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Sample Question paper and hints for solution*AE 1301 Flight dynamics

B.E./ B.tech Degree examination,November / December 2006.

Anna university.

Time : 3 hours Maximum:100 marks

Answer ALL questions

Part A- (10 x 2 = 20 marks)

1. What causes induced drag?

[For answ er see topic Drag coefficient of w ing in section 3.2 (Flight dynamics I )] .

2. Plot the variation of power available with flightspeed for a propeller powered airplane andindicate the effect of altitude on the curve.

* The hints are given in bold letters.

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[for answ er see Variation of THP w ith

velocity in section 4.2.4 (Flight dynamics I) ].3. Define service and absolute ceiling.

[ For answ er see section 5.3.4 Absolu te

ceiling & service ceiling( Flight dynamics I )]4. What are the conditions for maximum endurance of

a jet powered airplane?

[ For answ er see section 5.5.3 Breguetformulae (Flight dynamics I )] .

5. Define neutral point.

[ For answ er see section 2.3 Stick fixedneutral point and static margin (Flightdynamics II )].

6. What is the criterion for static longitudinal stability?

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[ For answ er see Criterion for longitudinal

static stability in section 2.1 (Flightdynamics II )].

7. What is meant by dihedral effect?

[ For answ er see section 3.3 Lateralstabili ty (Flight dynamics II )] .

8. Differentiate between yaw and sideslip angle.

[ For answ er see section3.2.1 Sideslip & yaw (Fl ight dynamics I I ] .

9. Graphically represents a system which isstatistically stable but dynamically unstable.

[ For answ er see section 1.2.2 Static

stabili ty and dynamic stabili ty (Flightdynamics II )].

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10. What is spiral divergence?[ For answ er see section 4.15.2 Analysis of

roots (Flight dynamics II )] .

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Part B - (5 x 16 = 80 marks)

11a) An aircraft weighing 2,50,000 N has a wing area of 80 m 2 and its drag equation is C D=0.016 +0.04 C L2 .Calculate (i) minimum thrust required (T min )(ii)minimum power required (P min ) for straight and levelflight and the corresponding true air speeds (V md & Vmp )at sea level and at an altitude where ( ) 1/2 =0.58.Assume sea level air density to be 1.226 kg/m 3 .

[ This problem is similar to example 5.1 of Flightdynamics I . The answ ers are :

At sea level : T min = 12649 N, V md =89.81 m/ s or323.3 kmph, P min = 996.6 kW , V mp =68.23 m/ s or245.62 kmph.

At altitude w here ( ) 1/ 2 =0.58: T min = 12649 N

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Vmd =154.8 m/ s or 557.5 kmph,

P min = 1717.7 kW , V mp =117.6 m/ s or423.1 kmph].

Or

(b)While flying straight and level at sea level at a speedof 100 m/s, the pilot causes his aircraft to enter ahorizontal , correctly banked circle of 1100m radius

while maintaining the same angle of incidence, theengine thrust being altered as necessary. Withoutaltering either the incidence or the engine thrust, the

pilot then brings the aircraft out of the turn and allowsit to climb. Estimate the rate of climb if, at the angleof incidence L/D ratio is 9.

[Solution to this problem is present at the end of this section .

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12 (a) Write short notes on:

(i) International standard atmosphere.

[For answ er see section 2.2 I SA (Flightdynamics I)] .

(ii) Various types of drag of an airplane.

[Answ er : the various types of drags are

skin friction drag, pressure drag, profiledrag, induced drag, w ave drag andparasite drag. See relevant portions inchapter 3 ( Flight dynamics I ) ].

Or

(b) Write short notes on:

(i) V-n diagram.

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[For answ er see section 5.10.3 V-n diagram

(Flight dynamics I) ].(ii) Methods to minimize airplane drag.

[Answ er : (a) P arasite drag is minimized by

smooth surface finish, low thickness ratioairfoil for w ing, high slenderness ratio forfuselage and smooth fillets at w ing-fuselage

junction.(b) W ave drag is reduced by low thickness ratioairfoil and w ing sw eep.

(c) I nduced drag is reduced by increasingaspect ratio.

See remark (ii i) at the end of section 5.2.4

(Flight dynamics I) ].

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13 (a) Discuss briefly the following:

(i) Aerodynamic balancing of control surfaces.[For answ er see section 3.4.1 Aerodynamicbalancing (Flight dynamics I I )] .

(ii) Determination of neutral point and maneuver point(X mp )from flight test.

[For answ er see section 2.5 Determination of neutral point from flight tests (Flight dynamicsI I ) . From discussion in section 2.9 Staticstability in accelerated flight suggest a methodfor determining X mp from flight test] .

Or

(b) Discuss in detail the power effects on staticlongitudinal stability for a jet powered airplane.

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[For answ er see section 2.2.4 Contribution of

pow er plant to C mcg and C m (Flightdynamics II ].

14 (a) Discuss in detail the contribution of various

components of the airplane to static directionalstability.

[For answ er see section 3.2.3 Contribution

of major components (f light dynamics I)] .Or

(b) Discuss briefly the following:

(i) Basic requirements of the rudder.

[ For answ er see section 3.2.6 Directionalcontrol (Flight dynamics I I )] .

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(ii) Aileron reversal

[For answ er see section 3.3.6 Role control (Flight dynamics I I )].

(iii) Adverse yaw.

[For answ er see Adverse yaw under section3.2.6 (Flight dynamics I I )] .

15 (a) discuss the following:

(i) Phugoid motion.

[For answ er see section 4.8 Modes of

longitudinal motion (flight dynamics I I )] .(ii) stability derivatives in longitudinal dynamics.

[For answ er see section 4.6 Estimation of

stabili ty derivatives (Flight dynamics I I )] .

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Or(b) Discuss in detail autorotation and spin and

procedure for recovery from these situations.

[For answ er see section 5.1 Stability afters tall (Flight dynamics II )] .

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Solution to question 11(b):

Solution :

This is a solved problem in Ref.1.5 chapter 13.However the solution needs careful understanding

Question : While flying straight and level at sealevel at a speed of 100 m/s, the pilot causes hisaircraft to enter a horizontal , correctly bankedcircle of 1100m radius while maintaining thesame angle of incidence, the engine thrust beingaltered as necessary. Without altering either theincidence or the engine thrust, the pilot thenbrings the aircraft out of the turn and allows it toclimb. Estimate the rate of climb if, at the angleof incidence L/D ratio is 9.

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of the question. We solve it in our own way. The

question says that initially the airplane is flying atsome velocity V 0 in steady level flight at sea level atan angle of attack 0 at which the lift drag ratio

(L/D) is prescribed to be 9. Then the pilot causesthe airplane to go into a steady level coordinatedterm of radius (r=1100m). The angle of attack in

turning flight is kept same as in the level flight( 0 ). Now in a turn the lift has to be more thanweight and to generate the extra lift, at the sameangle of attack, the airplane must fly at higher flightvelocity and would require more thrust. Let usdenote the flight velocity and thrust in turn by V tand T t . The prescribed radius of turn will decide theangle of bank( ) , the flight speed required (V t) and

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and the thrust (Tt). With this increased level of

thrust (T t) the pilot causes the airplane to climbkeeping the angle of attack again same ( 0). Theairplane would now climb at some angle of climb( ) and some resultant velocity (V Rc) . Thevalues of thrust (T t ) and the angle of attack ( 0 )will decide and V Rc . Knowing these we can getthe desired rate of climb.To get T t we proceed as follows.

Note from Eq.(5.92) that in a turn2

tanV

r g

=

Let,

L0 = Lift in level flight = W=1/2 V02

SC L0Lt=Lift in turning flight = W / cos =W sec

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CL0 = Lift coefficient in level flight corresponding to 0 .Lt = 1/2 Vt 2SC L0 = W sec = 1/2 V0 2SC L0 sec Hence for a turn at 0 or C L0, the velocity V t is given

by:

Vt 2 = V 0 2 sec or V t=V 0 (sec ) 1/2

Now or or

Noting V 0=100 m/s , r = 1100 m.sin = (100) 2 / (9.81 x 1100) = 0.927 or =67.9 0 .

Consequently sec = 2.666Vt= (2.666) 1/2 V0Since 0 is same, C D is same in both level flight and

turn. However V t is (2.66) 1/2 V0 . Hence thrust

20tan sec

V g r

=2

tant V r

g =

20sin

V gr

=

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required in turn (T t), in terms of that in level flightthrust (T 0) is :Tt = T 0 (V t /V 0) 2 = 2.666T 0 .As per the question the thrust setting in the climbis same as that in turning flight i.e. 2. 666T

0. The

angle of attack in climb is also 0 .In a steady climb at angle , the equations of motion are :

Tc=D c+W sin Lc = W cos Where T c = thrust in climb = T t = 2.666 T o ,Dc = drag in climbLc= lift in climb

Now L c =W cos = V0 2SC L0 cos But L c is also equal to VRc 2 SC L0

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Hence V rc = V 0 (cos ) 1/2

Since angle of attack in climb is same as in levelflight .

Lc / D c = L 0 /D 0 =

0

00

0

{1 }

cos (1 )cos

cos (1 tan )

c c

cc

c

Now T D W sin

W D sin D

W D sin

D L

D D

= +

= +

= +

= +

2

0( ) cos RcV V

=

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But L 0 /D 0 = lift drag ratio = 9 in this case.Since D 0 = T 0 .

0

0 0

(1 tan ) cos

cos sin

2.666

c

c t

T LT D

L D

T T

Now T T

= +

= +

= =

in this case

Consequently ,

2 22.666 cos 9sin

(2.666 cos ) 81(1 cos )

cos 0.982 0.92

or

or or

= +

=

=

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Ignoring the second value ,

0

cos 0.982. sin 0.187

cos 100 0.982 99.1 /

/ sin 99.1 0.187 18.5 /1110 / min

Rc

Rc

Hence

Then V V m s

R C V m sm

= =

= = =

= = = =

=

Rate of climb