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Aircraft Design - Initial Sizing Quiz(MCQ)
A)
aircraft Lofting
B)
aircraft sizing
C)
aircraft drafting
D)
aircraft drawing

Correct Answer :   aircraft sizing


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.

A)
remain fix
B)
cannot be scaled accordingly
C)
can be scaled accordingly
D)
will change as half always

Correct Answer :   can be scaled accordingly


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.

A)
Engine is natured rubber
B)
Engine made from rubber
C)
High strength rubber engine
D)
Engine can be rubberized for sizing

Correct Answer :   Engine can be rubberized for sizing


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.

4 .
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)
1.4
B)
2.0
C)
5
D)
30.5

Correct Answer :   2.0


Explaination : 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.

5 .
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

Correct Answer :   0.9821


Explaination : 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.

A)
2.5
B)
2.63 kg
C)
2.89 unit
D)
2.5*C*T unit

Correct Answer :   2.5*C*T unit


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.

A)
fixed engine
B)
unfixed engine
C)
rubber engine
D)
differential engine

Correct Answer :   fixed engine


Explanation : When existing engine is used for sizing of the aircraft then the engine is called fixed engine. Fixed engines are any existing engine which can almost fit to our requirements.

A)
zero at sizing
B)
fixed during sizing
C)
unfixed during sizing
D)
independent of engine

Correct Answer :   fixed during sizing


Explanation : Thrust will be fixed during sizing if fixed engine is used. Fixed engine is any existing engine that is available for use. However, when we use fixed engine we are bounded with some constraints such as fixed thrust. Hence, we cannot vary thrust as a rubber engine.

A)
Cost effective
B)
Can be stretched
C)
Can be rubberized
D)
Can provide variable thrust at sizing

Correct Answer :   Cost effective


Explanation : Fixed engine is more easily available than a rubber engine. Any existing engine can save millions of dollars which was going to be spend in developing the new engine. However, we cannot stretch it during sizing but they are more cost-effective than the rubber engine.

A)
True
B)
False
C)
Can Not Say
D)
None of the above

Correct Answer :   False


Explanation : Any existing engine has pre-defined size and diameter for their design. Fixed engine will produced finite thrust and has finite thrust loading. We cannot alter the size of the particular fixed engine.

11 .
Fixed engine sizing method is adopted for aircraft. If thrust per engine is 20000 unit and thrust loading T.L. is 0.8 then, find the take-off gross weight. Given aircraft has 2 engines.
A)
32589 unit
B)
50000 unit
C)
58000 unit
D)
60822 unit

Correct Answer :   50000 unit


Explaination : Given, number of engines N = 2, thrust per engine T = 20000 unit and thrust loading T.L. = 0.8
Now, take-off gross weight is given by,
W = N*T / (T.L.) = 2*20000/0.8 = 50000 unit.

A)
Has no disadvantage
B)
Availability is more than rubber engine
C)
Less costly compared to rubber engine
D)
Often requires compromise in mission range or performance

Correct Answer :   Often requires compromise in mission range or performance


Explanation : In general, for any reason if our weight needs to be alter during sizing then we need to provide optimization for such changes. In most of cases we need to alter the range or any other mission parameters.

A)
0.9365
B)
0.937
C)
0.956
D)
0.98052

Correct Answer :   0.956


Explanation : For an aircraft that accelerates from 0.1M to 0.8M, the required fuel fraction is 0.9805.
Similarly, for 0.1M to 2.0M we required a fuel fraction of 0.937 typically.
Now to find fuel fraction for aircraft from 0.8M to 2.0M is given by,
Fuel fraction = (required fuel fraction for 0.1M to 2.0M)/(required fuel fraction for 0.1M to 0.8M)
= 0.937 / 0.9805 = 0.956.

A)
to provide thrust
B)
to provide high lift
C)
to increase maximum lift
D)
to provide space for cargo and passenger

Correct Answer :   to provide space for cargo and passenger


Explanation : Main body of the aircraft is called fuselage. Fuselage is used to provide space for cargo, crew and passenger. Thrust is primarily provided by engines. High lift is generated through high lift devices.

A)
fuselage diameter
B)
length of fuselage and weight
C)
fuselage length and maximum diameter
D)
fuselage diameter and maximum length

Correct Answer :   fuselage length and maximum diameter


Explanation : Fineness ratio is defined as the length of the fuselage to the maximum diameter of the fuselage. If fuselage has a different cross section then, equivalent diameter is used for fineness ratio.

A)
0.1m
B)
0.25m
C)
5m
D)
20.5m

Correct Answer :   0.1m


Explanation : Given, fuselage fineness ratio F=15, length l=1.5m
Now, diameter D = l/F = 1.5/15 = 0.1m.

A)
2-5cm
B)
6m-10m
C)
5-9in
D)
11-20m

Correct Answer :   6m-10m


Explanation : Given a typical fuselage with length L = 20m.
Typical range for canard (control) moment arm = 30% to 50% of the length
= 30% of L to 50% of L = 30% of 20 to 50% of 20
= 0.3*20 to 0.5*20 = 6m – 10m.

A)
Cvt = Lvt*Svt
B)
Cvt = Lvt*Svt/(bwing*Swing)
C)
Cvt = Svt/(bwing*Swing)
D)
Cvt = Lvt/(bwing*Swing)

Correct Answer :   Cvt = Lvt*Svt/(bwing*Swing)


Explanation : For vertical tail Cvt represents tail volume coefficient. It is used for sizing purpose. By adequate selection of it will provide require surface area of the tail. Cvt is given by, Cvt = Lvt*Svt / (bwing*Swing).

A)
35 unit
B)
35.65
C)
34.5
D)
34 unit

Correct Answer :   34 unit


Explanation : Given, take-off gross weight is W0 = 10000 unit.
Now, length is given by,
L = a*W0c
Now, for jet fighter aircraft a=0.93, c=0.39
Hence, length of fuselage is,
L = 0.93*100000.39 = 33.76 = 34unit.

A)
drafting
B)
flap inversion
C)
aileron reversal
D)
independent from any disturbances

Correct Answer :   aileron reversal


Explanation : Typical high speed aircrafts are affected by ‘aileron reversal’. Aileron reversal is generated as a result of much complex air loads. Air loads are strong enough that they twits wing itself. This results in wrong way rolling at certain speeds.

A)
inboard fuel
B)
inboard aileron
C)
thrust reversal
D)
thrust decrement

Correct Answer :   inboard aileron


Explanation : To reduce the degree of aileron reversal we often use an inboard aileron. Inboard aileron will reduce the excessive twisting of wing. Thrust reversal are used to decelerate aircraft.

A)
1.8m
B)
2m
C)
2.5m
D)
2.8m

Correct Answer :   1.8m


Explanation : Given vertical length L = 2m.
For, typical aircraft rudder is extended up to 90% of the length of the vertical tail.
Hence, Rudder length = 90% of tail length = 90% of 2m = 0.9*2m = 1.8m.

A)
1.8m
B)
2.5m
C)
0.25m-0.5m
D)
1.5-2m

Correct Answer :   0.25m-0.5m


Explanation : Given, tail chord T.C. = 1m.
In typical aircraft, elevator chord is 25% to 50% of tail chord.
Elevator chord = 25% of T.C. to 50% of T.C. = 25% of 1 to 50% of 1
= 0.25*1 to 0.5*1 = 0.25m-0.50m.

A)
2.4m
B)
15m
C)
20m
D)
25m

Correct Answer :   2.4m


Explanation : For a typical wing, ailerons are extended from 50% of span to 90% of span.
Given, Span of wing b=12m. Since we need to find aileron length at one wing we will divide total wing span by 2.
Hence span of starboard wing = 12/2 = 6m.
Length of aileron at starboard wing (right wing) = |difference between 50% and 90% of span|
= |50% of 6m – 90% of 6m| = |0.5*6-0.9*6| = 2.4m.

25 .
Following diagram represents ______
A)
fluttered balance
B)
overhorn balance
C)
overhung balance
D)
notched aerodynamic balance

Correct Answer :   notched aerodynamic balance


Explaination : Above diagram is showing a typical notched aerodynamic balance. It is also called horn aerodynamic balance. We can reduce the force required for deflecting the control surface. However, it is not suitable for high speed aircrafts.

A)
control surface trim
B)
control surface flutter
C)
control surface span
D)
control surface effectiveness

Correct Answer :   control surface flutter


Explanation : Rapid oscillation of control surface is called control surface flutter. This rapid oscillation can tear off the surface or wing/tail itself. This will induces excessive vibration as well. Effectiveness of control surface is measure of how effectively we can control the aircraft.