# Aircraft Design Questions and Answers – Propulsion – Propeller-Engine Integration

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This set of Aircraft Design Multiple Choice Questions & Answers (MCQs) focuses on “Propulsion – Propeller-Engine Integration”.

1. What is the old thumb rule to design a propeller?
a) Keep it as long as possible
b) Keep it as short as possible
c) Keep it as short and wide as possible
d) Keep it as half of diameter always

Explanation: ‘Keep it as long as possible’ is old thumb rule for propeller. However, the length of the propeller is limited by the tip speed. Propeller tip speed should be below sonic speed.

2. Which of the following is correct?
a) (Vtip) static = π*n*d/60 fps
b) (Vtip) static = n*d feet
c) (Vtip) static = 2*n*d/360 s
d) (Vtip) static = d/60 fs

Explanation: Correct relation is given by,
(Vtip) static = π*n*d/60 fps
Where, n = rpm based on engine data
d = diameter
Units should be observed carefully and equation should be used accordingly.

3. Helical tip speed is given by ____________
a) $$\sqrt{Vtip^2+ V^2}$$
b) 2*ᴨ*d
c) CL*d
d) Drag*Lift

Explanation: The helical tip speed is given by, (Vtip) helical = $$\sqrt{Vtip^2+ V^2}$$. The tip of propeller is following a helical path through air. At sea level helical tip speed of wooden propeller should be below 850 fps typically.

4. Propeller has diameter of 25 unit. What will be the static tip speed of the propeller? (Given n = 1200rpm.)
a) 1570.8 unit
b) 3004.5
c) 2345.67 unit
d) 7643.23

Explanation: Given, diameter d = 25 unit, n = 1200rpm.
Now, static tip speed V = ᴨ*n*d/60 = ᴨ*1200*25/60 = 1570.8 unit.

5. If propeller has static tip speed of 600 ft. per second then, find diameter of propeller. Given n = 1250rpm.
a) 9.167
b) 19.8976
c) 23.96
d) 3.2876

Explanation: Given, Static tip speed V = 600 ft. per second, n = 1250rpm.
Now, diameter d = V*60/π*n = 600*60/π*1250 = 9.167 ft.

6. Which of the following is correct for two blade propeller?
a) Diameter d = 22*(Hp)0.25
b) Diameter d = 22*(Hp)0.56
c) Diameter d = 12.22*(Hp)0.25
d) Diameter d = 2.2 ft

Explanation: An initial estimation of propeller diameter is given by,
Diameter d = 22*(Hp)0.25. Where, Hp is horse power.
Above equation provides approximate value for 2 bladed propeller diameter.

7. If a propeller is designed to have 200 Hp then find the appropriate diameter for the propeller. Consider 2 blades.
a) 82.73 unit
b) 210
c) 34.56
d) 180

Explanation: Given, power = 200Hp, number of blades = 2
Now, diameter d = 22*(Hp)0.25 = 22*(200)0.25 = 82.73 unit.

8. Let’s consider we need to design a propeller engine which has 3 blades. What will be the typical value of propeller diameter? Engine is expected to produce 400 Hp of power.
a) 80.5 unit
b) 129
c) 128.94
d) 129.95

Explanation: Given, power = 400Hp, number of blades = 3
Now, diameter d = 18*(Hp)0.25 = 18*4000.25 = 80.5 unit.

9. Which of the following is an example of a cooling system?
a) Downdraft cooling
b) Downdraft lofting
c) Sideway lofting
d) Sideway circulation

Explanation: Downdraft cooling is a typical example of a cooling system which can be used in aircraft. Air flows in downward direction in the downdraft cooling configuration. Lofting is mathematical modelling of skin and aircraft. Circulation is related to vorticity in the flow.

10. Following diagram represents _____ a) updraft cooling
b) downdraft cooling
c) updraft lifting
d) downdraft lifting

Explanation: Downdraft and updraft cooling are typical cooling system used in aircraft. Above diagram is representing typical updraft cooling system. Here, in this method cooling air flows in upward direction. It creates more efficient cooling flow than the downdraft cooling.

11. Which of the following is correct?
a) Variable pitch propeller is used to improve thrust characteristics
b) Variable pitch should not be used at all
c) Always use variable pitch propeller
d) Variable pitch has no disadvantage

Explanation: Variable pitch propeller is used to improve thrust characteristics across broad speed range. Variable pitch can be used to alter accelerating force. However, variable pitch has some drawbacks as well. Hence, it should be used as per mission requirements.

12. Scaling factor of turboprop engine is 1.2. If, the actual length of an existing turboprop engine is 12 unit then, what will be the length of scaled engine?
a) 23.68
b) 56.6
c) 12
d) 32

Explanation: Given, scale factor SF = 1.2, Actual length L = 12 unit
Now, Length of the scaled engine l = L*SF3.73 = 12*1.23.73 = 23.68 unit.

13. A turboprop engine is designed to generate bhp of 2500. Estimate the appropriate weight of the engine based on statistical models.
a) 893.84 lb.
b) 21kg
c) 4555lb
d) 321 slug

Explanation: Given, bhp = 2500
Now, approximate value of weight is given by, W = 1.67*bhp0.803 = 1.67*25000.803 = 893.84 lb.

14. A turboprop engine is to be scaled from an existing engine. If scale factor is 0.85 then, what will be the value of diameter of our engine? Given actual diameter of selected existing engine is 8.56unit.
a) 8.4 unit
b) 23
c) 0.89
d) 9.123

Explanation: Given, scale factor SF = 0.85, actual diameter D = 8.56 unit
Now, scaled engine diameter d = D*SF0.12 = 8.56* 0.850.12 = 8.4 unit.

15. Find the ratio of weight to length of the turboprop engine which has bhp of 3000.
a) 12.60
b) 60.12
c) 60.012
d) 60.1206

Explanation: Given, bhp = 3000.
Now, ratio of weight to length is given by,
W/L = 0.403*(bhp)0.43 = 0.403*30000.43 = 12.60.

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