4 s A- 2: here Unit A - 1: List of...

AE301 Aerodynamics I

UNIT A: Fundamental Concepts

ROAD MAP . . .

A-1: Engineering Fundamentals Review

A-2: Standard Atmosphere

A-3: Governing Equations of Aerodynamics

A-4: Airspeed Measurements

A-5: Aerodynamic Forces and Moments


Unit A-1: List of Subjects

What is Aerodynamics?

Engineering Units Review

Flow Field Representations

Flow Field Properties

Equation of State

WHAT IS AERODYNAMICS?

Aerodynamics is a branch (means: a specific application) of Fluid Mechanics:

(i) Underlying assumption is the fluid as a continuum and,

(ii) The working fluid is standard air.

CLASSIFICATION OF AERODYNAMICS

1. Density: incompressible (constant density) or compressible

2. Viscosity: inviscid (zero viscousity) or viscous

=> If viscous: laminar, transitional, or turbulent

3. Velocity (Mach number): subsonic (M < 0.7), transonic (0.7 < M < 1), supersonic (1 < M < 5), or

hypersonic (5 < M)

=> If subsonic: incompressible (M < 0.3) or compressible (0.3 < M)

I. Analytical Solution:

=> Potential Flow Analysis (2-D & 3-D)

=> Panel Methods (2-D & 3-D)

=> Thin Airfoil (2-D) / Lifting Line & Surface Methods (3-D)

II. Experiment: Wind Tunnel (scale-model tests) / Prototype (full-scale tests)

III. Numerical Simulation: Computational Fluid Dynamics (CFD)

Unit A-1Page 1 of 8

What is Aerodynamics?

1. Density

2. Viscosity

3. Velocity

A. Space (1-D, 2-D, or 3-D)

I. Analytical Solution

II. Experiment

III. Numerical Simulation

B. Time (Steady, Periodic, or Unsteady)

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

CONSISTENT AND “INCONSISTENT” UNITS

• “Consistent” (or coherent) units allow physical relationships to be written without the need of

conversion factors in the basic formulas.

• “Inconsistent” units cannot be used against consistent units (important): inconsistent units must

be converted into consistent units, before executing numerical calculation.

SI UNIT SYSTEM

Kilogram or kg (mass), Meter or m (length), Second or s (time)

=> Force: Newton (N): 1 N = (1 kg)(1 m/s2)

“Inconsistent” force unit: Kilogram-Force: kgf (“Kilogram-Mass: kgm = kg” is consistent unit)

• Unit conversion: 9.8 N = 1 kgf

(Do you casually use “Newtons” as a unit of force, or weight . . .?)

US CUSTOMARY (ENGLISH) UNIT SYSTEM:

Slug (mass), foot or ft (length), Second or s (time)

=> Force: Pound (lb): 1 lb = (1 slug)(1 ft/s2)

“Inconsistent” mass unit: Pound-Mass: lbm (note: “Pound-Force: lbf = lb” is consistent unit)

• Unit conversion: 1 slug = 32.2 lbm

(Do you casually use “slugs” as a unit of mass . . .?)

Unit A-1Page 2 of 8

Engineering Units Review (1)

1 slug = 32.2 lbm9.8 N = 1 kgf

“Consistent” Unit:Slug / N

“Inconsistent” Unit:lbm / kgf

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

Both measures “weight” = 15 pounds force / 85 kilogram force

• Mass is not a measurable property (you can never directly determine the amount of mass, by any

measurement method) . . . one can only “estimate” the mass, by measuring the force (weight)

generated by the presence of mass, under a known gravitational acceleration (W mg= ).

WHAT IS THE BOTTOM LINE?

Under the gravitational acceleration on earth (9.8 m/s2 or 32.2 ft/s2):

• A substance of 15 pounds mass “weighs” 15 pounds force.

• A substance of 85 kilogram mass “weighs” 85 kilogram force.

Problems:

• Pound mass (lbm) is “inconsistent” mass unit, and cannot be used against pound force (lb).

• Kilogram force (kgf) is “inconsistent” unit, and cannot be used against kilogram mass (kg).

SO, “WHAT DO YOU NEED TO DO” IN ENGINEERING ANALYSIS?

In engineering, you should re-state “more accurately” as follows:

• A substance of 0.466 slugs (15 lbm / 32.2 ft/s2 = 0.466 slugs) “weighs” 15 pounds force.

( ) ( )( )215 lb 0.466 slugs 32.2 ft/sW mg= = and 32.2 lbm

0.466 slugs 15 lbm1 slug

=

“Pound (force) can only be created by the combination of slugs, ft, and second)”

“Pound (force) and pound (mass) cannot coexist within the same equation”

“Under the standard sea-level gravity on earth (32.2 ft/s2), 1 lbm of object “weighs” 1 lbf”

• A substance of 85 kilograms “weighs” 833 Newtons (85 kgf 9.8 m/s2 = 833 N).

( ) ( )( )2833 N 85 kg 9.8 m/sW mg= = and 1 kgf

833 N 85 kgf9.8 N

=

“Newton (force) can only be created by the combination of kilogram, meter, and second)”

“Kilogram (force) and kilogram (mass) cannot coexist within the same equation”

“Under the standard sea-level gravity on earth (9.8 m/s2), 1 kg of object “weighs” 1 kgf”

Unit A-1Page 3 of 8

Class Example Problem A-1-1

Related Subjects . . . “Engineering Units Review”

Suppose, if you go to “FRY’S” and measure a bunch

of fruits . . . The scale says, “15 pounds.” Does this

mean: 15 pounds mass, or 15 pounds force?

A standard health meter in Japan indicates everything

in “kilograms.” Suppose, if you step onto a Japanese

health meter and it indicates, “85 kilograms.” Does

this mean: 85 kilogram force, or 85 kilogram mass?

STANDARD UNITS FOR AERODYNAMIC PROPERTIES

• Pressure (SI Unit): N/m2 (Pascal, or “Pa”), 1,000 Pa = 1 kPa

• Pressure (US Customary Unit): lb/in2 (psi), lb/ft2 (psf)

• Pressure unit conversion: 1 psi = 144 psf

• Density: kg/m3 (SI Unit) / slugs/ft3 (US Customary Unit)

• Temperature (SI Unit): Celsius (C), Kelvin (K) / K = C + 273

• Temperature (US Customary Unit): Fahrenheit (F), Rankine (R) / R = F + 460

• Gas Constant (Standard Air): 287 J/(kgK), 1,716 (ftlb)/(slugR)

COMMON MISHAPS OF UNITS

• Angle of Attack: may be given in degrees or radians.

• Temperature: the problem may be given in relative scales (C / F), but usually need to use absolute

scales (K / R) in calculations.

• Pressure: you need to know either “absolute” or “gage” pressure.

• Do you use “inches” in your calculation?

Unit A-1Page 4 of 8

Engineering Units Review (2)

0 F = 460 R0 C = 273 K

1 atm (Standard Sea-Level) = 2,116 lb/ft2

1 atm (Standard Sea-Level) = 1.01105 N/m2

1 ft = 0.3048 m1 lb = 4.448 N

1 mile = 5,280 ft1 h = 3,600 s

60 mph = 88 ft/s

STREAKLINE AND PATHLINE

• A streakline is the line defined in a flow field by continuously injecting infinitesimally small

particles at a fixed location. The idea of streakline is the pure experimental method (Eulerian).

• A pathline is the line defined by tracing a path of an individual particle. A pathline can be visualized

by injecting small particles in the flow field. The idea of pathline is a particle trace (Lagrangian).

FLOW VELOCITY AND STREAMLINE

• For a steady flow, a moving fluid element traces out a fixed path in space, called a streamline.

• Streamlines are mathematical concept. Hence, streamlines can be formed by “equations of

streamlines” in a given flow field.

• A streamline is a line, at which the velocity vector is tangent along its path.

NEWTONIAN FLUID SHEAR STRESS

• Due to the viscosity, the shear stress will be developed between the adjacent streamlines.

• For Newtonian fluids: dV

dy = (: absolute or dynamic viscosity).

Unit A-1Page 5 of 8

Flow Field Representations

FLOW FIELD PROPERTIES AND STANDARD UNITS

Common properties in aerodynamics that can define flow field at a point:

(1) Pressure:

p = p (x,y,z,t): units => lb/ft2 (psf), lb/in2 (psi), N/m2, Pa

• Pressure in absolue value: psia, psfa

• Pressure in gage value: psig, psfg

(2) Density / Specific Volume:

= (x,y,z,t): units => slugs/ft3, kg/m3

• Specific Volume: 1v = (volume per unit mass: inverse of density)

(3) Temperature:

T = T (x,y,z,t): units => °F, °C, °R, K

Absolute Scale Temperature (conversion):

°R = °F + 460 / K = °C + 273

VELOCITY (A VECTOR PROPERTY)

(4) Velocity:

( , , , )x y z t=V V : vector property (= magnitude + direction)

Vector Notation: publication / handwritten V VV

Unit Vectors: ˆ ˆ ˆi j k (Cartesian) / ˆ ˆ ˆ r ze e e (cylindrical)

Unit A-1Page 6 of 8

Flow Field Properties

in "absolute" pressurepsia psi=

in "gage" pressurepsig psi=

1 (Specific Volume)v

=

EQUATION OF STATE

A perfect (ideal) gas is the one in which the intermolecular forces are negligible.

Specific (R) and universal ( ) gas constants for a standard air:

Note 1): Pressure must be in absolute pressure (not “gage” pressure).

Note 2): Temperature must be in absolute scale (K / R, not C / F).

IDEAL GAS MODELS IN AERODYNAMICS

Rule of Thumb (means “usually,” but not always true)

Calorically Perfect => Thermally Perfect => Non Ideal Gas

Ideal Gas of Air Ideal Gas of Air of Air

Subsonic Supersonic Hypersonic

The difference between thermally and calorically perfect gases will be discussed later!)

Unit A-1Page 7 of 8

Equation of State

= 8,314 J/(kg mole K)

= 4.97 104 (ft lb)/(slug mole ºR)

M = 28.96 kg/(kg mole) or slug/(slug mole)

“Calorically Perfect” Ideal Gas of Air“Thermally Perfect” Ideal Gas of Air“Non-Ideal” Gas of Air

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

The pressure of air on the wing = 3.25 psi (144) = 468 lb/ft2

The density of air on the wing: 0.00075 slugs/ft3

Using the equation of state: p RT= or p

TR

=

Note 1) Temperature must be in absolute scale temperature (R), not relative scale (F)

Note 2) Pressure must be in absolute pressure (not gage pressure)

2

3

468 lb/ft363.63 R

(0.00075 slugs/ft )[1,716 (ft lb)/(slug R)]

pT

R= = =

In Fahrenheit: 363.63 − 460 = o96.37 F−

Unit A-1Page 8 of 8


Related Subjects . . . “Equation of State”

The air is flowing on a wing of Boeing 787 Dreamliner at

cruise. The pressure and density of air measured were 3.25

psi (absolute) and 0.00075 slugs/ft3, respectively. What is the

temperature (in F) of the air at that point?



ROAD MAP . . .








What is Altitude ?

Hydrostatic Equation

The Standard Atmosphere

Gradient Region

Isothermal Layer

This is an ONLINE LECTURE material (YouTube)

• Not covered in class as a standard lecture of AE301

• WILL BE INCLUDED in exams

• Please feel free to ask any question (if you have any) to the instructor

DEFINITION OF ALTITUDE

• There are two definitions for an altitude.

Absolute altitude (ha): an altitude measured from the center of the earth.

Geometric altitude (hG): an altitude measured from the surface (called, “sea-level”) of the earth.

Geometric altitude of Prescott is: 5,000 ft, while Daytona Beach is 0 ft (sea-level).

DIFFERENCE FROM FLUID MECHANICS IN STANDARD ATMOSPHERIC MODEL

• The gravitational acceleration is NOT constant, but will change against altitude.

• The variation of gravitational acceleration: (altitude)g g= must be taken into account, when a

mathematical model of the earth’s atmosphere (standard atmosphere) is constructed. To do this, a

GEOPOTENTIAL altitude concept is employed.

Unit A-2Page 1 of 7

What is Altitude ?

: Absolute Altitude ah

: Geometric Altitude Gh

: Radius of Earthr

Gravitational Acceleration:

r = 6,378.137 km (Radius of Earth)

THE HYDROSTATIC EQUATION

Let’s take a close look at a small cubic element of an air (atmosphere). If we apply the equation of

equilibrium (statics) to this element:

0F

+= : (1)(1) ( )(1)(1) (1)(1) 0Gp p dp dh g− + − =

or ( ) Gp p dp gdh= + + => Gdp gdh= −

THE GEOPOTENTIAL ALTITUDE

We have a problem: g is a function of geometric altitude (hG) . . . To solve this problem, we need to

employ a “fictitious” altitude, called geopotential altitude (h), such that:

G

G

rh h

r h

=

+ (see handout)

Note: Earth radius is 6378.137 kmr =

Then the hydrostatic equation becomes: Gdp gdh= − => 0dp g dh= −

where 0g is the g at standard sea-level ( 0h = ): 9.8 m/s2 or 32.2 ft/s2

Unit A-2Page 2 of 7

Hydrostatic Equation

Geopotential Altitude:

THE STANDARD ATMOSPHERE MODEL

The baseline assumption of the U.S. standard atmosphere is the defined variation of temperature (T)

with altitude.

It consists of a series of linearly varying temperature (called gradient) regions and constant temperature

(called isothermal) layers.

The first gradient region is usually called, the “troposphere.”

The first isothermal layer is usually called, the “stratosphere.”

Unit A-2Page 3 of 7

The Standard Atmosphere

Sea-Level Standard Values:

Geopote

ntial Altitude (

km

)

HYDROSTATIC EQUATION

From the hydrostatic equation (with geopotential altitude): 0dp g dh= −

Divide both sides by the equation of state: 0 0g dh gdpdh

p RT RT

= − = −

Note that the temperature variation is: 1 1( )T T a h h= + −

GRADIENT REGION

The temperature variation is assumed to be linear, such that:

1

1

T TdTa

dh h h

−= =

− (a is called, the “lapse rate”) or

dTdh

a= (eqn. 1)

Substituting eqn. 1 into this equation yields: 0gdp dT

p aR T= −

Integrating this equation between the base of this gradient region (h1) and a point (within this gradient

region) at an altitude (h) yields:

1 1

0

p T

p T

gdp dT

p aR T= − => 0

1 1

ln lngp T

p aR T= − =>

0[ /( )]

1 1

g aR

p T

p T

−

=

Applying the equation of state:

0{[ /( )] 1}

1 1

g aR

T

T

− +

=

Unit A-2Page 4 of 7

Gradient Region

Linear Temperature Variation:

Geopote

ntial Altitude (

km

)

HYDROSTATIC EQUATION

Once again, from the hydrostatic equation (with geopotential altitude):

0dp g dh= −

Divide both sides by the equation of state:

0 0g dh gdpdh

p RT RT

= − = −

Note that the temperature is constant within the isothermal layer: constantT =

ISOTHERMAL LAYER

Integrating this equation between the base of this isothermal layer (h1) and a point (within this

isothermal layer) at an altitude (h) yields:

1 1

0

p h

p h

gdpdh

p RT= − => 0

1

1

ln ( )gp

h hp RT

= − − => 0 1[ /( )]( )

1

g RT h hpe

p

− −=

Applying the equation of state: 0 1[ /( )]( )

1

g RT h he

− −=

Unit A-2Page 5 of 7

Isothermal Layer

T = constant

Geopote

ntial Altitude (

km

)

The geopotential altitude of 11 km is the edge of the troposphere.

Let us start from the sea-level (0 km altitude). At the sea-level:

pS = 1.01325 105 N/m2, S = 1.2250 kg/m3, and TS = 288.16 K

Applying the equations for the gradient region:

T11km = 3

1 1( ) 288.16 ( 6.5 10 )(11,000 0)T a h h −+ − = + − − = 216.66 K

p11km =

30[ /( )] [9.8/( 6.5 10 )(287)]

5

1

1

216.661.01325 10

288.16

g aR

Tp

T

−− − −

= =

4 22.2650 10 N/m

11km =

30{[ /( )] 1} {[9.8/( 6.5 10 )(287)] 1}

1

1

216.661.2250

288.16

g aR

T

T

−− + − − +

= =

30.36420 kg/m

Unit A-2Page 6 of 7


Related Subjects . . . “The Standard Atmosphere”

Calculate the standard atmosphere values

of T, p, and at a geopotential altitude of

11 km.

Geopote

ntial Altitude (

km

)11 km

The geopotential altitude of 20 km is in the middle of the tropopause.

Let us start from the results from the previous class example problem A-2-1.

At 11 km altitude, we found that:

p11km = 2.2650 104 N/m2, 11km = 0.36420 kg/m3, and T11km = 216.66 K

Applying the equations for the isothermal layer:

T20km = T11km = 216.66 K

p20km = 0 1[ /( )]( ) 4 [9.8/(287)(216.66)](20,000 11,000)

11km 2.2650 10g RT h h

p e e− − − −= = 3 25.483 10 N/m

20km = 0 1[ /( )]( ) [9.8/(287)(216.66)(20,000 11,000)

11km 0.36420g RT h h

e e − − − −= = 30.088 kg/m

Unit A-2Page 7 of 7


Related Subjects . . . “The Standard Atmosphere”

Calculate the standard atmosphere values

of T, p, and at a geopotential altitude of

20 km.

Geopote

ntial Altitude (

km

)20 km



ROAD MAP . . .








Continuity Principle

Incompressible Flow

Momentum Principle

Bernoulli’s Equation

Energy Principle

Specific Heats

Isentropic Flow

Energy Equation

Summary of Equations

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

CONTINUITY PRINCIPLE

Continuity principle = conservation of mass

Density () will change as a small element of air is compressed. However, the mass will never change.

Specific volume will also be changed: 1

=

Let us define the mass flow rate: dm

m VAdt

= =

A STREAM TUBE

• A stream tube is an aerodynamic analytical concept: a tube within a flow field, defined by

streamlines (streamlines are the wall of the stream tube)

For a given stream tube, the continuity equation is defined by: 1 1 1 2 2 2AV A V =

Unit A-3Page 1 of 16

Continuity Principle

Time rate of change of mass

within the control volume+

“Flux” of mass

across the control surface= 0

1

1

m

V =

2

2

m

V =

COMPRESSIBLE & INCOMPRESSIBLE FLOW

Compressible flow is the flow in which the density of the fluid elements can change from point to point.

Incompressible flow is the flow in which the density of the fluid elements is always constant.

Let us look at a duct (converging or diverging, or combined). If the flow is incompressible, the

continuity principle can be applied as:

1 1 1 2 2 2AV A V = => 1 1 2 2AV AV= => 12 1

2

AV V

A

=

In aerodynamics, the low subsonic speed is considered “incompressible.”

The rule of thumb for incompressible flow is: M < 0.3 (Why?)

INCOMPRESSIBLE AERODYNAMICS

• You must understand that “air” (as a fluid substance), is not incompressible. This is fundamental

and called, material (fluid) property. However, if the flow field is sufficiently “low speed” (we

usually employ “Mach number” to measure that), we can ignore the effect of compressibility so that

our analysis can dramatically be simplified.

• Hence, under the assumption of M < 0.3 (we’ll discover the reasoning of this assumption later), we

purposefully ignore the change in density within the flow field. This is, in fact, flow field property.

• Equations we employ in aerodynamic analysis are built under assumptions: understanding the

assumptions means understanding the limitations of each equation (very important).


Incompressible Flow

M < 0.3

(Incompressible)1 2 =

1 2 =

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

At sea-level, the velocity 20 m/s and 25 m/s correspond to the Mach numbers of 0.059 and 0.073,

respectively (M << 0.3).

Therefore, we can assume that the flow is incompressible (constant density: 1 2 = ).

Applying the continuity principle:

1 1 2 2AV AV= => 12 1

2

VA A

V

=

Thus, 2 212 1

2

20(10)

4 4 25

VA d

V

= = =

262.832 m



Related Subjects . . . “Continuity Principle”

Consider a convergent circular duct with an inlet diameter d1 = 10 m. Air enters this

duct with a velocity V1 = 20 m/s and leaves the duct exit with a velocity V2 = 25 m/s.

What is the area of the duct exit?

Velocity Pressure Temperature


_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

MOMENTUM PRINCIPLE

• Momentum principle = conservation of momentum

• In aerodynamics, we consider three types of forces:

Pressure, Body , and Viscous forces

EULER’S EQUATION ALONG A STREAMLINE

Let us consider a small cubic element of air (with the volume: dx dy dz) moving steadily along the

streamline.

If we ignore body force and viscous force, the momentum equation will become Euler’s equation.

F ma= along the streamline: ( )dp dV

pdydz p dx dydz dxdydzdx dt

− + =

Note that: dV dV dx dV

Vdt dx dt dx

= = , thus: F ma= => ( ) ( )dp dV

dxdydz dxdydz Vdx dx

− =

Therefore, dp VdV= − (Euler’s equation along a streamline)


Momentum Principle

+“Flux” of momentum

across the control surface=

Sum of all the forces

in a given directionTime rate of change of momentum

within the control volume

BERNOULLI’S EQUATION

From Euler’s equation: dp VdV= −

If we assume that the density is constant along a streamline (incompressible), we can integrate this

equation along a streamline:

2 2

1 1

p V

p V

dp VdV= − => 2 2

2 12 1( )

2 2

V Vp p

− = − −

=>

2 2

1 21 2

2 2

V Vp p + = + or simply,

2

constant2

Vp + =

(along a streamline)

LIMITATIONS OF BERNOULLI’S EQUATION

The Bernoulli’s equation is the most convenient, but the most commonly abused (in other words,

“mistakenly used”) equation in aerodynamics.

Assumptions required for deriving Bernoulli’s equation includes:

• From Euler’s equation:

1. The flow is steady (streamline).

2. The body force is ignored (no body force: common assumption in aerodynamics, but NOT in

hydrodynamics).

3. The viscous force is ignored (inviscid).

• Added onto the Euler’s equation (specific for Bernoulli’s equation):

4. The density is assumed to be constant (incompressible).

Note: the assumption 4 (incompressible) is “usually not true” in aerodynamics . . . only if M < 0.3

(called, “incompressible subsonic” flow).


Bernoulli’s Equation

3

22 2

31 21 2 3

2 2 2

VV Vp p p + = + + (3 is not along the same streamline of 1 & 2)

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

First, let us check to see if we can really use Bernoulli’s equation to solve this problem.

1. Is this steady flow? => Yes, since the problem only provides a single value of properties in the flow

field (time independent: steady flow).

2. Is this inviscid flow? => Not really, but the viscosity of air is very low (if you compare against a

liquid). We can assume that the flow of air is inviscid, as long as we do not consider details of boundary

layer.

3. Is this incompressible flow? => The freestream airspeed is:

88 ft/s100 mph 146.667 ft/s

60 mph

=

At the sea-level, this is Mach 0.131 < 0.3, so we can assume incompressible flow.

So, let us apply Bernoulli’s equation:

2 2

2 2

AA

V Vp p

+ = +

2 22( ) 2(2,116.2 2,070)(146.667) 244.972 ft/s

0.0024

AA

p pV V

− −= + = + = = 167.026 mph



Related Subjects . . . “Bernoulli’s Equation”

Consider an aircraft wing in a flow of air, where far

ahead upstream of the wing (called a “freestream”),

the pressure, velocity, and density are measured as:

1 atm (absolute), 100 mph, and 0.0024 slugs/ft3,

respectively. At a given point A on the wing, the

measured pressure is 2,070 lb/ft2. What is the

velocity at point A?

A

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

From the problem A-3-1, recall: V1 = 20 m/s and V2 = 25 m/s.

Let us assume the standard sea-level air density: 1.225 kg/m3.

Using Bernoulli’s equation:

2 2

1 21 2

2 2

V Vp p + = +

2 2 5 2 2

2 1 1 2

1.225( ) 1.013 10 [(20) (25) ]

2 2p p V V

= + − = + − =

5 21.0116 10 N/m



Related Subjects . . . “Bernoulli’s Equation”

Consider the same convergent duct and conditions as in Class Example Problem A-3-1.

If the air pressure at the inlet is p1 = 1 atm, what is the pressure at the exit?



ENERGY PRINCIPLE

Energy principle = conservation of energy

Let us consider a system (a unit mass of air). The first law of thermodynamics can be applied to this

system (in thermodynamics, this is called the control mass) as:

q w de + =

This equation says that “energy added or subtracted by heat and work is balanced against the change of

internal energy: energy must be balanced.” This is the first law of thermodynamics.

d : “exact differential” (means “infinitesimally small” and process independent quantities = internal

energy / enthalpy)

: “infinitesimally small” quantities, but not “exact differential” (means process or path dependent =

heat/work energy transfer. Often upper case is also used: this is called “inexact differential”)


Energy Principle (1)

1st Law of Thermodynamics:

Energy added or subtracted

due to Heat

Energy added or subtracted

due to Work Change of Internal Energy

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

FIRST LAW OF THERMODYNAMICS

In aerodynamics, work due to pressure (in thermodynamics, this is called “pdV” or “moving boundary”

work) is the main component of energy transfer – consider work energy transfer per unit mass:

A

w p sdA pdv = = − (Note that: 1

v

= , called the “specific volume”)

The first law of thermodynamics (applied for aerodynamics) becomes:

q w de + = => q de w = − => q de pdv = + (eqn. 1)

ALTERNATIVE FORM OF FIRST LAW (IN TERMS OF ENTHALPY)

Let us define a new property, called the ehthalpy (per unit mass) as:

ph e e pv

= + = + (eqn. 2)

Let us differentiate the eqn. 2, such that:

dh de pdv vdp= + + or de dh pdv vdp= − − (eqn. 3)

Substituting the eqn. 3 into eqn. 1 yields:

( )q de pdv dh pdv vdp pdv = + = − − +

This results in the alternative form of the equation of the first law of thermodynamics: q dh vdp = −


Energy Principle (2)

1st Law of Thermodynamics (in terms of internal energy) :

1st Law of Thermodynamics (in terms of enthalpy):Moving boundary work:

SPECIFIC HEATS

Definition of specific heats: “heat energy added per unit change in temperature” of the system: q

cdT

Specific heats are the measure of energy that can be stored (per unit mass) of a substance.

There are two different kinds of specific heats:

Specific heat at constant volume: constant

v

v

qc

dT

=

=> vq c dT =

Specific heat at constant pressure: constant

p

p

qc

dT

=

=> pq c dT =

CONSTANT VOLUME SPECIFIC HEAT

Recall, the equation of the first law of thermodynamics: q de pdv = +

Since the volume is held constant during the heat addition process:

v = constant => 0dv = .

Therefore, q de pdv de = + =

From the definition of specific heat ( vq c dT = ): vde c dT=


Specific Heats (1)

Constant Volume Heat Addition

Thermally perfect: Calorically perfect:

( )vde c T dT= ve c T=

0

CONSTANT PRESSURE SPECIFIC HEAT

Recall, the first law of thermodynamics (alternative form): q dh vdp = −

Since the pressure is held constant during the heat addition process,

p = constant => 0dp =

Therefore, q dh vdp dh = − =

From the definition of specific heat ( pq c dT = ): pdh c dT=

AIR AS AN IDEAL GAS: TWO MODELS

Thermally perfect ideal gas: specific heats are not constant, but can be modeled as a simple functions

of temperature.

( )v vc c T= and ( )p pc c T=

Further, under a certain (limited) temperature range, the specific heats can be assumed constant. This is

called “calorically perfect” ideal gas.

constant(1)vc = and constant(2)pc = These constants can be found by the gas constant (R).

For a calorically perfect ideal gas, the ratio between pc and vc is constant.

This is called the specific heat ratio: p

v

c

c (for standard air, = 1.4)


Specific Heats (2)

v

p

c

c=Rcc vp +=

= 1.4 (Air at Standard Condition)

Constant Pressure Heat Addition

Thermally perfect: Calorically perfect:

( )pdh c T dT= ph c T=

Calorically perfect ideal gas relations:

0

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

(a) As we learned, there are two models for air as an ideal gas.

(i) An ideal gas means at least thermally perfect ideal gas in aerodynamics, means that specific heats are

functions of temperature only.

Thus, ( )vde c T dT= and ( )pdh c T dT=

(ii) A calorically perfect ideal gas is thermally perfect ideal gas and if the specific heats are constant.

Thus, ve c T= and ph c T=

For example, under the standard atmosphere (1 atm), the specific heats are constant in the range of from

−50C to 50 C (i.e., the room temperature): cp = 1,005 J/kgK = 6,006 ftlb/slugR.

(b) For calorically perfect ideal gas, ve c T= and ph c T= . (eqn. 1)

Recall, the definition of enthalpy: h e pv= + .

If we assume air as an ideal gas, pv RT= (equation of state) can be used, so: h e RT= + .

Let us plug-in eqn. 1 into this equation:

p vc T c T RT= + => p vc c R= +



Related Subjects . . . “Specific Heats”

(a) Explain the difference between: (i) thermally perfect

ideal gas and (ii) calorically perfect ideal gas.

(b) For calorically perfect ideal gas, derive the

relationship between two specific heats (cp and cv) as:

cp = cv + R

ISENTROPIC FLOW

Isentropic flow is the flow, in which the process is both adiabatic and reversible. As you know well at

this point, the Bernoulli’s equation cannot be used for compressible flow. Instead of Bernoulli’s

equation, you may be able to employ isentropic relationship, if the flow is still isentropic.

Isentropic flow assumption (adiabatic and reversible) is valid for wide variety of flow regimes, even

including supersonic flows.

However, typical non-isentropic flows include:

• High-speed flows across the shock waves (shock waves are non-isentropic compression)

• Flows with combustion and/or chemical reactions (example: rocket combustions)

• Flows under the viscous effects (i.e., boundary layer)

Isentropic flows will be discussed in-depth in AE302 (Aerodynamics II), so let us skip the derivation of

the equation at this point – the equation, called isentropic relations:


Isentropic Flow

Isentropic Relations:

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

FIRST LAW OF THERMODYNAMICS (IN TERMS OF ENTHALPY)

Staring from the alternative form of the first law of thermodynamics:

q dh vdp = −

If the flow is isentropic, there is no heat transfer (adiabatic or 0q = ):

0dh vdp− = => dh vdp= (eqn. 1)

ENERGY EQUATION

Recall, the Euler’s equation: dp VdV= − (eqn. 2)

Combining eqns. 1 & 2 yields:

( )dh v VdV= − => dh VdV= − (Note: 1v = )

Integrating this equation between any two points along the streamline: 2 2

1 1

h V

h V

dh VdV= − => 2 2

2 12 1( )

2 2

V Vh h

− = − −

2 2

1 21 2

2 2

V Vh h+ = + or

2

constant2

Vh + = (along the streamline)

For calorically perfect ideal gas: ph c T=

2 2

1 21 2

2 2p p

V Vc T c T+ = + or

2

constant2

p

Vc T + = (along the streamline)


Energy Equation

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

First, let’s check the flight Mach number. At 10 km altitude, the speed of sound is 299.5 m/s.

The flight Mach number is:

km 1,000 m 1 h750

h 1 km 3,600 s

299.5 m/s

VM

a

= = = 0.696 > 0.3

Bernoulli’s equation is not valid.

Let us apply energy equation, instead:

2 2

1 21 2

2 2p p

V Vc T c T+ = + (along the streamline, behind the

shockwave) => 2

2 1 1 22 ( )pV V c T T= + −

At 10 km altitude, the temperature is: 223.3 K.

Also, 1 750 km/hV = = 208.333 m/s

Therefore, 2

2 (208.333) 2[1,005 (J/kg K)][223.3 ( 45 273)]V = + − − + = 184.27 m/s (M = 0.615)



Related Subjects . . . “Energy Equation”

A high-speed business jet aircraft is flying at 10 km altitude with 750 km/h. The

temperature and pressure at a point on the wing is −45 C and 2.8 104 N/m2,

respectively. What is the velocity at this point?

Is Bernoulli’s equation still applicable to solve this problem?

Property of Air(p = 1 atm)

EQUATIONS AND RESTRICTIONS OF USE

1. Bernoulli’s equation (M < 0.3: Rule of Thumb)

= Euler’s equation + “incompressible” assumption

• Steady flow

• Incompressible ( = constant)

• Inviscid (no viscosity, no friction)

2. Energy equation (M > 0.3)

= First law of thermodynamics + Euler’s equation + Isentropic flow + Ideal Gas

• Steady flow

• Compressible ( ≠ constant), but isentropic (adiabatic and reversible)

• Inviscid (no viscosity, no friction: this is required for reversibility)


Summary of Equations



ROAD MAP . . .








Speed of Sound

Mach number

Measurement of Airspeed

Incompressible Flow

Compressible Flow

What’s “Incompressible” ?

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

APPLICATION OF CONTINUITY ON A SOUND WAVE

Let us consider a coordinate system attached to (and thus, moving with the same speed with) the sound

wave. Continuity equation (1 2m m= ) yields:

( ) ( )Aa d A a da = + +

( )( )a d a da a ad da d da = + + = + + +

Therefore, da

ad

= − (eqn. 1)

SPEED OF SOUND (1)

Recall, the Euler’s equation (in terms of speed of sound): dp ada= − => dp

daa

= − (eqn. 2)

Substituting eqn. 2 into eqn. 1 yields: dp

ad a

= => 2 dp

ad

=

The flow through a sound wave involves no heat addition, and the effect of friction is negligible: means,

it is isentropic flow).

Therefore: isentropic

dpa

d

=


Speed of Sound

0

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

ISENTROPIC FLOW

For isentropic flow: 2 2

1 1

p

p

=

=> 2 1

2 1

constantp p

c

= = = , or, simply: p

c

= (eqn. 1)

SPEED OF SOUND (2)

Starting from isentropic

dpa

d

=

From eqn. 1, the pressure of isentropic flow can be expressed as: p c =

Therefore, 1

isentropic

( )dp d

c cd d

− = =

(eqn. 2)

Substituting eqn. 1 back into eqn. 2, and simplifying: 1

isentropic

dp p p

d

− = =

Therefore, the speed of sound is: isentropic

dp pa

d

= =

For an ideal gas: p RT= => p

RT

= ; therefore, the speed of sound becomes: a RT=


Mach Number

Speed of Sound (Sea-Level Standard Value)

SI Units: 340.3 m/s or 1,225.08 km/hU.S. Customary Units: 1,116.5 ft/s or 761.25 mphor 661.508 knots

a RT=

AIRSPEED MEASUREMENT DEVICE

Pitot-static probe measures both stagnation (or total) pressure and static pressure: provides pressure

difference between them ( 0p p− )

STATIC, DYNAMIC, AND TOTAL (OR STAGNATION) PRESSURES

Static pressure (p) at a given point is the pressure we would feel if we were moving along with the flow

at that point.

Total pressure (p0) at a given point in a flow is the pressure that would exist if the flow was slowed

down “isentropically” to zero velocity: therefore, p < p0 (for a stagnant air: p = p0).

Dynamic pressure is a pressure due to the added energy into the moving fluid (air). The difference

between total and static pressures ( 0p p− ) is dynamic pressure. Dynamic pressure is zero for a stagnant

air (p = p0).

Stagnation point is where V = 0: so at stagnation point, the pressure becomes close to the total pressure:

stagnation pressure total pressure.


Measurement of Airspeed

Pitot-Static Probe

Pitot Tube: senses total pressure

Static Pressure Orifice: senses static pressure

− (subtract)

BERNOULLI’S EQUATION

For incompressible flow, we can employ Bernoulli’s equation.

Along a streamline: 21

constant2

p V+ = = p0

Let us define a dynamic pressure: 21

2q V

Then, the Bernoulli’s equation becomes: constantp q+ = = p0

AIRSPEED MEASUREMENT FOR SUBSONIC INCOMPRESSIBLE FLOW (M < 0.3)

Let us define: location ‘1’ being the flow far upstream (called, the “freestream”) and location ‘0’ being

the location of zero velocity, the ‘tip’ of the Pitot-Static tube (called, the “stagnation point”).

Applying Bernoulli’s equation between freestream (‘1’) and the tip of the Pitot-Static tube (‘tip’):

2 2

1 1 tip tip

1 1

2 2p V p V + = +

At the freestream: p1 = p, because V1 = V (this is freestream).

At the tip: ptip = p0, because Vtip = 0 (this is stagnation point).

Therefore, 2

0

1

2p V p+ = =>

0 12p p

V

−=


Incompressible Flow (1) (Subsonic: M < 0.3)

(p0)(p)

V

TRUE AND EQUIVALENT AIRSPEEDS

The air density is difficult to measure. For small (low subsonic and low altitude cruise) airplanes, often

the equivalent airspeed is indicated on its airspeed indicator:

Equivalent (or indicated) airspeed is the airspeed that uses the standard sea-level air density value for the

airspeed calculation:

0( )2e

s

p pV

−= (Equivalent Airspeed)

As long as the altitude is low (close to the sea-level), the equivalent (or indicated) airspeed is fairly

accurate.

The true airspeed is the airspeed that uses the actual air density value for a given flight altitude for the

airspeed calculation:

0true

( )2

p pV

−= (True Airspeed)

Have you heard about “KEAS” = “Knots in Equivalent AirSpeed”?

(1 knot = 1.15 mph)


Incompressible Flow (2) (Subsonic: M < 0.3)

Pitot-static probe

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

The equivalent airspeed can be calculated (by using the air density at standard sea-level).

Using s = 0.0023769 slugs/ft3 and 2

0 53.3 lb/ftp p− = :

02( ) 2(53.3)

0.0023769e

s

p pV

−= = = 211.774 ft/s

If the temperature is known, it is possible to calculate the true air density:

Using 21,896.7 lb/ftp = (pressure altitude 3,000 ft):

1,896.7

(1,716)(50 460)

p

RT = =

+ = 0.00216726 slugs/ft3

Using this true air density:

true

02( ) 2(53.3)

0.00216726

p pV

−= = = 221.780 ft/s

The error of equivalent airspeed: 221.780 211.774

100221.780

− =

4.5 %



Related Subjects . . . “Airspeed Measurement: M < 0.3”

The altimeter on a low-speed private aircraft (M < 0.3) reads 3,000

ft. If a Pitot-static probe (as shown in the figure) measures a

pressure of 53.3 lb/ft2, what is the equivalent airspeed of the

airplane? Suppose, if you know the outside air temperature

(through an independent measurement) is 50 ºF, what is the true

airspeed? Calculate the error of equivalent airspeed.

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

ENERGY EQUATION

For compressible flow, we can no longer use Bernoulli’s equation. Let us look at the energy equation

one more time.

Recall, the energy equation:

2

constant2

p

Vc T + = (along the streamline)

AIRSPEED MEASUREMENT FOR SUBSONIC COMPRESSIBLE FLOW (M > 0.3)

Applying the energy equation for a Pitot tube (freestream ‘1’ and stagnation point ‘0’): 2

11 0

2p p

Vc T c T+ = =>

2

0 1

1 1

12 p

T V

T c T= + (eqn. 1)

Also, the definition of specific heat can be given by: 1

p

Rc

=

−

Substituting this into eqn. 1:

2 2

0 1 1

1 1 1

11 1

2[ / ( 1)] 2

T V V

T R T RT

−= + = +

−

Note that from speed of sound, 2

1 1a RT= : thus the equation becomes,

2

20 112

1 1

1 11 1

2 2

T VM

T a

− −= + = + =>

201

1

11

2

TM

T

−= +

Using the isentropic relationship: 1

0 0 0

1 1 1

p T

p T

− = =

120

1

1

11

2

pM

p

−− = +

and

1

120

1

1

11

2M

−− = +


Compressible Flow (1)(Subsonic: 1 > M > 0.3)

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

TRUE AND CALIBRATED AIRSPEEDS (1)

Starting from:1

201

1

11

2

pM

p

−− = +

Solving this equation for M1:

1

2 01

1

21

1

pM

p

− = − −

(Note: V

Ma

= )

1 1

2 22 0 0 11 1

1

1 1

2 21 1 1

1 1

p p pa aV

p p

− − − = − = + − − −

(True Airspeed)

TRUE AND CALIBRATED AIRSPEEDS (2)

True airspeed requires information of a1 (i.e., T1) and p1. The static temperature and pressure in the air

surrounding the airplane is often very difficult to measure (in high-speed flight).

Therefore, all high-speed airspeed indicators are “calibrated.” For example, assuming that a1 and p1 are

both equal to the standard sea-level value (as = 340.3 m/s = 1,116.5 ft/s and ps = 1.013×105 N/m2 =

2,116.2 lb/ft2), the calibrated airspeed (based on the standard sea-level condition) becomes:

1

22 0 1

cal

21 1

1

s

s

a p pV

p

− − = + − −

(Calibrated Airspeed)


Compressible Flow (2) (Subsonic: 1 > M > 0.3)

( )12

2 0 111

1

21 1

1

p paV

p

− − = + −

−

( )12

2 0 1cal

21 1

1

s

s

a p pV

p

− − = + −

−

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

Pitot-static probe measurement:0 1p p− = 5.5 103 N/m2

At standard sea-level: ps = 1.013 105 N/m2 and as = 340.3 m/s

1 1.4 1

2 2 3 1.42 0 1

cal 5

2 2(340.3) 5.5 101 1 1 1

1 1.4 1 1.013 10

s

s

a p pV

p

− − − = + − = + − − −

= > Calibrated airspeed: calV = 93.878 m/s cal

s

93.8780.276

340.3

VM

a

= = =

Pressure altitude of 10 km: p1 = 2.65 104 N/m2

Measured temperature is: T1 = −45 C = −45+273 = 228 K

Speed of sound is: 1 1 (1.4)(287)(228) 302.672 m/sa RT= = =

1 1.4 1

2 2 3 1.42 0 11

1 4

1

2 2(302.672) 5.5 101 1 1 1

1 1.4 1 2.65 10

p paV

p

− − − = + − = + − − −

= > True airspeed: 1V = 159.243 m/s 1

1

159.2430.526

302.672

VM

a

= = =

Error of calibrated airspeed 159.243 93.878

100159.243

− = =

41.05 %



Related Subjects . . . “Airspeed Measurement: M > 0.3”

A jet aircraft is cruising high speed (high subsonic:

M > 0.3) at 10 km cruising altitude. If a Pitot-static

probe (as shown in the figure) measures a pressure

of 5.5 103 N/m2, what is the calibrated airspeed

(and associated Mach number) of the airplane?

Suppose, if you know that the outside air

temperature (through an independent measurement)

is −45 ºC, what is the true airspeed (and associated

Mach number)? Calculate the error of calibrated

airspeed.

DEFINITION OF INCOMPRESSIBLE FLOW

So far, we employed the rule of thumb (M < 0.3) as an indicator of incompressible flow.

But, why this is valid?

Recall, for isentropic flow, with calorically perfect ideal gas, the ratio of density between location ‘1’

(freestream) and location ‘0’ (stagnation point) can be given as: ( )1 1

201

1

11

2M

−−

= +

Note that the “freestream” is the location where the density is “lowest” within the flow field, while

“stagnation point” is the location where the density is “highest” (most compressed). Hence, this

equation is the “density variation” within the given flow field (from lowest to highest density).

For isentropic flows with Mach numbers less than about 0.3, the density variation within the flow field

is less than 5 percent. The variation is small, and thus the flow can be treated as incompressible.


What’s “Incompressible” ?



ROAD MAP . . .








Pressure and Shear Force Distributions

Lift, Drag, and Moment

Aerodynamic Coefficients

Aerodynamic Forces and Moments per Unit Span

2-D Aerodynamic Coefficients

Center of Pressure

Aerodynamic Center

SOURCES OF AERODYNAMIC FORCES AND MOMENTS

There are two sources for aerodynamic forces and moments.

(1) Pressure distribution over the body surface

(2) Shear stress distribution over the body surface

The resultant aerodynamic force (R) can be resolved into 2 components of forces:

(i) In the direction of chord (“body-fitted” coordinate system)

(ii) In the direction of freestream (V)

ANGLE OF ATTACK ( )

Angle of attack is defined as an angle between chord line & freestream.


Pressure and Shear Force Distributions

LIFT, DRAG, AND MOMENT

Lift (L) is defined as the component of the aerodynamic force perpendicular to the relative wind (called,

freestream: V∞). Non-dimensional lift coefficient (CL) is:

Sq

L

SV

LC

L

==2

2

1

Drag (D) is defined as the component of the aerodynamic force parallel to the relative wind (called,

freestream: V∞). Non-dimensional drag coefficient (CD) is:

Sq

D

SV

DC

D

==2

2

1

Pitching moment (M) is defined as a moment created by pressure differences between bottom and top

surfaces. Pitch up moment is positive and pitch down moment is negative. Non-dimensional moment

coefficient (CM) is:

Scq

M

ScV

MC

M

==2

2

1

Note: the moment depends on the location, about which we choose to take moments at:

MLE, Mc/4, and MTE, where: MLE, ≠ Mc/4 ≠ MTE.

(You MUST always specify the LOCATION (that you took the moment) with a subscript)


Lift, Drag, and Moment

CL

CD CM

AERODYNAMIC COEFFICIENTS

There are two different aerodynamic coefficients:

• Upper case: CL, CD, CM = coefficients for 3-D finite wings

L

LC

q S

= D

DC

q S

= M

MC

q Sc

=

S: projected surface area (characteristic surface)

For aircraft wings (lift) = “projected planform” area

Note: for circular cylinder (drag) = “projected frontal” area

• Lower case: cl, cd, cm = coefficients for 2-D airfoils (“unit span” of an infinite wing)

( )

( )' '

1l

L b L Lc

q S q c q c

= = = ( )

( )' '

1d

D b D Dc

q S q c q c

= = = ( )

( ) 2

' '

1m

M b M Mc

q Sc q c c q c

= = =

c: chord length

b: wing span

L’: lift per unit span

D’: drag per unit span

M’: moment per unit span


Aerodynamic Coefficients

S = frontal area

Unit D-1Page 1 of 18

Infinite v.s. Finite Wings

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

PRESSURE AND SHEAR STRESS DISTRIBUTIONS (1)

Let us consider a 2-D body: a cross section of a wing (airfoil). The normal and axial forces (per unit

span) will be developed due to the pressure and shear stress distributions over the surface. Let us

examine a small element ds (as shown in the figure) on upper (u) and lower (l) surfaces of this airfoil.

PRESSURE AND SHEAR STRESS DISTRIBUTIONS (2)

' ' 'u ldN dN dN= + ' ' 'u ldA dA dA= + ' ' 'LE u ldM dM dM= +

uuudspdN )sincos(' +−=

llldsp )sincos( −+

' ( sin cos )u u udA p ds = − +lll

dsp )cossin( ++

uuuuuLEdsypxpdM ])cossin()sincos[(' −−+=

lllll

dsypxp ])cossin()sincos[( +++−+


Aerodynamic Forces and Moments per Unit Span (1)

LIFT, DRAG, AND MOMENT PER UNIT SPAN (1)

Normal and axial forces (per unit span) as well as moment at the leading edge can be calculated by

integrating these equations over the entire surface of the 2-D body (airfoil): TE

LE

' 'N dN=

TE

LE

' 'A dA=

TE

LE LE

LE

' 'M dM=

LIFT, DRAG, AND MOMENT PER UNIT SPAN (2)

TE TE TE

LE LE LE

' ' ( cos sin ) ( cos sin )u u u l l lN dN p ds p ds = = − + + −

TE TE TE

LE LE LE

' ' ( sin cos ) ( sin cos )u u u l l lA dA p ds p ds = = − + + +

TE TE

LE LE

LE LE

' ' ( cos sin ) ( sin cos )u u u u uM dM p x p y ds = = + − −

TE

LE

( cos sin ) ( sin cos )l l l l lp x p y ds + − + + +

Lift and drag forces (per unit span) can be calculated from normal force (N’) and axial force (A’) and

angle of attack () as: ' 'cos 'sinL N A = − ' 'sin 'cosD N A = +


Aerodynamic Forces and Moments per Unit Span (2)

2-D AERODYNAMIC COEFFICIENTS

Lower case cl, cd, cm: 2-D airfoil

L’, D’, and M’: aerodynamic forces and moment per unit span.

'l

Lc

q c

= '

d

Dc

q c

= 2

'm

Mc

q c

=

SKIN FRICTION & PRESSURE COEFFICIENT

Skin friction coefficient (per unit length): fcq

=

: Shear stress developed on the surface (due to viscosity)

Skin friction drag coefficient (for a given surface S): ( )f

f

f D

DC C

q S

= =

fD : Skin friction drag (due to viscosity)

Pressure coefficient: p

p pC

q

−=


2-D Aerodynamic CoefficientsPressure coefficient:

Skin friction coefficient:

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

CENTER OF PRESSURE

The center of pressure is the location (along the chord line), where the resultant of a distributed load

effectively acts on the body. At the center of pressure, the pitching moment is equal to zero.

LOCATION OF THE CENTER OF PRESSURE

Taking moment at leading edge:

(1) Left & center figures: LE /4' ' '

4c

cM M L= − eqn. 1

(2) Center & right figures: /4 cp' ' '

4c

cM L x L− = − eqn. 2

Therefore, LE /4 cp' ' ' '

4c

cM M L x L= − = −

=> LEcp

'

'

Mx

L= − and /4

cp

'

4 '

cMcx

L= −


Center of Pressure

LE

cp

'

'

Mx

L= − / 4

cp

'

4 '

cMcx

L= −

Center of pressure:

or

M’

= =

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

The center of pressure can be given by:

LE c/4 cp' ' ' '4

cM L M x L= − + = −

or /4cp

'

4 '

cMcx

L= − =>

/4

cp

'

1

4 '

cMx c

c L= −

Further manipulation of this equation yields:

/4

2cp , /4

'

1 1 1 ( 0.1)

'4 4 4 0.8

c

m c

l

M

x cq c

Lc c

q c

−= − = − = − = 0.375

• Summary: Center of Pressure Locations

cp , /4

cp

1

4

m c

l

x cx

c c= = −

or

cp ,

cp

m le

l

x cx

c c= = −



Related Subjects . . . “Center of Pressure”

A set of experimental data for a series of conventional

airfoil is readily available. This is called “NACA

conventional airfoil data.” NACA 4412 is one of

these NACA conventional airfoils, and it shows the

following aerodynamic coefficients at an angle of

attack of 4 degrees: cl = 0.8 and cm,c/4 = −0.1.

Calculate the location of the center of pressure.

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

AERODYNAMIC CENTER

Aerodynamic center: the point on a body about which the aerodynamically generated moment is

independent of angle of attack.

LOCATION OF THE AERODYNAMIC CENTER

Once again, taking moment at leading edge:

ac ac 4' ' ' 4 'cM cx L M c L− = −

Hence, ( )ac ac 4' ' 4 'cM L cx c M= − +

Dividing this equation by 2q c yields:

( ) 4acac2 2

'' '0.25

cMM Lx

q c q c q c

= − + => ( ),ac ac , 40.25m l m cc c x c= − +

Differentiating this equation with respect to yields:

( ) ( ) ( ),ac ac , 40.25lm m c

dcd dc x c

d d d = − + => ( ) , 4

ac0 0.25m cl

dcdcx

d d = − +

Let 0

ldca

d and

, 4

0

m cdcm

d => ( )0 ac 00 0.25a x m= − +

Therefore, 0ac

0

0.25m

xa

= − +


Aerodynamic Center

=

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

Taking data points at ±8 degrees:

0

0.99 ( 0.74)

8 ( 8)a

− −=

− − = 0.1081 (/degree)

0.10.04

0.4

−

0.10.08

0.4

−

0

0.01 ( 0.02)

8 ( 8)m

− − −=

− − =

46.25 10− (/degree)

4

0ac

0

6.25 100.25 0.25

0.1081

mx

a

−= − + = − + = 0.244



Related Subjects . . . “Aerodynamics Center”

Determine the location of the aerodynamic center (in % chord) of NACA 23012 airfoil.

Center of pressure: a location where moments become zero: Mcp = 0.

Center of pressure will vary as angle of attack is changed.

Aerodynamic center (also, often called “Neutral Point” or “NP”): a location where moments do not

change with angle of attack (): Mac = const.

Aerodynamic center is independent of angle of attack (usually very close to the quarter chord

location, especially for “thin” airfoils).

The aerodynamic center is also often called the “Neutral Point” (NP), where it is used as a reference

location for aircraft control moments in three axes (“yaw,” “pitch,” and “roll”).

The aerodynamic center is independent of angle of attack (), while the center of pressure depends on

the angle of attack (): this makes the aerodynamic center an important (useful) reference location for

flight mechanics.

Imagine, when you “pitch up” an aircraft, using a control surface. The aircraft will rotate around a

center of gravity (CG). As an aircraft rotates around CG, the increased angle of attack will result in

increased lift force. This lift force will be resolved into two components: (i) a moment generated at the

aerodynamic center (which is “constant”) and (ii) a resultant force acting through the aerodynamic

center that will, in turn, generate control moment (“pitch” either up or down) around a pivot point (CG).

The center of pressure location keeps changing as the angle of attack changes . . . hence, this cannot

be used as a reference location for flight mechanics!



Related Subjects . . . “Center of Pressure” & “Aerodynamic Center”

For designing an aircraft, the location of

“aerodynamic center” (AC), not “center of pressure”

(CP) is usually used. What is the difference between

the center of pressure and aerodynamic center?

Apparently, the aerodynamic center is more important

(or “useful”) in flight mechanics than the center of

pressure. Why?

4 s A- 2: here Unit A - 1: List of...

Documents

Transcript of 4 s A- 2: here Unit A - 1: List of...