# Aircraft Design Questions and Answers – Rubber-Engine Sizing

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This set of Aircraft Design Multiple Choice Questions & Answers (MCQs) focuses on “Rubber-Engine Sizing”.

1. Process of determining the take-off gross weight and required fuel weight is called as ________
a) aircraft sizing
b) aircraft Lofting
c) aircraft drafting
d) aircraft drawing

Explanation: Aircraft sizing is nothing but the process of estimating and evaluating the total weight of an aircraft. The weight will be estimated based on requirements, specifications, fuel requirements etc. lofting is mathematical modelling of the skin.

2. We can provide sizing based on new design only.
a) True
b) False

Explanation: Aircraft sizing is process of weight determination based on different performance parameters. Aircraft can be sized in two different way; either using new design or using existing design of engine which meet all the required parameters.

3. Which of the following is correct?
a) Rubber engine can be stretched during A/C sizing
b) Rubber engine is fixed during sizing
c) Lift is always same as engine thrust
d) Engine sizing will be same for every A/C in the world

Explanation: A new engine used during sizing is flexible in terms of size, thrust etc. Such an engine is called rubber engine. They can be stretched during sizing as per requirements. Lift will not be same as thrust. Engine sizing will be different for different requirements of aircraft.

4. As weight varies, rubber engine _______
a) can be scaled accordingly
b) cannot be scaled accordingly
c) remain fix
d) will change as half always

Explanation: Rubber engine can be rubberized as per our requirement. If during the initial sizing process, we need to alter the weight then, we can scale the rubber engine accordingly. This is one of the advantages of the rubber engine.

5. The rubber engine can be sized for any thrust so we can have desired thrust loading.
a) True
b) False

Explanation: Fundamental of rubber engine sizing method is that we are using new design of engine. New design can be made as per our requirements. Rubber engine can be scaled according to our requirements and then the performance parameters are evaluated. Hence, rubber engine can be sized to meet our thrust requirements.

6. What is rubber engine?
a) Engine can be rubberized for sizing
b) Engine made from rubber
c) Engine is natured rubber
d) High strength rubber engine

Explanation: Rubber engine as name suggests can be rubberized for sizing. When we use new concept and new engine design, we use rubber engine for sizing. Rubber engines can be stretched during sizing to meet desired requirements.

7. For mission with payload drop, the take-off gross weight W0 is given by ____
a) W0 = Wcrew + Wdropped payload + Wfixed payload + Wfuel + Wempty
b) W0 = Wcrew + Wdropped payload + Wfixed payload + Wfuel
c) W0 = Wcrew + Wdropped payload + Wfixed payload + Wempty
d) W0 = Wcrew + Wdropped payload + Wfuel + Wempty

Explanation: Takeoff gross weight of the aircraft is the net or total weight of each and every possible components and systems. Typically, aircraft gross weight is divided into number of sub weight categories such as crew weight, fuel weight, empty weight etc. A typical takeoff gross weight can be expressed as W0 = Wcrew + Wdropped payload + Wfixed payload + Wfuel + Wempty.

8. A general aviation single engine aircraft has gross weight of 2800lb. The aircraft has maximum speed of 130mph. Determine the approximated value of empty weight fraction if, aspect ratio AR is 9, wing loading is 15 and power loading is 12.
a) 0.5476
b) 0.6457
c) 0.8962
d) 0.9028

Explanation: Given, a general aviation aircraft with single engine.
Maximum velocity Vmax =130mph
Gross weight W0 = 2800lb, wing loading = W.L. = 15, Aspect ratio AR = 9, Power loading P.L. = 12.
For, given parameters empty weight fraction is given by,
We/W0 = a + b*(W0c)*(AR d)*(P.L.-e)*(W.L.f)*(Vmaxg)
Now, for general aviation with single engine, a=-0.25, b = 1.14, c = -0.2, d = 0.08, e = 0.05, f = -0.05, g = 0.27.
Hence, empty weight fraction is given by,
We/W0 = -0.025 + 1.14*(W0-0.2)*(AR 0.08)*(P.L.-0.05)*(W.L.-0.05)*(Vmax0.27)
We/W0 = -0.025 + 1.14*(2800-0.2)*(9 0.08)*(12-0.05)*(15-0.05)*(1300.27)
We/W0 = -0.025 +.07976 = 0.5476.

9. For a given rubber engine, determine fuel burn at combat if combat time is 2.5min.
a) 2.5*C*T unit
b) 2.5
c) 2.89 unit
d) 2.63 kg

Explanation: Given, combat time d = 2.5 min.
Since SFC and thrust are not mentioned we will consider their notation as C and T respectively.
Now fuel burn = SFC * Thrust *Combat time = C*T*2.5 unit.

10. An aircraft is climbing and accelerating to reach the cruise mach number of 0.75. If aircraft is accelerating from 0.1M then, what will be the mission segment fuel fraction for this mission?
a) 0.9821
b) 0.9258
c) 0.5289
d) 0.28958

Explanation: Given, after acceleration mach number = 0.75
For given subsonic mission segment fuel fraction,
= 1.0065 – 0.0325M = 1.0065 – 0.0325*0.75 = 0.9821.

11. A jet fighter is designed to operate at maximum mach of 2.0. Which of the following is approximated value for empty weight fraction if rubber engine sizing is used? Given, W0 = 20000lb, AR = 5, thrust loading T.L. = 0.6 and wing loading W.L. is 25.
a) 0.8109
b) 0.8532
c) 0.526
d) 0.3562

Explanation: Given, jet fighter with maximum Mach number Mmax = 2.0
Gross weight W0 = 20000lb, wing loading = W.L. = 25, Aspect ratio AR = 5, Thrust loading T.L. = 0.6.
For, given parameters empty weight fraction is given by,
We/W0 = [a + b*(W0c)*(AR d)*(T.L.e)*(W.L.f)*(Mmax g)]*K
Now, for jet fighter a = -0.02, b = 2.16, c = -0.1, d = 0.2, e = 0.04, f = -0.1, g = 0.08, K = 1.0
We/W0 = [-0.02 + 2.16*(W0-0.1)*(AR0.2)*(T.L.0.04)*(W.L. -0.1)*(Mmax 0.08)]*1.0
We/W0 = [-0.02 + 2.16*(20000-0.1)*(50.2)*(0.60.04)*(25-0.1)*(20.08)]*1.0
We/W0 = -0.02+0.8309 = 0.8109.

12. If supersonic aircraft has fuel fraction of 0.937 as climb and accelerating segment fuel fraction then, find the cruise mach number. Consider aircraft is accelerating from 0.1M.
a) 2.0
b) 30.5
c) 5
d) 1.4

Explanation: Given, fuel fraction =0.937
For supersonic aircraft accelerating from 0.1M, fuel fraction is given by,
Fuel fraction = 0.991 – 0.007M -0.01M2.
0.937 = 0.991 – 0.007M -0.01M2.
0.01M2 + 0.007M -0.054 = 0.
Now, dividing by 0.01 then,
M2 + 0.7M -5.4 = 0.
Hence, by solving above equation,
M = 2.0 or M = -2.7. Since Mach number cannot be negative, answer is 2.0.

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