aero class #5

8/6/2019 Aero Class #5

1/27

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


2/27

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


3/27

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


4/27

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


5/27

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.


6/27

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.


7/27

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.


8/27

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


9/27

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.


10/27

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


11/27

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


12/27

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.


13/27


14/27

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.


15/27


16/27

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


17/27

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.


18/27

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


19/27

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,


20/27

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


21/27

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)


22/27

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


23/27

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.


24/27

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.


25/27

Stability & Control


26/27

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.


27/27

Acknowledgements

Slide 13 Hurt, Aerodynamics for Naval

Aviators

Slide 15 Serway, Physics for Engineers and

Scientists

Slide 25 Hurt, Ibid

aero class #5

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