# Aircraft Design Questions and Answers – Flight Mechanics – Steady Level Flight-1

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This set of Aircraft Design Multiple Choice Questions & Answers (MCQs) focuses on “Flight Mechanics – Steady Level Flight-1”.

1. Find climb angle of climb gradient is 0.6. Assume non accelerated flight.
a) 31°
b) 1.2
d) 9.94°

Explanation: Given, climb gradient G = 0.6.
Now, climb angle = arctan (G) = arctan (0.6) = 30.9 ∼ 31°.

2. If my aircraft is in steady level flight and has thrust of 100 unit then, what will be the approximate value of time Easter if change of aircraft weight. Consider SFC as C unit.
a) -100C
b) 1000
c) 1.25
d) 5.43

Explanation: Given, thrust T = 100 unit and SFC = C unit.
Now, time rate of change of aircraft weight = -C*T = -100*C = -100C.

3. Which of the following is correct for steady level flight?
a) T=D
b) T>D
c) D<T
d) L>W

Explanation: In steady level flight, flight path angle or climb angle is zero. Hence, conventional equation of motion reduces to the thrust T = Drag D. Steady level flight is unaccelerated and hence all the forces should give sum in respective direction as zero.

4. An aircraft has to be designed to operate at aerodynamic efficiency of 10 at cruise. Find the required thrust to weight ratio for maintaining this unaccelerated steady level flight.
a) 0.1
b) 0.6
c) 1.5
d) 10.45

Explanation: Given, aerodynamic efficiency L/D = 10.
Now, for given conditions, thrust to weight ratio T/W is given by,
T/W = D/L = 1/10=0.1.

5. Following diagram represents ______________

a) drag polar
b) weight polar
c) thrust polar
d) lift curve slope

Explanation: Above diagram represents the typical drag polar. A drag polar is simple graphical representation of variation of drag coefficient with respect to lift coefficient or AOA. Drag polar can be used to estimate drag characteristics. Lift curve slope is change in lift with respect to change in AOA.

6. An aircraft is designed with lift coefficient of 1.1 during steady level flight. If dynamic pressure is set to be 25.65 Pa then find weight of aircraft during this flight condition.
a) 28.215 N per unit area
b) 30 kg
c) 59 N
d) 125.25 N

Lift coefficient CL=1.1, q=25.65Pa.
Since, area is not mentioned we will find weight as per unit area.
Weight can be given by,
W = q*S*CL = 25.65*1.1 = 28.215 N per area.

7. If an aircraft is in steady level flight and weighs 2500 kg then, find lift at the given flight condition.
a) 24.525KN
b) 24525kg
c) 2500 N
d) 35000kg

Explanation: Given, a steady level flight and mass m = 2500kg
Lift = m*g = 2500*9.81=24525N = 24.525 KN.

8. An aircraft is designed to be in steady level flight with weight of 1500N and CL of 1.0. Determine at which speed we need to design this aircraft so that it can achieve this requirement. Consider density as 1.2 kg/m3 and reference area as 2m2.
a) 35.35m/s
b) 56m/min
c) 56m/s
d) 70km/s

Explanation: Given, steady level flight. Weight W=1500N, CL=1.0, density = 1.2kg/m3, reference area S=2m2.
Now, velocity is given by,
V = $$\sqrt{\frac{2*W}{density*S*CL}}$$
= $$\sqrt{\frac{2*1500}{1.2*1.0*2}}$$ = 35.35m/s.

9. Let’s consider our aircraft has to generate thrust of 150KN during cruise. If at the cruise thrust loading is 0.86 then, at which value of Lift our aircraft is flying?
a) 174.4KN
b) 450KN
c) 100N
d) 174N

Explanation: For cruise lift is given by,

10. Which of the following is correct?
a) Thrust required is minimum if L/D is maximum during cruise
b) Thrust required is minimum if L/D is minimum for cruise
c) Power required is minimum if L/D is maximum
d) Thrust required is maximum if L/D is maximum during steady state

Explanation: At steady level flight or in cruise, thrust loading or thrust to weight ratio is inversely proportional lift to drag ratio. Hence thrust required will be minimum of lift to drag ratio is maximum.

11. We can design our aircraft to cruise for 3 different lift to drag ratio. Lift to drag ratio are as follows: 10, 12 and 11.25. Now, if we want to reduce thrust required to fly then, at which value of lift to drag ratio our aircraft needs to be designed so that it can operate with minimum thrust required?
a) 12
b) 10
c) 11.25
d) 10.62

Explanation: Here, we’ve been asked to choose from 3 different lift to drag ratio so that our product can operate with minimum thrust required. To have thrust required minimum at cruise we should design our aircraft with maximum value of lift to drag ratio.
From given values, 12 is the highest value available. Hence, among given choices correct choice would be 12.

12. Find the velocity for minimum thrust required at steady level flight if wing loading is 75N/m2 and induced drag factor K is 0.0025. Consider CD0 as 0.02 and density as sea level.
a) 6.45m/s
b) 97m/s
c) 120m/s
d) 75m/s

Explanation: Given, CDO = 0.02, k = 0.0025. By solving these we get answer as 6.45m/s.

13. If lift coefficient at thrust required minimum is 0.85 and induced drag factor K is 0.0085 then, find the approximate value of drag coefficient at zero lift.
a) 0.0061
b) 0.061
c) 0.61
d) 6.1

Explanation: Given, CL as 0.85 and K as 0.0085.
Now, zero lift drag coefficient for given conditions,
CD0 = CL2/K = 0.85*0.85/0.0085 = 0.0061.

14. Which of the following is required condition of thrust required minimum?
a) CL = (CD0/K)0.5
b) CL = (CD0)*0.21
c) CL = (CD0/K)*1.5
d) CL = 0.5

Explanation: At cruise Thrust required is same as drag produced by the aircraft. Now, to operate the aircraft at Thrust required minimum then, aircraft should be operated with certain value of lift coefficient. Such value can be determined by using following relation: CL = (CD0/K)0.5 Where, CL is minimum Thrust required lift coefficient, CD0 is parasite drag coefficient and K is induced drag constant.

15. Find the drag at which thrust required is minimum. Consider CD0 as 0.015 and reference area is 5m2. Consider steady level flight with q of 15Pa.
a) 2.25N
b) 22.5N
c) 225N
d) 2250N

Explanation: Given, steady level flight with CD0 of 0
015, q as 15Pa. Reference area S = 5m2.
Now, drag at thrust required minimum is given by,
D = 2*CD0*q*S = 2*0.015*15*5=2.25N.

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