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Flight dynamics I - Airplaneperformance - Web course

COURSE OUTLINE

FLIGHT DYNAMICS - I - AIRPLANE PERFORMANCE

1. Introduction.

Definition and subdivisions of flight dynamics.

Forces and moments acting on vehicles in flight. .

Equations of motion and simplification for performanceanalysis.

2. Earth's atmosphere and International StandardAtmosphere.

3. Drag polar.

Various types of drags.

Methods of estimating drag polar.

Drag polar of vehicles from low speed to hypersonicspeeds.

High lift devices.

4. Review of the variations of thrust or power output andSFC with altitude and velocity for various air breathingengines.

5. Performance analysis.

Steady level flight - Maximum speed, minimum speedand their variations with altitude.

Steady climb - Maximum rate of climb, angle of climb andtheir variations with altitude; absolute ceiling and serviceceiling.

Range and endurance - Breguet formulae; range inconstant velocity flight; effect of wind on the range.

Accelerated level flight.

Accelerated climb.

NPTELhttp://nptel.iitm.ac.in

AerospaceEngineering

Pre-requisites:

The student is expected tohave undergone courses on:

1. Vectors.

2. Rigid body dynamics.

3. Aerodynamics

4. Aircraft engines.

Additional Reading:

1. Miele, A. "Flightmechanics Vol I"Addison Wesley(1962).

2. Hale, F.J., "Introduction toaircraft performance,selection and design",John Wiley (1984).

3. Anderson, Jr. J.D"Introduction to flight"Fifth edition, McGrawHill,(2005).

4. Roskam, J. "Methods for

Manoeuvres - flight in vertical plane (loop); turn(minimumradius of turn and maximum rate of turn and theirvariations with the altitude).

V - n diagram.

Flight limitations.

Estimations of take-off distance and landing distance.

6. Examples of estimation of the drag polar andperformance of a piston engined and a jet enginedairplane.

COURSE DETAIL

A Web course shall contain 40 or more 1 hour lectureequivalents.

S.No Topics No.ofHours

1 Chapter 1 : Introduction 3

2 Chapter 2 : Earth’s atmosphere 2

3 Chapter 3 : Drag polar 7

4 Chapter 4 : Engine characteristics 4

5 Chapter 5 :Performance analysis I –Steady level flight

4

6 Chapter 6 : Performance analysis II –Steady climb,descent and glide

3

7 Chapter 7 : Performance analysis III –Range and endurance

3

8 Chapter 8 : Performance analysis IV–Accelerated level flight and climb

1

4. Roskam, J. "Methods forestimating drag polarsof subsonic airplanes"published by author1973.

Coordinators:

Prof. E.G. TulapurkaraDepartment of AerospaceEngineeringIIT Madras

9 Chapter 9 : Performance analysis V –Manoeuvres

4

10 Chapter 10 : Performance analysis VI –Take-off and landing

3

11 Performance analysis of a piston-engined airplane

3

12 Performance analysis of a subsonic jettransport

3

Total 40

References:

1. Houghton and Carruthers, "Aerodynamics forengineering students", Edward Arnold (1982).

2. McCormick B.W, "Aerodynamics, aeronautics and flightmechanics", John Wiley (1995).

3. Anderson, Jr. J.D "Aircraft performance and design"McGraw Hill International edition (1999).

4. Eshelby , M.E."Aircraft performancetheory and practice" ,Butterworth-Heinemann, Oxford,U.K., (2001).

5. Pamadi, B., "Performance, stability, dynamics and controlof an airplane", AIAA (2004).

6. Phillips, W.F. "Mechanics of flight" 2nd Edition, John Wiley(2010).

A joint venture by IISc and IITs, funded by MHRD, Govt of India http://nptel.iitm.ac.in

http://nptel.iitm.ac.in

Flight dynamics-I Prof. E.G. Tulapurkara Chapter-1

Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1

Chapter 1 Introduction

(Lectures 1, 2 and 3)

Keywords: Definition and importance of flight dynamics; forces acting on an

airplane; degrees of freedom for a rigid airplane; subdivisions of flight dynamics;

simplified treatment of performance analysis; course outline.

Topics 1.1 Opening remarks

1.1.1 Definition and importance of the subject

1.1.2 Recapitulation of the names of the major components of the airplane

1.1.3 Approach in flight dynamics

1.1.4 Forces acting on an airplane in flight

1.1.5 Body axes system for an airplane

1.1.6 Special features of flight dynamics

1.2 A note on gravitational force

1.2.1 Flat earth and spherical earth models

1.3 Frames of reference

1.3.1 Frame of reference attached to earth

1.4 Equilibrium of airplane

1.5 Number of equations of motion for airplane in flight

1.5.1 Degrees of freedom

1.5.2 Degrees of freedom for a rigid airplane

1.6 Subdivisions of flight dynamics

1.6.1 Performance analysis

1.6.2 Stability and control analysis

1.7 Additional definitions

1.7.1 Attitude of the airplane

1.7.2 Flight path

1.7.3 Angle of attack and side slip



1.8 Simplified treatment of performance analysis

1.9 Course outline

1.10 Background expected

References

Exercises



Chapter 1

Lecture 1 Introduction – 1 Topics

1.1 Opening remarks



1.1.3 Approach in flight dynamics

1.1.4 Forces acting on an airplane in flight

1.1.5 Body axes system for an airplane

1.1.6 Special features of flight dynamics



1.3 Frames of reference


1.1 Opening remarks

At the beginning of the study of any subject, it is helpful to know its definition,

scope and special features. It is also useful to know the benefits of the study of

the subject, background expected, approach, which also indicates the limitations,

and the way the subject is being developed. In this chapter these aspects are

dealt with.


The normal operation of a civil transport airplane involves take-off, climb to

cruise altitude, cruising, descent, loiter and landing (Fig.1.1). In addition, the

airplane may also carry out glide (which is descent with power off), turning

motion in horizontal and vertical planes and other motions involving

accelerations.



Fig.1.1 Typical flight path of a passenger airplane

Apart from the motion during controlled operations, an airplane may also

be subjected to disturbances which may cause changes in its flight path and

produce rotations about its axes.

The study of these motions of the airplane – either intended by the pilot or

those following a disturbance – forms the subject of Flight dynamics.

Flight dynamics: It is a branch of dynamics dealing with the motion of an object

moving in the earth’s atmosphere.

The study of flight dynamics will enable us to (a) obtain the performance of the

airplane which is described by items like maximum speed, minimum speed,

maximum rate of climb, distance covered with a given amount of fuel, radius of

turn, take-off distance, landing distance etc., (b) estimate the loads on the

airplane, (c) estimate the power required or thrust required for desired

performance, (d) determine the stability of the airplane i.e. whether the airplane

returns to steady flight conditions after being disturbed and (e) examine the

control of the airplane.

Flight dynamics is a basic subject for an aerospace engineer and its

knowledge is essential for proper design of an airplane.

Some basic ideas regarding this subject are presented in this chapter. The topics

covered herein are listed in the beginning of this chapter.



In this course, attention is focused on the motion of the airplane. Helicopters,

rockets and missiles are not covered.


At this stage it may be helpful to recapitulate the names of the major

components of the airplane. Figures 1.2a, b and c show the three-view drawings

of three different airplanes.

Fig.1.2a Major components of a piston engined airplane

(Based on drawing of HANSA-3 supplied by

National Aerospace Laboratories, Bangalore, India)



Fig.1.2b Major components of an airplane with turboprop engine

(Based on drawing of SARAS airplane supplied by

National Aerospace Laboratories, Bangalore, India)



Fig.1.2c Major components of an airplane with jet engine

(Note: The airplane shown has many features, all of which may not be there in a

single airplane).

1.1.3 Approach

The approach used in flight mechanics is to apply Newton’s laws to the

motion of objects in flight. Let us recall these laws:

Newton’s first law states that every object at rest or in uniform motion

continues to be in that state unless acted upon by an external force.

The second law states that the force acting on a body is equal to the time

rate of change of its linear momentum.

The third law states that to every action, there is an equal and opposite

reaction.

Newton’s second law can be written as:

F = ma ; a = dV / dt ; V = dr / dt (1.1)



Where F = sum of all forces acting on the body, m = mass, a = acceleration,

V = velocity, r = the position vector of the object and t = time

(Note: quantities in bold are vectors).

Acceleration is the rate of change of velocity and velocity is the rate of

change of position vector.

To prescribe the position vector, requires a co-ordinate system with

reference to which the position vector/displacement is measured.

1.1.4 Forces acting on an airplane

During the analysis of its motion the airplane will be considered as a rigid

body. The forces acting on an object in flight are:

– Gravitational force

– Aerodynamic forces and

– Propulsive force.

The gravitational force is the weight (W) of the airplane.

The aerodynamic forces and moments arise due to the motion of the

airplane relative to air. Figure 1.3 shows the aerodynamic forces viz. the drag

(D), the lift (L) and the side force (Y).

The propulsive force is the thrust(T) produced by the engine or the engine-

propeller combination.



Fig.1.3 Forces on an airplane

1.1.5 Body axes system of an airplane

To formulate and solve a problem in dynamics requires a system of axes.

To define such a system it is noted that an airplane is nearly symmetric, in

geometry and mass distribution, about a plane which is called the ‘Plane of

symmetry’ (Fig.1.4a). This plane is used for defining the body axes system.

Figure 1.4b shows a system of axes (OXbYbZb) fixed on the airplane which

moves with the airplane and hence is called ‘Body axes system’. The origin ‘O’ of

the body axes system is the center of gravity (c.g.) of the body which, by

assumption of symmetry, lies in the plane of symmetry. The axis OXb is taken

positive in the forward direction. The axis OZb is perpendicular to OXb in the

plane of symmetry, positive downwards. The axis OYb is perpendicular to the

plane of symmetry such that OXbYbZb is a right handed system.



Fig.1.4a Plane of symmetry and body axis system

Fig.1.4b The forces and moments acting on an airplane and the components of

linear and angular velocities with reference to the body axes system



Figure 1.4b also shows the forces and moments acting on the airplane

and the components of linear and angular velocities. The quantity V is the

velocity vector. The quantities X, Y, Z are the components of the resultant

aerodynamic force, along OXb, OYb and OZb axes respectively. L’, M, N are the

rolling moment, pitching moment and yawing moment respectively about OXb,

OYb and OZb axes; the rolling moment is denoted by L’ to distinguish it from lift

(L). u,v,w are respectively the components, along OXb, OYb and OZb, of the

velocity vector (V). The angular velocity components are indicated by p, q, and r.

1.1.6 Special features of Flight Dynamics

The features that make flight dynamics a separate subject are:

i)During its motion an airplane in flight, can move along three axes and can

rotate about three axes. This is more complicated than the motions of machinery

and mechanisms which are restrained by kinematic constraints, or those of land

based or water based vehicles which are confined to move on a surface.

ii)The special nature of the forces, like aerodynamic forces, acting on the

airplane(Fig.1.3). The magnitude and direction of these forces change with the

orientation of the airplane, relative to its flight path.

iii)The system of aerodynamic controls used in flight (aileron, elevator, rudder).


In the case of an airplane, the gravitational force is mainly due to the

attraction of the earth. The magnitude of the gravitational force is the weight of

the airplane (in Newtons).

W = mg; where W is the gravitational force, m is the mass of the airplane and g

is the acceleration due to gravity.

The line of action of the gravitational force is along the line joining the

centre of gravity (c.g.) of the airplane and the center of the earth. It is directed

towards the center of earth.

The magnitude of the acceleration due to gravity (g) decreases with

increase in altitude (h). It can be calculated based on its value at sea level (go),

and using the following formula.

(g / g0) = [R / (R + h)]2 (1.2)



where R is the radius of the earth,

R = 6400 km (approx.) and g0 = 9.81ms-2

However, for typical airplane flights (h < 20 km), g is generally taken to be

constant.


In flight mechanics, there are two ways of dealing with the gravitational

force, namely the flat earth model and the spherical earth model.

In the flat earth model, the gravitational acceleration is taken to act

vertically downwards (Fig 1.5).

When the distance over which the flight takes place is small, the flat earth

model is adequate. Reference 1.1, chapter 4 may be referred to for details.

Fig.1.5 Flat earth model

In the spherical earth model, the gravitational force is taken to act along

the line joining the center of earth and the c.g. of the airplane. It is directed

towards the center of the earth (Fig.1.6).

The spherical earth model is used for accurate analysis of flights involving

very long distances.



Fig.1.6. Spherical earth model

Remarks:

In this course the flat earth model is used. This is adequate for the

following reasons.

i) The distances involved in flights with acceleration are small and the

gravitational force can be considered in the vertical direction by proper choice of

axes.

ii) In unaccelerated flights like level flight, the forces at the chosen instant of time

are considered and the distance covered etc. are obtained by integration. This

procedure is accurate as long as it is understood that the altitude means height

of the airplane above the surface of the earth and the distance is measured on a

sphere of radius equal to the sum of the radius of earth plus the altitude of

airplane.

iii) As mentioned in section 1.1.4, the forces acting on the airplane are the

gravitational force, the aerodynamic forces and the propulsive force. The first one

has been discussed in this section.The discussion on aerodynamic forces will be

covered in chapter 3 and that on propulsive force in chapter 4.

1.3 Frame of reference

A frame of reference (coordinate system) in which Newton’s laws of

motion are valid is known as a Newtonian frame of reference.



Since Newton’s laws deal with acceleration, a frame of reference moving

with uniform velocity with respect to a Newtonian frame is also a Newtonian

frame or inertial frame.

However, if the reference frame is rotating with an angular velocity (ω),

then, additional accelerations like centripetal acceleration {ω x (ω x r)} and

Coriolis acceleration (V x ω) will come into picture.

Reference 1.2,chapter 13 may be referred to for further details on non-Newtonian

reference frame.


In flight dynamics, a co-ordinate system attached to the earth is taken to

approximate a Newtonian frame (Fig.1.7).

The effects of the rotation of earth around itself and around the sun on this

approximation can be estimated as follows.

It is noted that the earth rotates around itself once per day. Hence

ω = 2 / (3600x24) = 7.27x10-5 s-1;

Since r roughly equals 6400 km; the maximum centripetal acceleration (ω2r)

equals 0.034 ms-2.

The earth also goes around the sun and completes one orbit in approximately

365 days. Hence in this case,

ω = 2 / (365 x 3600 x 24) = 1.99x10-7s-1;

Further, in this case, the radius would be roughly the mean distance between the

sun and the earth which is 1.5x1011m. Consequently, ω2 r = 0.006 ms-2.



Fig.1.7 Earth fixed and body fixed co-ordinate systems

Thus, it is observed that the centripetal accelerations due to rotation of earth

about itself and around the sun are small as compared to the acceleration due to

gravity.

These rotational motions would also bring about Coriolis acceleration

(V x ω). However, its magnitude, which depends on the flight velocity, would be

much smaller than the acceleration due to gravity in flights up to Mach number of

3. Hence, the influence can be neglected.

Thus, taking a reference frame attached to the surface of the earth as a

Newtonian frame is adequate for the analysis of airplane flight. Figure 1.7 shows

such a coordinate system.



Chapter 1



1.5 Number of equations of motion for airplane in flight




1.6.1 Performance analysis

1.6.2 Stability and control analysis


1.7.1 Attitude of the airplane

1.7.2 Flight path

1.7.3 Angle of attack and side slip


The above three types of forces (aerodynamic, propulsive and

gravitational) and the moments due to them govern the motion of an airplane in

flight.

If the sums of all these forces and moments are zero, then the airplane is

said to be in equilibrium and will move along a straight line with constant velocity

(see Newton's first law). If any of the forces is unbalanced, then the airplane will

have a linear acceleration in the direction of the unbalanced force. If any of the

moments is unbalanced, then the airplane will have an angular acceleration

about the axis of the unbalanced moment.

The relationship between the unbalanced forces and the linear

accelerations and those between unbalanced moments and angular



accelerations are provided by Newton’s second law of motion. These

relationships are called equations of motion.

1.5 Number of equations of motion for an airplane in flight

To derive the equations of motion, the acceleration of a particle on the

body needs to be known. The acceleration is the rate of change of velocity and

the velocity is the rate of change of position vector with respect to the chosen

frame of reference.


The minimum number of coordinates required to prescribe the motion is

called the number of degrees of freedom. The number of equations governing

the motion equals the degrees of freedom. As an example, it may be recalled

that the motion of a particle moving in a plane is prescribed by the x- and y-

coordinates of the particle at various instants of time and this motion is described

by two equations.

Similarly, the position of any point on a rigid pendulum is describe by just

one coordinate namely the angular position (θ) of the pendulum (Fig.1.8). In this

case only one equation is sufficient to describe the motion. In yet another

example, if a particle is constrained to move on a sphere, then its position is

completely prescribed by the longitude and the latitude. Hence, this motion has

only two degrees of freedom.

From the discussion in this subsection it is clear that the coordinates needed to

prescribe the motion could be lengths and/or angles.



Fig.1.8 Motion of a single degree of freedom system


To describe its motion, the airplane is treated as a rigid body. It may be

recalled that in a rigid body the distance between any two points is fixed. Thus

the distance r in Fig.1.9 does not change during the motion. To decide the

minimum number of coordinates needed to prescribe the position of a point on a

rigid body which is translating and rotating, one may proceed as follows.



Fig.1.9 Position of a point on a rigid airplane

A rigid body with N particles may appear to have 3N degrees of freedom,

but the constraint of rigidity reduces this number. To arrive at the minimum

number of coordinates, let us approach the problem in a different way. Following

Ref.1.3, it can be stated that to fix the location of a point on a rigid body one does

not need to prescribe its distance from all the points, but only needs to prescribe

its distance from three points which do not lie on the same line (points 1, 2 and 3

in Fig.1.10a). Thus, if the positions of these three points are prescribed with

respect to a reference frame, then the position of any point on the body is known.

This may indicate nine degrees of freedom. This number is reduced to six

because the distances s12, s23 and s13 in Fig.1.10a are constants.



Fig.1.10a Position of a point with respect to three reference points

Another way of looking at the problem is to consider that the three

coordinates of point 1 with respect to the reference frame are prescribed. Now

the point 2 is constrained, because of rigid body assumption, to move on a

sphere centered on point 1 and needs only two coordinates to prescribe its

motion. Once the points 1 and 2 are determined, the point 3 is constrained, again

due to rigid body assumption, to move on a circle about the axis joining points 1

and 2. Hence, only one independent coordinate is needed to prescribe the

position of point 3. Thus, the number of independent coordinates is six (3+2+1).

Or a rigid airplane has six degrees of freedom.

In dynamics the six degrees of freedom associated with a rigid body,

consist of the three coordinates of the origin of the body with respect to the

chosen frame of reference and the three angles which describe the angular

position of a coordinate system fixed on the body (OXbYbZb) with respect to the



fixed frame of reference (EXeYeZe) as shown in Fig.1.10b. These angles are

known as Eulerian angles. These are discussed in ch.7 of flight dynamics- II. See

also Ch.4 of Ref.1.3.

Fig.1.10b Coordinates of a point (P) on a rigid body

Remarks:

i) The derivation of the equations of motions in a general case with six degrees of

freedom (see chapter 7 of Flight dynamics-II or Ref.1.4 chapter 10, pt.3 or

Ref.1.5, chapter 10) is rather involved and would be out of place here.

ii) Here, various cases are considered separately and the equations of motion

are written down in each case.


The subject of flight dynamics is generally divided into two main branches viz.

(i) Performance analysis and (ii) Stability and control

1.6.1 Performance Analysis

In performance analysis, only the equilibrium of forces is generally

considered. It is assumed that by proper deflections of the controls, the moments



can be made zero and that the changes in aerodynamic forces due to deflection

of controls are small. The motions considered in performance analysis are steady

and accelerations, when involved, do not change rapidly with time.

The following motions are considered in performance analysis

- Unaccelerated flights,

• Steady level flight

• Climb, glide and descent

- Accelerated flights,

• Accelerated level flight and climb

• Loop, turn, and other motions along curved paths which are

called manoeuvres

• Take-off and landing.

1.6.2 Stability and control analyses

Roughly speaking, the stability analysis is concerned with the motion of

the airplane, from the equilibrium position, following a disturbance. Stability

analysis tells us whether an airplane, after being disturbed, will return to its

original flight path or not.

Control analysis deals with the forces that the deflection of the controls

must produce to bring to zero the three moments (rolling, pitching and yawing)

and achieve a desired flight condition. It also deals with design of control

surfaces and the forces on control wheel/stick /pedals. Stability and control are

linked together and are generally studied under a common heading.

Flight dynamics - I deals with performance analysis. By carrying out this

analysis one can obtain various performance characteristics such as maximum

level speed, minimum level speed, rate of climb, angle of climb, distance covered

with a given amount of fuel called ‘Range’, time elapsed during flight called

‘Endurance’, minimum radius of turn, maximum rate of turn, take-off distance,

landing distance etc. The effect of flight conditions namely the weight, altitude

and flight velocity of the airplane can also be examined. This study would also

help in solving design problems of deciding the power required, thrust required,



fuel required etc. for given design specifications like maximum speed, maximum

rate of climb, range, endurance etc.

Remark:

Alternatively, the performance analysis can be considered as the analysis

of the motion of flight vehicle considered as a point mass, moving under the

influence of applied forces (aerodynamic, propulsive and gravitational forces).

The stability analysis similarly can be considered as motion of a vehicle of finite

size, under the influence of applied forces and moments.


1.7.1 Attitude

As mentioned in section 1.5.2 the instantaneous position of the airplane,

with respect to the earth fixed axes system (EXeYeZe), is given by the

coordinates of the c.g. at that instant of time. The attitude of the airplane is

described by the angular orientation of the OXbY

bZ

b system with respect to

OXeYeZe system or the Euler angles. Reference 1.4, chapter 10 may be referred

to for details. Let us consider simpler cases. When an airplane climbs along a

straight line its attitude is given by the angle ‘γ’ between the axis OXb and the

horizontal (Fig.1.11a). When an airplane executes a turn, the projection of OXb

axis, in the horizontal plane, makes an angle Ψ with reference to a fixed

horizontal axis (Fig.1.11b). When an airplane is banked the axis OYb makes an

angle with respect to the horizontal (Fig.1.11c) and the axis OZb makes an

angle with respect to the vertical.

Fig.1.11a Airplane in a climb



Fig.1.11b Airplane in a turn - view from top

Fig.1.11c Angle of bank ( )



1.7.2 Flight path

In the subsequent sections, the flight path, also called the trajectory,

means the path or the line along which the c.g. of the airplane moves. The

tangent to this curve at a point gives the direction of flight velocity at that point on

the flight path. The relative wind is in a direction opposite to that of the flight

velocity.

1.7.3. Angle of attack and side slip

While discussing the forces acting on an airfoil, the chord of the airfoil is

taken as the reference line and the angle between the chord line and the relative

wind is the angle of attack (α). The aerodynamic forces viz. lift (L) and drag (D) ,

produced by the airfoil, depend on the angle of attack (α) and are respectively

perpendicular and parallel to relative wind direction (Fig.1.11 d).

Fig.1.11d Angle of attack and forces on a airfoil

In the case of an airplane the flight path, as mentioned earlier, is the line along

which c.g. of the airplane moves. The tangent to the flight path is the direction of

flight velocity (V). The relative wind is in a direction opposite to the flight velocity.

If the flight path is confined to the plane of symmetry, then the angle of attack

would be the angle between the relative wind direction and the fuselage

reference line (FRL) or OXb axis (see Fig.1.11e). However, in a general case the

velocity vector (V) will have components both along and perpendicular to the



plane of symmetry. The component perpendicular to the plane of symmetry is

denoted by ‘v’. The projection of the velocity vector in the plane of symmetry

would have components u and w along OXb and OZb axes (Fig.1.11f). With this

background the angle of sideslip and the angle of attack are defined as follows.

Fig.1.11e Flight path in the plane of symmetry



Fig.1.11f Velocity components in a general case and definition of angle of attack

and sideslip

The angle of sideslip (β) is the angle between the velocity vector (V) and the

plane of symmetry i.e.

β = sin-1 (v/ |V|); where |V| is the magnitude of V.

The angle of attack (α) is the angle between the projection of velocity vector (V)

in the Xb - Zb plane and the OXb axis or

-1 -1 -1

2 2 2 2

w w wα = tan = sin = sin

u | | -v u +wV

Remarks:

i) It is easy to show that, if V denotes magnitude of velocity (V), then

u = V cos α cos β, v = V sin β; w = V sin α cos β.



ii) By definition, the drag (D) is parallel to the relative wind direction. The lift force

lies in the plane of symmetry of the airplane and is perpendicular to the direction

of flight velocity.



Chapter 1


1.8 Simplified treatment of performance analysis

1.9 Course outline

1.10 Background expected

1.8 Simplified treatment in performance analysis

In a steady flight, there is no acceleration along the flight path and in a

level flight; the altitude of the flight remains constant. A steady, straight and level

flight generally means a flight along a straight line at a constant velocity and

constant altitude.

Sometimes, this flight is also referred to as unaccelerated level flight. To illustrate

the simplified treatment in performance analysis, the case of unaccelerated level

flight is considered below.

The forces acting on an airplane in unaccelerated level flight are shown in the

Fig.1.12.

They are: Lift (L), Thrust (T), Drag (D) and Weight (W) of the airplane.

It may be noted that the point of action of the thrust and it’s direction depend on

the engine location. However, the direction of the thrust can be taken parallel to

the airplane reference axis.



Fig.1.12 Forces acting in steady level flight

The lift and drag, being perpendicular to the relative wind, are in the

vertical and horizontal directions respectively, in this flight. The weight acts at the

c.g. in a vertically downward direction.

In an unaccelerated level flight, the components of acceleration in the

horizontal and vertical directions are zero.

Hence, the sums of the components of all the forces in these directions

are zero. Resolving the forces along and perpendicular to the flight path (see

Fig.1.12.), gives the following equations of force equilibrium.

T cos α – D = 0 (1.3)

T sin α + L – W = 0 (1.4)

Apart from these equations, equilibrium demands that the moment about

the y-axis to be zero, i.e.,

Mcg = 0

Unless the moment condition is satisfied, the airplane will begin to rotate

about the c.g.

Let us now examine how the moment is balanced in an airplane. The

contributions to Mcg come from all the components of the airplane. As regards the

wing, the point where the resultant vector of the lift and drag intersects the plane

of symmetry is known as the centre of pressure. This resultant force produces a

moment about the c.g. However, the location of the center of pressure depends

on the lift coefficient and hence the moment contribution of wing changes with

the angle of attack as the lift coefficient depends on the angle of attack. For



convenience, the lift and the drag are transferred to the aerodynamic center

along with a moment (Mac). Recall, that moment coefficient about the a.c. (Cmac)

is, by definition, constant with change in angle of attack.

Similarly, the moment contributions of the fuselage and the horizontal tail

change with the angle of attack. The engine thrust also produces a moment

about the c.g. which depends on the thrust required.

Hence, the sum of the moments about the c.g. contributed by the wing,

fuselage, horizontal tail and engine changes with the angle of attack. By

appropriate choice of the horizontal tail setting (i.e. incidence of horizontal tail

with respect to fuselage central line), one may be able to make the sum of these

moments to be zero in a certain flight condition, which is generally the cruise

flight condition. Under other flight conditions, generation of corrective

aerodynamic moment is facilitated by suitable deflection of elevator (See

Fig.1.2a, b and c for location of elevator). By deflecting the elevator, the lift on the

horizontal tail surface can be varied and the moment produced by the horizontal

tail balances the moments produced by all other components.

The above points are illustrated with the help of an example.

Example 1.1

A jet aircraft weighing 60,000 N has it’s line of thrust 0.15 m below the line

of drag. When flying at a certain speed, the thrust required is 6000 N and the

center of pressure of the wing lift is 0.45 m aft of the airplane c.g. What is the lift

on the wing and the load on the tail plane whose center of pressure is 7.5 m

behind the c.g.? Assume unaccelerated level flight and the angle of attack to be

small during the flight.

Solution:

The various forces and dimensions are presented in Fig.1.13. The lift on

the wing is LW and the lift on the tail is LT. Since the angle of attack (α) is small, it

may be considered that cos α = 1 and sin α = 0. Thus, the force equilibrium (Eqs.

1.3 and 1.4), yields :

T – D = 0

LW + LT – W = 0



i.e. D = T = 6000 N and LT + LW = 60000 N

From Fig. 1.13., the moment equilibrium about the c.g. gives:

Mcg = T (zd + 0.15) – D.zd – 0.45.LW – 7.5.LT = 0 where zd is the distance of drag

below the c.g; not shown in figure as it is of no significance in the present

context.

Fig.1.13 Forces acting on an airplane in steady level flight

Solving these equations, gives :

LW = 63702.13 N and LT = -3702.13 N

Following observations can be made.

A) The lift on the wing is about 63.7 kN. The lift on the tail is only 3.7 kN and is in

the downward direction.

B) The contribution of tail to the total lift is thus small, in this case, about 6% and

negative. This negative contribution necessitates the wing lift to be more than the

weight of the airplane. This increase in the lift results in additional drag called trim

drag.

C) The distance zd is of no significance in this problem as the drag and thrust

form a couple whose moment is equal to the thrust multiplied by the distance

between them.

D) Generally, the angle of attack (α) is small. Hence, sin α is small and cos α is

nearly equal to unity. Thus, the equations of force equilibrium reduce to

T – D = 0 and L – W = 0.

E) It is assumed that the pitching moment equilibrium i.e. Σ Mcg = 0 is achieved

by appropriate deflection of the elevator. The changes in the lift and drag due to



elevator deflections are generally small and in performance analysis, as stated

earlier, these changes are ignored and the simplified picture as shown in Fig.1.14

is considered adequate.

Fig.1.14. Simplified picture of the forces acting on an airplane in level flight.

1.9 Course outline

Let us consider the background material required to carry-out the

performance analysis. It is known that :

L = (1/2) ρ V2 S CL

D = (1/2) ρ V2 S CD

where CL and CD are the lift and drag coefficients; S is the area of the wing.

The quantities CL and CD depend on α , Mach number (M = V / a) and Reynolds

number (Re = ρ V l /µ); where l is the reference length. Thus

CD = f (CL, M, Re) (1.6)

The relation between CL and CD at given M and Re is known as the drag

polar of the airplane. This has to be known for carrying the performance

analysis. The density of air (ρ) depends on the flight altitude. Further the Mach

number depends on the speed of sound, which in turn depends on the ambient

air temperature. Thus, performance analysis requires the knowledge of the



variations of pressure, temperature, density, viscosity etc. with altitude in earth’s

atmosphere.

The evaluation of performance also requires the knowledge of the engine

characteristics such as, variations of thrust (or power) and fuel consumption with

the flight speed and altitude.

Keeping these aspects in view, following will be the contents of this course.

Earth’s atmosphere (chapter 2)

Drag polar (chapter 3)

Engine characteristics (chapter 4)

Performance analysis. ( chapters 5 to 10)

These topics will be taken up in the subsequent chapters.

The Appendices ‘A’ and ‘B’ present the performance analyses of piston-engined

and jet airplane respectively.

1.10 Back ground expected

The student is expected to have undergone courses on (a) Vectors (b)

Rigid body dynamics (c) Aerodynamics and (d) Aircraft engines.

Remark: References 1.5 to 1.14 are some of the books dealing with airplane

performance. They can be consulted for additional information.



Chapter 1

References

1.1 Miele, A. “Flight mechanics Vol I” Addison Wesley (1962).

1.2 Shames, I.H. and Krishna Mohana Rao, G. “Engineering mechanics – statics

and dynamics”, 4th Edition, Dorling Kindersley (India), licensees of Pearson

Education (2006).

1.3 Goldstein H. “Classical mechanics “Second edition Addison Wesley (1980).

1.4 Davies, M. (Editor) “The standard handbook for aeronautical and

astronautical engineers” McGraw Hill (2003).

1.5 Perkins, C.D. and Hage, R. E. “Airplance performance, stability and

control” John Wiley (1963).

1.6 Dommasch, D.O. Sherby, S.S. and Connolly, T.F. “Airplane

aerodynamics” Pitman (1967).

1.7 Houghton E.L. and Carruthers N.B. “Aerodynamics for engineering

students”, Edward Arnold (1982).

1.8 Hale, F.J. “Introduction to aircraft performance, selection and design”,

John Wiley (1984).

1.9 McCormick B.W. “Aerodynamics, aeronautics and flight mechanics”, John

Wiley (1995).

1.10 Anderson, Jr. J.D. “Aircraft performance and design” McGraw Hill

International edition (1999).

1.11 Eshelby, M.E. ”Aircraft performance-theory and practice”, Butterworth-

Heinemann, Oxford, U.K., (2001).

1.12 Pamadi, B. “Performance, stability, dynamics and control of an

airplane”, AIAA (2004).

1.13 Anderson, Jr. J.D. “Introduction to flight” Fifth edition, McGraw-Hill,

(2005).

1.14 Phillips, W.F. “Mechanics of flight” 2nd Edition John Wiley (2010).



1.15 Jackson, P. (Editor) “Jane’s all the world’s aircraft” Published annually

by Jane’s information group Ltd., Surrey, U.K..



Chapter 1 Exercises

1. Sketch the three views of an airplane and show it’s axes systems.

2. Define, with neat sketches, the following terms.

(a) flight path

(b) flight velocity

(c) body axes system

(d) angle of attack

(e) angle of slide slip and

(f) bank angle.

3.“Jane’s All the World Aircraft” (Ref.1.15) is a book published annually and

contains details of airplanes currently in production in various countries. Refer to

this book and study the three view drawings, geometrical details and

performance parameters of different types of airplanes.



Chapter 2 Earth’s atmosphere (Lectures 4 and 5)

Keywords: Earth’s atmosphere; International standard atmosphere;

geopotential altitude; stability of atmosphere.

Topics

2.1 Introduction

2.2 Earth’s atmosphere

2.2.1 The troposphere

2.2.2 The stratosphere

2.2.3 The mesosphere

2.2.4 The ionosphere or thermosphere

2.2.5 The exosphere

2.3 International standard atmosphere (ISA)

2.3.1 Need for ISA and agency prescribing it.

2.3.2 Features of ISA

2.4 Variations of properties with altitude in ISA

2.4.1 Variations of pressure and density with altitude

2.4.2 Variations with altitude of pressure ratio, density ratio speed of

sound, coefficient of viscosity and kinematic viscosity.

2.5 Geopotential altitude

2.6 General remarks

2.6.1 Atmospheric properties in cases other than ISA

2.6.2 Stability of atmosphere

References

Exercises



Chapter 2

Lecture 4 Earth’s atmosphere – 1 Topics

2.1 Introduction






2.2.5 The exosphere

2.3 International standard atmosphere (ISA)

2.3.1 Need for ISA and agency prescribing it.


2.1 Introduction

Airplanes fly in the earth’s atmosphere and therefore, it is necessary to

know the properties of this atmosphere.

This chapter, deals with the average characteristics of the earth’s

atmosphere in various regions and the International Standard Atmosphere (ISA)

which is used for calculation of airplane performance.


The earth’s atmosphere is a gaseous blanket around the earth which is

divided into the five regions based on certain intrinsic features (see Fig.2.1).

These five regions are: (i) Troposphere, (ii) Stratosphere, (iii) Mesosphere,

(iv) Ionosphere or Thermosphere and (v) Exosphere. There is no sharp

distinction between these regions and each region gradually merges with the

neighbouring regions.



Fig.2.1 Typical variations of temperature and pressure in the earth’s atmosphere


This is the region closest to the earth’s surface. It is characterized by

turbulent conditions of air. The temperature decreases linearly at an approximate

rate of 6.5 K / km. The highest point of the troposphere is called tropopause. The

height of the tropopause varies from about 9 km at the poles to about 16 km at

the equator.


This extends from the tropopause to about 50 km. High velocity winds

may be encountered in this region, but they are not gusty. Temperature remains

constant up to about 25 km and then increases. The highest point of the

stratosphere is called the stratopause.


The mesosphere extends from the stratopause to about 80 km. The

temperature decreases to about -900C in this region. In the mesosphere, the



pressure and density of air are very low, but the air still retains its composition as

at sea level. The highest point of the mesosphere is called the mesopause.


This region extends from the mesopause to about 1000 km. It is

characterized by the presence of ions and free electrons. The temperature

increases to about 00C at 110 km, to about 10000C at 150 km and peak of about

17800C at 700 km (Ref.2.1). Some electrical phenomena like the aurora borealis

occur in this region.

2.2.5 The exosphere

This is the outer fringe of the earth’s atmosphere. Very few molecules are

found in this region. The region gradually merges into the interplanetary space.

2.3 International Standard Atmosphere (ISA)

2.3.1 Need for ISA and agency prescribing it

The properties of earth’s atmosphere like pressure, temperature and

density vary not only with height above the earth’s surface but also with the

location on earth, from day to day and even during the day. As mentioned in

section 1.9, the performance of an airplane is dependent on the physical

properties of the earth’s atmosphere. Hence, for the purpose of comparing

(a) the performance of different airplanes and (b) the performance of the same

airplane measured in flight tests on different days, a set of values for atmospheric

properties have been agreed upon, which represent average conditions

prevailing for most of the year, in Europe and North America. Though the agreed

values do not represent the actual conditions anywhere at any given time, they

are useful as a reference. This set of values called the International Standard

Atmosphere (ISA) is prescribed by ICAO (International Civil Aviation

Organization). It is defined by the pressure and temperature at mean sea level,

and the variation of temperature with altitude up to 32 km (Ref.1.11, chapter 2).

With these values being prescribed, it is possible to find the required physical

characteristics (pressure, temperature, density etc) at any chosen altitude.



Remark:

The actual performance of an airplane is measured in flight tests under

prevailing conditions of temperature, pressure and density. Methods are

available to deduce, from the flight test data, the performance of the airplane

under ISA conditions. When this procedure is applied to various airplanes and

performance presented under ISA conditions, then comparison among different

airplanes is possible.


The main features of the ISA are the standard sea level values and the

variation of temperature with altitude. The air is assumed as dry perfect gas.

The standard sea level conditions are as follows:

Temperature (T0) = 288.15 K = 150C

Pressure (p0) = 101325 N/m2 = 760 mm of Hg

Rate of change of temperature:

= - 6.5 K/km upto 11 km

= 0 K/km from 11 to 20 km

= 1 K/km from 20 to 32 km

The region of ISA from 0 to 11 km is referred to as troposphere. That

between 11 to 20 km is the lower stratosphere and between 20 to 32 km is the

middle stratosphere (Ref.1.11, chapter 2).

Note: Using the values of T0 and p0 , and the equation of state, p = ρRT, gives the

sea level density (ρ0) as 1.225 kg/m3.



Chapter 2

Lecture 5 Earth’s atmosphere – 2 Topics



2.4.2 Variations with altitude of pressure ratio, density ratio speed of

sound, coefficient of viscosity and kinematic viscosity.


2.6 General remarks



Atmospheric properties of ISA (Table 2.1)


For calculation of the variations of pressure, temperature and density with

altitude, the following equations are used.

The equation of state p = ρ R T (2.1)

The hydrostatic equation dp/dh = - ρ g (2.2)

Remark:

The hydrostatic equation can be easily derived by considering the balance of

forces on a small fluid element.

Consider a cylindrical fluid element of area A and height Δh as shown in Fig.2.2.



Fig.2.2 Equilibrium of a fluid element.

The forces acting in the vertical direction on the element are the pressure forces

and the weight of the element.

For vertical equilibrium of the element,

pA – {p + (dp /dh) Δh} A – ρ g A Δh = 0

Simplifying, dp /dh = - ρ g


Substituting for ρ from the Eq.(2.1) in Eq.(2.2) gives:

dp / dh = -(p/RT) g

Or (dp/p) = -g dh/RT (2.3)

Equation (2.3) is solved separately in troposphere and stratosphere, taking into

account the temperature variations in each region. For example, in the

troposphere, the variation of temperature with altitude is given by the equation

T = T0 – λ h (2.4)

where T0 is the sea level temperature, T is the temperature at the altitude h and λ

is the temperature lapse rate in the troposphere.

Substituting from Eq.(2.4) in Eq.(2.3) gives:

(dp /p) = - gdh /R (T0 – λ h) (2.5)



Taking ‘g’ as constant, Eq.(2.5) can be integrated between two altitudes h1 and

h2. Taking h1 as sea level and h2 as the desired altitude (h), the integration gives

the following equation, the intermediate steps are left as an exercise.

(p/p0) = (T/T0)(g/λR) (2.6)

where T is the temperature at the desired altitude (h) given by Eq.(2.4).

Equation (2.6) gives the variation of pressure with altitude.

The variation of density with altitude can be obtained using Eq.(2.6) and

the equation of state. The resulting variation of density with temperature in the

troposphere is given by:

(ρ/ρ0) = (T/T0)(g/λR)-1 (2.7)

Thus, both the pressure and density variations are obtained once the

temperature variation is known.

As per the ISA, R = 287.05287 m2sec-2 K and g = 9.80665 m/s2.

Using these and λ = 0.0065 K/m in the troposphere yields (g/Rλ) as 5.25588.

Thus, in the troposphere, the pressure and density variations are :

(p/p0) = (T/T0)5.25588 (2.8)

(ρ/ρ0) = (T/T0)4.25588 (2.9)

Note: T= 288.15 - 0.0065 h; h in m and T in K.

In order to obtain the variations of properties in the lower stratosphere (11

to 20 km altitude), the previous analysis needs to be carried-out afresh with λ = 0

i.e., ‘T’ having a constant value equal to the temperature at 11 km (T = 216.65 K).

From this analysis the pressure and density variations in the lower stratosphere

are obtained as :

(p / p11) = (ρ / ρ11) = exp { -g (h - 11000) / RT11 } (2.10)

where p11, ρ11 and T11 are the pressure, density and temperature respectively at

11 km altitude.

In the middle stratosphere (20 to 32 km altitude), it can be shown that (note in

this case λ = -0.001 K / m):

(p / p20) = (T / T20)- 34.1632 (2.11)

(ρ / ρ20) = (T/ T20)- 35.1632 (2.12)



where p20, ρ20 and T20 are pressure, density and temperature respectively at

20 km altitude.

Thus, the pressure and density variations have been worked out in the

troposphere and the stratosphere of ISA. Table 2.1 presents these values.

Remark:

Using Eqs.(2.1) and (2.2) the variations of pressure and density can be worked

out for other variations of temperature with height (see exercise 2.1).

2.4.2. Variations with altitude of pressure ratio, density ratio, speed of

sound, coefficient of viscosity and kinematic viscosity

The ratio (p/p0) is called pressure ratio and is denoted by δ. Its value in ISA can

be obtained by using Eqs.(2.8),(2.10) and (2.11). Table 2.1 includes these

values.

The ratio (ρ / ρ0) is called density ratio and is denoted by σ. Its values in ISA can

be obtained using Eqs.(2.9),(2.10) and (2.12). Table 2.1 includes these values.

The speed of sound in air, denoted by ‘a’, depends only on the temperature and

is given by:

a = (γ RT)0.5 (2.13)

where γ is the ratio of specific heats; for air γ = 1.4. The values of ‘a’ in ISA can

be obtained by using appropriate values of temperature. Table 2.1 includes these

values.

The kinematic viscosity ( ) is given by:

= μ / ρ where μ is the coefficient of viscosity.

The coefficient of viscosity of air (μ) depends only on temperature. Its variation

with temperature is given by the following Sutherland formula.

3/2-6 T

μ = 1.458X10 [ ]T+110.4

, where T is in Kelvin and μ is in kg m-1 s-1 (2.14)

Table 2.1 includes the variation of kinematic viscosity with altitude.

Example 2.1

Calculate the temperature (T), pressure (p), density (ρ ), pressure ratio

(δ ) , density ratio (σ ), speed of sound (a) , coefficient of viscosity (μ ) and

kinematic viscosity ( ) in ISA at altitudes of 8 km, 16 km and 24 km.



Solution:

It may be noted that the three altitudes specified in this example, viz.

8 km, 16 km and 24 km, lie in troposphere, lower stratosphere and middle

stratosphere regions of ISA respectively.

(a) h = 8 km

Let the quantities at 8 km altitude be denoted by the suffix ‘8’.

In troposphere: 0T = T -λh

where, T0 = 288.15 K, λ = 0.0065 K /m

Hence, 8T = 288.15 - 0.0065 8000 = 236.15K

From Eq.(2.8)

5.25588 5.2558888 0

0

p= δ = T/T = 236.15/288.15 = 0.35134

p

Or 28p = 0.35134 × 101325 = 35599.5 N/m

38 8 8

35599.5ρ = p / RT = = 0.52516 kg/m

287.05287×236.15

8 8 0σ = ρ /ρ = 0.52516/1.225 = 0.42870

a8 = (γ RT8)0.5 0.5

= 1.4×287.05287×236.15 = 308.06 m/s

From Eq.(2.14):

1.5 1.5-6 -6 -5 -1 -18

88

T 236.15μ = 1.458×10 = 1.458×10 = 1.5268×10 kg m s

T +110.4 236.15+110.4

-5 -5 28 8 8= μ /ρ = 1.5268×10 / 0.52516 = 2.9072×10 m /s

Remarks:

(i) The values calculated above and those in Table 2.1 may differ from each

other in the last significant digit. This is due to the round-off errors in the

calculations.

(ii) Consider an airplane flying at 8 km altitude at a flight speed of 220 m/s.

The Mach number of this flight would be: 220/308.06 = 0.714



(iii) Further if the reference chord of the wing (cref) of this airplane be 3.9 m,

the Reynolds number in this flight, based on cref, would be:

6refe -5

V c 220×3.9R = = = 29.51×10

2.9072×10

(iv) For calculation of values at 16 km altitude, the values of temperature,

pressure and density are needed at the tropopause viz. at h=11 km.

Now 11T = 288.15-0.0065×11000 = 216.65 K

5.25588 211p = 101325 216.65/288.15 = 22632 N/m

311ρ = 22632/ 287.05287×216.65 = 0.36392 kg/m

(b) h = 16 km

In lower stratosphere Eq.(2.10) gives :

1111 11

p ρ= = exp -g h-11000 /RT

p ρ

Consequently,

16 16

11 11

p ρ= = exp -9.80665 16000-11000 / 287.05287×216.65 = 0.45455

p ρ

Or 216p = 22632×0.45455 = 10287 N/m

316ρ = 0.36392×0.45455 = 0.16541kg/m

16δ = 10287 /101325 = 0.10153

16σ = 0.16541/1.225 = 0.13503

0.5

16a = 1.4×287.05287×216.65 = 295.07m/s

1.5

-6 -5 -1 -116

216.65μ = 1.458×10 = 1.4216×10 kg m s

216.65+110.4

-5 -5 216 = 1.4216×10 / 0.16541 = 8.594×10 m /s

Remark :

To calculate the required values at 24 km altitude, the values of T and p are

needed at h = 20 km. These values are :

T20 = 216.65



20

11

p= exp -9.80665 20000-11000 / 287.05287×216.65 = 0.24191

p

Or 220p = 22632 0.24191 = 5474.9 N/m

(c) h = 24 km

24T = 216.65+0.001 24000-20000 = 220.65K

From Eq.(2.11):

-34.16322424 20

20

p= T /T

p

Or -34.1632 224p = 5474.9 220.65/216.65 = 2930.5N/m

24ρ = 2930.5/ 287.05287×220.65 = 0.04627

Hence, 24δ = 2930.5/101325 = 0.02892

and 24σ = 0.04627/1.225 = 0.03777

0.5

24a = 1.4×287.05287×220.65 = 297.78 m/s

1.5-6 -5 -1 -1

24

220.65μ = 1.458×10 = 1.4435×10 kg m s

220.65+110.4

-5 -4 224 = 1.4435×10 / 0.04627 = 3.12×10 m /s

Answers:

h (km) 8 16 24

T (K) 236.15 216.65 220.65

p (N/m2) 35599.5 10287.0 2930.5

0δ = p/p 0.35134 0.10153 0.02892

3ρ kg/m 0.52516 0.16541 0.04627

0σ = ρ/ρ 0.42870 0.13503 0.03777

a (m/s) 308.06 295.07 297.78

-1 -1μ kg m s 1.5268 x 10-5 1.4216 x 10-5 1.4435 x 10-5

2m /s 2.9072 x 10-5 8.594 x 10-5 3.12 x 10-4




The variations of pressure, temperature and density in the atmosphere

were obtained by using the hydrostatic equation (Eq.2.2). In this equation ‘g’ is

assumed to be constant. However, it is known that ‘g’ decreases with altitude.

Equation (1.1) gives the variation as:

0

G

Rg = g ( )

R+h

where ‘R ’ is the radius of earth and ‘hG’ is the geometric altitude above earth’s

surface.

Thus, the values of p and ρ obtained by assuming g = 0

g are at an

altitude slightly different from the geometrical altitude (hG). This altitude is called

geopotential altitude, which for convenience is denoted by ‘h’. Following Ref.1,

the geopotential altitude can be defined as the height above earth’s surface in

units, proportional to the potential energy of unit mass (geopotential), relative to

sea level. It can be shown that the geopotential altitude (h) is given, in terms of

geometric altitude (hG), by the following relation. Reference 1.13, chapter 3 may

be referred to for derivation.

GR

h = hR-h

It may be remarked that the actual difference between h and hG is small

for altitudes involved in flight dynamics; for h of 20 km, hG would be 20.0627 km.

Hence, the difference is ignored in performance analysis.

2.6 General remarks:


It will be evident from chapters 4 to 10 that the engine characteristics and

the airplane performance depend on atmospheric characteristics. Noting that ISA

only represents average atmospheric conditions, other atmospheric models have

been proposed as guidelines for extreme conditions in arctic and tropical regions.

Figure 2.3 shows the temperature variations with altitude in arctic and tropical

atmospheres along with ISA. It is seen that the arctic minimum atmosphere has

the following features. (a) The sea level temperature is -500C (b) The



temperature increases at the rate of 10 K per km up to 1500 m altitude. (c) The

temperature remains constant at -350C up to 3000 m altitude. (d) Then the

temperature decreases at the rate of 4.72 K per km up to 15.5 km altitude (e)

The tropopause in this case is at 15.5 km and the temperature there is -940c.

The features of the tropical maximum atmosphere are as follows.

(a) Sea level temperature is 450 C.

(b) The temperature decreases at the rate of 6.5 K per km up to 11.54 km

and then remains constant at -300 C.

Fig.2.3 Temperature variations in arctic minimum, ISA and tropical maximum

atmospheres (Reproduced from Ref.1.7, Chapter 3 with permission of author)

Note:

(a) The local temperature varies with latitude but the sea level pressure (p0)

depends on the weight of air above and is taken same at all the places i.e.

101325 N/m2. Knowing p0 and T0, and the temperature lapse rates, the pressure,

temperature and density in tropospheres of arctic minimum and tropical



maximum can be obtained using Eqs. (2.4), (2.6) and (2.7). (see also exercise

2.1).

(b) Some airlines/ air forces may prescribe intermediate values of sea level

temperature e.g. ISA +150C or ISA +200C. The variations of pressure,

temperature and density with altitude in these cases can also be worked out from

the aforesaid equations.


It is generally assumed that the air mass is stationary. However, some

packets of air mass may acquire motion due to local changes. For example, due

to absorption of solar radiation by the earth’s surface, an air mass adjacent to the

surface may become lighter and buoyancy may cause it to rise. If the

atmosphere is stable, a rising packet of air must come back to its original

position. On the other hand, if the air packet remains in the disturbed position,

then the atmosphere is neutrally stable. If the rising packet continues to move up

then the atmosphere is unstable.

Reference 1.7, chapter 3 analyses the problem of atmospheric stability

and concludes that if the temperature lapse rate is less than 9.75 K per km, then

the atmosphere is stable. It is seen that the three atmospheres, representing

different conditions, shown in Fig.2.3 are stable.



Altit-

ude

(m)

Tempe-

rature

(K)

Pressure

(N/m2)

δ

(p/po)

Density

(kg/m3)

σ

(ρ/ρo)

speed

of

sound

(m/s)

Kinematic

viscosity

(m2/s)

0 288.15 101325.0 1.00000 1.22500 1.00000 340.29 1.4607E-005

200 286.85 98945.3 0.97651 1.20165 0.98094 339.53 1.4839E-005

400 285.55 96611.0 0.95348 1.17864 0.96216 338.76 1.5075E-005

600 284.25 94321.6 0.93088 1.15598 0.94365 337.98 1.5316E-005

800 282.95 92076.3 0.90872 1.13364 0.92542 337.21 1.5562E-005

1000 281.65 89874.4 0.88699 1.11164 0.90746 336.43 1.5813E-005

1200 280.35 87715.4 0.86568 1.08997 0.88977 335.66 1.6069E-005

1400 279.05 85598.6 0.84479 1.06862 0.87234 334.88 1.6331E-005

1600 277.75 83523.3 0.82431 1.04759 0.85518 334.10 1.6598E-005

1800 276.45 81489.0 0.80423 1.02688 0.83827 333.31 1.6870E-005

2000 275.15 79494.9 0.78455 1.00649 0.82162 332.53 1.7148E-005

2200 273.85 77540.6 0.76527 0.98640 0.80523 331.74 1.7432E-005

2400 272.55 75625.4 0.74636 0.96663 0.78908 330.95 1.7723E-005

2600 271.25 73748.6 0.72784 0.94716 0.77319 330.16 1.8019E-005

2800 269.95 71909.7 0.70969 0.92799 0.75754 329.37 1.8321E-005

3000 268.65 70108.2 0.69191 0.90912 0.74214 328.58 1.8630E-005

3200 267.35 68343.3 0.67450 0.89054 0.72697 327.78 1.8946E-005

3400 266.05 66614.6 0.65744 0.87226 0.71205 326.98 1.9269E-005

3600 264.75 64921.5 0.64073 0.85426 0.69736 326.18 1.9598E-005

3800 263.45 63263.4 0.62436 0.83655 0.68290 325.38 1.9935E-005

4000 262.15 61639.8 0.60834 0.81912 0.66867 324.58 2.0279E-005

4200 260.85 60050.0 0.59265 0.80197 0.65467 323.77 2.0631E-005

4400 259.55 58493.7 0.57729 0.78510 0.64090 322.97 2.0990E-005

4600 258.25 56970.1 0.56225 0.76850 0.62735 322.16 2.1358E-005

4800 256.95 55478.9 0.54753 0.75217 0.61402 321.34 2.1734E-005

Table 2.1 Atmospheric properties in ISA (Cont..)



5000 255.65 54019.4 0.53313 0.73611 0.60091 320.53 2.2118E-005

5200 254.35 52591.2 0.51903 0.72031 0.58801 319.71 2.2511E-005

5400 253.05 51193.7 0.50524 0.70477 0.57532 318.90 2.2913E-005

5600 251.75 49826.4 0.49175 0.68949 0.56285 318.08 2.3324E-005

5800 250.45 48488.8 0.47855 0.67446 0.55058 317.25 2.3744E-005

6000 249.15 47180.5 0.46564 0.65969 0.53852 316.43 2.4174E-005

6200 247.85 45900.9 0.45301 0.64516 0.52666 315.60 2.4614E-005

6400 246.55 44649.5 0.44066 0.63088 0.51501 314.77 2.5064E-005

6600 245.25 43425.9 0.42858 0.61685 0.50355 313.94 2.5525E-005

6800 243.95 42229.6 0.41677 0.60305 0.49229 313.11 2.5997E-005

7000 242.65 41060.2 0.40523 0.58949 0.48122 312.27 2.6480E-005

7200 241.35 39917.1 0.39395 0.57617 0.47034 311.44 2.6974E-005

7400 240.05 38799.9 0.38292 0.56308 0.45965 310.60 2.7480E-005

7600 238.75 37708.1 0.37215 0.55021 0.44915 309.75 2.7998E-005

7800 237.45 36641.4 0.36162 0.53757 0.43884 308.91 2.8529E-005

8000 236.15 35599.2 0.35134 0.52516 0.42870 308.06 2.9073E-005

8200 234.85 34581.2 0.34129 0.51296 0.41875 307.21 2.9629E-005

8400 233.55 33586.9 0.33148 0.50099 0.40897 306.36 3.0200E-005

8600 232.25 32615.8 0.32189 0.48923 0.39937 305.51 3.0784E-005

8800 230.95 31667.6 0.31254 0.47768 0.38994 304.65 3.1383E-005

9000 229.65 30741.9 0.30340 0.46634 0.38069 303.79 3.1997E-005

9200 228.35 29838.2 0.29448 0.45521 0.37160 302.93 3.2627E-005

9400 227.05 28956.1 0.28577 0.44428 0.36268 302.07 3.3272E-005

9600 225.75 28095.2 0.27728 0.43355 0.35392 301.20 3.3933E-005

9800 224.45 27255.2 0.26899 0.42303 0.34533 300.33 3.4611E-005

10000 223.15 26435.7 0.26090 0.41270 0.33690 299.46 3.5307E-005

10200 221.85 25636.2 0.25301 0.40256 0.32862 298.59 3.6020E-005

10400 220.55 24856.4 0.24531 0.39262 0.32050 297.71 3.6752E-005

10600 219.25 24096.0 0.23781 0.38286 0.31254 296.83 3.7503E-005




10800 217.95 23354.4 0.23049 0.37329 0.30473 295.95 3.8274E-005

11000 216.65 22631.5 0.22336 0.36391 0.29707 295.07 3.9065E-005

11200 216.65 21929.4 0.21643 0.35262 0.28785 295.07 4.0316E-005

11400 216.65 21248.6 0.20971 0.34167 0.27892 295.07 4.1608E-005

11600 216.65 20588.9 0.20320 0.33106 0.27026 295.07 4.2941E-005

11800 216.65 19949.7 0.19689 0.32079 0.26187 295.07 4.4317E-005

12000 216.65 19330.4 0.19078 0.31083 0.25374 295.07 4.5736E-005

12200 216.65 18730.2 0.18485 0.30118 0.24586 295.07 4.7202E-005

12400 216.65 18148.7 0.17911 0.29183 0.23823 295.07 4.8714E-005

12600 216.65 17585.3 0.17355 0.28277 0.23083 295.07 5.0275E-005

12800 216.65 17039.4 0.16817 0.27399 0.22366 295.07 5.1886E-005

13000 216.65 16510.4 0.16294 0.26548 0.21672 295.07 5.3548E-005

13200 216.65 15997.8 0.15789 0.25724 0.20999 295.07 5.5264E-005

13400 216.65 15501.1 0.15298 0.24925 0.20347 295.07 5.7035E-005

13600 216.65 15019.9 0.14823 0.24152 0.19716 295.07 5.8862E-005

13800 216.65 14553.6 0.14363 0.23402 0.19104 295.07 6.0748E-005

14000 216.65 14101.8 0.13917 0.22675 0.18510 295.07 6.2694E-005

14200 216.65 13664.0 0.13485 0.21971 0.17936 295.07 6.4703E-005

14400 216.65 13239.8 0.13067 0.21289 0.17379 295.07 6.6776E-005

14600 216.65 12828.7 0.12661 0.20628 0.16839 295.07 6.8916E-005

14800 216.65 12430.5 0.12268 0.19988 0.16317 295.07 7.1124E-005

15000 216.65 12044.6 0.11887 0.19367 0.15810 295.07 7.3403E-005

15200 216.65 11670.6 0.11518 0.18766 0.15319 295.07 7.5754E-005

15400 216.65 11308.3 0.11160 0.18183 0.14844 295.07 7.8182E-005

15600 216.65 10957.2 0.10814 0.17619 0.14383 295.07 8.0687E-005

15800 216.65 10617.1 0.10478 0.17072 0.13936 295.07 8.3272E-005

16000 216.65 10287.5 0.10153 0.16542 0.13504 295.07 8.5940E-005

16200 216.65 9968.1 0.09838 0.16028 0.13084 295.07 8.8693E-005

16400 216.65 9658.6 0.09532 0.15531 0.12678 295.07 9.1535E-005




16600 216.65 9358.8 0.09236 0.15049 0.12285 295.07 9.4468E-005

16800 216.65 9068.2 0.08950 0.14581 0.11903 295.07 9.7495E-005

17000 216.65 8786.7 0.08672 0.14129 0.11534 295.07 1.0062E-004

17200 216.65 8513.9 0.08403 0.13690 0.11176 295.07 1.0384E-004

17400 216.65 8249.6 0.08142 0.13265 0.10829 295.07 1.0717E-004

17600 216.65 7993.5 0.07889 0.12853 0.10492 295.07 1.1060E-004

17800 216.65 7745.3 0.07644 0.12454 0.10167 295.07 1.1415E-004

18000 216.65 7504.8 0.07407 0.12068 0.09851 295.07 1.1780E-004

18200 216.65 7271.9 0.07177 0.11693 0.09545 295.07 1.2158E-004

18400 216.65 7046.1 0.06954 0.11330 0.09249 295.07 1.2547E-004

18600 216.65 6827.3 0.06738 0.10978 0.08962 295.07 1.2949E-004

18800 216.65 6615.4 0.06529 0.10637 0.08684 295.07 1.3364E-004

19000 216.65 6410.0 0.06326 0.10307 0.08414 295.07 1.3793E-004

19200 216.65 6211.0 0.06130 0.09987 0.08153 295.07 1.4234E-004

19400 216.65 6018.2 0.05939 0.09677 0.07900 295.07 1.4690E-004

19600 216.65 5831.3 0.05755 0.09377 0.07654 295.07 1.5161E-004

19800 216.65 5650.3 0.05576 0.09086 0.07417 295.07 1.5647E-004

20000 216.65 5474.9 0.05403 0.08803 0.07187 295.07 1.6148E-004

20200 216.85 5305.0 0.05236 0.08522 0.06957 295.21 1.6694E-004

20400 217.05 5140.5 0.05073 0.08251 0.06735 295.34 1.7257E-004

20600 217.25 4981.3 0.04916 0.07988 0.06521 295.48 1.7839E-004

20800 217.45 4827.1 0.04764 0.07733 0.06313 295.61 1.8440E-004

21000 217.65 4677.9 0.04617 0.07487 0.06112 295.75 1.9060E-004

21200 217.85 4533.3 0.04474 0.07249 0.05918 295.89 1.9701E-004

21400 218.05 4393.4 0.04336 0.07019 0.05730 296.02 2.0363E-004

21600 218.25 4257.9 0.04202 0.06796 0.05548 296.16 2.1046E-004

21800 218.45 4126.8 0.04073 0.06581 0.05372 296.29 2.1752E-004

22000 218.65 3999.7 0.03947 0.06373 0.05202 296.43 2.2480E-004

22200 218.85 3876.7 0.03826 0.06171 0.05038 296.56 2.3232E-004




22400 219.05 3757.6 0.03708 0.05976 0.04878 296.70 2.4009E-004

22600 219.25 3642.3 0.03595 0.05787 0.04724 296.83 2.4811E-004

22800 219.45 3530.5 0.03484 0.05605 0.04575 296.97 2.5639E-004

23000 219.65 3422.4 0.03378 0.05428 0.04431 297.11 2.6494E-004

23200 219.85 3317.6 0.03274 0.05257 0.04291 297.24 2.7376E-004

23400 220.05 3216.1 0.03174 0.05091 0.04156 297.38 2.8287E-004

23600 220.25 3117.8 0.03077 0.04931 0.04026 297.51 2.9228E-004

23800 220.45 3022.6 0.02983 0.04776 0.03899 297.65 3.0198E-004

24000 220.65 2930.4 0.02892 0.04627 0.03777 297.78 3.1200E-004

24200 220.85 2841.1 0.02804 0.04482 0.03658 297.92 3.2235E-004

24400 221.05 2754.6 0.02719 0.04341 0.03544 298.05 3.3302E-004

24600 221.25 2670.8 0.02636 0.04205 0.03433 298.19 3.4404E-004

24800 221.45 2589.6 0.02556 0.04074 0.03325 298.32 3.5542E-004

25000 221.65 2510.9 0.02478 0.03946 0.03222 298.45 3.6716E-004

25200 221.85 2434.7 0.02403 0.03823 0.03121 298.59 3.7927E-004

25400 222.05 2360.9 0.02330 0.03704 0.03024 298.72 3.9178E-004

25600 222.25 2289.4 0.02259 0.03589 0.02929 298.86 4.0468E-004

25800 222.45 2220.1 0.02191 0.03477 0.02838 298.99 4.1800E-004

26000 222.65 2153.0 0.02125 0.03369 0.02750 299.13 4.3174E-004

26200 222.85 2087.9 0.02061 0.03264 0.02664 299.26 4.4593E-004

26400 223.05 2024.9 0.01998 0.03163 0.02582 299.40 4.6056E-004

26600 223.25 1963.9 0.01938 0.03064 0.02502 299.53 4.7566E-004

26800 223.45 1904.7 0.01880 0.02969 0.02424 299.66 4.9124E-004

27000 223.65 1847.3 0.01823 0.02878 0.02349 299.80 5.0732E-004

27200 223.85 1791.8 0.01768 0.02788 0.02276 299.93 5.2391E-004

27400 224.05 1737.9 0.01715 0.02702 0.02206 300.07 5.4102E-004

27600 224.25 1685.8 0.01664 0.02619 0.02138 300.20 5.5868E-004

27800 224.45 1635.2 0.01614 0.02538 0.02072 300.33 5.7690E-004

28000 224.65 1586.2 0.01565 0.02460 0.02008 300.47 5.9569E-004

Table 2.1 Atmospheric properties in ISA (Cont…)



28200 224.85 1538.7 0.01519 0.02384 0.01946 300.60 6.1508E-004

28400 225.05 1492.6 0.01473 0.02311 0.01886 300.74 6.3508E-004

28600 225.25 1448.0 0.01429 0.02239 0.01828 300.87 6.5572E-004

28800 225.45 1404.8 0.01386 0.02171 0.01772 301.00 6.7700E-004

29000 225.65 1362.9 0.01345 0.02104 0.01718 301.14 6.9896E-004

29200 225.85 1322.2 0.01305 0.02040 0.01665 301.27 7.2161E-004

29400 226.05 1282.8 0.01266 0.01977 0.01614 301.40 7.4497E-004

29600 226.25 1244.7 0.01228 0.01916 0.01564 301.54 7.6906E-004

29800 226.45 1207.6 0.01192 0.01858 0.01517 301.67 7.9391E-004

30000 226.65 1171.8 0.01156 0.01801 0.01470 301.80 8.1954E-004

30200 226.85 1137.0 0.01122 0.01746 0.01425 301.94 8.4598E-004

30400 227.05 1103.3 0.01089 0.01693 0.01382 302.07 8.7324E-004

30600 227.25 1070.6 0.01057 0.01641 0.01340 302.20 9.0136E-004

30800 227.45 1038.9 0.01025 0.01591 0.01299 302.33 9.3035E-004

31000 227.65 1008.1 0.00995 0.01543 0.01259 302.47 9.6026E-004

31200 227.85 978.3 0.00966 0.01496 0.01221 302.60 9.9109E-004

31400 228.05 949.5 0.00937 0.01450 0.01184 302.73 1.0229E-003

31600 228.25 921.4 0.00909 0.01406 0.01148 302.87 1.0557E-003

31800 228.45 894.3 0.00883 0.01364 0.01113 303.00 1.0895E-003

32000 228.65 867.9 0.00857 0.01322 0.01079 303.13 1.1243E-003

Table 2.1 Atmospheric properties in ISA

Note: Following values / expressions have been used while preparing ISA table.

2 -2

2

R=287.05287m sec K

g= 9.80665m/s

Sutherland formula for viscosity:

3/2-6 T

μ = 1.458X10 [ ]T+110.4



In troposphere (h = 0 to 11000 m): T= 288.15 - 0.0065 h.

p = 101325 [1-0.000022588h] 5.25588

ρ = 1.225 [1-0.000022588h]4.25588 . In lower stratosphere (h = 11000 to 20000 km): T=216.65 K. p = 22632 exp {-0.000157688 (h-11000)} ρ = 0.36391 exp {-0.000157688 (h-11000)} In middle stratosphere (h = 20000 to 32000 km): T = 216.65 + 0.001h p = 5474.9 [1+0.000004616(h-20000)]-34.1632

ρ = 0.08803 [1+0.000004616(h-20000)]-35.1632



Chapter 2

Reference

2.1 Gunston, B, “The Cambridge aerospace dictionary” Cambridge University

Press (2004).



Chapter 2

Exercises

2.1 On a certain day the pressure at sea level is 758 mm of mercury

(101059 N / m2) and the temperature is 25oC. The temperature is found to fall

linearly with height to -55oC at 12km and after that it remains constant upto

20 km. Calculate the pressure, density and kinematic viscosity at 8km and 16km

altitude.

(Hint : When the temperature variation is linear, Eqs. (2.6) and (2.7) can be used

to obtain the pressure and density at a chosen altitude by using appropriate

values of p0, T0, 0ρ and λ . As regards the constant temperature region, an

equation similar to Eq (2.10) can be used; note that, in this exercise, the

tropopause is at 12 km altitude)

[Answers:

p8 = 36,812 N/m2, 8 = 0.5238 kg/m3, 8 = 3.002 x 10-5 m2/sec,

p 16 = 10897 N/m2, 16 = 0.1740 kg/m3,16 = 8.218 x 10-5 m2/sec]

Remark : Due to round off errors in calculations, the student may get the

numerical values which are slightly different from those given as answers. Values

within 0.5% of those given as answers can be regarded as correct.

2.2 If the altimeter in an airplane reads 5000m, on the day described in exercise

2.1, what is the altitude of airplane above mean sea level? What would be the

indicated altitude after landing on aerodrome at sea level?

(Hint: An altimeter is an instrument which senses the ambient pressure and

indicates height in ISA corresponding to that pressure. It does not read the

correct altitude when the atmospheric conditions differ from ISA.

To solve this exercise, obtain the pressure corresponding to 5000 m altitude in

ISA. Then find the altitude corresponding to this pressure in the atmospheric

conditions prevailing as in exercise 2.1. As regards the second part of this

exercise, the pressure at the sea level on that day is 101059 N/m2. When the

airplane lands at sea level, the altimeter would indicate altitude, in ISA,

corresponding to this pressure. In actual practice, the air traffic control would



inform the pilot about the local ambient pressure and the pilot would adjust zero

reading of his altimeter.)

[Answers: 5152 m, 22.3 m].

2.3 An altimeter calibrated according to ISA reads an altitude of 3,600 m. If the

ambient temperature is –60 C, calculate the ambient density.

[Answer: 0.847 kg/m3].

2.4 During a flight test for climb performance, the following readings were

observed at two altitudes:

Record Number 1 2

Indicate altitude (m) 1,300 1,600

Ambient temperature (0C) 16 14

The altimeter is calibrated according to ISA. Obtain the true difference of height

between the two indicated altitudes.

(Hint: Note that the ambient temperatures are different from those in ISA at 1300

and 1600 m altitudes. Hence the actual altitudes are different from the indicated

altitudes. To get the difference between these two altitudes (Δh), obtain

pressures at 1300 and 1600 m heights in ISA. Let the difference in pressures be

Δp. Calculate density at the two altitudes using corresponding pressures and

temperature. Take average of the two densities (avg). Using Eq. (2.2) :

Δh ≈ -Δp / {avg x g} )

[Answer: 311 m]

Remark:

The difference between the actual altitudes (311 m) and the indicated

altitudes (300 m) is small. Since altimeters of all the airplanes are calibrated

using ISA, the difference between indicated altitudes and actual altitudes of two

airplanes will be small. To take care of any uncertainty, the flight paths of two

airplanes are separated by several hundred meters. However, with the

availability of Global Positioning System (GPS) the separation between two

airplanes can be reduced.

2.5 A light airplane is flying at a speed of 220 kmph at an altitude of 3.2 km.

Assuming ISA conditions and the mean chord of the wing to be 1.5 m, obtain the



Reynolds number, based on wing mean chord, and the Mach number in this

flight.

[Answers: Re = 4.83 x 106, M = 0.186]



Chapter 2




Altit-

ude

(m)

Tempe-

rature

(K)

Pressure

(N/m2)

δ

(p/po)

Density

(kg/m3)

σ

(ρ/ρo)

speed

of

sound

(m/s)

Kinematic

viscosity

(m2/s)

0 288.15 101325.0 1.00000 1.22500 1.00000 340.29 1.4607E-005

200 286.85 98945.3 0.97651 1.20165 0.98094 339.53 1.4839E-005

400 285.55 96611.0 0.95348 1.17864 0.96216 338.76 1.5075E-005

600 284.25 94321.6 0.93088 1.15598 0.94365 337.98 1.5316E-005

800 282.95 92076.3 0.90872 1.13364 0.92542 337.21 1.5562E-005

1000 281.65 89874.4 0.88699 1.11164 0.90746 336.43 1.5813E-005

1200 280.35 87715.4 0.86568 1.08997 0.88977 335.66 1.6069E-005

1400 279.05 85598.6 0.84479 1.06862 0.87234 334.88 1.6331E-005

1600 277.75 83523.3 0.82431 1.04759 0.85518 334.10 1.6598E-005

1800 276.45 81489.0 0.80423 1.02688 0.83827 333.31 1.6870E-005

2000 275.15 79494.9 0.78455 1.00649 0.82162 332.53 1.7148E-005

2200 273.85 77540.6 0.76527 0.98640 0.80523 331.74 1.7432E-005

2400 272.55 75625.4 0.74636 0.96663 0.78908 330.95 1.7723E-005

2600 271.25 73748.6 0.72784 0.94716 0.77319 330.16 1.8019E-005

2800 269.95 71909.7 0.70969 0.92799 0.75754 329.37 1.8321E-005

3000 268.65 70108.2 0.69191 0.90912 0.74214 328.58 1.8630E-005

3200 267.35 68343.3 0.67450 0.89054 0.72697 327.78 1.8946E-005

3400 266.05 66614.6 0.65744 0.87226 0.71205 326.98 1.9269E-005

3600 264.75 64921.5 0.64073 0.85426 0.69736 326.18 1.9598E-005

3800 263.45 63263.4 0.62436 0.83655 0.68290 325.38 1.9935E-005

4000 262.15 61639.8 0.60834 0.81912 0.66867 324.58 2.0279E-005

4200 260.85 60050.0 0.59265 0.80197 0.65467 323.77 2.0631E-005

4400 259.55 58493.7 0.57729 0.78510 0.64090 322.97 2.0990E-005

4600 258.25 56970.1 0.56225 0.76850 0.62735 322.16 2.1358E-005




4800 256.95 55478.9 0.54753 0.75217 0.61402 321.34 2.1734E-005

5000 255.65 54019.4 0.53313 0.73611 0.60091 320.53 2.2118E-005

5200 254.35 52591.2 0.51903 0.72031 0.58801 319.71 2.2511E-005

5400 253.05 51193.7 0.50524 0.70477 0.57532 318.90 2.2913E-005

5600 251.75 49826.4 0.49175 0.68949 0.56285 318.08 2.3324E-005

5800 250.45 48488.8 0.47855 0.67446 0.55058 317.25 2.3744E-005

6000 249.15 47180.5 0.46564 0.65969 0.53852 316.43 2.4174E-005

6200 247.85 45900.9 0.45301 0.64516 0.52666 315.60 2.4614E-005

6400 246.55 44649.5 0.44066 0.63088 0.51501 314.77 2.5064E-005

6600 245.25 43425.9 0.42858 0.61685 0.50355 313.94 2.5525E-005

6800 243.95 42229.6 0.41677 0.60305 0.49229 313.11 2.5997E-005

7000 242.65 41060.2 0.40523 0.58949 0.48122 312.27 2.6480E-005

7200 241.35 39917.1 0.39395 0.57617 0.47034 311.44 2.6974E-005

7400 240.05 38799.9 0.38292 0.56308 0.45965 310.60 2.7480E-005

7600 238.75 37708.1 0.37215 0.55021 0.44915 309.75 2.7998E-005

7800 237.45 36641.4 0.36162 0.53757 0.43884 308.91 2.8529E-005

8000 236.15 35599.2 0.35134 0.52516 0.42870 308.06 2.9073E-005

8200 234.85 34581.2 0.34129 0.51296 0.41875 307.21 2.9629E-005

8400 233.55 33586.9 0.33148 0.50099 0.40897 306.36 3.0200E-005

8600 232.25 32615.8 0.32189 0.48923 0.39937 305.51 3.0784E-005

8800 230.95 31667.6 0.31254 0.47768 0.38994 304.65 3.1383E-005

9000 229.65 30741.9 0.30340 0.46634 0.38069 303.79 3.1997E-005

9200 228.35 29838.2 0.29448 0.45521 0.37160 302.93 3.2627E-005

9400 227.05 28956.1 0.28577 0.44428 0.36268 302.07 3.3272E-005

9600 225.75 28095.2 0.27728 0.43355 0.35392 301.20 3.3933E-005

9800 224.45 27255.2 0.26899 0.42303 0.34533 300.33 3.4611E-005

10000 223.15 26435.7 0.26090 0.41270 0.33690 299.46 3.5307E-005

10200 221.85 25636.2 0.25301 0.40256 0.32862 298.59 3.6020E-005

10400 220.55 24856.4 0.24531 0.39262 0.32050 297.71 3.6752E-005




10600 219.25 24096.0 0.23781 0.38286 0.31254 296.83 3.7503E-005

10800 217.95 23354.4 0.23049 0.37329 0.30473 295.95 3.8274E-005

11000 216.65 22631.5 0.22336 0.36391 0.29707 295.07 3.9065E-005

11200 216.65 21929.4 0.21643 0.35262 0.28785 295.07 4.0316E-005

11400 216.65 21248.6 0.20971 0.34167 0.27892 295.07 4.1608E-005

11600 216.65 20588.9 0.20320 0.33106 0.27026 295.07 4.2941E-005

11800 216.65 19949.7 0.19689 0.32079 0.26187 295.07 4.4317E-005

12000 216.65 19330.4 0.19078 0.31083 0.25374 295.07 4.5736E-005

12200 216.65 18730.2 0.18485 0.30118 0.24586 295.07 4.7202E-005

12400 216.65 18148.7 0.17911 0.29183 0.23823 295.07 4.8714E-005

12600 216.65 17585.3 0.17355 0.28277 0.23083 295.07 5.0275E-005

12800 216.65 17039.4 0.16817 0.27399 0.22366 295.07 5.1886E-005

13000 216.65 16510.4 0.16294 0.26548 0.21672 295.07 5.3548E-005

13200 216.65 15997.8 0.15789 0.25724 0.20999 295.07 5.5264E-005

13400 216.65 15501.1 0.15298 0.24925 0.20347 295.07 5.7035E-005

13600 216.65 15019.9 0.14823 0.24152 0.19716 295.07 5.8862E-005

13800 216.65 14553.6 0.14363 0.23402 0.19104 295.07 6.0748E-005

14000 216.65 14101.8 0.13917 0.22675 0.18510 295.07 6.2694E-005

14200 216.65 13664.0 0.13485 0.21971 0.17936 295.07 6.4703E-005

14400 216.65 13239.8 0.13067 0.21289 0.17379 295.07 6.6776E-005

14600 216.65 12828.7 0.12661 0.20628 0.16839 295.07 6.8916E-005

14800 216.65 12430.5 0.12268 0.19988 0.16317 295.07 7.1124E-005

15000 216.65 12044.6 0.11887 0.19367 0.15810 295.07 7.3403E-005

15200 216.65 11670.6 0.11518 0.18766 0.15319 295.07 7.5754E-005

15400 216.65 11308.3 0.11160 0.18183 0.14844 295.07 7.8182E-005

15600 216.65 10957.2 0.10814 0.17619 0.14383 295.07 8.0687E-005

15800 216.65 10617.1 0.10478 0.17072 0.13936 295.07 8.3272E-005

16000 216.65 10287.5 0.10153 0.16542 0.13504 295.07 8.5940E-005

16200 216.65 9968.1 0.09838 0.16028 0.13084 295.07 8.8693E-005




16400 216.65 9658.6 0.09532 0.15531 0.12678 295.07 9.1535E-005

16600 216.65 9358.8 0.09236 0.15049 0.12285 295.07 9.4468E-005

16800 216.65 9068.2 0.08950 0.14581 0.11903 295.07 9.7495E-005

17000 216.65 8786.7 0.08672 0.14129 0.11534 295.07 1.0062E-004

17200 216.65 8513.9 0.08403 0.13690 0.11176 295.07 1.0384E-004

17400 216.65 8249.6 0.08142 0.13265 0.10829 295.07 1.0717E-004

17600 216.65 7993.5 0.07889 0.12853 0.10492 295.07 1.1060E-004

17800 216.65 7745.3 0.07644 0.12454 0.10167 295.07 1.1415E-004

18000 216.65 7504.8 0.07407 0.12068 0.09851 295.07 1.1780E-004

18200 216.65 7271.9 0.07177 0.11693 0.09545 295.07 1.2158E-004

18400 216.65 7046.1 0.06954 0.11330 0.09249 295.07 1.2547E-004

18600 216.65 6827.3 0.06738 0.10978 0.08962 295.07 1.2949E-004

18800 216.65 6615.4 0.06529 0.10637 0.08684 295.07 1.3364E-004

19000 216.65 6410.0 0.06326 0.10307 0.08414 295.07 1.3793E-004

19200 216.65 6211.0 0.06130 0.09987 0.08153 295.07 1.4234E-004

19400 216.65 6018.2 0.05939 0.09677 0.07900 295.07 1.4690E-004

19600 216.65 5831.3 0.05755 0.09377 0.07654 295.07 1.5161E-004

19800 216.65 5650.3 0.05576 0.09086 0.07417 295.07 1.5647E-004

20000 216.65 5474.9 0.05403 0.08803 0.07187 295.07 1.6148E-004

20200 216.85 5305.0 0.05236 0.08522 0.06957 295.21 1.6694E-004

20400 217.05 5140.5 0.05073 0.08251 0.06735 295.34 1.7257E-004

20600 217.25 4981.3 0.04916 0.07988 0.06521 295.48 1.7839E-004

20800 217.45 4827.1 0.04764 0.07733 0.06313 295.61 1.8440E-004

21000 217.65 4677.9 0.04617 0.07487 0.06112 295.75 1.9060E-004

21200 217.85 4533.3 0.04474 0.07249 0.05918 295.89 1.9701E-004

21400 218.05 4393.4 0.04336 0.07019 0.05730 296.02 2.0363E-004

21600 218.25 4257.9 0.04202 0.06796 0.05548 296.16 2.1046E-004

21800 218.45 4126.8 0.04073 0.06581 0.05372 296.29 2.1752E-004

22000 218.65 3999.7 0.03947 0.06373 0.05202 296.43 2.2480E-004




22200 218.85 3876.7 0.03826 0.06171 0.05038 296.56 2.3232E-004

22400 219.05 3757.6 0.03708 0.05976 0.04878 296.70 2.4009E-004

22600 219.25 3642.3 0.03595 0.05787 0.04724 296.83 2.4811E-004

22800 219.45 3530.5 0.03484 0.05605 0.04575 296.97 2.5639E-004

23000 219.65 3422.4 0.03378 0.05428 0.04431 297.11 2.6494E-004

23200 219.85 3317.6 0.03274 0.05257 0.04291 297.24 2.7376E-004

23400 220.05 3216.1 0.03174 0.05091 0.04156 297.38 2.8287E-004

23600 220.25 3117.8 0.03077 0.04931 0.04026 297.51 2.9228E-004

23800 220.45 3022.6 0.02983 0.04776 0.03899 297.65 3.0198E-004

24000 220.65 2930.4 0.02892 0.04627 0.03777 297.78 3.1200E-004

24200 220.85 2841.1 0.02804 0.04482 0.03658 297.92 3.2235E-004

24400 221.05 2754.6 0.02719 0.04341 0.03544 298.05 3.3302E-004

24600 221.25 2670.8 0.02636 0.04205 0.03433 298.19 3.4404E-004

24800 221.45 2589.6 0.02556 0.04074 0.03325 298.32 3.5542E-004

25000 221.65 2510.9 0.02478 0.03946 0.03222 298.45 3.6716E-004

25200 221.85 2434.7 0.02403 0.03823 0.03121 298.59 3.7927E-004

25400 222.05 2360.9 0.02330 0.03704 0.03024 298.72 3.9178E-004

25600 222.25 2289.4 0.02259 0.03589 0.02929 298.86 4.0468E-004

25800 222.45 2220.1 0.02191 0.03477 0.02838 298.99 4.1800E-004

26000 222.65 2153.0 0.02125 0.03369 0.02750 299.13 4.3174E-004

26200 222.85 2087.9 0.02061 0.03264 0.02664 299.26 4.4593E-004

26400 223.05 2024.9 0.01998 0.03163 0.02582 299.40 4.6056E-004

26600 223.25 1963.9 0.01938 0.03064 0.02502 299.53 4.7566E-004

26800 223.45 1904.7 0.01880 0.02969 0.02424 299.66 4.9124E-004

27000 223.65 1847.3 0.01823 0.02878 0.02349 299.80 5.0732E-004

27200 223.85 1791.8 0.01768 0.02788 0.02276 299.93 5.2391E-004

27400 224.05 1737.9 0.01715 0.02702 0.02206 300.07 5.4102E-004

27600 224.25 1685.8 0.01664 0.02619 0.02138 300.20 5.5868E-004

27800 224.45 1635.2 0.01614 0.02538 0.02072 300.33 5.7690E-004




28000 224.65 1586.2 0.01565 0.02460 0.02008 300.47 5.9569E-004

28200 224.85 1538.7 0.01519 0.02384 0.01946 300.60 6.1508E-004

28400 225.05 1492.6 0.01473 0.02311 0.01886 300.74 6.3508E-004

28600 225.25 1448.0 0.01429 0.02239 0.01828 300.87 6.5572E-004

28800 225.45 1404.8 0.01386 0.02171 0.01772 301.00 6.7700E-004

29000 225.65 1362.9 0.01345 0.02104 0.01718 301.14 6.9896E-004

29200 225.85 1322.2 0.01305 0.02040 0.01665 301.27 7.2161E-004

29400 226.05 1282.8 0.01266 0.01977 0.01614 301.40 7.4497E-004

29600 226.25 1244.7 0.01228 0.01916 0.01564 301.54 7.6906E-004

29800 226.45 1207.6 0.01192 0.01858 0.01517 301.67 7.9391E-004

30000 226.65 1171.8 0.01156 0.01801 0.01470 301.80 8.1954E-004

30200 226.85 1137.0 0.01122 0.01746 0.01425 301.94 8.4598E-004

30400 227.05 1103.3 0.01089 0.01693 0.01382 302.07 8.7324E-004

30600 227.25 1070.6 0.01057 0.01641 0.01340 302.20 9.0136E-004

30800 227.45 1038.9 0.01025 0.01591 0.01299 302.33 9.3035E-004

31000 227.65 1008.1 0.00995 0.01543 0.01259 302.47 9.6026E-004

31200 227.85 978.3 0.00966 0.01496 0.01221 302.60 9.9109E-004

31400 228.05 949.5 0.00937 0.01450 0.01184 302.73 1.0229E-003

31600 228.25 921.4 0.00909 0.01406 0.01148 302.87 1.0557E-003

31800 228.45 894.3 0.00883 0.01364 0.01113 303.00 1.0895E-003

32000 228.65 867.9 0.00857 0.01322 0.01079 303.13 1.1243E-003


Note: Following values / expressions have been used while preparing ISA table.

2 -2

2

R=287.05287m sec K

g= 9.80665m/s

Sutherland formula for viscosity:

3/2-6 T

μ = 1.458X10 [ ]T+110.4



In troposphere (h = 0 to 11000 m): T= 288.15 - 0.0065 h.

p = 101325 [1-0.000022588h] 5.25588

ρ = 1.225 [1-0.000022588h]4.25588 . In lower stratosphere (h = 11000 to 20000 km): T=216.65 K. p = 22632 exp {-0.000157688 (h-11000)} ρ = 0.36391 exp {-0.000157688 (h-11000)} In middle stratosphere (h = 20000 to 32000 km): T = 216.65 + 0.001h p = 5474.9 [1+0.000004616(h-20000)]-34.1632

ρ = 0.08803 [1+0.000004616(h-20000)]-35.1632



Chapter 3

Drag polar

(Lectures 6 to 12)

Keywords: Various types of drags; streamlined body and bluff body; boundary

layers; airfoil characteristics and designations; drags of airplane components;

drag polars at subsonic, transonic, supersonic and hypersonic speeds; high lift

devices

Topics

3.1. Introduction- Need and definition of drag polar

3.1.1 Contributions to airplane drag

3.1.2 Interference drag

3.1.3 Contributions to airplane lift

3.1.4 Contributions to airplane pitching moment

3.1.5 Drag coefficient, lift coefficient and pitching moment coefficient of the

airplane

3.1.6 Categorization of airplane components

3.2 Estimation of drag polar at low subsonic speeds

3.2.1 Angle of attack of airplane, wing incidence and tail incidence

3.2.2 Skin friction drag, pressure drag and profile drag of an airfoil

3.2.3 Summary of lift coefficient, drag coefficient, pitching moment

coefficient, centre of pressure and aerodynamic centre of an airfoil

3.2.4 Examples of pressure coefficient distributions

3.2.5 Introduction to boundary layer theory

3.2.6 Boundary layer over a flat plate – height of boundary layer,

displacement thickness and skin friction drag

3.2.7 Boundary layer separation, adverse pressure gradient and

favourable pressure gradient

3.2.8 Boundary layer transition

3.2.9 Turbulent boundary layer over a flat plate



3.2.10 General remarks on boundary layers

3.2.11 Presentation of aerodynamic characteristics of airfoils

3.2.12 Geometric characteristics of airfoils

3.2.13 Airfoil nomenclature\designation

3.2.14 Induced drag of wing

3.2.15 Drag coefficient of fuselage

3.2.16 Drag coefficients of other components

3.2.17 Parabolic drag polar, parasite drag, induced drag and Oswald

efficiency factor

3.2.18 Parasite drag area and equivalent skin friction coefficient

3.2.19 A note on estimation of minimum drag coefficients of wings and

bodies

3.2.20 Typical values of CDO, A, e and subsonic drag polar

3.2.21 Winglets and their effect on induced drag

3.3 Drag polar at high subsonic, transonic and supersonic speeds

3.3.1 Some aspects of supersonic flow - shock wave, expansion fan

and bow shock

3.3.2 Drag at supersonic speeds

3.3.3 Transonic flow regime - critical Mach number and drag

divergence Mach number of airfoils, wings and fuselage.

3.3.4 Parabolic drag polar at high speeds

3.3.5 Guidelines for variations of CDo and K for subsonic jet transport

airplanes

3.3.6 Variations of CDo and K for a fighter airplane

3.3.7 Area ruling

3.4 Drag polar at hypersonic speeds

3.5 Lift to drag ratio

3.6 Other types of drags

3.6.1 Cooling drag

3.6.2 Base drag



3.6.3 External stores drag

3.6.4 Leakage drag

3.6.5 Trim drag

3.7 High lift devices

3.7.1 Need for increasing maximum lift coefficient (CLmax)

3.7.2 Factors limiting maximum lift coefficient

3.7.3 Ways to increase maximum lift coefficient viz. increase in camber,

boundary layer control and increase in area

3.7.4 Guidelines for values of maximum lift coefficients of wings with

various high lift devices

References

Exercises



Chapter 3

Lecture 6 Drag polar – 1 Topics

3.1. Introduction- Need and definition of drag polar


3.1.2 Interference drag




airplane


3.2 Estimation of drag polar at low subsonic speeds

3.2.1 Angle of attack of airplane, wing incidence and tail incidence

3.2.2 Skin friction drag, pressure drag and profile drag of an airfoil

3.1 Introduction- Need and definition of drag polar

As mentioned in section 1.9, to obtain the performance of an airplane

requires the value of the drag coefficient of the airplane (CD) when the lift

coefficient (CL) and Mach number (M) are given. The relationship between the

drag coefficient and the lift coefficient is called ‘Drag polar’. It may be pointed out

that aerodynamics generally deals with the drag, lift and pitching moment of

individual components like wing, fuselage etc. Whereas, for the estimation of the

airplane performance the knowledge of the drag, lift and pitching moment of the

entire airplane is required.

Equation (1.6) indicates that the drag coefficient is a function of lift

coefficient (CL), Mach number (M) and Reynolds number (Re). However, for a

given airplane a single drag polar can be used for flights upto critical Mach



number (Ref.1.4 section 10.14); see sections 3.3.3 to 3.3.5 for details of critical

Mach number. For airplanes flying at transonic and supersonic speeds, the drag

polar depends on Mach number. Hence, the usual practice is to obtain the drag

polar of subsonic airplanes at a suitable flight speed (generally the cruising

speed) and for a high speed airplane, the drag polars are obtained at suitable

values of Mach numbers spread over the range of operating Mach numbers.

In this chapter the estimation of the drag polar at subsonic, transonic and

supersonic speeds is discussed. The topic of drag polar at hypersonic speed is

also touched up on.


The usual method to estimate the drag of an airplane is to add the drags

of the major components of the airplane and then apply correction for the

interference effects.

The major components of the airplane which contribute to drag are wing,

fuselage, horizontal tail, vertical tail, nacelle(s) and landing gear.

Thus,

D = Dwing

+ Dfuse

+ Dht + D

vt + D

nac + D

lg + D

etc + D

int (3.1)

where Dwing

, Dfuse

, Dht, D

vt , D

nac and D

lg denote drag due to wing, fuselage,

horizontal tail, vertical tail , nacelle(s) and landing gear respectively.

Detc

includes the drag of items like external fuel tanks, bombs, struts etc.

Dint

is the drag due to interference which is described in the next section.

3.1.2 Intereference drag

While presenting the data on the drag of wing or fuselage or any other

component of the airplane, the data generally refers to the drag of that

component when it is alone in the airstream and free from the influence of any

other component. Whereas, in an airplane, the wing, the fuselage and the tails

are present in close proximity of each other and the flow past one component is

influenced by the others. As a result, the drag of the airplane as a combination of

different components is different from the sum of the drags of individual

components. To appreciate this, let us consider the case examined in Ref.3.1.



Flow past a rather thick airfoil section, shown in Fig.3.1a, is examined at a

Reynolds number of 420,000. The maximum thickness and the chord of the

airfoil are denoted respectively by ‘d’ and ‘c’. The thickness ratio (d/c) for the

airfoil in Fig.3.1a is 33.3%.

The drag coefficient is defined as :

d 212

DC =

ρV cb ; b = span of the airfoil model

The drag coefficient (Cd) is found to be 0.0247.

Subsequently, another identical airfoil is placed side by side with a spacing(s)

as shown in Fig.3.1b. The tests were carried out for different values of s/d. It is

found that for large values of s/d, say s/d > 5, the flows past the two sections do

not interfere and the total drag coefficient of the combination is equal to the sum

of the drags of each airfoil namely (Cd)combination = 0.0494.However, at closer

spacings the results presented in table 3.1 are obtained.

s/d 1.16 1.4 1.8 2.0 2.6 4 5

(Cd)combination 0.1727 0.1194 0.0824 0.0761 0.0627 0.0527 0.0494

Cdint 0.2233 0.070 0.033 0.0267 0.0133 0.0033 0.0

Table 3.1 Interference drag coefficient for different spacings between two airfoils

Note:

(Cd)combination = (Cd)airfoil1 + (Cd)airfoil2 + Cdint

It is evident that Cdint depends on the relative positions and could be very large.



(a) Single airfoil

(b) Configuration with airfoils placed side by side as seen in plan view

(c) Circular cylinder with splitter plate at rear

Fig.3.1 Interference effects



Remarks:

(i)The drag coefficient of the individual airfoil in this example is large as the airfoil

is thick and Reynolds number is rather low. Airfoils used on airplanes would have

thickness ratio (t/c) of 12 to 18% and the values of Cd, for Reynolds number of

6 x 106, would be around 0.006.

(ii) Ways to reduce interference drag

A large number of studies have been carried out on interference drag and

it is found that Dint

can be brought down to 5 to 10% of the sum of the drags of all

components, by giving proper fillets at the junctions of wing and fuselage and

tails and fuselage ( Fig.3.2 ).

Fig.3.2 Reduction of interference drag using fillets



(iii) Favorable interference effect

The interference effects need not always increase the drag. As an

example the drag of a circular cylinder with a splitter plate (Fig.3.1c) is lower than

the drag of a cylinder without it at certain Reynolds numbers (Ref.3.2). In an

another example, the birds flying in formation flight experience lower drag than

when flying individually.

(iv) Chapter VIII of Ref.3.3 can be consulted for additional information on

interference drag.


The main contribution to the lift comes from wing-fuselage combination

and a small contribution from the horizontal tail i.e.

L = Lwing + fuselage + Lht (3.2)

For airplanes with wings having aspect ratio greater than six, the lift due to the

wing-fuselage combination is roughly equal to the lift produced by the gross wing

area. The gross wing area (S) is the planform area of the wing, extended into the

fuselage, up to the plane of the symmetry.


The pitching moment of the airplane is taken about its center of gravity

and denoted by Mcg.

Main contributions to Mcg are from wing, fuselage, nacelle(s) and

horizontal tail i.e.

Mcg

= Mwing

+ Mfuselage

+ Mnac

+ Mht (3.3)


airplane

To obtain the non-dimensional quantities namely drag coefficient (CD), lift

coefficient (CL) and pitching moment coefficient (Cmcg) of the airplane, the

reference quantities are the free stream dynamic pressure (½ ρ 2V ), the gross

wing area (S) and the mean aerodynamic chord of the wing ( c ). Consequently,

cgD L mcg2 2 21 1 1

2 2 2

MD LC = ; C = ; C =

ρV S ρV S ρV Sc (3.4)



However, the drag coefficient and lift coefficient of the individual

components are based on their own reference areas as given below.

(a) For wing, horizontal tail and vertical tail the reference area is their planform

area.

(b)For fuselage, nacelle, fuel tanks, bombs and such other bodies the reference

area is either the wetted area or the frontal area. The wetted area is the area of

the surface of the body in contact with the fluid. The frontal area is the maximum

cross-sectional area of the body.

(c) For other components like landing gear the reference area is given along with

the definition of CD.

Remarks:

(i)The reference area, on which the CD and C

L of an individual component is

based, is also called proper area and denoted by S; the drag coefficient based

on S is denoted by CD

.

(ii)The reference areas for different components are different for the following

reasons. The aim of using non-dimensional quantities like CD is to be able to

predict the characteristics of many similar shapes by carrying out computations

or tests on a few models. For this to be effective, the phenomena causing the

drag must be taken into account while specifying the reference qualities. In this

context the drag of streamline shapes like wing and slender bodies is mainly due

to the skin friction and depends on the wetted area. Whereas, the drag of bluff

bodies like the fuselage of a piston-engined airplane, is mainly the pressure drag

and depends on the frontal area. It may be added that for wings, the usual

practice is to take the reference area as the planform area because it (planform

area) is proportional to the wetted area.

(iii)At this stage the reader is advised to the revise the background on

aerodynamics (see for examples References 1.9, 1.10 and 1.12).

(iv) Following the above remarks, the total drag of the airplane can be expressed

as:



2 2 2 2Dwing fuse Dfuse nac Dnac ht Dht

2 2 2vt Dvt lg Dlg etc Detc int

1 1 1 1D = ρV SC + ρV S C + ρV S C + ρV S C

2 2 2 21 1 1

+ ρV S C + ρV S C + ρV S C +D (3.5)2 2 2

It may be recalled that Setc and CDetc

refer to areas and drag coefficients of other

items like external fuel tanks, bombs, struts etc..

Or D 212

DC =

ρV S

lgfuse ht vt nac etcDwing Dfuse Dht Dvt Dnac Dlg Detc Dint

SS S S S S= C + C + C + C + C + C + C + C (3.6)

S S S S S S

The data on drag, lift and pitching moment, compiled from various sources, is

available in references 1.9, 1.10, 1.12 and 3.3 to 3.9.


During the discussion in the previous section it was mentioned that (a) for

wing, horizontal tail and vertical tail, the planform area is taken as the reference

area, (b) for fuselage, the wetted area or the frontal area is taken as the

reference area. The reason for these specifications lies in the fact that in

aerodynamics the airplane components are categorised as (a) wing type

surfaces, (b) bodies and (c) others. This categorisation, described below, is

based on common geometrical features of certain airplane components.

Figure 3.3 shows the geometric parameters of a wing. It is observed that the

span (b) of the wing is much larger than the chord (c) of the wing section (or the

airfoil) and in turn the chord is much larger than the thickness (t) of the airfoil. For

wings of subsonic airplanes the ratio (b/c) is between 5 to 12 and the ratio (t/c)

for the commonly used profiles is 0.10 to 0.18 or 0.1t/c and 0.1c/b . This

separation of sizes ( or scales in more technical terms) enables the simplification

that the flow past a wing can be analysed as a study of flow past an airfoil and

then applying correction for the effect of finite wing span. It may be recalled that



in aerodynamics an airfoil is treated as a wing of infinite span or a two-

dimensional problem.

Fig.3.3 Geometric parameters of a wing

Hence, subsections 3.2.2 to 3.2.13 deal with various aspects of flow past airfoils

which are relevant to the estimation of drag polar. The subsequent subsection

deals with the induced drag which is the result of finite span. It may be added

that in aerodynamics, the quantity finite aspect ratio (A) is employed instead of

the finite span. The aspect ratio is defined as :

A = b2/S; b = wing span, S = wing planform area

Remarks :

(i)When the aspect ratio is less than about 5, which is characteristic of wings of

high speed airplanes, the flow past the wing has to be treated as three-

dimensional.

(ii) Horizontal tail, vertical tail and streamlined struts, seen on some low speed

airplanes, come under the category of wing type surfaces.

Figure 3.4a shows the fuselage of a jet airplane. Here the length (lf) is much

larger than the height (h) and width (w), but ‘h’ and ‘w’ are generally not very

different in their dimensions. Hence, the flow past a fuselage cannot be

considered as two-dimensional. However, for jet airplanes, lf/h is around 6 to 10

and the analysis of flow past fuselage can be simplified by assuming the fuselage

to be a slender/streamlined body.



Fig.3.4 Fuselage parameters

Figure 3.4 b shows the fuselage of a low speed airplane. Here lf/h is rather low

and the fuselage is treated as a bluff body.

Precise definitions of the streamlined body and bluff body are given in the

subsequent sections.

Remarks:

(i) As regards the analysis of flow is concerned, the fuselage, nacelle, external

fuel tanks, bombs, and antenna masts have common geometric features and are

categorised as “bodies”.

(ii) Components of airplane like landing gear, which do not fall under the above

two categories, are designated as ‘others’.

3.2. Estimation of drag polar at low subsonic speeds

As mentioned in the previous section, the drag polar of an airplane can be

obtained by summing-up the drags of individual components and then adding 5

to 10% for the interference drag. As the drag coefficient depends on the angle of

attack, this exercise has to be carried-out at different angles of attack. The

definition of the angle of attack of the airplane and brief descriptions of the drag



coefficients of the airplane components are presented before discussing the drag

polar.

3.2.1 Angles of attack of the airplane, wing incidence and tail incidence

For defining the angle of attack of an airplane, the fuselage reference

line(FRL) is taken as the airplane reference line (Figs.1.9 and 3.5).The angle

between the free stream velocity and FRL is the angle of attack of the airplane.

However, the angles of attack of the wing and tail are not the same as that of the

fuselage.

The wing is fixed on the fuselage such that it makes an angle, iw, to the

fuselage reference line (Fig.3.5). This angle is called wing incidence. The angle

iw is generally chosen such that during the cruising flight the wing can produce

enough lift when fuselage is at zero angle of attack. This is done because the

fuselage produces least drag when it is at zero angle of attack and that is ideal

during the cruising flight. In other words, during cruise the wing produces the lift

required to balance the weight whereas the fuselage, being at zero angle of

attack, produces least drag.

The horizontal tail is set on fuselage at an angle it (Fig.3.5). This angle is

called tail incidence. It is generally chosen in a manner that during cruise the lift

required from the tail, to make the airplane pitching moment zero, is produced by

the tail without elevator deflection. This is because, the drag, at low angles of

attack, is least when the required lift is produced without elevator deflection.

Remark :

The angles iw and it are measured clockwise from FRL. The angle iw is

positive but the angle it is generally negative.

Fig.3.5 Wing incidence (iw) and tail incidence (it)



3.2.2 Skin friction drag, pressure drag and, profile drag of an airfoil

The drag coefficient of a wing consist of the (i) the profile drag due to

airfoil (Cd) and (ii) the induced drag due to the finite aspect ratio of the wing (CDi).

The symbols dC and Cl with lower case suffices refer to the drag coefficient and

lift coefficient of the airfoil. The profile drag of the airfoil consists of the skin

friction drag and the pressure drag. It may be added that an element on the

surface of an airfoil, kept in a flow, experiences shear stress tangential to the

surface and pressure (p) normal to it (Fig.3.6). The shear stress multiplied by the

area of the element gives the tangential force. The component of this tangential

force in the free stream direction when integrated over the profile gives the skin

friction drag. Similarly, the pressure distribution results in normal force on the

element whose component in the free stream direction, integrated over the profile

Fig.3.6 Shear stress ( ) and pressure(p) on an airfoil

gives the pressure drag. The pressure drag is also called form drag. The sum of

the skin friction drag and the pressure drag is called ‘Profile drag’. The profile

drag depends on the airfoil shape, Reynolds number, angle of attack and surface

roughness.



Chapter 3


3.2.3 Summary of lift coefficient, drag coefficient, pitching moment

coefficient, centre of pressure and aerodynamic centre of an airfoil



3.2.6 Boundary layer over a flat plate – height of boundary layer,


3.2.3 Summary of the lift coefficient, drag coefficient, pressure coefficient,

pitching moment coefficient, centre of pressure and aerodynamic centre of

an airfoil

In order to understand the dependence of pressure drag and skin friction

drag on various factors, it is appropriate, at this stage, to present brief

discussions on (I) generation of lift, drag and pitching moment from the

distributions of pressure (p) and shear stress ( ) and (II) outline of boundary

layer theory. These and the related topics are covered in this subsection and in

the subsections 3.2.4 to 3.2.10. In subsections 3.2.11 to 3.2.13 the airfoil

characteristics and their nomenclature are dealt with. Subsequently, the

estimation of the drags of wing, fuselage and the entire airplane at subsonic

speeds are discussed(sections 3.2.14 to 3.2.21).

Figure 3.7 shows an airfoil at an angle of attack (α )kept in a stream of

velocity V . The resultant aerodynamic force (R) is produced due to the

distributions of the shear stress( ) and the pressure (p). The distributions also

produce a pitching moment (M). By definition, the component of R perpendicular

to the free stream direction is called lift (L) and the component along the free

stream direction is called drag (D). The resulant aerodynamic force (R) can also



be resolved along and perpendicular to the chord of the airfoil. These

components can be denoted by C and N respectively(Fig.3.7). From the

subsequent discussion in this section, it will be evident that it is more convenient

to evaluate N and C from the distributions of shear stress ( ) and pressure (p)

and then evaluate L and D.

Fig.3.7 Aerodynamic forces and moment on an airfoil

From Fig.3.7 it can be deduced that :

L = Ncosα - Csinα (3.7)

D = Nsinα+Ccosα (3.8)

Figure 3.8 shows elements of length and dsu and dsl at points Pu and Pl

on the upper and lower surfaces of the airfoil respectively. The cartesian

coordinates of points Pu and Pl are (xu,yu) and ( xl , yl ) respectively. Whereas su

and sl are respectively the distances along the airfoil surface, of the points Pu

and Pl measured from the stagnation point (Fig.3.8).



Fig.3.8 Pressure and shear stress at typical points on upper

and lower surfaces of an airfoil

To obtain the forces at points Pu and Pl , the local values of p and are

multiplied by the local area. Since the flow past an airfoil is treated as two-

dimensional, the span of the airfoil can be taken as unity without loss of

generality. Hence, the local area is (ds x 1) and the quantities, L, D, N and C, on

the airfoil, are the forces per unit span. Keeping these in mind, the local

contributions, dNu and dCu, to N and C respectively, from the element at point Pu

are obtained as:

u u u u u u udN = -p ds cosθ - ds sinθ (3.9)

u u u u u u udC = -p ds sinθ + ds cosθ (3.10)

Note that the suffix ‘u’ denotes quantities at point Pu and the positive direction of

the angle uθ is as shown in Fig.3.8 .

Expressions similar to Eqs.(3.9) and (3.10) can be written down for the

contributions to N and C from element at point Pl .

Integrating over the entire airfoil yields :

u u u u u

upper surface lower surface

N = - p cosθ + sinθ ds + p cosθ - sinθ dsl l l l l (3.11)

u u u u u

upper surface lower surface

C = -p sinθ + cosθ ds + p sinθ + cosθ dsl l l l l (3.12)



Proceeding in a similar manner, it can be shown that Mle, the pitching

moment about leading edge of the airfoil, per unit span, is :

le u u u u u u u u u u u

upper surface

M = p cosθ + sinθ x - p sinθ - cosθ y ds

lower surface

+ -p cosθ + sinθ x + p sinθ + cosθ y dsl l l l l l l l l l l (3.13)

Note: Once N and C are known, the lift per unit span (L) and drag per unit

span (D) of the airfoil can be obtained using Eqs.(3.7) and (3.8).

It is convenient to work in terms of lift coefficient (Cl ) and drag coefficient

(Cd). The definitions of these may be recalled as :

l

2

LC =

1ρV c

2

(3.14)

and

d2

DC =

1ρV c

2

(3.15)

It may be pointed out, that integration of a constant pressure, say p ,

around the body would not give any resultant force i.e.

p ds = 0 (3.16)

Hence, instead of ‘p’ the quantity p-p can be used in Eqs.(3.11), (3.12)

and (3.13). At this stage the following quantities are also defined.

pressure coefficient : p2

p-pC =

1ρV

2

(3.17)

skin friction drag coefficient :

f2

c =1ρV

2

(3.18)



:

n2

c2

lemle

2 2

NNormalforcecoefficient C =

1ρV c

2C

Chordwiseor axial forcecoefficient: C = (3.19)1ρV c

2M

Pitchingmomentcoefficient: C =1ρV c

2

It may be noted that dx = ds cos θ and dy = -ds sin θ, where “ds” is an

elemental length around a point P on the surface and θ is the angle between the

normal to the element and the vertical (Fig.3.8). Note that θ is measured positive

in the clockwise sense. It can be shown that :

n p pu fu f

0 upper surface lower surface

c

c fu f pu p

0 upper surface lower surface

c1C = C -C dx + c dy + c dy

c(3.20)

1C = c +c dx + C dy - C dy

c

l l

l l

Following section 10.2 of Ref.1.4, the expressions for Cn, Cc and Cmle can be

rewritten as:

2

1

1

1

un p pu fu f

0 0

uc pu p fu f

0 0

umle pu p fu f

0 0

upu fu u p2

0 0

c c

c c

c c

c c

dy dyC = C -C dx+ c +c dx

c dx dx

dy dy1C = C -C dx + c -c dx

c dx dx

dy dyC = C -C x dx - C +c x dx

c dx dx

dy d+ C +c y dx+ -C

c dx

ll l

ll l

ll l

l

f

(3.21)

y+c y dx

dxl

l l



Remarks:

(i) From Cn and Cc the lift coefficient (Cl ) and drag coefficient (Cd) are obtained

as :

n cC = C cosα - C sinαl (3.22 )

d n cC = C sinα + C cosα (3.23)

(ii) Centre of pressure : The point on the airfoil chord through which the

resultant aerodynamic force passes is the centre of pressure. The aerodynamic

moment about this point is zero. It may be noted that the location of centre of

pressure depends on the angle of attack or the lift coefficient.

(iii) Aerodynamic centre: As the location of the centre of pressure depends on

lift coefficient (Cl ) the pitching moment coefficient about leading edge (Cmle) also

changes with Cl . However, it is found that there is a point on the airfoil chord

about which the pitching moment coefficient is independent of the lift coefficient.

This point is called ‘Aerodynamic centre‘. For incompressible flow this point is

close to the quarter chord point of the airfoil.

(iv) If the distributions of Cp and cf are obtained by analytical or computational

methods, then the pressure drag coefficient (Cdp) and the skin friction drag

coefficient(Cdf) can be evaluated.

In experimental work the pressure distribution on an airfoil at different angles of

attack can be easily measured. However, measurement of shear stress on

an airfoil surface is difficult.The profile drag coefficient (Cd) of airfoil, which is the

sum of pressure drag coefficient and skin friction drag coefficient, is measured in

experiments by ‘Wake survey technique’ which is described in Chapter 9, section

‘f’ of Ref.3.10. In this technique, the momentum loss due to the presence of the

airfoil is calculated and equated to the drag (refer section 7.5.1 of Ref.3.11 for

derivation).


Though the expression for lift coefficient (Cl ) involves both the pressure

coefficient (Cp) and the skin friction drag coefficient (cf), the contribution of the



former i.e. Cp is predominant to decide Cl . On the other hand, the pressure drag

coefficient (Cdp) is determined by the distribution of Cp and the skin friction drag

coefficient (Cdf) is decided by the distribution of shear stress .

In this subsection the distributions of CP in typical cases and their implications for

Cl and Cdp are discussed.

The distribution of the pressure coefficient is generally plotted on the outer side

of the surface of the body (Fig.3.9a). The length of the arrow indicates the

magnitude of Cp. As regards the sign convention, an arrow pointing towards the

surface indicates that Cp is positive or local pressure is more than the free stream

pressure p . An arrow pointing away from the surface indicates that Cp is

negative i.e. the local pressure is lower than p .

(a) Ideal fluid flow (b) Real fluid flow

Fig.3.9 Distribution of Cp around a circular cylinder

Figure 3.9 shows distributions of Cp in ideal fluid flow and real fluid flow past a

circular cylinder. It may be recalled that an ideal fluid is inviscid and

incompressible whereas a real fluid is viscous and compressible. From the

distribution of Cp in ideal fluid flow (Fig.3.9a) it is seen that the distribution is

symmetric about X-axis and Y-axis. It is evident that in this case, the net forces in

vertical and horizontal directions are zero. This results in Cl = 0, Cdp = 0. These

results are available in books on fluid mechanics and aerodynamics. In the real



fluid flow case, shown in Fig.3.9b, it is seen that the flow separates from the body

(see description on boundary layer separation in section 3.2.7) and the pressure

coefficient behind the cylinder is negative and nearly constant. However, the

distribution is still symmetric about horizontal axis. Thus in this case Cl = 0 but

dpC > 0 .

The distributions of Cp over symmetrical and unsymmetrical foils at Cl = 0

and Cl > 0 are shown in Figs.3.10 a to d. Note also the locations of centre

pressure and the production of pitching moment for the unsymmetrical airfoil.

Flow visualization pictures at three angles of attack(α ) are shown in Figs.3.36 a,

b and c. An attached flow is seen at low angle of attack. Some separated flow is

seen at moderate angle of attack and large separated flow region is seen near α

close to the stalling angle ( stallα ). It may be pointed out that theoretical calculation

of skin friction drag using boundary layer theory can be done, when flow is

attached. This topic is discussed in the next subsection.

(a)Distribution of pressure coefficient on symmetrical airfoil at Cl = 0 and α= 0

Note : Lu = Ll



(b) Distribution of pressure coefficient on symmetrical airfoil at Cl > 0 and α> 0

(c) Distribution of pressure coefficient on cambered airfoil at Cl = 0, α< 0 ;

Note: Lu and Ll form a couple; centre of pressure is at infinity, Cmac < 0,



(d) Distribution of pressure coefficient on cambered airfoil at Cl > 0, α> 0

Note : Cmac same as in Fig.(c)

Fig.3.10 Distributions of pressure coefficient on symmetrical and unsymmetric

airfoils at Cl = 0 and Cl > 0


Under conditions of normal temperature and pressure a fluid satisfies the

‘No slip condition’ i.e. on the surface of a solid body the relative velocity between

the fluid and the solid wall is zero. Thus, when the body is at rest the velocity of

the fluid layer on the body is zero. In this and the subsequent subsections, the

body is considered to be at rest and the fluid moving past it. Though the velocity

is zero at the surface, a velocity of the order of free stream velocity is reached in

a very thin layer called ‘Boundary layer’. The velocity gradient normal to the

surface

U

yis very high in the boundary layer. Hence even if the coefficient of

viscosity μ is small, the shear stress,

Uμ

y,in the boundary layer may be

large or comparable to other stresses like pressure. Outside the boundary layer



the gradient / U y is very small and viscous stress can be ignored and flow

treated as inviscid. It may be recalled from text books on fluid mechanics, that in

an inviscid flow the Bernoulli’s equation is valid.

Features of the boundary layer over the surface of a streamlined body are shown

in Fig.3.11a. On the surface of a bluff body the boundary layer develops upto a

certain extent and then separates (Fig.3.11b). The definitions of the streamlined

body and bluff body are presented at the end of this subsection.

(a) boundary layer over a streamlined body

(b) Boundary layer over a bluff body

Fig.3.11 Boundary layer over different shapes (not to scale)



The features of the flow are as follows.

1.Near the leading edge (or the nose) of the body the flow is brought to rest.This

point is called the ‘Stagnation point’. A laminar boundary layer develops on the

surface starting from that point. It may be recalled, from topics on fluid

mechanics, that in a steady laminar flow the fluid particles move downstream in

smooth and regular trajectories; the streamlines are invariant and the fluid

properties like velocity, pressure and temperature at a point remain the same

with time. In an unsteady laminar flow the fluid properties at a point may vary but

are known functions of time. In a turbulent flow, on the other hand, the fluid

properties at a point are random functions of time. However, the motion is

organized in such a way that statistical averages can be taken. In a laminar

boundary layer the parameter which mainly influence its development is the

Reynolds number x eR = ρU x/μ ; x being distance along the surface, from the

stagnation point.

2.Depending on the Reynolds number (RX), the pressure gradient and other

parameters, the boundary layer may separate or become turbulent after

undergoing transition. The turbulent boundary layer may continue till the trailing

edge of the body (Fig.3.11a) or may separate from the surface of the body (point

‘S’ in Fig 3.11b). It may be added that the static pressure across the boundary

layer at a station ‘x’, is nearly constant with ‘y’. Hence the pressure gradient

referred here is the gradient (dp/dx) in the flow outside the boundary layer.

3.Nature of boundary layer decides the drag and the heat transfer from the body.

If the boundary layer is separated, the pressure in the rear portion of the body

does not reach the freestream value resulting in a large pressure drag (Fig.3.9b).

Incidently a streamlined body is one in which the major portion of drag is skin

friction drag. For a bluff body the major portion of drag is pressure drag. A

circular cylinder is a bluff body. An airfoil at low angle of attack is a streamlined

shape. But, an airfoil at high angle of attack like stallα is a bluff body.

Remark:

General discussion on boundary layer is a specialised topic and the

interested reader may consult Ref.3.11 for more information. Here, the features



of the laminar and turbulent boundary layers on a flat plate are briefly described.

While discussing separation, the boundary layer over a curved surface is

considered.

3.2.6 Laminar boundary layer over flat plate – height of boundary layer,


The equations of motion governing the flow of a viscous fluid are called

‘Navier-Stokes (N-S) equations’. For derivation of these equations refer to

chapter 15 of Ref.3.12. Taking into account the thinness of the boundary layer,

Prandtl simplified the N-S equations in 1904. These equations are called

‘Boundary layer equations’ (Chapter 16 of Ref.3.12). Solution of these equations,

for laminar boundary layer over a flat plate with uniform external stream, was

obtained by Blasius in 1908. Subsequently many others obtained the solution.

The numerical solution by Howarth, presented in Ref.3.10, chapter 7, is given in

Table 3.2. In this table U is the local velocity, Ue is the external velocity (which in

this particular case is V ), and η is the non-dimensional distance from the wall

defined as :

eUη = y

x (3.24)



Table 3.2 Non-dimensional velocity profile in a laminar boundary layer over a

flat plate

Height of boundary layer

It is seen from table 3.2 that the external velocity (Ue) is attained very

gradually. Hence the height at which U/Ue equals 0.99 is taken as the height of

the boundary layer and denoted by 0.99δ . From table 3.2, eU/U 0.99 is attained

at η = 5 . Noting the definition of η in Eq.(3.24) gives :

e

0.99

U5 = δ

x

Or 0.99 ex

e x

δ U x5 5= = ; R =

x U x R

(3.25)



Figure 3.12 shows a typical non-dimensional velocity profile in a laminar

boundary layer. While presenting such a profile, it is a common practice to plot

U/Ue on the abscissa and ( 0.99y/δ ) on the ordinate.

Fig.3.12 Non-dimensional velocity profile in laminar and turbulent boundary

layers on a flat plate

It is seen from Eq.(3.25) that 0.99δ grows in proportion to 1

2x (see Fig.3.13). It

may be added that in this special case of laminar boundary layer on flat plate, the

velocity profiles are similar at various stations i.e. the non-dimensional profiles of

U/Ue vs (y/ 0.99δ ) are same at all stations.



Fig.3.13 Schematic growth of boundary layer

Displacement thickness and skin friction drag coefficient

The presence of boundary layer causes displacement of fluid and skin

friction drag. The displacement thickness 1δ is defined as :

1e0

Uδ = 1- dy

U

(3.26)

The local skin friction coefficient ( fC or cf)is defined as :

wallf f wall wall

2 y=0e

uC = c = ; = μ ;Note: is a function of 'x'.

1 yρU2

(3.27)

If the length of the plate is L, then the skin friction drag per unit span of the

plate (Df) is :

f wall

0

L

D = dx

Hence, skin friction drag coefficient Cdf is given by:

fdf

2

DC =

1ρV L

2

(3.28)

From the boundary layer profile (table 3.2) it can be shown that for a flat

plate of length, L, the expressions for 1δ and Cdf are:

1

L

δ 1.721=

L R (3.29)



df L

L

V L1.328C = ; R =

R (3.30)

Remark :

Reference 1.11, chapter 6 may be consulted for additional boundary

layer parameters like momentum thickness ( 2δ ), shape parameter (H = 1δ / 2δ )

and energy thickness ( 3δ ) of a boundary layer.

Example 3.1

Consider a flat plate of length 500 mm kept in an air stream of velocity 15 m/s.

Obtain (a) the boundary layer thickness 0.99δ and the displacement thickness

1δ at the end of the plate (b) the skin drag coefficient. Assume -6 2= 15×10 m /s

and the boundary layer to be laminar.

Solution:

L = 0.5 m, V = 15 m/s , -6 2= 15×10 m /s

Hence, 5L -6

0.5×15R = = 5×10

15×10

Consequently, from Eq.(3.25):

-30.99

5L

δ 5 5= = = 7.07×10

L R 5×10

Or -3 -30.99δ = 7.07×10 ×0.5 = 3.54×10 m = 3.54mm

From Eq.(3.29):

-31

5L

δ 1.721 1.721= = = 2.434×10

L R 5×10

Or -5 -31δ = 2.434×10 ×0.5 = 1.217×10 m = 1.217 mm

From Eq.(3.30):

df 5L

1.328 1.328C = = = 0.00188

R 5×10

Remark:

0.99δ /L is found to be 7.07 x 10-3. Hence the assumption of the thinness of

boundary layer is confirmed by the results.



Chapter 3


3.2.7 Boundary layer separation, adverse pressure gradient and

favourable pressure gradient




3.2.7 Boundary layer separation, adverse pressure gradient and favourable

pressure gradient

When the flow takes place around airfoils and curved surfaces, the velocity

outside the boundary layer is not constant. From Bernoulli’s equation it can be

deduced that when the velocity decreases the pressure increases and vice-

versa. When the velocity is decreasing i.e. dp/dx is positive, the pressure

gradient is called ‘Adverse pressure gradient’. When dp/dx is negative it is called

‘Favourable pressure gradient’.

Figure 3.14 shows the development of a boundary layer in an external stream

with adverse pressure gradient (dp/dx > 0). Such a flow may occur on the upper

surface of an airfoil beyond the point of maximum thickness. Since the static

pressure at a station remains almost constant across the boundary layer, the

pressure inside the boundary layer at stations separated by distance Δx also

increases in the downstream direction.



Fig.3.14 Flow in boundary layer before and after point of separation (not to

scale)

Figure 3.14 also shows a small element ABCD in the boundary layer. The

pressure on the face AD is p whereas that on the face BC is p+ dp/dx Δx .

Since dp/dx is positive in this case, the net effect causes a deceleration of the

flow, in addition to that due to viscosity. The effect is more pronounced near the

surface and the velocity profile changes as shown in Fig.3.14. Finally at point S

the slope of the velocity profile at the wall, wallU / y , becomes zero. Besides

the change in shape, the boundary layer also thickens rapidly in the presence of

adverse pressure gradient. Downstream of the point S, there is a reversal of the

flow direction in the region adjacent to the wall. A line can be drawn (indicated as

dotted line in Fig.3.14) in such a way that the mass flow above this line is the

same as that ahead of point S. Below the dotted line, there is a region of

recirculating flow and the value of the stream function ψ for the dotted line is

zero. However, ahead of the point S, the ψ = 0 line is the surface of the body.

Thus, after the point S, it is observed that between the main flow (i.e. region

above ψ = 0 line) and the body surface lies a region of recirculating flow. When

this happens the flow is said to be ‘Separated’ and S is referred to as the ‘Point

of separation’. Due to separation, the pressure recovery, which would have taken



place in an unseparated flow, does not take place and the pressure drag of the

body increases.

Remarks:

(i) If the adverse pressure gradient is very gradual then separation may

not take place (Refer to Ch. 8 of Ref.3.11 for dp/dx needed for

separation).

(ii) Separation does not take place when the pressure gradient is

favourable.

(iii) In the two-dimensional case shown in Fig.3.14 the gradient / U y is

zero at the point of separation. Hence cf is zero at this point. This

behaviour is used in computations to determine the location of the

separation point.


In a laminar boundary layer, either the flow variables at a point have

constant values or their values show a definite variation with time. However, as

Reynolds number increases, it is found that the flow variables inside the

boundary layer show chaotic variation with time. Such a boundary layer is called

‘Turbulent boundary layer’. The change over from laminar to turbulent boundary

layer is called ‘Transition’ and takes place over a distance called ‘Transition

length’.

Initiation of transition in a boundary layer, can be studied as an instability

phenomenon. In this study, the flow is perturbed by giving a small disturbance

and then examining whether the disturbance grows. Details of the analysis are

available in chapter 15 of Ref.3.11 and chapter 5 of Ref.3.13. The salient

features can be summarized as follows.

1. In a boundary layer on a flat plate with uniform external subsonic stream (Ue =

constant), the flow becomes sensitive to some disturbances as Rx exceeds

5 x 105 . This is called ‘Critical Reynolds number (Rcrit)’. For boundary layers in

other cases, Rcrit depends on Mach number, surface curvature, pressure gradient

in external stream and heat transfer from wall.



2. After Rcrit is exceeded, some disturbances grow. These are called ‘Tollmien –

Schlichting (T-S) waves’.

3. The T-S waves lead to three-dimensional unstable waves and formation of

isolated large scale vortical structures called turbulent spots.

4. The turbulent spots grow and coalesce to form fully turbulent flow.

Remarks:

(i) As Rcrit is exceeded only some disturbances grow and hence in flows with very

low free stream turbulence level, Rcrit as high as 2.8 x 106 has been observed in

experiments. It may be recalled from fluid mechanics that the flow in pipe can

become turbulent when Reynolds number, based on pipe diameter (Red),

exceeds 2000. But laminar flow has been observed, in very smooth pipes, even

at Red = 40,000.

(ii) Transition process takes place over a length called transition length.

Reference 3.11, chapter 15 gives some guidelines for estimating this length.

Surface roughness reduces this length.

(iii) In flows with external pressure gradient, the transition is hastened by adverse

pressure gradient. It is generally assumed that transition does not take place in

favourable pressure gradient.


When the flow is turbulent, one of its dominant features is that the velocity at

a point is a random function of time(Fig.3.15). When a quantity varies in a

random manner, one cannot say as to what the value would be at a chosen time,

though the values may lie within certain limits. In such a situation, the flow

features are described in terms of statistical averages. For example, the average

Uof a fluctuating quantity U is given by :

0

0

T +T

0

T -T

1U T = Udt

2T (3.31)



Fig.3.15 Typical turbulence signal

If the quantity U is independent of T0 then the phenomenon is called ‘Stationary

random phenomenon’. The discussion here is confined to this type of flow. In

such a case, the instantaneous value, U t , is expressed as :

U t = U+u t ; u = U-U .

By definition u = 0 . Hence, to distinguish different turbulent flows the root mean

square (r.m.s.) of the fluctuating quantity is used.

2

rmsu = 2u' =T

2

-T

1u dt

2T (3.32)

T

r.m.s. value of u is 2u'

Another feature of turbulent flows is that even if the mean flow is only in one

direction, the fluctuations are in all three directions i.e. the instantaneous velocity

vector ( V ) at a point would be

V = U+u i+v j+w k; u ,v andw are the components of the

fluctuating velocity along x, y and z directions.

Because of the random fluctuations, the transfer of heat, mass and momentum is

many times faster in turbulent flows than in laminar flows. However, the part of



the kinetic energy of the mean motion which gets converted into the random

fluctuations is finally dissipated into heat and as such losses are higher when the

flow is turbulent.

Characteristics of turbulent boundary layer:

Analysis of turbulent boundary layer is more complicated than that of laminar

boundary layer. Reference 3.11, chapters 16 to 22 can be referred to for details.

A few results are presented below.

Velocity profile:

Velocity profile of a turbulent boundary layer on a flat plate with zero

pressure gradient is also shown in Fig.3.12. The profile can be approximated by

a power law like:

1/7 5 7

e

U= y/δ ; 5×10 < Re < 10

U (3.33)

This approximation is called ‘1/7th power law profile’.

Boundary layer thickness 0.99δ :

Though the velocity gradient U/ y near the wall is much higher for

turbulent boundary layer than for the laminar case, the gradient is lower away

from the wall and 0.99δ is much higher for a turbulent boundary layer. Reference

3.13, chapter 6, gives the following expression for 0.99δ .

0.99 turb

1/7x

δ 0.16=

x R (3.34)

Skin friction:

The value of wall

U/ y is higher for turbulent boundary layer than for

laminar boundary layer (Fig.3.12). Hence, the skin friction drag for turbulent

boundary layer is much higher than that for a laminar boundary layer. Reference

3.13, chapter 6 gives the following expression for dfC .

df 1/7L

0.031C =

R (3.35)



Remarks:

(i) In Eqs.(3.34) and (3.35) it is assumed that the boundary layer is turbulent

from the leading edge. Corrections to these expressions can be applied by

taking the start of the transition region as the origin of the turbulent boundary

layer. However, at the values of RL obtained in actual airplanes the error in Cdf,

by ignoring the laminar region is small.

(ii) In certain references following expressions are found for 0.99δ and Cdf.

1

5/ 0.37 /0.99 xδ x = R and 1

50.072 /df LC = R .

However, Ref.3.13 chapter 6 shows that Eqs. (3.34) and (3.35) are more

accurate.

(iii) For the 1/7th power law profile of the turbulent boundary layer (Eq.3.33),

it can be shown using Eqs.(3.26) and (3.33) that :

1

δδ =

8 for turbulent boundary layer (3.36)

Example 3.2

Consider a case with L = 0.5 m, V = 30m/s , -6 2= 15×10 m /s which gives

RL = 106. Obtain the values of 0.99δ , 1δ and Cdf in the following cases.

(i) Assume that the boundary layer is laminar throughout even at RL = 106

(ii) Assume that the boundary layer is turbulent from landing edge of plate.

Solution:

(i) Laminar flow

0.99

6L

δ 5 5= = = 0.005

L R 10 or 0.99δ = 2.5 mm

1

6L

δ 1.721 1.721= = = 0.001721

L R 10 or 1δ = 0.86 mm

df 6L

1.328 1.328C = = = 0.001328

R 10

(ii) Turbulent flow

0.991/7L

δ 0.16= = 0.02223

L R or 0.99δ = 11.12 mm



1

δδ = 1.39 mm

8

df 1/7 67L

0.031 0.031C = = = 0.00431

R 10

Remark :

The comparison of the above results points out that the values of 0.99δ and 1δ are

larger when the boundary layer is turbulent than when it is laminar. The value of

Cdf in the former case is nearly three times of that in the later case.


In this subsection the following four topics are briefly touched upon.

(i) Calculation of boundary layer, (ii) Separation of turbulent boundary layer,

(iii) Laminar flow airfoil and (iv) Effect of roughness on transition and skin friction

(i) Calculation of boundary layer

To calculate the boundary layer over an airfoil the first step is to obtain the

velocity distribution using potential flow theory. It may be recalled from

aerodynamics, that in potential flow analysis the velocity on the surface of the

body is not zero. It is assumed that this velocity distribution, given by potential

flow, would roughly be the distribution of velocity outside the boundary layer (Ue).

From this velocity distribution and using Bernoulli’s equation, the first estimate of

dp/dx is obtained. Based on this data the growth of laminar boundary layer and

the location of transition point are determined. After the transition, the growth of

turbulent boundary layer is calculated. After obtaining the boundary layers the

displacement thickness 1δ is added to the airfoil shape and calculations are

repeated till the displacement thickness assumed at the beginning of an iteration

is almost same as that obtained after calculation of the boundary layer.

Subsequently, the skin friction drag can be calculated. Section 18.4 of Ref.3.11

may be consulted for details. Presently, Computational Fluid Dynamics (CFD) is

used for these calculations.



(ii) Separation of turbulent boundary layer

A turbulent boundary layer may also separate from the surface when it is

subjected to adverse pressure gradient. However, due to turbulent mixing the

value of wU/ y for separation to take place is much higher than that in the

case of laminar boundary layer. Hence, a turbulent boundary layer has a higher

resistance to separation. This behaviour is used in bluff bodies to delay the

separation and reduce their pressure drag. For example, in the case of a circular

cylinder the laminar boundary layer separates at around 800 leaving a large

Fig.3.16 Schematic of flow past circular cylinder (a) Laminar

separation (b) Turbulent separation

separated region (Fig.3.16a). However, if the transition to turbulent flow takes

place before separation of laminar boundary layer, the separation is delayed. A

turbulent layer separates at around o108 , giving a smaller separated region

(Fig.3.16b). Since the drag of a bluff body is mainly pressure drag, the total drag

decreases significantly when the flow is turbulent before separation.

For example, the drag coefficient of a circular cylinder is around 1 when the

separation is laminar and it is 0.3 when the separation is turbulent (Refer chapter

1 of Ref.3.11)

(iii) Laminar flow airfoils

For a streamlined body, like an airfoil at low angle of attack, the drag is mainly

skin friction drag. Figure 3.17 and Eqs.(3.30) and (3.35) indicate that Cdf is much

higher when boundary layer is turbulent. Hence, to reduce the drag of the airfoil,



its shape is designed in such a way that the transition to turbulence is delayed

and the flow remains laminar over a longer portion of the airfoil.

These airfoils are called ‘Laminar flow or low drag airfoils’. Presently, efforts are

in progress to delay the transition by boundary layer control (see remark in

section 3.7.2).

Fig.3.17 Skin friction drag coefficient at various Reynolds numbers and

levels of roughness

(iv) Effect of roughness on transition

It was mentioned that the critical Reynolds number (Rcrit) depends on factors like

pressure gradient, Mach number, surface curvature and heat transfer. However,

the onset of transition may be delayed when disturbance like free stream

turbulence is low. However, if the surface is rough, this delay may not be

observed when roughness exceeds a certain value (see chapter 15 of Ref.3.11)

(v) Effect of roughness on skin friction in turbulent boundary layer

Equation (3.35) indicates that Cdf is proportional to -1/7LR i.e. Cdf decreases with

RL. However, when the surface is rough it is observed that the decrease in Cdf

stops after a certain Reynolds number(Fig.3.17). This Reynolds number is called

‘Cut-off Reynolds number’ and is denoted by (Re)cut-off.



The roughness is quantified by the parameter ( l /k), where

l = characteristic length e.g. the length (L) in case of a flat plate and chord (c) in

case of an airfoil.

k = height of roughness referred to as equivalent sand roughness.

Following Ref.3.6, chapter 3, typical values of k are given in table 3.3.

Type of surface Equivalent sand roughness (m)

Natural sheet metal 4.06 x 10-6

Smooth paint 6.35 x 10-6

Standard camouflage paint 1.02 x 10-5

Mass production paint 3.048 x 10-5

Table 3.3 Equivalent sand roughness for typical surfaces

Based on Ref.3.11, chapter 18, Fig.3.17 shows typical plots of Cdf vs Re for

turbulent boundary layer with l /k as parameter. For example, when l /k = 105 , Cdf

remains almost constant at 0.0032 beyond Re = 7 x 106. Reference 3.6 section

3.1 may be seen for plot of (Re)cutoff vs ( l /k) with Mach number as parameter.

These plots are based on section 4.1.5, of Ref.3.5.



Chapter 3


3.2.11 Presentation of aerodynamic characteristics of airfoils

3.2.12 Geometric characteristics of airfoils

3.2.13 Airfoil nomenclature\designation





efficiency factor

3.2.11. Presentation of aerodynamic characteristics of airfoils

As mentioned in the beginning of subsection 3.2.3, the ways of presenting

the aerodynamic and geometric characteristics of the airfoils and the

nomenclature of the airfoils are discussed in this and the next two subsections.

Figure 3.18 shows typical experimental characteristics of an aerofoil. The

features of the three plots in this figure can be briefly described as follows.

(I) Lift coefficient (Cl ) vs angle of attack (α). This curve, shown in Fig.3.18a, has

four important features viz. (a) angle of zero lift ( 0α l ), (b) slope of the lift curve

denoted by dCl / dα or a0 or αCl , (c) maximum lift coefficient ( maxCl ) and (d)

angle of attack (αstall

) corresponding to maxCl .



Fig.3.18 Aerodynamic characteristics of an airfoil

(a) Cl vs α (b) Cl vs Cd (c) Cmc/4 vs α

(II) Drag coefficient (Cd) vs Cl . This curve, shown in Fig.3.18b, has two

important features viz. (a) minimum drag coefficient (Cdmin

) and (b) lift coefficient

( optCl ) corresponding to Cdmin

. In some airfoils, called laminar flow airfoils or low-

drag airfoils, the minimum drag coefficient extends over a range of lift coefficients

(Fig.3.18b). This feature is called ‘Drag bucket’. The extent of the drag bucket

and the lift coefficient at the middle of this region are also characteristic features

of the airfoil. It may be added that the camber decides optCl and thickness ratio

decides the extent of the drag bucket.

(III) Pitching moment coefficient about quarter-chord Cmc/4 vs α . This curve is

shown in Fig.3.18c. Sometimes this curve is also plotted as Cmc/4 vs Cl . From

this curve, the location of the aerodynamic center (a.c.) and the moment about it

(Cmac

) can be worked out. It may be recalled that a.c. is the point on the chord

about which the moment coefficient is independent of Cl .

(IV) Stall pattern : Variation of the lift coefficient with angle of attack near the stall

is an indication of the stall pattern. A gradual pattern as shown in Fig.3.18a is a

desirable feature. Some airfoils display abrupt decrease in Cl after stall. This

behaviour is undesirable as pilot does not get adequate warning regarding



impending loss of lift. Airfoils with thickness ratio between 6 – 10% generally

display abrupt stall while those with t/c more than 14% display a gradual stall. It

may be added that the stall patterns on the wing and on the airfoil are directly

related only for high aspect ratio (A > 6) unswept wings. For low aspect ratio

highly swept wings three-dimensional effects may dominate.

3.2.12 Geometrical characteristics of airfoils

To describe the geometrical characteristics of airfoils, the procedure given in

chapter 6 of Ref.3.14 is followed. In this procedure, the camber line or the mean

line is the basic line for definition of the aerofoil shape (Fig.3.19a). The line

joining the extremities of the camber line is the chord. The leading and trailing

edges are defined as the forward and rearward extremities, respectively, of the

mean line. Various camber line shapes have been suggested and they

characterize various families of airfoils. The maximum camber as a fraction of the

chord length (ycmax/c) and its location as a fraction of chord (xycmax/c) are the

important parameters of the camber line.

Various thickness distributions have been suggested and they characterize

different families of airfoils Fig.3.19b. The maximum ordinate of the thickness

distribution as fraction of chord (ytmax/c) and its location as fraction of chord

(xytmax/c) are the important parameters of the thickness distribution.

Airfoil shape and ordinates:

The aerofoil shape (Fig.3.19c) is obtained by combining the camber line

and the thickness distribution in the following manner.

a) Draw the camber line shape and draw lines perpendicular to it at various

locations along the chord (Fig.3.19c).

b) Lay off the thickness distribution along the lines drawn perpendicular to the

mean line (Fig.3.19c).

c) The coordinates of the upper surface (xu, yu) and lower surface (xl, yl) of the

airfoil are given as :



u t

u c t

l t

l c t

x = x - y sinθ

y = y + y cosθ

x = x + y sinθ

y = y - y cosθ

3.37

where yc and yt are the ordinates, at location x, of the camber line and the

thickness distribution respectively; tan θ is the slope of the camber line at location x (see also Fig.3.19d). d) The leading edge radius is also prescribed for the aerofoil. The center of the

leading edge radius is located along the tangent to the mean line at the

leading edge (Fig.3.19c).

e) Depending on the thickness distribution, the trailing edge angle may be zero

or have a finite value. In some cases, thickness may be non-zero at the

trailing edge.



Fig.3.19 Airfoil geometry

3.2.13 Airfoil nomenclature/designation

Early airfoils were designed by trial and error. Royal Aircraft

Establishment (RAE), UK and Gottingen laboratory of German establishment

which is now called DLR(Deutsches Zentrum fϋr Luft-und Raumfahrt – German



Centre for Aviation and Space Flight) were pioneers in airfoil design. Clark Y

airfoil shown in Fig.3.20a is an example of a 12% thick airfoil with almost flat

bottom surface which has been used on propeller blades.

Taking advantage of the developments in airfoil theory and boundary

layer theory, NACA (National Advisory Committee for Aeronautics) of USA

systematically designed and tested a large number of airfoils in 1930’s. These

are designated as NACA airfoils. In 1958 NACA was superseded by NASA

(National Aeronautic and Space Administration). This organization has

developed airfoils for special purposes. These are designated as NASA airfoils.

Though the large airplane companies like Boeing and Airbus, design their own

airfoils the NACA and NASA airfoils are generally employed by others. A brief

description of their nomenclature is presented below. The description of NACA

airfoils is based on chapter 6 of Ref.3.14.

NACA four-digit series airfoils

Earliest NACA airfoils were designated as four-digit series. The thickness

distribution was based on successful RAE & Gottigen airfoils. It is given as :

2 3 4t

ty = 0.2969 x - 0.1260 x - 0.3516 x +0.2843 x -0.1015 x

20 (3.38)

where t = maximum thickness as fraction of chord.

The leading radius is : rt = 1.1019 t2

Appendix I of Ref.3.14 contains ordinates for thickness ratios of 6%, 9%, 10%,

12%, 15%, 18%, 21% and 24%. The thickness distributions are denoted as

NACA 0006, NACA 0009,……..,NACA 0024. Figure 3.20b shows the shape of

NACA 0009 airfoil. It is a symmetrical airfoil by design. The maximum thickness

of all four-digit airfoils occurs at 30% of chord. In the designation of these airfoils

the first two digits indicate that the camber is zero and the last two digits indicate

the thickness ratio as percentage of chord.



a) Clark Y – Airfoil with flat bottom surface, used on propeller blades

b) NACA 0009 – Symmetrical airfoil used on control surfaces

c) NACA 23012 – Airfoil with high Clmax , used on low speed airplanes

d) NACA 662 – 215 – Laminar flow or low drag airfoil

e) NASA GA(W) -1 or LS(1) - 0417 – Airfoil specially designed for general aviation airplanes

f) NASA SC(2)-0714 – Supercritical airfoil with high critical Mach number, specially designed for high subsonic airplanes

Fig.3.20 Typical airfoils



The camber line for the four-digit series airfoils consists of two parabolic arcs

tangent at the point of maximum ordinate. The expressions for camber(yc) are :

2c ycmax2

2ycmax2

my = 2px - x ; x x

p3.39

m= 1-2p +2px - x ; x > x

1-p

m = maximum ordinate of camber line as fraction of chord

p = chordwise position of maximum camber as fraction of chord

The camber lines obtained by using different values of m & p are denoted by two

digits, e.g. NACA 64 indicates a mean line of 6% camber with maximum camber

occuring at 40% of the chord. Appendix II of Ref.3.14 gives ordinates for NACA

61 to NACA 67 mean lines. The ordinates of other meanlines are obtained by

suitable scaling. For example, NACA 24 mean lines is obtained by multiplying the

ordinates of NACA 64 mean line by (2/6).

A cambered airfoil of four-digit series is obtained by combining meanline and

the thickness distribution as described in the previous subsection. For example,

NACA 2412 airfoil is obtained by combining NACA 24 meanline and NACA 0012

thickness distribution. This airfoil has (a) maximum camber of 2% occurring at

40% chord and (b) maximum thickness ratio of 12%.

Refer appendix III of Ref.3.14, for ordinates of the upper and lower surfaces of

several four-digit series airfoils. Appendix IV of the same reference presents the

low speed aerodynamic characteristics at M = 0.17 and various Reynolds

numbers. Chapter 7 of the same reference gives details of experimental

conditions and comments on the effects of parameters like camber, thickness

ratio, Reynolds number and roughness on aerodynamic characteristics of airfoils.

NACA five-digit series airfoils

During certain tests it was observed that maxCl of the airfoil could be

increased by shifting forward the location of the maximum camber. This finding

led to development of five-digit series airfoils. The new camber lines for the five-

digit series airfoils are designated by three digits. The same thickness distribution



was retained as that for NACA four-digit series airfoils. The camber line shape is

given as :

3 2 2c 1

31

1y = k x -3mx +m 3-m x , 0 < x m

6 (3.40)1

= k m 1- x ; m<x<16

The value of ‘m’ decides the location of the maximum camber and that of k1 the

design lift coefficient( iCl or optCl ). A combination of m = 0.2025 and k1 = 15.957

gives iCl = 0.3 and maximum camber at 15% of chord. This meanline is

designated as NACA 230. The first digit ‘2’ indicates that iCl = 0.3 and the

subsequent two digits (30) indicate that the maximum camber occurs at 15% of

chord.

A typical five-digit cambered airfoil is NACA 23012. Its shape is shown in

Fig.3.20c. The digits signify :

First digit(2) indicates that iCl = 0.3.

Second & third digits (30) indicate that maximum camber occurs at 15% of chord.

Last two digits (12) indicate that the maximum thickness ratio is 12%.

Remarks:

(i) Refer Appendices II, III and IV of Ref.3.14 for camber line shape,

ordinates and aerodynamic characteristics of five-digit series airfoils.

(ii) Modified four and five digit series airfoils were obtained when leading

edge radius and position of maximum thickness were altered. For

details Ref.3.14, chapter 6 may be consulted.

Six series airfoils

As a background to the development of these airfoils the following points may

be mentioned.



(i) In 1931 T.Theodorsen presented ’Theory of wing sections of arbitrary

shape’ NACA TR 411 which enabled calculation flow past airfoils of

general shape .

(ii) Around the same time the studies of Tollmien and Schlichting on

boundary layer transition, indicated that the transition process, which

causes laminar boundary layer to become turbulent, depends

predominantly on the pressure gradient in the flow around the airfoil.

(iii) A turbulent boundary layer results in a higher skin friction drag

coefficient as compared to when the boundary layer is laminar. Hence,

maintaining a laminar boundary layer over a longer portion of the airfoil

would result in a lower drag coefficient.

(iv) Inverse methods, which could permit design of meanline shapes and

thickness distributions, for prescribed pressure distributions were also

available.

Taking advantage of these developments, new series of airfoils called low drag

airfoils or laminar flow airfoils were designed. These airfoils are designated as

1-series, 2-series,…….,7-series. Among these the six series airfoils are

commonly used airfoils. Refer Ref.3.14, chapter 6 for more details.

When the airfoil surface is smooth. These airfoils have a Cdmin which is lower

than that for four-and five-digit series airfoils of the same thickness ratio. Further,

the minimum drag coefficient extends over a range of lift coefficient. This extent

is called drag bucket (see Fig.3.18b).

The thickness distributions for these airfoils are obtained by calculations which

give a desired pressure distribution. Analytical expressions for these distributions

are not available. Appendix I of Ref.3.14 gives symmetrical thickness

distributions for t/c between 6 to 21%.

The camber lines are designated as : a = 0, 0.1, 0.2 …., 0.9 and 1.0. For

example, the camber line shape with a = 0.4 gives a uniform pressure distribution

from x/c = 0 to 0.4 and then linearly decreasing to zero at x/c = 1.0. If the camber

line designation is not mentioned, ‘a’ equal to unity is implied.



An airfoil with a designation as NACA 662-215 is shown in Fig.3.20d. It is

obtained by combining NACA 662 – 015 thickness distribution and a = 1.0 mean

line. The digits signify :

1st digit ‘6’ indicates that it is a 6 series airfoil

2nd digit ‘6’ denotes the chordwise position of the minimum pressure in tenths of

chord for the symmetrical airfoil at Cl = 0. i.e. the symmetrical section

(NACA 662 - 015) would have the minimum pressure at x/c = 0.6 when producing

zero lift.

The suffix ‘2’ indicates that the drag bucket extends ±0.2 around optCl .

The digit ‘2’ after the dash indicates that optCl is 0.2. Thus in this case, drag

bucket extends for Cl = 0.0 to 0.4.

The last two digits ”15” indicate that the thickness ratio is 15%.

Since the value of ‘a’ is not explicitly mentioned, the camber line shape

corresponds to a = 1.0.

Remarks:

(i) Refer appendices I, II, III and IV of Ref.3.14 for details of thickness

distribution, camber distribution, ordinates and aerodynamic

characteristics of various six series airfoils.

(ii) The lift coefficient at the centre of the drag bucket ( optCl ) depends on

the camber. The extent of drag bucket depends on the thickness ratio

and the Reynolds number. The value given in the designation of the

airfoil is at Re = 9 x 106. The extent is about ±0.1 for t/c of 12%,±0.2

for t/c of 15% and ±0.3 for t/c of 18%. When the extent of the drag

bucket is less than ±0.1 , the subscript in the designation of the airfoil

is omitted, e.g. NACA 66-210

NASA airfoils

NASA has developed airfoil shapes for special applications. For

example GA(W) series airfoils were designed for general aviation airplanes. The

‘LS’ series of airfoils among these are for low speed airplanes. A typical airfoil of



this category is designated as LS(1)-0417. In this designation, the digit ‘1’ refers

to first series, the digits ‘04’ indicate optCl of 0.4 and the digits ‘17’ indicate the

thickness ratio of 17%. Figure 3.20c shows the shape of this airfoil. For the

airfoils in this series, specifically designed for medium speed airplanes, the

letters ‘LS’ are replaced by ‘MS’.

NASA NLF series airfoils are ‘Natural Laminar Flow’ airfoils.

NASA SC series airfoils are called ‘Supercritical airfoils’. These airfoils have a

higher critical Mach number. Figure 3.20f shows an airfoil of this category.

Chapter 3 of Ref.1.9 may be referred to for further details.

Remarks:

(i)Besides NACA & NASA airfoils, some researchers have designed airfoils for

specialized applications like (a) low Reynolds number airfoils for micro air

vehicles, (b) wind mills, (c) hydrofoils etc. These include those by Lissaman,

Liebeck, Eppler and Drela. Reference 3.9, chapter 4, and internet

(www.google.com) may be consulted for details.

(ii)The coordinates of NACA, NASA and many other airfoils are available on the

website entitled ‘UIUC airfoil data base’.


In the beginning of section 3.2.2 it was mentioned that the drag of the

wing consists of (i) the profile drag coefficient due to airfoil (Cd) and (ii) the

induced drag coefficient (CDi) due to finite aspect ratio of the wing. Subsections

3.2.3 to 3.2.13 covered various aspects of profile drag. In this subsection the

induced drag of the wing is briefly discussed.

For details regarding the production of induced drag and derivation of the

expression for the induced drag coefficient, the books on aerodynamics can be

consulted e.g. Ref.3.12, chapter 5. Following is a brief description of the induced

drag.

Consider a wing kept at a positive angle of attack in an air stream. In this

configuration, the wing produces a positive lift. At the wing root, the average

pressure on the upper surface is lower than the free stream pressure p and



the average pressure on the lower surface is higher than p . Since the span of

the wing is finite it has the wing tips and at these tips there cannot be a pressure

discontinuity or the pressure at the wing tips would be the same on the upper

side and the lower side. The pressure at the wing tips is expected to be mean of

the pressures on the upper and lower sides at the root section. Because of the

difference of pressures between the root and the tip, the pressure on the upper

surface of the wing increases from root to the tip in the spanwise direction.

Similarly, the pressure on the lower surface of the wing, decreases from the root

to the tip in the spanwise direction. These pressure gradients on the upper and

lower surfaces would lead to cross flows on these surfaces. Thus, at a given

spanwise station, the airstreams from the upper and lower surfaces would meet,

at the trailing edge, at an angle. This would cause shedding of vortices from the

trailing edge. Viewed from the rear, the vortices would appear rotating clockwise

from the left wing and anticlockwise from the right wing. These vortices soon roll

up to form two large vortices springing from positions near the wing tips. As a

consequence of these vortices the air stream in the vicinity and behind the wing

acquires a downward velocity component called induced downwash.

This downwash tilts the aerodynamic force rearwards resulting in a component in

the free stream direction called induced drag. The induced drag coefficient (CDi)

is given as :

2 2

L LDi

wing

C 1+δ CC =

A Ae (3.41)

Where A is the wing aspect ratio (A = b2/S) and δ is a factor which depends on

wing aspect ratio, taper ratio, sweep and Mach number. The quantity ‘ewing ‘ is

called Oswald efficiency factor for wing.

It may be added that a wing with elliptic chord distribution has the minimum

induced drag i.e. δ = 0 in Eq.(3.41).

Reference 3.6 section 3.3, gives the following expression for ewing which is

based on Ref.3.5 section 4.1.5.2.



LαW

wingLαW

1.1 C /Ae =

CR + 1-R

A

(3.42)

where,

LαW2

12 22

2 2

2 AC =

tan ΛA β

2+ 1+ +4β

(3.43)

LαWC = slope of lift curve of wing in radians

A = aspect ratio of wing

R = a factor which depends on (a) Reynolds number based on leading

edge radius, (b) leading edge sweep ( LE ), (c) Mach number (M), (d) wing

aspect ratio (A) and (e) taper ratio (λ ).

2β = 1-M

1/2 = sweep of semi-chord line

= ratio of the slope of lift curve of the airfoil used on wing divided by 2 .

It is generally taken as unity.

Remarks:

(i)Example 3.3 illustrates the estimation of ewing for an unswept wing. Section 2.5

of Appendix ‘B’ illustrates the steps for estimating ewing of a jet airplane.

(ii) When a flap is deflected, there will be increments in lift coefficient and also in

profile drag coefficient and induced drag coefficient. Refer section 2.9 of

Appendix ‘A’.

(iii) The drags of horizontal and vertical tails, can be estimated by following a

procedure similar to that for the wing. However, contributions to induced drag

from the tail surfaces are generally neglected.


The drag coefficient of a fuselage (CDf

) consists of (a)the drag of the fuselage at

zero angle of attack (CDo

)f plus (b) the drag due to angle of attack. Following



Ref.3.7, section 19.3 it can be expressed as:

2

(3.44)Df DofC = C 1+K15

where α is the angle of attack of fuselage in degrees.

For a streamlined body (CDo

)f is mainly skin friction drag and depends on

(i) Reynolds number, based on length of fuselage ( fl ), (ii) surface roughness

and (iii) fineness ratio (Af). The fineness ratio is defined as:

Af = fl /de (3.44a)

‘de‘ is the equivalent diameter given by:

( /4)de2 = A

fmax

where Afmax

equals the area of the maximum cross-section of the fuselage.

When the fineness ratio of the fuselage is small, for example, in case of

general aviation airplanes, the fuselage may be treated as a bluff body. In such a

case the term CDof is mainly pressure drag and the drag coefficient is based on

the frontal area (Afmax

). However, the expression for (CD0)f given in Ref.3.6,

section 3.1.1 includes the effect of pressure drag and is also valid for general

aviation airplanes (refer section 2 of Appendix A).

The quantity ‘K’ in Eq.(3.44) has a value of 1 for a circular fuselage and 4 to 6 for

a rectangular fuselage. However, the general practice is to include the increase

in drag of fuselage, due to angle of attack, by adding a term fuselage

1

e

to

wing

1

e

.

Remark:

The drag coefficients of other bodies like engine nacelle, external fuel tanks and

bombs suspended from the wing, can also be estimated in a manner similar to

that of fuselage.




The drag coefficients of other components like landing gear are based on

areas specific to those components. They should be obtained from the sources

of drag data mentioned earlier. The change in drag of these components, with

angle of attack, is included by adding a term other

1

e

to (fuselage

1

e +

wing

1

e ) i.e.

for the entire airplane, wing fuselage other

1 1 1 1= + +

e e e e (3.44b)

Reference 3.6, section 3.2 recommends other

1

e

as 0.05.


efficiency factor

It was mentioned earlier that the drag polar can be obtained by adding the

drag coefficients of individual components at corresponding angles of attack.

This procedure needs a large amount of detailed data about the airplane

geometry and drag coefficients. A typical drag polar obtained by such a

procedure or by experiments on a model of the airplane has the shape as shown

in Fig.3.21a. When this curve is replotted as CD vs C

L2 (Fig.3.21b), it is found

that over a wide range of 2LC the curve is a straight line and one could write:

CD = C

D0 + KC

L2 (3.45)

CD0 is the intercept of this straight line and is called zero lift drag coefficient or

parasite drag coefficient (Fig.3.21b).



Fig.3.21a Typical drag polar of a piston – engined airplane

Fig.3.21b Drag polar replotted as CD vrs.

2

LC

The term 2LKC is called induced drag coefficient or more appropriately lift

dependent drag coefficient. K is written as:

1

K =A e

(3.46)



The quantity ‘e’ in Eq.(3.46) is called ‘Oswald efficiency factor’. It includes the

changes in drag due to angle of attack of the wing, the fuselage and other

components as expressed by Eq.(3.44b). It may be added that in the original

definition of Oswald efficiency factor, only the contribution of the wing was

included. However, the expression given by Eq.(3.44b) is commonly

employed(Ref.1.12, Chapter 2 and Ref.3.6, Chapter 2).

The drag polar expressed by Eq.(3.45) is called ‘Parabolic drag polar’.

Remarks:

i) A parabolic expression like Eq.(3.45) fits the drag polar because the major

contributions to the lift dependent drag are from the wing and the fuselage and

these contributions are proportional to the square of the angle of attack or CL.

ii) Rough estimate of CDo :

Based on the description in Ref.1.9, chapter 4 and Ref.3.7, chapter 14, the

parasite drag (Dparasite or D0) of an airplane can be approximately estimated as the

sum of the minimum drags of various components of the airplane plus the

correction for the effect of interference.

Modifying Eq.(3.1), the parasite drag can be expressed as:

Dparasite = D0 = (Dmin)wing + (Dmin)fuse + (Dmin)ht + (Dmin)vt + (Dmin)nac + (Dmin)lg +

(Dmin)etc + Dint

(3.46a)

Modifying Eq.(3.5), the above equation can be rewritten as :

( )

2 2 20 Dmin fuse Dmin nac Dminwing fuse nac

2 2ht Dmin vt Dminht vt

2 2lg Dmin etc Dmin intlg etc

1 1 1D = ρV S C + ρV S C + ρV S C +

2 2 21 1ρV S C + ρV S C +

2 21 1ρV S C + ρV S C +D 3.46b

2 2

Dividing Eq.(3.46b) by 21ρV S

2 yields:



( )

fuse nacD0 Dmin Dmin Dminwing fuse nac

lght vtDmin Dmin Dminht vt lg

etcDmin Dintetc

S SSC = C + C + C +

S S SSS S

C + C + C +S S SS

C +C 3.46cS

To simplify Eq.(3.46c) the minimum drag coefficient of each component is

denoted by CD

and the area on which it is based is called ‘Proper drag area’ and

denoted by S. Thus, when the contribution of fuselage to CDo

is implied, then

CD

refers to Dmin fuseC and S refers to fuseS . With these notations Eq.(3.46c)

simplifies to :

D0C = DintD1

C S + CS

(3.46d)

The product D0C S is called ‘Parasite drag area’.

Note :

1)See example 3.3 for estimation of D0C of a low speed airplane.

2) In Appendices A and B the parasite drag coefficient ( D0C ) is estimated using

the procedure given in Ref.3.6, chapter 3, which in turn is based on Ref.3.5,

section 4.5.3.1. In this procedure the contributions of the wing and fuselage to

D0C are estimated togather as wing – body combination and denoted by D0WBC

iii) The parabolic polar is an approximation. It is inaccurate near CL= 0 and

CL= C

Lmax (Fig.3.21b). It should not be used beyond CLmax

.



Chapter 3



3.2.19 A note on estimation of minimum drag coefficients of wings and

bodies

3.2.20 Typical values of CDO, A, e and subsonic drag polar



3.3.1 Some aspects of supersonic flow - shock wave, expansion fan

and bow shock


3.3.3 Transonic flow regime - critical Mach number and drag

divergence Mach number of airfoils, wings and fuselage


As mentioned in remark (ii) of the previous subsection, the product CDo

x S is

called the parasite drag area. For a streamlined airplane the parasite drag is

mostly skin friction drag. Further, the skin friction drag depends on the wetted

area which is the area of surface in contact with the fluid. The wetted area of the

entire airplane is denoted by Swet

. In this background the term ‘Equivalent skin

friction coefficient (Cfe)’ is defined as:

CDo

x S = Cfe x Swet

Hence, Cfe = CDo x

wet

S

S and wet

DO f e

SC = C

S (3.47)

Reference 3.9, Chapter 12 gives values of Cfe for different types of airplanes.



Example 3. 3

A quick estimate of the drag polar of a subsonic airplane is presented in this

example which is based Ref.3.7, section 14.8. However, modifications have been

incorporated, keeping in view the present treatment of the drag polar.

An airplane has a wing of planform area 51.22 m2 and span 20 m. It has a

fuselage of frontal area 3.72 m2 and two nacelles having a total frontal area of

3.25 m2. The total planform area of horizontal and vertical tails is 18.6 m2. Obtain

a rough estimate of the drag polar in a flight at a speed of 430 kmph at sea level,

when the landing gear is in retracted position.

Solution:

Flight speed is 430 kmph = 119.5 m/s.

Average chord of wing( wingc ) = S / b = 51.22/20 = 2.566 m.

The value of kinematic viscosity ( ) at sea level = -614.6× 10 m2

Reynolds number (Re) based on average chord is:

6-6

119.5× 2.566= 21× 10

14.6× 10

It is assumed that NACA 23012 airfoil is used on the wing. From Ref.3.14,

Appendix IV, the minimum drag coefficient, (Cd)min, of this airfoil at Re = 9 x 106 is

0.006. However, the value of drag coefficient is required at Re = 621× 10 .

Assuming the flow to be turbulent (Cd)min can be taken proportional to 1-7

eR (Eq.

3.35). Thus, Cdmin at wRe = 21 x 106 would roughly be equal to:

6 61 17 70.006 21 10 / 9 10

= 0.0053

As regards the fuselage and nacelle, the frontal areas are specified. Hence, they

are treated as a bluff bodies. The value of (CDmin

)fuselage

can be taken as 0.08

(Ref.3.4). The nacelle generally has a lower fineness ratio and (CDmin

)nac

can be

taken as 0.10.



As regards the horizontal and vertical tails, the Reynolds number based on their

average chords (Retail) can be calculated if the areas and spans of these were

given. The following is suggested to obtain a rough estimate of Retail.

2ht vtS S 18.6/2 9.3 m . Then

tail tail

wing wing

c S

c S

and tail

w

tail

wing

cRe 9.3= 0.426

Re c 51.22

Hence, Re tail 6 6 621×10 ×0.426 = 8.95×10 9×10

At this Reynolds number (Cdmin)tail can be assumed to be 0.006

The calculation of the parasite drag coefficient (CDo) is presented in Table E 3.3.

Part S (m2) CD

CD

S (m2)

Wing 51.22 0.0053 0.271

Fuselage 3.72 0.080 0.298

Nacelles 3.25 0.1 0.325

Tail surfaces 18.6 0.006 0.112

Total 1.006

Table E3.3 Rough estimate of CDo

Adding 10% for interference effects, the total parasite drag area (CD

S ) is:

1.006 + 0.1006 = 1.1066 m2.

Hence,- CD0 = 1.1066/51.22 = 0.0216

Wing aspect ratio is 202 / 51.22 = 7.8

To obtain Oswald efficiency factor for the airplane (e) , the quantities ewing,

efuselage and eother are obtained below.

Equations (3.42) and (3.43) give :



LαW

WingLαW

1.1 C /Ae =

CR + 1-R

A

222 4

LαW

12 2

2

2 AC =

tan2ΛA β

1+ +β

Here

A = 7.8, M = V/a = 119.5/340 = 0.351

Hence, 2 2β = 1-M = 1-0.351 = 0.936

For the purpose of calculating ewing, the taper ratio (λ ), the quarter chord

sweep ( 14

) and the quantity , are taken as 0.4, 0 and 1 respectively.

Consequently, oLEΛ = 3.14

Hence, Lα 2 2

2 ×7.8C = = 5.121rad

7.8 ×0.9362+ +4

1

-1

From Ref.3.14, chapter 6, the leading edge radius, as a fraction of chord, for

NACA 23012 airfoil is :

1.109 t2 = 1.019 x 0.122 = 0.016

Rle = 0.016 x c = 0.016 x 2.566 = 0.041 m

Reynolds number, based leading edge radius ( eLERR ), is :

eLER5

-6

0.041×119.5R = = 3.35×10

14×10

Hence, 2eLER LE LER cotΛ 1-M cosΛ 5 2= 3.35×10 ×18.22× 1-0.351 ×0.998

= 57.16 x 105

Further, LE

Aλ 7.8×0.4= = 3.13

cosΛ 0.998



Corresponding to the above values of ( 2eLER LE LER cotΛ 1-M cosΛ ) and

(LE

Aλ

cosΛ), Fig 3.14 of Ref.3.6, gives R = 0.95.

Hence,

wing

1.1× 5.121/7.8e = = 0.925

0.95 × 5.121 / 7.8 + 0.05×

To obtain efuselage , it is assumed that the fuselage has a round cross section.

In this case, Fig.2.5 of Ref 3.6 gives: fuselagefuselage

1/ S /S

e

= 0.75 when A = 7.8.

Consequently,

fuselage

1= 0.75×3.72/51.22 = 0.054

e

others

1

e is recommended as 0.05(Ref.3.6, section 2.2)

Thus, wing fuselage other

1 1 1 1= + +

e e e e =

1+0.054+0.05 = 1.185

0.925

Or e = 0.844

Hence, 1 1

= = 0.0484Ae × 7.8 × 0.844

Hence, a rough estimate of the drag polar is:

2LDC = 0.0216+0.0484C

Answer: A rough estimate of the drag polar is : 2LDC = 0.0216+0.0484C

Remark:

i) A detailed estimation of the drag polar of Piper Cherokee airplane is

presented in appendix A.

3.2.19 Note on estimation of minimum drag coefficients of wings and

bodies

Remark (ii) of section 3.2.17 mentions that the parasite drag coefficient of

an airplane (CD0

) is given by :



D0C = DintD1

C S + CS

where the values of DC represent the minimum drag coefficients of various

components of the airplane.

In example 3.3 the minimum drag coefficients of wing, fuselage, nacelle,

horizontal tail and vertical tail were estimated using experimental data. However,

the minimum drag coefficients of shapes like the wing, the horizontal tail, the

vertical tail and the streamlined bodies can be estimated using the background

presented in subsections 3.2.5 to 3.2.10. The procedure for such estimation are

available in Ref.3.6 which in turn is based on Ref.3.5. The basis of this procedure

is that the minimum drag coefficient of a streamlined shape can be taken as the

skin friction drag coefficient of a flat plate of appropriate characteristic length,

roughness and area.

With these aspects in view, the procedure to estimate the minimum drag

coefficient of the wing can be summerised as follows. It is also illustrated in the

sections on drag polar in Appendices A & B.

(a) The reference length ( l ) is the mean aerodynamic chord of the exposed wing

i.e. the portion of wing outside the fuselage. This chord is denoted by ec . Obtain

roughness parameter ( l /k) with ec as ‘ l ’ and value of k from Table 3.3.

(b) The flow is assumed to be turbulent over the entire wing.

(c)Choose the flight condition. Generally this is the cruising speed (Vcr) and the

cruising altitude (hcr). Obtain the Reynolds number (Re) based on Vcr, kinematic

viscosity cr at hcr and the reference length as ec i.e. e cr

cr

c VRe =

.

Obtain (Re)cutoff corresponding to ( l /k) using Fig.3.2 of Ref.3.6. Obtain Cdf

corresponding to lower of Re and (Re)cut-off. Following Ref.3.6 this value is

denoted by Cfw in Appendices A & B.

(d) Apply correction to Cfw for type of airfoil and its thickness ratio. Multiply this

value by (Swet/Sref), where Swet is the wetted area of the exposed wing and Sref is

the reference area of the wing. Refer to section 3.1 of Ref.3.6 for estimation of



Swet and correction for airfoil shape. When the shape of the airfoil changes along

the wingspan, a representative section is taken for estimation of Swet.

Similar procedure can be used to estimate the minimum drag coefficients of the

horizontal tail and vertical tail.

As regards estimation of the minimum drag coefficient of fuselage, the reference

length is taken as the length of fuselage ( fl ) and the roughness factor is taken as

( fl /k). Correction is applied for fineness ratio ( fl /de) of the fuselage. Where ‘de’ is

the equivalent diameter of the fuselage (see section 3.2.15). The wetted area in

this case is the wetted area of the fuselage.

Finally, correction is applied for wing-body interference effect (see Appendices A

& B for details).

Similar procedure can be used to estimate the minimum drag coefficients of

bodies like nacelle, external fuel tanks, bombs etc.

3.2.20 Typical values of CDO, A, e and subsonic drag polar.

Based on the data in Ref.3.9, chapter 4 , Ref.3.18 vol. VI , chapter 5 and

Ref.3.15 , chapter 6, the typical values of CD0

, A, e and the drag polar for

subsonic airplanes are given in Table 3.4.

Type of

airplane

CD0

A

e

Typical polar

Low speed

(M <0.3)

0.025 to

0.04

6 to 8

0.75 to

0.85

0.025 + 0.06CL2

Medium speed

(M around 0.5)

0.02 to

0.024

10 to 12

0.75 to

0.85 0.022 + 0.04CL

2

High subsonic

(M around 0.8,

Swept wing)

0.014 to

0.017

6 to 9

0.65 to

0.75 0.016 +0.045CL

2

Table 3.4 Typical values of CD0

, A, e and subsonic drag polar



Remarks:

(i) Table 3.4 shows that CD0

for low speed airplanes is higher than other

airplanes. This is because these airplanes have exposed landing gear, bluff

fuselage (see Fig.1.2a) and struts when a high wing configuration is used. The

CD0

for high subsonic airplanes is low due to smooth surfaces, thin wings and

slender fuselage. It may be added that during the design process, the values of

airfoil thickness ratio, aspect ratio and angle of sweep for the wing are obtained

from considerations of optimum performance.

(ii) The low speed airplanes have a value of K (=1/ Ae ) higher than the other

airplanes. One of the reasons for this is that these airplanes have only a

moderate aspect ratio (6 to 8) so that the wing-span is not large and the hanger-

space needed for parking the plane is not excessive.

(iii) See section 2 of Appendix A for estimation of the drag polar of a subsonic

airplane in cruise and take-off conditions.


According to Ref.2.1, a Winglet is an upturned wing tip or added axialliary airfoil

above and / or below the wing tips. Figure 1.2c shows one type of winglets at

wing tips. The winglets alter the spanwise distribution of lift and reduce the

induced drag. Reference 1.9, chapter 4 can be referred for a simplified analysis

of the effect of winglets. However, along with reduction in induced drag, the

winglets increase the weight of the wing and also the parasite drag. After trade-

off studies which take into account the favourable and unfavourable effects of the

winglets, the following approximate dimensions are arrived at for the winglets.

Root chord of about 0.65 ct, tip chord of about 0.2 ct and height of about ct ;

where ct is the tip chord of the wing. As regards the effect on induced drag,

Ref.3.15, chapter 5 suggest that the effect of winglets can be approximately

accounted for by increasing the wing span by an amount equal to half the height

of the winglet. The procedure is illustrated in example 3.4

Example 3.4

Consider a wing, with the following features. Area (S) = 111.63 m2,

Aspect ratio (A) = 9.3, span (b) = 32.22 m, root chord (cr) = 5.59 m,



tip chord (ct) = 1.34 m

Further, the airplane has (a) parasite drag coefficient (CDO) = 0.0159 ; (b) Oswald

efficiency factor (e) = 0.8064 (c) lift coefficient during cruise (CLcr)=0.512.

Examine the benefits of fitting winglets to this wing.

Solution :

The drag polar of the existing airplane is :

CD = 2 2L L

10.0159+ C = 0.0159+0.04244C

π×9.3×0.8064

When CL = 0.512, CD = 0.0159 + 0.04244 x 0.5122 = 0.02703

With winglets, the wing span effectively increases to :

te

c 1.34b = b+ = 32.22+ = 32.89 m

2 2

Hence, the effective aspect ratio (Ae) = 2 2eb 32.89

= = 9.691S 111.63

Consequently, the drag polar approximately changes to :

2 2L L

10.0159+ C = 0.0159+0.0407C

×9.691×0.8064

At CL = 0.512, the CD of the wing with winglet is :

0.0159 + 0.04074 x 0.5122 = 0.02658

Reduction in drag coefficient is 0.02703 – 0.02658 = 0.00045 or 1.7%

Note :

(CL/CD)existing wing = 0.512/0.02703 = 18.94

(CL/CD)modified wing = 0.512/0.02658 = 19.26


At this stage, the reader is advised to revise background on compressible

aerodynamics and gas dynamics. References.1.9 & 1.10 may be consulted.

Before discussing the drag polar at high subsonic, transonic and supersonic

speeds, the relevant features of supersonic and transonic flows are briefly

recapitulated in the following subsections.



3.3.1 Some aspects of supersonic flow – shock wave, expansion fan and

bow shock

When the free stream Mach number roughly exceeds a value of 0.3, the

changes in the fluid density, within the flow field, become significant and the flow

needs to be treated as compressible. In a compressible flow, the changes of

temperature in the flow field may be large and hence the speed of sound

(a = γRT ) may vary from point to point. When the free stream Mach number

exceeds unity, the flow is called supersonic. When a supersonic flow

decelerates, shock waves occur. The pressure, temperature, density and Mach

number change discontinuously across the shock. The shocks may be normal or

oblique. The Mach number behind a normal shock is subsonic; behind an oblique

shock it may be subsonic or supersonic. When supersonic flow encounters a

concave corner, as shown in Fig.3.22a, the flow changes the direction across a

shock. When such a flow encounters a convex corner, as shown in Fig.3.22b, the

flow expands across a series of Mach waves called expansion fan. A typical flow

past a diamond airfoil at supersonic Mach number is shown in Fig.3.23. If the

free stream Mach number is low supersonic (i.e. only slightly higher than unity)

and the angle θ, as shown in Fig.3.23, is high then instead of the attached shock

waves at the leading edge, a bow shock wave may occur ahead of the airfoil. A

blunt-nosed airfoil can be thought of an airfoil with large value of ‘θ’ at the leading

edge and will have a bow shock at the leading edge as shown in Fig.3.24.

Behind a bow shock there is a region of subsonic flow (Fig.3.24).



(a) (b)

Fig.3.22 Supersonic flow at corners

(a) Concave corner (b) Convex corner

Fig.3.23 Supersonic flow past a diamond airfoil



Fig.3.24 Bow shock ahead of blunt-nosed airfoil


At supersonic Mach numbers also the drag of a wing can be expressed as

sum of the profile drag of the wing section plus the drag due to effect of finite

aspect ratio. The profile drag consists of pressure drag plus the skin friction drag.

The pressure drag results from the pressure distribution caused by the shock

waves and expansion waves (Fig.3.23) and hence is called ‘Wave drag’.

It is denoted by Cdw. Figures 3.25a and b show the distributions of pressure

coefficients (Cp) on an airfoil at angles of attack (α ) of 00 and 20.



(b) oα = 2

Fig.3.25 Pressure distributions over a diamond airfoil (a) oα = 0 (b) oα = 2

From the distributions of pC at oα = 0 , on various faces of the diamond airfoil, it is

observed that the distributions are symmetric about the X-axis but not symmetric

about the Y-axis. This indicates Cl = 0 but Cdw > 0. From the distributions of Cp at

oα = 2 , it is seen that the distributions are unsymmetric about both X- and



Y-axes. Thus in this case, Cl > 0 and Cdw > 0. It may be added that a leading

edge total angle of 10would give a thickness ratio of 8.75%, which is rather high.

Supersonic airfoils would have (t/c) between 3 to 5%.

At supersonic speed the skin friction drag is only a small fraction of the wave

drag. The wave drag of a symmetrical airfoil (Cdw) can be expressed as (Ref.1.9,

chapter 5):

2 2

2dW

4C = [α +(t/c) ]

M -1

(3.48)

where α = angle of attack in radians and

t / c = thickness ratio of the airfoil

The wave drag of a finite wing at supersonic speeds can also be expressed as

KCL2 (refer Ref.1.9, chapter 5 for details). However, in this case K depends on

the free stream Mach number (M∞), aspect ratio and leading edge sweep of the

wing (refer Ref.1.9 chapter 5 for details).

The estimation of the wave drag of a fuselage at supersonic speeds is more

involved than that of the wing. It is considered as flow past a combination of a

nose cone, a cyclindrical mid-body and a conical after body. It may be added that

the supersonic airplanes generally have low aspect ratio wings and the wave

drag of the wing-body is analysed as a combination. Reference 1.9, chapter 5

deals with some of these aspects. Reference 3.5 is generally used to estimate

the drag of wing-body-tail combination at desired values of Mach numbers.

3.3.3 Transonic flow regime, critical Mach number and drag divergence

Mach number of airfoil , wing and fuselage

A transonic flow occurs when the free stream Mach number is around one.

The changes in the flow and hence in the drag occurring in this range of Mach

numbers can be better understood from the following statements.

I) In the subsonic flow past an airfoil the flow velocity is zero at the stagnation

point. Subsequently, the flow accelerates, it reaches a maximum value (Vmax) and

later attains the free stream velocity (V∞). The ratio (Vmax /V∞) is greater than

unity and depends on (a) the shape of airfoil (b) the thickness ratio (t/c) and



( c ) the angle of attack (α). As (Vmax / V∞ ) is greater than unity, the ratio of the

maximum Mach number on the airfoil ( M max) and free stream Mach number

(M∞) would also be more than unity. However, (Mmax/ M∞) would not be equal to

(Vmax /V∞) as the speed of sound varies from point to point in the flow.

II) Critical Mach number: As M∞ increases, Mmax also increases. The free stream

Mach number for which the maximum Mach number on the airfoil equals unity is

called the critical Mach number (Mcrit

).

III) The changes in flow patterns when the free stream Mach number changes

from subcritical (i.e. critM M ) to supersonic (M > 1 ) are highlighted below .

(A) When M is less than or equal to Mcrit then the flow is subsonic everywhere

i.e. in the free stream, on the airfoil and behind it (Fig.3.26a).

(B) When M∞ exceeds Mcrit

, a region of supersonic flow occurs which is

terminated by a shock wave. The changes in flow pattern are shown in

Figs.3.26b and c.

(C) As free stream Mach number increases further the region of supersonic flow

enlarges and this region occurs on both the upper and lower surfaces of the

airfoil (Figs.3.26c, d & e).

(D) At a free stream Mach number slightly higher than unity, a bow shock is seen

near the leading edge of the airfoil (Fig.3.26f).

(E) At a still higher Mach numbers, the bow shock approaches the leading edge

and if the leading edge is sharp, then the shock waves attach to the leading edge

as shown in Fig.3.23.

Fig.3.26 (a) Mach number subsonic everywhere



Fig.3.26 (b) M∞ only slightly higher than Mcrit ; shock waves are not discernible

Fig.3.26 (c) M∞ greater than Mcrit ; shock wave seen on the upper surface

Fig.3.26 (d) M∞ greater than Mcrit ; shock waves seen on both the upper and

lower surfaces



Fig.3.26 (e) M∞ greater than Mcrit; shock waves seen on both the upper

and lower surfaces at the trailing edge

Fig.3.26 (f) M∞ greater than unity; bow shock wave seen ahead of the airfoil;

shock waves also seen at the trailing edge on both upper and lower surfaces

Fig.3.26 Flow past airfoil in transonic range at α=20

(Adapted from Ref.3.16, chapter 9 with permission from author). The angle of

attack (α) being 20 is mentioned in Ref.3.17 chapter 4.

(IV) Transonic flow regime

When M∞ is less than Mcrit the flow every where i.e. in the free stream,

and on the body and behind it, is subsonic. It is seen that when Mcrit

< M∞ < 1,



the free stream Mach number is subsonic but there are regions of supersonic

flow on the airfoil (Figs.3.26c, d & e). Further, when M∞ is slightly more than

unity i.e. free stream is supersonic; there is bow shock ahead of the airfoil

resulting in subsonic flow near the leading edge (Fig.3.24). When the shock

waves are attached to the leading edge (Fig.3.23) the flow is supersonic every-

where i.e. in the free stream and on the airfoil and behind it.

Based on the above features, the flow can be classified into three regimes.

(a) Sub-critical regime - when the Mach number is subsonic in the free stream

as well as on the body (M∞ < Mcrit

).

(b) Transonic regime - when the regions of both subsonic and supersonic flow

are seen within the flow field.

(c) Supersonic regime - when the Mach number in the free stream as well as

on the body is supersonic.

The extent of the transonic regime is commonly stated as between 0.8 to

1.2. However, the actual extent of this regime is between Mcrit

and the Mach

number at which the flow becomes supersonic everywhere. The extent depends

on the shape of the airfoil and the angle of attack. In the transonic regime the lift

coefficient and drag coefficient undergo rapid changes with Mach number

(Figs.3.27a, b and c). It may be recalled that Cd and Cl refer to the drag

coefficient and lift coefficient of an airfoil respectively.



Fig.3.27a Variation of lift coefficient (Cl ) in transonic range for the airfoil in

Fig.3.26 (α=20). (Adapted from Ref. 3.16, chapter 9 with permission from author)

Note: The points A, B, C, D, E and F corresponds to those in Figs.3.26a, b, c, d,

e and f respectively.

Fig.3.27b Typical variations of drag coefficient (Cd) in transonic region for

airfoils of different thickness ratios (Adapted from Ref.3.17,

chapter 4 with, permission of McGraw-Hill book company)



Figure 3.27a shows the variation of the lift coefficient (Cl ) with Mach number at a

constant value of angle of attack. It is seen that at sub critical Mach numbers, Cl

increases with Mach number. This is due to the effect of compressibility on

pressure distribution. However, as the critical Mach number is exceeded the

formation of shocks changes the pressure distributions on the upper and lower

surfaces of the airfoil and the lift coefficient decreases (points C & D in

Fig.3.27a). This phenomenon of decrease in lift due to formation of shocks is

called ‘Shock stall’. For a chosen angle of attack the drag coefficient begins to

increase near Mcrit

and reaches a peak around M∞ = 1 (Fig.3.27b).

(V) Drag divergence Mach number (MD)

The critical Mach number (Mcrit

) of an airfoil has been defined in statement (II) of

this subsection. It is the free stream Mach number (M∞) for which the maximum

Mach number on the airfoil equals one. The critical Mach number is a theoretical

concept. It is not possible to observe this (Mcrit

) in experiments as the changes in

flow, when M∞ just exceeds Mcrit

, are very gradual. Hence, a Mach number

called ‘Drag divergence Mach number (MD)’ is used in experimental work. The

basis is as follows.

When the change in Cd with Mach number is studied experimentally, the effects

of changes in flow, due to the appearance of shock waves, are noticed in the

form of a gradual increase in the drag coefficient. The Mach number at which the

increase in the drag coefficient is 0.002 over the value of Cd at sub-critical Mach

numbers is called ‘Drag divergence Mach number’ and is denoted by MD.

Figure 3.27c shows a typical variation of Cd with M and also indicates MD.

The following may be added. (a) For a chosen angle of attack the value of Cd

remains almost constant when the Mach number is sub-critical. (b) The drag

divergence Mach number of an airfoil depends on its shape, thickness ratio and

the angle of attack. (c) The increase in the drag coefficient in the transonic region



is due to the appearance of shock waves. Hence, this increment in Cd is called

wave drag.

Fig.3.27c Definition of drag divergence Mach number

(The curve corresponds to t / c =0.12 in Fig.3.27b)

Remark:

Supercritical airfoil

For airplanes flying at high subsonic speeds the lift coefficient under

cruising condition (CLcr) is around 0.5. At this value of lift coefficient, the older

NACA airfoils have drag divergence Mach number (MD) of around 0.68 for a

thickness ratio (t/c) of around 15%.

With the advancements in computational fluid dynamics (CFD) it was

possible, in 1970’s to compute transonic flow past airfoils. This enabled design of

improved airfoils, called supercritical airfoils, which have MD around 0.75 for t/c of

15% (Ref.3.18 part II, chapter 6). For comparison, the shapes of older airfoil

(NACA 662 – 215) and a supercritical airfoil are shown in Fig.3.20d and f. Note



the flat upper surface of the supercritical airfoil (refer Ref. 1.9 chapter 3 for

additional information).

(VI) Drag divergence Mach number of a wing

The drag divergence Mach number of an unswept wing depends on the

drag divergence Mach number of the airfoil used on the wing and its aspect ratio.

The drag divergence Mach number of the wing can be further increased by

incorporating sweep (Λ) in the wing. Figure 3.3 shows the geometrical

parameters of the wing including the sweep. The beneficial effects of sweep on

(a) increasing MD, (b) decreasing peak value of wave drag coefficient (C

Dpeak)

and (c) increasing the Mach number, at which CDpeak

occurs, are evident from

Fig.3.28.

Fig.3.28 Effect of wing sweep on variation of CD with Mach number.

(Adapted from Ref.3.3, chapter 16 with permission)

(VII) Drag divergence Mach number of fuselage

It can be imagined that the flow past a fuselage will also show that the

maximum velocity (Vmax

) on the fuselage is higher than V∞. Consequently, the



fuselage will also have critical Mach number (Mcritf

) and drag divergence Mach

number. These Mach numbers depend on the fineness ratio of the fuselage. For

the slender fuselage, typical of high subsonic jet airplanes, Mcritf

could be around

0.9. When Mcritf

is exceeded the drag of the fuselage will be a function of Mach

number in addition to the angle of attack.



Chapter 3




airplanes

3.3.6 Variations of CDo and K for a fighter airplane

3.3.7 Area ruling



3.6 Other types of drags

3.6.1 Cooling drag

3.6.2 Base drag


3.6.4 Leakage drag

3.6.5 Trim drag


The foregoing sections indicate that the drag coefficients of major airplane

components change as the Mach number changes from subsonic to supersonic.

Consequently, the drag polar of an airplane, being the sum of the drag

coefficients of major components, will also undergo changes as Mach number

changes from subsonic to supersonic. However, it is observed that the

approximation of parabolic polar is still valid at transonic and supersonic speeds,

with CD0

and K becoming functions of Mach number i.e.:

CD = C

D0 (M) + K (M)CL2 (3.49)

Detailed estimation of the drag polar of a subsonic jet airplane is presented in

section 2 of Appendix B.




airplanes

Subsonic jet airplanes are generally designed in a manner that there is no

significant wave drag up to the cruise Mach number (Mcruise). Further, the drag

polar of the airplane for Mach numbers upto Mcruise can be estimated, using the

methods for subsonic airplanes. Section 2 of Appendix B illustrates the

procedure for estimation of such a polar. However, to calculate the maximum

speed in level flight (Vmax

) or the maximum Mach number Mmax

, guidelines are

needed for the increase in CD0 and K beyond Mcruise. Such guidelines are

obtained in this subsection by using the data on drag polars of B727-100 airplane

at Mach numbers between 0.7 to 0.88.

Reference 3.18 part VI, chapter 5, gives drag polars of B727-100 at M = 0.7,

0.76, 0.82, 0.84, 0.86 and 0.88. Values of CD and C

L corresponding to various

Mach numbers were recorded and are shown in Fig.3.29 by symbols. Following

the parabolic approximation, these polars were fitted with Eq.(3.49) and CD0

and

K were obtained using least square technique. The fitted polars are shown as

curves in Fig.3.29. The values of CD0

and K are given in Table 3.5 and presented

in Figs.3.30 a & b.

Fig.3.29 Drag polars at different Mach numbers for B727-100



M CD0

K

0.7

0.76

0.82

0.84

0.86

0.88

0.01631

0.01634

0.01668

0.01695

0.01733

0.01792

0.04969

0.05257

0.06101

0.06807

0.08183

0.103

Table 3.5 Variations of CD0

and K with Mach number

Fig.3.30a Parameters of drag polar - CD0

for B727-100



Fig.3.30b Parameters of drag polar - K for B727-100

It is seen that the drag polar and hence CD0

and K are almost constant up to

M = 0.76. The variations of CD0

and K between M = 0.76 and 0.86, when fitted

with polynomial curves, give the following equations (see also Figs.3.30 a & b).

CD0

= 0.01634 -0.001( M-0.76)+0.11 (M-0.76)2 (3.50)

K= 0.05257+ (M-0.76)2 + 20.0 (M-0.76)3 (3.51)

Note: For M ≤ 0.76, CD0

= 0.01634, K = 0.05257

Based on these trends, the variations of CD0

and K beyond Mcruise but upto

Mcruise + 0.1 are expressed by the following two equations.

CD

= DOcrC - 0.001 ( M-Mcruise) + 0.11 (M-Mcruise)2 (3.50 a)

K = Kcr + (M-Mcruise)2 + 20.0 (M-Mcruise)3 (3.51 a)

where DOcrC and Kcr are the values of CD0

and K at cruise Mach number for the

airplane whose Vmax

or Mmax

is required to be calculated. It may be pointed out

that the value of 0.01634 in Eq.(3.50) has been replaced by DOcrC in Eq.(3.50a).



This has been done to permit the use of Eq.(3.50a) for different types of

airplanes which may have their own values of DOcrC (see section 4.2 of Appendix

B). For the same reason the value of 0.05257 in Eq.(3.51) has been replaced by

Kcr in Eq.(3.51a).

Section 4.2 of Appendix B illustrates the application of the guidelines given in this

subsection.

3.3.6 Variations of CD0

and K for a fighter airplane

Reference 1.10, chapter 2 has given drag polars of F-15 fighter airplane at

M = 0.8, 0.95, 1.2, 1.4 and 2.2.These are shown in Fig.3.31. These drag polars

were also fitted with Eq.(3.49) and CD0

and K were calculated. The variations of

CD0

and K are shown in Figs.3.32a & b. It is interesting to note that CD0 has a

peak and then decreases, whereas K increases monotonically with Mach

number. It may be recalled that the Mach number, at which CD0

has the peak

value, depends mainly on the sweep of the wing.

Fig.3.31 Drag polars at different Mach numbers for F15 (Reproduced from

Ref.1.10, chapter 2 with permission from McGraw-Hill book company)

Please note: The origins for polars corresponding to different Mach numbers are

shifted.



Fig.3.32a Typical variations of CD0

with Mach number for a fighter airplane

Fig.3.32b Typical variations of K with Mach number for a fighter airplane



3.3.7 Area ruling

The plan view of supersonic airplanes indicates that the area of cross section

of fuselage is decreased in the region where wing is located. This is called area

ruling. A brief note on this topic is presented below.

It was observed that the transonic wave drag of an airplane is reduced when the

distribution of the area of cross section of the airplane, in planes perpendicular to

the flow direction, has a smooth variation. In this context, it may be added that

the area of cross section of the fuselage generally varies smoothly. However,

when the wing is encountered there is an abrupt change in the cross sectional

area. This abrupt change is alleviated by reduction in the area of cross section of

fuselage in the region where the wing is located. Such a fuselage shape is called

‘Coke-bottle shape’. Figure 3.33c illustrates such a modification of fuselage

shape.



Fig.3.33 Design for low transonic wave drag

(a) Abrupt change in cross sectional area at wing fuselage junction

(b) Coke-bottle shape

Figure 3.34, based on data in Ref.1.9 , chapter 5, indicates the maximum wave

drag coefficient, in transonic range, for three configurations viz (i) a body of

revolution (ii) a wing-body combination without area ruling and (iii) a wing-body

combination with area ruling (Ref.1.9, chapter 5 may be referred to for further

details). Substantial decrease in wave drag coefficient is observed as a result of

area ruling. Figure 3.35. presents a practical application of this principle.



CDWave = 0.0035 0.008 0.0045

Fig.3.34 Maximum transonic wave drag coefficient of three different shapes

(a) body of revolution (b) wing-body combination without area ruling (c) wing-

body combination with area ruling

Fig.3.35 An example of area ruling - SAAB VIGGEN

(Adapted from http://upload.wikimedia.org)




When the free stream Mach number is more than five, the changes in

temperature and pressure behind the shock waves are large and the treatment of

the flow has to be different from that at lower Mach numbers. Hence, the flows

with Mach number greater than five are termed hypersonic flows. Reference 3.19

may be referred to for details. For the purpose of flight mechanics it may be

mentioned that the drag polar at hypersonic speeds is given by the following

modified expression (Ref.1.1, chapter 6).

CD= C

D0 (M) + K (M)CL3/2 (3.52)

Note that the exponent of the CL term is 1.5 and not 2.0.


The ratio CL/ C

D is called lift to drag ratio. It is an indicator of the aerodynamic

efficiency of the design of the airplane. For a parabolic drag polar CL/ C

D can be

worked out as follows.

CD= C

D0 +KC

L2

Hence, CD / CL = (C

D0 / C

L) +KC

L (3.53)

Differentiating Eq.(3.53) with CL and equating to zero gives CLmd

which

corresponds to minimum of (CD / CL

) or maximum of (CL / C

D).

CLmd = (C

D0 / K)1/2 (3.54)

CDmd = C

D0 + K (C

Lmd)2 = 2 C

D0 (3.55)

(L/D)max = (CLmd / CDmd

) = D0

1

2 C K (3.56)

Note:

To show that CLmd

corresponds to minimum of (CD / CL

), take the second

derivative of the right hand side of Eq.(3.53) and verify that it is greater than zero.



3.6 Other types of drag

Subsections 3.1.1, 3.2.2, 3.2.14, 3.2.17 and 3.3.2. dealt with the skin

friction drag, pressure drag (or form drag), profile drag , interference drag ,

parasite drag, induced drag, lift dependent drag and wave drag. Following

additional types of drags are mentioned briefly to conclude the discussion on this

topic.

3.6.1Cooling drag

The piston engines used in airplanes are air cooled engines. In such engines, a

part of free stream air passes over the cooling fins and accessories. This causes

some loss of momentum and results in a drag called cooling drag.

3.6.2 Base drag

If the rear end of a body terminates abruptly, the area at the rear is called a

base. An abrupt ending causes flow to separate and a low pressure region exists

over the base. This causes a pressure drag called base drag.


Presence of external fuel tank, bombs, missiles etc. causes additional parasite

drag which is called external stores drag. Antennas, lights etc. also cause

parasite drag which is called protuberance drag.

3.6.4 Leakage drag

Air leaking into and out of gaps and holes in the airplane surface causes

increase in parasite drag called leakage drag.

3.6.5 Trim drag

In example 1.1 it was shown that to balance the pitching moment about c.g.

(Mcg), the horizontal tail which is located behind the wing produces a lift (- LT) in

the downward direction. To compensate for this, the wing needs to produce a lift

(L W

) equal to the weight of the airplane plus the downward load on the tail i.e. LW

= W + LT. Hence, the induced drag of the wing, which depends on Lw, would be

more than that when the lift equals weight. This additional drag is called trim drag

as the action of making Mcg equal to zero is referred to as trimming the airplane.



It may be added that a canard surface is located ahead of the wing and the lift on

it, to make Mcg equal to zero, is in upward direction. Consequently, the lift

produced by the wing is less than the weight of the airplane. SAAB Viggen

shown in Fig.3.35, is an example of an airplane with canard. Reference1.15 and

internet (www.google.com) may be consulted for details of this airplane.



Chapter 3





3.7.3 Ways to increase maximum lift coefficient viz. increase in camber,


3.7.4 Guidelines for values of maximum lift coefficients of wings with




An airplane, by definition, is a fixed wing aircraft. Its wings can produce lift only

when there is a relative velocity between the airplane and the air. The lift (L)

produced can be expressed as :

2L

1L= ρV SC

2 (3.57)

In order that an airplane is airborne, the lift produced by the airplane must be

atleast equal to the weight of the airplane i.e.

2L

1L = W = ρ V S C

2 (3.58)

Or L

2WV =

ρSC (3.59)

However, CL has a maximum value, called LmaxC , and a speed called ‘Stalling

speed (Vs)’ is defined as :

sLmax

2WV =

ρSC (3.59a)



The speed at which the airplane takes-off ( T0V ) is actually higher than the

stalling speed.

It is easy to imagine that the take-off distance would be proportional 2T0V and in

turn to 2SV . From Eq.(3.59a) it is observed that to reduce the take-off distance (a)

the wing loading (W/S) should be low or (b) the CLmax

should be high. Generally,

the wing loading of the airplane is decided by considerations like minimum fuel

consumed during cruise. Hence, it is desirable that CLmax

should be as high as

possible to reduce the take-off and landing distances. The devices to increase

the CLmax

are called high lift devices.


Consider an airfoil at low angle of attack (α). Figure 3.36a shows a flow

visualization picture of the flow field. Boundary layers are seen on the upper and

lower surfaces. As the pressure gradients on the upper and lower surfaces of the

airfoil are low at the angle of attack under consideration, the boundary layers on

these surfaces are attached. The lift coefficient is nearly zero. Now consider the

same airfoil at slightly higher angle of attack (Fig.3.36b). The velocity on the

upper surface is higher than that on the lower surface and consequently the

pressure is lower on the upper surface as compared to that on the lower surface.

The airfoil develops higher lift coefficient as compared to that in Fig.3.36a.

However the pressure gradient is also higher on the upper surface and the

boundary layer separates ahead of the trailing edge (Fig.3.36b). As the angle of

attack approaches about 15o the separation point approaches the leading edge

of the airfoil (Fig.3.36c). Subsequently, the lift coefficient begins to decrease

(Fig.3.36d) and the airfoil is said to be stalled. The value of α for which Cl equals

maxCl is called stalling angle (αstall). Based on the above observations, the stalling

should be delayed to increase maxCl .



Fig.3.36a Flow past an airfoil at low angle of attack. Note: The flow is from left to

right (Adapted from Ref.3.20, chapter 6 with permission of editor)

Fig.3.36b Flow past an airfoil at moderate angle of attack.

Note: The flow is from right to left

(Adapted from Ref. 3.21, part 3 section II B Fig.200 with permission from

McGraw-Hill book company)



Fig.3.36c Flow past an airfoil at angle of attack near stall. Note: The flow is from

left to right (Adapted from Ref.3.12, chapter 6 with permission of editor)

Fig.3.36d Typical Cl vrs α curve

Remark:

Since stalling is due to separation of boundary layer, many methods have

been suggested for boundary layer control. In the suction method, the airfoil

surface is made porous and boundary layer is sucked (Fig.3.37a). In the blowing



method, fluid is blown tangential to the surface and the low energy fluid in the

boundary layer is energized (Fig.3.37b). Blowing and suction require supply of

energy and are referred to as active methods of control. The energizing of the

boundary layer can be achieved in a passive manner by a leading edge slot

(Fig.3.37c) and a slotted flap which are described in section 3.7.3. Reference

3.11, chapter 11 may be referred for other methods of boundary layer control and

for further details.

a. Suction

b. Blowing

c. Blowing achieved in a passive manner

Fig.3.37 Boundary layer control with suction and blowing



3.7.3. Ways to increase maximum lift coefficient viz. increase in camber,


Beside the boundary layer control, there are two other ways to increase

the maximum lift coefficient of an airfoil ( maxCl ) viz. increase of camber and

increase of wing area. These methods are briefly described below.

I) Increase in maximum lift coefficient due to change of camber

It may be recalled that when camber of an airfoil increases, the zero lift

angle ( 0α l ) decreases and the Cl vs α curve shifts to the left (Fig.3.38). It is

observed that αstall does not decrease significantly due to the increase of

camber and a higher maxCl is realized (Fig.3.38). However, the camber of the

airfoil used on the wing is chosen from the consideration that the minimum drag

coefficient occurs near the lift coefficient corresponding to the lift coefficient

during cruise. One of the ways to achieve a temporary increase in the camber

during take-off and landing is to use flaps. Some configurations of flaps are

shown in Fig.3.39. In a plain flap the rear portion of the airfoil is hinged and is

deflected when maxCl is required to be increased (Fig.3.39a). In a split flap only

the lower half of the airfoil is moved down (Fig.3.39b). To observe the change in

camber brought about by a flap deflection, draw a line in-between the upper and

lower surfaces of the airfoil with flap deflected. This line is approximately the

camber line of the flapped airfoil. The line joining the ends of the camber line is

the new chord line. The difference between the ordinates of the camber line and

the chord line is a measure of the camber.



Fig.3.38 Increase in Clmax

due to increase of camber



Fig.3.39 Flaps, slot and slat

II) Increase in maximum lift coefficient due to boundary layer control

In a slotted flap (Fig.3.39c) the effects of camber change and the

boundary layer control (see remark at the end of section 3.7.2) are brought

together. In this case, the deflection of flap creates a gap between the main

surface and the flap (Fig.3.39c). As the pressure on the lower side of airfoil is

more than that on the upper side, the air from the lower side of the airfoil rushes

to the upper side and energizes the boundary layer on the upper surface. This

way, the separation is delayed and maxCl increases (Fig.3.40). The slot is

referred to as a passive boundary layer control, as no blowing by external source

is involved in this device. After the success of single slotted flap, the double

slotted and triple slotted flaps were developed (Figs.3.39d and e).



Fig.3.40 Effects of camber change and boundary layer control on maxCl

III) Increase in Clmax due to change in wing area

Equation (3.57) shows that the lift can be increased when the wing area

(S) is increased. An increase in wing area can be achieved if the flap, in addition

to being deflected, also moves outwards and effectively increases the wing area.

This is achieved in a Fowler flap (Fig.3.39f). Thus a Fowler flap incorporates

three methods to increase maxCl viz. change of camber, boundary layer control

and increase of wing area. It may be added that while defining the maxCl , in case

of Fowler flap, the reference area is the original area of the wing and not that of

the extended wing.

A zap flap is a split flap where the lower portion also moves outwards as

the flap is deflected.

IV) Leading edge devices

High lift devices are also used near the leading edge of the wing. A slot

near the leading edge (Fig.3.39g) also permits passive way of energizing the

boundary layer. However, a permanent slot, in addition to increasing the lift, also

increases the drag and consequently has adverse effects during cruise. Hence, a



deployable leading edge device called ‘Slat’ as shown in Fig.3.39h is used. When

a slat is deployed it produces a slot and increases maxCl by delaying separation.

On high subsonic speed airplanes, both leading edge and trailing edge

devices are used to increase maxCl (Fig.1.2c).

Remarks:

i) References 1.9, 1.10, 1.12 and 3.9 may be referred for other types of high

lift devices like Kruger flap, leading edge extension, blown flap etc.

ii) Reference1.10, chapter 1 may be referred for historical development of

flaps.

3.7.4 Guide lines for values of maximum lift coefficients of wings with


An estimate of the maximum lift coefficient of a wing is needed to calculate

the stalling speed of the airplane. It may be added that the maximum lift

coefficient of an airplane depends on (a) wing parameters (aspect ratio, taper

ratio and sweep) (b) airfoil shape, (c) type of high lift device(s), (d) Reynolds

number , (e) surface finish , (f) the ratio of the area of the flap to the area of wing

and (g) interference from nacelle and fuselage.

Table 3.6 presents the values of CLmax

which are based on (a) Ref.1.10,

chapter 5, (b) Ref.3.9 chapter 5 and (c) Ref.3.15 chapter 5. These values can be

used for initial estimate of CLmax

for subsonic airplanes with unswept wings of

aspect ratio greater than 5.

The quarter chord sweep( 1/4 ) has a predominant effect on CLmax

. This effect,

can be roughly accounted for by the following, cosine relationship:

(CLmax

)Λ = (C

Lmax)Λ=o

cos 1/4

For example, when the unswept wing without flap has CLmax

of 1.5, the same

wing with 30o sweep would have a CLmax

of 1.5 x cos 30o or 1.3. Similarly, an

unswept wing with Fowler flap has CLmax

of 2.5. The same wing with 30o sweep



would have CLmax

of 2.5 x cos 30o or 2.17. With addition of leading edge slat, this

can go upto 2.43.

Type of flap Guideline for CLmax

in landing configuration

No flap 1.5

Plain flap 1.8

Single slotted flap 2.2

Double slotted flap 2.7

Double slotted flap with slat 3.0

Triple slotted flap 3.1

Triple slotted flap with slat 3.4

Fowler flap 2.5

Fowler flap with slat 2.8

Table 3.6 Guidelines for CLmax

of subsonic airplanes with unswept wings of

moderate aspect ratio

Figure 3.41 shows CLmax

for some passenger airplanes. The solid lines

correspond to the cosine relation given above.



Fig.3.41 Maximum lift coefficient of passenger airplanes operating at high

subsonic Mach numbers

(Adapted from Ref.3.22, Chapter 8 with permission of authors)

Remarks:

i) The value of CLmax

shown in Table 3.6 can be used in landing configuration.

The flap setting during take-off is lower than that while landing. The maximum lift

coefficient during take-off can be taken approximately as 80% of that during

landing.

ii) The values given in Table 3.6 should not be used for supersonic airplanes

which have low aspect ratio wings and airfoil sections of small thickness ratio.

Reference 3.5, section 4.1.3.4 may be referred to for estimating CLmax

in these

cases.

iii) As the Mach number (M) increases beyond 0.5, the maxCl of the airfoil section

decreases due to the phenomena of shock stall (see item IV in section 3.3.3).

Hence CLmax of the wing also decreases for M > 0.5. The following relationship

between CLmax

at M between 0.5 to 0.9, in terms of CLmax

at M = 0.5, can be

derived based on the plots of CLmax

vs M in Ref.3.23, chapter 9, and Ref.3.9

chapter 12.



Lmax M

Lmax M=0.5

(C )= - 0.418M + 1.209 , 0.5 M 0.9

(C )

For example at M = 0.9, CLmax

would be about 0.833 of CLmax

at M = 0.5.

Note: The maximum lift coefficient (CLmax

) in transonic Mach number range is not

likely to be monotonic as seen in Fig.3.27a. At transonic and supersonic Mach

numbers, CLmax

must be estimated at each Mach number. Reference 3.5,

section 4.1.3.4 may be consulted for this estimation.



Chapter 3

References

3.1 Biermann, D. and Herrnitein Jr., W.H.“The interference between struts in

various combinations” NACA TR 468, (1934). Note: This report can be

downloaded from the site “NASA Technical Report Server(NTRS)”.

3.2 Apelt, C.J. and West, G.S. “The effects of wake splitter plates on bluff body

flow in the range of 104 < Re < 5 x 104 part-2” J.Fluid Mech. Vol.71, pp 145-160,

(1975).

3.3. Hoerner, S.F. “Fluid dynamic drag” Published by author (1965).

3.4. Royal Aeronautical Society data sheets – Now known as Engineering

Sciences Data Unit (ESDU).

3.5. Hoak, D.E. et al. “USAF stability and control DATCOM,” Air Force Wright

Aeronautical Laboratories Technical Report 83-3048, October 1960. (Revised

April 1978). Note: USAF Digital DATCOM can be accessed from net.

3.6 Roskam, J. “Methods for estimating drag polars of subsonic airplanes”

Roskam aviation and engineering (1973).

3.7. Wood K.D. “Aerospace vehicle design Vol.I” Johnson Publishing Co.,

Boulder, Colarado (1966).

3.8. Torenbeek, E. “Synthesis of subsonic airplane design” Delft University Press

(1982).

3.9. Raymer D.P.“Aircraft design: A conceptual approach” AIAA Educational

Series, Fourth Edition (2006).

3.10 Schlichting,H. “Boundary layer theory” McGraw-Hill (1968).

3.11 Schlichting, H. and Gersten , K. “Boundary layer theory” 8th Edition,

Spinger-Verlag, (2000).

3.12 Anderson,Jr. J.D. “Fundamentals of aerodynamics” McGraw-Hill,

International Edition (1988).

3.13 White,F.M. “Viscous fluid flow” 2nd Edition, McGraw-Hill (1991).

3.14 Abbott, I.H and Von Doenhoff, A.E. “Theory of wing sections” Dover (1959).



3.15 Howe, D. “Aircraft conceptual design synthesis” Professional Engineering

Publishing Limited, London (2000).

3.16 Duncan, W.J., Thom, A.S., and Young, A.D. “Mechanics of fluids” E.L.B.S.

and Edward Arnold, (1975).

3.17 Schlichting, H.and Truckenbrodt, E.D. “Aerodynamics of the airplane”

translated by H.J. Ramm, McGraw Hill, (1979).

3.18 Roskam, J “Airplane design volume I-VIII” Roskam Aviation and

engineering (1990).

3.19 Anderson, Jr, J.D. “Hypersonic and high temperature gas dynamics”

McGraw Hill (1989).

3.20 Oertel, H. (Editor) “Prandtl’s essentials of fluid mechanics” Second edition

Springer-Verlag,(2004).

3.21 Kaufmann, W. “Fluid mechanics” McGraw Hill ,(1963).

3.22 Jenkinson L. R., Simpkin P. and Rhodes D. “Civil jet aircraft design” Arnold,

(1999).

3.23 Huenecke. K, “Combat aircraft design” Airlife Pub. Co. (1987).



Chapter 3

Exercises

3.1 Following data relate to a light airplane.

W =11000 N

Wing: S = 15 m2, C

D0 = 0.007, A.R. = 6.5, taper ratio () = 1.0, e = 0.9.

Fuselage: Has a drag of 136N at V = 160 km/hr at sea level when the angle of

attack is zero.

Horizontal tail: CD0

= 0.006, St = 2.4 m2

Vertical tail: CD0

= 0.006, Sv = 2.1 m2

Other components: CD0

based on wing area = 0.003

Estimate the drag polar of the airplane assuming the contribution of the fuselage

to the lift dependent drag as small.

[Answer: CD = 0.0193 + 0.0544 C

L2 ]

[Remark : The CDO of wing in this exercise appears higher than CDO of tails. It is

likely that the airfoil section used on Wing may be thicker (say 15 to 18%) and

that on tail be thinner (say 9%).]

3.2 A drag polar is given as:

CD = C

D0 + KC

Ln

Show that:

CLmd

= 1/nD0C{ }K(n-1)

, CDmd

=n

n-1C

D0

(CL/C

D) max = (n-1)/n 1/n

1/n D0

1 1{ }

n C K(n-1)n-1

Verify that when n = 2, the above expressions reduced to those given by

Eqs. (3.54),(3.55) and (3.56).

3.3 Based on data in Ref.1.1, chapter 6, the drag polar of a hypersonic glider is

given in the table below.



CL 0 0.05 0.1 0.15 0.2

CD 0.028 0.0364 0.05 0.07 0.0907

Fit Eq.(3.52) to this data and obtain CD0

and K. Also obtain CLmd

, CDmd and

(CL/C

D)max

. The expressions mentioned in exercise 3.2 can be used.

[Answers: CD0

= 0.0283, K = 0.703, CD = 0.0283+0.0703 C

L3/2

CLmd

= 0.1865, CDmd

= 0.0849, (CL/C

D)max

= 2.197].

Flight dynamics-I Prof. E.G.Tulapurkara Chapter IV

Indian Institute of Technology, Madras 1

Chapter 4

Engine characteristics

(Lectures 13 to 16)

Keywords: Engines for airplane applications; piston engine; propeller

characteristics; turbo-prop, turbofan and turbojet engines; choice of engine for

different applications.

Topics

4.1 Introduction

4.1.1 Engines considered for airplane applications

4.2 Piston engine-propeller combination

4.2.1 Operating principle of a piston engine

4.2.2 Effect of flight speed on the output of a piston engine

4.2.3 Effect of altitude on the output of a piston engine

4.2.4 Specific fuel consumption (SFC)

4.2.5 The propeller

4.2.6 Propeller efficiency

4.2.7 Momentum theory of propeller

4.2.8 Parameters for describing propeller performance and typical

propeller characteristics

4.2.9 Selection of propeller diameter for chosen application

4.2.10 Procedure for obtaining propeller efficiency for given h,V, BHP

and N

4.2.11 Variations of THP and BSFC with flight velocity and altitude

4.2.12 Loss of propeller efficiency at high speeds

4.3 Gas turbine engines

4.3.1 Propulsive efficiency

4.3.2 Why turboprop, turbo fan and turbojet engines?

4.3.3 Characteristics of a typical turboprop engine

4.3.4 Characteristics of a typical turbofan engine



4.3.5 Characteristics of a typical turbojet engines

4.4 Deducing output and SFC of engines where these characteristics are

not available directly

4.5 A note on choice of engines for different range of flight speeds

References

Excercises



Chapter 4

Lecture 13 Engine characteristics – 1 Topics

4.1 Introduction







4.2.5 The propeller



4.1. Introduction

To evaluate the performance of an airplane we need to know the

atmospheric characteristics, the drag polar and the engine characteristics like

variations of thrust (or power) output and specific fuel consumption with flight

speed and altitude. In this chapter the engine characteristics are briefly reviewed.


Following power plants are considered for airplane applications.

(a) Piston engine-propeller combination.

(b) Gas turbine engines - turboprop, turbofan and turbojet.

(c) Ramjets.

(d) Rockets.

(e) Combination power plants like ramrocket and turboramjet.

At present, piston engine-propeller combination and gas turbine engines are the

power plants used on airplanes. Ramjets offer simplicity of construction and have



been proposed for hypersonic airplanes. However, a ramjet cannot produce any

thrust when flight speed is zero. Hence, it is proposed to use a rocket or turbojet

engine to bring it (ramjet) to a flight speed corresponding to Mach number (M) of

2 or 3 and then the ramjet engine would take over. Consequently, the

combination power plants viz. ramrocket or turboramjet have been proposed.

Rockets have sometimes been used on airplanes as boosters to increase

the thrust for a short duration e.g. during take-off.


In this case the output of the engine viz. brake horse power (BHP) is

available at the engine shaft and is converted into thrust by the propeller.


A few relevant facts about the operation of piston engines, used on

airplanes, are mentioned here. In these engines a certain amount for fuel-air

mixture is taken in, it is compressed, then ignition, due to a spark, takes place

which is followed by the power stroke and the exhaust stroke (Fig.4.1).

Remarks:

i ) The piston engine in which the ignition is caused by a spark from the spark

plug is called a spark-ignition engine. There are other types of piston engines in

which the pressure and temperature at the end of the compression stroke are

high enough to cause spontaneous ignition. Such engines do not need a spark

plug and are known as compression-ignition engines.



Fig.4.1 Four stroke cycle of a spark-ignition engine

ii) The volume of the air-fuel mixture taken in, is almost equal to the swept

volume i.e., product of the area of cross-section of the engine cylinder and the

length of the piston stroke. The mass of fuel taken in per power stroke is thus

approximately equal to:

(swept volume) X (density of air) / (air-fuel ratio).


For a given altitude and r.p.m. (N) the power output changes only slightly

with flight speed. This is because the piston engines are generally used at low

speeds (M < 0.3) and at these low Mach numbers, the increase in manifold

pressure due to the deceleration of air in the engine manifold is negligible. Hence

power output increases only slightly with flight speed. This increase is generally

ignored.


To understand the effect of altitude on the output of the piston engine, the

following three facts need to be noted. (a) As stated at the end of the subsection

4.2.1, the mass of fuel taken in per stroke is equal to the product of swept volume

and density of air divided by air-fuel ratio. (b) For complete combustion of fuel,



the air-fuel ratio has a definite value (around 15, the stoichiometric ratio). (c) As

the flight altitude increases, the density of air decreases.

Thus, for a given engine r.p.m. and air-fuel ratio, the mass of air and

consequently, that of the fuel taken in decreases as the altitude increases. Since,

the power output of the engine depends on the mass of the fuel taken in, it

(power output) decreases with altitude. The change in power output (P) with

altitude is roughly given as (Ref.3.7,Appendix 1 A-5 and Ref.4.3, chapter 14):

(P / P0) =1.13σ – 0.13 (4.1)

where P0 is the power output at sea level under ISA conditions and σ is the

density ratio.

Remark:

(i) Reference 3.15, chapter 3, gives the following alternate relationship for

decrease of power output with altitude :

(P / P0) = σ1.1 (4.1a)

(ii) Figure 4.2 shows the performance for a typical piston engine. To prepare

such a performance chart, the engine manufacturer carries out certain tests, on

each new engine. During these tests the engine is run at a chosen RPM and

different loads are applied. The throttle setting is adjusted to get steady

conditions. The quantities like (a) engine RPM(N), (b) torque developed, (c)

manifold air pressure(MAP) and (d) the fuel consumed in a specific interval of

time, are measured.These tests are conducted at different RPM’s. From these

test data the power output and the fuel flow rate per hour are calculated. The

data are also corrected for any difference between the ambient conditions during

the test, and the sea level standard conditions. The left side of Fig.4.2 presents

the sea level performance of a Lycoming engine. The upper part of the figure



Fig.4.2 Typical piston engine performance (Lycoming 0-360-A)

(with permission from Lycoming aircraft engines )

shows the power output at different MAP’s with RPM as parameter. The lower

part of the figure shows the fuel flow rate in US gallons per hour.

To obtain the effect of altitude on the engine output, the power output is

measured at different RPM’s and MAP’s during flight tests at different altitudes.

Typical altitude performance of Lycoming engine is presented in the right side of

Fig.4.2.

From such a chart, the output of the engine and the fuel flow rate can be

obtained for a chosen combination of altitude, RPM, MAP and ambient

temperature. The steps to obtain these are explained with the help of examples

4.1 and 4.2



It may be added that the units used in Fig.4.2, which is reproduced from

manufacturer’s catalogue, are in FPS system. However, SI units are used in this

and the subsequent chapters.


In engine performance charts, the fuel consumption is presented as fuel

flow rate per hour. However, in engineering practice the fuel consumption is

expressed as specific fuel consumption (SFC). It is defined as :

Fuel flow rate in Newton per hour

SFC = BHP in kW

(4.1b)

Remarks :

(i) The output of a piston engine or turboprop engine is available as power at the

engine shaft. It is called BHP and measured in HP when FPS system is used. In

SI units the output is measured in kW. On the other hand, the output of a

turbofan or a turbojet engine is available as thrust, which is measured in ‘lb’ in

FPS system and in Newton in SI units.

The specific fuel consumption of a jet engine is defined as:

SFC Fuel flow rate in Newton per hour

=Thrust in Newton

(4.1c)

(ii) To distinguish the specific fuel consumption of a piston or a turboprop engine,

from that of a jet engine, the SFC defined by Eq.(4.1b), is denoted as BSFC i.e.


BSFC =BHP in kw

with units of N/kW-hr (4.1d)

The specific fuel consumption of a turbofan or a turbojet engine is denoted by

TSFC i.e.


TSFC =Thrust in Newton

with units of hr-1 (4.1e)

(iii) BSFC in metric units is also expressed as mg/W-s



Example 4.1

Obtain the power output and BSFC for the Lycoming engine when

operating at sea level at an RPM(N) of 2400 and MAP of 24 of mercury (Hg).

Solution :

From plots in the left side of Fig.4.2, for N = 2400 and MAP = 24 of Hg

the power output is 136 HP and the fuel flow rate is 10.7 US gallons/hr.

Taking 1 US gallon = 3.78 litre and density of petrol as 0.76 kg/m3 gives:

1 gallon per hour of petrol = 3.78 x 0.76 kg/hr

= 3.78 x 0.76 x 9.81 N/hr

= 28.18 N/hr of petrol

Hence, the fuel flow rate in the case under study is :

10.7 x 28.18 = 301.5 N/hr.

Noting that 1 lb/hr = 4.45 N/hr, The fuel flow rate in this case is 67.75 lbs/hr.

Further, 1 HP is 0.7457 kW. Hence, power output of 136 HP equals 101.4 kW.

Hence, BSFC in SI units is: 301.5/101.4 = 2.973 N/kW-hr

In FPS units it is: 67.75/136 = 0.498 lb/HP-hr

Answers:

For the given engine, the power output, fuel flow rate and BSFC at N = 2400 and

MAP = 24 of Hg under sea level standard conditions are :

(i)Power output = 101.4 kW = 136 HP, (ii) Fuel flow rate = 10.7 US gallons/hr

or 301.5 N/hr or 67.75 lb/hr of petrol (iii) BSFC = 2.973 N/kW-hr = 0.498 lb/HP-hr

Example 4.2

Obtain the power output and BSFC for the Lycoming engine when

operating at 8000 altitude, RPM (N) of 2200 and MAP of 20 of Hg.

Solution :

Reference 1.9 chapter 6, gives the following procedure to obtain the

output and fuel flow rate using left and right sides of Fig.4.2.

(i)At sea level for N = 2200 and MAP of 20of Hg the output would be 97.5 BHP.

This is indicated by point ‘B’ in the left hand side of Fig.4.2. This side of the

diagram is also called sea level performance.



(ii)Transfer this point to the right hand side of Fig.4.2 at sea level which is

indicated by point ‘C’. The right side of the diagram is also called altitude

performance.

(iii)Locate a point on the altitude curve corresponding to N = 2200 and MAP of

20 of Hg. This point is indicated by ‘A’.

(iv)Join points C and A by a dotted line. The value at 8000 on this line (the point

‘D’) is the output at h = 8000 corresponding to N = 2200 and MAP = 20 of Hg.

It is seen that the value is 107 HP.

(v)To get the fuel flow rate, mark a point ‘F’ on the sea level performance at 107

HP and N = 2200. The MAP at this point is observed to be 21.2 of Hg. The fuel

flow rate corresponding to N = 2200 and MAP of 21.2 of Hg, from the lower part

of figure in the left side is 8.25 gallons per hour. This point is indicated by ‘G’

Hence, at h = 8000 , N = 2200 and MAP of 20 . The output is 107 HP (79.79

kW) and the fuel flow rate is 8.25 gallons / hr (232.5 N/hr or 52.2 lbs/hr of petrol).

Consequently, BSFC = 232.5

= 2.914 N/kW-hr79.79

or in FPS units, BSFC = 52.2

= 0.488 lb/HP-hr107

Answers :

At h = 8000 , N = 2200 and MAP = 20 of Hg :

Output = 79.79 kW =107 HP and BSFC =2.914 N/kW-hr = 0.488 lb/BHP-hr

Note : Reference 1.9, chapter 6 may be referred to obtain the correction to the

output if the ambient temperature is different from that in ISA.

4.2.5 The propeller

The output of the engine is converted into thrust by the propeller. A typical

engine with a two bladed propeller is shown in Fig.4.3. Depending on the engine

power and the operating conditions, the propeller may have two to four blades.

Special propellers with five or six blades have also been used in practice when

required.



Fig.4.3 Typical engine-propeller combination

(Source: www.flickr.com)

The propeller blade, as seen in Fig.4.3, is like a wing with significant amount of

twist. Refer Fig.3.3 for geometric parameters of a wing. The geometry of the

propeller is defined by the following features. (a) The variation of the chord,

shape and thickness of the airfoil section (also called blade element) over the

span of the blade. (b) The angle between the chord of the blade element and the

plane of rotation. This is also one of the definitions of the pitch angle (β).

The pitch angle (β) varies along the span of the blade for the following reason.

Since the propeller blade moves forward as it rotates, the blade element has a

forward velocity of V and a circumferential velocity of 2πrn ; where ‘r’ is the

radius of the blade element and ‘n’ is the revolutions per second of the propeller.

The blade element experiences a relative wind which is resultant of the forward

and circumferential velocities. As ‘r’ varies from the root to the tip, the blade

elements at various spanwise locations of the propeller are subjected to a

relative wind which varies significantly, in magnitude and direction, along the

span. Further, each blade element being an airfoil, must operate at a moderate

angle of attack. These two considerations require that the blade elements along

the span of the blade make different angles to the plane of rotation or have



different pitch angles (β). The pitch of the blade is generally the pitch of the blade

element at r/R = 0.75, where R is the radius of the blade.

For other definitions of pitch Ref.2.1 and chapter 6 of Ref.1.9, be

consulted. For details of the geometry of propellers refer chapter 6 of Ref.1.9,

chapter 16 of Ref.3.7 and Ref.4.1.


Consider that an engine located in an airplane is developing certain output

indicated as BHP. The propeller attached to this engine produces a thrust T

when the airplane moves with a speed V . In this situation, the power output

called ‘Thrust horse power (THP)’, is (TV / 1000) in kW. The efficiency of the

propeller is therefore defined as:

ηρ = THP / BHP = T V /(1000 x BHP) ( 4.2 )

Note: T is in Newton, V is in m/s and THP and BHP are measured in kW.

The efficiency of a propeller can be estimated by analysing the flow through a

propeller. The momentum theory of propeller is briefly discussed in the next

subsection. Subsequent subsections deal with determination of propeller

efficiency from experimental results.


As the name suggests, this theory is based on the idealization that the

thrust produced by the propeller is the result of the increase in momentum

imparted to the airstream passing through the propeller. It is assumed that the

propeller can be thought of as an actuator disc. This disc is an idealised device

which produces a sudden pressure rise in a stream of air passing through it. This

pressure rise integrated over the disc gives the thrust developed by the propeller.

Figure 4.4 shows the actuator disc and the flow through it. It is assumed that : (i)

the flow is incompressible and inviscid, (ii) the increase in pressure is constant

over the disc (iii) there is no discontinuity in flow velocity across the disc (iv) in

the flow behind the disc, called slipstream, there is no swirl.



Fig.4.4 Flow through an actuator disc

(a) Stream tube (b) Pressure variation

In Fig.4.4 the actuator disc is located at plane AA. Far upstream, the velocity is

V and the pressure p is the atmospheric pressure. The velocity V equals the

forward speed of the airplane on which the propeller is mounted. A stream tube

enclosing the disc is also shown in Fig.4.4. As the stream approaches the front

face of the disc the fluid velocity reaches a value V1 at the disc. As the flow is

assumed to be inviscid and incompressible, Bernoulli’s equation is valid till the

front face of the disc and the pressure decreases, to a value p1. At the disc,

energy is added in the form of increase in pressure by an amount Δp while the

velocity remains the same as V1 through the disc (Fig.4.4a). After the disc the

pressure gradually returns to the atmospheric value of p . Bernoulli’s equation is

again valid behind the disc and the fluid velocity increases to a value Vj. The

changes in pressure and velocity are shown in Fig.4.4a.

Applying Bernoullis equation ahead and behind the disc gives :



Total head ahead of disc = H = 1

2 21

1 1p + ρV = p + ρV

2 2 (4.3)

Total head behind the disc = H1 = 2 21 1 j

1 1p +Δp + ρV = p + ρV

2 2 (4.4)

Consequently, 2 21 1

2 22 2

1 j j

1Δp = H - H = p ρV p ρV = ρ V - V

2 (4.5)

Since Δp is the change in pressure over the disc, the thrust acting on the disc is:

2 2j

ρT = AΔp=A V - V

2 (4.6)

where, A = area of disc = 2d4

; d = diameter of the propeller

Alternatively, the thrust produced can also be obtained as the rate of change of

momentum of the stream i.e.

jT = m V - V (4.7)

where, m = rate of mass flow through the disc = 1ρA V (4.8)

Hence, 1 jT = ρA V V - V (4.9)

Equating Eqs.(4.6) and (4.9) yields :

2 21 j j

ρρAV V - V = A V - V

2

Or j1

V + VV =

2 (4.10)

Thus, the momentum theory shows that the velocity at the disc (V1) is the

average of Vj & V . In other words, half of the increase in velocity takes place

ahead of the disc and the remaining half behind it.

The efficiency of the actuator disc can be obtained by considering the ratio of

power output to the power input.

The power output = work done = jT V = m V V V (4.11)

The power input is the energy imparted to the fluid stream. This is the energy of

the stream far behind the disc minus the energy of the stream far ahead of the

disc. i.e.



Power input = 2 2j

1 1mV - mV

2 2 (4.12)

Hence, propeller efficiency is:

2

2

1

j

p2 jjj

mV V V 2Vpower outputη = = = =

m Venergy input V VV V2 V

(4.13)

Remarks:

(i)Equation (4.13) gives the propeller efficiency under ideal conditions and

represents an upper limit on efficiency obtainable. In practical situations, the

efficiency would be lower due to losses associated with (a) profile drag of blades,

(b) swirl in slip stream and (c) the pressure at the blade tips being the same

ahead and behind the disc.

(ii)For production of thrust, Vj must be greater than V . But for high propeller

efficiency Vj must be only slightly higher than V . Hence to get adequate amount

of thrust with high propeller efficiency a large mass of air should be given a small

velocity increment.

(iii)Propeller theories like blade element theory, and vortex theory take into

account effects of drag of blades, finite span of blade etc. For details of these

theories refer to chapter 6 of Ref.1.9.

Example 4.3

A propeller of diameter 1.8 m is mounted on an airplane. When

moving at a speed of 200 kmph it produces a thrust of 2070 N under standard

sea level conditions. Calculate the velocity of slip stream far behind the propeller

and the ideal efficiency of the propeller.

Solution :

Diameter of propeller = d = 1.8 m

Free stream velocity = V = 200 kmph = 55.56 m/s

Slipstream velocity = Vj

Consequently, Thrust = T =

jm V - V



j2 V +Vm = ρ d

4 2

Hence, j2j

V +55.562070 = 1.225× ×1.8 V -55.56

4 2

Or 1328.1 = 2 2jV -55.56

Or jV = 66.45 m/s

Ideal propeller efficiency = j

2 2= = 0.9107 = 91.07 %

V 66.451+1+

55.56V

Answers :

Velocity of slip stream far behind propeller = 66.45 m/s = 239.22 kmph

Ideal propeller efficiency = 91.07 %



Chapter 4


4.2.8 Parameters for describing propeller performance and typical

propeller characteristics


4.2.8 Parameters for describing propeller performance and typical propeller

characteristics

As pointed out at the end of the previous subsection, the momentum

theory of propeller has limitations. Though the refined theories are helpful in

design of propeller blades, the propeller characteristics obtained from the wind

tunnel tests are used for estimation of airplane performance. These

characteristics are presented in terms of certain parameters. First these

parameters are defined and then typical characteristics of propellers are

presented. The procedures for (a) selection of the propeller diameter and (b)

obtaining the propeller efficiency for given h, v, BHP and N, are given in the next

two subsections.

Following Ref.4.1 and Ref.3.7 chapter 6, the propeller performance is

expressed in terms of the following coefficients. It may be pointed out that FPS

units are used in these references whereas SI units are used here.

Advance ratio : J = V/nd (4.14)

Power coefficient: PC = P/ρn3d5; P in Watts (4.15)

Thrust coefficient: CT = T/ρn2d4 (4.16)

Speed power coefficient: Cs = V (ρ/ Pn2)1/5 = 5PJ/ C (4.17)

Propeller efficiency: pη =TV / P; P in Watts

= J (CT / PC ) (4.18)



Torque coefficient: 2 5Q

Q=ρn d

C (4.19)

Torque speed coefficient: 3S QQ = J/ C = V ρd /Q (4.20)

where, P = Power in watts, T = thrust (N); V = flight velocity (m/s), n = rotational

speed (rev/s),

d = diameter of propeller (m)

Q = Torque (Nm) = P / 2 n

In FPS units:

T = thrust (lbs); P = power (ft lbs/s) = 550 BHP

V = velocity (ft / s), BHP = brake horse power

The performance of a propeller is indicated by thrust coefficient (CT), power

coefficient ( PC ) and efficiency ( pη ). These quantities depend on advance ratio

(J) and pitch angle β . Based on Ref.4.1, the experimental characteristics of a

two bladed propeller are presented in Figs.4.5a to d.

Figure 4.5a presents the variation of pη vs J with β as parameter. It is seen that

pη is zero when V is zero; J is also zero in this case by virtue of its

definition(Eq.4.14). Equation (4.2) also indicates that pη is zero when V is zero.

This is because even though the engine is working and producing thrust, no

useful work is done when V is zero. This is like a person pressing an immovable

wall. He spends muscular energy to push the wall but the output and hence the

efficiency is zero as the wall does not move and no useful work is done.



Fig.4.5a Propeller efficiency ( pη ) vs advance ratio (J) with pitch angle (β) as

parameter.

For a chosen value of β , the efficiency ( pη ) increases as J increases. It reaches

a maximum for a certain value of J and then decreases (Fig.4.5a). The maximum

value of pη is seen to be around 80 to 85%. However, the value of J at which the

maximum of pη occurs, depends on the pitch angleβ . This indicates that for a

single pitch or fixed pitch propeller, the efficiency is high (80 to 85%) only over a

narrow range of flight speeds (Fig.4.5a). Keeping this behaviour in view, the

commercial airplanes use a variable pitch propeller. In such a propeller the entire

blade is rotated through a chosen angle during the flight and the pitch of all blade

elements changes. Such propellers have high efficiency over a wide range of

speeds. However, propellers with variable pitch arrangements are expensive and

heavy. Hence, personal airplanes, where cost of the airplane is an important

consideration, employ a fixed pitch propeller. As a compromise, in some designs,

propellers with two or three pitch settings are employed.

Figure 4.5b presents the variation of power coefficient ( PC ) vs J with β and CT

as parameters. This chart is useful to obtain pη for given values of altitude,

velocity, RPM and BHP (see subsection 4.2.10).



Fig.4.5b Power coefficient ( PC ) vs advance ratio (J) with pitch angle (β) and

thrust coefficient (CT) as parameters.

Figure 4.5c presents the variations of CS vs J and CS vs pη with βas parameter.

This figure is designated as ‘Design chart’ and is used for selection of the

diameter of the propeller. A brief explanatory note on this topic is as follows.

Using defintions of J and PC , the parameter sC , defined below, is obtained. It is

observed that this parameter does not involve the diameter (d) of the propeller.

2 1/51/5P

sJ

C = = V (ρ / Pn )C

(4.21)



It is also observed that the parameter sC depends on V, ρ , P and N.

Consequently, this parameter can be evaluated when the power output (P),

engine RPM(N) and flight condition viz. V and h are specified.

The design problem involves obtaining the value of J which would give the

maximum value of pη for a specified value of sC . This is arrived at in the

following manner.

Fig.4.5c Design chart

Using the data in Figs.4.5b & a , the values of sC can be obtained for constant

values of J or β . For example, for β= 15o the values given in Table 4.1 are

obtained.



J PC

From Fig.4.5b

sC

From Eq.(4.21)

pη

From Fig.4.5a

0 0.04 0 0

0.2 0.04 0.381 0.43

0.4 0.037 0.773 0.69

0.6 0.025 1.255 0.805

0.8 0.005 3.685 0.35

Table 4.1 variation of sC with J for β = o15

Similar calculations at , , ,o o o o o oβ = 20 25 30 ,35 40 and 45 yield additional values.

From these values the curves for sC vs pη and sC vs J at different values of β

can be plotted. These are shown in the upper and lower parts of Fig.4.5c. Based

on these plots, the dotted line in the lower part of Fig.4.5c gives the values of J

and β which would give maximum pη . This line is designated as ‘Line of

maximum efficiency for sC ’. For example, corresponding to a value of sC = 1.4,

the dotted line gives J = 0.74 and β = o20 . The upper part of the Fig.4.5c gives

pη = 82% for the chosen value of sC = 1.4.

From the value of J, the propeller diameter is obtained as d = V/(nJ) ; note that

the values of V and n are already known. Subsection 4.2.9 gives additional

details and example 4.4 illustrates the procedure to select the propeller diameter.

Figure 4.5d presents the variation of thrust coefficient (CT) vs J with β as

parameter. It is observed that when J is zero, CT is not zero as the propeller

produces thrust, even when ‘V’ is zero. The curves in Fig.4.5d are useful to

estimate the thrust developed by the propeller especially during the take-off flight.



Fig.4.5d Thrust coefficient (CT) vs advance ratio (J) with pitch angle β as

parameter.

Fig.4.5 Typical characteristics of a two bladed propeller

(Adapted from Ref. 4.1)

Remark :

Reference 4.1 contains information on propellers with three and four

blades. Reference 3.7 chapter 16 contains information on six bladed propellers.

Additional information can be obtained from Ref.4.2 which is cited in chapter 17

of Ref.4.3.


A propeller is selected to give the best efficiency during a chosen flight

condition which is generally the cruising flight for transport airplanes. Some

companies may design their own propellers but it is an involved task. Hence, the

general practice is to use the standard propellers and the charts corresponding to

them. As a first step, the number of blades of the propeller is decided depending

on the amount of power to be absorbed by the propeller.

The designer of a new airplane generally chooses the diameter of the propeller

using the design chart (e.g. Fig.4.5c) appropriate to the propeller. Let us consider

a two bladed propeller. Following steps are used to select the diameter of a

propeller.



(a) Choose a level flight condition i.e. altitude ( ch ) and speed ( cV ).

(b) Obtain lift coefficient (CL) in this flight using :

CL = W/ 2c0.5ρV S . Obtain the corresponding CD from the drag polar of the

airplane.

(c) Obtain THP required during the flight using : THP = 3DC0.5ρV SC /1000

(d) Assume pη = 0.8 .

(e) Obtain BHP = THP/0.8. Then RPM (N) which will give this power output at the

chosen hc with low BSFC is known from the engine curves e.g. Fig.4.2.

Calculate n = N/60.

(f) Calculate 2 1/5

SC = V ρ /Pn .

(g) From the design chart like Fig.4.5c, obtain the value of J on the dotted line,

corresponding to the value of CS in step (f). Also obtain the value of β from the

same curve. Obtain the value of pη from the upper part of the design chart.

(h) Since V, n and J are known, obtain propeller diameter (d) using : d = V/n J

(i) If the value of pη obtained in step (g) is significantly different from the value of

0.8 assumed in step (d), then iterate by using the value of pη obtained in step (g).

Finally round-off the propeller diameter to nearby standard value.

Remark :

The choice of the parameters of the propeller like, diameter, pitch, blade

size are also influenced by factors like noise level of the propeller, ground

clearance, and natural frequency of the blade. Refer chapter 6 of Ref.1.9.

Example 4.4

Consider the case of Piper Cherokee airplane dealt with in Appendix A

and obtain the diameter of the propeller for this airplane. According to chapter 6

of Ref.1.9, the chosen speed and altitude for propeller design are 132 mph

(212.4 kmph or 59 m/s) and sea level standard conditions respectively. The

engine operates at 75% of the maximum power at an RPM of 2500.



Solution :

From Appendix ‘A’ the following data are obtained on Piper Cherokee

airplane.

Weight of airplane = W = 10673.28 N

Drag polar : CD = 0.0349 + 0.0755 2LC

Wing area = S = 14.864 m2 .

At sea level ρ= 1.225 kg/m3

CL under chosen flight condition is = 1. 2

10673.28= 0.3368

1× 225 × 59 ×14.864

2

CD = 0.0349 + 0.0755 x 0.33682 = 0.04346

Hence, thrust horse power required (THPr) is :

THPr =

31× 1.225 × 59 × 14.864 × 0.04346

2 = 81.26 kW1000

As a first step, assume pη = 0.8 .

Consequently, the required BHP is :

BHPr = 81.26/0.8 = 101.6 kW = 101600 W

Noting that at sea level the maximum power is 135 kW, the BHPr of 101.6 kW is

close to 75% of that value which is prescribed in the exercise.

N = 2500. Hence, n = revolutions per second = 2500/60 = 41.67

Consequently, CS = V 1 1/52 25ρ / pn = 59 1.225 /101600 × 41.67 = 1.38

The airplane has a two bladed propeller of standard design and hence Fig.4.5c is

applicable. From this figure, corresponding to CS of 1.38, the dotted line gives

J = 0.74 , oβ = 20 , pη = 0.83.

Consequently, the first estimate of propeller diameter is :

V 59d = = = 1.91m

nJ 41.67×0.74



Since, the value of pη obtained is somewhat different from the value of 0.8

assumed earlier, the steps are repeated with pη = 0.83.

BHPr = 81.26/0.83 = 97.90 kW = 97900 W

CS = 59 (1.225/97960 x 41.472)1/5 = 1.390

From Fig.4.5c corresponding to CS of 1.39, the dotted line gives:

J = 0.75 and oβ = 20 and pη = 0.83.

Consequently, the second estimate of propeller diameter is :

d = 59

= 1.89 m41.67×0.75

Since the latest value of pη is same as the value with which the steps were

repeated, the propeller diameter is taken as 1.89 m.

Remark:

The value of the propeller diameter obtained above is very close to the value of

1.88 m in the actual airplane.



Chapter 4


4.2.10 Procedure for obtaining propeller efficiency for given h,V, BHP

and N



4.3 Gas turbine engines


4.3.2 Why turboprop, turbo fan and turbojet engines?

4.2.10 Procedure for obtaining THP for given h, V, BHP and N

For calculating the performance of the airplane, the thrust horse power (THP)

is needed at different values of engine RPM(N), break horse power (BHP), flight

speed (V) and flight altitude (h). In this context the following may be noted.

(a) The engine output (BHP) depends on the altitude, the RPM (N) and the

manifold air pressure (MAP).

(b) The propeller absorbs the engine power and delivers THP; pTHP = η BHP

(c) The propeller efficiency depends, in general, on BHP, V, N and β.

(d) The three quantities viz. d, V and n can be combined as advance ratio

(J = V/nd).

(e) Once pη is known :

THP = pη x BHP and T = THP×1000 / V .

The steps required to obtain pη depend on the type of propeller viz. variable pitch

propeller, constant speed propeller and fixed pitch propeller. The steps in the

three cases are presented below.



I) Variable pitch propeller

In this type of propeller the pitch of the propeller is changed during the flight

so that the maximum value of pη is obtained in various phases of flight. The steps

are as follows.

(a) Obtain the ambient density ρ for the chosen altitude.

(b) Obtain 3 5PC = P / ρ n d ; P is BHP in watts

(c) Obtain J = V/nd

(d) Calculate 1/5PSC = J/C

(e) From the design chart for the chosen propeller (e.g. Fig.4.5c for a two

bladed propeller), obtain β which will give maximum efficiency. Obtain

corresponding pη . Consequently,

THP = pη x BHP and T = THP x 1000 / V

(f) To get the thrust (T) at V = 0, obtain CT at J = 0 for a pitch setting which

would give high value of CT. In Fig.4.5d, it is seen that for a two bladed

propeller, CT at J = 0, is nearly maximum for oβ 30 . Having known CT, the

thrust(T) is given by :

T = 2T

4ρ n d C

II) Constant speed propeller

The variable pitch propellers were introduced in 1930’s. However, it was

noticed that as the pilot changed the pitch of the propeller, the engine torque

changed and consequently the engine RPM deviated from its optimum value.

This rendered, the performance of the engine-propeller combination,

somewhat suboptimal. To overcome this problem, the constant speed

propeller was introduced. In this case, a governor mechanism alters the fuel

flow rate so that the required THP is obtained even as rpm remains same.

The value of β is adjusted to give maximum possible pη .

The steps to obtain pη are the same as mentioned in the previous case.



III) Fixed pitch propeller

From Fig.4.5b it is observed that a fixed pitch propeller has a definite

value of CP for a chosen value of advance ratio (J). Consequently, the

propeller can absorb only a certain amount of power for a given value of J.

Thus when the flight speed changes, the power absorbed by the propeller

also changes. However, for the engine-propeller combination to be in

equilibrium i.e. run at a constant r.p.m, the power absorbed by the propeller

and that produced by the engine must be the same. This would render the

problem of determining power output as a trial and error procedure. However,

it is observed that the torque of a piston engine remains nearly constant over

a wide range of r.p.m’s. Using this fact, the torque coefficient (CQ) and torque

speed coefficient (Qs) are deduced in Ref.3.7, chapter 16, from the data on

CP & CT . Further a procedure is suggested therein to obtain pη at different

flight speeds.

Herein, the procedure suggested in the Appendix of Ref.4.1 is presented. It is

also illustrated with the help of example 4.5.

It is assumed that the propeller is designed for a certain speed, altitude, rpm

and power absorbed.

Let V0 = design speed (m/s)

N0 = design rpm ; n0 = N0 / 60

BHP0 = BHP of the engine under design condition (kW)

d = diameter of propeller (m)

J0 = Advance ratios under design condition = V0 / n0 d

0β = design blade angle; this angle is fixed

0η = efficiency of propeller under design condition

The steps, to obtain the THP at different flight speeds, are as follows.

1. Obtain from propeller charts, CT and CP corresponding to J0 and 0β .

These values are denoted by CTO and CPO.



2. Choose values of J from 0 to a suitable value at regular intervals. Obtain

from the relevant propeller charts, the values of CT and CP at these values

of J’s and the constant value of 0β .

3. Calculate J/J0, CT/CP and CP0/CP from values obtained in step 2.

4. Calculate:

T0 = 0 0 0η BHP ×1000 / V and (4.22)

0 0 PO TOK = T C /C (4.23)

5. The assumption of constant torque (Q0) gives that N and P are related.

Note: 0 0 0Q = P / 2πn

This yields:

PO

0 P

CN=

N C (4.24)

00 0

J NV = V × ×

J N (4.25)

and P0 T T0 0

T0 P P

C C CT = T = K

C C C (4.26)

Consequently, THP = TV/1000 and BHP = THP/ pη

The procedure is illustrated with the help of example 4.5.

Example 4.5

Obtain the thrust and the thrust horse power at sea level for V upto 60 m/s

for the propeller engine combination of example 4.4

Solution:

From example 4.4 it is noted that the propeller is designed to absorb

97.9 kW at 2500 rpm at V = 59 m/s.The propeller diameter is 1.88 m and β = 20o.

Hence, V0 = 59 m/s, N0 = 2500, n0 = 41.67, 0β = 20o

BHP0 = 97.9 kW, 0η = 0.83

00

0

V 59J = = = 0.753

n d 41.67×1.88

From Fig.4.5d, CTO = 0.046

From Fig.4.5b, CPO = 0.041



Hence, CTO/CPO = 0.046/0.041 = 1.122

0

97.9×1000×0.83T = = 1377.24N

59

PO0

TO

C 0.041K = T = 1377.24× = 1227.54

C 0.046

The remaining calculations are presented in Table E 4.5

J

J/J0

CT

*

PC

$

T

P

C

C

CP0/CP

N/N0

£

V

€

T

(N)

€€

N

#

pη

**

THP

$$

BHP

££

0 0 0.104 0.066 1.576 0.621 0.788 0 1927 1971 0 0 -

0.1 0.133 0.104 0.065 1.589 0.629 0.793 6.21 1951 1983 0.17 12.15 71.23

0.2 0.266 0.104 0.065 1.606 0.636 0.792 12.49 1971 1993 0.33 24.61 74.60

0.3 0.398 0.102 0.062 1.631 0.657 0.811 19.05 2002 2027 0.49 38.14 77.83

0.4 0.531 0.093 0.060 1.545 0.683 0.827 25.91 1897 2067 0.62 49.15 79.28

0.5 0.664 0.082 0.058 1.420 0.712 0.844 33.05 1743 2109 0.70 57.61 82.29

0.6 0.797 0.070 0.059 1.306 0.765 0.875 41.12 1603 2187 0.77 65.91 85.60

0.7 0.930 0.055 0.046 1.185 0.884 0.900 51.55 1455 2350 0.81 75.00 92.60

0.8 1.062 0.040 0.036 1.099 1.126 1.061 66.50 1349 2653 0.83 89.71 108.1

*From Fig.4.5d ; $ From Fig.4.5b; £ From Eq.(4.24); € From Eq.(4.25);

€€ From Eq.(4.26); # N = (N/N0)x N0 ; ** From Fig.4.5a; $$ THP = TV/1000 ;

££ BHP = THP/ pη

Table E4.5 Thrust and power output of an engine-propeller combination with

fixed pitch propeller



The results are shown in Figs.E4.5a and b

Fig.E 4.5 Variations of thrust (T) and thrust horse power (THP) with velocity(V)

(a) T vs V (b) THP vs V



Answers :

The variations of T and THP with V are given in table below.

V(m/s) 0 6.21 12.49 19.05 25.91 33.05 41.12 51.55 66.50

T (N) 1927 1951 1971 2002 1897 1743 1603 1455 1349

THP(kW) 0 12.15 24.61 38.14 49.15 57.61 65.91 75.00 89.71

BHP(kW) - 71.23 74.10 77.83 79.28 82.29 85.60 92.60 108.1

N (RPM) 1971 1983 1993 2027 2067 2109 2187 2350 2653


As mentioned earlier,THP equals pη × BHP . Thus, the variations of THP with V

and h depends on variations of pη and BHP with V and h. In this context, the

following may be recalled.

(i)At a given altitude and RPM, the engine output (BHP) is almost constant with

flight velocity. (ii) BHP decreases with altitude as given by Eqs.(4.1) or (4.1a).

(iii) The propeller efficiency pη depends on BHP, h, V, n and β . For a variable

pitch propeller pη remains nearly constant over a wide range of flight speeds.

Thus for an airplane with variable pitch propeller, the THP vs V curve for a

chosen RPM and h remains flat over a wide range of flight speeds. A typical

variations of THP with V, at chosen ‘RPM(N)’ and with ‘h’ as parameter are

shown in Fig.4.6.

From the engine charts the fuel flow rate and BSFC are known at chosen

MAP & N. From these values the BSFC at the chosen MAP & N, can be

calculated using Eq.(4.1d) . See section 6 of Appendix A for typical calculations.



Fig.4.6 Schematic variation of THP with flight speed for an engine-propeller

combination with variable pitch propeller


As noted earlier, the propeller blade is like a rotating wing with forward

motion. The resultant velocity at the propeller tip (VRtip

) would be the highest. It is

equal to:

VRtip

= { V2 + (2 π n R)2}1/2, where R is the radius of the propeller.

When the Mach number corresponding to VRtip

exceeds the critical Mach number

for the airfoil used on the propeller, the drag coefficient of the airfoil would

increase and the lift coefficient would decrease (see subsection 3.3.3).

Consequently, the efficiency of the propeller would decrease. This loss of

efficiency can be delayed to higher flight Mach numbers by use of advanced

propellers. These propellers have swept blades and are being used on turboprop

airplanes up to flight Mach number of 0.7. Figure 4.7a shows one such propeller

placed in a wind tunnel and Fig.4.7b shows another propeller mounted on

ATR 72 airplane.



Fig.4.7a Advanced propeller being tested in a Wind tunnel

(Adapted from Ref 4.4)

Fig.4.7b Advanced propeller mounted on ATR72 airplane

(Source : www.fspilotshop.com)

Fig.4.7 Features of an advanced propeller



4.3 Gas Turbine Engines

A gas turbine engine consists of a diffuser to decelerate the air stream

entering the engine, a compressor, a combustion chamber, a turbine and a

nozzle (Fig.4.8a). In some turbojet engines, an afterburner is incorporated

between the exit of the turbine and the entry of the nozzle (Fig.4.8b).The hot

gases leaving the combustion chamber expand partly in the turbine and partly in

the nozzle. The need for three variants of gas turbine engines viz. turboprop,

turbofan & turbojet can be explained by considering their propulsive efficiencies.

Fig.4.8 Turbojet engine

(Source : http://www.aerospaceweb.org)


Propulsive efficiency is the ratio of useful work done by the air stream and

the energy supplied to it.

In a gas turbine engine, the velocity of the air stream ( V ) is augmented

to Vj,the velocity of the jet stream, thereby imparting kinetic energy at the rate of :

(m /2) [ Vj 2 - V

2] (4.27)

where m is the mass flow rate.



The engine develops a thrust T and hence results in a useful work of T V .

Noting that:

T = m (Vj - V ), (4.28)

the propulsive efficiency (ηpropulsive) is:

m(V -V )(V ) 2j

η = =propulsive m V2 2 j(V -V ) 1+j2 V

(4.29)

4.3.2 Why turboprop, turbofan and turbojet engines?

The overall efficiency of a gas turbine engine is the product of items like

cycle efficiency, combustion efficiency, mechanical efficiency and propulsive

efficiency. The cycle efficiency depends on the engine cycle and in turn on the

maximum temperature / pressure in the engine. The combustion efficiency and

mechanical efficiency are generally of the order of 95%. Thus propulsive

efficiency finally decides the overall efficiency of a gas turbine engine as a

propulsive system.

Remark:

The action of a propeller is also similar to that of a jet engine i.e. it also enhances

velocity of the free stream from V to Vj, In this case, Vj is the velocity of the

stream far behind the propeller(see subsection 4.2.7). Hence, the propulsive

efficiency of a propeller which was called ideal efficiency of propeller, is also

given by Eq.(4.29), which is same as given by Eq.(4.13).

The variation of propulsive efficiency with flight speed provides the reason for

use of turboprop, turbofan and turbojet engines in airplanes operating at different

range of flight speeds.Consider the variation of propulsive efficiency with flight

speed. For this purpose, a subsonic jet engine with convergent nozzle is

considered. In this case, the Mach number at the exit, would be unity and the

temperature of the exhaust gases would be around 600 K. Under these

conditions, Vj, the velocity of jet exhaust would be around 500 m/s. Using

Eq.(4.29), the values of propulsive efficiency obtained at different flight speeds

( V ) are given in the Table 4.2.



V (m/s) 100 125 166.7 250 333.3 400

Vj / V 5 4 3 2 1.5 1.25

ηρ % 33.3 40.0 50.0 66.7 80.0 88.9

Table 4.2 Variation of propulsive efficiency with flight speed for Vj = 500 m/s

Remarks:

i) Turboprop engine

It is seen from Table 4.2 that ηp will be low if a pure jet engine is used at

low speeds. An analysis of Eqs.(4.28 and 4.29) points out that for having

adequate thrust and high propulsive efficiency at low flight speeds, a small

increment in velocity should be given to a large mass of air. This is effectively

done by a propeller. Thus for airplanes with flight Mach number less than about

0.5, a turbo-prop engine is used (Fig.4.9). In this case, the turbine drives the

compressor and also the propeller through a gearbox (Fig.4.9). The gear box is

needed because the turbine r.p.m. would be around 15000-20000 whereas, the

propeller rotates at about 3000 r.p.m.

For practical reasons, the expansion of the gases coming out of the combustion

chamber is not allowed to take place completely in the turbine and a part of the

expansion is carried out in the nozzle. Hence, in a turboprop engine, about 80 to

90% of the total output is produced through the propeller and the rest 20 to 10%

as output from the jet coming out of the nozzle.



Fig.4.9 Turboprop engine

(Source: www.aircraftenginedesign.com)

ii) Turbofan engine

As the flight Mach number increases beyond 0.7, the propeller efficiency

decreases rapidly due to the formation of shock waves at the tip of the propeller blade. Hence, for airplanes flying near Mach number of unity, a turbo-fan engine

is used (Fig.4.10).In this engine a major portion of the power output (about 60%)

is obtained as jet thrust and the rest as thrust from the fan. A fan has a smaller

diameter as compared to the propeller and it is generally placed inside a duct. A

ducted fan has a higher propulsive efficiency than a propeller.

It is observed in Fig.4.10 that all the air taken in by the fan does not go

through the turbine. Incidentally the part of the engine consisting of the

compressor, combustion chamber, turbine and nozzle is called ‘Gas generator’.

The ratio of the mass of the air that passes through the fan to the mass of air that

passes through the gas generator is called ‘Bypass ratio’.

Early turbofan engines had bypass ratio of 1:1. At present, it is around 6.5:1 and

is likely to increase in future.



Fig.4.10 Turbofan engine

(Source : http://www.aerospaceweb.org)

iii) Turbojet engine

At supersonic Mach numbers, up to three, a turbo-jet engine is used. In this

engine entire power output is through the jet thrust.



Chapter 4


4.3.3 Characteristics of a typical turboprop engine

4.3.4 Characteristics of a typical turbofan engine

4.3.5 Characteristics of a typical turbojet engines




4.3.3 Characteristics of a turboprop engine

As noted earlier, in this engine, a major portion of the output is available at

the propeller shaft (SHP) and a small fraction through the jet thrust (Tj). Hence,

the output is represented as:

THP = ηp SHP + (Tj V /1000) (4.30)

where SHP = shaft horse power available at propeller shaft in kW,ηp = propeller

efficiency and Tj = jet thrust

The total output of a turbo-prop engine, also called ‘Equivalent shaft horse

power (ESHP)’, is defined as :

ESHP = SHP + {Tj V / (0.8x1000) } (4.31)

Note : (i) For the purpose of defining ESHP, the value of ηρ is taken as 0.8 in

Eq.(4.31). The ESHP and SHP are in kW.

(ii) Equation (4.31) would not be able to account for the contribution, to ESHP, of

the thrust produced when the flight velocity (V) is zero or the static condition. For

this case and when V < 100 knots (or 185 kmph), the convention is to define

ESHP as follows (Ref.4.3, chapter 14).

ESHP = SHP + (Tj / 14.92) (4.31a)

where ESHP and SHP are in kW and Tj is in N.



For example a turboprop engine developing SHP of 746 kW and jet thrust of

503 N, under sea level static condition, would have :

ESHP = 746 + (503/14.92) = 780 kW.

Characteristics of a typical turbo-prop engine are shown in Fig.4.11. It is

observed that the power output increases with flight speed. This increase is due

to two factors viz. (a) the mass flow through the engine ( ;i im = ρA V Ai and Vi

being the area of intake, and the velocity at the intake) increases with flight

speed and (b) the pressure rise due to the deceleration of the flow in the inlet

diffuser also increases with flight Mach number.

Figure 4.11 also shows the influence of ambient temperature on power output. It

is observed that there is a significant fall in ESHP as the ambient temperature

rises.

From the curves regarding fuel flow rate in Fig.4.11, the BSFC can be obtained

at various speeds and altitudes as:

BSFC = (Fuel flow/hr) / ESHP

Remark:

Reference 3.9 Appendix E.3 gives performance curves for a large turboprop

engine with sea level static power of 6500 HP. It may be noted that the ‘Sea level

static power’ is the engine output at sea level at zero velocity. Reference 1.9,

chapter 6 gives characteristics of an engine of around 1700 HP.



Fig.4.11 Characteristics of PT6A-25 turboprop engine

(Adapted from Brochure of Pratt & Whitney Canada Corp. 1000, Marie-Victorin,

Longueuil Quebec J4G 1A1, Canada © Pratt & Whitney Canada Corp.

Reproduced with permission)

4.3.4 Characterisitcs of typical turbofan engine

In the early turbofan engines the thrust output used to remain fairly

constant with flight speed. In the modern turbofan engines the performance at

low speeds and low altitudes (up to about 5 km) has been improved so that the

ratio of the sea level static thrust and that (thrust) in high speed-high altitude

flight is much higher than the early turbofan engines. The ‘Sea level static thrust’

is the engine output at M=0 at sea level. Higher sea level static thrust helps in

reducing the distance required for take-off. Figure 4.12 shows the variations of

thrust with Mach number at different altitudes for an engine with bypass ratio of

4.9. The figure also shows the values of the specific fuel consumption (TSFC).



Remark:

Chapter 9 of Ref.3.22 gives the performance, in terms of non-dimensional

parameters, for engines with bypass ratios of 3, 6.5, 8 and 13. The curves are

also presented for take-off rating, climb rating and cruise rating. It may be added

that the ‘Take-off rating’ is the engine output which can be availed for about 5

min. The engine can be run at ‘Climb rating’ for about half an hour and at ‘Cruise

rating’ for long periods.

Fig.4.12 Characteristics of Pratt and Whitney PW4056 turbofan engine -

maximum cruise thrust

(With permission from Pratt and Whitney, East Hartford)

4.3.5 Characterisitcs of typical turbojet engine

The characteristics of a supersonic turbojet engine are shown in Figs.4.13a to d.

It is observed that at subsonic speeds the thrust is fairly constant, but it increases

considerably at supersonic speeds. This rise is due to increased ram pressure

in the intake, as a result of the deceleration of the supersonic flow. The Mach



number at which the peak value of thrust occurs depends on the design of the

engine.

Fig.4.13a Characteristics of Pratt and Whitney

JT4A-3 turbojet engine (estimated thrust, TSFC, and airflow) under standard

atmospheric condition and 100% RAM recovery. h = sea level




Fig.4.13b Characteristics of engine in Fig.4.13a, h = 15000 ft


Fig.4.13c Characteristics of engine in Fig.4.13a, h = 30000 ft




Fig.4.13d Characteristics of engine in Fig.4.13a, h = 45000 ft


Remarks:

i)In Fig.4.13a to d the true airspeed is given in knots;one knot is equal to

1.852 kmph. Further, the speed of sound at h = 0, 15000’ , 30000’ and 45000’ is

respectively 661, 627, 589 and 574 knots.

ii) Bypass supersonic turbofan engines are also being considered for supersonic

flight. Reference 3.9, gives, in Appendix E, typical curves for an engine with sea

level static thrust of 30000 lb (133 kN). Similarly Ref.4.5, chapter 8 also presents

curves for an engine with 33000 lb (146.3 kN) sea level static thrust. Figures 4.13

a to d also indicate the values of specific fuel consumption (TSFC) and the air

flow rate.

iii) Figure 4.8b shows an afterburner duct between the turbine exit and the entry

of the nozzle. The same figure also shows the fuel spray bars and the flame

holder. An afterburner is used to increase the thrust output for a short duration.

When the fuel is burnt in the afterburner, the temperature of the gases goes up

and the thrust increases when these gases subsequently expand in the nozzle.



However, the specific fuel consumption also goes up considerably and the

afterburner operation is resorted to only for a short duration like during take-off or

transonic acceleration.



The detailed information about engine performance (i.e. variations with

altitude and flight velocity of the thrust (or power) and TSFC (or BSFC) is

generally available only in a limited number of cases. To get the performance of

an engine with other rating, scaling of the available data is carried out. For this

purpose, the values of thrust(or power) of the engine, whose characteristics are

known, are multiplied by a suitable factor which will bring the output of the

existing engine equal to the output of the desired engine. It is assumed that the

SFC values will be the same for the two engines. This kind of scaling is generally

applicable for outputs within ± 25% of the output of the known engine (Ref.4.5,

chapter 8).


The topic of choice of engine for different types of airplanes is generally covered

in airplane design. Here some salient points are mentioned to conclude the

discussion on engines.

The following five criteria are used to select a power plant for a specific

application.

1.Overall efficiency 0η : This quantity is the product of (a) thermodynamic

cycle efficiency tη (b) Combustion efficiency cη (c) mechanical efficiency

mη and (d) propulsive efficiency pη . The thermodynamic efficiency depends

on the thermodynamic cycle on which the engine operates. The details regarding

estimation of tη are available in books on thermodynamics. However, it is of the

order of 40 to 50%. The combustion efficiency and mechanical efficiency would

be around 95%. The propulsive efficiency of the propeller and gas turbine

engines have been described in subsections 4.2.7, 4.2.8 and 4.3.2. It has been

pointed out there that pη depends on flight speed or Mach number.



The specific fuel consumption (SFC) is an indication of the overall efficiency.

Based on Ref.3.9 chapter 3, it can be mentioned that the piston engine-propeller

combination would have lowest SFC for Mach number (M) upto about 0.3. The

turboprop engine would have lowest SFC in the range of Mach number from 0.3

to 0.6 which may extend to M 0.7 with the use of a transonic propeller. The

high bypass ratio turbofans have lowest SFC between for M 0.7 to 1.0 and the

low by-pass ratio ones between M 1 to1.6 . Turbojets are more suited for

1M .6 to about 3.5 and ramjets later upto M 8 . It may be recapitulated that a

ramjet engine requires another powerplant to bring it to Mach number of about

1.5.

2. Variation of thrust (or power) with flight speed and altitude:

The shaft horse powers of piston engine and turboprop engine do not change

significantly with flight speed. Consequently, the thrust outputs of these engines

decrease significantly with flight speed or Mach number. The output of a turbofan

engine decreases with Mach number, especially at low altitudes (Fig.4.12). The

thrust of a jet engine is fairly constant at subsonic speeds but increases

considerably at supersonic speeds (Fig.4.13 c & d). As regards the effect of flight

altitude Eq.(4.1a) shows that for a piston engine 1.1slP /P = σ where σ is the

density ratio and the suffix ‘sl’ denotes a quantity at sea level.

For a turbo-prop engine (from Ref 1.10 chapter 3), 0.7slP/P σ . From

Ref.3.15, chapter 3, (T/Tsl) for turbofan and turbojet engines is also roughly

proportional to 0.7σ

3. Weight of the engine:

The weight of the engine contributes to the gross weight of the airplane and

hence it should be as low as possible.This quantity is indicated by the ratio Wpp/T

or Wpp/BHP, where WPP is the weight of the power plant. This ratio depends on

the type of engine and the engine rating; it (ratio) decreases as the rating

increases. Based on data in Ref.1.15, it can be mentioned that the weight per

unit BHP for a piston engine is around 9N/kW for an engine with a rating of

around 150 kW and about 6N/kW for a rating of around 500 kW. For a turboprop



engine WPP/ESHP is around 2.9 N/kW for rating of 500 kW, 2.3 N/kW for a rating

of 2500 kW and 1.4 N/kW for a rating of 7500 kW. For a turbofan engines the

ratio WPP/T could be around 0.25 N/N for a rating of around 100 kN and about

0.15 N/N for a rating of about 250 kN.

4. Frontal area:

The frontal area of an engine contributes to the parasite drag of the airplane.

Hence, a lower frontal area is a desirable feature of the engine. For a given

output the piston engine-propeller combination generally has the highest frontal

area. Turboprop, turbofan and turbojet follow in the decreasing order of the

frontal area.

5.Other considerations :

Gas turbine engines have mechanical simplicity as compared to a piston

engine. However, gas turbine engines are costlier than the piston engines as

some of the components of the gas turbine engines operate at higher

temperature and RPM. This requires special materials and fabrication

techniques.

Keeping these factors in view the different types of engine are used in

the speed range/application as given in Table 4.3



Type of engine Speed / Mach number

range

Application – airplanes

in the following

categories

Piston engine-propeller

combination

Upto 300 kmph

General aviation, trainer,

agricultural and sports.

Turboprop

250 to 600 kmph;

upto 750 kmph with

advanced propeller

Short and medium range

transport/cargo, aerial

survey, feeder liner and

executive transport.

Turbofan

M from 0.7 to 1.0

Medium and long range

transports, cargo,

maritime patrol, executive

transport, jet trainer.

Turbojet

M from 1 to 3

Trainers, supersonic

transport, fighter,

interceptor, bomber.

Ramjet

M from 2 to 8

Intended for hypersonic

transport.

Table 4.3 Speed range and applications of different types of engines



Chapter 4

References

4.1 Hartman, E.P. and Biermann, D. “The Aerodynamic characteristics of full-

scale propellers having 2, 3, and 4 blades of clark Y and R.A.F. 6 airfoil sections”

NACA TR 640, Nov.1937. This report can be downloaded from the site “NASA

Technical Report Server (NTRS)”.

4.2 “Generalized method for propeller performance estimation” Hamilton

Standard Division, Hamilton Standard Publication PDB6101A, United Aircraft

Corp., 1963.

4.3 Nicholai, L.M. and Carichner, G.E “Fundamentals of aircraft and airship

design – Vol I – Aircraft design” AIAA educational series (2010).

4.4 Mikkelson D.C. and Mitchell G.A. “High speed turboprop for executive aircraft

– potential and recent test results“ NASA TM 31482, Jan 1980. This report can

be downloaded from the site “NASA Technical Report Server (NTRS)“.

4.5 Jenkinson L.R., Marchman III J.F. “Aircraft design projects” Butterworth-

Heinemann (2003).



Chapter 4

Exercises

4.1) What are the different types of engines used on airplanes? State the

speed/Mach number range in which they are used.

4.2) Sketch a typical BHP vrs altitude curve for a piston engine. Why does the

power output of a piston engine decrease rapidly with altitude? Supercharger is

needed to delay this loss of power to higher altitudes. Look for information on

supercharger from books (e.g. Ref.1.9) and internet (www.google.com).

4.3) What are the essential differences between turboprop, turbofan and turbojet

engines? Derive an expression for the propulsive efficiency and justify the range

of flight Mach numbers in which these engines are used.

4.4) A propeller of 2 m diameter is mounted on an airplane flying at a speed of

216 kmph. If the velocity of air far behind the propeller be 81 m/s, calculate the

propulsive efficiency and the thrust developed by the propeller.

[Answers: ηp = 85.1%, T = 5695 N]

4.5) Neatly sketch the following:

(a) variation of propeller efficiency vs flight velocity with propeller pitch angle as

parameter.

(b) Variation of thrust vs. Mach number with altitude as parameter for a

turbofan engine.



Chapter 5

Lecture 19 Performance analysis I – Steady level flight – 3 Topics

5.8 Influence of level flight analysis on airplane design

5.9 Steady level flight performance with a given engine

5.10 Steady level flight performance with a given engine and parabolic

polar

5.10.1 Airplane with jet engine


The significant manner in which the performance analysis helped in

evolution of the airplane configuration can be appreciated from the following

discussion.

(a) The low speed airplanes are powered by engines delivering BHP or ESHP. In

this case, the major portion of the power required is induced power, which

depends on the factor K in drag polar (Eq.5.10). This factor is given as 1 / A e

where A is the aspect ratio of the wing and ‘e’ is the Oswald’s efficiency factor

(see Eq.3.46). Hence the low speed airplanes and gliders have high aspect ratio

wings. It may be added that personal airplanes have aspect ratio between 6 to 8

as hanger space is also an important consideration. However, medium speed

commercial airplanes have aspect ratio between 10 to 12. Gliders have aspect

ratio as high as 16 to 20.

(b) For high subsonic airplanes most of the drag is parasite drag which depends

on DOC (see Eq.5.12). Hence, high speed airplanes have features like smooth

surfaces, thin wings, streamlined fuselage, smooth fairings at wing-fuselage joint

and retractable landing gear. These features reduce DOC . Manufacturing

techniques have also been improved to achieve smooth surface finish. High



speed airplanes also have high wing loading (W/S) to reduce the wing area.

Table 3.4 may be referred to for typical values of DOC , A and e of different types

of airplanes. The reciprocal of (CD / CL

) is (CL / C

D). It is called lift-drag ratio

(L / D). The maximum value of this ratio, (L / D)max

, is an indication of the

aerodynamic efficiency of the airplane. (L / D)max

lies between 12 to 22 for a

subsonic airplanes and between 5 to 8 for supersonic airplanes.

(c) When the weight of an airplane increases the thrust required increases in

proportion to W and the power required increases in proportion to W3/2

(Eqs.5.3

and 5.4). Hence, airplane design bureaus have a group of engineers which

keeps a close watch on any increase in the weight of the airplane.


At the outset the following three points may be noted.

(I)In steady level flight the thrust must be equal to drag (Eq.5.1).

(II) The thrust is provided by the engine or the engine-propeller combination and

from chapter 4, it is noted that the thrust or power output varies with engine RPM,

flight speed and altitude.

(III) For airplanes with piston engine or turboprop engine, the output is the power

available at the engine shaft. Hence, to estimate the performance of such

airplanes the calculations are carried-out in terms of BHP or THP. For airplanes

with turbofan or turbojet engines, the output is in terms of thrust and to estimate

the performance of such airplanes the calculations are carried-out in terms of

thrust.

Typical variations, with altitude and flight speed, of the maximum thrust available

(Ta) and the maximum thrust horse power available (THP)a are shown in

Figs.5.5 and 5.6a respectively. The thrust required and power required curves

are also shown in same figures.

Consider the curves of Ta and Tr corresponding to sea level conditions. It is seen

that the power or thrust available is much more than the minimum power or thrust

required. Hence, flights over a wide range of speeds are possible by controlling

the engine output with the help of throttle and ensuring thrust equals drag.



However, as the speed increases above the speed for minimum power or thrust

( Vmp or Vmd), the power or thrust required increases and at a certain speed the

power or thrust required is equal to the maximum available engine output (point

A in Figs.5.5 & 5.6a). This speed is called the ‘Maximum speed(Vmax

)’. Similar

intersections between power available and power required curves or thrust

available and thrust required curves are seen at higher altitudes (points B, C and

D in Fig.5.5, point B in Fig.5.6a and point C in Fig.5.6b).

Similarly, when the flight speed decreases below Vmp or Vmd the power or

thrust required increases and there is a speed at which the power or thrust

required is equal to the available power or thrust - point D’ in Fig.5.5 and point C’

in Fig.5.6b. Figure 5.6b is drawn separately from Fig.5.6a to show the points C

and C’ clearly.

Thus, the minimum speed can be limited by available thrust or power output. It is

denoted by (Vmin

)e. However, in level flight the requirement of lift equal to weight

should also be satisfied(Eq.5.1). Hence, level flight is not possible below stalling

speed. Thus, two factors viz. the thrust or power available and the stalling, limit

the minimum flight speed of an airplane. Satisfying both these requirements, the

minimum speed of the airplane at an altitude will be the higher of the two speeds

viz. (Vmin

)e and V

S.

Typical variations of Vmax

, (Vmin

)e and V

S are shown for a jet engined airplane in

Fig.5.9. The details of the calculations are given in Appendix B. Similarly, typical

variations of these speeds in case of a piston engined airplane are shown in

Fig.5.10 with details of calculation given in Appendix A. The following

observations are made.

(i)For a jet airplane Vmax may slightly increase initially with altitude and then

decrease. However, there is an altitude at which the thrust required curve is

tangential to the thrust available curve and flight is possible only at one speed.

This altitude is called ‘Ceiling’ and denoted by hmax

. Above hmax

the thrust

available is lower than the minimum thrust required and level flight is not possible

as the requirement of T = D cannot be satisfied.



Fig.5.9 Vmax

and Vmin for jet airplane

Fig.5.10 Vmax

, (Vmin

)e and Vs for airplane with engine-propeller combination

(ii)The minimum speed of a jet airplane is the stalling speed (Vs) at low altitudes.

However, near the ceiling, the minimum speed is that limited by the thrust

available i.e. (Vmin

)e.

0

1000

2000

3000

4000

5000

6000

0 10 20 30 40 50 60 70 80

Velocity (m/s)

Vs Vmax

(Vmin)e



(iii)In the case of a piston engined airplane, the maximum speed seems to

decrease with altitude. In this case also there is a ceiling altitude beyond which

the power available is lower than the minimum power required and hence level

flight is not possible. The ceiling in this case, is lower than in the case of a jet

airplane because the power output of a piston engine decreases rapidly with

altitude. As regards the minimum speed, it is also limited by stalling at low

altitudes and by power available near the ceiling altitude.

5.10 Steady level flight with a given engine and parabolic polar

If the drag polar is parabolic and the engine output can be assumed to be

constant with speed, then Vmax

and (Vmin

)e from the engine output consideration,

can be calculated analytically. i.e. by solving an equation. It may be noted from

Figs.5.5 & 5.6 that the assumption of Ta or Pa as constant with V appears

reasonable near the speeds where Vmax occurs.

5.10.1 Airplane with jet engine:

The steps to calculate Vmax

and (Vmin

)e are as follows.

(1) Choose an altitude ‘h’. Let Ta be the thrust available in the range of speeds

where Vmax is likely to occur.

(2) Tr =Ta= W(CD / CL

)

Hence,

DOT CCa D= = +KCLW C CL L

Or DOT2 aKC - C + C = 0LL W

(5.25)

Equation (5.25) is a quadratic in CL. Its solution gives two values of C

L at which

level flight with the given thrust is possible. Let these values of CL be denoted as

CL1

and CL2

. Then, the corresponding flight speeds, V1 and V

2, are given as:

1 12 22W 2W

V = and V =1 2ρSC ρSCL1 L2

(5.26)



It may be pointed out that the same results can be obtained by using Eq.(5.12),

i.e.

1

2 DO

22W2T = T = ρV S C + Ka r 2ρV S

Or AV4 – BV

2 + C = 0 (5.27)

where, 1

2 DO

22KWA = ρSC , B = T and C =a ρS

For given value of thrust (Ta), Eq.(5.27) also gives two solutions for level flight

speeds V1 and V

2.

Let V1 be the higher among V

1 and V

2.Then, V

1 is the maximum speed and V2 is

the minimum speed, based on engine output i.e. (Vmin

)e. The higher of (V

min)e

and the stalling speed (Vs) will be the minimum speed at the chosen altitude.

The example 5.2 illustrates the procedure.

Remarks:

i) Calculate the Mach number corresponding to V1. If it is more than the critical

Mach number then DOC and K would need correction and revised calculation,

would be required.

ii) Obtain, from the engine charts, the thrust available at V1 . Let it be denoted by

Ta1. If the thrust available (Ta), assumed at the start of the calculation(step 1), is

significantly different from Ta1, then the calculations would have to be revised

with new value of Ta. However, it is expected that the calculations would

converge to the correct answer in a few iterations.



Chapter 5

Performance analysis I – Steady level flight

(Lectures 17 to 20)

Keywords: Steady level flight – equations of motion, minimum power required,

minimum thrust required, minimum speed, maximum speed; stalling speed;

equivalent airspeed.

Topics

5.1 Introduction

5.1.1 Subdivisions of performance analysis

5.1.2 Importance of performance analysis

5.1.3 Approach in performance analysis

5.2 Equations of motion for steady level flight

5.3 Stalling speed

5.4 Equivalent airspeed

5.4.1 Airspeed indicator

5.5 Thrust and power required in steady level flight – general case

5.6 Thrust and power required in steady level flight when drag

polar is independent of Mach number

5.7 Thrust and power required in steady level flight – consideration of

parabolic polar



5.10 Steady level flight performance with a given engine and parabolic

polar

5.10.1 Airplane with jet engine

5.10.2 Parameters influencing Vmax of a jet airplane

5.10.3 Airplane with engine-propeller combination

5.11 Special feature of steady level flight at supersonic speeds

References

Exercises



Chapter 5

Lecture 17 Performance analysis I – Steady level flight – 1 Topics 5.1 Introduction





5.3 Stalling speed




5.1 Introduction:

During its normal operation an airplane takes –off, climbs to the cruising

altitude, cruises at almost constant altitude, descends and lands. It may also fly

along curved paths like turns, loops etc. The flights along curved paths are also

called manoeuvres. Analyses of various flights are the topics under the

performance analysis. A revision of section 1.6 would be helpful at this stage.


Performance analysis covers the following aspects.

I) Unaccelerated flights:

(a) In a steady level flight an airplane moves with constant velocity at a constant

altitude. This analysis would give information on the maximum level speed and

minimum level speed at different altitudes.

(b) In a steady climb an airplane climbs at constant velocity. This analysis would

provide information on the maximum rate of climb, maximum angle of climb and

maximum attainable altitude (ceiling).



(c) In a steady descent an airplane descends with constant velocity. A glide is a

descent with zero thrust. This analysis would give the minimum rate of sink and

time to descend from an altitude.

(d) Range is the horizontal distance covered, with respect to a given point on the

ground, with a given amount of fuel. Endurance is the time for which an airplane

can remain in air with a given amount of fuel.

II) Accelerated flights:

(a) In an accelerated level flight an airplane moves along a straight line at

constant altitude and undergoes change in flight speed. This analysis provides

information about the time required and distance covered during acceleration

over a specified velocity range.

(b) In an accelerated climb, an airplane climbs along a straight line accompanied

by a change in flight speed. This analysis gives information about the change in

the rate of climb in an accelerated flight as compared to that in a steady climb.

(c) Loop is a flight along a curved path in a vertical plane whereas a turn is a

flight along a curved path in a horizontal plane. This analysis would give

information about the maximum rate of turn and minimum radius of turn. These

items indicate the maneuverability of an airplane.

(d) During a take-off flight an airplane starts from rest and attains a specified

height above the ground.This analysis would give information about the take-off

distance required.

(e) During a landing operation the airplane descends from a specified height

above the airport, lands and comes to rest. This analysis would provide

information about the distance required for landing.


The performance analysis is important to asses the capabilities of an

airplane as indicated in the previous subsection. Moreover, from the point of

view of an airplane designer, this analysis would give the thrust or power

required, maximum lift coefficient required etc. to achieve a desired performance.

This analysis would also point out the new developments required, in airplane

aerodynamics and engine performance, to achieve better airplane performance.




As mentioned in subsection 1.1.3 the approach here is to apply the

Newton’s laws and arrive at the equations of motion. The analysis of these

equations would give the performance.

Remarks:

i) References 1.1, 1.5 to 1.13 may be referred to supplement the analysis

described in this and the subsequent five chapters.

ii) It would be helpful to recapitulate the following points.

(a) A ‘Flight path’ is the line along which the centre of gravity (c.g.) of the airplane

moves. Tangent to the flight path gives the direction of the ‘Flight velocity’ (see

Fig.5.1).

Fig.5.1 Flight path

(b) The external forces acting on a rigid airplane are:

(I) Aerodynamic forces (lift and drag)

(II) Gravitational force

(III) Propulsive force (thrust)



(c) The forces produced due to control deflection, needed to balance the

moments, are assumed to be small as compared to the other forces. With this

assumption all the forces acting on the airplane are located at the centre of

gravity (c.g.) of the airplane (Fig.5.2) and its motion is simplified to that of a point

mass moving under the influence of aerodynamic, propulsive and gravitational

forces.

Fig.5.2 Steady level flight


In this flight the c.g. of the airplane moves along a straight line at a

constant velocity and at a given altitude. The flight path, in this case, is a

horizontal line. The forces acting on the airplane are shown in Fig.5.2. ‘T’ is

Thrust, ‘D’ is Drag, ’L’ is lift and ‘W’ is the weight of the airplane.The equations of

motion are obtained by resolving, along and perpendicular to the flight direction,

the forces acting on the airplane. In the present case, the following equations are

obtained.

T - D = m ax

L - W = m az

where, m is the mass of airplane and ax, and az, are the components of the

acceleration along and perpendicular to the flight path respectively.



As the flight is steady i.e. no acceleration along the tangent to the flight path,

implies that ax = 0. Further, the flight is straight and at constant altitude, hence,

az= 0.

Consequently, the equations of motion reduce to:

T – D = 0, L – W = 0 (5.1)

Noting that, L = (1/2)ρV2SCL and L = W in level flight, gives :

W = (1/2)ρV2SCL

Or V = (2W / ρSCL)1/2 (5.2)

Further, (1/2)ρV2S = W / CL

Noting that, D = (1/2)ρV2SCD and T = D in level flight, gives

the thrust required (Tr) as :

Tr = D =(1/2) ρV2SCD

Substituting for (1/2) ρV2S as W / CL, yields:

Tr = W (CD/ CL) (5.3)

The power required (Pr), in kiloWatts, is given by:

Pr = Tr V/1000 (5.3a)

where Tr is in Newton and V in m/s.

Substituting for V and Tr from Eqs. (5.2) and (5.3) in Eq.(5.3a) yields:

CW 2WDP = × ×r 1000 C ρS CL L

Or 3 C1 2W DPr 3/21000 ρ S CL

(5.4)

Remarks:

i) Equations (5.1) to (5.4) are the basic equations for steady level flight and would

be used in subsequent analysis of this flight.

ii) To fly in a steady level flight at chosen values of h and V, the pilot should

adjust the following settings.

(a) The angle of attack of the airplane to get the desired lift coefficient so that the

lift(L) equals the weight(W).



(b) The throttle setting of the engine, so that thrust equals drag at the desired

angle of attack. He (pilot) will also have to adjust the elevator so that the airplane

is held in equilibrium and the pitching moment about c.g. is zero at the required

angle of attack. As noted earlier, the forces (lift and drag) produced due to the

elevator deflection are neglected.

5.3 Stalling speed:

Consider that an airplane which has weight (W) and wing area (S), is flying at

an altitude (h). From Eq.(5.2) it is observed that, the flight velocity (V) is

proportional to 1/CL

1/2. Thus, the value of C

L required would increase as the flight

speed decreases. Since CL cannot exceed C

Lmax, there is a flight speed below

which level flight is not possible. The flight speed at which CL equals C

Lmax is

called ‘Stalling speed’ and is denoted by Vs. Consequently ,

Vs= (2W / ρSCLmax

)1/2 (5.5)

It is evident from Eq.(5.5) that Vs increases with altitude since the density (ρ)

decreases with height.The variations of Vs with h for a typical piston engined

airplane and a typical jet airplane are presented in Figs.5.3a and b respectively.

Appendices A & B give the details of calculations.

Remark:

The maximum lift coefficient (CLmax

) depends on the flap deflection (δf). Hence,

Vs will be different for the cases with (a) no flap (b) flap with take-off setting (c)

flap with setting for landing. Figure 5.3a presents the variations of stalling speed,

with altitude, for four cases viz. with no flap and with three different flap settings.



Fig.5.3a Variations of stalling speed with altitude for a low speed airplane

Fig.5.3b Variations of stalling speed with altitude for a jet transport


Equivalent airspeed (Ve) is defined by the following equation.



,

2 21 1ρV = ρ Vo e2 2Noting σ = ρ/ρ V can be expressed as :o e

2W1/2V = Vσ =e ρ SCo L (5.6)

Remarks:

i) From Eq.(5.6) it is evident that for a given wing loading (W/S), the equivalent

airspeed in steady level flight is proportional to 1/CL1/2 and is independent of

altitude. Thus the stalling speed, for a given airplane configuration, when

expressed as equivalent airspeed is independent of altitude.

ii) To avoid confusion between equivalent airspeed ( Ve ) and the actual speed of

the airplane relative to the free stream (V), the latter is generally referred to as

true airspeed.


The equivalent airspeed is also significant from the point of view of

measurement of speed of the airplane using Pitot-static system. It may be

recalled from the topics studied in fluid machanics that a Pitot-static tube senses

the Pitot (or total) pressure (pt) and the static pressure ( sp ). The difference

between pt and sp is related to the velocity of the stream ( V ) by the following

equation.

1;

2

2 42

stM M

p -p = ρV 1+ + +.... M = V /a,4 40

a = speed of sound (5.6a)

Thus, at low speeds (M < 0.2),

1

22

stp - p ρV

It may be pointed out that, in the case of an airplane, the air is stationary and the

airplane is moving. Hence, the quantity V in the above expressions, equals the

speed of the airplane(V). Hence, at low speeds:

2 2s et 0

1 1p -p ρV = ρ V

2 2 (5.6b)



In an airplane the Pitot pressure is sensed by a Pitot tube mounted on the

airplane and static pressure is sensed by a hole located at a suitable point on the

airplane.These two pressures are supplied to the airspeed indicator mounted in

the cockpit of the airplane. The mechanism of the airspeed indicator in low speed

airplanes is such that it senses t sp -p and indicates eV . Note that 0ρ is a

constant value.

At subsonic speeds, when the compressibility effects become significant,

the airspeed indicator mechanism is calibrated to indicate ‘Calibrated airspeed

(Vcal)’, based on the following equation which is a simplified form of Eq.(5.6a).

1 11

2 4

22

2cal

st 0 cal0

Vp -p = ρ V

a

(5.6c)

where, a0 = speed of sound under sea level standard conditions.

For further details like construction of airspeed indicators and measurement of

airspeed at supersonic Mach numbers, Refs. 5.1, 5.2 and 5.3 may be consulted.

Information is also available on the internet. However, it may be added that the

static pressure sensed by the static pressure hole may be influenced by the flow

past the airplane. It may be slightly different from the free stream static pressure

and hence the speed indicated by the airspeed indicator may be slightly different

from eV or calV . The speed indicated by the airspeed indicator is called ‘Indicated

airspeed’ and denoted by Vi.

Remark:

On high speed airplanes the speed with respect to ground called ‘Ground

speed’ is deduced from the coordinates given by the global positioning system

(GPS). However, airspeed indicator based on Pitot static system is one of the

mandatory instruments on the airplane.


From Eqs.(5.3) and (5.4) it is noted that :

3C C1 2WD DT = W and P = r r 3/2C 1000 ρSL CL

.



The drag coefficient (CD)depends on the lift coefficient (C

L) and the Mach

number. The relationship between CD and C

L, the drag polar, is already known

from the estimation of the aerodynamic characteristics of the airplane.

Thus, when the drag polar, the weight of the airplane and the wing area are

prescribed, the thrust required and the power required in steady level flight at

various speeds and altitudes can be calculated for any airplane using the above

equations. The steps are as follows.

i) Choose an altitude (h).

ii) Choose a flight velocity (V).

iii) For the chosen values of V and h, and given values of the weight of

airplane (W) and the wing area(S) calculate CL as :

L 2

2WC =

ρSV

where corresponds to the density at the chosen ‘h’.

iv) Calculate Mach number from M = V/a; ‘a’ is the speed of sound at the chosen

altitude.

v) For the values of CL and M, calculated in steps (iii) and (iv), obtain C

D from the

drag polar. It (drag polar) may be given in the form of Eqs.(3.45) or (3.49). The

drag polar can also be given in the form of a graph or a table.

vi) Knowing CD, The thrust required (Tr) and power required (Pr) can now be

calculated using Eqs.(5.3) and (5.4).



Chapter 5


5.6 Thrust and power required in steady level flight when drag

polar is independent of Mach number


parabolic polar

5.6 Thrust and power required in steady level flight when drag polar is

independent of Mach number

When the Mach number is less than about 0.7, the drag polar is generally

independent of Mach number. In this case, CD / C

L and C

D / CL

3/2 can be

calculated for different values of CL. The curves shown in Figs.5.4a and b are

obtained by plotting CD / CL

and CD / CL

3/2 as functions of C

L. From these curves

it is observed that CD / CL

is minimum at a certain value of CL. This C

L is denoted

by CLmd

as the drag is minimum at this CL. The power required is minimum when

CD / CL

3/2 is minimum. The C

L at which this occurs is denoted by C

Lmp . Thus in

steady level flight:

Trmin = W (C

D / CL)min

(5.7)

3 C2W DP =rmin 3/2ρS CL min

(5.8)



Fig.5.4a Variation of CD / CL

with CL

Fig.5.4b Variation of CD / CL

3/2 with C

L



The speeds, at which the drag and the power required are minimum, are denoted

by Vmd and Vmp respectively. The expressions for Vmd and Vmp are:

2W 2WV = , V =md mpρSC ρSCLmd Lmp

(5.9)

Note:

i) CLmd

and CLmp

are not equal and the corresponding speeds are different. As the

density occurs in the denominator of Eq.(5.9), it implies that Vmd and Vmp

increase with altitude.

ii) Since for Mach number is lower than about 0.7, the drag polar is assumed to

be independent of Mach number, the values of CLmd

, CLmp

, (CD / C

L)min

and

(CD / C L

3/2)min

are also independent of Mach number. From Eqs.(5.7) and (5.8) it

is seen that Trmin

is independent of altitude whereas Prmin

increases with altitude

in proportion to 1/ σ1/2

.

iii) It is also observed in Fig.5.4a that a line drawn parallel to the X-axis cuts the

curve at two points A and B. This shows that for the same value of CD / CL

or the

thrust {Tr = W(CD / CL

) }, an airplane can have steady level flight at two values of

lift coefficients viz. CLA

and CLB

. From Eq.(5.2) each value of CL corresponds to a

velocity. Hence for the same amount of thrust, in general, flight is possible at two

speeds (VA and V

B). These speeds are:

VA= (2W / ρSC

LA)1/2

, VB = (2W / ρSC

LB)1/2 (5.9a)

Similarly, from Fig.5.4b it is observed that with the same power, in general, level

flight is possible at two values of lift coefficient viz. CLA

and CLB and

correspondingly at two flight speeds viz. VA and V

B .

iii) Typical variations of thrust required with flight speed and altitude are shown in

Fig.5.5. Following interesting observations are made in this case where the drag

polar is independent of Mach number. From Eq.(5.7) the minimum drag depends



only on W and (CD / CL

)min

and hence is independent of altitude. However, the

speed corresponding to minimum drag ( Vmd) increases with altitude (Eq.5.9).

Hence, the thrust required curves at various altitudes have the same minimum

thrust at all altitudes and the curves have a horizontal line, corresponding to

T = Trmin

, as a common tangent (see Fig.5.5). This feature should be kept in

mind when thrust required curves for subsonic airplanes are plotted.

Fig.5.5 Thrust required and thrust available for subsonic jet airplane

iv) Typical variations of power required with flight speed and altitude are shown

in Fig.5.6a. Interesting observations are made in this case also. From Eq.(5.8)

the minimum power required (Prmin) depends on W3/2 , (CD/CL3/2)min and ρ-1/2 .

From Eq.(5.9) it is observed that Vmp depends on ρ-1/2 . Noting that for airplanes

with piston engine or turboprop engine, the flight Mach number is less than 0.7,

the drag polar is independent of Mach number. However, due to dependence on

ρ-1/2 , the Prmin and Vmp increase with altitude (Fig.5.6a) . It may be added that

the slope of a line, joining a point on the Pr vs V curve and the origin, is Pr / V or

Tr. However, as pointed out earlier, Tr has a minimum value (Trmin) which is



independent of altitude. Hence, all Pr vs. V curves have a common tangent

passing through the origin. Such a tangent is shown in Fig.5.6a. This feature

should be pointed out when Pr vs. V curves are plotted at different altitudes. Note

that the common tangent to Pr vs. V curves does not touch at Vmp but at Vmd.

Fig.5.6a Power required and power available curves

Fig.5.6b Power required and power available at an altitude near ceiling




parabolic drag polar

The discussion in section 5.6, was with reference to a general drag polar

which may be given in tabular form or a plot. Consider the parabolic polar given

by :

CD = C

D0 + KCL2 (5.10)

Since an equation is available for the drag polar, it is possible to obtain

mathematical expressions for the power required and thrust required. In this

section it is assumed that CDO and K are constant with Mach number.

Substituting for CD in expression for thrust required gives:

Tr = D = (1/2)ρV2SCD

= (1/2) ρV2 S (C

D0+KC

L2) (5.11)

Substituting for CL as W /{(1/2)ρV2S} in Eq.(5.11) yields:

2

1 2W2T = ρ V S C + Kr Do 12 2ρV S2

Or 2 2 2= + / ( )r DOT V SC KW ρV S1ρ 2

2 (5.12)

In Eq.(5.12) the first term (½) ρ V2 S CD0 is called ‘Parasite drag’. The second

term 2 K W2 / (ρV

2S) is called ‘Induced drag’. Typical variations of the parasite

drag, induced drag and total drag are shown in Fig.5.7.



Fig.5.7 Variation of drag with flight speed

It is observed that from Fig.5.7 that the parasite drag, being proportional to V2,

increases rapidly with speed. The induced drag being proportional to 1/ V2 is high

at low speeds but decreases rapidly as speed increases. The total drag, which is

the sum of the induced drag and the parasite drag, is approximately equal to

induced drag at low speeds and approaches parasite drag at high speeds. It has

a minimum value at a speed (Vmd

) where the parasite drag and induced drag are

equal to each other (Fig.5.7). This can be verified by differentiating Eq.(5.12) with

respect to V and equating it to zero i.e.

2dT 2KW (-2)r = ρ V S C + = 0md D0 3dV ρS Vmd

Or

12 1/4

2W KV =md ρS CD0

(5.13)



Substituting Vmd in Eq.(5.12) gives minimum thrust required i.e.

Trmin = W (C

D0 K)1/2+W(CD0

K)1/2 = 2W(CD0

K)1/2 (5.14)

From Eq.(5.14) it is observed that when V equals Vmd , the parasite drag and

induced drag both are equal to W (CD0 K)1/2. This is also shown in Fig.5.7.

Expression for power required in the present case is given by :

T V 1 1 3rP = = ρV SCr D1000 1000 2

Substituting for CD from Eq.5.10 gives:

1 1 3 2P = ρ V S [C + KC ]r D0 L1000 21 1 W3 2Or P = ρ V S [C + K ( ) ]r D0 211000 2 ρV S

2

Or 21 1 KW3P = ρV S C +r D02000 500 ρVS

(5.15)

The first term in Eq.(5.15) is called ‘Parasite power’ and the second term is called

‘Induced power’. The variations with flight velocity (V) of induced power, parasite

power and the total power required are shown in Fig.5.8.

It is observed that the minimum power occurs at a speed, Vmp , at which the

induced power is three times the parasite power. This can be verified by

differentiating Eq.(5.15) with respect to V and equating it to zero. The verification

is left as an exercise to the student.

1/2

1/42W K

V =mp ρS 3CDo (5.16)

1/2

1/431 2W 256 3P = C Krmin Do1000 ρS 27 (5.17)



Fig.5.8 Variation of power required with flight speed

Remarks:

i) The expressions given in Eqs.(5.13) and (5.14) can be obtained in the following

alternate way.

Tr = W (CD / CL

)

Hence, Trmin = W (C

D / CL)min (5.18)

But, for a parabolic polar

CC D0D = + K CLC CL L (5.19)

The value of CL at which (C

D / CL) is minimum i.e. (C

Lmd) is given by :



Cd(C /C ) D0D L = 0 or - + K = 02dCL CLmd

This gives CLmd

as:

CLmd = (C

Do / K)1/2

(5.20)

The corresponding drag coefficient, CDmd

is

Dmd DODO

DOKC

C = C + = 2CK

(5.21)

Equation (5.21) shows that when Tr equals Tmin

, both parasite drag coefficient

and induced drag coefficient are equal to DOC . Hence under this condition, the

parasite drag and induced drag both are equal to (1/2)ρ V2 S DOC .

Further,

12

12Dmd DO

Lmd DO

DDO

L

C 2CC= = = 2 (C K)

C C C / Kmin

(5.22)

Hence, Trmin and Vmd are:

DO DO2W1/2 1/2 1/4T = 2 W (C K) and V = ( ) (K / C )rmin md ρS

,

which are the same as Eqs.(5.14) & (5.13).

(ii) Exercise 5.4 gives expressions for Tr in terms of V/Vmd and Trmin.

(iii) Similarly, expressions given in Eqs.(5.16) and (5.17) can be obtained in the

following alternate manner.

1/23 C1 2W DP = r 3/21000 ρS CL

Hence, Prmin occurs when CD/C

L

3/2 is minimum. For a parabolic polar

1/2DOCCD = + K CL3/2 3/2C CL L

Therefore,



12

DO5/2L L

3/2d C / CD CL 3 1 K = - +

dC 2 2L C C

Equating the R.H.S. to zero, the value of CL at which the power required is

minimum ( LmpC ) is given as:

LmpC = (3 DOC / K)1/2

(5.23)

Then the drag coefficient, corresponding to LmpC is given by:

Dmp DO DODO3KC

C = C + = 4 CK

(5.24)

Equation (5.24) shows that when Pr equals P

rmin the parasite drag coefficient is

equal to DOC and the induced drag coefficient is equal to 3 DOC . Consequently,

the parasite power is (1/2) ρ V3 S DOC and induced power is 3 times of that.

Hence,

DODO

DO

1/44CC 256 3D = = C K3/2 3/4 27C (3C / K)L min

(5.24a)

1/2 1/2 1/42W 2W K

V = =mp ρSC ρS 3 CLmp DO

1= V 0.76 Vmd md1/43

(5.24b)

The above expression for Vmp is the same as in Eq.(5.16).

Example 5.1

An airplane weighing 100,000 N is powered by an engine producing

20,000 N of thrust under sea level standard conditions. If the wing area be

25 m2 , calculate (a) stalling speeds at sea level and at 10 km altitude,

(b) (CD / C

L)min

, (CD / 3/2

LC )min, T

rmin, P

rmin, Vmd and Vmp under sea level

conditions.

Assume CLmax = 1.5, C

D = 0.016 + 0.064 2

LC .



Solution:

The given data are : W = 100,000 N, T = 20,000 N, CD = 0.016 + 0.064 C

L

2 ,

S = 25 m2, C

Lmax =1.5

a) SLmax

2WV = ,

ρSC

at s.l. ρ = 1.225 kg/m3,

at 10 km = 0.413 kg / m3

Hence, at sea level,2 × 100000

V =S 1.225 × 25 × 1.5 = 66 m/s = 237.6 kmph

At 10 km altitude,2 × 100000

V =S 0.413 × 25× 1.5 = 113.6 m/s = 409.0 kmph.

b) CD0C = = 0.016/0.064 = 0.5Lmd K

DmdC 2= = 0.032DOC

Hence, (CD / C

L)min

= 0.032/0.5 = 0.064 and

Trmin

= W (CD / C

L)min

= 100000 x 0.064 = 6400 N

C = 3C /K = 0.866Lmp DO

C = 4C = 0.064Dmp DO

3/2 3/2(C / C ) = 0.064/0.866 = 0.0794D minL

2W 2 ×100000V = = = 114.5 m/s = 412.2 kmphmd ρ S C 1.225× 25 ×0.5Lmd

2 ×100000V = = 86.30 m/s = 310.7 kmphmp 1.225× 25 ×0.866

Note: V mp = Vmd / 31/4

3 31 2W 1 2 × 1000003/2P = C / C = × 0.0794 = 641.5 kW.rmin D L1000 ρS 1000 1.225 × 25min



Answers :

a) SV at sea level = 237.6 kmph

SV at 10 km altitude = 409.0 kmph

b) (CD / C

L)min

= 0.064 ; 3/2(C / C ) = 0.0794D minL ; Trmin

= 6400 N

At sea level : P = 641.5 kW;rmin V = 412.2 kmph;md V = 310.7 kmphmp



Chapter 5




5.11 Special feature of steady level flight at supersonic speeds


From Eq.(5.27), an analytical expression for Vmax can be deduced when it is

assumed that the thrust available (Ta) ,CDO and K remain constant with flight

speed. The derivation is as follows.

22

2DOa1 2W

T = ρV S C K2 ρV S

(5.27)

or AV4 – BV2 + C = 0

where, DO

1A = ρSC

2, B = Ta and

22KWC =

ρS. (5.27a)

When Ta, CDO and K have constant values, Eq.(5.27a) gives :

122B ± B -4AC

V =2A

Consequently, Vmax being the larger of the two solutions, is :

2

max

12B+ B -4AC

V =2A

Substituting for A, B and C from Eq.(5.27a) yields :

2 2

max 2 2 2 2 2

12

DO DO

a a

DO

T T W KV = + - 4

ρSC ρ S C S ρ C

(5.27b)



Multiplying and dividing some of the terms in Eq.(5.27b) by ‘W’ gives:

12

max

2 2 2DO

2 2 2 2DO DO DO

a aT /W W/S T /W W/S K CWV = + - 4

ρC Sρ C ρ C

Simplifying yields :

Or DO

maxDO

22

1/2

a aW

T /W W/S + T /W -4C KS

V =ρC

(5.27c)

Equation (5.27c) shows that Vmax depends on thrust to weight ratio ( aT /W), wing

loading (W/S), CDO , K and ρ .The maximum speed (Vmax) increases with increase

of ( aT /W) and (W/S) and decreases with increase of CDO and K. The term ‘ρ ’ in

the denominator of Eq.(5.27c) indicates that Vmax would be higher at higher

altitudes because ρ decreases with altitude. In section 4.5 it is pointed out that

the thrust output decreases as 0.7σ . Taking this into account, Eq.(5.27c) indicates

that Vmax would increase slightly upto a certain altitude as shown in Fig.5.9.

The trend of Vmax, decreasing after a certain altitude, observed in Fig.5.9, can be

explained as follows.

From atmospheric characteristics (Chapter 2), it is observed that, with the

increase of altitude the speed of sound decreases. Thus for a given Vmax the

Mach number corresponding to it would increase with altitude. When the Mach

number exceeds the critical Mach number, CDO & K would no longer be constant

but actually increase. This would result in lowering of Vmax as compared to that

obtained with constant values of CDO and K. In section 4.2 of Appendix ‘B’ the

values of Vmax at different altitudes are obtained by a graphical procedure which

takes into account the changes in CDO and K when Mach number is greater than

0.8.


The steps to calculate Vmax

and (Vmin

)e in this case, are as follows.

(1) Assume an altitude ‘h’. Let Pa be the THP available in kW at this altitude.



(2) From Eq.(5.15) :

21 1 KW3P = P = ρ V S +r a DO2000 500 ρVSC

or A1V

4 – B

1V + C

1 = 0 (5.28)

where, 2

, = 1DO1 1 KW

A = ρSC , B = P C1 1 a2000 500 ρS.

Equation (5.28) is not a quadratic. An iterative method of solving Eq.(5.28) is

given in example 5.3. Equation (5.28) has two solutions V1 and V

2. The higher of

these two gives Vmax

and the lower value gives (Vmin

)e. The minimum speed at

the chosen altitude is higher of (Vmin

)e and Vs (see example 5.3).

Remark:

Obtain power available at V1 calculated above and denote it by Pa1. If Pa

assumed at the beginning of the calculation in step (1), is significantly different

from Pa1, then the calculations would need to be revised with the new value of

Pa1. However, it is expected that the calculations would converge in a few

iterations.

Example 5.2

For the airplane in example 5.1 obtain the maximum and minimum speed in

steady level flight at sea level.

Solution:

The given data are :

W = 100,000N, T = 20,000N, S = 25 m2, 2C = 0.016 + 0.064 CD L , CLmax = 1.5

In this case, T / W = 20000 / 100000 = 0.2 = CD / CL

0.0160.2 = + 0.064CLCL

2Or 0.064 C - 0.2 C + 0.016 = 0LL

Solving the above equation gives: CL = 3.04 and 0.0822. The corresponding

speeds are :



2 ×100000V = = 281.8 m/smax 1.225× 25 × 0.0822

and 2 ×100000

( V ) = = 45.4 m/smin e 1.225× 25 × 3.04

Since VS , as calculated in example 5.1, is 66.0 m/s, the minimum speed is

decided by VS and equals 66.0 m/s.

The Mach number corresponding to Vmax

is :

281.8 / 340.29 = 0.828.

This value of Mach number is likely to be greater than Mcrit

. As a possible

assumption let us assume Mcruise

= 0.8 and obtain DOΔC and ΔK from Eqs.3.50a

and 3.51a. Consequently,

DOΔC = - 0.001 (M - 0.8) + 0.11 (M - 0.8)2 and ΔK = (M - 0.8)2 + 20(M - 0.8)3

For M = 0.828, DOΔC = 0.000055 and ΔK = 0.00122

Hence, the drag polar at M = 0.828 is likely to be:

CD = (0.016 + 0.000055) + (0.064+0.00122)C

L

2 = 0.016055 + 0.06522 C

L

2

Using this polar and revising the calculations, gives: Vmax = 281.3 m/s

This revised value of Vmax

is very close to the value of 281.8 m/s obtained earlier

and hence further revision is not needed.

(Answers: Vmax = 281.3 m/s =1012.7 kmph, Vmin = 66.00 m/s = 237.6 kmph)

Example 5.3

A piston-engined airplane has the following characteristics.

W = 11,000 N, S = 11.9 m2, CD = 0.032 + 0.055 C

L2, C

Lmax = 1.4.

Obtain the maximum and minimum speeds in level flight at an altitude of 3 km

assuming that the engine BHP is 103 kW and the propeller efficiency is 83%.

Solution :

W = 11,000 N, S = 11.9 m2, CD = 0.032 + 0.055 C

L

2

CLmax

= 1.4, ρ at 3km altitude = 0.909 kg/m3,



Pa = η x BHP = 0.83 x 103 = 85.5 kW

From Eq.(5.15):

21 2K W3P = P = ρV S C +a r Do2000 1000 ρSV

Or 21 2 0.055 ×11000385.5 = × 0.909 × 11.9× 0.032×V + ×

2000 1000 0.909 × 11.9 × V

= 1.731 x 10-4 V3 + 1230.5

V (5.28a)

Equation (5.28a) is not a quadratic. However, it can be solved for Vmax

and (Vmin

)e

by an iterative procedure.

Solution for Vmax:

When solving for V max

, by an iterative procedure, it is assumed that the first

approximation (Vmax1

) is obtained by retaining only the term containing the

highest power of V in Eq.(5.28a) i.e.

1st approximation: 85.5 = 1.731 x 10-4 V3

max1

This gives Vmax1

= 79.05 m/s

To obtain the 2nd approximation, substitute Vmax1

in the second term on RHS of

Eq.(5.28a). Note that this term was ignored in the first approximation.

3max2

1230.5-485.5 = 1.731 x 10 V + 79.05

Or V max2

= 73.93 m/s

To obtain the 3rd approximation, substitute Vmax2


Eq.(5.28a), i.e.

3max3

1230.5-485.5 = 1.731× 10 V + 73.93

Or Vmax3

= 73.54 m/s

To obtain the 4th approximation, substitute Vmax3


Eq.(5.28a), i.e.



3max4

1230.5-485.5 = 1.731× 10 V +73.54

Or Vmax4

= 73.51 m/s

Since the 3rd and 4th approximations are close to each other, Vmax

is taken as

73.51 m/s.

Solution for (Vmin

)e:

When solving for (Vmin

)e, by an iterative procedure, it is assumed that the first

approximation 1min e

V , is obtained by retaining only the term containing the

lowest power of V in Eq.(5.28a) i.e.

1min e

1230.585.5 =

V

Or 1min e

V = 14.4 m/s.

To obtain the 2nd approximation, substitute (Vmin

)e1 in the first term on RHS of

Eq.(5.28a). Note that this term was ignored in the first approximation.

2min e

1230.5-4 385.5 = 1.731× 10 × 14.4 +(V )

Or 2min e(V ) = 14.48 m/s

Since the second approximation is very close to the first one,

(Vmin

)e is taken as 14.48 m/s

The stalling speed at 3 km altitude is :

2W 2 × 11000V = = = 38.2m/ss ρSC 0.909×11.9 ×1.4Lmax

Since Vs is greater than (Vmin

)e, the minimum speed is 38.2 m/s.

Answers:

At 3 km altitude:

Vmax = 73.51 m/s = 265.0 kmph , V

min = 38.20 m/s = 137.4 kmph



5.11 Special features of steady level flight at supersonic speeds

At transonic and supersonic speeds the variations of CD0

, K and Ta with

Mach number do not permit simple mathematical treatment of the performance

analysis. The thrust required (Tr) increases rapidly as the Mach number

approaches unity (Fig.5.11).

Fig.5.11 Level flight performance at high speeds

The thrust available also increases but the increase is not as fast as that

of Tr and the thrust available and thrust required curves may intersect at many

points (points A, B, and C in Fig.5.11). It is interesting to note that if the airplane

can go past Mach number represented by point B in Fig.5.11, then it can fly up to

Mach number represented by point C with the same engine. To overcome the

rapid drag rise in transonic region (Fig.5.11), the afterburning operation of the

engine is resorted to. It may be mentioned that an afterburner duct is located

between the turbine and the nozzle (Fig.4.8b). When the afterburner is on,



additional fuel is burnt in the afterburner duct. This gives additional thrust.

However, the specific fuel consumption is very high with afterburner on and

hence this operation is resorted to only for a short duration.

The thrust with afterburner on is shown schematically by a dotted line in Fig.5.11.

It is observed that the thrust available is more than the thrust required and

airplane can accelerate beyond point B. When the Mach number is close to that

represented by point C, the afterburner can be shut down and the airplane runs

with normal engine operation.



Chapter 5

References

5.1 Pallett, E.H.J. “Aircraft instrument integrated systems” 3rd Edition, Longman

Science & Technology, (1992).

5.2 Illman, P.E. “The pilot’s handbook of aeronautical knowledge“ 3rd Edition,

Tab books division of McGraw Hill (1995).

5.3 Perkins, C.D. (Editor) “AGARD Flight test manual, Vol.I – Performance”

Pergamon Press (1959).



Chapter 5

Exercises

5.1 Obtain the maximum speed and minimum speed in steady level flight at sea

level for the following airplane:

W = 36,000 N; S = 26.0 m2; CD = 0.032 + 0.043C

L2 BHP = 503 kW; Propeller

efficiency = 82%; CLmax = 1.5

[Answers: Vmax = 324.6 kmph; Vmin = 139.8 kmph]

5.2 A jet engined airplane has a weight of 64,000 N and wing area of 20 m2. If

the engine output at 5 km altitude be 8000 N, calculate the maximum and

minimum speeds in level flight. Given that

CDO

= 0.017, A = 6.5, e = 0.80, CLmax

= 1.4.

[Answers: Vmax

= 877 kmph,Vmin = 283.6 kmph]

5.3 An airplane stalls at M=0.2 at sea level. What will be the Mach number and

equivalent airspeed when it stalls at 5 km altitude? Compare the thrust required

to maintain level flight near stall at the two altitudes. Assume the weight of the

airplane to be same at the two altitudes.

[Answers: M = 0.274, Ve = 68.06 m/s, (Tr)s.l = (Tr)5 km as CL is same]

5.4 (a) Show that the thrust required in steady level flight at a speed V for an

airplane with parabolic drag polar is given by:

V AW2T = D = AW ( ) +r 2Vmd (V / V )md

where, Vmd = speed for minimum drag, W = weight of airplane and A = (CD0

K)1/2

.

(b) Also show that if V = m Vmd, then the thrust required (Tr) in terms of the

minimum thrust required (Trmin) is given by :

2 -2r

rmin

T 1= m +m

T 2



Chapter 6

Performance analysis II – Steady climb, descent and glide

(Lectures 21,22 and 23 )

Keywords: Steady climb – equations of motion, thrust and power required;

maximum rate of climb; maximum angle of climb; absolute ceiling; service ceiling;

glide – equations of motion, minimum angle of glide, minimum rate of sink;

hodographs for climb and glide.

Topics

6.1 Introduction

6.2 Equations of motion in steady climb

6.3 Thrust and power required for a prescribed rate of climb at a given

flight speed

6.4 Climb performance with a given engine

6.4.1 Iterative procedure to obtain rate of climb

6.5 Maximum rate of climb and maximum angle of climb

6.5.1 Parameters influencing (R/C)max of a jet airplane

6.5.2 Parameters influencing (R/C)max of an airplane with engine-propeller

combination

6.6. Climb hydrograph

6.7. Absolute ceiling and service ceiling

6.8 Time to climb

6.9 Steady descent

6.10 Glide

6.10.1 Glide performance – minimum angle of glide, minimum rate of sink

and maximum range and endurance in glide.

6.11 Glide hodograph

Exercises



Chapter 6

Lecture 21 Performance analysis II – Steady climb, descent and glide – 1 Topics

6.1 Introduction

6.2 Equations of motion in steady climb


flight speed



6.1. Introduction

In this chapter the steady climb, descent and glide are dealt with. A glide

is a descent with thrust equal to zero. The approach in this chapter is as follows.

(a) Present the forces acting on the airplane in the chosen flight,

(b) Write down equations of motion using Newton’s second law,

(c) Derive expressions for performance items like rate climb, angle of climb.

(d) Obtain variation of these with flight velocity and altitude.

6.2 Equations of motion in a steady climb

During a steady climb the center of gravity of the airplane moves at a

constant velocity along a straight line inclined to the horizontal at an angle γ

(Fig.6.1). The forces acting on the airplane are shown in Fig.6.1.



Fig.6.1 Steady climb

Since the flight is steady, the acceleration is zero and the equations of motion in

climb can be obtained by resolving the forces along and perpendicular to the

flight path and equating their sum to zero i.e.

T – D – W sin = 0 (6.1)

L – W cos = 0 (6.2)

Hence, sin = (T- D / W) (6.3)

From the velocity diagram in Fig.6.1, the vertical component of the flight velocity

(Vc) is given by:

Vc = V sin = (T- D / W) V (6.4)

The vertical component of the velocity (Vc) is called rate of climb and also

denoted by R/C. It is also the rate of change of height and denoted by (dh / dt).

Hence,

Vc = R/C = dh/dt = T-D

Vsin = VW

(6.5)

Rate of climb is generally quoted in m/min.

Remarks :

i) Multiplying Eq.(6.1) by flight velocity V, gives:

T V = D V + W V sin = D V + W Vc dh d= DV + mg = DV + mgh

dt dt (6.6)

In Eq.(6.6) the terms ‘TV’, ‘DV’ and ‘ dmgh

dt’ represent respectively, the power

available, the energy dissipated in overcoming the drag and the rate of increase



of potential energy. Thus, when the airplane climbs, its potential energy

increases and a part of the engine output is utilized for this gain of potential

energy.

Two facts may be pointed out at this juncture. (a) Energy supply to the airplane

comes from the work done by the engine which is represented by the term‘TV’ in

Eq.(6.6). (b) The drag acts in a direction opposite to that of the flight direction.

Hence, energy has to be spent on overcoming the drag which is represented by

the term ‘DV’ in Eq.(6.6). This energy (DV) is ultimately lost in the form of heat

and is appropriately termed as ’Dissipated’. Continuous supply of energy is

needed to overcome the drag. Thus, a climb is possible only when the engine

output is more than the energy required for overcoming the drag.

It may be recalled from section 5.9 that in a level flight, at speeds equal to Vmax

and (Vmin)e , the power (or thrust ) available is equal to the power (or thrust)

required to overcome the drag (see points D and D’ in Fig.5.5 and points C and

C’ in Fig.5.6b). Hence, the rate of climb will be zero at these speeds. The climb is

possible only at flight speeds in between these two speeds viz. Vmax and (Vmin)e.

It is expected that there will be a speed at which the rate of climb is maximum.

This flight speed is denoted by V(R/C)max

and the maximum rate of climb is

denoted by (R/C)max

. The flight speed at which the angle of climb () is maximum

is denoted by V max

.

ii) In a steady level flight, the lift is equal to weight but in a climb, the lift is less

than weight as cos is less than one, when is not zero. Note that when an

airplane climbs vertically, its attitude is as shown in Fig.6.2. It is observed that in

this flight, the resolution of forces along and perpendicular flight direction gives:

L = 0, T = D + W

These expressions are consistent with Eqs.(6.1) and (6.2) when = 90o is

substituted in them. Note that in this flight the thrust is more than the weight.



Fig.6.2 Airplane in vertical climb


flight speed

Here it is assumed that the weight of the airplane (W), the wing area (S) and the

drag polar are given. The thrust required and power required for a chosen rate of

climb (Vc) at a given altitude (h) and flight speed (V) can be obtained, for a

general case, by following the steps given below. It may be pointed out that the

lift and drag in climb are different from those in level flight. Hence, the quantities

involved in the analysis of climb performance are, hereafter indicated by the

suffix ‘c’ i.e. lift in climb is denoted by Lc

i) Since Vc and V are prescribed, calculate the angle of climb γ from:

γ = sin-1 (Vc / V)

ii) From Eq.(6.2) the lift required in climb (Lc ) is :

Lc = W cos γ (6.7)



iii) Calculate the lift coefficient in climb (CLc ) as:

CLc = L W cosc =

1 212 ρV SρV S 22

(6.8)

iv) Obtain the flight Mach number; M = V/a ; a = speed of sound at the chosen

altitude.

v) Corresponding to the values of CLc and M, obtain the drag coefficient in climb

(CDC

) from the drag polar. Hence, drag in climb (Dc) is given by:

Dc = (1/2 ρV2 S C

DC) (6.9)

vi) The thrust required in climb (Trc) is then given by:

Trc = W sin γ + Dc (6.10)

and the power required in climb (Prc) is :

Prc = T Vrc in kW1000

(6.11)

Example 6.1

An airplane weighing 180,000N has a wing area of 45 m2 and drag polar given

by CD = 0.017 + 0.05 2

LC . Obtain the thrust required and power required for a

rate of climb of 2,000 m/min at a speed of 540 kmph at 3 km altitude.

Solution:

The given data are:

W = 180,000 N, S = 45 m2, CD = 0.017 + 0.05 2

LC

Vc = 2,000 m/ min = 33.33 m/s, V = 540 kmph = 150 m/s.

at 3 km altitude = 0.909 kg/m3

sin = Vc / V = 33.33/150 = 0.2222 or = 12o-50’, cos = 0.975

Lc = W cos = 180000 x 0.975

Or CLc = 2

2W cos

ρ V S

=

180000 × 0.975 × 2= 0.381

0.909 × 150× 150 × 45



Hence, CDC

= 0.017 + 0.05 X0.3812 = 0.02426

Dc = (1/2 ρ V2 S) C

DC

= (1/2) X 0.909 X 150 X 150 X 45 X0.02426 = 11163 N

Hence, Trc = W sin + Dc = 180000 X 0.2222 + 11163 = 51160 N

Prc = TrcV/1000 = 51160 X 150/1000 = 7674 kW

Answers:

Thrust required in climb (Trc) = 51,160 N

Power required in climb (Prc) = 7,674 kW


In this case, the engine output is prescribed at a certain altitude and flight

speed. This is in addition to the data on weight of the airplane (W), the wing area

(S) and the drag polar.The rate of climb (Vc) and the angle of climb(γ) are

required to be determined at the prescribed altitude and flight speed.

The solution to this problem is not straightforward as sin γ depends on

(T- Dc) and the drag in climb (Dc) depends on the lift in climb (Lc ), which in turn

depends on W cos γ. Hence, the solution is obtained in an iterative manner. This

is explained later in this section. However, if the drag polar is parabolic with

constant coefficients, an exact solution can be obtained using Eqs. (6.1) to (6.4).

The procedure is as follows.

From Eq.(6.4), sin = Vc / V.

Using Eq.(6.7), the lift during climb (Lc) = W cos = W (1-sin2 )1/2

122= W 1-(V /V)c

(6.12)

Hence, Lift coefficient during climb

122W 1-(V /V)cLcC = =Lc 1 212 ρV SρV S 22

(6.13)

By its definition, D = (1/2) ρ V2 SC

D.



When the polar is parabolic, the drag in climb (Dc) can be expressed as :

Dc = (1/2) ρ V2 S (C

DO+K 2CLc ) DO

22 V1 KW2 c= ρ V S C + 1-212 VρV S

2

(6.14)

From Eq.(6.10), the thrust required in climb (Trc) is given by :

Trc = W sin γ + Dc =W Vc + DcV

Substituting for Dc , yields :

Trc DO

22 V WV1 KW2 c c = ρ V S + 1- +212 V VρV S

2

C

(6.15)

or 2V Vc cA - B + C =0

V V

(6.16)

where,2KW

A = , B = W21 ρV S

2

and 21 2KW2C = T - ρV SC -DO 22 ρV S

(6.17)

Equation (6.16) is a quadratic in (Vc / V), and has two solutions. The solution

which is less than or equal to one, is the valid solution because Vc / V equals

sin γ and sin γ cannot be more than one. Once (Vc / V) is known, the angle of

climb and the rate of climb can be immediately calculated. This is illustrated in

example 6.2.

Example 6.2.

For the airplane in example 6.1, obtain the angle of climb and the rate of climb at

a flight speed of 400 kmph at sea level, taking the thrust available as 45,000 N.

Solution:

In this case, W = 1,80,000 N, S = 45 m2 , CD = 0.017 + 0.05 2CL

V = 400 kmph = 111.1 m/s, T = 45,000 N

From Eq.(6.15), Trc DO

22 V WV1 KW2 c c = ρ V S + 1- +212 V VρV S

2

C



Substitution of various quantities yields:

2145000 = × 1.225 × 111.1 × 45 × 0.0172

22 V V0.05 × 180000 c c + × 1 - 180000 × 2 21 V× 1.225 × 111.1 × 45 V

2

Simplifying, 2V Vc c- 37.82

V V

+ 7.24 = 0

Solving the above quadratic gives : ( Vc / V) = 37.62, 0.192.

Since sin cannot be larger than unity, the first value is not admissible.

Hence, Vc / V = sin = 0.192 or = 11o 4

Vc = 0.192 111.1 = 21.33 m/s = 1280 m/min.

Answers:

Angle of climb () = 11o 4 ; Rate of climb (Vc) = 1280 m/min


When the drag polar is not given by a mathematical expression, an

iterative procedure is required to obtain the rate of climb for a given thrust ( aT ) or

thrust horse power (THPa). The need for an iterative solution can be explained as

follows.

From Eq.(6.10), sin = T - Da c

W (6.18)

To calculate sin , the drag in climb (Dc) should be known. The term Dc

depends on the lift in climb (Lc). In turn Lc is W cos , but cos is not known in

the beginning.

To start the iterative procedure, it is assumed that the lift during climb (Lc) is

approximately equal to W. Using this approximation, calculate the first estimate

of the lift coefficient (CL1

) as :

(CL1

) = W / (1/2)ρ V2 S



From CL1

and the flight Mach number obtain CD1

from the drag polar. Calculate

the first approximation of drag (D1) as:

D1 = (1/2) ρ V

2 S C D1

Hence, the first approximation to the angle of climb (1) is given by:

sin 1

T - Da 1=W

(6.19)

In the next iteration, put L = W cos 1 and carry out the calculations and get a

second approximation to the angle of climb (2). The calculations are repeated till

the values of after consecutive iterations are almost the same. Once the angle

is known, Vc is given as V sin .

It is observed that the convergence is fast and correct values of and Vc are

obtained within two or three iterations. This is due to the following two reasons.

(a)When is small (i.e. less than 10o), cos is almost equal to one, and the

approximation, L = W, is nearly valid. (b) When is large the lift dependent part

of the drag, which is affected by the assumption of L W , is much smaller than

Ta . Consequently, the value of given by Eq.(6.18) is close to the final value.

Example 6.3 illustrates the procedure.

Example 6.3

An airplane weighing 60,330 N has a wing area of 64 m2 and is equipped

with an engine-propeller combination which develops 500 kW of THP at 180

kmph under standard sea-level conditions. Calculate the rate of climb at this flight

speed. The drag polar is given in the table below.

CL 0.0 0.1 0.2 0.3 0.4 0.5 0.6

CD 0.022 0.0225 0.024 0.026 0.030 0.034 0.040

CL 0.7 0.8 0.9 1.0 1.2

CD 0.047 0.055 0.063 0.075 0.116



Solution:

The given data are: W = 60,330 N, S = 64 m2, V=180 kmph = 50 m/s,

THPa = 500 kW. Hence, Ta = (THPa x 1000)/V = 500 x 1000 / 50 = 10,000 N

The values of and Vc are obtained by the iterative procedure explained in

section 6.4.1.

1st approximation: L W = 1

22ρ V S CL1

Hence, CL1

60330 × 2= = 0.615

1.225× 50 × 50 × 64

CD1

: By interpolating between the values given in the above table, the value of

C D1

is 0.041, corresponding to CL1

of 0.615.

Hence, D1 = (1/2) 1.225 50 50 64 X0.041= 4030 N

From Eq.(6.19) : sin 1

T - Da 1=W

Or sin 1=

10000-4030= 0.099

60330

Or 1 = 5o 41

Hence, Vc1

= 50 x 0.099 = 4.95 m/s

cos 1 = 0.995

2nd approximation:

L = W cos 1 = 60330 0.995 = 60036 N

L260036 × 2

C = = 0.6121.225 × 50 × 50 × 64

From above table CD2

is 0.0408 corresponding to CL2

of 0.612.

Hence, D2 = (1/2) 1.225 50 50 64 0.0408 = 4010 N

sin 2=

10000-4010= 0.0993

60330



Hence, V

c2 = 50 x 0.0993 = 4.965 m/s

The two approximations, Vc1 and V

c2 are fairly close to each other. Hence, the

iteration process is stopped.

Vc = 4.965 m/s = 298 m/min.

Remark:

In the present example, is small (5041’) hence 2nd iteration itself gives

the correct value. For an interceptor airplane which has very high rate of climb

(about 15000 m/min) few more iterations may be needed.



Chapter 6





combination


Using the procedure outlined above, the rate of climb and the angle of

climb can be calculated at various speeds and altitudes. Figures 6.3a to 6.3f

present typical climb performance of a jet transport. Figure 6.4a to 6.4d present

the climb performance of a piston engined airplane. Details of the calculations for

these two cases are presented in Appendices B and A respectively.

Fig.6.3a Climb performance of a jet transport - rate of climb



Fig.6.3b Climb performance of a jet transport - angle of climb

Fig.6.3c Climb performance of a jet transport - V(R/C) max



Fig.6.3d Climb performance of a jet transport-Vmax

Fig.6.3e Climb performance of a jet transport - variation of (R/C)max with altitude



Fig.6.3f Climb performance of a jet transport - variation of max with altitude

Fig.6.4a Climb performance of a piston engined airplane- rate of climb



Fig.6.4b Climb performance of a piston engined airplane- angle of climb

Fig.6.4c Climb performance of a piston engined airplane - V(R/C)max, and Vmax



Fig.6.4d Climb performance of a piston engined airplane - Variation of (R/C)max

with altitude

Remarks:

i) At V = Vmax

the available thrust or THP is equal to the thrust required or power

required in level flight. Hence climb is not possible at this speed. Similar is the

case at (Vmin)e limited by engine output (Figs.6.3a and 6.4a). For the same

reasons, at Vmax

and (Vmin)e the angle of climb () is also zero (Figs.6.3b and

6.4b). It may be recalled from subsection 5.9, that at low altitudes the minimum

speed is decided by stalling and hence the calculations regarding the rate of

climb and the angle of climb are restricted to flight speeds between Vmin and

Vmax.

ii) The speed at which R/C is maximum is denoted by V(R/C)max

, and the speed at

which γ is maximum is called Vmax . Figures 6.3c and d and Fig.6.4c show the

variation of these speeds with altitudes for a jet transport and a piston engined

airplane respectively. It may be noted that V(R/c)max

and Vmax

are different from

each other. For a jet airplane V(R/c)max

is higher than Vmax at low altitudes . The

two velocities approach each other as the altitude increases. For a piston



engined airplane V(R/C)max

is lower than Vmax at low altitudes . The two velocities

approach each other as the altitude increases. These trends can be explained as

follows.

From Eqs.(6.3) and (6.4), it is observed that is proportional to the excess thrust

i.e. (Ta - Dc) and the rate of climb is proportional to the excess power i.e.

(TaV – DcV). It may be recalled that for a piston engined airplane the power

available remains roughly constant with velocity and hence, the thrust available

(Ta = Pa / V) will decrease with velocity. On the other hand, for a jet airplane the

thrust available is roughly constant with velocity and consequently the power

available increases linearly with velocity (see exercise 6.3). The differences in

the variations of Ta and Pa with velocity, in the cases of jet engine and piston

engine, decide the aforesaid trends.

iii) As the excess power and the excess thrust decrease with altitude, (R/C)max

and max also decrease with altitude.


In subsection 5.10.2, the parameters influencing Vmax were identified by

simplifying the analysis with certain assumptions. In this subsection the

parameters influencing (R/C)max are identified in a similar manner. The limitations

of the simplified analysis are pointed out at the end of this subsection.

The case of a jet airplane is considered in this subsection.

From Eq.(6.5) it is noted that :

C

T -DV = R/C = V

W

Following simplifying assumptions are made to identify the parameters

influencing (R/C)max.

(a)Drag polar is parabolic with constant coefficients i.e. CDO and K are constants.

(b) Though the angle of climb is not small, for the purpose of estimating the

induced drag, the lift (L) is taken equal to weight. See comments at the end of

section 6.4.1 for justification of this approximation.



(c) At a chosen altitude, the thrust available (Ta) is constant with flight speed.

With these assumptions, the expression for drag simplifies to that in the level

flight i.e.

2

2 22D DO

1 1 2WD = ρV SC = ρV S C + K

2 2 ρSV

Hence,

CaT -D

V = VW

Or

2

2DOa

c

1ρVT 2K W2V = V - C -

W W/S ρV S

Or -13C DO

aT 1 2K WV = V - ρV W/S C -

W 2 ρV S

(6.20)

To obtain the flight speed corresponding to (Vc)max , Eq.(6.20) is differentiated

with respect to V and equated to zero i.e.

-122DO

ac dV T 3 2K W= - ρV W/S C + = 0

dV W 2 ρV S

(6.21)

Simplifying Eq.(6.21) yields:

2(R/C)max (R/C)max

2

24

DO DO

a2 T /W W/S 4K W/SV - V - = 0

3ρC 3ρ C (6.22)

Noting from Eq.(3.56) that 2maxDO

1L/D =

4C K, yields:

2(R/C)max (R/C)max

max

2

22

42

DO DO

a2 T /W W/S W/SV - V - = 0

3ρC 3ρ C L/D

Thus,

(R/C)max 2

2

max

12

DO

a

a

T /W W/S 3V = 1 ± 1 +

3ρC LT /W

D



The negative sign in the above equation, would give an imaginary value

for R/C maxV and is ignored.

Hence,

R/C max 2DO max

12

2a

a

T /W W/S 3V = 1 + 1 +

3ρC L/D T /W

1

2

DO

aT /W W/S Z= ,

3ρC

(6.23)

where, 22

max a

3Z = 1 + 1 +

L/D T /W (6.24)

Substituting V(R/C)max from Eq.(6.23) in Eq.(6.5) yields:

12

DODOmax

DO DO

a aa

a

2 W/S K 3ρCT /W W/S Z T /W W/S ZCT 1R/C = × - ρ -

3ρC W 2 3ρC W/S ρ T /W W/S Z

Or DO

DO

12

maxa a

aa

T /W W/S Z T 6KCZR/C = - T /W -

3ρC W 6 T /W Z

=

DO

1/2 3/2

2 2max

a

a

W/S Z T Z 31- -

3ρC W 6 2 T /W L/D Z

(6.25)

Remarks:

The following observations can be made based on Eqs.(6.23) to (6.25)

(i) In Eq.(6.24), the quantity Z appears to depend on (L/D)max and ( aT /W). In this

context it may be noted that for jet airplanes (a) the value of (L/D)max would be

around 20 and (b) the value of ( aT /W) would be around 0.25 at sea level and

around 0.06 at tropopause. With these values of (L/D)max and ( aT /W), Z would be

around 2.1 at sea level and 2.7 at tropopause.



However, in Eqs.(6.23) and (6.25) the terms involving Z, appear as 1/2Z or Z/6.

Hence, the dependence of R/C maxV and maxR/C on Z does not appear to be of

primary importance. The important parameters however, are ( aT /W), (W/S),ρ

and CDO. It may be recalled from section 4.5 that for a turbofan engine, aT

decreases with altitude in proportion to 0.7σ ; σ being the density ratio.

(ii) From Eq.(6.25) it is observed that for given values of W/S and CDO ,

maxR/C decreases with altitude. Hence, suitable value of ( aT /W) is required to

achieve the specified rates of climb at different altitudes.

The same equation also indicates that the rate of climb increases when wing

loading increases and CDO decreases. However, the performance during cruise

and landing generally place a limit on the value of (W/S).

(iii) From Eq.(6.23) it is observed that the flight speed for maximum rate of

climb(V(R/Cmax), increases with ( aT /W), (W/S) and altitude. In this context it may

be pointed out that the Mach number corresponding to V(R/C)max, should always

be worked out and corrections to the values of CDO and K be applied when this

Mach number exceeds Mcruise. Without these corrections, the values of

maxR/C obtained may be unrealistic.

(iv) Figure 4.12 shows typical variations of thrust vs Mach number with altitude as

parameter. It is observed, that the thrust decreases significantly with Mach

number for altitudes equal to or less than 25000 (7620 m). Thus, the

assumption of thrust being constant with flight speed is not a good approximation

for h 25000 .


combination

In this subsection the simplified analysis is carried out for climb performance of

an airplane with engine-propeller combination.

From Eq.(6.5)

C

V T-D TV - DVV = R/C = =

W W



TV = 1000 p aη P ; aP = power available in kW

DV = Power required to overcome drag = 3D

1ρV SC

2

Following assumptions are made to simplify the analysis and obtain parameters

which influence (R/C)max are V(R/C)max in this case.

(a) Drag polar is parabolic with CDO and K as constants.

(b) L = W for estimation of induced drag.

(c) Power available is constant with flight speed (V).

Consequently,

2

3DO 2

1 2WDV = ρV S C +K

2 ρSV

(6.26)

Since Pa is assumed to be constant, the maximum rate of climb would be

obtained when DV is minimum. This occurs at the flight speed corresponding to

minimum power mpV .

Hence, in this case : R/C max mpV = V

The expression for mpV is given in Eq.(5.24b), consequently :

R/C max

1142

mpDO

2W KV = V =

ρS 3C

(6.27)

Substituting V(R/C)max from Eq.(6.27) in Eq.(6.5) gives:

R/C max

R/C maxR/C max

2

22DO

pmax

a V1000η P 1 2WR/C = - ρV SC + K

W W 2 ρSV

-1

R/C max

p aDO

DODO

1000η P 2K W/S1 2 K W= - V ρ W/S C +

2W 2 ρ 3C S ρ K/ 3C W/Sρ

(6.28)

Noting that ,max

DO

1L/D =

2 C K and



1/ 3 + 3= 1.155, yields

2:

R/C maxp a

maxmax

1000η P 1.155R/C = - V

W L/D (6.29)

Substituting for V(R/C)max from Eq.(6.27) yields :

p 1/2a

maxDO max

1000η P 2 K 1.155R/C = - W/S

W ρ 3C L/D (6.30)

Remarks:

(i) From Eq.(6.27) it is observed that V(R/C)max increases with wing loading (W/S).

(ii) From Eq.(6.30) it is observed that (R/C)max increases as pη , Pa and (L/D)max

increase. However, the second term on the right hand side of this equation

indicates that (R/C)max decreases with increase of wing loading. This trend is

opposite to that in the case of jet airplanes. Thus, for a specified (R/C)max , the

wing loading for an airplane with engine-propeller combination should be rather

low, to decrease the power required.

(iii) The first term in Eq.(6.30) involves pη and Pa. From subsection 4.2.2 it is

noted that Pa is nearly constant with flight speed (V). However, the assumption

of pη being constant with V is roughly valid only when the airplane has a variable

pitch propeller. For a fixed pitch propeller pη varies significantly with V (Fig.4.5a)

and the assumption of Pa being constant with V is not appropriate in this case.



Chapter 6


6.6. Climb hydrograph

6.7. Absolute ceiling and service ceiling

6.8 Time to climb

6.9 Steady descent

6.10 Glide

6.10.1 Glide performance – minimum angle of glide, minimum rate of sink

and maximum range and endurance in glide.


6.6 Climb hodograph

From Fig.6.1 it is observed that in a climb, the vertical velocity is the rate of climb

(Vc) and the horizontal velocity is Vh. From the discussion in section 6.5 it is

observed that for a chosen altitude, the vertical velocity (Vc) and the horizontal

velocity (Vh) change with the flight speed (V). A plot of the values of Vc and Vh at

a particular altitude, in which Vc is plotted on y-axis and Vh is plotted on the x-axis

is called ‘Climb hodograph’. Figure 6.5 shows a hodograph, based on the sea

level climb performance (Fig.6.3) of a jet airplane.



Fig.6.5 Climb hodograph

In a hodograph the line, joining the origin to a point on the curve, has the length

proportional to the flight velocity (V) and the angle this line makes to the

horizontal axis (Vh - axis) is the angle of climb ( ). This becomes evident when it

is noted that Vc and Vh are the components of the flight velocity (V) (see

Fig.6.1).

A line from the origin which is tangent to the hodograph gives the value of max

and also the velocity corresponding to it (Fig.6.5). Actually, a climb hodograph

gives complete information about the climb performance at the chosen altitude

especially max

, Vmax

, (R/C)max

, (R/C)max

, V(R/C)max

, (R/C)max

and Vmax

. These

quantities are marked in Fig.6.5

6.7 Absolute ceiling and service ceiling

Figures 6.3e and 6.4d present the variations of maximum rate of climb,

(R/C)max

, with altitude. It is observed that (R/C)max

decreases as the altitude

increases. The altitude at which the (R/C)max is zero is called ‘Absolute ceiling’. It

is denoted by h max

. At this altitude level flight is possible only at one speed(see

sec. 5.9).

Near absolute ceiling, the rate of climb is very small and the time to climb

becomes very large.It is not possible to reach the absolute ceiling (see remark (ii)



in section 6.8). Hence, for practical purposes an altitude at which the maximum

rate of climb is 100 ft /min(30.5 m/min) is used as ‘Service ceiling’.

To obtain the absolute ceiling and service ceiling the values of (R/C)max

at

different altitudes are plotted as shown in Figs.6.3e and 6.4d. Subsequently, the

(R/C)max vs h curve is extrapolated till (R/C)

max= 0. The altitude at which (R/C)

max

equals zero is the absolute ceiling. The altitude at which (R/C)max

equals 100

ft/min (30.5 m/min) is the service ceiling. From Fig.6.3e and Appendix ‘B’ the

absolute ceiling and service ceiling for the jet transport are 11.95 and 11.71 km

respectively. From Fig.6.4d and Appendix A the values of these ceilings for a

piston engined airplane are 5.20 and 4.61 km respectively.

6.8 Time to climb:

From the knowledge of the variation of rate of climb with altitude, the time

required (t) to climb from an altitude h1 to h2 can be calculated as follows.

dh dhV = or dt =c dt Vc

Hence, t = 2

1

hdh

Vch (6.31)

The rate of climb (Vc) in Eq.(6.31) depends on the speed and altitude at which

the climb takes place. The appropriate values of Vc can be taken from plots

similar to those given in Figs.6.3e or 6.4d.

Remarks:

i) It may be noted that in a climb which attempts to fly at (R/C)max at each altitude,

the flight velocity, V(R/C)max , increases with altitude (Figs.6.3c and 6.4c).

Consequently, such a flight is an accelerated climb and the values of Vc obtained

using steady climb analysis will need to be appropriately corrected for the

acceleration (see section 8.3.2 on accelerated climb).

ii) As an exercise the student should plot the height (h) on y-axis and the time to

climb (t) on x-axis. It is observed that this curve reaches the absolute ceiling

(hmax) in an asymptotic manner. In other words, the time taken to reach absolute



ceiling would be infinite and not practically attainable. Hence, service ceiling is

used for practical purposes.

6.9 Steady descent

Figure 6.6 shows an airplane in a steady descent. In such a flight thrust is

less than drag. The equations of motion are as follows.

T + W sin - D = 0 (6.32)

L – W cos = 0 (6.33)

Fig.6.6 Descent or glide

Hence, sin = D - T

W (6.34)

Rate of descent ( Vd ) = D-T

VW

(6.35)

The rate of descent is also called rate of sink and denoted by (R/S).

6.10 Glide

In a glide the thrust is zero. This may happen in a powered airplane due

to failure of engine while in flight. In a class of airplanes called gliders there is no

engine and the thrust is always zero. With thrust equal to zero, the following

equations of motion for glide, are obtained from Eqs.(6.32) and (6.33).

W sin - D =0 (6.36)

L - W cos =0 (6.37)

Hence,

sin = D / W and (6.38)



Vd = V sin = D V

W (6.39)

The angle of glide, , is generally small. Hence, L W and one can write,

sin = = CD D D

W L CL (6.40)

1/2 1/22L 2W

V ρ SC ρSCL L

and 1/2

3/2

CDV DV 2W DV = d W L ρS CL

(6.41)

Remarks:

i) Multiplying Eq.(6.31) by V gives:

W V sin -D V = 0

Or W Vd – D V = 0

Noting that Vd is the rate of descent and equals dh / dt,

dhW - DV = 0

dt (6.42)

From Fig.6.6 it is to be noted that ‘V’ is along the glide path and hence in the

downward direction. Consequently in Eq.(6.42) dh/dt is negative as the altitude

is decreasing. As a result, the potential energy of the glider decreases with time.

This loss of potential energy is utilized to provide for the energy required to

overcome the drag (the second term in Eq.6.42). Hence, for a glider to stay aloft,

it must be brought to a certain height and speed before it can carry out the glide.

This is done by launching the glider by a winch or by towing the glider by another

powered airplane.

6.10.1 Glide performance – minimum angle of glide and minimum rate of

sink and maximum range and endurance in glide

The performance in a glide is stated in terms of the following four quantities.

(a) Minimum angle of glide (γmin) (b) Minimum rate of climb ((R/S)min or Vdmin).



(c) The distance covered in glide from a certain height or ’range in glide (Rglide)’

(d) The time elapsed in descending from a given height or ‘Endurance in glide

(Eglide)’.

From Eq.(6.40) the minimum angle of glide (γmin), occurs when CD / CL is

minimum or at CL = CLmd . From Eq.(6.41) the minimum rate of sink (R/S)min or

Vdmin occurs when CD / 3/2CL is minimum or at CL = CLmp . This can be

understood from the following alternate explanation. When a glider sinks, it is

expending energy to overcome the drag, which comes from the potential energy

initially imparted to it. Thus, the rate of sink would be minimum when the rate of

power consumption is minimum and this occurs when V equals Vmp .

Gliders with very low rate of sink (around 0.5 m/s) are called ‘Sail planes’. From

Eq.(6.41) it is observed that a low rate of sink is achieved by (a) low wing loading

(b) low CDO with smooth surface finish and (c ) large aspect ratio (16 to 20) to

reduce K. Note from Eq.(5.24a ) that (CD / 3/2CL )min depends on 1/4CDO

and K3/4.

If a glider is left at a height ‘h’ above the ground, then the horizontal distance

covered in descending to the ground is called ‘Range in glide’ and denoted by

Rglide. Assuming γ to be constant during the glide, the range in glide can be

expressed as:

L

D

Ch hR = = hglide tan C

(6.43)

Thus, the range in glide would be maximum when the flight is at LC

corresponding to CLmd or at V= Vmd.

The time to descend from a height h is called Eglide. Assuming Vd to be constant,

Eglide equals (h / Vd). The quantity Eglide would be maximum when the descent

takes place at CL = CLmp or V = Vmp.

It is evident from the above discussion that the flight speeds for min

and (R/S) min

are different .




In section 6.6, the climb hodograph was discussed. Similarly, a glide hodograph

is obtained when horizontal velocity (Vh) is plotted on the x-axis and the rate of

sink (Vd) is plotted on the y-axis. A typical diagram is shown in Fig.6.7. Such a

diagram gives complete information about glide performance at an altitude

especially, min

, Vmin , (R/S)min, V(R/S)min and (R/S)min .

Fig.6.7 Glide hodograph

Example 6.4

A glider weighing 4905 N has a wing area of 25 m2, DOC = 0.012, A = 16

and e = 0.87. Determine (a) the minimum angle of glide, minimum rate of sink

and corresponding speeds under sea level standard conditions (b) the greatest

duration of flight and the greatest distance that can be covered when glided from

a height of 300 m. Neglect the changes in density during glide.

Solution:

(a) D DO2L

2C 1LC = + = 0.012 + CA e 3.14 × 16 ×0.87

C

= 0.012 + 0.023 CL2



12

DO

DO

L min

D min

L(R/S)min

D(R/S)min

C = C = (0.012/0.023) = 0.721Lmd

C = 2C = 0.024

C = C = 3 C = 1.25Lmp Lmd

C = 4C = 0.048

Hence,

= (C /C ) = 0.024 / 0.721min D L min =0.0332 radians or 1.9o

12

3/22W(R/S) = (C /C )min D minLρS

Noting that the density ()has been assumed to be constant and equal to that at

sea level i.e. = 1.225 kg/m3 , the above equation gives :

122 × 4905 0.048

(R/S) = = 0.615 m/smin 3/21.225 × 25 1.25

12

1 12 2

2 × 4905V = = 21.05 m/s.min 1.225 × 25 × 0.721

2W 2 × 4905V = = = 16m/s(R/S)min ρ S C 1.225 × 25 × 1.25Lmp

(b) The greatest distance, in descending from 300 m to sea level, ( Rglide)max,

is (note is assumed constant during glide) :

( Rglide)max = 300 /0.0332 = 9040 m = 9.04 km.

Longest time taken in descending from 300 m to sea level (Eglide

)max

is (note R/S

is assumed constant during glide) :

(Eglide

)max

= 300/0.615 = 487 s = 8 min 7s.

Note:

The rate of sink, in a flight when the greatest distance is covered, is higher than

the minimum rate of sink. Hence, the time of flight will be shorter in this case than

in a flight for longest endurance in glide. From the above data, the student may



show that (R/S) in the flight corresponding to (Rglide)max is 0.7015 m/s and the

endurance in this flight is 427 s.

Remarks:

i) If the glide takes place from a sufficiently high altitude (as may happen for an

airplane having an engine failure in cruise), the rate of sink (R/S) cannot be taken

as constant during the descent. Equation (6.41) should be used to calculate the

rates of sink at various altitudes.

ii) The time elapsed during glide (Eglide), in a general case is given by:

2

1

hdh

E =glide Vdh ; (6.44)

where Vd is the rate of descent corresponding to an altitude ‘h’.



Chapter 6

Exercises

6.1 An airplane powered by a turbojet engine weighs 180,000 N, has a wing area

of 50 m2 and the drag polar is CD = 0.016 + 0.048C

L2. At sea level a rate of

climb of 1200 m/min is obtained at a speed of 150 m/s. Calculate the rate of

climb at the same speed when a rocket motor giving an additional thrust of

10,000 N is fitted to the airplane.

(Answer: 1702 m /min.)

6.2 A glider having a wing loading of 185 N / m2 has the following drag polar.

CL

0.0

0. 1

0.2

0.4

0.6

0.8

1.0

1.2

1.4

C

D

0.0145

0.014

0.0155

0.0183

0.0231

0.0299

0.0385

0.0491

0.062

Obtain the minimum rate of sink, minimum angle of glide and corresponding

speeds at sea level.

(Hint: Obtain CD / CL

and CD / CL

3/2 from the given data, plot them, obtain

(CD / CL

) min , (C

D / CL3/2)min and proceed.)

(Answers: (R/S)min

= 0.647 m/s, min

= 2.13o, V(R/S)min

= 54.2 kmph,

Vmin

= 71.35 kmph)

6.3 Consider a subsonic jet airplane. Assume that (a) thrust available (Ta) is

roughly constant, (b) L ≈ W in climb or the drag in climb (D) is roughly equal to

the drag in level flight and (c) the drag polar is parabolic. With these

assumptions and from exercise 5.4 which gives:

DOV A W2D = A W( ) + , A = C K,

2Vmd (V/V )md



show that (V / Vmd

) for (R/C)max

i.e. (V / Vmd

)(R/C)max is given by:

T T 2 2a a± ( ) +12AV W W=

V 6Amd (R/C)max

Further taking DOC = 0.016 and K = 0.05625 or A = 0.03 obtain the following

table.

Ta / W

0.2

0.15

0.1

0.06

V

Vmd (R/C)max

1.54

1.36

1.16

1.0

Note that, as the altitude increases, Ta / W decreases and as a consequence

V

Vmd (R/C)max

tends to 1. At absolute ceiling V

Vmd (R/C)max

= 1 but

(R/C)max is zero !.



Chapter 7

Performance analysis III – Range and endurance

(Lectures 24-26)

Keywords: Range; endurance; safe range; gross still air range; Breguet

formulae; cruising speed and altitude; cruise climb; effect of wind on range.

Topics

7.1 Introduction

7.2 Definitions of range and endurance

7.2.1 Safe range

7.2.2 Head wind, tail wind, gust and cross wind

7.2.3 Gross still air range (GSAR)

7.3 Rough estimates of range and endurance

7.4 Accurate estimates of range and endurance

7.4.1 Dependence of range and endurance on flight plan and remark on

optimum path

7.4.2 Breguet formulae for range and endurance of airplanes with engine-

propeller combination and jet engine

7.4.3 Discussion on Breguet formulae – desirable values of lift coefficient

and flight altitude

7.4.4 Important values of lift coefficient

7.4.5 Influence of the range performance analysis on airplane design

7.5 Range in constant velocity - constant altitude flight (Rh,v)

7.6 Cruising speed and cruising altitude

7.7 Cruise climb

7.8 Effect of wind on range and endurance

References

Excercises



Chapter 7

Lecture 24 Performance analysis III – Range and endurance – 1 Topics

7.1 Introduction


7.2.1 Safe range


7.2.3 Gross still air range (GSAR)




optimum path

7.1 Introduction Airplane is a means of transport designed to carry men and materials

safely over a specified distance. Hence, the fuel required for a trip or the distance

covered with a given amount of fuel are important items of performance analysis.

Similarly, airplanes used for training, patrol and reconnaissance would be

required to remain in air for a certain period of time. Thus, the fuel required to

remain in air for a certain length of time or the time for which an airplane can

remain in air with a given amount of fuel are also important aspects of

performance analysis. These two aspects viz. distance covered and the time for

which an airplane can remain in air are discussed under the topic of range and

endurance and are the subject matter of this chapter.


Range (R) is the horizontal distance covered, with respect to a given point

on the ground, with a given amount of fuel. It is measured in km. Endurance (E)



is the time for which an airplane can remain in air with a given amount of fuel. It

is measured in hours. The above definition of range is very general and terms

like safe range and gross still air range are commonly used. These terms include

details of the flight plan and are explained in the subsequent subsections.

7.2.1 Safe range

It is the maximum distance between two destinations over which an

airplane can carry out a safe, reliably regular service with a given amount of fuel.

This flight involves take-off, acceleration to the speed corresponding to desired

rate of climb, climb to the cruising altitude, cruise according to a chosen flight

plan, descent and landing. Allowance is also given for the extra fuel requirement

due to factors like (i) head winds (see next subsection) normally encountered en-

route (ii) possible navigational errors (iii) need to remain in air before permission

to land is granted at the destination and (iv) diversion to alternate airport in case

of landing being refused at the scheduled destination.


Generally the performance of an airplane is carried out assuming that the flight

takes place in still air. However the air mass may move in different directions.

Following three cases of air motion are especially important.

(a) Head wind and tail wind: In these two cases the direction of motion of air (Vw)

is parallel to the flight direction. If Vw is opposite to that of the flight direction, it is

called ‘Head wind’. When Vw is in the same direction as the flight direction, it is

called ‘Tail wind’ (Fig.7.1a). In the presence of wind, the velocitiy of the airplane

with respect to air (Va) and that with respect to ground (Vg) will be different. For

the head wind case, Vg = Va - Vw, and for the tail wind case, Vg = Va + Vw.

(b) Gust: When the velocity of the air mass is perpendicular to flight path and

along the vertical direction, it is called gust. Here the velocity of gust is denoted

by Vgu (Fig.7.1b). This type of air movement would result in a change of the

angle of attack of the airplane.



(c ) Cross wind: When the velocity of the air mass is perpendicular to flight path

and parallel to the sideward direction, it is called ‘Cross wind’. Here it is denoted

by ‘v’ (Fig.7.1c).

Fig.7.1a Head wind and tail wind

Fig.7.1b Gust

Fig.7.1c Side wind

7.2.3 Gross still air range (G.S.A.R.)

The calculation of safe range depends on the route on which the flight

takes place and other practical aspects. It is not a suitable parameter for use

during the preliminary design phase of airplane design. For this purpose, gross

still air range (G.S.A.R.) is used. In this case, it is assumed that the airplane is

already at the cruising speed and cruising altitude with desired amount of fuel



and then it carries out a chosen flight plan in still air, till the fuel is exhausted.

The horizontal distance covered in this flight is called ‘Gross still air range’. In the

subsequent discussion the range will mean gross still air range.

Remark:

As a guideline G.S.A.R. is roughly equal to one and a half times the safe range.


If the weight of the fuel available (Wf in N) and the average rate of fuel

consumption during the flight are known, then the rough estimates of range (R)

and endurance (E) are given as follows.

R = Wf x (km / N of fuel)average

(7.1)

E = Wf x (hrs / N of fuel) average

(7.2)

The estimation procedure is illustrated with the help of example 7.1.

Example 7.1

An airplane has a weight of 180,000 N at the beginning of the flight and

20% of this is the weight of the fuel. In a flight at a speed of 800 kmph the lift to

drag ratio (L/D) is 12 and the TSFC of the engine is 0.8. Obtain rough estimates

of the range and endurance.

Solution:

W1 = Weight at the start of the flight = 180,000N

Wf = Weight of the fuel = 0.2x180,000 = 36,000N

W2 = Weight of the airplane at the end of the flight

= 180,000 - 36,000= 144,000N.

Hence, the average weight of the airplane during the flight is :

Wa =

180000 + 144000= 162000N

2

Consequently, the average thrust (Tavg

) required during the flight is:

Tavg

= Wa / (L / D) = 162000/12 = 13500 N



The average fuel consumed per hour is:

Tavg x TSFC = 13500 x 0.8 = 10800 N

Since the average speed is 800 kmph, the distance covered in 1 hr is 800 km.

Noting that the fuel consumed in 1 hr is 10,800 N, gives:

(km / N of fuel)average

= 800/10800.

Consequently, 800

R = 36000 × = 2667km10800

and

the endurance E = 36000 x1

= 3.33 hrs.10800


For accurate estimates of range and endurance, the continuous variation

of the weight of the airplane during the flight and consequent changes in the

following quantities are considered.

(a) The thrust required (or power required),

(b) TSFC (or BSFC) and

(c) Flight velocity and lift coefficient.

It may be recalled from subsection 4.2.4, that is the specific fuel consumption

(SFC) of an engine delivering shaft horse power to a propeller is denoted by

BSFC and the SFC of a jet engine is denoted by TSFC. The units of BSFC and

TSFC are respectively N/kW-hr and N/N-hr (or hr -1).

The steps to accurately estimate the range and endurance are as follows.

Let W be the weight of the airplane at a given instant of time and Wfi be the

weight of the fuel consumed from the beginning of the flight up to the instant

under consideration.

Then, W = W1 - Wfi (7.3)

where, W1 = weight of the airplane at the start of the flight.

Let dR and dE be the distance covered in km and the time interval in hours

respectively, during which a small quantity of fuel dWf is consumed. Then,

dR = dWf x (km/N of fuel) (7.4) and dE = dWf x (hrs/ N of fuel) (7.5)



Following Ref.1.5, chapter 4, the Eqs.(7.4) and (7.5) are rewritten as:

dR = dWf km/hr

N of fuel /hr

(7.6)

and dE = dWf x { 1 / (N of fuel / hr)} (7.7) It may be pointed out that (a) km/hr = 3.6 x V, where V is the flight speed in m/s.

(b) the fuel / hr in Newtons is equal to BSFC x BHP for an airplane with engine-

propeller combination and equal to TSFC x T for a jet airplane.

Hereafter, the airplane with engine-propeller combination is referred to as “E.P.C”

and the jet airplane as “J.A” Note that in the case of an engine-propeller

combination, the engine could be a piston engine or a turboprop engine and in

the case of a jet airplane the engine could be a turbofan or a turbojet engine.

Equations (7.6) and (7.7) can be rewritten as :

3.6 VdR = dWf BSFC × BHP

For E.P.C. (7.8)

and3.6 V

dR = dWf TSFC × T For J.A. (7.8a)

dWfdE = BSFC × BHP

For E.P.C. (7.9)

and dWfdE =

TSFC × T For J.A. (7.9a)

Recall from section 5.2 that in a level flight,

C 1 2DT = D = W , L = W = ρV SCLC 2L ,

1122

L 0 L

2W 2WV = =

ρS C σ ρ S C

(7.10)

Using, ρ0 = 1.225 kg/m3 yields:

W 1/2V = 1.278 ( )σ SCL

. (7.11)

Substituting for T and V from Eqs.(7.10) and (7.11), the expression for BHP is:



12 3/2

D L1 T V 1 3/2BHP = = W / η (σ S) C / Cpη 1000 782.6p

(7.12)

where p is the propeller efficiency.

During the analysis of range, the rate of change of weight of the airplane is only

due to the consumption of fuel. Hence,

dWf = - dW

Substituting for V, BHP, T and dWf in Eqs. (7.8),(7.8a),(7.9) and (7.9a) gives:

-3600 η dWpdR =

BSFC × W (C /C )D L For E.P.C. (7.13)

and1/2 1/2

-4.6 dWdR =

TSFC (σ S W) (C /C )D L

For J.A. (7.13a)

1/2782.6 η (σS) dWp

dE = -3/2 3/2BSFC × W C /CD L

For E.P.C. (7.14)

and

-dWdE =

TSFC × W C /CD L For J.A (7.14a)

Let W2 be the weight of the airplane at the end of the flight. Integrating

Eqs.(7.13), (7.13a), (7.14) and (7.14a), the range and endurance are given as:

2 2

1 1

W W3600 η dWp

R = dR = -BSFC × W (C /C )D LW W

For E.P.C. (7.15)

and 2

1122

1

W-4.6 dW

R =TSFC (σ S W) (C /C )DW L

For J.A (7.15a)

1/22 2

1 1

W W782.8 η (σS) dWp

E = dE = -3/2 3/2BSFC × W (C /C )DW W L

For E.P.C. (7.16)

and 2

1

W-dW

E =TSFC × W (C /C )D LW

For J.A. (7.16a)




optimum path

The set of Eqs.(7.15),(7.15a),(7.16) and (7.16a) or (7.8), (7.8a), (7.9) and (7.9a)

when integrated, give the range and endurance. However, while doing this, it

should be noted that the weight of the aircraft decreases continuously as the fuel

is consumed. Further, the flight is treated as steady level flight and hence, T = D

and L= W must be satisfied at each instant of time. Consequently, the thrust and

power required and the flight speed may change continuously. Hence, it is

necessary to prescribe the flight plan i.e., the manner in which the velocity

changes with time during the flight. The following three types of flight plans can

be cited as examples.

(a) Level flight at a constant velocity. In this flight, the lift coefficient decreases

gradually as the weight of the airplane decreases (Eq.7.10). Simultaneously, the

thrust required also decreases continuously.

(b) Level flight with constant lift coefficient (or constant angle of attack) . In this

flight, in accordance with Eq.(7.10), the flight velocity and the thrust required

decrease continuously as the weight of the airplane decreases.

(c) Level Flight with constant thrust. In this case, the continous decrease in the

airplane weight during the flight, requires that the flight velocity and the lift

coefficient (CL) be adjusted so that at each instant of time, the thrust balances

the drag and the lift balances the weight.

As mentioned earlier, the airplanes are commercial means to transport men and

materials. Hence, maximization of range and endurance are important

requirements. However, the right hand sides of Eqs.(7.15),(7.15a),(7.16) and

(7.16a) involve integrals. The optimization of an integral is different from the

optimization of an expression. The latter is done by taking the derivative of the

expression and equating it to zero. Whereas, in the case of an integral, it is to be

noted that the value of the integral depends on how the integrand varies with the

independent variable. This variation, in mathematical terms, is called a path. For

example, as mentioned above, the range will depend on the flight plan viz.

constant angle of attack flight or constant velocity flight or constant thrust flight.



The problem of optimization is to find out the path that will maximize the integral.

The branch of Mathematics which deals with optimization of integrals is called

‘Calculus of variation’. This topic is outside the scope of the present introductory

course. Interested reader may refer, chapter 20 of Ref.7.1.

Remark:

It can be shown, using calculus of variation, that if the specific fuel consumption,

propeller efficiency and altitude are assumed constant, then the maximum range

is obtained in a flight with constant lift coefficient. With these assumptions

Eqs.(7.15),(7.15a),(7.16) and (7.16a) become easy to integrate. The expressions

for range and endurance, obtained with these assumptions, are called ‘Breguet

formulae’. These are derived in the next subsection. It may be pointed out that

Breguet was a French pioneer in aeronautical engineering.



Chapter 7




7.4.3 Discussion on Breguet formulae – desirable values of lift coefficient

and flight altitude




The derivations of these formulae are based on the assumptions that during

the flight:

(i) BSFC or TSFC is constant

(ii) p is constant for engine propeller combination (E.P.C).

(iii) altitude is constant

(iv) CL is constant and

(v) flight Mach number is below critical Mach number so that the drag polar is

independent of Mach number.

With, these assumptions, certain terms in Eqs.(7.15) and (7.15a) can be taken

outside the integral and the equations reduce to:

2

1

w3600 η dWp

R = -WBSFC (C /C )D wL

For E.P.C

and 2

1

W-4.6 dW

R =1/2 1/2 1/2TSFC(σS) (C /C ) WD WL

For J.A.

Hence,



8289.3 η Wp 1R = log10BSFC (C /C ) WD L 2

For E.P.C. (7.17)

and 1/2

W W9.2 1/21 2R = ( ) 1-1/2 σS W1TSFC(C /C )D L

For J.A. (7.17a)

Similarly, from the above assumptions, Eqs.(7.16) and (7.16a) reduce to :

1/2 2

1

W782.6 η (σS) dWp

E = -3/23/2 WBSFC × C / CD WL

For E.P.C

and 2

1

W1 dW

E = -TSFC(C / C ) WD L W

For J.A.

Hence,

1/2 1/21565.2 η σSp 1/2E = [ ] W - W2 13/2 W1BSFC × C /CD L

For E.P.C. (7.18)

and 10W2.303 1E = log

TSFC (C / C ) WD L 2

For J.A. (7.18a)

7.4.3 Discussion of Breguet formulae – desirable values of lift coefficient

and flight altitude

The following conclusions can be drawn from the above expressions for

range and endurance viz. Eqs.7.17, 7.17a, 7.18 and 7.18a.

(1) For range and endurance to be high, it is evident that p should be high and

the TSFC and BSFC should be low.

(2) Desirable values of lift coefficients for an airplane with engine-propeller

combination: The endurance is maximum (Eq.7.18) when the lift coefficient is

such that CD/C

L

3/2 is minimum, i.e., C

L = C

Lmp. This can be understood from the

fact, that with BSFC being assumed constant, the rate of fuel consumption per

hour would be minimum, in this case, when the power required is minimum.



From Eq.(7.17), for range to be maximum, in this case, CD/C

L should be

minimum or CL = C

Lmd. This can be understood from the fact that the range, in

this case is proportional to V/THP or L / D. Hence, range is maximised when CL

corresponds to minimum drag.

(3) Desirable values of lift coefficients for a jet airplane: From Eq.(7.18a) it is

observed that the endurance is maximum when CD / CL

is minimum or CL= C

Lmd.

This can be understood from the fact that with TSFC being assumed constant,

the fuel flow rate per hour would be minimum when the thrust required is

minimum.

From Eq.(7.17a) the range, in this case is maximum when CD/C

L

1/2 is minimum.

This can be understood from the fact that the range, in this case, is proportional

to (V / T) or 1/2L D C /C . The C

L corresponding to (C

D/C

L

1/2)min

is denoted here by

CLmrj

.

(4) Desirable values of flight altitude : Equation (7.17a), also shows that for a jet

airplane, the range would be high, when (a) the wing loading (W/S) is high and

(b) density ratio () is low or the altitude is high. Hence, the jet airplanes have

wing loading of the order of 4000 to 6000 N/m2, which is much higher than that

for the low speed airplanes which have a wing loading of 1000 to 2500 N/m2. The

jet airplanes also cruise at high altitude (10 to 12 km) which is not much below

the ceiling altitude of 12 to 14 km for these airplanes.

From Eq.(7.18) it is observed that the endurance of an airplane with engine-

propeller combination is high when (a) the wing loading is low and (b) is high or

flight takes place near sea level.

It may be added that the final wing loading chosen for an airplane is a

compromise between requirements of cruise, climb, take-off and landing. The

take-off and landing distances increase in direct proportion to the wing loading

(subsections 10.4.5 and 10.5.3), and hence, a high wing loading is not desirable

from this point of view.



Remarks:

i) If the drag polar is parabolic, an expression for CLmrj

can be derived as follows.

1/2 1/2

1/2

2

3/2

C = C + K CD DO LCC DODHence, = + K CL

C CL L

d(C / C ) C 3D -3/2 1/2DOL = C + K C = 0Lmrj LmrjdC 2 2L

Or CLmrj = (C

DO / 3K)1/2


The points on the drag polar at which CL is equal to C

Lmax, C

Lmp, C

Lmd and C

Lmrj

are shown in Fig.7.2. The importance of these values of lift coefficient can be

reemphasized as follows.

(i) The maximum lift coefficient (CLmax

) decides the stalling speed which is one of

the criterion for the minimum speed of the airplane. It also affects the minimum

radius of turn (see subsection 9.3.3) and the take-off and landing distances (see

subsections 10.4.5 and 10.5.3)

(ii) The lift coefficient corresponding to minimum power required (CLmp) influences

the performance of airplanes with engine-propeller combination. It decides the

flight speeds corresponding to maximum rate of climb, minimum rate of sink and

maximum endurance of these airplanes.

(iii) The lift coefficient corresponding to minimum thrust required ( LmdC ) is also

the value of CL at which (L/D) is maximum. From Fig.7.2 it is observed that the

slope of a line joining the origin to a point on the curve, is equal to (CL/ C

D). At,

CL = LmdC this line, from the origin, is tangent to the drag polar and has the

maximum slope (Fig.7.2). The value of CLmd decides the flight speed for

maximum range of an airplane with engine-propeller combination and the

maximum endurance of a jet airplane.



(iv) The lift coefficient corresponding to max1/2L D C /C or C

Lmrj decides the flight

speed for maximum range of jet airplanes.

Fig.7.2 Important points on a drag polar

Example 7.2

An airplane having an engine-propeller combination weighs 88,290 N and

has a wing area of 45 m2. Its drag polar is given by: C

D = 0.022 + 0.059CL2.

Obtain the maximum range and endurance at sea level in a steady level flight at

a constant angle of attack from the following additional data.

Weight of fuel and oil = 15,450 N, BSFC = 2.67 N/kW-hr,

propeller efficiency (p) = 85%.

Note: Along with the fuel, the lubricating oil is also consumed and this fact is

taken into account in this example, by specifying the weight of the oil along with

the weight of fuel.

Solution:

W1 = 88290 N, W2 = 88290 – 15450 = 72840 N



Since BSFC, p and CL are constant, the maximum range and endurance occur

when CL has the values of C

Lmd and CLmp respectively.

C Lmd = (CDO/K)

1/2 = (0.022/0.059)1/2 = 0.6106,

CDmd = 2 CDO = 0.044

C Lmp = (3CDO/K)

1/2 = (3 x 0.022/0.059)1/2 = 1.058, C

Dmp = 4 CDO = 0.088

Hence, (CD /C

L)min

= 0.044/0.6106 = 0.0721

and 3/2 3/2(C /C ) = 0.088/(1.058)D minL = 0.0808

From Eq. (7.17):

8289.3 η W 8289.3 × 0.85 88290p 1R = log = × log10 10BSFC × (C /C ) W 2.67 × 0.0721 72840D L min 2

= 3058 km.

Remark:

Since CL is constant during the flight, the flight velocity and the power required

change as the fuel is consumed. In the present case, the following results

illustrate the changes.

Velocity at the beginning of the flight:

V1= 1/2 1/22W /ρ S C = (2 × 88290/1.225 ×45 × 0.6106)1 Lmd = 72.41 m/s.

= 260.7kmph.

Velocity at the end of flight:

= 1/222W /ρ S CLmd

= (2 x 72840/1.225 x 45 x 0.6106)1/2 = 65.8 m/s = 236.8 kmph.

Power required in the beginning of the flight:

1 D L 11 1

W × C /C ×VT × V 88290 × 0.044= = = ×72.41 = 460.7kW

1000 1000 1000 × 0.6106

Power required at the end of flight:



2 22 2 D LW × C /C ×VT × V 72840 × 0.044

= = = × 65.8 = 345.5kW1000 1000 1000 × 0.6106

Maximum endurance:

From Eq.(7.18) the maximum endurance is :

1/2 1/2

1/2 1/2

η W1565.2 σSp 1E = -1max 3/2BSFC W W1 2C /CD L min

1565.2 ×0.85 1.00 ×45 88290= -1 = 14.06hrs

2.67 × 0.0808 88290 72840

In this flight CL equals 1.058. Proceeding in a manner similar to the remark

above, it can be shown that the speeds at the beginning and end of the flight for

Emax

are 197.8 kmph and 179.7 kmph respectively. The power outputs required

at the beginning and the end of this flight are 402.8 kW and 302.0 kW

respectively.

Example 7.3

A jet airplane has a weight of 922,140 N and wing area of 158 m2. The

weight of the fuel and oil together is 294,300 N. The drag polar is given by:

2C = 0.017 + 0.0663 CD L

Obtain the maximum range in constant CL flight at an altitude of 10 km assuming

the TSFC to be 0.95 hr-1.

Solution:

In a flight with constant C L the maximum range occurs when

1/2C = C = (C / 3K)L Lmrj D0

1/20.017

C = = 0.292 Lmrj 3 × 0.0663

CDmrj

= 0.017 + 0.0663 (0.292)2 = 0.02265

1/2 1/2( )C /C = 0.02265 / 0.292 = 0.04192Dmrj Lmrj



σ at 10 km altitude is: 0.3369

From Eq.(7.17a):

1/21/2

1/2

1/2 1/2

W W9.2 1 2R = 1 -max σS W1TSFC × C /CD L

9.2 922140 922140-294300= × 1 - = 5317 km

0.95 × 0.04192 0.3369 × 158 922140

Remarks:

i)The flight velocity corresponding to a CL of 0.292 at an altitude of 10 km is equal

to: [2 x 922140 / (158 x 0.413 x 0.292)]1/2

= 311.1 m/s.

The speed of sound at 10 km is 299.5 m/s. Thus the Mach number at this speed

would be 311.1/299.5 = 1.04. This value is definitely higher than the critical Mach

number of the airplane. Consequently, the prescribed drag polar is not valid. The

CD will actually be much higher and the range much lower.

As an alternative, let the critical Mach number be taken as 0.85 and the range be

calculated in a flight at constant CL which begins at this Mach number.

Consequently, V = 0.85 x 299.5 = 254.5 m/s.

Hence,CL = (2 x 922140/0.413 x 158 x 254.52) = 0.436

Consequently, CD = 0.017 + 0.0663 x (0.436)2 = 0.0296

and 1/2 1/2C /C = 0.0296/0.436 = 0.0448D L

The range in a constant CL flight with CL=0.436 would be:

1/2 1/24975

9.2 922140 922140-294300= × 1- km

0.95 × 0.0448 0.3369×158 922140

.

(ii) The data given in this example, roughly corresponds to that of Boeing 727,

the famous jetliner of 1970’s. The value of TSFC corresponds to engines of that

period. The value of K equal to 0.0663 includes the change in K, when Mach

number lies in the transonic range.



Chapter 7





7.7 Cruise climb



In section 5.8 it was pointed out that the analysis of level flight

performance led to improvements in design of airplanes. Similarly, the analysis of

range also helped in improvements in airplane design in the following way.

The high speed airplanes are jet airplanes and for these airplanes the range

(R) is proportional to :

1 11/2TSFC C /CD L

or C1 1L

1/2TSFC CD CL

Noting that 1/ 1/2LC

is proportional to flight speed (V),

C1 LR VTSFC CD

Since, high speed airplanes fly in lower stratosphere, where speed of sound is

constant,

C1 LR MTSFC CD

(7.19)

The quantity C1 L M

TSFC CD can be referred to as figure of merit (FM) for the

following reasons.



(a) A low value of TSFC in an indicator of high engine efficiency and (b) A high

value of (CL/CD) is an indicator of high aerodynamic efficiency.

The figure of merit provided guidelines when the supersonic airplane

Concorde was being designed in early 1960’s.The subsonic jets of that period

like Boeing 707 would fly around M = 0.8, have (L / D)max

around 16 and TSFC

around 0.9. These values would give the FM of 0.8x16/0.9 or 14.2. If Concorde

were to compete with subsonic jets, it needed to have a similar value of FM. The

fighter airplanes of that period flying at Mach number of two had TSFC of 1.5 and

(L/D) max

of 5. This would give FM of (2 x 5) /1.5 = 6.66 which was far too low as

compared to that for subsonic airplanes. Hence the targets for Concorde, which

was being designed for a Mach number of 2.2, were fixed at (L/D) max

of 7.5 and

TSFC of 1.2. This would give FM of 2.2(1/1.2) x 7.5 = 13.75, which was

comparable to the FM of subsonic airplanes. However, to achieve a TSFC of 1.2

at M =2.2, a large amount of research was carried out and the Olympus engine

used on Concorde was developed jointly by Rolls-Royce of U.K. and SNECMA of

France. Similarly, to achieve an (L/D) max of 7.5 at M = 2.2 needed a large amount

of computational and experimental effort. A picture of Concorde, a technological

marvel, is shown in Fig.7.3.

It may be added that for Concorde the Mach number was limited to 2.2 as

the designers had chosen to use aluminum as structural material. At M = 3 the

FM could be greater than that of subsonic airplanes but the aerodynamic heating

would cause surface temperatures of around 300oC at which the strength and

modulus of elasticity of aluminum will be significantly reduced.



Fig.7.3 Concorde

(Source: www.airplane-pictures.net)

The B787 (Fig.7.4) being brought out by Boeing and called ‘Dream liner’ has

M = 0.85, (L/D) max

of 22 and TSFC of 0.54 hr-1. These values of (L/D) max and

TSFC indicate steady improvements in aerodynamics and engine performance

over the last five decades .



Fig.7.4 Boeing 787 Dream liner

(Source: www.lotz.com)


The assumption of constant CL during cruise gives the longest range(R).

However, it is more convenient for the pilot to fly the airplane at a constant speed

or Mach number. He just needs to keep an eye on the airspeed indicator or

Machmeter and adjust other parameters like the angle of attack and engine

setting.

To derive an expression for range in level flight at constant speed (Rh,v),

an airplane with jet engine is considered and it is assumed that TSFC is

constant. Equation (7.8a) is the basic equation for range of a jet airplane. When

V is constant, the equation takes the following form.

2

1

1

W3.6 V dW

R =h v TSFC TrW (7.20)

Tr = thrust required

Assuming a parabolic polar,

Tr = 2 21 2K W KW2 21ρV S C + = q S C + ; q = ρVDO DO2 22 qSρV S



Note: The dynamics pressure (q), is constant in a constant velocity and constant

altitude flight.

Substituting for Tr in Eq.(7.20) gives:

2

1

1

w3.6 V -dW

R =h v 2TSFC q S CD0 1+aWw where, a =

K2 2q S CD0

1

3.6 V -1 -1Or R = [tan a W - tan a W ]h v 1 2qS C TSFC aD0

(7.20a)

where W2 = weight of the airplane at the end of the flight.

Let Wfζ =W1

, where Wf = weight of fuel

Hence, W2 = W

1(1-) ;

Further, let E1 = W1/D1 = initial lift to drag ratio,

21

1 D1 DO 2 2

KWD = qSC = qS C +

q S

2

1DO

KW= qSC +

qS

CL1

= CL at start of flight =

W121 ρV S

2

= 1W

qS

Emax

= 1

2 K CD0

Noting that ,

-1 -1 -1 1 21 2

1 2

θ -θtan θ - tan θ = tan

1 + θ θ

,

Equation (7.20a) can be rewritten as :

-1 1 2h,v

1 2DO

DO

aW - aW3.6 VR = tan

1+aW WKTSFC qSC

qS C



1 2-1

21DO

2 2DO

a W - W3.6V= tan

WKTSFC KC 1 + 1-ζC q S

1

DO-1max2

12 2

DO

WKζ

C qS7.2E V= tan

WKTSFC1+ 1-ζ

C q S

Multiplying the denominator and numerator of the terms in square brackets by

qSCDO gives :

DO 1-1maxh,v 2 2

1 1DO

KC W ζ7.2E VR = tan

KW KWTSFCqSC + - ζ

qS qS

DO 1-1max2

11

K C W ζ7.2E V= tan

WTSFCD - K ζ

qS

Dividing the numerator and denominator of the term in square brackets by D1,

gives :

1DO

-1max 1h,v

1 1

1

WKC ζ

7.2E V DR = tan

TSFC W W1 - K ζ

qS D

Or

-1max 1h,V

max L1 1

7.2E V E ζR = tan

TSFC 2E 1-K C E ζ

(7.21)

For an airplane with an engine-propeller combination, the range at constant

speed and constant altitude (Rh,v) is given as:

2

1

1

W 3600 η dWpR =h v BSFC T

W (7.22)



Assuming BSFC and ηp to be constant and the drag polar as parabolic i.e.

1

2

22KW2T = ρV S C +DO 2ρSV and substituting in Eq.(7.22) gives:

1

7200 η E ζp -1 1R = E tanh v maxBSFC 2E (1-KC E ζ)max L1 1

(7.23)

Remarks:

i) Comparing the ranges in the constant velocity and constant CL flights, Ref.1.1,

chapter 9, shows that the maximum range in a constant velocity flight is only

slightly lower than that in a constant CL flight.

ii) In actual practice BSFC (or TSFC) and ηp may vary during the cruise. If

detailed information about their variations is available, then better estimates of

range and endurance can be obtained by numerical integration of Eqs.(7.8),

(7.8a),(7.9) and (7.9a).

iii) Appendix ‘A’ section 6 considers the range and endurance performance of a

piston engined airplane at an altitude of 8000 feet (2438 m) in constant velocity

flights at different speeds. The variations in propeller efficiency and fuel

consumption are taken into account. It is seen that the endurance is maximum

around flight speed of 135 kmph. The range is maximum for flight speeds

between 165 to 185 kmph.

iv) Section 6 of Appendix ‘B’ considers the range and endurance performance of

a jet transport at an altitude of 36000 feet (10973 m) in constant velocity flights at

different speeds. The endurance is near its maximum value in the speed range of

684 to 828 kmph. The maximum range occurs around 240 m/s (864 kmph). The

corresponding Mach number is 0.82, which is slightly higher than the Mach

number beyond which the CDOand K begin to increase due to compressibility

effects.


The cruising speed (Vcr

) and the cruising altitude (hcr

) together constitute

the combination at which the maximum range is obtained. To arrive at the values



of Vcr

and hcr

the range is calculated at various speeds at a number of altitudes

and the plots as shown in Fig.7.5 are obtained. The dotted line in Fig.7.5 is the

envelop of all the curves. The speed and altitude at which the maximum of this

envelop occurs is called the most economical cruising speed and altitude. In

some cases this speed is rather low and a higher cruising speed may be chosen

from other considerations like, shorter flight time and speed appeal. i.e. a faster

airplane may be more appealing to the passengers even if it consumes more fuel

per kilometer of travel.

Fig.7.5 Determination of cruising speed and cruising altitude

7.7 Cruise climb

To prepare the back ground for the analysis of the cruise climb, consider

Eq.(7.8a) which gives the range of a jet airplane. i.e.



2

1

W

W D

L

-3.6 V dWR =

CTSFC W

C

(7.24)

The TSFC is generally assumed to be constant during the flight. Further

simplifications are needed to carry out the integration in Eq.(7.24). In the

constant altitude - constant CL flight considered in subsection 7.4.2, as the name

suggests, the lift coefficient (CL) is assumed constant during the flight. In this

case, to satisfy the requirement of L = W = 2L

1ρV SC

2, the flight velocity is

decreased as the weight of the airplane decreases due to consumption of fuel

(see example 7.2). In the constant altitude – constant velocity flight, considered

in section 7.5, the flight speed (V) is held constant during the flight. In this case,

CL decreases as the fuel is consumed.

Equation (7.24) suggests a third possibility, other than the above two cases, of

both V and CL being held constant during the flight.In this case, to satisfy

L = W = L21

ρV SC2

, it has been suggested that the airplane be allowed to climb

slowly such that the decrease of atmospheric density ρ with altitude

compensates for the decrease of airplane weight due to consumption of fuel.

With these simplifications Eq.(7.24) gives :

1 2D L D L

2

1

W

W

-3.6V dW 3.6VR = = ln W /W

TSFC C /C W TSFC C /C (7.25)

The flight is called ’Cruise climb’ as the altitude continuously increases during the

flight.

Remarks:

(i) Exercise 7.3 would show that for a jet airplane with Wf / W1 = 0.2 and starting

the cruise climb at h = 11 km, the range would be 5141 km and the change of

flight altitude between the end and the start of cruise climb would be only



1.415 km. Thus, it is observed that the change in the altitude between the start

and end of cruise climb is very small as compared to the distance covered and

the level flight equations (L = W and T = D) are valid.

(ii) It can be shown (Ref.1.1, chapter 5) that the range in a cruise climb is higher

than that in level flight at the altitude where the cruise climb begins.

(iii) In actual practice continuous increase in altitude may not be permitted by Air

Traffic Regulations. As an alternative, a stepped climb approximation may be

used i.e. the flight path is divided into segments of constant altitude flights with

stepped increase in altitude after certain distance.

(iv)In a cruise climb the thrust required would be

T = D = (1/2) ρV2 SC

D

Since, the flight velocity and CL (and hence C

D) are held constant, the thrust

required will be proportional to ambient density (ρ). It may be pointed out that in

lower stratosphere the engine output (thrust available) is also proportional to the

ambient density. Thus, in a cruise climb in lower stratosphere the thrust setting

required is also constant and it becomes a very convenient flight – the pilot has

just to set the Mach number and then the autopilot will take care of the flight.


In the foregoing discussion, it was assumed that the airplane moves in a

mass of air which is stationary with respect to the ground. However, in many

situations the air mass has a velocity with respect to the ground and the airplane

encounters head wind or tail wind. (see subsection 7.2.2 for definition of head

wind and tail wind). The wind velocity is denoted by VW . When VW is non-zero,

the velocity of the airplane with respect to the ground (Vg) and that with respect

to air (Va) are different. To analyze the effect of wind on airplane performance, it

may be pointed out that the aerodynamic characteristics of the airplane (lift, and

drag) and the engine characteristics depend on the velocity with respect to air

(Va), whereas the distance covered in the flight depends on the velocity with

respect to the ground (Vg). In the presence of head wind the velocity of the



airplane with respect to the ground will be lower than its velocity with respect to

air and the range decreases. For example, in a hypothetical case of head wind

being equal to the stalling speed, the airplane, in principle, can remain airborne

without moving with respect to the ground. The fuel will be consumed as engine

would produce thrust to overcome the drag, but no distance will be covered as

the airplane is hovering! When there is tail wind the range increases.

An expression for range with effect of wind can be derived as follows.

Consider a jet airplane. Let Rg be the range in the presence of wind.

Equation(7.8a) can be used to calculate Rg, but the quantity ‘V’ in that equation

should be replaced with Vg i.e. :

2 2

1 1

W

W W3.6 V dW 3.6 (V - V ) dWg aR = - =g TSFC × T TSFC × T

W W , VW in m/s

2 2

1 1

W W3.6 V dW dWaR = - - 3.6 V = R - 3.6 V Eg w a wTSFC × T TSFC × T

W W (7.26)

where Ra is the range in still air = -2

1

W3.6 V dWaTSFC × T

W

and E is the duration of flight in hours. Thus, with head wind the range decreases

by 3.6 Vw E. In example 7.1 the range is 2667 km and the endurance is 3.33

hours. If a head wind of 15 m/s is encountered then the range would decrease

by 15 x 3.6 x 3.33 = 180 km.

Remarks:

i) Before a flight takes- off, the information about head wind, likely to occur on

the route is gathered from weather reports, and adequate amount of fuel is

provided to take care of the situation.

ii) The maximum endurance (Emax) is not affected by the presence of wind,

because Emax depends on airspeed only. The airspeed indicator in the cockpit, as



the name suggests, indicates airspeed and the pilot only needs to fly at airspeed

corresponding to Emax.



Chapter 7

Reference

Riley , K.F., Hobson, M.P. and Bence, S.J. “Mathematical methods for physics

and engineering” Cambridge University press Cambridge, U.K. (1998).



Chapter 7

Exercises

7.1 A jet airplane is flying in level flight at a constant velocity (V). Show that when

the drag polar is parabolic the endurance (E) is given by :

2E E ζ-1max 1E = tanTSFC 2E 1 - K C E ζmax L1 1

where, = Wfuel

/ W1

W1= Weight of airplane at the beginning of the flight; W2 = W1 (1 - ζ )

E1 = W1/D1, D1 = drag at the beginning of the flight

CL1 = Lift coefficient at the beginning of the flight.

TSFC = Specific fuel consumption (assume as constant).

7.2 Define safe range and gross still air range. Obtain the gross still air range

in steady level flight for a turboprop airplane flying at a constant speed of

400 kmph at an altitude where = 0.65, given that:

CD = 0.021 + 0.06C

L2 ; W

1 = 176, 600 N, W

fuel = 35, 300 N,

S = 90 m2, p = 0.82, BSFC = 3.90 N/kW - hr.

(Answer: R = 2104 Km).

7.3 Consider a jet airplane with 20% of its weight as fuel fraction. It starts the

cruise climb at an altitude of 11km. What will be the altitude at the end of cruise

climb (hf)? Assuming V = 240 m/s, TSFC = 0.6 and CL/ C

D = 16, estimate the

range in cruise climb (Rcc

). What is the angle of climb ( γcc

) in cruise climb?

(Answers: hf = 12415 m, Rcc = 5141 m, γcc = 0.0157o).



Chapter 8

Performance analysis IV – Accelerated level flight and climb

(Lecture 27)

Keywords: Accelerated level flight; accelerated climb; energy height.

Topics

8.1 Introduction

8.2 Accelerated level flight

8.2.1 Equations of motion in accelerated level flight

8.2.2 Time taken and distance covered in accelerated level flight

8.3 Accelerated climb

8.3.1 Equations of motion in accelerated climb

8.3.2 Effect of acceleration on rate of climb

8.3.3 Performance in accelerated climb from energy point of view

8.3.4 Energy height

Exercise



Chapter 8

Lecture 27 Performance analysis IV – Accelerated level flight and climb Topics

8.1 Introduction


8.2.1 Equations of motion in accelerated level flight

8.2.2 Time taken and distance covered in accelerated level flight

8.3 Accelerated climb




8.3.4 Energy height

8.1 Introduction

The last three chapters dealt with the performance airplane in steady

flights. The flights with acceleration are considered in this and the next two

chapters. The accelerated flights could be along a straight line e.g. accelerated

level flight and accelerated climb or along curved paths like loops and turn. In

this chapter the accelerated level flight and climb are discussed.


When an airplane moves along a straight line at a constant altitude but its

velocity changes with time, then it is said to execute an accelerated level flight.

This type of flight occurs in the following situations.

(i) The take-off speed of an airplane is about 1.15 to 1.3 times the stalling speed.

However, the speed corresponding to the best rate of climb is generally much

higher than this speed (see Figs.6.3a and c). Hence the airplane may accelerate

from the take-off speed to the speed corresponding to the desired rate of climb.

Similarly, the speed , at the end of the climb to the cruising altitude, is lower than



the cruising speed (Figs.6.3a and c) and an airplane would accelerate at the

cruising altitude to attain the desired cruising speed.

(ii) The airplane may also accelerate in the transonic flight range to quickly pass-

over to the supersonic speeds (see Fig.5.11)

(iii) The airplane may decelerate during a combat or when the pilot notices the

possibility of over-shooting a target.

8.2.1 Equation of motion in accelerated level flight

The forces acting on an airplane in an accelerated level flight are shown in

Fig.8.1. It may be recalled that the equations of motion are obtained by applying

Newton’s second law. For this purpose, the forces acting on the airplane are

resolved along and the perpendicular to the flight path. Sum of the components

of the forces in each of these directions, is equated to the product of the mass of

the airplane and the component of the acceleration in that direction.

The flight path in this case is a horizontal line. Hence, the equations of motion

are :

WT - D = m a = a

g (8.1)

L - W = 0 (8.2)

where ‘a’ is the acceleration.

Fig.8.1 Forces in accelerated level flight



8.2.2 Time taken and distance covered in an accelerated level flight

As regards the analysis of performance in an accelerated flight, it is of

interest to obtain the time taken and the distance covered for a given change in

velocity. The accelerated or decelerated flights last only for a short duration and

the weight of the airplane can be assumed to remain constant during such flights.

However, in a level flight L = W = (1/2) ρ V2 S C

L, should be satisfied. Hence, the

value of CL and consequently of C

D change continuously as the flight velocity

changes. From Eq.(8.1), the acceleration ‘a’ is given by:

a = g (T-D) / W

Substituting for D as (1/2)ρ V2 S C

D , gives:

1

2

g 2a = (T - ρ V S C )DW (8.3)

Note that :ds dV dV ds dV

V = and a = = = Vdt dt ds dt ds

dV V dVConsequently, dt = and ds = 8.3a

a a

Let the distance covered and the time taken for velocity to change from V1 to V

2

be denoted by ‘s’ and ‘t’ respectively, Integrating expressions in Eq.(8.3a) gives:

2 2

1 1

V VVdV dV

s = and t =a a

V V (8.4)

Substituting for ‘a’ from Eq.(8.3) yields:

2 2

1 1

V VW V dV W dV

s = and t =2 21 1g (T- ρ V S C ) g (T- ρ V S C )D DV V2 2

(8.5)

The expressions in Eq.(8.5) can be directly integrated if T and D are simple

functions of velocity. Otherwise a numerical integration as illustrated in the

following example can be carried out.



Example 8.1

An airplane with a weight of 156,960 N and a wing area of 49 m2 has a

drag polar given by CD = 0.017+0.06C

L2. It accelerates under standard sea level

conditions from a velocity of 100 m/s to 220 m/s. Obtain the distance covered

and the time taken during the acceleration, assuming the thrust output to remain

roughly constant at 53,950 N.

Solution:

1

2

1 1

2 2

1

2

2L = W = ρ V SCL

22 KW2 2D = ρ V SC = ρ V S C +D D0 2ρ SV

22× 0.06 × 1569602Or D = × 1.225 ×49×0.017 × V +21.225 × 49 × V

74.9225 × 102Or D = 0.5102 V +2V

To carry out the numerical integration, the integrands in Eq.(8.5) are evaluated

for several values of V and the methods like trapezoidal rule or Simpson’s rule

are used. Books on numerical analysis be consulted for further details of these

methods. Simpson’s rule gives accurate results with a small number of points

and is used here. For this purpose the range between V1 and V

2 is divided into

six intervals, each of 20 m/s. The values are tabulated below:

V (m/s) 100 120 140 160 180 200 220

D (N) 10042 10771 12518 14999 18050 21660 25731

W

g(T-D) 0.3644 0.3705 0.3861 0.4107 0.4456 0.4954 0.5669

W V

g(T-D) 36.44 44.46 54.06 65.72 80.21 99.09 124.72



Using Simpson’s rule,

s = (20/3) {36.44 + 4 (44.46 + 65.72 + 99.09 ) + 2(54.06 + 80.21) + 124.72}

= 8445 m = 8.445 km

t = (20/3) {0.3644 + 4 (0.3705 + 0.4107+0.4954) + 2 (0.3861 + 0.4456 ) + 0.5669}

= 51.34 s.

Answers:

Distance covered = 8.445 km ; time taken = 51.39 s.

8.3 Accelerated Climb

In this case, the flight takes place along a straight line inclined to the

horizontal at an angle as shown in Fig.8.2. The flight velocity increases or

decreases along the flight path. Figure 8.2 also shows the forces acting on the

airplane.

Fig.8.2 Accelerated climb


The equations of motion are:

WT - D - Wsin = a

g (8.6)

L - W cos = 0 (8.7)


From Eq.(8.6), the acceleration can be expressed as:

g (T - D - W sin )a =

W

(8.8)



Note that : ;dV dV dh dh

a = = . but = V = R/CCdt dh dt dt

dVConsequently, a = Vc dh

(8.9)

Substituting for ‘a’ in Eq.(8.6) and noting that sin = Vc / V,

V W dV (T-D)VcT - D - W - V = 0 or V =c cV g dh V dVW 1+

g dh

(8.10)

From Eq.(6.4), (T-D) V / W is the rate if climb in steady flight. Denoting it by Vco

,

Eq.(8.10) reduces to:

VcoV =cV dV

1+g dh

(8.11)

Remark:

The term (dV/dh) in Eq.(8.11) represents the rate of change of velocity with

altitude. This quantity would be positive if the flight velocity increases with

altitude. Thus, in an accelerated climb, the rate of climb, for given values of

thrust, speed and altitude, will be lower than that in a steady climb. This has

relevance to the flight with shortest time to climb, i.e., to calculate the shortest

time required to achieve desired altitude.

From Fig.6.3c it is observed that the flight speed for maximum rate of climb

(VR/Cmax) increases with altitude. Thus, in a climb which attempts to fly the

airplane at speeds corresponding to the maximum rate of climb (V(R/C)max) at

different altitudes, would not be a steady climb but an accelerated climb.

Consequently, the values of (R/C)max given in Fig.6.3e may need to be corrected

for the effect of acceleration.


The performance of an airplane in an accelerated flight can also be viewed from

the energy point of view. Multiplying Eq.(8.6) by V gives:

W dVT V - DV - W V sin = V

g dt



Or 2dh W d V

TV = DV + W +dt g dt 2

(8.12)

In Eq.(8.12) the term ‘TV’ represents the available energy provided by the

propulsive system. The term ‘DV’ represents the energy dissipated in overcoming

the drag. The term ‘W (dh / dt)’ represents the rate of change of potential energy

and (W/g) {d(V2

/ 2) / dt} represents the rate of change of kinetic energy. Thus,

the total available energy can be utilized in three ways viz. overcoming drag,

change of potential energy and change of kinetic energy. If the flight takes place

at Vmax

or (Vmin

)e in level flight, then entire energy is used in overcoming the

drag and no energy is available for climb or acceleration. Only at speeds in

between (Vmin

)e and V

max, can an airplane climb or accelerate and the excess

power (T-D)V has to be shared for increase of potential energy or kinetic energy

or both. If climb takes at V(R/C)max

then no acceleration is possible.

8.3.4 Energy height

Equation (8.12) can be rewritten as:

2(T-D)V d V= h +

W dt 2g

(8.13)

The term (h + V2/2g) is denoted by he and is called ‘Specific energy or Energy

height’. It is called specific energy because it is equal to the sum of potential

energy and kinetic energy divided by the weight. It is called energy height

because this term has the dimensions of height. It may be noted that

(dhe / dt) = (T-D)V/ W (8.14)

The energy height concept is used in optimization of climb performance.

Reference 1.9 chapter 7 and Ref.1.12 chapter 2 may be referred to for details.

The quantity (dhe/ dt) is called specific excess power and denoted by ‘Ps’.



Example 8.2

An airplane climbs at constant equivalent air speed in troposphere. Obtain an

expression for the correction to be applied to the value of rate of climb calculated

with the assumption of the steady climb(the denominator in Eq.8.11).

Solution:

In a climb with Ve as constant, the true air speed (V) is given by:

1

2

12V = V / σ ,e

dV dσ-3/2Consequently, = - V σedh dh

In troposphere the variation of with h is given as follows (Eq.2.7):

g-1T -λho λRσ =

To

where, TO = Temperature at sea level,

λ = Temperature lapse rate and

R = gas constant.

1

2

-(g+λR)2(g-λR)

eo

dV λ g-λRHence, = V σ

dh T λR

(8.15)

In I.S.A., λ = 0.0065 K/m. Using g = 9.81 m/s and R = 287.05 m/s2 K, the

correction factor in Eq.(8.11) is:

V dV -6 2 -1.2351+ = 1 + 4.894 × 10 V σeg dh (8.16)

It is seen that the correction required depends on Ve and . Typical values of the

correction factor at sea level ( = 1) and at 11 km altitude ( = 0.2971) are given

in Table E8.1.



Ve (m/s)

50

100

200

V dV1+

g dh

at s.l

1.01224

1.0489

1.1958

V dV1+

g dh

at 11 km

1.0548

1.2191

1.8766

Table E8.1 Correction factor in climb at constant equivalent air speed in

troposphere.

It is worth noting that at 11 km altitude the actual rate of climb, in constant Ve

flight at 200 m/s, is reduced to about half of its value in a steady climb.

Remark:

In a constant Mach number flight in troposphere, the flight velocity

decreases with altitude. Hence, the term (dV / dh) is negative and the rate of

climb in constant Mach number flight is more than that in a steady climb. See

exercise 8.1.



Chapter 8

Exercise

8.1 A jet trainer is climbing in troposphere at a constant Mach number of 0.6.

Obtain the rate of climb when it is climbing at an altitude of 5 km. The airplane

has the following data.

W = 54,000 N, S = 17 2m , CD = 0.017 + 0.055 2

LC , and thrust available at 5 km

altitude = 13,000 N.

[Hint: Show that in a constant Mach number flight :

dV λ R V dV 1 λ 2 = - M and 1+ = 1- RMdh 2 T -λh g dh g 2o

where, = ratio of specific heats,

λ = temperature lapse rate

R = Gas constant

Taking = 1.4, λ = 0.0065 K / m , R = 287.05 2m -2s K and g = 9.81 m / 2s

gives:

V 1c =2Vco 1-0.1331M

]

(Answers: Vco = 1798 m/min, Vc = 1888 m/min)



Chapter- 9

Performance analysis – V- Manoeuvres

(Lectures 28 to 31)

Keywords : Flights along curved path in vertical plane – loop and pull out ;

load factor ; steady level co-ordinated-turn - minimum radius of turn, maximum

rate of turn; flight limitations ; operating envelop; V-n diagram.

Topics

9.1 Introduction

9.2 Flight along a circular path in a vertical plane (simplified loop)

9.2.1 Equation of motion in a simplified loop

9.2.2 Implications of lift required in a simplified loop

9.2.3 Load factor

9.2.4 Pull out

9.3 Turning flight

9.3.1 Steady, level, co-ordinated-turn

9.3.2 Equation of motion in steady, level, co-ordinated-turn

9.3.3 Factors limiting radius of turn and rate of turn

9.3.4 Determination of minimum radius of turn and maximum rate of turn

at a chosen altitude

9.3.5 Parameters influencing turning performance of a jet airplane

9.3.6 Sustained turn rate and instantaneous turn rate

9.4 Miscellaneous topics – flight limitations, operating envelop and V-n

diagram

9.4.1 Flight limitations

9.4.2 Operating envelop

9.4.3 V-n diagram

Exercises



Chapter 9

Lecture 28 Performance analysis V – Manoeuvres – 1 Topics

9.1 Introduction

9.2 Flight along a circular path in a vertical plane (simplified loop)

9.2.1 Equation of motion in a simplified loop

9.2.2 Implications of lift required in a simplified loop

9.2.3 Load factor

9.2.4 Pull out

9.3 Turning flight


9.3.2 Equation of motion in steady, level, co-ordinated-turn

9.1 Introduction

Flight along a curved path is known as a manoeuvre. In this flight the

radial acceleration is always present even if the tangential acceleration is zero.

For example, from particle dynamics (Ref.1.2) we know that when a body moves

with constant speed along a circle it is subjected to a radial acceleration equal to

(V2 / r) or 2ω r where, V is the speed, r is the radius of curvature of the path and

is the angular velocity ( = V / r). In a general case, when a particle moves

along a curve it has an acceleration along the tangent to the path whose

magnitude is equal to the rate of change of speed ( V ) and an acceleration along

the radius of curvature whose magnitude is (V2 / r). Reference 1.1, chapter 1 may

be referred to for details. In order that the body has these accelerations a net

force, having components along these directions, must act on the body. For

example, in the simpler case of a body moving with constant speed along a



circle, there must be a centripetal force of magnitude m 2ω r in the radially inward

direction; m is the mass of the body.

For the sake of simplicity, the motions of an airplane along curved paths

confined to either the vertical plane or the horizontal plane, are only considered

here. The flight along a closed curve in a vertical plane is refered to as loop and

that in the horizontal plane as turn. Reference 2.1 and Ref.1.12, chapter 2, may

be referred to for various types of loops and turns. However, the simpler cases

considered here illustrate important features of these flights.

9.2 Flight along a circular path in vertical plane (simplified loop)

Consider the motion of an airplane along a circular path of radius r with

constant speed V. The forces acting on the airplane at various points of the flight

path are shown in Fig.9.1. Note also the orientation of the airplane at various

points and the directions in which D and L act; in a flat earth model W always

acts in the vertically downward direction.



Fig.9.1 Flight along a loop with constant radius and speed

(Note: The quantity 2W V

g r is the magnitude of the inertia force at various points)

9.2.1 Equations of motion in a simplified loop

The equations of motion, when the airplane is at specified locations, can be

written down as follows.

At point A : 2WV

T - D = 0 ; L - W = g r

(9.1)

At Point B: 2W V

T - D - W = 0 ; L = g r

(9.2)

At point C : 2W V

T - D = 0 ; L + W = g r

(9.3)



At point D : 2WV

T - D + W = 0 ; L = gr

(9.4)

At a general point G the equations of motion are:

2WVT - D - W sin = 0 ; L + W cos =

gr (9.5)

Note that the Eqs.(9.1) to (9.4) for points A, B, C and D can be obtained from

Eqs.(9.5) by substituting as 180o, 90o, 0o and 270o respectively.

Remarks:

i) If the tangential velocity is not constant during the loop then the first equation of

Eqs.(9.5) would become:

T - D - W sin = (W / g) a, where a = dV / dt (9.6)

ii) From Eqs.(9.1 to 9.5) it is observed that the lift required and the thrust required

during a loop with constant ‘r’ and ‘V’ change rapidly with time. It is difficult for the

pilot to maintain these values and the actual flight path is somewhat like the one

shown in Fig.9.2.

Fig.9.2 Shape of a normal loop



9.2.2 Implications of lift required during simplified loop

It is observed, that at the bottom of the loop i.e. point ‘A’ in Fig.9.1, the lift

required is equal to 2WV

W + gr

or 2V

L = W 1+gr

. The term (V2

/ gr) could be

much larger than 1 and the lift required in a manoeuvre could be several times

the weight of the airplane. As an illustration, let the flight velocity be 100 m/s and

the radius of curvature be 200 m, then the term (V2/ gr) is equal to 5.1. Thus the

total lift required at point ‘A’ is 6.1 W. In order that an airplane carries out the

manoeuvres without getting disintegrated, its structure must be designed to

sustain the lift produced during manoeuvres. Secondly, when lift produced is

high, the drag would also be high and the engine must produce adequate output.

Further, lift coefficient cannot exceed CLmax, and as such no manoeuvre is

possible at V= Vstall.

9.2.3 Load factor

The ratio of the lift to the weight is called ‘Load factor’ and is denoted by ‘n’ i.e.

n = (L / W) (9.7)

A flight with a load factor of n is called ‘ng’ flight. For example, a turn (see

example 9.2) with load factor of 4 is referred to as a 4g turn. In level flight, n

equals 1 and it is a 1g flight.

Higher the value of n, greater would be the strength required of the structure and

consequently higher structural weight of the airplane. Hence, a limit is prescribed

for the load factor to which an airplane can be subjected to. For example, the civil

airplanes are designed to withstand a load factor of 3 to 4 and the military

airplanes to a load factor of 6 or more. The limitation on the military airplane

comes from the human factors namely, a pilot subjected to more than 6g may

black out during the manoeuvre which is an undesirable situation.

To monitor the load factor, an instrument called ‘g-meter’ is installed in the

cockpit.

9.2.4 Pull out

The recovery of an airplane from a dive or a glide is called a pull out



(Fig.9.3). The dive is an accelerated descent while the pull out phase can be

regarded as a flight along an arc of a circle (See example 9.1).

Fig.9.3 Pull out from dive

Example 9.1

An airplane with a wing area of 20 m2 and a weight of 19,620 N dives with

engine switched off, along a straight line inclined at 60o to the horizontal. What is

the acceleration of the airplane when the flight speed is 250 kmph? If the airplane

has to pull out of this dive at a radius of 200 m, what will be the lift coefficient

required and the load factor? Drag polar is given by: CD = 0.035 + 0.076 2

LC and

the manouevre takes place around an altitude of 2 km.

Solution:

From Fig.9.3 the equations of motion in the dive can be written as follows.

WL- Wcos = 0; Wsin - D = a

g

= 60o, Hence, cos = 0.5 and sin = 0.866



Consequently, L = 19620 x 0.5 = 9810 N.

The drag of the airplane(D) can be obtained by knowing CD which depends on

CL .

2LC =L 2ρS V

V = 250 kmph = 69.4 m/s, at 2 km = 1.0065 kg / m3

Hence,

CL =

2 × 981021.0065 × 20 × 69.4

= 0.2024

Consequently, CD = 0.035 + 0.076 0.20242 = 0.03811

The drag D = LC 0.03811D = 9810 × = 1847.3NC 0.2024L

Hence, (W/g) a = W sin - D = 19620 x 0.866 - 1847.3 = 15144.1 N

Or a = 15144.1 9.81/ 19620 = 7.57 m/s2.

To obtain the lift required during the pull out, let us treat the bottom part of the

flight path during the pull out as an arc of a circle.

From Eqs.(9.1) to (9.5), the lift required is maximum at the bottom of the loop and

is given by:

2 2WV 1 69.4L= W+ or L = 19620 × 1+ ×

gr 9.81 200

Or L = 19620 x 3.45 Then,

19620 × 3.45 × 2C = = 1.396L 21.0065 × 20 × 69.4

Remarks:

i) The maximum load factor in the above pull out is 3.45. The value of lift

coefficient required is 1.396. This value may be very close to CLmax and the

parabolic drag polar may not be valid.



ii) Since CL cannot exceed C

Lmax, a large amount of lift cannot be produced at low

speeds. Thus maximum attainable load factor (nmaxattainable

) at a speed is:

nmaxattainable

= (1/2) ρ V2 S CLmax / W

At stalling speed the value of n is only one.

9.3 Turning flight

When an airplane moves along an arc of a circle about a vertical axis then

the flight is called a turning flight. When the altitude of the airplane remains

constant in such a flight, it is called a level turn. In order that a turning flight is

possible, a force must act in the direction of the radius of curvature. This can be

done by banking the airplane so that the lift vector has a component in the

horizontal direction. It may be added that the side force produced by deflecting

the rudder is not large. It also causes considerable amount of drag, which is

undesirable.


If there is no tangential acceleration i.e. the flight speed is constant, then

the flight is called a steady turn. If the altitude remains constant then the flight is

called a level turn. When the airplane executes a turn without sideslip, it is called

co-ordinated-turn. In this flight the X-axis of the airplane always coincides with

the velocity vector. The following two aspects may also be noted regarding the

steady, level, co-ordinated-turn.

(a) The centripetal force needed to execute the turn is provided by banking the

wing. The horizontal component of the lift vector provides the centripetal force

and the vertical component balances the weight of the airplane. Hence, the lift in

a turn is greater than the weight.

(b) An airplane executing a turn, does produce a sideslip.

Because of the aforesaid two factors, a pilot has to apply appropriate deflections

of elevator and rudder to execute a co-ordinated-turn.

A co-ordinated-turn is also called ‘Correctly banked turn’. In this chapter,

the discussion is confined to the steady level, co-ordinated-turn.



9.3.2 Equations of motion in steady level co-ordinated-turn

The forces acting on an airplane in steady, level, co-ordinated-turn are

shown in Fig.9.4. The equations of motion in such a flight can be obtained by

resolving the forces in three mutually perpendicular directions.

Fig.9.4 Turning flight

As the turn is a steady flight: T – D = 0 . (9.8)

As the turn is a level flight: W – L cos = 0. (9.9)

As the turn is co-ordinated which implied that, there is no unbalanced sideforce.

2W VL sin =

g r (9.10)

where is the angle of bank and r is the radius of turn.

Remarks:

i) From the above equations it is noted that L = W / cos . Hence, in a turn L is

larger than W. Consequently, drag will also be larger than that in a level flight at

the same speed. The load factor n is equal to 1/ cos and is higher than 1.

ii) From Eqs.(9.9) and (9.10), the radius of turn r is given by:



2 2W V Vr = =

g Lsin g tan (9.11)

Noting that, 1

cos =n

gives 2tan = n -1 and

2

2

Vr =

g n -1 (9.11a)

The rate of turn, denoted by (ψ ), is given by:

2V V g tanψ = = V / =

r gtan V

(9.12)

Noting 2tan = n -1 gives :

2g n -1

ψ =V

(9.12a)

(iii) In some books, the radius of turn is denoted by ‘R’. However, herein the letter

‘R’ is used to denote range, and to avoid confusion, the radius of turn is denoted

by ‘r’.

Example 9.2

An airplane has a jet engine which produces a thrust of 24,525 N at sea

level. The weight of the airplane is 58,860 N. The wing has an area of 28 m2,

zero-lift angle of – 2.2o and a slope of lift curve of 4.6 per radian. Find (a) the

radius of a correctly banked 4g level turn at the altitude where = 0.8 and the

wing incidence is 8o, (b) time required to turn through 180o and (c) thrust

required in the manoeuvre if the drag coefficient at this angle of attack be 0.055.

Solution:

The given data are: W = 58860 N, S = 28 m2, = 8o,0L = -2.2o,

dCLdα

= 4.6 per radian = 4.6

180 x 2 per degree = 0.083 per degree,

allowable n = 4 and T = 24525 N at sea level.



Consequently, dCLC = (α - α )L oLdα

= 0.0803 (8 + 2.2) = 0.82

In a 4g turn L = 4W = 1/2 ρV2 S CL

Hence, V = 1/2

2 ×4 × 588601/2(2L /ρ SC ) =L 1.225 × 0.8 × 28 × 0.82

= 144.6 m/s.

0 '1 1cos = = or = 75 31

n 4

Hence, tan = 3.873

Consequently, 22 144.6V

r = = = 550.3mgtan 9.81× 3.873

Rate of turn = V 144.6

ψ = =r 550.3

= 0.2627 rad /s

Hence, time to turn through 180o is equal to = 11.95s0.2627

The thrust required = Tr = 1/2 ρ V2 S CD

= (1/2) x 1.225 x 0.8 x 144.62 x 28 x 0.055 = 15786 N

Answers : (a) Radius of correctly banked turn = 550.3 m, (b) time required to turn

through 1800 = 11.95 s and (c) thrust required during turn = 15,786 N

Remark:

The thrust available is given as 24525 N at sea level. If the thrust available is

assumed to be roughly proportional to (σ0.7), the thrust available at the chosen

altitude would be 24525 x 0.80.7 = 20978 N. This thrust is more than the thrust

required during the turn and the flight is possible.



Chapter 9



9.3.4 Determination of minimum radius of turn and maximum rate of turn

at a chosen altitude


Turning flight is a very important item of performance evaluation,

especially for the military airplanes. Minimum radius of turn and maximum rate of

turn are important indicators of the manoeuverability of an airplane. From

Eqs.(9.11) and (9.12) it is observed that, at a given altitude and flight velocity, a

small radius of turn and a high rate of turn are achieved when the bank angle ( )

has the highest possible value. Equations (9.11a) and (9.12a) indicate that at a

given altitude, the minimum radius of turn (rmin) and the maximum rate of turn

( maxψ ) are obtained when ‘V’ is low and ‘n’ is high. The following considerations

limit the achievable values of rmin and maxψ .

(I)Limitation due to CLmax

: From the above discussion we observe that the lift

coefficient in a turning flight is higher than the lift coefficient required at the same

speed in level flight. Let CLT

be the lift coefficient in the turning flight and CLL

be

the lift coefficient in the level flight at the same speed.

Then, CLT = n W / ( ½ V2 S ) = n C

LL

However, CLT

cannot be more than CLmax

and this imposes limitations on the

attainable values of load factor (n) and the bank angle ( ). Let these two values

be denoted by max CLmaxn and max CLmax . They can be expressed as :



(nmax

)CLmax = /Lmax LLC C (9.13)

,-1Noting that = cos 1/n max CLmax =cos-1( /LL LmaxC C ) (9.14)

It may be noted that, at stalling speed (Vs), the value of CLL

equals CLmax

or n =

1. Hence, turn is not possible at stalling speed .

(II) Limitation due to allowable load factor from structural consideration : The

bank angle and the load factor in a turn are related by:

cos = 1/n .

However, n cannot exceed the value permitted by the structural design of the

airplane. Let this value be denoted by (nmax)str. Hence, max is limited to

cos-1 {1/(nmax)str}.

(III) The drag coefficient in a turning flight is higher than that in a level flight at the

same speed. However, in a steady turn the thrust required cannot exceed the

thrust available (Ta). This also imposes limitations on the attainable values of

and n. Let these two values be denoted as (max

)Ta

and (nmax

)Ta

.

It may be noted that, at V = Vmax

and (Vmin

)e the entire engine output is used in

overcoming the drag in level flight. Hence, the steady level turn is not possible at

these two speeds.

The lowest of the above three values viz max CLmaxn , (nmax)str and (nmax

)Ta

is

the permissible value of nmax

. Let this value be denoted by (nmax)perm .

Substituting this value in Eqs.(9.11a) and (9.12a) gives r and ψ .

9.3.4 Determination of minimum radius of turn and maximum rate of turn at

a chosen altitude

In a general case, the drag polar and the thrust available are functions of

Mach number. In such a case, the minimum radius of turn (rmin) and the

maximum rate of turn (ψmax

) at an altitude, can be obtained by using the



following steps. The limitations stated in the previous subsection, are taken into

account during the procedure.

(i) Choose an altitude. Obtain Vmax

and Vmin

at this altitude. Note that a steady

level, co-ordinated-turn is possible only within this speed range.

(ii) Choose a flight speed (V) in between Vmax

and Vmin and obtain C

LL as:

CLL

= 2W / ( SV2)

Obtain Mach number (M) corresponding to the chosen V and the speed of sound

at chosen altitude.

(iii) Obtain the CLmax at the chosen flight Mach number. It may be recalled from

subsection 3.7.4, that for airplanes flying at high speeds, the CLmax depends on

Mach number. Obtain the ratio CLmax / CLL

.

The ratio CLmax

/ CLL

gives the quantity max CLmaxn defined above. If this value

is smaller than the allowable load factor from structural consideration viz.

(nmax)str, then the turn may be limited by CLmax

. In this situation, choose

CLT1

= CLmax

. It may be mentioned that the procedure presented here, aims at

obtaining the value of lift coefficient in the turn (CLT

) which satisfies all the three

limitations on the turn mentioned above. The quantity CLT1

is the value of CLT

as

limited by CLmax

. This will be modified in the subsequent steps.

If CLmax

/ CLL

is more than (nmax)str, then the turn may be limited by (nmax)str. In

this situation, choose CLT1 as (nmax)str x C

LL.

(iv) Obtain from the drag polar, the drag coefficient CDT1

, corresponding to CLT1

and the chosen Mach number. Calculate the drag DT1

from:

DT1

= 1/2 V2 S CDT1



If D T1

is greater than the available thrust (Ta), then the turn is limited by engine

output. In this situation, obtain the maximum permissible value of drag coefficient

in turning flight (CDT

) as limited by Ta . It is given as : C

DT = Ta / (1/2 V

2 S)

Corresponding to this value of CDT

, obtain the lift coefficient CLT

by referring to

the drag polar.

If DT1

is smaller than Ta , then the turn is not limited by the engine output. In this

situation, the turn is limited by CLmax

or (nmax)str. Consequently, CLT

is the

smaller of the two values obtained in step (iii).

(v) Once CLT

is known, is given by:

= cos-1 (CLL

/CLT

). Knowing and V, the radius of turn (r) and rate of turn

(ψ ), at the chosen speed, can be calculated using Eqs.(9.11) and (9.12).

(vi) The previous steps should be repeated at various values of flight speeds

ranging between Vmin

and Vmax

. Plotting these results, the values of rmin and

ψ max and the corresponding speeds Vrmin and Vψ max

can be determined at the

chosen altitude.

(vii) Repeat steps (i) to (vi) at different altitudes.

The procedure is illustrated, at a chosen altitude, in example 9.3. Example 9.3

A passenger airplane has a gross weight of 176,400 N and a wing area of

45 2m . Obtain the variations of r and ψ with velocity at an altitude of 8 km from

the following data.

CLmax

= 1.4, (nmax)str = 3.5, CD = 0.017 + 0.05 2

LC

Vmin

= 103 m/s, Vmax

= 274 m/s and the thrust output (Ta) varies as given in the

table below.

V (m/s) 105 115 125 145 165 185 205

Ta (N) 21100 21125 21150 21480 21580 21980 22270



Solution:

At 8 km altitude the value of ρ is 0.525 kg/m3. The minimum radius of turn and

ψmax

at various speeds are worked out in a tabular manner using the procedure

outlined above.

V (m/s) 105 115 125 145 165 185 205

CLL

1.354 1.129 0.955 0.710 0.548 0.436 0.355

CLmax / CLL

1.034 1.240 1.466 1.972 2.553 3.21 3.94

CLT1

1.4 1.4 1.4 1.4 1.4 1.4 1.243*

CDT1

0.115 0.115 0.115 0.115 0.115 0.115 .0942

DT1

(N) 15000 17993 21258 28601 37042 46568 46852

Ta (N) 21100 21125 21150 21480 21580 21980 22270

CDT

** ** 0.1114 0.0864 0.067 0.0543 0.0448

CLT

1.4$ 1.4$ 1.396£ 1.178£ 1.08£ 0.863£ 0.745£

LT

LL

C=n

C 1.034 1.240 1.461 1.659 1.824 1.98 2.10

(degrees) 14.75 36.25 46.9 52.93 56.76 59.63 61.6

r (m) 4273 1838 1491 1619 1819 2043 2321

ψ (rad/s) 0.0246 0.0626 0.0838 0.0896 0.0907 0.0906 0.0883

The symbols in the above table have the following meanings:

* Turn is limited by load factor (nmax)str hence CLT1

= (nmax)str CLL.

** Thrust available is more than thrust required. Hence, CLT = C

LT1

$ Turn is limited by CLmax

£ Turn is limited by Ta

Table E9.3 Variations of radius of turn (r) and rate of turn ψ with

flight velocity (V) for airplane in example 9.3



Fig.E9.3 Variations of radius of turn (r) and rate of turn ψ with flight

velocity (V) for the airplane in example 9.3

The plots of r vs V and ψ vs V are shown in Fig.E9.3. From these plots

rmin = 1.490 m and maxψ = 0.0907 rad/s, Vrmin = 124 m/s and ψmaxV = 165 m/s

Answers:

Minimum radius of turn (rmin) = 1490 m at Vrmin = 124 m/s

Maximum rate of turn ( maxψ ) = 0.090 rad/s at ψmaxV = 165 m/s

Remarks: i) Turning performance of a jet airplane :

Section 7 of Appendix B presents the turning performance of a jet airplane.

Figures 9.5a and b show the variations of ψ and r with velocity at different



altitudes for that airplane. Figures 9.5c and d present the variations of ψmaxV

and Vrmin with altitude.

Fig.9.5a Turning performance of a jet transport – rate of turn (ψ )



Fig.9.5b Turning performance of jet transport – radius of turn (r)

Fig.9.5c Turning performance of jet transport - variation of Vψ max



Fig.9.5d Turning performance of jet transport - variation of Vrmin

Note:

Some curves in Figs.9.5a,c and d show discontinuity in slope at certain points.

This occurs when the criterion limiting the turning performance changes from

(nmax)str to (nmax

)Ta

.

ii) Turning performance of a piston engined airplane :

Section 7 of Appendix A presents the turning performance of a piston engined

airplane. Figures 9.6a and b show the variations of r and maxψ with velocity at

different altitudes for that airplanes. Figures 9.7c and d present the variations of

rmin and maxψ with altitude. Figure 9.6e presents the variations of ψmaxV and

Vrmin with altitude. Both these speeds increase with altitude. The two speeds

come close to each other as absolute ceiling is approached. Minimum radius of

turn (rmin) increases with altitude and maxψ decreases with altitude. At absolute

ceiling, the rate of turn becomes zero and the radius becomes infinite.



Fig.9.6a Turning performance of a piston engined airplane

- variation of rate of turn (ψ )

Fig.9.6b Turning performance of a piston engined airplane –

variation of radius of turn ( r)



Fig.9.6c Turning performance of a piston engined airplane –

variation of rmin with altitude

Fig.9.6d Turning performance of piston engined airplane –

variation of maxψ with altitude



Fig.9.6e Turning performance of piston engined airplane –

variations of Vrmin and ψmaxV with altitude

iii) In many situations the minimum radius of turn in level flight is limited by the

available engine output. This can be overcome and a smaller radius of turn can

be obtained by allowing the airplane to descend during the turn. In this manner a

loss of potential energy is used to increase the available energy during turn.

Reference 1.12, chapter 2 may be consulted for additional details. See also

subsection 9.3.6 for further information.



Chapter 9





The steps, described in subsection 9.3.4 , to determine rmin, maxψ , Vrmin and

ψmaxV constitute a general procedure which is applicable to all types of airplane.

However, the influence of the wing loading (W/S) and the thrust loading (Ta/W)

can be examined by the following simplified analysis. It is based on the following

two assumptions.

(a)Thrust available (Ta) is constant.

(b)The drag polar is parabolic with DOC and K as constants.

The following relationships are observed in a steady, level, co-ordinated-turn.

T = D, L = nW ; n = 1

cos ,

2

2

Vr =

g n -1,

2g n -1ψ =

V

Hence, 2 2 2a D LDO

1 1T = ρV SC = ρV S C + K C

2 2

Or 2

2a DO 2

1 2nWT = ρV S C +K

2 ρSV

(9.15)

Solving Eq.(9.15) for n2, gives :

2 2DO2 a

1 1ρV ρV CT2 2n = -

K W/S W W/S

(9.16)



Let the free stream dynamic pressure be denoted by ‘q’ :

Or q = 21ρV

2 (9.17)

Consequently, Eq.(9.16) can be rewritten as :

2 a DOqCTqn = -

K(W/S) W W/S

(9.18)

From Eq.(9.11a)

2

2Vr =

g n -1

Or 2

2qr =

gρ n -1 (9.19)

From Eq.(9.19) it is observed that ‘r’ is a function of q and n. However, when the

constraint of thrust available is taken into account, then n and q are related by

Eq.(9.18).

The value of q which would give minimum radius of turn (rmin ), can be obtained in

two stages.

(a) Substitute the expression for ‘n’ as given by Eq.(9.18) in Eq.(9.19).

(b) Differentiate the resultant equation for ‘r’ obtained in step (a), with respect to

‘q’, and equate it to zero.However, the resulting expression is complicated. An

alternate way is as follows.

(i) Differentiate Eq.(9.19) with respect to q and equate it to zero.

-1/22 2

2 2 2

2gρ n -1 - 2gρqn n -1 dn/dqdr=

dq g ρ n - 1= 0

Or 2 dnn - 1 - qn = 0

dq (9.20)

(ii)The quantity (dn/dq) is obtained by differentiating Eq.(9.18) with respect to q

i.e.



DO2

aT /W qCdnn = -

dq 2K(W/S) K W/S (9.21)

(iii)Substituting for n2 and n (dn/dq) in Eq.(9.20) yields :

2 2aa DO DO

2 2

q C q Cq T /WTq- - 1 - + = 0

K W/S W 2K W/SK W/S K W/S

Simplifying :

aq T /W

= 12K W/S

(9.22)

Equation (9.22) yields the value of q which gives minimum radius of turn. This

value is denoted by ‘qrmin ‘ i.e. :

qrmin = a

2K W/S

T /W (9.23)

Using Eq.(9.17), Vrmin is given as :

Vrmin =

a

4K W/S

ρ T /W (9.24)

Substituting qrmin in Eq.(9.18) gives :

222 DOrmin 2 2

a

a a

2K W/S T /W 4K W/S Cn = -

T /W K W/S T /W K W/S=

DO

2a

4 K C2 -

T /W

Or 2

DOrmin

a

4K Cn = 2 -

T /W (9.25)

Substituting from Eqs.(9.24) and (9.25) in Eq.(9.11a) gives :

2rmin

min 2rmin

Vr =

g n -1

DO2

a

a

4K(W/S) 1=ρ T /W 4KC

g 2- -1T /W

2a aDO

4K W/S=

gρ T /W 1 - 4K C / T / W (9.26)



Proceeding in a similar manner, the values of ψmaxV , ψmaxn and maxψ , which

take into account the constraint of thrust available, can be derived. The final

expressions are given below.

DO

1/21/4

ψmax2(W/S)

V = K/Cρ

(9.27)

DO

1/2

ψmaxaT /W

n = - 1KC

(9.28)

1/2

DOmax

1/2a CT /Wρ

ψ = g -W/S 2K K

(9.29)

Remarks:

(i)From Eqs.(9.26) and (9.29) it is observed that for a jet airplane to have a low

value of rmin and a high value of ψmaxV , the value of (Ta/W) should be high and

that of (W/S) should be low. However, as stated in section 7.4.3 the wing loading

(W/S) is a compromise between various considerations like range, take-off and

landing. Consequently, the general practice is to select (Ta/W) to give the desired

value of maxψ , taking into account the wing loading chosen from other

considerations.

(ii) The constraints of (nmax)str and CLmax have not been taken into account in

the above analysis. Also the variation of thrust available with flight speed has

been ignored.

Equation (9.25) shows that the load factor for minimum radius of turn (nrmin) is

less than 2 . However, the load factor for maximum rate of turn ( ψmaxn ), as

given by Eq.(9.28), could be high, especially near the sea level where (Ta/W) is

at its highest. In this situation the constraint of (nmax)str needs to be taken into

consideration.



(iii)The constraint of CLmax is likely to affect the value of rmin. Example 9.4

illustrates such a situation.

(iv) A simplified analysis of the turning performance of an airplane with engine

propeller combination can be carried out by assuming that (a) THP in constant

with flight velocity and (b) DOC and K are constants. However, the resulting

expression has the following form.

4rmin r minA V +B V +C = 0

This equation does not have an analytical solution and a graphical or numerical

procedure is needed. Reference 1.12 chapter 2 can be consulted for details.

It can be inferred from the analysis of Ref.1.12, that if it is desired to increase

maxψ or decrease rmin of a given airplane, then the wing loading (W/S) should

be reduced and / or the ratio (BHP/W) should be increased.

Example 9.4

Consider the airplane in example 9.3 with the simplification that the thrust

remains constant with flight velocity and has the value of 21685 N. Obtain the

values of Vrmin, ψmaxV , nrmin, ψmaxn , rmin and maxψ as given by the analysis in

subsection 9.3.5.

Solution :

The given data are :

W = 176, 400 N, S = 45 m2 , DOC = 0.017, K = 0.05, h = 8000 m or

3ρ = 0.525 kg/m , Ta = 21685 N .

The constraints are : CLmax = 1.4, (nmax)str = 3.5.

Consequently, W/S = 176400 /45 = 3920 N/m2 &Ta/W = 21685/176400 = 0.1229.

Based on the analysis of subsection 9.3.5, which considers only the constraint of

thrust available, the following expressions are obtained.

rmina

4K(W/S)V =

ρ T /W



2DO

rmin

a

4K Cn = 2 -

T /W

minr

2DOa a

4K W/S=

gρ T /W 1-4K C / T /W

DO

1/21/4

ψmax2(W/S)

V = K/Cρ

DO

1/2

ψmaxaT /W

n = - 1KC

1/2

DOmax

1/2a CT /Wρ

ψ = g -W/S 2K K

Accordingly ,

rmin4 × 0.05 × 3920

V = = 110.23 m/s0.525 × 0.1229

rmin 24 ×0.05 × 0.017

n = 2 - = 1.3320.1229

2

min 2 2

2rminV 110.23

r = = = 1407.6 mg n -1 9.81 1.332 -1

1/2 1/4

ψmax2×3920 0.05

V = = 160.04 m/s0.525 0.017

1/2

ψmax0.1229

n = - 1 = 1.7930.05 × 0.017

2ψmax

maxψmax

g n - 1ψ =

V

29.81× 1.793 -1= = 0.0912 rad/s

160.04

The values of lift coefficients corresponding to Vrmin and ψmaxV are:



2rmin

Lrmin2rmin

n W 1.332 × 176400C = = = 1.637

1 0.5 × 0.525 × 110.23 × 45ρV2

S

2

ψmaxLψmax 2

ψmax

n W 1.793 × 176400C = = = 1.045

1 1ρV S × 0.525 × 160.04 × 45

2 2

It is observed that in case of maxψ the values of n and CL are 1.793 and 1.045.

These values are lower than the prescribed values of (nmax)str and CLmax.

Hence, this turn is possible and maxψ of 0.0912 rad/s at V = 160.04 m/s is

possible. However, the value of CLrmin is 1.637 which is higher than CLmax and

this turn is not possible. In this situation, a new value of flight velocity (V) is to be

obtained at which the values of load factor (n) given by the two constraints of

thrust available and CLmax, are equal.

The value of n from the constraint of thrust available can be denoted by ‘ Tan ’. It

is given by Eq.(9.16):

2

2 DO

1/2

Taa

1ρV CT 12n = - ρV

K W/S W 2 W/S

(9.30)

The value of n from the constraint of CLmax can be denoted by ‘ CLmaxn ’. It is

given by :

L = 2LmaxCLmax

1n W = ρV SC

2

Or 2 LmaxCLmax

C1n = ρV

2 W/S (9.31)

Equating Eqs.(9.30) and (9.31) gives the value of ‘V’ which satisfies both the

constraints i.e.



2

2DO Lmax

1/2

2a1ρV CT C1 12 - ρV = ρV

K W/S W 2 W/S 2 W/S

Simplifying yields :

DO

22Lmax

2 2aT /W CC 1

= + ρVK W/S 2K W/S W/S

Substituting various values gives :

22

2 2

0.1229 0.017 1.4 1= + ×0.525×V

0.05×3920 20.05×3920 3920

Or V = 126.32 m/s

Consequently,

22Lmax

1ρV C 0.5×0.525×126.32 ×1.42n = = = 1.496

W/S 3920

2

min 2

126.32r = = 1461.9 m

9.81 1.496 -1

The value of V = 126.32 m/s satisfies the constraints of Ta and CLmax. The

corresponding value of n = 1.496 is also less than (nmax)str of 3.5. Hence, all

constraints are satisfied.

Answers : Based on the simplified analysis at 8000 m altitude the following

values are obtained.

Vrmin = 126.32 m/s, nrmin = 1.496, rmin = 1461.9 m,

ψmaxV = 160.04 m/s , ψmaxn = 1.793 , maxψ = 0.0912 rad/s

Remark :

The values by exact analysis are :

Vrmin = 124 m/s, nrmin = 1.451 , rmin = 1490 m ,



ψmaxV = 165 m/s , ψmaxn = 1.824 maxψ = 0.0907 rad/s .

The agreement between the two results is seen to be reasonable. The reasons

are that (Ta/W) is rather low and the variation of Ta with V is not large.


The maximum rate of turn in a steady level co-ordinated-turn is called

‘Maximum sustained turn rate(MSTR)’ (Ref. 1.12 chapter 2). An airplane can

maintain this turn rate continuously for some time. However, as explained in

subsections 9.3.3 and 9.3.4 this turn rate is generally limited by the thrust

available. A rate of turn higher than MSTR can be obtained if the airplane is

allowed to descend or slow down. In this manner, the loss of potential energy or

kinetic energy can be utilized to increase the available energy during turn and

increase the rate of turn. This rate of turn is called ‘Instantenous rate of turn’. The

maximum instantenous rate of turn will be limited by other two factors viz. CLmax

and (nmax)str. See also item (iv) in subsection 9.4.3.

General Remark:

In the foregoing sections various types of flight situations of practical

interest have been analyzed. To analyze any other flight situation one can begin

by writing down the equations of motion along and perpendicular to the flight

path. From these equations, the lift required, thrust required and accelerations in

tangential and radial directions can be worked out.



Chapter 9


9.4 Miscellaneous topics – flight limitations, operating envelop and V-n

diagram

9.4.1 Flight limitations


9.4.3 V-n diagram

9.4 Miscellaneous topics

A flight is called free flight when the airplane is away from the influence of

the ground i.e. it is at a height more than a few wing spans above the ground.

The performance in level flight, climb, turn etc. come under this category. In

contrast, the analysis of take-off and landing requires consideration of the

influence of proximity of ground. The discussion of performance in free flight is

concluded in this section by describing aspects like flight limitations, operating

envelop and V-n diagram. Chapter 10 describes the performance in take-off and

landing.

9.4.1 Flight Limitations:

In chapters 5 to 8 and the previous subsections of this chapter, the

performance of an airplane in free flight has been discussed under categories of

level flight, climb, accelerated flights and manoeuvres. The important aspects of

these analyses are generally brought out in a diagram which is called here as

‘Variations of characteristics velocities’. Figure 9.7a shows the variations of Vmax

,

(Vmin

)e, Vs, V

(R/C) max V

max, Vrmin and ψmaxV with altitude for a typical subsonic jet

airplane. Figure 9.7b presents similar plots for a piston engined airplane.



It has been pointed out earlier that (a) the maximum lift coefficient limits the

minimum speed in level flight (Vs), the minimum radius of turn (rmin

) and the

maximum rate of turn ( maxψ ) , (b) the power output limits the maximum speed

(Vmax

), the minimum speed (Vmin

)e , the maximum angle of climb (max

), the

maximum rate of climb (R/C)max

, rmin and maxψ , (c) the maximum allowable load

factor, (nmax)str, limits rmin and maxψ . In addition to these, the performance of the

airplane may also be limited by considerations like buffeting, sonic boom,

maximum dynamic pressure (q) limit and aerodynamic heating . These limitations

are briefly described below.

Fig.9.7a Variations of characteristic velocities – Jet transport



Fig.9.7b Variations of characteristic velocities – Piston engined airplane

(i)Buffeting is an irregular oscillation of a part of an airplane, caused by the

passing of separated flow from another component. For example, the horizontal

tail experiences buffeting when the separated flow from the wing passes over it

(horizontal tail). This happens when the wing is at a high angle of attack or the

shock stall takes place on it in the transonic flow regime. To prevent buffeting,

the permissible value of CLmax

may be limited. This in turn would affect Vs, rmin

and ψmax

.

(ii)The sonic boom problem is encountered when an airplane flies at supersonic

speed at low altitudes. The shock waves created by an airplane, when it is flying

at a supersonic speed, coalesce and form two waves across which there is a

finite pressure rise (overpressure). When these waves reach the ground each of

them is perceived as an explosive like sound called sonic boom or sonic bang.

The intensity of the boom depends on the size and shape of the airplane, its flight

altitude and the atmospheric conditions. It increases with the increase in the size

of the airplane and decreases with the increase of the altitude of the flight. An

overpressure in excess of about 100 N/m2 is quite annoying and may cause

0

1000

2000

3000

4000

5000

6000

0 20 40 60

Velocity (m/s)

Alt

itu

de

(m)

V for minimum radius of turn

V for maximum rate of turn

Stalling speed

Vmin from engine output

V max

V for maximum angle of climb

V for maximum rate of climb



vibrations of buildings and rattling of window panes. To keep the overpressure on

the ground within socially acceptable limits, the supersonic transport airplanes

are generally not permitted to cross Mach number of one below tropopause and

they cruise at altitudes of 16 to 20 km.

(iii)The airplanes are generally designed for a dynamic pressure (q = 21ρV

2) of

100,000 N/m2. This limit would not permit attainment of high supersonic Mach

number at low altitudes.

(iv)As the flight Mach number increases, the stagnation temperature (Ts) on the

surface increases. It is given by: -1 2T = T (1+ε M )s amb 2

(9.32)

where, Tamb is the ambient temperature and ε is the recovery factor which has a

value of around 0.9 for turbulent boundary layer on the surface. The maximum

stagnation temperature (Ts) may be limited from the consideration of material

used for the fabrication of the airplane. This would limit the maximum permissible

Mach number.

(v) Reference 3.9, chapter 17, mentions about other limits like engine relight limit,

pilot ejection altitude and duct pressure limit. The minimum speed from engine

relight limit is encountered in some cases at high altitudes where enough air may

not be available to restart the engine in the event of flame-out. The highest

altitude may be limited to about 15 kms which is the the highest altitude at which

ejection by the pilot is permitted.


The maximum speed and the minimum speed of the airplane can be

calculated from the level flight analysis. However, the attainment of maximum

speed may be limited by the considerations mentioned in the previous

subsection. A diagram which indicates the range of flight speeds permissible for

an airplane at different altitudes is called ‘Operating envelope’. Typical operating



envelope for a military airplane is shown in Fig.9.8. Explanation of the curves in

this figure is as follows.

(i)The curve ABCDE is the level flight boundary based on the engine output. The

portion ABC is the Vmax

or (Mmax

) boundary. The portion CDE is the (Vmin

)e or

(Mmin

) boundary, limited by the engine output. It may be mentioned that for these

Fig.9.8 Operating envelope of a military airplane - Schematic

curves (ABC and CDE) the engine output is with the afterburner on. On this

boundary (ABCDE) the specific excess power (Ps) is zero.

(ii)The curve FG is the line representing stalling speed (Vs).

2WV = ; C without flaps Lmaxρ S CLmax

Recalling that when Mach number exceeds 0.5, the maximum lift coefficient

(CLmax

) decreases due to shock stall or buffetting. The line FG includes this effect

when Mach number corresponding to Vs is more than 0.5.



(iii)The line HJK represents the dynamic pressure (q) limit corresponding to q of

100,000 N/m2.

(iv)The line LMNOP represents the boundary corresponding to stagnation

temperature (Ts) of 400K. It may be pointed out that Tamb and hence the speed

of sound change with altitude in troposphere. They are constant in lower

stratosphere. Hence, the allowable flight Mach number, for stagnation

temperature to be below allowable value, changes with altitude.

The flight envelope taking into account the above limits is the curve FDCONMJH.

Remark:

Figure 9.8 also shows zones marked as : (I) advantageous for interceptor role,

(II) advantageous for aerial combat and (III) suitable for high speed low altitude

flight.

It may be added that for the interceptor role, it is advantageous if the airplane

flies at high altitude and high speed (zone I in Fig.9.8).

For aerial combat the manoeuverability, which is measured mainly by the rate of

turn, is important. It may be recalled from subsection 9.3.3 that the rate of turn is

low at (a) altitudes near the ceiling and (b) flight speeds close to Vmax and Vmin.

Further, the aerial combat cannot take place at very low altitudes. Hence, the

aerial combat zone is the region marked as (II) in Fig.9.8.

For airplanes used as ground attack fighter, the ability to fly at high speed and at

low altitude is important. Zone (III) in Fig.9.8 is appropriate for these airplanes.

9.4.3 V-n diagram

The load factor (n) has already been defined as the ratio of lift and weight

i.e. n = L / W. In level flight n = 1. However, as pointed out in subsections 9.2.3

and 9.3.3 the value of ‘n’ during a manoeuver is greater than one. Hence, the

structure of the airplane must be designed to withstand the permissible load

factor. Further, when an airplane encounters a gust of velocity Vgu (see Fig.7.1b)

the angle of attack of the airplane would increase by Δα = Vgu /V. This increase

in angle of attack, would increase the lift by ΔL, given by :



ΔL = ½ ρV2 S C

Lα Δα = ½ ρV

2 S CLα

Vgu / V

= ½ ρVSCLα

Vgu (9.33)

Hence, Δn = ΔL / W = ½ ρVSCLα

Vgu / W (9.34)

From Eqs.(9.33) and (9.34), ΔL increases with Vgu. Further, for a given Vgu, the

values of ΔL & Δn increase with flight velocity. An airplane must be designed to

withstand the gust loads also.

In aeronautical engineering practice, the load factors due to manoeuver

and gust are indicated by a diagram called ‘Velocity-load factor diagram or V-n

diagram’. A typical V-n diagram is shown in Fig.9.9. This diagram can be

explained as follows.

(i) Curves OIA and OHG : The lift (L) produced by an airplane is given by

L = ½ ρV2S CL. It should be noted that (i) C

L ≤ C

Lmax and (ii) at stalling

speed(Vs), L = W and n = 1. However, if the airplane is flown with CL = C

Lmax at

speeds higher than Vs, then (a) L will be more than W and (b) L or n would be

proportional to V2. This variation is a parabola and is shown by curve OIA in

Fig.9.9. In an inverted flight the load factor will be negative and the V vs n curve

in such a flight is indicated by the curve OHG in Fig.9.9. It may be mentioned that

an airplane can fly only at V ≥ Vs and hence the portions OI and OH in Fig.9.9

are shown by chain lines.



Fig.9.9 A typical V-n diagram

(ii) Positive and negative manoeuver load factors : An airplane is designed to

withstand a certain permissible load factor. Higher the permissible load factor,

heavier will be the weight of airplane structure. Hence, for actual airplanes the

manoeuver load factor is limited depending on its intended use. Federal Aviation

Administration (FAA) in USA and similar agencies in other countries, prescribe

the values of permissible manoeuver limit load factors (npositive and nnegative) for

different categories of airplanes. Table 9.1 gives typical values. A limit load is

obtained by multiplying the limit load factor with the weight (W). The airplane

structure is designed such that it can withstand the limit load without yielding.

The ultimate load factor, in aeronautical practice, is 1.5 times the limit load factor.

The ultimate load is obtained by multiplying the ultimate load factor with the

weight (W). The airplane structure is designed such that it can withstand the

ultimate load without failing, though there may be permanent damage to the

structure.



Type of airplane npositive nnegative

General aviation-non aerobatic 2.5 to 3.8 -1

Transport 3 to 4 -1

Fighter 6 to 9 -3

Table 9.1 Typical limit load factors

In Fig.9.9, npositive

= 3 and nnegative

= -1.2 have been chosen; the actual values

depend on the weight of the airplane and its category. Reference 3.18 part

V,chapter 4 may be consulted for details.

(iii) Lines AC, GF and FD : The positive manoeuver load factor is prescribed to

be constant upto the design diving speed (Vd); line AC in Fig.9.9. According to

Ref.3.9, chapter 14, the design diving speed could be 40 to 50% higher than the

cruising speed (Vc) for subsonic airplanes. For supersonic airplanes, the Mach

number corresponding to Vd could be 0.2 faster than the maximum level flight

Mach number. The negative manoeuver load factor is prescribed to be constant

upto design cruising speed (line GF in Fig.9.9) and then increases linearly to zero

at V = Vd (line FD in Fig.9.9).

(iv) Manoeuvre load diagram : The diagram obtained by joining the points

OACDFGO is called ‘Manoeuvre load diagram’.

(v) Manoeuvre point and Corner speed : The point ‘A’ in Fig.9.9 is called

’Manoeuvre point’. The flight speed at this point is denoted by V* and is called

’Corner speed’. At point ‘A’ the lift coefficient equals CLmax and the load factor

equals npositive. This combination would result in the maximum instanteneous

turn rate at the speed V*. See subsection 9.3.6 for definition of instanteneous

turn rate.

(vi) Positive and negative gust load factors : From Eq.(9.33) it is observed that

the gust load factor varies linearly with velocity. The regulating agencies like FAA



prescribe that an airplane should be able to withstand load factors corresponding

to Vgu = 30 ft/s (9.1 m/s) upto cruising speed (Vc) and Vgu = 15 ft/s (4.6 m/s)

upto design diving speed (Vd). Lines JB and JF’ in Fig.9.9 show the gust lines

corresponding to Vgu = 30 ft/s (9.1 m/s) and Lines JC’ and JE in the same figure

show the gust lines corresponding to Vgu = 15 ft/s (4.6 m/s).

It may be pointed out that a gust in real situation, is not a sharp edged gust as

shown in Fig.7.1b and the velocity Vgu is attained in a gradual manner. This

causes reduction in the gust load factor. To take care of this reduction the gust

load factor is multiplied by a quantity called ‘Gust alleviation factor’. Reference

3.9 chapter 14 may be referred for details.

(vii) Gust load diagram : The diagram obtained by joining the points JBC’EF’J is

called ‘Gust load diagram’.

(vi) Final V-n diagram : For its safe operation, an airplane must be designed to

withstand load factors occuring at all points of the gust and manouever load

diagrams. Hence, the final V-n diagram is obtained by joining the parts of these

two diagrams representing the higher of the manoeuver and gust load factors.

The final V-n diagram in the case presented in Fig.9.9, is given by the solid curve

obtained by joining the points IAB’BB’’CEF’’FGHI.

It may be pointed out that the gust load line JB’ is above the curve IA in the

region IK. However, along the curve IK the airplane is already operating at CLmax

and any increase in angle of attack due to gust cannot increase CL beyond C

Lmax.

Hence, the portion JK of the line JB is not included in the final V-n diagram.

It may also be pointed out that the angles of attack of the airplane are different at

various points of the V-n diagram. Consequently, the components of the resultant

aerodynamic force along and perpendicular to the chord of the wing (N and C in

Fig.3.7) would be different at different angles of attack. The structural analysis

needs to take this into account. For example, the angle of attack is positive and

high at point A and it is positive and low at point C. At points G and E the angles

of attack are negative. Books on Airplane structures may be consulted for details.



Chapter 9

Exercises

9.1 Define steady level co-ordinated-turn. An airplane having a weight of 11,000

N has a wing area of 15 m2 and drag polar of CD = 0.032 + 0.06C

L2. Obtain the

radius of turn in a steady level coordinated turn at a speed of 160 kmph at sea

level from the following data.

CLmax

= 1.4, (THP)available

= 90 kW, maximum load factor = 3.5.

What is the time taken to turn through 180o?

[Answers: r = 124.6 m; t = 8.81 s]

9.2 Define load factor. What are its values in (a) level flight (b) free fall (c) in a

turn of radius 200 m at a speed of 100 m/s and (d) at the bottom of a loop of

radius 200 m at a speed of 100 m/s?

[Answers: (a) 1 (b) 0 (c) 5.19 (d) 6.097]



Chapter 10

Performance analysis VI – Take-off and landing

(Lectures 32-34)

Keywords: Phases of take-off flight — take-off run, transition and climb; take-

off distance; balanced field length; phase of landing flight; landing distance.

Topics

10.1 Introduction

10.2 Definitions of take-off run and take-off distance

10.3 Phases of take-off flight

10.3.1 Take-off ground run

10.3.2 Transition and climb phases

10.4 Estimation of take-off performance

10.4.1 Distance covered and time taken during ground run

10.4.2 Various speeds during take-off run

10.4.3 Distance covered and time taken during transition phase

10.4.4 Distance covered and time taken during climb phase

10.4.5 Parameters influencing take-off run

10.4.6 Effect of wind on take-off run

10.4.7 Guidelines for estimation of take-off distance

10.4.8 Balanced field length, its estimation and effect of number of

engines on it.

10.5 Landing performance

10.5.1 Definition of landing distance

10.5.2 Phases of landing flight

10.5.3 Estimation of landing distance

10.6 Flap settings during take-off and landing

References

Exercises



Chapter 10

Lecture 32 Performance analysis VI – Take-off and landing –1 Topics

10.1 Introduction

10.2 Definitions of take-off run and take-off distance

10.3 Phases of take-off flight






10.1 Introduction

An airplane, by definition, is a fixed wing aircraft. Its wings can produce lift

only when there is a relative velocity between the airplane and the air. In order to

be airborne, the lift produced by the airplane must be at least equal to the weight

of the airplane. This can happen when the velocity of the airplane is equal to or

greater than its stalling speed. To achieve this velocity called ‘Take-off

velocity(VTO

)’ the airplane accelerates along the runway. Thus, an airplane

covers a certain distance before it can take-off. Similarly, when an airplane

comes in to land, the lift produced must be nearly equal to the landing weight.

Hence, the airplane has a velocity, called ‘Touch down speed (VTD)’, when it

touches the ground. It then covers a certain distance before coming to halt.

The estimation of take-off distance and landing distance are the topics

covered in this chapter.



10.2. Definitions of take-off run and take off distance

The horizontal distance covered along the ground, from the start of take-

off till the airplane is airborne is called the take-off run. However, to decide the

length of the runway required for an airplane, it is important to ensure that the

airplane is above a certain height before it leaves the airport environment. This

height is called ‘Screen height’ and is equal to 15 m (sometimes 10 m), which is

above the height of common obstacles like trees and electricity poles. The take-

off distance is defined as the horizontal distance covered by an airplane from the

start of the run till it climbs to a height equal to the screen height. It is assumed

that the weight of the airplane during take-off is the gross weight for which it is

designed and that the take-off takes place in still air.

10.3 Phases of take off flight

The take-off flight is generally divided into three phases namely (i) ground

run (ii) transition (or flare) and (iii) climb (see Fig.10.1a).

Fig.10.1a Phases of take-off flight


During the ground run the airplane starts from rest and accelerates to the

take-off speed (VT0

or V1). The flaps and engine(s) are adjusted for their take-off



settings. In the case of an airplane with tricycle type of landing gear, all the three

wheels remain in contact with the ground till a speed of about 85% of the VT0

is

reached. This speed is called ‘Nose wheel lift off speed’. At this speed the pilot

pulls the stick back and increases the angle of attack of the airplane so as to

attain a lift coefficient corresponding to take-off (CLT0

). At this stage, the nose

wheel is off the ground (Fig.10.1b) and the speed of the airplane continues to

increase. As the speed exceeds the take off speed the airplane gets airborne and

the main landing gear wheels also leave the ground.

When the airplane has a tail wheel type of landing gear, the angle of attack is

high at the beginning of the take-off run (Fig.10.1c). However, the tail wheel is

lifted off the ground as soon as some elevator effectiveness is gained

(Fig.10.1d). This action reduces the angle of attack and consequently the

drag of the airplane during most of the ground run. As the take-off speed is

approached the tail wheel is lowered to get the incidence corresponding to CLT0

.

When VT0

is exceeded, the airplane gets airborne.

The point at which all the wheels have left the ground is called ‘Unstick point’

(Fig.10.1a).

Fig.10.1b Nose wheel lift-off



Fig.10.1c Tail wheel type airplane at start of take-off run

Fig.10.1d Tail wheel type of airplane during middle part of take-off run


During the transition phase the airplane moves along a curved path

(Fig.10.1a) and the pilot tries to attain a steady climb. As soon as the airplane

attains an altitude equal to the screen height, the take-off flight is complete. For

airplanes with high thrust to weight ratio the screen height may be attained during

the transition phase itself.




From the point of view of performance analysis, the following two

quantities are of interest.

(i) The take-off distance (s) (ii) The time (t) taken for it.

Since the equations of motion are different in the three phases of take-off flight,

they (phases) are described separately in the subsequent subsections.


The forces acting on the airplane are shown in Fig.10.1a. It is observed

that the ground reaction (R) and the rolling friction, μR, are the two additional

forces along with the lift, the drag, the weight and thrust ; μ is the coefficient of

rolling friction between the runway and the landing gear wheels. The equations of

motion are :

WT- D - μ R = a

g (10.1)

L + R – W = 0 (10.2)

Hence, R = W - L and

T - D - μ (W-L)a =

W/ g (10.3)

Further,

dV dV ds dVa = = = V

dt ds dt ds

Hence, ground run (s1) is given by:

1 1V V

V dV W V dVs = =1 a g T- D - μ(W-L)

o o (10.4)

The time taken during ground run (t1) is given by:

1 1V V

dV W dVt = =1 a g T - D - μ (W-L)

o o (10.5)



Equations (10.4) and (10.5) can be integrated numerically, when the variations

of T, D and L are prescribed and µ is known. The value of µ depends on the type

of surface. Typical values are given in Table 10.1.

Type of surface Coefficient of

rolling friction (µ)

Concrete, wood or asphalt 0.02

Hard turf 0.04

Average field-short grass 0.05

Average field-long grass 0.1

Soft ground 0.1-0.3

Table 10.1 Coefficient of rolling friction

The thrust during take-off run can be approximated as T = A1 – B

1V

2. The angle

of attack and hence, the lift coefficient (CL ) and the drag coefficient (CD ) can be

assumed to remain constant during the take-off run. With these assumptions, the

left-hand side of Eq.(10.1) becomes :

12 2T - D -μ (W -L) = A - B V - μ W - ρV S (C - μ C )1 1 D L2

= A – BV2 where A = A

1 - µ W and

1B = B + ρS (C - μC )1 D L2

Substituting in Eqs.(10.4) and (10.5) gives :

1V

W V dV W 2s = = ln A/(A - BV )1 12g 2gBA - BVo (10.6)



and 1

VA + B VW dV W 1t = = ln1 2g 2g AB A - B VA-BV 1o

(10.7)

Remarks:

i) The denominator in the integrands of Eqs.10.4 and 10.5, i.e. [T- D - µ (W - L)],

is the accelerating force during the take-off run. A good approximation to s1 and

t1 is obtained by taking an average value of the accelerating force (Fa) to be its

value at V = 0.7 V1 i.e.

Fa = [T- D - µ (W - L)]V = 0.7 V1

Consequently,

1V 2W VW V dV 1s = =1 g F 2g Fa ao (10.8)

and 1

VWVW dV 1t = =1 g F g Fa ao

(10.9)

ii) Generally the flaps are kept in take-off setting (partial flaps) right from the

beginning of the take-off run. Hence, CD during the take-off run should include

the drag due to flaps and landing gear.

Reference 3.6, section 3.4.1 may be consulted for increase in CDO due to the flap

deflection and the landing gear. See also section 2.9 of Appendix A. The

proximity of the ground reduces the induced drag. As a rough estimate, the

induced drag with ground effect can be taken to be equal to 60% of that in free

flight at the same CL.

(iii) The take-off speed (VTO or V1) is (1.1 to 1.2) Vs ; where Vs is the stalling

speed with W = WTO and CL = CLTO . As mentioned in subsection 3.7.4, CLTO is

0.8 times CLland.




In the subsection 10.3.1 the nose wheel lift-off speed and take-off speed have

been explained. Section 6.7 of Ref.1.10 mentions additional flight speeds

attained during the ground run. A brief description of the speeds, in the sequence

of their occurance, is as follows.

(a) Stalling speed (Vs) : It is the speed in a steady level flight at W = WTO and

CL = CLTO.

(b) Minimum control speed on ground (Vmcg): At this speed, the deflection of full

rudder should be able to counteract the yawing moment due to failure of one

engine of a multi-engined airplane when the airplane is on ground.

(c) Minimum control speed in air (Vmca) : At this speed, the deflection of full

rudder should be able to counteract the yawing moment, due to failure of one

engine of a multi-engined airplane if the airplane was in air.

(d) Decision speed (Vdecision) : This speed is also applicable to a multi-engined

airplane. In the event of the failure of one engine, the pilot has two options. (I) If

the engine failure takes place during the initial stages of the ground run, the pilot

applies brakes and stops the airplane. (II) If the engine failure takes place after

the airplane has gained sufficient speed, the pilot continues to take-off with one

engine inoperative.

If the engine failure takes place at decision speed (Vdecision), then the distance

required to stop the airplane is the same as that required to take-off with one

engine inoperative. See subsection 10.4.8 for additional details.

(e) Take-off rotation speed (VR): At this speed the elevator is powerful enough to

rotate the airplane to attain the angle of attack corresponding to take-off.

(f) Lift-off speed (VLO) : This is the same as unstick speed mentioned in

subsection 10.3.1. This speed is between (1.1 to 1.2) VS.



It is mentioned in Ref.1.10, chapter 6, that Vmcg, Vmca, Vdecision, VR lie

between VS and VLO.



Chapter 10

Lecture 33 Performance analysis VI – Take-off and landing –2 Topics







From Fig.10.1a it is observed that during the transition phase the airplane

changes the direction of flight and its speed would increase from V1 to V

2. The

height attained during this phase and the horizontal distance traversed can be

obtained by treating the flight path as part of a circle. However, according to the

procedure given in Royal Aeronautical Data sheets (now called Engineering

Sciences Data Unit, ESDU for short), the increase in height during the transition

phase is small and the horizontal distance (s2) can be obtained by assuming that

the work done by the engine is used in overcoming the drag and in increasing the

kinetic energy of the airplane i.e.

)2 22 2 2 1

WT s = D s + (V - V

2g

Or 2 2

2 1

2

V - VWs =

2g T -D (10.10)

T and D in Eq.(10.10) are evaluated at the mean speed during this phase i.e., at

(V2 + V

1) / 2. The time taken (t

2) in transition is given by:



2 1

22

st =

0.5 (V + V ) (10.11)

V1 generally lies between (1.15 to 1.2) Vs and V

2 is (1.05 to 1.1) V

1.


Since the vertical height covered during the transition has been ignored,

the horizontal distance covered in climb phase (s3) is the distance covered while

climbing to screen height i.e.

s3 = (Screen height) / tan (10.12)

where, is the angle of climb at velocity V2 :

sin γ = T -D

W

where, T and D are evaluated at V2.

The time taken in climb phase (t3) is:

t3 = (Screen height) / V

2 sin (10.13)

Hence, the take-off distance (s) and the time taken for it (t) are given by :

s = s1 + s

2 + s3 (10.14)

t = t1 + t

2 + t

3 (10.15)

Example 10.1

A jet airplane with a weight of 441, 450 N and wing area of 110 m2 has a

tricycle type landing gear. Its CLmax with flaps is 2.7. Obtain the take-off distance

to 15 m screen height and the time taken for it. Given that:

(i) V1 = 1.16 Vs

(ii) V2 = 1.086 V

1

(iii) CL during ground run is 1.15

(iv) Drag polar with landing gear and flaps deployed is CD = 0.044 +0.05C

L2

(v) Thrust variation during take-off can be approximated as :

T = 128,500 – 0.0929 V2 ; where V is in kmph and T is in Newton



(vi)Take-off takes place from a level, dry concrete runway (µ = 0.02) at sea level.

Solution:

0Lmax LmaxTC = 0.8 C = 0.8 ×2.7 = 2.16

1/2 1/22W 2 × 441450

V = = = 55.08 m/ss ρS C 1.225 × 110 × 2.16Lmax

Hence,

V1 = 1.16 x 55.08 = 63.89 m/s

and V2 = 1.086 x 63.89 = 69.38 m/s.

For CL = 1.15,

CD = 0.044 + 0.05 x 1.152 = 0.1101

Hence, 2D L

1T - D - μ (W-L) = T- μW - ρV S {C - μC }

2

= 128500 - 0.0929 (3.6V)2 - 0.02 x 441450

– 0.5 x 1.225 x V2 x 110 (0.1101 - 0.02 x 1.15)

= 119671 – 7.0752 V2

Using Eqs.(10.6) and (10.7) the ground run (s1) and time taken for it (t

1) are:

s1 =

441450

2 × 9.81× 7.0752 ln [119671/(119671-7.0752 x 63.89 x 63.89)] = 878.32 m

1 12 2

1 1 12 2 2

1441450 (119671) + (7.0752) × 63.89

t = ln2 × 9.81(119671× 7.0752) 119671 - 7.0752 × 63.89

= 26.34 s.

The distance covered during transition (s2) is obtained as follows.

63.89 + 69.38Average speed during transition = m/s = 66.635 m/s = 239.9 kmph

2

Hence, thrust during this phase = 128500 – 0.0929 x 239.92 = 123,153 N

To get the drag during this phase it is assumed that CL equals C

LTO and it is given

by :

CLTO = C

Lmax (Vs / V1

)2 = 2.16 / (1.16)

2 = 1.605



Assuming the same drag polar as in the ground run gives:

CD = 0.044 + 0.05 x 1.6052 = 0.1728

Hence, D = 0.5 x1.225 x (66.635)2 x 110 x 0.1728 = 51695 N

Using Eqs.(10.10) and (10.11) gives :

2

2 2441450 69.38 - 63.89s = = 230.4 m

2 × 9.81 123159 - 51695

and t2 = 230.4 / 66.635 = 3.46 s.

During the climb phase, V = 69.38 m/s = 249.77 kmph.

Hence, T = 128500 - 0.0929 x 249.772 = 122704 N

To get the drag in the climb phase the lift coefficient should be known. For this

purpose L is taken roughly equal to W.

Hence, 1

122

L4414502C = W/ ρV S = = 1.36

2× 1.225 × 110 × 69.38

Consequently, CD = 0.044 + 0.05 x 1.362 = 0.1365

and D = 0.5 x 1.225 x (69.38)2 x 110 x 0.1365 = 44269 N

Hence, 122704 - 44269

sin = = 0.1777441450

Consequently, tan = 0.1805.

Using Eqs.(10.12) and (10.13) gives:

s3 = 15/0.1805 = 83.1 m

and t3 = 15/(69.38 x0.1805) = 1.20 s.

Finally, s = s1 + s

2 + s

3 = 878.32 + 230.4 + 83.1 = 1192 m

and t = t1 + t

2 + t3 = 26.34 + 3.46 + 1.20 = 31.0 s.

Answers:

Take off distance = 1192 m ; Time taken for take-off = 31 s.




The major portion of the take-off distance is the ground run. Hence if

ground run is reduced, the take-off distance is also reduced. From Eq.(10.8), it is

observed that the distance s1 is given by :

2

11

avg

VWs =

g T-D-μ W -L (10.16)

Let V1 = 1.1 VS. Recalling, SLmax

2WV =

ρSC, Eq.(10.16) can be rewritten as :

1

2

Lmax avg

1.21× 2Ws =

2gρSC T -D-μ W -L

=

Lmax avg

1.21 W/S

gρC T/W - D/W -μ 1-L/W (10.17)

The following observations can be made from Eq.(10.17).

i) The ground run increases when the wing loading (W / S) increases.

ii)The ground run also increases when decreases. Since decreases with

altitude, the take-off distance will be more when the altitude of the airport

increases.

iii)The ground run decreases as CLmax

increases. Hence, the high speed

airplanes which have high wing loading from consideration of cruise, employ

elaborate high lift devices to increase CLmax

.

iv)The take-off run decreases by increasing the accelerating force which mainly

depends on (T/W). It may be recalled from subsection 4.3.5 that the thrust of a jet

engine can be increased temporarily by using an afterburner. The thrust can also

be augmented by using an auxiliary rocket fired during the take-off run. In

shipboard airplanes a catapult is used to augment the accelerating force.


While discussing the range performance it was shown, with the help of a

derivation in section 7.8, that the distance covered with respect to the ground



decreases when the flight takes place in the presence of head wind. Same effect

occurs during the take-off and the take-off distance reduces in the presence of

head wind. In a hypothetical case of head wind being equal to the stalling speed

(Vs), the airplane can get airborne without having to accelerate along the ground.

A quantitative estimate of the effect of wind velocity (Vw) on s

1, can be obtained

from Eq.(10.4), by replacing the limits of integration from (0 to V1) by (V

w to V

1)

i.e. :

1

w

1 with wind

VW V dV

s =g T - D - μ(W -L)

V

Thus, the head wind, though bad for range, is beneficial during take-off as it

reduces the take-off distance.

Airports have a device to indicate the direction of wind. The take-off flight takes

place in such a manner that the airplane experiences a head wind. This is

referred to as ‘Take-off into the wind’.


In subsections 10.4.1, 10.4.3 and 10.4.4, a procedure to estimate the

take-off distance has been presented. However, it is based on several

assumptions and consequently has significant amount of uncertainty. In actual

practice, there would be further uncertainty due to factors like condition of the

runway surface (wet or dry), and piloting technique. Hence, for the purpose of

preliminary design of airplane, the following guidelines can be used.

For airplanes with engine-propeller combination, the Federal Aviation

Regulations designated as FAR-23 (Ref.10.1) are used. Under these regulations,

the take-off distance to attain 50 feet (or 15 m) is obtained under certain

prescribed conditions. This distance is denoted here by ’sto23

’. Reference 10.2

has estimated sto23

for several airplanes. It is observed (Ref.10.2) that sto23

is

related to the following parameter.



T0 T0

LT0

W W( ) ( )

S Pσ C

where,

(W/S)T0

= wing loading based on take-off weight.

(W/P) T0

= power loading based on take-off weight and sea level static power

output.

σ = density ratio = ρ/ρ0

CLT0 = Lift coefficient in take off configuration (about 80% of C

Lmax in landing

configuration)

The above quantity is called take-off parameter for FAR-23 and denoted by

‘TOP23’ i.e.

T0 T0

23LT0

W W( ) ( )

S PTOP =σ C

(10.18)

Based on the data of Ref.10.2, the following relationship has been deduced in

Ref.3.18, pt.I, chapter 3.

sto23

(in ft) = 8.134 TOP23 + 0.0149 2

23TOP (10.19)

where, W/S is in lbs / ft2, W in lbs and P in hp.

When SI units are used this relationship takes the following form.

sto23 (in m) = 8.681x10-3 TOP

23+5.566x10-8 2

23TOP (10.20)

where W / S is in N / m2 , W in N and P in kW.

Example 10.2

Consider an airplane with the following features.

W/S = 2400 N / m2 , W/P = 24 N /kW , C

LTO = 1.6 and σ = 1.

Estimate the take-off distance for this airplane.

Solution :

The parameter ‘TOP23

’ in this case is :



TOP23

= 2400 x 24/(1 x 1.6) = 36000 N2/( m2 kW)

Using Eq.(10.20) gives,

sto23

= 8.681x10-3 x 36000 +5.566x10-8 x 360002 = 385.9 m or 1260 ft.

Answer : Take-off distance = 385.9 m or 1260 ft.

As regards the airplanes with jet engines, the take-off parameter (TOP) is defined

as :

T0

LT0 T0

W( )

STOP = T

σC ( )W

(10.21)

where, T = sea level static thrust.

Reference 3.9 chapter 5, gives a curve as guideline for sto in feet and TOP in

lbs / ft2. However, when a second order equation is fitted to that curve, the

relationship can be expressed in SI units in the following form.

sto (in m) = 0.1127 TOP +1.531 x 10-6 TOP2 (10.22)

Example 10.3

Consider a jet airplane with the following features.

W/S = 5195 N/m2, T / W = 0.3, C

LT0= 2.16 and σ = 1.

Estimate the take-off distance.

Solution :

In this case TOP is :

TOP = 25195= 8017 N/m

1×2.16×0.3

From Eq.(10.22) :

sto = 0.1127x8017 +1.531 x 10-6 x 80172 = 1002 m.

Answer : Take-off distance = 1002 m.



Chapter 10

Lecture 34 Performance analysis VI – Take-off and landing – 3 Topics

10.4.8 Balanced field length, its estimation and effect of number of

engines on it.



10.5.2 Phases of landing flight



10.4.8 Balanced field length and its estimation

Take-off is a critical phase of flight operation and various eventualities are

taken into account to arrive at the length of the runway required for the operation

of the airplane. In the case of multi-engined airplane, the possibility of the failure

of one of the engines during take-off is an important consideration. If the engine

failure takes place during initial stages of ground run, then the pilot can apply the

brakes and bring the airplane to halt. If the engine failure takes place after the

airplane has gained sufficient speed, then the following two alternatives are

available.

(a) Apply brakes and stop the airplane, but this may need much longer runway

length than in the case of take-off without engine failure.

(b) Instead of applying brakes, continue to fly with one engine inoperative and

take-off; but the take-off distance would be longer than when there is no engine

failure.

These two alternatives indicate the possibility of a speed, called “Decision

speed”. If the engine failure occurs at the decision speed, then the distance



required to stop the airplane is the same as that required to take-off with one

engine inoperative. The take-off distance required when engine failure takes

place at the decision speed is called ‘Balanced field length (BFL)’. It is estimated

as follows.

FAR 25 (see Ref.10.1) is used as a set of regulations for obtaining the take-off

distance of jet airplanes. The regulations also prescribe a procedure to calculate

the balanced field length (BFL). Reference 10.2 has estimated BFL for many jet

airplanes and observed that BFL is a function of TOP defined in Eq.(10.21).

Based on this data, the BFL in feet, when W/S in lbs / ft2 is given as (Ref.3.18,

Pt.I, chapter 3):

BFL (in ft) = 37.5 TOP (in lbs / ft2) (10.23)

When SI units are used, Eq.(10.23) takes the following form.

TO

LT0 TO

W( )

SBFL (inm) = 0.2387T

σ C ( )W

(10.24)

where W / S is in N / m2.

Remark :

(i) Effect of number of engines on BFL :

The data in Ref.10.2, on which Eq.(10.23), is based, shows some scatter

(Fig.3.7 of Ref.10.2). However, the data for airplanes with two, three and four

engines show some definite trend; the BFL is more as the number of engines

decrease. This is expected, as for a two engined airplane, when one engine is

inoperative, the thrust available would decrease to half of the full thrust, whereas

for an airplane with four engines, with one engine inoperative, the thrust available

would be three fourth of the full thrust. Consequently, BFL would be less for a

four engine airplane as compared to that for a two engined airplane. Perhaps,

based on this argument, Ref.3.9, chapter 5, suggests three different lines for BFL

vs TOP curve for airplane with two three and four engines. In SI units these lines

can be expressed as:



For two engined airplane: BFL (in m) = 0.2613 TOP (in N / m2) (10.25)

For three engined airplane: BFL (in m) = 0.2387 TOP (in N / m2) (10.26)

For four engined airplane: BFL (in m) = 0.2196 TOP (in N / m2) (10.27)

Example 10.4

Consider the airplane of example 10.3 and obtain the balance field length.

Solution:

In this case :

W / S = 5195 N/m

2, σ = 1.0 , C

LTO = 2.16 and T/ W = 0.3.

Consequently, TOP is 8017 N/m2.

Using Eqs (10.25) to (10.27) the BFL would be (a) 2095 m for an airplane

configuration with two engines, (b) 1914 m for three engine configuration and

(c) 1761 m for four engine configuration. Comparing sto and BFL in examples

10.3 and 10.4, it is seen that is BFL is nearly twice of sto .

(ii) See Appendices A and B for calculation or take-off distance for a piston

engined airplane and a jet airplane respectively.



While describing the take-off distance it was mentioned that the airplane should

clear the screen height before it leaves the airport environment. For the same

reason, the landing flight begins when the airplane is at the screen height. The

landing distance is defined as the horizontal distance that the airplane covers in

descending from the screen height and to come to halt. In actual practice, the

airplane does not halt on the runway. After reaching a sufficiently low speed the

pilot takes the airplane to the allotted parking place.

10.5.2 Phases of landing

Figure 10.2 shows the phases of landing flight for an airplane with tricycle type

landing gear.



Fig.10.2 Phases of landing flight

During the final approach phase, the airplane performs a steady descent. The

flight velocity in this phase is called approach speed and denoted by VA. During

the flare, the pilot makes the flight path almost horizontal. In the float phase the

pilot gently touches the main wheels to the ground. This is done gradually so that

the vertical velocity of the airplane is not more than about 4 m/s. The flight speed

at the point of touch down is denoted by VT. It is about 90% of V

A. After the

touch down, the airplane rolls for a period of about 3 seconds during which the

nose wheel is gently lowered to touch the ground. Brakes are not applied in this

phase as their application would produce a large decelerating force which would

cause a large nose down moment and the nose wheel may hit the ground with a

bang. After the three wheels have touched the ground, the brakes are applied

as well as other devices like reverse thrust or reversed pitch of propeller are

deployed. The ground run is said to be over when the airplane comes to halt or

attains a low speed when it can turn off the runway and go to the parking place.


This can be done in a way similar to the estimation of the take-off distance

i.e., by writing down equations for each phase of the flight. However, the



estimation cannot be done accurately as the flare and float phases depend very

much on the judgment of the pilot.

Royal Aeronautical Society Data sheets (presently called Engineering Science

Data Unit or ESDU) have given a simple method which amounts to assuming a

constant deceleration and calculating the distance to decelerate from VA and to

come to a halt i.e.

sland = - (V

A)2 / 2a (10.28)

where, a = -1.22 m/s2 (or 4ft/s2) for simple braking system

= -1.52 m/s2 (or 5 ft/s2) for average braking system.

= - 1.83 m/s2 (or 6 ft/s2) for modern braking system and

= - 2.13 to 3.0 m/s2 (or 7 to 10 ft /s

2) for airplane with modern braking

system and reverse thrust or reverse pitch propellers.

The approach speed (VA) depends on factors like stalling speed under

approach conditions, minimum speed at which adequate control is possible and

the type of approach viz. visual landing or instrumented landing system or aircraft

carrier deck approach. As a first estimate VA can be taken as 1.3 Vs.

Example 10.5

Obtain the landing distance for the airplane in example 10.1. Assume that

the airplane has modern braking system with reverse thrust and that VA = 1.3 Vs.

Solution:

From example 10.1, W = 441, 450 N, S = 110 m2 ,

CLmax during landing = 2.7.

Hence, 1/2

s2 ×441450

V =1.225×110×2.7

= 49.24 m/s

Consequently, VA = 1.3 x 49.24 = 64.01 m/s.

Taking a = - 2.13 m/ 2s , the estimate of landing distance is :



land

264.01s = - = 961.9 m

2 × (-2.13)

Answer : Landing distance = 961.9 m

Remarks:

i) Appendix A also estimates the landing distance using Eq.(10.28). Appendix B

uses a different formula.

ii) The landing distance is proportional to (VA)2 and consequently it is proportional

to (Vs)2

. The following observations can be made by noting that (Vs)2

equals

2W/(SCLmax

).

(a) The landing distance increases with increase of (W/S) and the altitude of

landing field. (b) The landing distance decreases with increase of CLmax

.

iii) The use of reverse thrust and reverse pitch propeller to reduce the landing

distance has been mentioned earlier. The landing run can also be decreased by

using (a) arresting gear, (b) drag parachute and (c) spoilers.

The arresting gear is used for airplane landing on the deck of a ship.

The drag parachute, when opened, increases the drag significantly and reduces

the landing run.

The spoilers are located on the upper surface of the wing. When deflected up,

the spoiler disturbs the flow, resulting in reduction of lift and increase of drag.

Spoiler ailerons are shown in Fig.1.2c. When used as a device to produce a

rolling moment, the spoiler aileron is deflected only on the left or the right wing

half. The lift on that wing half is reduced and the airplane rolls. Whereas, during

landing, the spoiler ailerons on both the wing halves are deployed

simultaneously. This results in a large reduction in lift and increase in drag. Both

these effects help in reducing the landing run.

iv) Like take-off distance the landing distance is also reduced by head wind.




It is mentioned in subsection 10.4.1, that the LmaxC during take-off is 80% of that

during landing. The flap setting during take-off is lower than the setting during

landing. The reasons for this difference are as follows.

Equation (10.17) shows that the take-off run depends on ambient density ρ ,

wing loading (W/S), maximum lift coefficient (CLmax) and the average accelerating

force. Out of these parameters, as pointed out earlier, the values of (W/S) and

(T/W) are chosen based on considerations of cruise, maximum speed etc. In this

situation, the choices available to reduce the take-off distance are (a) CLmax and

(b) average accelerating force during the take-off.

It may be pointed out at this juncture that a high value of CLTO would reduce V1

and hence the take-off run (Eq.10.17). However, the high value of CLTO would

also result in high value of CD and consequently high value of drag and a lower

accelerating force. This would tend to increase the take-off run (Eq.10.17). On

account of these two opposing effects, there is an optimum value of C LTO and

the corresponding flap setting, that would result in lowest take-off run.

On the other hand, during landing the approach speed and the touch down

speed would be lowest when the CLmax is highest. Further, the high value of CD

associated with high value of CLmax would also increase the decelerating force

during landing run and consequently reduce it. Thus a high value of CLmax is

beneficial for reducing the landing run & distance.

Keeping these two aspects in view, the flap setting during the take-off is

lower than that during the landing. As a guideline it is mentioned in Ref.3.15,

chapter 5, that the flap deflection for take-off f TOδ is about half of that during

landing f Landδ .The deflection of the leading edge slat during take-off, is about

two-thirds of that during landing.

It may be further added that during landing run, after all the landing gear

wheels have touched ground, the lift is not needed. Hence, in airplanes with



provision of spoilers, they (spoilers) are deployed during the landing run to

reduce the lift and increase the drag.

Acknowledgements

The major portion of the lecture material was prepared when the author was an

AICTE Emeritus fellow at IIT Madras. Support of AICTE and IIT Madras is

greatfully acknowledged. He is also grateful to Prof.J.Kurian, Prof.P.Sriram and

Prof.K.Bhaskar, the Heads of the department of Aerospace engineering, IIT

Madras and to Prof. K. Mangala Sunder, Co-ordinator NPTEL, and Prof.S.R.

Chakravarthy, Co-ordinator for Aerospace Engineering, NPTEL, IIT Madras for

providing facilities to carry-out the work.

The lecture material in powerpoint format was reviewed by Prof. K.

Sudhakar, Dept.of Aerospace Engg. , IIT Bombay, Prof.C.V.R. Murti, formerly of

IIT Kanpur and now at Institue of Aeronautical Engg. near Hyderabad,

Prof. B.S.M. Augustine, Sathyabama University, Prof.K.Elangovan ,Dept.of

Aeronautical Engg., M.I.T., Chennai, Prof. R.Rajasekhar, Park college of

Engg.&Technology, Coimbatore and Mr.K.Ibrahim , former chief deisgner, HAL.

The author is indebted to them for their comments which helped in considerably

improving the text. Prof.C.V.R. Murti made detailed comments and even went

through the revised draft. Special thanks are due to him.

The lecture material in the running matter format was reviewed by two reviewers

selected by NPTEL. The comments by the reviewers, helped in adding new

topics and giving explanatory notes. Author’s wife, Mrs. Suniti, also went through

the lecture material and her comments helped in refining the text. The author is

thankful to these persons.

The help of Mr. Amudan Arun Kumar and Mr.S.Ananth former B.Tech

students, Mr.Aditya Sourabh, Dual Degree student, Mr.M.Mahendran, M.S.

scholar, Mr.S.Gurusideswar, Ph.D. scholar and Sandip Chajjed, Project staff

Department of Aerospace Engineering and Ms. K. Sujatha of NPTEL Web studio,

IIT Madras is gratefully acknowledged.



The author would like to record appreciation of Mr. G.Manikandasivam of

NPTEL Web studio, IIT Madras for painstakingly keying in several revisions of

the lecture material and also for preparing figures suitable for conversion to PDF

format.



Chapter 10

References

10.1 Federal Aviation Regulations (FAR), Federal Aviation Administration,

Washington D.C. USA.

10.2 Loftin , Jr. L.K. “Subsonic aircraft evolution and the matching of size to

performance” NASA Reference publications ,1060, August 1980. This report can

be downloaded from the site “NASA Technical Report Server (NTRS)“ .



Chapter 10

Exercises

10.1 Describe the various phase of take-off flight, Write down the equations of

motion during take-off run. Taking CD, C

L and T as constant during take-off run

show that the ground run (s1) is given by:

W Γs = ln1 S g ρ (C -μ C ) Γ -qD L 1

where W, S, g, and have the usual meanings, q1 = dynamic pressure at the

unstick point and

T -μWΓ =

S (C -μ C )D L.

10.2 A rocket motor firing for a short duration of say 10 s is proposed to be used

to reduce the take of run. Explain that a larger reduction in the take-off distance

would be achieved by using the rocket motor in the later part of the take-off run

than in the beginning of the take-off run.

10.3 Describe various methods to reduce the take-off distance and landing

distance.

APPENDIX - A

PERFORMANCE ANALYSIS OF A PISTON ENGINED AIRPLANE – PIPER CHEROKEE PA-28-180

(Lectures 35 - 37)

E.G. TULAPURKARA

S. ANANTH TEJAS M. KULKARNI

REPORT NO: AE TR 2007-1

FEBRUARY 2007 (REVISED OCTOBER 2011)

1

Performance Analysis of a piston engined airplane – Piper Cherokee PA-28-180

E.G.Tulapurkara*, S Ananth$ and Tejas M Kulkarni$

ABSTRACT

The report is intended to serve as an example of performance calculation of a typical piston

engined airplane.

Problem statement: Obtain the following for the prescribed airplane:

Information about the airplane.

Drag Polar at cruising speed and during take-off condition.

Engine Characteristics.

Variation of stalling speed with altitude for flaps up and flaps down

conditions.

Variations of the maximum speed (Vmax) and minimum speed (Vmin) with altitude.

Variations of maximum rate of climb (R/C)max and maximum angle of climb (γmax) with

speed and altitude. Variation of VR/Cmax and Vγmax with altitude. Values of absolute

ceiling and service ceiling.

Variations of range and endurance with flight speed in constant velocity

flights at cruising altitude. Speeds corresponding to Rmax and Emax .

Variation of minimum radius of turn (rmin ) and maximum rate of turn

( maxψ ) at selected altitudes and variations of (Vrmin ) and ( ψmaxV ) with

altitude.

Take-off and landing distances.

* AICTE Emeritus Fellow, Department of Aerospace Engineering, IIT Madras

$ Third year B.Tech students, Department of Aerospace Engineering, IIT Madras

2

Contents

1 Information about the airplane

1.1 Overall dimensions

1.2 Power plant

1.3 Weights

1.4 Wing geometry

1.5 Fuselage geometry

1.6 Horizontal tail geometry

1.7 Vertical tail geometry

1.8 Landing gear

1.9 Flight condition

1.10 Performance of PA-28-181 as given in Ref.3*

2 Estimation of drag polar

2.1 Estimation of DOWBC

2.2 Estimation of DOHC

2.3 Estimation of DOVC

2.4 Estimation of DLGC and DMISCC

2.5 Cooling drag and leakage drag

2.6 Estimation of parasite drag coefficient DOC

2.7 Estimation of induced drag coefficient DiC

2.8 Expression for drag polar during cruise

2.8.1 Slight modification of drag polar

2.9 Expression for drag polar during take-off condition

* Reference numbers in this Appendix relate to those given at the end of this appendix.

3

3 Engine characteristics

3.1 Variation of engine BHP

3.2 Thrust horsepower available

4 Steady level flight

4.1 Variation of stalling speed with altitude

4.2 Variations of Vmax and Vmin with altitude

5 Steady climb performance

6 Range and endurance

6.1 Estimation of range in constant velocity flight

6.2 Calculation of BHP and fuel flow rate at different RPM’s and MAP’s at 8000

6.3 Sample calculations for obtaining optimum N and MAP for a chosen flight

velocity (V)

7 Turning performance

8 Take-off and landing distance estimates

8.1 Distance covered during take-off run

8.2 Distance covered during transition

8.3 Distance covered during climb phase

8.4 Landing distance estimate

9 Concluding remarks

Acknowledgements

References

4

Appendix A

Lecture 35

Performance analysis of a piston engined airplane –1

Topics

1 Information about the airplane

1.1 Overall dimensions

1.2 Power plant

1.3 Weights

1.4 Wing geometry




1.8 Landing gear


1.10 Performance of PA-28-181 as given in Ref.3


2.1 Estimation of DOWBC

2.2 Estimation of DOHC

2.3 Estimation of DOVC

2.4 Estimation of DLGC and DMISCC


2.6 Estimation of parasite drag coefficient DOC

2.7 Estimation of induced drag coefficient DiC


2.8.1 Slight modification of drag polar

5


1. Information about the airplane

Airframe: Piper Cherokee PA-28-180

Type: Piston-engined propeller driven low speed recreational airplane.

Manufacturer and country of origin: The Piper Airplane Corporation, USA.

1.1 Overall dimensions*

Length : 7.148 m

Wing span : 9.144 m

Height above ground : 2.217 m

Wheel base : 1.897 m

Wheel track : 3.048 m

1.2 Power plant

Name : Lycoming O-360-A3A

Rating : 180BHP (135 kW) at 2700 RPM

Weight : 129 kgf (1265.5 N)

Number : 1

Propeller : 1.88 m diameter, fixed pitch.

1.3 Weights

Maximum take-off weight : 1088 kgf (10673.28 N)

Empty weight : 558 kgf (5473.98 N)

Fuel capacity : 50 US gallons (189 litres) usable 178.63 litres

Payload : 468.1 kgf (4592.06 N)

Maximum wing loading : 73.2 kgf/m2 (718.1 N/m2)

Maximum power loading (P/W) : 0.1241 kW/kgf (0.01265 kW/N)

1.4 Wing geometry

Planform shape : Trapezoidal near root, rectangular afterwards

and elliptical fillets at the tip.

Span (b) : 9.144 m

Reference area (S or SRef) : 14.864 m2

* The dimensions / areas are based on Fig.1 and the additional details given in Ref.2.

6

Flap area : 1.384 m2

Aileron area : 1.003 m2

Airfoil : NACA 652– 415, t/c = 15 %, Clopt = 0.4

Root chord : 2.123 m

Tip chord : 1.600 m

Quarter chord Sweep : 1.480

Dihedral : 60

Twist : -20

Incidence : 4.620 at root, 2.620 at tip

High lift devices : Simple flaps having 3 different settings : 100 ,

250 and 400

Derived parameters of wing:

(i) Aspect ratio (A ) :

A = b2/ S = 9.144 2 / (14.864) = 5.625

(ii) Root chord of equivalent tropazoidal wing (creq) :

req t

bS= (c +c )

2

Or 14.864 = req

9.144(c +1.60)

2

creq= 1.651 m

(iii) Root chord of exposed wing (cre):

From Fig.1, the maximum fuselage width is 1.168 m. Hence semi span of the exposed wing

(be / 2) is:

eb 1= (9.144-1.168)=3.988m

2 2

(iv) The root chord of exposed equivalent wing (cre) is obtained as follows.

An expression for the chord of the equivalent wing is

yc =1.651- (1.651-1.600)

b/2

Hence,

re

0.584c =1.651- (1.651-1.600)=1.644m

9.144/2

7

(v) Taper ratio of the exposed wing (λe) is:

λe= 1.6 / 1.644 = 0.9732

(vi) Mean aerodynamic chord of the exposed wing ( ec )

2 2e e

e ree

(1+λ +λ )2 2 (1+0.9732+0.9732 )c = c = ×1.644

3 1+λ 3 1+0.9732

=1.622 m

(vii) Planform area of the exposed wing (Se) is:

Se = 3.988 (1.644+1.6) = 12.937 m2

(viii) Wetted area of exposed wing (Swet)e is :

(Swet)e = 2 Se {1+1.2 x (t/c) } = 2 x 12.937 { 1 + 1.2 x 0.15} = 30.53 m2


Length (lb) : 6.547 m (measured from Fig.1)

Frontal area (Sb) : 1.412 m2 (Ref.2 p.179)

Maximum width : 1.168 m

Derived parameters for fuselage:

(i) Equivalent diameter (de) of fuselage :

2e e

πd =1.412 or d =1.341m

4

(ii) Height of maximum cross section (hmax)

hmax = 1.412 / 1.168 = 1.209 m.

(iii) Rough estimate of wetted area of fuselage (Ss)e is : (Ss)e = 0.75 x ( perimeter of the maximum cross section ) x lb = 0.75 (1.209 + 1.168) x 2 x 6.547 = 23.34 m2. (iv) Fineness ratio of fuselage (Af) : Af = lb/ de = 6.547 / 1.341 = 4.882


Plan-form shape : Rectangular with elliptical fillets at tips.

Span : 3.048 m

Area : 2.267 m2

Root chord and tip chord : 0.762 m

Airfoil : NACA 0012.

8

Derived parameters of horizontal tail: (i)Aspect ratio = At = 3.0482 / 2.267 = 4.098 (ii)Exposed area of horizontal tail = area of h.tail – area inside fuselage ≈ 2.15 m2 Hence wetted area of h.tail (Swet)h is : (Swet)h : = 2 x 2.15 [1+1.2 x 0.12] = 4.919 m2 1.7 Vertical tail geometry

Span : 1.219 m

Area : 1.059 m2

Root chord : 1.182 m

Tip chord : 0.517 m

Quarter chord sweep : 21.80

Airfoil : NACA 0010.

Derived parameters of vertical tail: (i) Taper ratio : 0.4374

(ii) Aspect ratio : 1.403

(iii)Exposed area of vertical tail : same as area of v.tail = 1.059 m2

(iv) Wetted area of v.tail (Swet)v is :

(Swet)v = 2 x 1.059 { 1+ 1.2 x 0.1} = 2.372 m2

(v) Mean aerodynamic chord of vertical tail is :

Vtc = (2/3) x 1.182 x (1+0.4374 + 0.43742 ) /(1+0.4374) = 0.893 m.

1.8 Landing gear

Type : Non-retractable, nose wheel type with fairing.

Number of wheels : Nose 1, main 2, all same size.

Thickness : 0.135 m

Diameter : 0.4547 m

Wheel base : 1.897 m

Wheel track : 3.048 m

9


Altitude : 2438 m (8000 )

Density : 0.9629 kg/m3

Speed of sound : 330.9 m/s

Kinematic viscosity ( υ ) : 40.17792 10 (m2/s)

Flight speed : 237 km/hr (65.83 m/s)

Mach number : 0.1992

Weight of the airplane : 1088 kgf (10673.28 N)

1.10 Performance of PA-28-181$ as given in Ref.3.

Maximum take-off weight : 1157 kgf (2550 lbf)

Power plant rating : 135 kW (180 BHP)

Wing loading : 73.3kgf/ m2

Maximum level speed : 246 kmph

Cruising speed : 237 kmph

Stalling speed : 86 kmph, with flaps down condition

Maximum rate of climb : 203 m/min at sea level

Service ceiling : 4035 m

Take-off run : 350 m

Take-off to 15m : 488 m

Landing run : 280 m

Landing distance from 15m : 427 m

Range with allowance for taxi, take-off, climb, descent and 45 min reserves at 6000 feet

(1830 m) : 924 km at 55 % power ; 875 km at 65 % power ; 820 km at 75 % power. $Remark: The performance calculations are being done for PA-28-180 as a large

amount of data on the airplane, the engine and the propeller are available in Ref.2. However,

information on actual performance of this airplane is not given there. Ref.3 (which is easily

accessible) contains information about PA-28-181 which is only slightly different from

PA-28-180.

10

Fig.1. Three-view drawing of Piper Cherokee PA-28-180

Dimensions in m

11

2. Estimation of drag polar Following Ref.1, the drag polar is assumed to be of the following form.

2

2LD Do Do L

CC =C + = C +KC

πAe (1)

Do DoWB DoV DoH DoMiscC =C +C +C +C (2)

where suffixes WB, V, H and Misc denote wing-body combination, vertical tail, horizontal tail

and miscellaneous items respectively.

2.1 Estimation of CDOWB

From Ref.1, section 3.1.1, at low subsonic Mach number, DoWBC is given by the following

expression.

4 wet eDoWB fw LS

Ref

S eB BfB WB Db3

f Ref Ref

(S )t tC ={C [1+L( )+100( ) ]R +

c c S

(S )l S60C [1+ +0.0025( )] }R +C

(l /d) d S S

(3)

Cfw = skin friction drag coefficient of wing (see below).

L = 1.2 when (t/c)max of the airfoil used on the wing is located at (x/c) 0.3, which is the case

here.

t/c = 0.15.

RLS = 1.07 from Fig. 3.3 of Ref.1; note M < 0.25 and = 0.

(Swet)e = 30.53 m2

Sref = 14.864 m2

CfB = skin friction drag of fuselage (see below)

lb/de= 4.882

(Ss)e = 23.34 m2

RWB = wing - body interference correction factor (see below)

CDb = base drag coefficient.Base drag contribution is neglected as the fuselage

gradually tapers down to zero width.

12

Skin friction drag of wing (Cfw):

This quantity depends on the lower of the two Reynolds numbers viz.

(i) Reynolds number based on mean aerodynamic chord of exposed wing e(c ) and (ii) cut-off

Reynolds number (Recut-off) based on the roughness of the surface.

Reynolds number based on ec is :

Re = (1.622 x 65.83) / (0.17792 x 10-4) = 6 x 106

The roughness parameter is (l/k) where l is the reference chord, here 1.662 m.

The value of k corresponding to mass production point, from Ref.1, is :

3.048 x 10-5 m. Hence, l/k = 1.622/(3.048 x 10-5) = 53215

Corresponding to this value of (l/k), Recut-off from Fig 3.2 of Ref.1 is 4 x 106.

Since Recut-off is lower, Cfw depends on it. Corresponding to Recutoff , the value of Cfw from

Fig .3.1 of Ref.1 is 0.00348

Skin friction coefficient of fuselage (CfB):

The Reynolds number based on length of the fuselage (RlB) is:

RlB = 6.547 x 65.83 / (0.17792 x 10-4) = 24.22 x 106

In this case l/k = 6.547/(3.048 x 10-5) = 2.14 x 105

Recut-off in this case,Fig 3.2 of Ref.1, is : 18 x 106

Hence CfB, from Fig 3.1 of Ref.1, is 0.00272

RWB : From Fig 3.5 of Ref.1, for M < 0.25 and RlB = 24.22 x 106, RWB = 1.06.

13

Hence,

4DWB

3

30.53C ={0.00348[1+1.2(0.15)+100(0.15) ]×1.07×

14.86460 23.34

+0.00272[1+ +0.0025×4.882] }1.06+04.882 14.864

30.53 23.34

={0.00348[1+0.18+0.051]×1.07× +0.00272[1+0.5156+0.0122]× }1.0614.864 14.864

={0.00941+0.006525}×1.06=0.009975+0.006917=0.01689

2.2 Estimation of CDoH

The drag coefficient of horizontal tail is given by (Ref.1) as:

wet4 h

DoH fH LSref

(S )t tC = C [1+L( )+100( ) ]R

c c S (4)

The tail has NACA 0012 airfoil. Hence, t/c = 0.12

The wetted surface area of horizontal tail (Swet)h = 4.919 m2

Sref = 14.864 m2

The mean aerodynamic chord of exposed horizontal tail is taken equal to the root chord of the

horizontal tail i.e. etc = 0.762 m.

Reynolds number based on etc is :

0.762 x 65.83 / (0.17792 x 10 -4) = 2.52 x 106

The value of l/k in this case is:

0.762/(3.048 x 10-5) =25000

Hence, Recutoff =1.5 x 106

Consequently, CfH = 0.00414

For = 0 and M< 0.25, RLS =1.07

Finally, CDoH = 0.00414[1+1.2 x 0.12 + 100(0.12)4]1.07x4.919/14.864 = 0.00171

14

2.3 Estimation of CDoV

The drag coefficient CDoV is given by:

wet4 v

DoV fV LSref

(S )t tC = C [1+L( )+100( ) ]R

c c S (5)

The vertical tail has NACA 0010 airfoil; Hence, t/c=0.10

Wetted surface area of vertical tail = 2.372 m2

Sref = 14.564 m2

Reynolds number based on Vtc is:

0.893 x 65.83 /(0.17792 x 10-4)=3.30 x 106

The value of l/k is 0.893/(3.048 x 10-4) = 29298

Recutoff = 1.9 x 106

Consequently, CfV = 0.00394

Corresponding to M < 0.25 & = 21.8°, RLS=1.07

Finally, CDOV = 0.00394[1+1.2 x 0.1 + 100(0.1)4] x1.07x2.372/14.864

= 0.00076

2.4 Estimation of CDOLG and CDOMisc

The landing gear drag coefficient can be obtained from Ref.1. However, the values for Piper

Cherokee given in Ref.2 are used as guidelines. Table 4.3 of Ref.2 indicates that parasite area of

landing gears components would be (a) wheel strut 0.19 ft2, (b) wheels 0.44 ft2 (c) wheel pants

0.40 ft2 (see remark on p.180 of Ref.2). Thus, parasite drag area of landing gear would be:

0.19 + 0.44 + 0.4 = 1.03 ft2 = 0.0957 m2

Again from Table 4.3 of Ref.2 The sum of the parasite drag areas of miscellaneous items like

beacon, antennas etc is 0.52 ft2 or 0.0483 m2. Thus,

CDOLG + CDOMisc = (0.0957 + 0.0483)/14.864 = 0.00645+0.00325 = 0.00970

15

Remarks:

i) Reference 7, chapter 5 mentions that the drag of landing gear (CDLG)without fairing is

about 35% of the sum of the drags of major components viz. wing-body, horizontal

tail and vertical tail. For landing gear with fairings, CDLG would be about 25% of the

aforesaid sum. In the present case :

CDWB + CDHT + CDVT is (0.01689 + 0.00171 + 0.00076 = 0.01936).Thus CDLG of

0.00645, estimated above appears reasonable.

ii) The value of CDmisc of 0.00325 is about 17% of the aforesaid sum and appears

reasonable.


These drags are important for piston engined airplanes. Appendix A of Ref.7 gives some

guidelines. However, Ref.2, p.179 mentions that the sum of the two drags could be

approximately taken into account by multiplying the sum of all the other drags by a factor of 1.2.

2.6 Estimation of parasite drag coefficient (CDO)

In light of the above discussion CDo can be expressed as:

CDO = 1.2 (CDOWB + CDOHT + CDOVT + CDOLG + CDOMisc) (7)

In the present case,

CDO = 1.2(0.01689+0.00171+0.00076+0.00645+0.00325)

= 1.2 x 0.02905 = 0.0349 (8)

Remark:

For comments on the above value of CDO see remark at the end section 2.8

2.7 Estimation of induced drag coefficient (CDi)

The quantity K in Eq.(1) is given by:

1

K =πAe

16

where, A = Aspect ratio of the wing, e = Oswald efficiency factor

Following Ref.1 section 2.3.1 the Oswald efficiency factor is expressed as:

wing fuselage other

1 1 1 1= + +

e e e e (9)

Figure 2.4 of Ref.1 presents ewing for unswept wings of rectangular and tapered planforms. In the

present case the taper ratio ( ) is almost unity. The value of ewing for rectangular wing of

A = 5.625 is 0.845.

Further, for a fuselage of rectangular cross section and wing of aspect ratio 5.625, Fig.2.5 of

Ref.1 gives:

bb

fuse ref

S1( ) =1.6 ; S = frontalarea of fuselage

e S

Or fuse

1 1.412=1.6× = 0.152

e 14.864

Again from Ref 1, other

1= 0.05

e

Consequently,

1 1

= +0.152+.05=1.3854e 0.845

e = 0.722

Hence, 1

K = = 0.0784π×5.625×0.722

CDi = 0.0784 2LC (10)


Combining Eqs.(8) and (10) gives the drag polar in cruise condition as:

CD = 0.0349 + 0.0784 2LC (11)

17

The value of (L/D)max is given by

max

DO

1(L/D) =

2 C K

Substituting the values of CDO and K from Eq.(11) gives:

max

1(L/D) = = 9.56

2 0.0349×0.0784

2.8.1 Slight modification of the expression for drag polar

The value of (L/D)max is an indication of the aerodynamic efficiency of the airplane.

From Ref.7 chapter 3 it is observed that the value of (L/D)max for Piper Cherokee is slightly more

than 10. Thus, the estimated value of 9.56 is lower than that of the actual airplane and suggests

need for slight modification. References 8 and 9 give the values of CDO and e for similar

airplanes, with non-retracted landing gear, made by manufacturers of Piper, Cessna and Beech

aircraft. These values are presented in Table 1.

Airplane A CDO L/D e

Piper Cherokee 6.02 0.0358 10 0.758

Piper J-3 cub 5.81 0.0373 9.6 0.75

Cessna Skyhawk 7.32 0.0317 11.6 0.747

Beechcraft D17S 6.84 0.0348 10.8 0.76

Table 1 Values of A, CDO, (L/D) and e for similar airplane

From Table 1 it is seen that the estimated value of CDO in the present case appears reasonable.

However, the value of e should perhaps be higher, say 0.76 . With CDO of 0.0349 and e = 0.76

the drag polar becomes :

2D L

1C = 0.0349+ C

π×5.625×0.76

Or 2D LC = 0.0349+0.0755C (12)

The expression in Eq.(12) would give (L/D)max of 9.81.

18

Remarks:

i) It may be added that Piper Cherokee is an airplane famous in its class but is of older design. The current trend is to have (a) smoother surfaces which would reduce CDO to about 0.032 and (b) wing of larger aspect ratio of 8 and above, which would give K of around 0.053. These would give (L/D)max of in excess of 12.

ii) For subsequent calculations, the following expression for drag polar is used.

2D LC = 0.0349+0.0755C


To obtain the drag polar under take-off condition, the flight velocity is taken as 1.2 Vs, where Vs

is the stalling speed with flaps in take-off condition (δf =100). In the present case, CLmax with 100

flap deflection, from Ref.2 is 1.42. Hence,

s

2×10673.28V = = 28.73m/s

1.42×1.225×14.864

Consequently, VTo = 1.2 28.73 = 34.47m/s

Reynolds number based on mean aerodynamic chord of the exposed wing in take-off condition

is:

6-6

1.622×34.47 = 3.83×10

14.6×10

We notice that this Reynolds number is very close to the cutoff Reynolds number for the wing

(4 106) obtained in Section 2.1. Thus, the value of Cf and other calculations will remain almost

the same. Hence, DoC for the airplane in take-off condition, without the flap, can be taken as

0.0349.

Similarly K, without the flap, can be taken as 0.0755.

The correction to the drag polar for flap deflection, is carried-out using the following steps.

The flap type is plain flap.

From Fig.1, the ratio of flap chord to wing chord is 0.16 and flap deflection is 100 .

The ratio of the area of the flapped portion of the wing to the wing plan-form area is 0.4827.

The ratio of the span of the flapped portion of the wing (including the fuselage width) to the total

span is 0.604.

The ratio of the fuselage width to the wing span is 0.127; the wing aspect ratio is 5.625.

Following Ref.1, section 3.4.1

19

Dflap = Dp Di DintΔC ΔC + ΔC + ΔC ,

where, DpΔC = increase in profile drag coefficient due to flaps,

DiΔC = increase in induced drag coefficient due to flaps and

DintΔC = increase in interference drag due to flaps.

The increment in CLmax due to 100 flap deflection, Δ CLmax, as noted earlier, is 0.09.

Using these data and interpolating the curves given in Ref.1, section 3.4.1, gives dpC , the

increment in the drag coefficient of airfoil due to flap deflection, as 0.008. Hence,

DpΔC = dpΔC x (area of flapped portion of the wing/ wing area)

= 0.008 x 0.4827 = 0.0038

According to Ref.1, the increase in induced drag coefficient ( DiΔC ) due to flap deflection is

2 2LmaxfΔK ×ΔC . Using Ref.1, section 3.4.1 fΔK is estimated as 0.163.

Consequently, 2 2Di ΔC = 0.163 ×0.09 = 0.00022

The interference drag due to deflection, of plain flaps is negligible.

Thus, the parasite drag coefficient in take-off condition is

DoC = 0.0349+0.0038+0.00022= 0.03892 0.0389

Hence, the drag polar in take-off condition is given by:

2D LC =0.0389+0.0755C (13)

Remarks:

i) In the approach just presented, to estimate the drag polar in take-off condition, the change in

the induced drag coefficient is included in the parasite drag coefficient. When the flap

deflections are large, the change in the induced drag can be accounted for by reducing the value

of the Ostwald efficiency factor (e) by 0.05 for take-off condition and 0.1 for landing condition

(Ref.4 section 3.4.1). Equations 12 and 13 are the drag polars for cruise condition and take-off

condition respectively. The polars are presented in Fig.2.

ii) It may be pointed out that the parabolic drag polar is not valid beyond CLmax. It is only

approximate near CL = 0 and CL = CLmax.

20

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0 0.05 0.1 0.15 0.2

Drag coefficient

Lif

t co

effi

cien

t

Cruise conditionTake-off condition

Fig.2 Drag polars at cruise and take-off conditions

Appendix A

Lecture 36

Performance analysis of a piston engined airplane – 2

Topics




4 Steady level flight

4.1 Variation of stalling speed with altitude

4.2 Variations of Vmax and Vmin with altitude



6.1 Estimation of range in constant velocity flight


6.3 Sample calculations for obtaining optimum N and MAP for a chosen flight

velocity (V)

1


Model : Lycoming O-360-A3A.

Type : Air-cooled, carbureted, four-cylinder, horizontally opposed

piston engine.

Sea level power : 180 BHP (135 kW)

Propeller : 74 inches (1.88 m) diameter

The variations of power output and fuel consumption with altitude and rpm are shown in Fig.3.

For the present calculations, the values will be converted into SI units.

Fig.3 Characteristics of Lycoming O-360-A (with permission from Lycoming aircraft engines )

2


The variation of available engine output (BHPa) with altitude is assumed to be of the form:

aBHP = sealevelBHP (1.13σ -0.13)

where σ is the density ratio = sLρ/ρ .

The power outputs of the engine at select altitudes are given in Table 2 and plotted in Fig.4.

Note: At a given altitude, the variation of engine BHP with flight speed is very slight and is

generally neglected.

h(m) σ BHPa (kW)

0 1 135.00

1000 0.9075 120.89

2000 0.8217 107.80

3000 0.7423 95.69

4000 0.6689 84.49

5000 0.6012 74.16

5500 0.5691 69.27

6000 0.538 64.52

6500 0.5093 60.14

7000 0.4812 55.86

Table 2 Variation of BHP with altitude

3

Fig.4 Variation of BHP with altitude at maximum power condition


The available thrust horsepower is obtained as product of a pBHP × η , where pη is the propeller

efficiency. The propeller efficiency ( pη ) depends on the flight speed, rpm of the engine and the

diameter of the propeller. It can be worked out at different speeds and altitudes using the

propeller charts. However, chapter 6 of Ref.2 gives an estimated curve of efficiency as a function

of the advance ratio V

(J = )nD

for the fixed pitch propeller used in the present airplane. This

variation is shown as data points in Fig.5.

It may be added that this variation of pη with J is used in chapter 6 of Ref.2, to estimate the drag

of Piper Cherokee airplane from measurements in flight. In another application, in Ref.10,

chapter 17, the same variation is used to compare the performance of fixed pitch and variable

pitch propellers. Based on these two applications, it is assumed here that the variation of pη with

J shown in Fig.5, can be used at all altitudes and speed relevant to this airplane.

For the purpose of calculating the airplane performance, an equation can be fitted to the pη vs J

curve in Fig.5. A fourth degree polynomial for pη in terms of J is as follows.

4 3 2pη (J) = -2.071895J +3.841567J -3.6786J +2.5586J-0.0051668 (14)

4

It is seen that the fit is very close to the data points. The dotted portions are extrapolations.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 0.2 0.4 0.6 0.8 1 1.2

Advance ratio (J)

Pro

pel

ler

effi

cien

cy

Fig.5 Variation of propeller efficiency with advance ratio For the calculation of maximum speed, maximum rate of climb and maximum rate of turn, it is

convenient to have maximum power available a p(THP =η ×BHP) as a function of velocity. The

maximum power occurs at 2700 rpm (45 rps). Noting the propeller diameter as 1.88m, the ηP vs J

curve can be converted to pη vs V curve (Fig.6).

The expression for ηP in terms of velocity is as follows.

-8 4 -6 3 -4 2 -2pη = -4.0447×10 V + 6.3445 ×10 V - 5.1398×10 V +3.0244× 10 V - 0.0051668 (15)

Curve fit (Eq.14)

Data from Ref.2

5

Fig.6 Variation of propeller efficiency with velocity at 2700rpm

Making use of the power available at different altitudes as given in Table 2 and the values of the

propeller efficiency at different speeds given by Eq.(15), the maximum available thrust

horsepower a p(THP = η ×BHP) can be obtained at different speeds and altitudes. The variations

are plotted in Fig.7.

6

0.00

20.00

40.00

60.00

80.00

100.00

120.00

0 10 20 30 40 50 60 70 80

Velocity (m/s)

TH

Pa

(k

W)

Sea level

1000 m

2000 m

3000 m

4000 m

5000 m

5500 m

Fig.7 Variations of THPa with altitude

4 Steady level flight 4.1 Variation of stalling speed with altitude

Fig.8 Forces on an airplane in steady level flight In steady level flight, the equations of motion are:

T - D = 0 (16)

L - W = 0 (17)

7

Further,

2L

1L= ρV SC =W

2 (18)

2D

1T = D = ρV SC

2 (19)

L

2WV=

ρSC

Since CL cannot exceed CLmax , there is a flight speed below which the level flight is not

possible. The flight speed at which CL equals CLmax is called the stalling speed and is denoted

by Vs.

Hence,

sLmax

2WV =

ρSC

Since density decreases with altitude, the stalling speed increases with height.

In the present case, W = 1088 9.81 = 10673.28 N and S = 14.864 m2.

As regards LmaxC , Reference 2 gives the values of LmaxC as 1.33, 1.42, 1.70 and 1.86 for flap

deflections of o0 , o10 , o25 and 40o respectively.

Using these data, the variations of stalling speeds with altitude are presented in Table 3 and

plotted in Fig.9.

H(m) σ Vs (δf = 0o)

(m/s)

Vs (δf =10o)

(m/s)

Vs (δf =25o)

(m/s)

Vs (δf = 40o)

(m/s)

0 1.000 29.69 28.73 26.26 25.10

1000 0.908 31.16 30.16 27.57 26.35

2000 0.822 32.75 31.70 28.97 27.69

3000 0.742 34.46 33.35 30.48 29.14

4000 0.669 36.30 35.13 32.11 30.70

4500 0.634 37.28 36.08 32.97 31.52

5000 0.601 38.29 37.06 33.87 32.38

5500 0.569 39.36 38.09 34.81 33.28

6000 0.538 40.46 39.16 35.79 34.22

Table 3 Stalling speeds for various flap settings

8

Fig.9 Variations of stalling speed with altitude for different flap settings

4.2 Variations of Vmax and Vmin with altitude With a parabolic drag polar and the engine output given by an analytical expression, the

following procedure gives Vmax and Vmin. Available power is denoted by Pa and power required

to overcome drag is denoted by Pr. At maximum speed in steady level flight, required power

equals available power.

a pP = BHP×η (20)

3r D

D×V 1P = = ρV SC

1000 2000

The drag polar expresses CD in terms of CL. Writing CL as 2

2W

ρSV and substituting in the above

equations we get:

23

p DO

1 KWBHP×η = ρV SC +

2000 500ρSV. (21)

The propeller efficiency has already been expressed as a fourth order polynomial function of

velocity and at a chosen altitude, BHP is constant with velocity. Their product ( pη BHP) gives

0

1000

2000

3000

4000

5000

6000

0 10 20 30 40 50

Stalling speed (m/s)

No flap

Flap deflection 10 degrees



Alt

itu

de

(m)

9

an analytical expression for power available. Substituting this expression on the left hand side of

Eq.(21) and solving gives maxV and min e(V ) at at the chosen altitude. Repeating the procedure at

different altitudes, we get maxV and min e(V ) at various heights. Sample calculations and the plot

for sea level conditions are presented in Table 4 and Fig.10.

V (m/s)

ηp Pa

(kW) Pr(kW)

0 0.000 0.000 -

5 0.134 18.086 188.983

10 0.252 33.995 94.789

15 0.352 47.549 64.053

20 0.438 59.185 49.778

25 0.513 69.259 42.753

30 0.578 78.045 40.069

35 0.635 85.735 40.615

40 0.685 92.438 43.953

45 0.727 98.184 49.947

50 0.762 102.918 58.611

55 0.789 106.503 70.040

60 0.805 108.724 84.376

65 0.809 109.280 101.792

70 0.798 107.790 122.480

Table 4 Steady level flight calculations at sea level

10

0

20

40

60

80

100

120

140

160

180

200

0 10 20 30 40 50 60 70 80

Velocity (m/s)

Po

wer

(k

W)

PaPr

Fig.10 Sample plot for Pa and Pr at sea level

It may be noted that

The minimum speed so obtained corresponds to that limited by power min e(V ) .

If this minimum speed is less than the stalling speed, a level flight is not possible at this

speed. The minimum velocity is thus higher of the stalling speed and min e(V ) .

The results for VS , min e(V ) , Vmin and max(V ) at various altitudes are tabulated in Table 5 and

plotted in Fig.11. It may be noted that at h = 5200 m, maxV and min e(V ) are same. This altitude is

the maximum height attainable by the airplane and will be referred later as absolute ceiling.

A

B

Point A: (Vmin)e

Point B: Vmax

11

Table 5 VS , (Vmin)e, Vmin and maxV at various altitudes

Fig.11 Variations of maximum and minimum flight velocities with altitude

Remark:

The calculated value of maxV of 240.6 kmph at sea level is fairly close to the value of

246 kmph of the actual airplane quoted in section 1.10.

h

(m)

Vs(no flap)

(m/s)

(Vmin)e

(m/s)

Vmin

(m/s)

Vmax

(m/s)

Vmax

(kmph)

0 29.7 18 29.7 66.84 240.624

1000 31.2 20.4 31.2 65.75 236.7

2000 32.75 23.3 32.75 64.3 231.48

3000 34.46 27 34.46 62.3 224.28

4000 36.3 32 36.3 59.15 212.94

5000 38.29 41 41 52.7 189.72

5200 38.73 46.5 46.5 46.5 167.4

12


Fig.12 Forces on an airplane in steady climb

Calculation of rate of climb:

In this flight, the C.G of the airplane moves along a straight line inclined to the horizontal at an

angle γ . The velocity of flight is assumed to be constant during the climb.

Since the flight is steady, acceleration is zero and the equations of motion can be written as:

T -D-W sin γ = 0 (22) L - W cos γ = 0 (23)

Noting that CL = 2L/ρV2S = 2

2 W cos γ

ρSV, gives:

2

D Do 2

2 W cos γC = C +K ( )

ρSV

Also cV = Vsin γ , or cVsin γ =

V

2

c2

Vcos γ = 1-

V

Substituting in Eq.(22) gives :

222 c c

a DO2

V V1 KWT = ρV S C + 1- + W

12 V VρV S2

13

Or 2

c cV VA + B( ) + C = 0

V V

(24) where,

22

a Do2

KW 1A = , B = - W and C = T - ρV SC -A

1 2ρV S2

;

Ta = available thrust = 1000 x Pa/V. The available thrust horsepower is given by the following expression:

Pa = sealevel pBHP (1.13σ -0.13)η

Equation 24 gives 2 values of cV

V. The value which is less than 1.0 is chosen as appropriate.

Consequently,

-1 cVγ= sin

V (25)

cV = Vsin γ (26)

The climb performance is calculated using following steps.

(i) Choose an altitude.

(ii) Choose a velocity between Vmin and Vmax and obtain A, B and C in Eq.(24).

(iii) Solve for cV

V, obtain γ and cV .

(iv) Repeat calculations, at chosen altitude, at various velocities in the range of Vmin to

Vmax.

(v) Repeat steps (i) to (iv) at various altitudes.

Sample calculations at sea level are presented in Table 6.

14

V (m/s)

ηp

THPa

(kW)

T

(N)

A C Vc/V γ

(deg.)

Vc

(m/s)

Vc

(m/min)

30 0.578 78.04 2601.49 1049.68 1265.85 0.120 6.894 3.60 216.03

35 0.635 85.73 2449.56 771.19 1289.14 0.122 7.000 4.26 255.89

40 0.685 92.43 2310.96 590.44 1212.13 0.114 6.563 4.57 274.29

45 0.727 98.18 2181.86 466.52 1071.92 0.101 5.790 4.53 272.36

50 0.762 102.91 2058.35 377.88 886.123 0.083 4.777 4.16 249.80

55 0.789 106.50 1936.42 312.30 662.971 0.062 3.568 3.42 205.35

60 0.805 108.72 1812.06 262.42 405.797 0.038 2.181 2.28 137.00

65 0.809 109.28 1681.23 223.60 115.193 0.011 0.619 0.70 42.10

Note: B = - W = -10673.28 N

Table 6 Steady climb calculations at sea level.

Repeating similar calculations at various altitudes gives the variations of and Vc with velocity

at different altitudes. The results are plotted in Figs.13 and 14. From these figures the variations

of maxγ , Vcmax or (R/C)max, γmaxV and R/CmaxV at various altitudes are obtained. The results are

presented in Table 7 and in Figs.15, 16 and 17.

15

Fig.13 Variations of angle of climb with flight velocity at different altitudes

Fig.14 Variations of rate of climb with flight velocity at different altitudes

16

h (m) maxγ (deg) Vcmax (m/min)

γmaxV (m/s) R/CmaxV (m/s)

0 7 276 34.1 41.7

1000 5.4 219.7 35 42.6

2000 3.83 165.8 38 43.6

3000 2.5 111.7 40.9 45

4000 1.28 60.5 44 45.9

5000 0.2 10 46 46.5

5200 0 0 46.5 46.5

Table 7 Climb performance

Fig.15 Variation of maximum angle of climb with altitude

17

Fig.16 Variation of maximum rate of climb with altitude

Fig.17 Variations of Vγmax and V(R/C)max with altitude

Remark:

It is observed that the maximum rate of climb and maximum angle of climb decrease with

altitude, but the velocity at which the rate of climb and angle of climb are maximum increase

slightly with height.

18

Service ceiling and absolute ceiling

The altitude at which the maximum rate of climb becomes 100 ft/min (30.5 m/min) is called the

service ceiling and the altitude at which the maximum rate of climb becomes zero is called the

absolute ceiling of the airplane. These can be obtained from Fig.16. It is observed that the

absolute ceiling is 5200 m and the service ceiling is 4610 m. It may be pointed out that the

absolute ceiling obtained from maxR/C consideration and that from maxV consideration are same

(as they should be). Further, the service ceiling of 4610 m is close to the value of 4035 m for the

actual airplane quoted in section 1.10.


6.1 Estimation of range in a constant velocity flight It is convenient for the pilot to cruise at constant velocity. Hence, the range performance in

constant velocity flights is considered here. In such a flight at a given altitude, the range (R) of a

piston-engined airplane is given by the following expression (Eq.7.23 of the main text of the

course).

p p-1 -1 -11 1 2max

max L1 1 1 2 1 2 1 2

7200 η 3600ηE ζ W WR = E tan = [ tan - tan ] (27)

BSFC 2E (1-KC E ζ) BSFC k k k /k k /k

where, 21 Do 2 1 22

1 2Kk = ρV SC ,k = , W and W

2 ρSV are the weights of the airplane at the start and

end of the cruise, L1 1 2max 1 L1 2

D1 1DO

C 2W W1E = , E = , C = , ζ =1-

C ρSV W2 C K,

D1C = DC corresponding to L1C .

From this expression the range and endurance in constant velocity flights, can be obtained at

different flight speeds, at the cruising altitude. From this information, the flight speeds which

would give the maximum range and endurance can be arrived at. It may be pointed that the

above expression gives the gross still air range as defined in section 7.2.3 of the main text of the

course.

The following values are taken as the common data for the subsequent calculations.

Cruising altitude = 8000 (2478 m)

W1 = Weight at the start of range flight = maximum gross weight = 10673.28 N

Usable fuel = 178.63 litre = 1331.78 N of petrol

19

W2 = Weight at the end of the flight = 10673.28 – 1331.78 = 9341.5 N

Wing area (S) = 14.864 m2, CDO = 0.0349, K = 0.0755

ρ= density at 8000 = 0.9629 kg/m3

The power required during a constant velocity flight varies as the fuel is consumed. However,

for the purpose of present calculations the power required is taken as the average of power

required at the start and end of cruise. It is denoted as THPavg . It is noted that the power required

(THPavg) can be delivered by the engine operating at different settings of RPM (N) and manifold

air pressure (MAP). But, for each of these settings the propeller efficiency ( pη ) and fuel flow

rate would be different. The optimum setting, which would give the maximum range, can be

arrived at by using the following steps.

(a) Select a value of N and calculate J (= V/nd); n = N/60 .

(b) Obtain pη corresponding to this value of J from Eq.(14).

(c) Then, BHP required (BHPr) = THPavg / pη

(d) The left hand side of Fig 4.2 of the main text, shows the BHP vs MAP and fuel flow rate vs

MAP curves with rpm as parameter. Similar curves are generated for h = 8000 .

(e) From the curves in step (d) the sets of N and MAP values which would give desired BHPr can

be obtained.

(f) Obtain fuel flow rate for each set of MAP and N. Calculate BSFC. Subsequently Eq.(27)

gives the range for chosen set of N and MAP.

(g) Repeat calculations at different value of N.

(h) The combination of N and MAP which gives longest range is the optimum setting.

The aforesaid steps are carried-out in the next three subsections.


Example 4.2 of the main text illustrates the procedure to obtain BHP and fuel flow rate at

N = 2200 and MAP of 20 of Hg. Similar calculations are repeated at N = 2700, 2600, 2400,

2200 and 2000 and at MAP = 15, 16, 17, 18, 19, 20, 21 and 21.6 of Hg. It may be pointed out

that the atmosphere pressure at 8000 is 21.6 of Hg. (see also right hand side of the engine

characteristics shown in Fig.3 of this Appendix). The values so obtained are plotted and

smoothed. Figure 18 shows the calculated values by symbols and the smoothed variations by

curves.

20

Fig.18 Variations of BHP and fuel flow rate with MAP

6.3 Sample calculations for obtaining optimum N and MAP for a chosen flight velocity (V)

For the purpose of illustration V is chosen as 50 m/s or 180 kmph

I) Calculation of THPavg :

21

CL1 = Lift coefficient at start of range 12

10673.280.5966

0.5 0.9629 50 14.8642

W1ρV S

2

CL2 = Lift coefficient at the end of cruise 12

9341.50.5220

0.5 0.9629 50 14.8642

W1ρV S

2

CD1 = CD corresponding to CL1 = 0.0349 + 0.0755 20.5966 = 0.06177

CD2 = CD corresponding to CL2 = 0.0349 + 0.0755 20.5220 = 0.05547

CDavg = (0.06177 + 0.05547)/2 = 0.05862

THPavg = 2Davg

1ρV SC /1000

2= 0.5 x 0.9629 x 503 x 14.864 x 0.05862/1000 = 52.43 kW

= 70.31 HP

II) Steps to obtain highest pη /BSFC or the range

(a) Choose ‘N’ from 2700 to 2000

(b) Calculate, J = V/nd ; n = N/60 ; d = 1.88 m

(c) Corresponding to the value of J in step (b), obtain pη from Eq.(14)

(d) Obtain BHPr = THPavg/ pη ; THPavg in HP

(e) From upper part of Fig.18, obtain MAP which would give BHPr at chosen N. For these values

of N and MAP obtain the fuel flow rate (FFR) in gallons/hr , from the lower part of Fig.18.

(f) Convert FFR in gallons per hour to that in N/hr and BHPr in HP to kW.

Obtain BSFC = FFR in N/hr

BHPin kW

(g) Obtain pη /BSFC and also range from Eq.(27).

The above calculations at different values of N are presented in the table below.

N (RPM)

J ηp BHP (HP)

MAP FFR (gal/hr)

FFR (N/hr)

BHP (kW)

BSFC (N/kW-hr)

ηp/BSFC Range (km)

2700 0.591 0.762 92.22 15.90 8.32 234.58 68.77 3.410 0.223 1023.82600 0.613 0.773 90.88 16.10 7.92 223.44 67.77 3.297 0.234 1074.82400 0.664 0.794 88.54 16.47 7.38 208.07 66.02 3.151 0.251 1154.22200 0.725 0.807 87.03 17.04 6.95 196.06 64.90 3.020 0.267 1224.92000 0.797 0.806 87.22 18.25 6.97 196.56 65.04 3.021 0.266 1221.8

Table 8 Sample calculation at V = 180 kmph

22

It is observed from the above table that at the chosen value of V =180 kmph, the range is

maximum for the combination of N = 2200 and MAP of 17.04 of Hg. The value of R is

1224.9 km.

III Obtaining range and endurance at different flight speeds

Repeating the calculations indicated in item (II), at different values of flight speeds in the

range of speeds Vstall from Vmax at 8000 , yield the results presented in Table 8a. Since the flight

speed is constant, the endurance (E) is given by the following expression.

E (in hours) =

Range in km

V in kmph

V

(m/s)

V

(kmph)

THPavg

(kW)

RPMopt

MAP

FFR

(gal/hr)

FFR

(N/hr)

ηp

BHP

(kW)

BSFC

(N/kW

‐hr)

Range

(km)

Endur‐

nce(hr)

34 122.4 41.01 2000 16.61 6.254 176.26 0.734 55.859 3.155 929.6 7.59

36 129.6 41.13 2000 16.39 6.16 173.62 0.753 54.597 3.18 999.2 7.66

38 136.8 41.64 2000 16.29 6.12 172.52 0.77 54.069 3.19 1061.1 7.76

40 144 42.51 2000 16.32 6.13 172.79 0.784 54.198 3.19 1114.3 7.74

43 154.8 44.53 2000 16.58 6.24 175.81 0.8 55.643 3.16 1176.5 7.6

46 165.6 47.37 2000 17.1 6.46 182.07 0.808 58.573 3.11 1214.5 7.33

50 180 52.43 2200 17.04 6.95 196.06 0.807 64.904 3.02 1225 6.81

52 187.2 55.53 2200 17.7 7.25 204.4 0.81 68.576 2.98 1222.1 6.52

54 194.4 58.98 2200 18.49 7.63 215.12 0.808 72.97 2.95 1205.5 6.2

56 201.6 62.81 2200 19.43 8.13 228.99 0.803 78.23 2.93 1174.1 5.82

58 208.8 67.03 2200 20.56 8.77 247.02 0.793 84.5 2.92 1127.1 5.4

60 216 71.63 2400 20.42 9.37 264.14 0.806 88.87 2.97 1090.2 5.05

Table 8a Range and endurance in constant velocity flights at 8000 (2438 m)

The results are plotted in Figs.19 and 20.

23

Fig.19 Range in constant velocity flights at h = 8000 (2438 m)

Fig.20 Endurance in constant velocity flights at h = 8000 (2438 m)

Remarks:

i) It is seen that the maximum endurance of 7.7 hours occurs in the speed range of

125 to 145 kmph.

24

ii) The range calculated in the present computation is the Gross Still Air Range (GSAR).

The maximum range is found to be around 1220 km which occurs in the speed range of

165 to 185 kmph.

iii) The range quoted in Section 1.10 for Cherokee PA – 28 - 181 accounts for taxi, take-off,

climb, descent and reserves for 45 min. This range can be regarded as safe range. This

value is generally two-thirds of the GSAR. Noting that two-thirds of GSAR (1220 km) is

813 km, it is seen that the calculated value is within the range of performance given in

Section 1.10.

Appendix A

Lecture 37

Performance analysis of a piston engined airplane – 3

Topics



8.1 Distance covered during take-off run

8.2 Distance covered during transition

8.3 Distance covered during climb phase

8.4 Landing distance estimate


Acknowledgements

1


Fig.21 Forces on an airplane in turning flight

In this section, the performance of the airplane in a steady level co-ordinated-turn is studied.

The forces acting on the airplane are shown in Fig.21.

The equations of motion in this flight are:

T – D = 0, as it is a steady flight (28)

W – L cos = 0, as it is a level flight (29)

2W VLsin =

g r , as it is a co-ordinated-turn (30)

These equations give:

Radius of turn = 2 2W V V

r = =g Lsin g tan

(31)

Rate of turn = 2V V g tan

ψ = = V =r g tan V

(32)

2

Load Factor = n = L 1

=W cos

(33)

In the following calculations, LmaxC =1.33 and nmax = 3.5 are assumed ; where maxn is the

maximum load factor for which the airplane is designed. The following procedure is used to

obtain minr and maxψ .

1. The flight speed and altitude are chosen. The lift coefficient in level flight ( LLC ) is

obtained as :

LL 2

2(W/S)C =

ρV

2. Obtain Lmax

LL

C

C. If Lmax

maxLL

Cn

C , then the turn is limited by LmaxC and LT1 LmaxC = C .

However, if Lmax LL maxC /C > n , then the turn is limited by maxn , and LT1C = maxn LLC .

3. From the drag polar, DT1C is obtained corresponding to LT1C . Then,

2T1 DT1

1D = ρV SC

2

If T1D > aT , where aT is the available thrust at chosen speed and altitude, then the turn is

limited by the engine output. The maximum permissible value of DC in this case is found

from:

aDT 2

2TC =

ρV S

From the drag polar, the value of LTC is calculated as:

DT DOLT

C - CC =

K

However, if T1D < aT , then the turn is not limited by the engine output and the value of CLT1

calculated in step (2) is taken as CLT.

4. Once LTC is known, the load factor n, which satisfies the three constraints namely of

LmaxC , maxn and aT , is given by:

3

LT

LL

Cn =

C

5. Knowing n, the values of the radius of turn (r) and the rate of turn (ψ ) can be calculated

from Eqs.(31), (32) and (33).

6. The above steps are repeated for various speeds at the same altitude and subsequently the

procedure is repeated at various altitudes.

Sample calculations of turning performance at sea level are represented in Table 9. Figures

22 and 23 present turning performance at various altitudes.

V (m/s)

CLL

Lmax

LL

C

C CLT1

CDT1

DT1 (N)

THP1

(kW) ηp

THPa

(kW) CLT

n

(deg) r (m)

ψ

(rad/s)

30 1.30 1.021 1.33 0.168 1380 41.4 0.578 78.0 1.33 1.02 11.6 445 0.067

35 0.96 1.390 1.33 0.168 1879 65.8 0.635 85.7 1.33 1.39 44.0 129 0.270

38 0.81 1.638 1.33 0.168 2215 84.2 0.666 89.9 1.33 1.64 52.4 113 0.335

40 0.73 1.815 1.33 0.168 2454 98.2 0.685 92.4 1.28 1.75 55.1 114 0.351

45 0.58 2.297 1.33 0.168 3106 139.8 0.727 98.2 1.05 1.82 56.6 136 0.330

50 0.47 2.836 1.33 0.168 3834 191.7 0.762 102.9 0.86 1.83 56.9 166 0.300

55 0.39 3.432 1.33 0.168 4639 255.2 0.789 106.5 0.69 1.77 55.5 212 0.260

60 0.33 4.084 1.14 0.133 4359 261.5 0.805 108.7 0.52 1.60 51.2 295 0.203

65 0.28 4.793 0.97 0.106 4082 265.3 0.809 109.3 0.34 1.23 35.7 600 0.108

Table 9 Turning performance calculations at sea level

4

Fig.22 Variations of radius of turn with velocity at various altitudes

Fig.23 Variations of rate of turn with velocity at various altitudes From Figs.22 and 23 the values of rmin , maxψ , Vrmin , ψmaxV can be obtained at various

altitudes. The variations are presented in Table 10 and Figs.24, 25 and 26.

5

h (m) rmin(m) maxψ (rad/s) Vrmin(m/s)

ψmaxV (m/s)

0 110 0.351 38 40.0

1000 135 0.301 39 41.2

2000 163 0.248 39.5 41.5

3000 198 0.194 40.5 41.7

4000 324 0.128 41 44.0

5000 918 0.048 45.7 46.0

Table 10 Turning performance

Fig.24 Variation of minimum radius of turn with altitude

6

Fig.25 Variation of maximum rate of turn with altitude

Fig.26 Variations of rminV and ψmaxV with altitude

Remark:

The minimum radius of turn at sea level is about 110 m at flight speed of about 38 m/s. The

maximum rate of turn at sea level is about 0.35 rad/sec at flight speed of about 40 m/s.

7


Take-off flight can be divided into three phases: take-off run or ground run, transition and

climb (Fig.27).

Fig.27 Phases of take-off flight

8.1 Distance covered during take-off run (s1)

The equations of motion during the take off run are:

W dV

T -D-μR =g dt

and R = W - L (34)

where R is the ground reaction. The acceleration can be written as:

dV g

= ×[T-D-μ(W-L)]dt W

Writing dV

dtas

dV ds×

ds dt, gives :

W VdV

ds =g T -D-μ(W -L)

Further, at sea level, BHP = constant = 135kW at 2700 rpm. Thrust is given by :

T = BHP x pη / V .

The distance covered during the take-off run (s1) can be expressed as:

TOV

1

0

W Vs = dV

g F (35)

where F is the accelerating force given by:

8

pBHP ηF = - D-μ (W-L)

V

Since pη is a function of velocity, an accurate way to estimate 1s is to evaluate the integrand in

Eq.(35) at several values of V and carry-out a numerical integration. Simpson’s rule is used for

this purpose. Various quantities needed for the purpose are estimated below.

TOV =1.2 sV , where sV is stalling speed, given by :

sLmax

2WV =

ρSC

During the take-off, flap deflection ( fδ ) is10 , hence LmaxC =1.42 . It is assumed that the

coefficient of friction is 0.02.

The take-off weight is W =10673.28 N

S = wing planform area =14.864 2m

Density ρsl =1.225 kg/m3

Thus, sV = 28.73m/s and TOV = 34.48 m/s

To estimate CL and CD during take -off run it is noted that the airplane has a nose wheel type of

landing gear and hence the airplane axis can be considered as horizontal and the wing produces

lift corresponding to the wing setting angle (see Section 10.3.1 of the main text of the course

material).

From Sec. 1.4, the average wing incidence is the average of incidence at root (4.62o) and that at

tip (2.62o) i.e. 3.62o. The slope of the lift curve of the wing ( LαWC ) is approximately given by:

LαW

A 5.625C = 2π = 2π = 4.63/ rad = 0.0808/ deg

A+2 7.625

The angle of zero lift ( 0Lα ) for the airfoil NACA 652 – 415, from Ref. 5, is -2.6o.

Hence, lift coefficient during take-off run due to wing lift is :

0.0808(3.62+ 2.6) = 0.502

Since, the flaps are deflected during the take-off run the lift coefficient will be increased by (1.42

– 1.33 = 0.09). Hence, CL during take-off run ( LtrC ) is :

LtrC = 0.502+0.09= 0.592

The drag coefficient during take-off run (CDtr) , using the drag polar corresponding to take-off,

is:

9

2DtrC = 0.0389+0.0755×0.592 = 0.0654

For applying Simpson’s rule in this case, the various quantities are evaluated at seven points in

the speed range of 0 to 34.48. The calculations are shown in Table 11.

V

(m/s) pη T

(N)

D

(N)

L

(N)

F

(N)

WV/gF

(s)

0 0 * 0 0 - 0.000

5.75 0.153 3589.94 19.68 132.9 3364.45 1.859

11.49 0.283 3329.77 78.60 711.5 3051.9 4.096

17.24 0.392 3072.83 176.9 1601.3 2714.5 6.910

22.99 0.484 2843.74 314.7 2848.7 2372.5 10.54

28.73 0.562 2642.65 491.4 4448.2 2026.7 15.42

34.48 0.629 2464.68 707.8 6407.0 1671.6 22.42

* The value of thrust (T) at V = 0 is not zero. It can be evaluated using propeller charts. However, it is not needed in the present calculation, as the integrand is zero when V is zero.

Table 11 Evaluation of integrand in Eq.(35).

Using the values of integrand in Table 11 and employing Simpson’s rule the ground run (s1) is

given by :

1

5.747s = 0+ 4(1.859+6.91+15.42) +2(4.096+10.54) + 22.42 = 284.4

3 m

8.2 Distance covered during transition (s2)

The entire power of the engine is assumed to be used to overcome the drag and to accelerate to a

velocity V2 given by V2 = 1.1 VT0. The height attained during the transition phase is ignored.

Hence, 2 22 2 2 TO

WTs = Ds + (V -V )

2g

2 22 TO

2

(V -V )Ws =

2g T -D

where T and D are evaluated at a speed which is mean (Vavg) of 2V and TOV

V2 = 1.1 x 34.48 = 37.93 m/s

TO 2avg

V + V 34.48+37.93V = = = 36.71m/s

2 2

10

p

avg

η ×BHP×1000T =

V

From Eq.(15), pη at a speed of 36.71 m/s is 0.65285

Hence, 0.65285×135×1000

T = = 2400.5 N36.71

L 2

2×10673.28Further, C = = 0.87

1.225×14.864×36.71

2DC = 0.0389+0.0755×0.87 = 0.096

21D = ×1.225×36.71 ×14.864×0.096=1178.4 N

2

Hence, 2 2 2 22 TO

2

V -VW 10673.28(37.93 -34.48 )s = = =111.3m

2g T-D 2×9.81(2400.5-1178.4)

8.3 Distance covered during climb phase (s3)

The airplane is assumed to climb to screen height (15m) at an angle of climb γ,

where the climb angle γ is given by:

T-D

γ = ( )W

For the climb phase, T and D are evaluated at V2 which is equal to 37.93 m/s

From Eq.(15), pη at a speed of 37.93 m/s is 0.665. Hence,

0.665×135×1000

T = = 2366.86 N37.93

2×10673.28

C = = 0.82L 21.225×14.864×37.93

2DC = 0.0389+0.0755×0.82 = 0.0897

21D = ×1.225×36.71 ×14.864×0.0897 =1174.5 N

2

oT -D 2366.86 - 1174.5sinγ = = = 0.1117 or γ = 6.41

W 10673.28

3

15 15Hence, s = = = 133.4m

tan γ 0.1124

11

Total takeoff distance is given by:

1 2 3s = s +s +s = 284.4+111.3+133.4 = 529.1 m

Remarks:

i) The above estimation of take off distance is based on several assumptions. Reference 8 has

compiled data on take-off distances of many propeller driven airplanes. This take-off distance is

based on FAR 23 specifications and can be denoted by sto23 . Based on this data the following

formula is obtained for sto23 in terms of a parameter called take-off parameter and denoted by

TOP23 . In SI units the relationship is given as (See Guidelines for take-off distance in

Section 10.4.7 of the main text of the course) .

STo23 = 8.681 x 10-3 x TOP23 + 5.566 x 10-8 x 223TOP

where, 223

LTO

W W( )×( )

S PTOP = ; (W/S) is in N/m , W in N and Pin kW.σC

σ is density ratio at the altitude of take-off.

In the present case:

W/S = 10673.28 / 14.864 = 718.1 N/m2, W/P = 10673.28 / 135 = 79.06 N/ kW

σ = 1.0 and CLTO = 1.42.

Consequently,

23

718.1×79.06TOP = =39981

1.42

-3 -8 2to23 s = 8.681 x 10 x 39981+ 5.566 x 10 x 39981 =347 +89= 436 m

ii) The estimated take-off distance of 530 m is somewhat higher than the actual take-off distance

of 488 m (section 1.10). This may be because the height attained during the transition phase has

been ignored.

12

8.4 Estimation of landing distance

The landing distance can be calculated in a manner similar to that for take-off distance.

However, due to uncertainty associated with piloting techniques during landing, the following

formula is used.

2

aland

Vs = -

2a

where, aV = 1.3 s×V in landing configuration

The weight of the airplane during the landing is taken same as that during the take-off. However,

LmaxC with landing flap setting is 1.86. The stalling speed in this configuration is 25.1 m/s.

Hence, approach speed is 32.6 m/s.Taking a = -1.22m/ s2 for a simple braking system yields:

sland = 436 m,

which is close to the value of 426 m given in Section 1.10.


1. The performance of a piston-engined airplane has been estimated for stalling speed, maximum

speed, minimum speed, steady climb, range, endurance, turning, take-off and landing.

2. A reasonable agreement has been observed between the calculated performance and the actual

performance of the airplane (PA – 28 – 181).

3. Figure 28 presents the variations, with altitude, of the characteristic velocities corresponding

to:

Stalling speed Vs

Maximum speed Vmax

Minimum speed as dictated by power (Vmin)e

Maximum rate of climb VR/C max

Maximum angle of climb Vγmax

Maximum rate of turn ψmaxV

Minimum radius of turn Vr min

13

0

1000

2000

3000

4000

5000

6000

0 20 40 60

Velocity (m/s)

Alt

itu

de

(m)

V for minimum radius of turn

V for maximum rate of turn

Stalling speed

Vmin from engine output

V max

V for maximum angle of climb

V for maximum rate of climb

Fig.28 Variations of characteristic velocities with altitude

Acknowledgements

The first author (EGT) thanks AICTE for the fellowship which enabled him to carry out the work

at IIT Madras. He is grateful for the support given by Prof.J.Kurian, Prof.P.Sriram,

Prof.K.Bhaskar, the Heads of the department of Aerospace engineering, IIT Madras. The help

rendered, while carrying out the revision, by Mr.Aditya Sourabh, Dual Degree student,

Mr.S.Gurusideswar, Ph.D. scholar and Sandip Chajjed, Project staff Department of Aerospace

Engineering and Ms. K. Sujatha and Mr. G. Manikandasivam of NPTEL Web studio is gratefully

acknowledged.

Appendix A

References

1.Roskam, J. “Methods for estimating drag polars of subsonic airplanes” Roskam aviation

engineering corporation, Kansas,U.S.A,(1983).

2. McCormick B.W. “Aerodynamics, aeronautics and flight mechanics”, John Wiley,

New York, (1979 First edition, 1995 Second edition).

3. Jackson, P. (editor-in-chief) “Jane’s all the world’s airplane (1999-2000)” Jane’s

information group ltd, Surrey, U.K.

4. Roskam, J. “Airplane design Vol. I”, Roskam aviation engineering corporation, Kansas,

U.S.A, (1989).

5. Perkins C.D. and Hage R.E. “Airplane performance stability and control”, John Wiley,

(1960).

6. Torenbeek. E. “Synthesis of subsonic airplane design” Delft University Press (1981).

7. Raymer, D.P. “Aircraft design: a conceptual approach” AIAA` educational series fourth

edition, (2006).

8. Loftin, Jr. L.K. “Subsonic aircraft evolution and the matching of size to performance”

NASA Reference publications, 1060, August 1980. This report can be downloaded from

the site “NASA Technical Report Server (NTRS)”.

9. Samoylovitch, O. and Strelets, D. “Determination of the Oswald efficiency factor at

airplane design preliminary stage”, Aircraft Design, Vol. 3, pp. 167-174, (2000).

10. Nicholai, L.M. and Carichner, G.E. “Fundamentals of aircraft and airship design Vol. I –

Aircraft design” AIAA educational series (2010).

APPENDIX- B

PERFORMANCE ESTIMATION OF A TYPICAL

SUBSONIC JET TRANSPORT AIRPLANE

(Lectures 38 – 40)

E.G.TULAPURKARA

V.GANESH

REPORT NO: AE TR 2007-2

FEBRUARY 2007

(REVISED OCTOBER 2011)

1

Performance estimation of a typical subsonic

jet transport airplane

E.G.Tulapurkara*V.Ganesh $

February 2007

(Revised November 2009)

Abstract

This report contains details of the performance estimation of a medium range jet

airplane similar to B737. The following aspects are considered.

Drag polar estimation

Engine characteristics

Level flight performance - stalling speed, maximum and minimum speeds

Steady climb performance – maximum rate of climb, maximum angle of climb,

service ceiling and absolute ceiling

Range and endurance

Steady level co-ordinated turn - minimum radius of turn, maximum rate of turn

Take-off and landing distances

The report is intended to serve as an example of performance calculation of a

typical jet airplane.

* AICTE Emeritus Fellow, Department of Aerospace Engineering, IIT Madras $ Dual Degree Student, Department of Aerospace Engineering, IIT Madras

2

Contents 1 Airplane details 1.1 Overall dimensions 1.2 Engine details 1.3 Weights 1.4 Wing geometry 1.5 Fuselage geometry 1.6 Nacelle geometry 1.7 Horizontal tail geometry 1.8 Vertical tail geometry 1.9 Other details 1.10 Flight condition Three-view drawing of the airplane

2 Estimation of drag polar 2.1 Estimation of (CDo)WB 2.2 Estimation of (CDo)V and (CDo)H 2.3 Estmation of misc drag – nacelle 2.4 CDo of the airplane 2.5 Induced drag 2.6 Final drag polar

3 Engine characteristics 4 Level flight performance 4.1 Stalling speed 4.2 Variations of Vmin and Vmax with altitude

5 Steady climb 6 Range and endurance 7 Turning performance 8 Take-off distance 9 Landing distance 10 Concluding remarks Acknowledgements References

3

Appendix B

Lecture 38 Performance analysis of a subsonic jet transport –1 Topics 1 Airplane details 1.1 Overall dimensions 1.2 Engine details 1.3 Weights 1.4 Wing geometry 1.5 Fuselage geometry 1.6 Nacelle geometry 1.7 Horizontal tail geometry 1.8 Vertical tail geometry 1.9 Other details 1.10 Flight condition Three-view drawing of the airplane

2 Estimation of drag polar 2.1 Estimation of (CDo)WB 2.2 Estimation of (CDo)V and (CDo)H 2.3 Estmation of misc drag – nacelle 2.4 CDo of the airplane 2.5 Induced drag 2.6 Final drag polar

4

1 Airplane Details

1.1 Overall Dimensions

Length : 34.32 m

Wing span : 32.22 m

Height above ground : 11.17 m

Wheel base : 13.2 m

Wheel track : 5.8 m

1.2 Engine details

Similar to CFM 56 - 2B

Seal level static thrust : 97.9 kN per engine

By pass ratio : 6.5 (For which the engine characteristics are given

in Ref.3*)

SFC : 0.6 hr-1 at M = 0.8 and h = 10973 m (36000 ft)

1.3 Weights

Gross weight : 59175 kgf (580506.8 N)

Empty weight : 29706 kgf (291415.9 N)

Fuel weight : 12131 kgf (119005.1 N)

Payload : 17338 kgf (170085.8 N)

Maximum landing weight : 50296 kgf (493403.8 N)

1.4 Wing Geometry

Planform shape : Cranked wing

Span : 32.22 m

Area (Sref) : 111.63 m2

Airfoil : NASA - SC(2) series, t/c = 14%,

Clopt = 0.5

Root chord : 5.59 m (Equivalent trapezoidal wing)

Tip chord : 1.34 m (Equivalent trapezoidal wing)

Root chord of cranked wing : 7.44 m

Portion of wing with straight

trailing edge : 11.28 m

* Reference numbers in this Appendix relate to those given on page 40.

5

Mean aerodynamic chord : 3.9 m

Quarter chord sweep : 27.69o

Dihedral : 5 o

Twist : 3 o

Incidence : 1.4 o

Taper ratio : 0.24 (Equivalent trapezoidal wing)

Aspect ratio : 9.3


Length : 33 m

Maximum diameter : 3.59 m

1.6 Nacelle geometry

No. of nacelles : 2

Nacelle diameter : 1.62 m

Cross sectional area : 2.06 m2

Length of nacelle : 3.3 m (based on B737 Nacelle)


Span : 11.98 m

Area : 28.71 m2


Quarter chord sweep : 32 o

Root chord : 3.80 m

Tip chord : 0.99 m

Taper ratio : 0.26

Aspect ratio : 5


Span : 6.58 m

Area : 25.43 m2

Root chord : 5.90 m

Tip chord : 1.83 m


Quarter chord sweep : 37 o

6

Taper ratio : 0.31

Aspect ratio : 1.70

1.9 Other details

CLmax without flap : 1.4

CLmax with landing flaps : 2.7

CLmax with T.O flaps : 2.16

Maximum load factor (nmax ) : 3.5


Altitude : 10973 m (36000 ft)

Mach number : 0.8

Kinematic viscosity : 3.90536 x10-5 m2/s

Density : 0.3639 kg/m3

Speed of sound : 295.07 m/s

Flight speed : 236.056 m/s

Weight of the airplane : 59175 kgf (580506.8 N)

7

Fig.1 Three-view drawing of the airplane

8


The drag polar is assumed to be of the form:

2L

D Do

CC =C +

πAe

The quantity CDO is assumed to be given by:

Do Do WB Do V Do H Do MiscC = (C ) + (C ) + (C ) + (C ) (1)

where suffices WB, V, H, Misc denote wing-body combination, vertical tail,

horizontal tail, and miscellaneous contributions respectively.

2.1 Estimation of (CDo)WB

Initially, the drag polar is obtained at a Mach number of 0.6 as suggested in

Ref.5, section 3.1.2. (CDo)WB is given as :

BDo WB Do w Do B

ref

S(C ) = (C ) + (C )

S

The suffix B denotes fuselage and SB is the maximum frontal area of fuselage.

(CDO)W is given as :

wetDo W fW wing

ref

St(C ) =C 1+L( ) ( )

c S

where, Cfw is the turbulent flat plate skin friction coefficient. The Reynolds number used

to determine it (Cfw) is lower of the two Reynolds numbers viz. Reynolds number based

on the mean aerodynamic chord of the exposed wing (Re) and Recuttoff based on surface

roughness. Further, (Swet)e is the wetted area of the exposed wing.

In the present case, cr = 5.59m, ct= 1.34m, b/2 = 16.11m and dfus = 3.59m. Hence,

Root chord of exposed wing = cre = 5.59 1.34 3.59

5.5916.11 2

= 5.116 m

e

1.34λ = =0.262

5.116

Hence, mean aerodynamic chord of exposed wing ( ec ) is :

2

e

2 1+0.262+0.262c = [5.116( )]

3 1+0.262 = 3.596 m

Span of exposed wing = (b/2)e = 16.11 – 1.795 = 14.315m

9

Further, M = 0.6, a = 295.07m/s. Hence, V = 177.12m/s.

Also ν = 3.90536 X 10-5 m2/s.

Hence,

Re 5

177.12 3.596

3.90536 10

16.31 x 106

The height of roughness corresponding to the standard camouflage paint, average

application, is k = 1.015 x 10-5m (Ref.5, table 3.1). Hence, l/k in this case is:

55

l 3.5963.543 10

k 1.015 10

The Recutoff corresponding to the above l/k is 30 x 106. Consequently, fwC corresponding

to Re = 16.31 x 106 is obtained from Fig.3.1 of Ref.5, as :

fwC = 0.00265.

(t/c)avg = 14% and (t/c)max occurs at x/c > 0.3 Hence, L = 1.2 and

Sexposedplanform = 5.116 1.341

14.314( ) 22

= 92.41m2

wet WS = 2 x 92.41(1+1.2 x 0.14) = 215.8m2

Hence,

(CDf)w = 0.00265 (1+1.2 x 0.14)215.8

111.63 = 0.00598

(CDo)B is given as:

(CDo)B = (CDf)B + (CDp)B + CDb

(CDo)B = b wet basefB fus Db3

b B ref

l S S60C [1+ +0.0025( )]( ) +C

(l /d) d S S

In the present case , lf = 33.0m , dmax = 3.59m ,

Reb = 5

177.12 33

3.905 10

= 149.6 x 106

5

l 33

k 1.015 10

= 32.51x105

The Recutoff corresponding to the above l/k is 2.6 x 108. The Cfw corresponds to

Reb = 149.6 x 106 measured from the graph in Ref.5, Fig.3.1 is:

Cfw = 0.0019

10

(Swet)fus = 0.75 x x 3.59 x 33 = 279m2

SB = 4

x 3.592 = 10.12m2

Hence,

(CDf)B = 0.0019 x 279

10.12 = 0.0524

Dp B 3

60 279(C ) =0.0019[ +0.0025×(33/3.59)]

(33/3.59) 10.12=0.00524

Since, base area is almost zero, CDb is assumed to be zero. Hence,

(CDo)B = 0.0524 + 0.00524 + 0 = 0.0576

D canopy(ΔC ) is taken as 0.002. Hence, (CDo)B = 0.0596

Finally,

(CDo)WB = 0.00598 + 0.059610.12

111.63 = 0.01138

2.2 Estimation of (CDo)V and (CDo)H

The estimation of (CDo)H and (CDo)V can be done in a manner similar to that for the wing.

However, the details regarding the exposed tail area etc. would be needed. In the absence

of the detailed data on the shape of fuselage at rear, a simplified approach given in Ref.5,

section 2.2 is adopted, wherein CDf = 0.0025 for both horizontal and vertical tails.

SW = 2(Sh + Sv)

Hence,

(CDo)HV = 0.0025(28.71 + 25.43) 2

111.63 = 0.0024 (2)

2.3 Estmation of misc drag - nacelle

For calculating drag due to the nacelles the short cut method is used i.e.:

(CDo)nacelle = 0.006 x wet

ref

S

S

where, Swet is the wetted area of nacelle. Here, Swet = 16.79m2. Since, there are two

nacelles, the total drag will be twice of this. Finally,

(CDo)nacelle = 0.006 x 16.79

111.63 x 2 = 0.0018

11

2.4 CDo of the airplane

Taking 2% for miscellaneous roughness and protuberances(Ref.5, section 3.4.6 ), the CDo

of the airplane is:

CDo = 1.02 [0.01138 + 0.0024 + 0.0018] = 0.0159 (3)

2.5 Induced drag

The induced drag component has the Oswald's efficiency factor e which is estimated by

adding the effect of all the airplane components on induced drag (Ref.5, section 2.3).The

rough estimate of e can be obtained as follows :

Figure 2.4 of Ref.5 is useful only for estimating ewing of unswept wings of low speed

airplanes. For the present case of swept wing, the following expression given in Ref.2 ,

chapter 7 is used.

ewing = (ewing)Λ=0 cos(Λ - 5)

where Λ is the quarter chord sweep. Ref.1, chapter 1 is used to estimate (ewing)Λ=0. In

the present case, with A = 9.3 and = 0.24, the value of (ewing)Λ=0 is 0.97.

Hence, ewing = 0.97 x cos (27.69 - 5) = 0.8948.

From Ref.5, section 2.3, fus

f

1/e

(S /S) = 0.8 for a round fuselage. Hence,

fus

1 10.122= 0.8 ×

e 111.63= 0.0725

Further, from Ref.5, section 2.3, other

1= 0.05

e

Finally,

e = -1

1= 0.8064

0.8948 + 0.0725 + 0.05

Hence,

1 1K = =

πAe π × 9.3 × 0.8064 = 0.04244

Remark:

Based on Ref.7, a detailed estimates of ewing and efuselage are given in Ref.5, section 3.3.

For an untwisted wing the value of ewing is given as:

12

Lαwwing

Lαw

1.1(C /A)e =

CR( )+(1-R)π

A

where,

LαW2

12 22

2 2

2πAC =

tanA β

2+ 1+ +4κ β

LαWC = slope of lift curve of wing per radian

A = aspect ratio of wing

R = a factor which depends on (a) Reynolds number based on leading edge radius, (b)

leading edge sweep (ΛLE), (c) Mach number (M), (d) wing aspect ratio (A) and (e) taper

ratio (λ).

2β = 1-M

Λ1/2 = sweep of semi-chord line

= ratio of the slope of lift curve of the airfoil used on wing divided by 2π . It is

generally taken as unity.

In the present case,

M= 0.6, h= 10973 m (36000 ft), V= 177.12 m, -5 2=3.90536×10 m /s , S = 111.63 m2

b = 32.22 m, cre = 5.59 m, ct = 1.34 m, Λ1/4 = 27.69 deg,

Hence, A = 9.3, =0.24, β =0.8 , 1/ 2tan = 0.4589, LEcos 0.8609 ,

Average chord = 3.615 m The airfoil is NASA – SC(2) with 14 % thickness. From Ref.8 the leading edge radius is

3 % of the chord.

From these data:

LαWC = 5.404

RLer = Reynolds number based on leading edge radius = 4.974 x 104

RLer x cot ΛLE x 2 2

LE1-M cos = 7.198 x 105

LE

Aλ

cos= 2.592

13

Corresponding to these data, R = 0.943 is obtained from Ref.5, Fig.3.14. Consequently,

wing

1.1 5.404 / 9.3e

5.4040.943 1 0.943

9.3

= 0.8793

This value of ewing is close to the value of 0.8948 obtained by the simpler approach.

However, detailed approach is recommended for wings with sweep of above 35o.

Reference 7, section 4.5.3 contains guidelines for estimating drag of wing-body-tail

combination with allowance for trim drag.

2.6 Final drag polar

CD = 0.0159 + 0.04244 2LC (4)

The drag polar is presented in Fig.2.

Fig.2 Drag polar at sub-critical Mach numbers

Remarks:

i) The polar given by Eq.(4) is valid at subcritical Mach numbers. The increase in CDO

and K at higher Mach numbers is discussed in section 4.2.

14

ii) The maximum lift to drag ratio ((L/D)max) is given by:

max

Do

1(L/D) =

2 C K

Using CDO and K from Eq.(4), (L/D)max is 19.25, which is typical of modern jet

transport airplanes.

iii) It may be noted that the parabolic polar is an approximation and is not valid beyond

CLmax. It is also not accurate close to CL = 0 and CL = CLmax.

Appendix B

Lecture 39 Performance analysis of a subsonic jet transport –2 Topics 3 Engine characteristics 4 Level flight performance 4.1 Stalling speed 4.2 Variations of Vmin and Vmax with altitude

5 Steady climb


To calculate the performance, the variations of thrust and SFC with speed and altitudes

are needed. Chapter 9 of Ref.3 contains these variations for turbofan engines with various

bypass ratios. The thrust variations versus Mach number with altitude as parameter are

given, in non-dimensional form, for take-off, cruise and climb ratings. The values were

read from those curves, interpolated and later smoothed. The values multiplied by

97.9 kN, the sea level static thrust rating for the chosen engine, are shown in Figs.3 and 4.

Figure 3 also contains (a) the variation of thrust with Mach number at sea level with take-

off rating and (b) variations of climb thrust with Mach number at various altitudes. The

values at h = 38000 ft and 39000 ft are obtained by interpolating the values at 36000 ft

and 40000 ft and are used for computation of performance.

The SFC variation is also given in Ref.3, but is taken as 0.6 hr-1 under cruise

conditions based on the trend shown in Fig.3.3 of Ref.4.

1

Fig.3 Output for single engine – take-off thrust at sea level and climb

thrust at various altitudes.

Fig.4 Output of single engine – cruise thrust at various altitudes

2

4 Level flight performance

Forces on an airplane in steady level flight

In steady level flight, the equations of motion, in standard notation, are:

T - D = 0 (5)

L - W = 0 (6)

2L

1L = W = ρV SC

2 (7)

2D

1D = ρV SC =T

2 (8)

4.1 Stalling speed

In level flight,

L

2WV =

ρSC (9)

Since, CL cannot exceed CLmax, there is a flight speed below which level

flight is not possible. The flight speed at CL = CLmax is called the stalling

speed and is denoted by Vs

smax

2WV =

ρSC (10)

Since, ρ decreases with altitude, Vs increases with height. It may be noted that

3

W/S = 5195 N/m2, CLmax = 2.7 with landing flaps and CLmax = 1.4 without flaps. The

values of stalling speed at different altitudes and flap settings are tabulated in Table 1 and

shown in Fig.5.

h

(m)

ρ

(kg/m3)

Vs

(CLmax = 1.4)

(m/s)

Vs

(CLmax = 2.7)

(m/s)

0

2000

4000

6000

8000

10000

11000

12000

1.225

1.006

0.819

0.659

0.525

0.412

0.363

0.310

77.83

85.86

95.18

106.06

118.87

134.09

142.80

154.52

56.04

61.83

68.54

76.37

85.59

96.56

102.83

111.27

Table 1 Variation of stalling speed with altitude

Fig.5 Stalling speed vs altitude

4

4.2 Variations of Vmin and Vmax with altitude

To determine the Vmin and Vmax at each altitude, the following procedure is adopted. The

engine thrust as a function of velocity at each altitude is obtained from the smoothed data.

The drag at each altitude is obtained as a function of velocity using the drag polar and the

level flight formulae given below.

L 2

2 (W/S)C =

ρV

(11)

2D Do LC C K C (12)

Thrust required = Drag = 2D

1ρV SC

2 (13)

Thrust available = Ta = f(M) (14)

where, CDo = 0.0159 and K = 0.04244.

However, the cruise Mach number (Mcruise) for this airplane is 0.8. Hence, CDo and K are

expected to become functions of Mach number above Mcruise. To get some guidelines

about variations of CDo and K, the drag polars of B-727 given in Volume VI, Chapter 5 of

Ref.6 are considered. These drag polars are shown in the Fig.6 as discrete points.

Fig.6 Drag polars at different Mach numbers for B727-100; Symbols are data from Ref.6

and various lines are the parabolic fits.

5

These polars were approximated by the parabolic polar expression namely

2D Do LC = C + KC . The values of CDo and K at various Mach numbers, obtained by least

square method, are given in the Table 2. The parabolic fits are also shown in Fig.6.

M CDo K

0.7

0.76

0.82

0.84

0.86

0.88

0.01631

0.01634

0.01668

0.01695

0.01733

0.01792

0.04969

0.05257

0.06101

0.06807

0.08183

0.10300

Table 2 Variations of CDo and K with Mach number (Parabolic fit)

The variations of CDo and K with Mach number are plotted in Figs.7 and 8. It is seen that

there is no significant increase in CDo and K upto M = 0.76. This is expected to be the

cruise Mach number for the airplane (B727-100). Following analytical expressions have

been found to closely represent the changes in CDo and K from M = 0.76 to M = 0.86.

CDo = 0.01634 – 0.001 x (M – 0.76) + 0.11x (M – 0.76)2 (15)

K = 0.05257 + (M – 0.76)2 + 20.0 x (M – 0.76)3 (16)

Fig.7 Variation of CDo with Mach number

6

Fig.8 Variation of K with Mach number

In the case of the present airplane, the cruise Mach number is 0.8. The variations of CDo

and K above Mcruise and upto M = 0.9, based on the B727-100 data are taken as follows.

CDo = 0.0159 – 0.001 x (M – 0.80) + 0.11 x (M – 0.80)2 (17)

K = 0.04244 + (M – 0.80)2 + 20.0 x (M – 0.80)3 (18)

The thrust available and thrust required curves are plotted at each altitude as a function of

velocity. The points of intersection give the (Vmin)e and Vmax at each altitude from thrust

available consideration (Figs.9 – 14).

However, to arrive at the minimum speed (Vmin), the stalling speed (Vs) also needs to be

taken in to account. Since, the drag polar is not valid below Vs, in the Figs.9 to 14, the

thrust required curves are plotted only for V ≥ Vs. Stalling speed is taken for CLmax

without flaps.

The calculations are carried out for h = 0, 10000, 15000, 25000, 30000 and 36000 ft, i.e

S.L, 3048, 4572, 7620, 9144 and 10972.8 m using Ta as both climb thrust (Tclimb) and as

cruise thrust (Tcr). Results in Figs.9 – 14 are presented only for climb thrust case. The

variations of Vs, (Vmin)e and Vmax are tabulated in Table 3 and presented in Fig.15.

7

Fig.9 Available and required thrust at S.L

Fig.10 Available and required thrust at h = 3048 m

8



9



10

h (in ft)

h (in m)

Vs (m/s)

(Vmin)e (m/s)

T = Tcr

(Vmin)e (m/s)

T=Tclimb

Vmax(m/s)T = Tcr

Vmax(m/s) T=Tclimb

Vmax(kmph)T=Tclimb

S.L 10000 15000 25000 30000 36000 38000 38995 39220

0 3048 4572 7620 9144 10973 11582 11884 11954

77.833 90.579 98.131 116.292 127.278 142.594 149.557 153.159 153.950

< Vs

< Vs

< Vs

< Vs

< Vs 176.054 217.386 235.48 ----

< Vs

< Vs

< Vs

< Vs

< Vs

169.071 200.896 229.865 236.40

258.711 272.060 275.613 272.929 267.854 253.671 243.676 235.48 ------

269.370 280.595 283.300 279.291 271.755 258.154 248.630 238.649 236.40

969.7 1010.1 1019.9 1005.4 978.3 929.4 895.1 859.1 851.04

Table 3 Variations of Vs, (Vmin)e , Vmin and Vmax

Fig.15 Variations of Vmin and Vmax with altitude

11

5 Steady climb

Forces on an airplane in a steady climb

In this flight, the C.G of the airplane moves along a straight line inclined to the horizontal

at an angle γ . The velocity of flight is assumed to be constant during the climb. Since,

the flight is steady, the acceleration is zero and the equations of motion can be written as:

T - D - W sin = 0 (19)

L - W cos = 0 (20)

To calculate the variation of rate of climb with flight velocity at different altitudes, the

following procedure is adopted.

Choose an altitude.

Choose a flight speed.

Noting that CL = 2W cos γ / SV2, gives :

2

D Do 2

2Wcos γC = C + K

ρSV

Also,

Vc = V sin

Hence, 2c2

Vcos γ = 1-

V

Substituting various quantities in Eq.(19) yields :

12

22

2 c ca DO

2

V V1 KWT = ρV S C + 1- + W

12 V VρV S2

Or 2c cV VA( ) + B( ) +C = 0

V V (21)

2

2

KWA =

1ρV S

2

; B = -W; 2

2a Do a2

1 2KWC = T - ρV SC - ,T Thrust available

2 ρV S (22)

Since, altitude and flight velocity have been chosen, the thrust available is read from the

climb thrust curves in Fig.3. Further, the variation of CDo and K with Mach number is

taken as in Eqs.17 and 18. Equation 21 gives 2 values of Vc/V . The value which is less

than 1.0 is chosen, as sin γ cannot be greater than unity. Hence ,

γ = sin-1(Vc/V) (23)

and Vc = V sin γ (24)

This procedure is repeated for various speeds between Vmin and Vmax. The entire

procedure is then repeated for various altitudes. The variations of (R/C) and γ with

velocity and with altitude as parameter are shown in Figs.16 and 18. The variations of

(R/C)max and γmax with altitude are shown in Figs.17 and 19. The variations of V(R/C)max

and maxV with altitude are shown in Figs.20 and 21. A summary of results is presented in

Table 4.

13

h

(ft)

h

(m)

(R/C)max

(m/min)

V(R/C)max

(m/s)

γmax

(degrees)

maxV

(m/s)

0

10000

15000

25000

30000

36000

38000

38995

39220

0.0

3048.0

4572.0

7620.0

9144.0

10972.8

11582.4

11885.7

11954.0

1086.63

867.34

738.16

487.41

313.43

115.57

41.58

1.88

0

149.7

167.5

174.0

198.2

212.2

236.1

236.9

236.5

236.40

8.7

6.0

4.7

2.6

1.5

0.5

0.2

0.0076

0

88.5

111.6

125.7

164.1

188.0

230.2

234.0

236.0

236.40

Table 4 Climb performance

Fig.16 Rate of climb vs velocity for various altitudes

14

Fig.17 Maximum rate of climb vs altitude

Fig.18 Angle of climb vs velocity for various altitudes

15

Fig.19 Maximum angle of climb vs altitude

Fig.20 Velocity at maximum rate of climb vs altitude

16

Fig.21 Velocity at maximum angle of climb vs altitude

Remarks:

i) The discontinuity in slope in Figs.20 and 21 at high velocities are due to the change in

drag polar as the Mach number exceeds 0.8.

ii) From Fig.17, the absolute ceiling (at which (R/C)max is zero) is 11.95 km. The service

ceiling at which (R/C)max equals 100 ft /min (30.5 m/min) is 11.71 km.

Appendix B

Lecture 40 Performance analysis of a subsonic jet transport – 3 Topics 6 Range and endurance 7 Turning performance 8 Take-off distance 9 Landing distance 10 Concluding remarks Acknowledgements 6 Range and endurance

In this section, the range of the airplane in a constant altitude and constant

velocity cruise is studied. The range is given by the following formula.

1

-1max 1

max L 1

7.2 E V E ζR = tan

TSFC 2E (1-KC E ζ)

(25)

where, O

max

D

1E =

2 K C; K and CDo are at Mach number corresponding to V.

f 2

1 1

W Wζ = = 1-

W W

L11

D1

CE =

C , 1

L12

WC =

1ρV S

2

,

CD1 = Drag coefficient at CL1 and Mach number corresponding to V.

W1 is the weight of the airplane at the start of the cruise and W2 is the weight of the

airplane at the end of the cruise.

The cruising altitude is taken as h = 10973 m (36000 ft). TSFC is taken to be constant as

0.6hr-1. The variation of drag polar above M = 0.8 is given by Eqs.17 and 18.

W1 = Wo = 59175 x 9.81 = 580506.8 N , Wf = 0.205 x W1

Allowing 6% fuel as trapped fuel, W2 becomes

W2 = W1 – 0.94 x Wf or ζ = 0.94 x 0.205 = 0.1927

1

The values of endurance (in hours) are obtained by dividing the expression for range by

3.6V where V is in m/s. The values of range (R) and endurance (E) in flights at different

velocities are presented in Table 5 and are plotted in Figs.22 and 23.

M V

(m/s)

CDo K Emax CL1 CD1 E1 R

(km)

E

(hr)

0.50

0.55

0.60

0.65

0.70

0.75

0.80

0.81

0.82

0.83

0.84

0.85

0.86

0.87

0.88

147.53

162.29

177.04

191.79

206.54

221.30

236.05

239.00

241.95

244.90

247.85

250.80

253.75

256.71

259.66

0.0159

0.0159

0.0159

0.0159

0.0159

0.0159

0.0159

0.0159

0.01592

0.01597

0.01604

0.01613

0.01624

0.01637

0.01652

0.04244

0.04244

0.04244

0.04244

0.04244

0.04244

0.04244

0.04256

0.04300

0.04388

0.04532

0.04744

0.05036

0.05420

0.05908

19.25

19.25

19.25

19.25

19.25

19.25

19.25

19.22

19.11

18.89

18.54

18.08

17.48

16.79

16.00

1.312

1.085

0.911

0.777

0.670

0.583

0.513

0.500

0.488

0.476

0.465

0.454

0.444

0.433

0.424

0.089

0.066

0.051

0.041

0.035

0.030

0.027

0.02654

0.02616

0.02591

0.02584

0.02591

0.02617

0.02653

0.02714

14.75

16.48

17.82

18.72

19.17

19.23

18.95

18.84

18.65

18.37

18.00

17.52

16.97

16.32

15.62

2979.0

3608.0

4189.6

4691.7

5095.6

5396.5

5599.8

5619.7

5621.6

5597.7

5544.1

5460.4

5349.3

5210.1

5051.1

5.61

6.18

6.57

6.80

6.85

6.77

6.59

6.53

6.45

6.35

6.21

6.05

5.86

5.64

5.40

Table 5 Range and endurance in constant velocity flights at h = 10973 m (36000 ft)

2

Fig.22 Range in constant velocity flights at h = 10973 m

Fig.23 Endurance in constant velocity flights at h = 10973 m

3

Remarks:

i) It is observed that the maximum range of 5620 km is obtained around a velocity of

240 m/s (864 kmph). Corresponding Mach number is 0.82 which is slightly higher than

the Mach number beyond which CDo and K increase. This can be explained based on two

factors namely (a) the range increases as the flight speed increases and (b) after Mcruise is

exceeded, CDo and K increase thus, reducing (L/D)max.

ii) The range calculated above is the gross still air range. The safe range would be about

two-thirds of this. In the present case, the safe range would be around 3750 km.

iii) The maximum endurance of 6.85 hours occurs in a flight at V = 206 m/s. (742 kmph).

It is observed that the endurance is roughly constant over a speed range of 190 m/s to 230

m/s (684 to 828 kmph).


Forces acting on an airplane in turning flight

In this section, the performance of the airplane in a steady level, co-ordinated-turn is

studied. The equations of motion in this case are:

T – D = 0

W - L cos = 0

4

L sin = W

g

2V

r

where is the angle of bank.

These equations give:

r = 2V

gtan

V g tan

ψ = =r V

Load factor = n = L 1

=W cos

where, is the rate of turn and r is the radius of turn.

The following procedure is used to obtain rmin andψ max .

1) A flight speed and altitude are chosen and the level flight lift coefficient

CLL is obtained as :

CLL = 2

2(W/S)

ρV

2) If CLmax/CLL < nmax, where nmax is the maximum load factor for which the airplane is

designed, then the turn is limited by CLmax and CLT1 = CLmax. However, if

CLmax/CLL > nmax, then the turn is limited by nmax, and CLT1 = nmaxCLL.

3) From the drag polar, CDT1 is obtained corresponding to CLT1 . Then,

DT1 = 2DT1

1ρV SC

2 .

If DT1 > Ta, where Ta is the available thrust at that speed and altitude, then the turn is

limited by the engine output. In this case, the maximum permissible value of CD in

turning flight is found from

aDT

2

TC =

1ρV S

2

From drag polar, the value of CLT is calculated as

5

DT DoLT

C -CC =

K

However, if DT1 < Ta, then the turn is not limited by the engine output and the value of

CLT calculated in step (2) is retained.

4. Once CLT is known, the load factor during the turn is determined as

LT

LL

Cn =

C

Once n is known, the values of , r and can be calculated using the equations given

above.

The above steps are repeated for various speeds and altitudes. A typical turning flight

performance estimation is presented in Table 6. In these calculations, CLmax = 1.4 and

nmax = 3.5 are assumed. The variation of turning performance with altitude is shown in

Table 7. Figures 24, 25, 26 and 27 respectively present (a) radius of turn vs velocity with

altitude as parameter, (b) Vrmin vs altitude, (c) rate of turn vs velocity with altitude as

parameter and (d) ψmaxV vs altitude.

V

(m/s)

CLL

Lmax

LL

C

C

CLT1

CDT1

T1

(N)

Ta

(N)

CDT

CLT

n

r

(m)

ψ (rad/s)

78.8 1.365 1.026 1.4 0.0991 42106 126250 0.0991 1.4 1.026 12.9 2768 0.0285 98.8 0.868 1.612 1.4 0.0991 66182 118125 0.0991 1.4 1.612 51.7 787 0.1255118.8 0.602 2.331 1.4 0.0991 95678 113750 0.0991 1.4 2.331 64.6 684 0.1738138.8 0.440 3.181 1.4 0.0991 130595 106611 0.0809 1.238 2.813 69.2 747 0.1858158.8 0.336 4.164 1.177 0.0747 128778 101539 0.0589 1.006 2.993 70.5 912 0.1742178.8 0.265 5.279 0.928 0.0525 114709 97041 0.0444 0.819 3.089 71.1 1115 0.1603198.8 0.215 6.527 0.751 0.0398 107635 92606 0.0343 0.661 3.080 71.1 1384 0.1437218.8 0.177 7.905 0.620 0.0322 105461 89483 0.0273 0.519 2.930 70.0 1772 0.1235238.8 0.149 9.415 0.521 0.0274 106860 86229 0.0221 0.383 2.573 67.1 2452 0.0974241.8 0.145 9.655 0.508 0.0268 107282 85779 0.0215 0.362 2.494 66.4 2609 0.0927

Table 6 A typical turning flight performance at sea level

6

Fig.24 Radius of turn vs velocity at various altitudes

Fig.25 Velocity at rmin vs altitude

7

Fig.26 Rate of turn (ψ ) vs speed at various altitudes

Fig.27 Velocity at maxψ vs altitude

8

h

(m)

rmin

(m)

Vrmin

(m/s)

ψ max

(rad/s)

Vψ max

(m/s)

0.0

3048.0

4572.0

7620.0

9144.0

10972.8

666

945

1155

1971

3247

8582

126.8

132.6

135.1

138.3

151.3

211.0

0.1910

0.1410

0.1170

0.0731

0.0513

0.0256

127.8

133.6

136.1

165.3

187.3

231.0

Table 7 Turning flight performance

Remarks:

i) The maximum value of is 0.191 and occurs at a speed of 127.8m/s at sea level.

ii) The minimum radius of turn is 666 m and occurs at a speed of 126.8m/s at sea level.

iii) The various graphs show a discontinuity in slope when the criterion which limits the

turn, changes from nmax to thrust available.

8 Take-off distance

In this section, the take-off performance of the airplane is evaluated. The take-off

distance consists of take-off run, transition and climb to screen height. Rough estimates

of the distance covered in these phases can be obtained by writing down the appropriate

equations of motion. However, the estimates are approximate and Ref.4 chapter 5

recommends the following formulae for take-off distance and balance field length based

on the take-off parameter.

This parameter is defined as:

Take-off parameter = LTo

W/S

σC (T/W) (26)

9

where W/S is wing loading in lb/ft2, CLTO is 0.8 x CLand and is the density ratio at

take-off altitude. In the present case:

W

S = 5195N/m2 = 108.2lb/ft2; CLTO = 0.8 x 2.7 = 2.16; = 1.0(sea level)

and T 2×97900

=W 59175×9.81

= 0.3373

Hence, take-off parameter = 108.2

1.0 2.16 0.3373 = 148.86 (27)

From Ref.4, chapter 5, the take-off distance, over 50', is 2823' or 861m. The balance

field length for the present case of two engined airplane is 6000' or 1829m.

Remark:

It may be noted that the balance field length in this case, is more than twice the take-off

distance.

9 Landing distance

In this section the landing distance of the airplane is calculated. From Ref.4, chapter 5,

the landing distance for commercial airliners is given by the formula:

landLmax

W 1s = 80 ( ) +1000 ft

S σC (28)

where W/S is in lbs/ft2. In the present case:

(W/S)land = 0.85 x (W/S)takeoff = 0.85 x 108.5 = 92.225 lb/ft2

CLmax = 2.7 , = 1.0

Hence,

sland = 80 x 92.225 1

10001.0 2.7

= 3732 ft =1138 m (29)

10


1. Performance of a typical commercial airliner has been estimated for stalling speed,

maximum speed, minimum speed, steady climb, range, endurance, turning, take-off

and landing.

2. The performance approximately corresponds to that of B737-200.

3. Figure 28 presents the variations with altitude of the characteristic velocities

corresponding to :

stalling speed, Vs

maximum speed, Vmax

minimum speed as dictated by thrust, (Vmin)e

maximum rate of climb, V(R/C)max

maximum angle of climb, V max

maximum rate of turn, V max

minimum radius of turn, Vrmin

11

Fig.28 Variations of characteristic velocities with altitude

11 Acknowledgements

The first author (EGT) thanks AICTE for the fellowship which enabled him to carry out

the work at IIT Madras. He is grateful for the support given by Prof.J.Kurian,

Prof.P.Sriram, Prof.K.Bhaskar, the Heads of the department of Aerospace engineering,

IIT Madras. The help rendered, while carrying out the revision, by Ms. K. Sujatha and

Mr. G. Manikandasivam of NPTEL, Web studio is gratefully acknowledged.

Appendix B

References

1. Abbot,I.H. and von Doenhoff, A.E. “Theory of wing sections” Dover publications,

(1959).

2. Hoerner,S.F. “Fluid dynamic drag” Hoerner Fluid Dynamics, Bricktown, U.S.A,

(1965).

3. Jenkinson,L.R., Simpkin,P. and Rhodes,D. “Civil jet airplane design” Elsevier –

Butterworth Heinemann, (1999).

4. Raymer,D.P. “Airplane design : a conceptual approach” AIAA Education series,

AIAA, (2006).

5. Roskam, J. “Methods for estimating drag polars of subsonic airplanes” Roskam

Aviation and Engineering Corporation, Kansas, (1983).

6. Roskam,J. “Airplane design Vol. I to VIII” Roskam Aviation and Engineering

Corporation, Kansas, (1990).

7. Hoak, D.E. et al. “USAF stability and control DATCOM,” Air force Wright

aeronautical laboratories Technical Report 83-3048, October 1960. (Revised April

1978). Note: Digital DATCOM can be accessed from net.

8. Harris, C.D. “NASA supercritical airfoils” NASA TP 2969, March 1990. This report

can be downloaded from the site “NASA Technical Report Server (NTRS) ”.