Aero Class #5
Transcript of Aero Class #5
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MidCourse Review
Streamline flow low velocity, little change indirection, also called potential flow, constantmass flow between streamlines, i.e. no flow
across streamlines Forces on a body in a stream are due to
pressure forces and friction forces.
Incompressible flow: =con., AU = Q = con.,good for liquids & gas flow U
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Review
p + U2/2 = constant (along a streamline)
Bernoullis eqn: p1 + U12/2 = p2 + U2
2/2
(frictionless, incompressible flow ONLY)
(NO GOOD for compressible flow)
Isentropic flow: frictionless, adiabatic (Q=0)
(p2/p1) = (2/1) = (T2/T1)
/(-1)
T0/T1 = 1 + (-1)(M12)/2; T0, p0, 0 are constant
throughout the isentropic flow field
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Review
Equivalent Airspeed an airplane flying at
some true airspeed at altitude its equivalent
airspeed would be its velocity at sea level to
experience the same dynamic pressure,
q=U2/2, i.e. U(EAS) = U(/SL)1/2 where:
=density at altitude, SL=density at sea level
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Review
Viscous Flow (flow with friction) The flow
field can be split into two regions, 1) inviscid
region, frictionless, potential flow and 2)
viscous region, boundary layer, shear layers in
the flow hence friction. This friction creates a
shear stress at the wall which results in skin
friction drag. The wall shear stress equals theviscosity times the velocity gradient at the
wall. wall = (dU/dy)wall
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Review
The skin friction drag can be reduced by
keeping the boundary layer laminar by
maintaining a decreasing pressure in the
streamwise direction (accelerating flow like in
a nozzle). For example, a laminar airfoil,
however they are very sensitive to
disturbances in the boundary layer whicheasily cause transition to turbulent flow.
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Review
Flow separation produces another source of
drag called pressure drag, promoted by an
increasing pressure in the streamwise
direction (decreasing velocity like in a diffuser)
A high angle of attack also promotes
separation. Not only does the drag increase
drastically but the lift also decreasesdrastically as the wing stalls.
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Review
Critical Mach No.: Mcr = value of M for whichMpeak = 1 (somewhere on the surface whereM->1). Also at this location p = pmin, i.e. Cp will
have the highest negative value for M > Mcr ,& the drag will drastically increase due towave drag. For supersonic flight, the wavedrag is significantly decreased by sweeping
the wings inside the Mach cone so that thevelocity is subsonic normal to the wingsleading edge.
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Review
To determine TR (thrust reqd for level flight),
pick U and h. CL = W /[(U2/2)S];
CD = CD,0 + CL2 /( e AR); L/D = CL / CD ;
TR = W/(L/D); repeat for selected values of Uand plot TR vs U or make a table.
TR
U
Tavail (dep. on engine)
U,max
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Review
Power reqd, PR=TRU=[2W3CD
2/(S CL3)]1/2
To generate the PR(U) curve, take the
preceding values from the table for TR
(U
)U= PR. Power available, PA = P; : propeller eff.
PA
U
recip. eng.
+ prop.
PA
U
Jet eng.
PA=TAU
TA
U
Jet eng.
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Review
One can establish a family of PR(U, altitude)
by Ualt=USL(SL/)1/2; PR,alt= PR,SL(SL/)
1/2 ;
Umax also decreases with altitude
Rate of climb, R/C=(PA PR) / W
Gliding flight: Rmax= h/tan min = h(L/D)max
Absolute Ceiling: R/C = 0; Service Ceiling: R/C
= 100 ft/min; calculate R/C (h), plot and
extrapolate to 0 and 100 ft/min
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Review
Range: total distance on a tank of fuel
Endurance: total time in the air on a tank of
fuel
Turning flight: n = L/W (load factor)
Radius of a horizontal turn, r = U2/*g(n2 -1)]
Turning rate, = U
/r = g(n2-1) / U
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Aero Class #5
Up to this juncture we have considered an
airplane to be a point mass concentrated at its
c.g. To discuss stability and control, we need
to treat the airplane as an extended body not
only capable of translation in the x-, y-, z-
directions but also capable of rotation about
the x-, y-, z-axes. The next slide illustrates thesix degrees of freedom.
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Rotation
We need to review a little more elementaryPhysics. This has to do with rotational motionabout an axis. In the case of an airplane the
axes pass through the c.g. of the airplane. Consider the simple case of a disk or pulley
mounted on an axle and we will neglect thefriction in the bearings of the pulley. The nextslide illustrates such a system with a singleweight suspended from the pulley.
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Rotation
Nomenclature:
Angular displacement: (radians)[2 rad/revolution]
Angular velocity: (radians/sec), ex. suppose a
helicopter rotor is turning at 1.5 rev/sec, this gives
= 1.5 rev/sec (2 rad/rev)= 3 rad/sec. Further,
if the rotor radius = 6 ft, this gives a tip speed of
v=r=3 rad/sec x 6 ft=18 ft/sec=56.5 ft/sec
Angular acceleration: (rad/sec2), this is related to
the tangential acceleration at radius r, at = r
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Rotation
Nomenclature:
The resistance to changes in the rotationalmotion is called the moment of inertia which
is dependent on the mass distribution aboutthe axis of rotation. If a body is thought of as anumber of discrete masses, mi, located atrespective distances from the axis of rotation,ri, then the moment of inertia is, I = mi ri
2.
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Rotation
To change the rotational motion requires the
application of a torque, a force x a lever arm.
Note: only the component of the force at right
angles to the lever arm is effective inproducing a torque. = Fr. This applied
torque now produces an angular acceleration
of = / I. Similar to linear motion, for =con,and 0=0, 0=0 at t=0; =t, =t
2/2, 2=2
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Rotation
Continuing the analogy with linear motion,
The rotational momentum, L = I = con.
The rotational kinetic energy, = I
2
/2The rotational power, P = .
And now back to the pulley on the axle,
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Rotation
We need to consider two free bodies: the mass
suspended by a string and the pulley constrained
to rotate about the fixed axle.
The forces on the mass lie in the vertical directionso Fy = mg T = may ; T: tension in the string.
The torque on the pulley is due to the tension in
the string so 0 = T R = I , I: moment of inertiaof the pulley, : angular acceleration of the
pulley, ay = R
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Rotation
We want to solve for the angular accelerationof the pulley, . The moment of inertia of thepulley about the axle is I = MR2/2. We have 2
eqns and 2 unknowns. Solving both eqns for Tand equating them gives mg - may = I/R, ormg - mR = (MR2/2)(/R) = MR/2 ormg = MR/2 + mR = (MR/2 + mR) or
= mg/(MR/2 + mR) = g/[R(M/2m +1)] orR = g/(1 + M/2m) = ay . We can now calculatethe position & velocity of the mass as fct(t)
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Stability & Control
An equilibrium position is reached where thereare no unbalanced torques on the airplane
Static Stability if this position is disturbed,
the airplane initially starts toward theequilibrium position, like a spring-mass system
Dynamic Stability disturbed from
equilibrium and oscillating, the motion isdamped and returns to equilibrium, like aspring-mass-damper system
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Stability & Control
The airplane can rotate in pitch, roll, and yaw.
The most important rotation to stabilize and
control is pitch, the rotation of the
longitudinal axis about the lateral axis. Wehave seen that the Lift acts through the
aerodynamic center, a.c., of the wing which is
located at the quarter chord point, c/4. One ofthe requirements for static stability is that the
c.g. is located forward of the a.c.
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Stability & Control
Since any rotation of the airplane is about the
c.g., a perturbation which increases the lift
would generate an increased force through
the a.c. which would result in a pitch-downmoment and a resulting decrease in and a
decrease in lift returning the airplane to its
original position. This effect is illustrated inthe next slide which compares stable and
unstable configurations.
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Stability & Control
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Stability & Control
It should be noted that stability comes at theexpense of controllability. The more stable,the less responsive the airplane.
Anderson goes into some detail with aplethora of partial derivatives which are fine ifyou are so inclined but I sense that we shouldleave that for the next course in Aero. Nohomework for next week. Catch up on yourreading.
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Acknowledgements
Slide 13 Hurt, Aerodynamics for Naval
Aviators
Slide 15 Serway, Physics for Engineers and
Scientists
Slide 25 Hurt, Ibid