# Aircraft Design MCQ – Propulsion – Propeller-Engine Integration

1 - Question

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 - Question

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 - Question

Helical tip speed is given by ____________
a) Vtip2+V2−−−−−−−−−√
b) 2*ᴨ*d
c) CL*d
d) Drag*Lift
Explanation: The helical tip speed is given by, (Vtip) helical = Vtip2+V2−−−−−−−−−√. 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 - Question

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 - Question

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 - Question

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 - Question

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 - Question

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 - Question

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 - Question

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 - Question

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 - Question

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 - Question

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 - Question

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 - Question

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