Aircraft Design Mcqs - Set 21

1)   Total energy of an aircraft is defined as ____________

a. sum of the potential and kinetic energy
b. total kinetic energy only
c. total potential energy only
d. product of kinetic and potential energy
Answer  Explanation 

ANSWER: sum of the potential and kinetic energy

Explanation:
At any given point in time, the total energy of the aircraft or energy state of an aircraft is the sum of the potential energy and kinetic energy. Kinetic energy is equal to half of product of mass and square of velocity.


2)   Total energy divided by the aircraft weight is term as ____________

a. Specific energy
b. KE
c. PE
d. Total energy
Answer  Explanation 

ANSWER: Specific energy

Explanation:
When total energy is divided by the aircraft weight then, it is termed as specific energy. Total energy of an aircraft is summation of kinetic and potential energy. It is often known as energy state of the aircraft at a given point in time.


3)   If Total energy of an aircraft is found to be 1200 unit and kinetic energy is 1100 unit then, determine the potential energy.

a. 100 unit
b. 213 unit
c. 456 unit
d. 500 unit
Answer  Explanation 

ANSWER: 100 unit

Explanation:
Potential energy = Total energy – kinetic energy = 1200 – 1100 = 100 unit.


4)   Find the total energy of an aircraft if kinetic energy and potential energy are 3600 and 1200 units respectively.

a. 4800 unit
b. 2800 unit
c. 1234 unit
d. 5789 unit
Answer  Explanation 

ANSWER: 4800 unit

Explanation:
Total energy = kinetic energy + potential energy
= 3600 + 1200 = 4800 unit.


5)   If sum of potential and kinetic energy of an aircraft is 5000 unit and aircraft weighs around 1200 unit then, determine specific total energy.

a. 4.16 unit
b. 2.89 unit
c. 40
d. 10.8
Answer  Explanation 

ANSWER: 4.16 unit

Explanation:
Specific total energy = total energy/weight
= 5000/1200 = 4.16 unit.


6)   Unit of specific energy is __________

a. meter
b. meter per second
c. unit less
d. newton
Answer  Explanation 

ANSWER: meter

Explanation:
Specific energy is defined as the ratio of total energy to the weight of the aircraft. Specific energy is defined by distance unit. It can be meter, feet etc. Meter per second is the unit of velocity or speed. Newton is standard unit of force.


7)   If my aircraft has access power off 520 KW and it weighs around 1200 N then, determine the value of specific excess power.

a. 433.33 m/s
b. 123.88 ft. s
c. 119
d. 345.768m/s
Answer  Explanation 

ANSWER: 433.33 m/s

Explanation:
Specific excess power = excess power/weight
= 520*1000/1200 = 433.33 m/s.


8)   Find the approximate value of time required to change energy height. Consider change in energy height is 250m and specific power for the operation is 100m/s.

a. 2.5s
b. 4.5min
c. 2.5hr
d. 2.59min
Answer  Explanation 

ANSWER: 2.5s

Explanation:
Time required to change energy height can be approximated as follows,
Time required = change in energy height/specific power = 250/100 = 2.5 s.


9)   If a Maneuver requires specific power of 1200 unit then, find the approximate value of fuel specific energy. Given thrust as 1000 unit and tsfc of 0.00042 unit.

a. 2857.14 unit
b. 1568.45 unit
c. 4500 unit
d. 6789 unit
Answer  Explanation 

ANSWER: 2857.14 unit

Explanation:
Fuel specific energy = specific power / (thrust*tsfc)
= 1200/1000*0.00042 = 2857.14 unit.


10)   The level flight envelope is determined by ________

a. stall limit lines and from zero specific power
b. wing aileron size
c. spoilers area
d. tail area
Answer  Explanation 

ANSWER: stall limit lines and from zero specific power

Explanation:
The level flight operating envelope is determined by using stall limit lines and zero specific power. The zero specific power lit is typically shown for both maximum Thrust and for military thrust. Tail area is based on tail sizing.


11)   The highest altitude where Ps is zero is called ________

a. absolute ceiling
b. drag polar
c. geometric height
d. geometric altitude
Answer  Explanation 

ANSWER: absolute ceiling

Explanation:
Absolute ceiling is defined as the highest altitude at where Ps is zero. Geometric height or geometric altitude is based on height measured. Drag polar is graphical representation of drag properties.


12)   Which of the following is incorrect?

a. Operating envelope will give same values for all the aircraft
b. Load factor at cruise is unity
c. Lift is same as weight during cruise
d. Wing loading is not same as thrust
Answer  Explanation 

ANSWER: Operating envelope will give same values for all the aircraft

Explanation:
Load factor is defined as ratio of lift to weight. At cruise lift and weight both are equal. Hence, at cruise load factor is unity. Wing loading and Thrust are two different things. Wing loading is ratio of weight by reference area.


13)   Operating envelope is also known as ________

a. flight envelope
b. fly by wire
c. stall speed diagram
d. lift curve
Answer  Explanation 

ANSWER: flight envelope

Explanation:
Operating envelope is also known as flight envelope. Fly by wire is a typical system used for converting pilot’s input to signal. Lift curve is used to illustrate the lift variation with respect to the angle of attack.


14)   If static pressure of flow is 1200psf and operating Mach number is 1.19 then, determine the external flow dynamic pressure.

a. 1189.52 psf
b. 12.89 psf
c. 112 psf
d. 2587 psf
Answer  Explanation 

ANSWER: 1189.52 psf

Explanation:
No explanation is available for this question!


15)   Determine static pressure if external flow dynamic pressure is 1800psf and flow has Mach number of 2.

a. 642.857psf
b. 1000psf
c. 2400 psf
d. 4673 psf
Answer  Explanation 

ANSWER: 642.857psf

Explanation:
No explanation is available for this question!


16)   Consider external flow dynamic pressure as 2000psf and static pressure as 800psf. find the approximate value of Mach number.

a. 1.9
b. 5
c. 4.5
d. 2.78
Answer  Explanation 

ANSWER: 1.9

Explanation:
No explanation is available for this question!


17)   Determine total pressure if static pressure is 1600psf and Mach number is 1.8. Consider flow of air.

a. 9193.2 psf
b. 1000 psf
c. 3450 psf
d. 7892 psf
Answer  Explanation 

ANSWER: 9193.2 psf

Explanation:
No explanation is available for this question!


18)   If flow has Mach number of 1.2 then, determine the ratio of total and external flow dynamic pressure.

a. 2.4
b. 1
c. 3.8
d. 5.6
Answer  Explanation 

ANSWER: 2.4

Explanation:
No explanation is available for this question!


19)   Which of the following is not part of takeoff?

a. Ground roll
b. Transition
c. Climb
d. Descending
Answer  Explanation 

ANSWER: Ground roll

Explanation:
Takeoff segment is subdivided into number individual segments or sections. These segments are ground roll, transition, climb etc. Descending is not the part of takeoff.


20)   Determine the acceleration of an aircraft which is in ground roll segment. Consider weight of the aircraft as 150kN, thrust as 110kN and drag as 10kN. Assume weight on the wheels as 60kN and rolling friction as 0.05.

a. 5.68 m/s2
b. 100kN
c. 100 m
d. 3.5689 Kg*m/s
Answer  Explanation 

ANSWER: 5.68 m/s2

Explanation:
Given, weight W=150KN, thrust T=100KN, drag D = 10KN, rolling friction f= 0.05, weight on wheels w = 60KN.
Acceleration a = g*(T- D – f*w)/W
= 9.81(100 -10-0.05*60)/150
= 9.81*87/150 = 5.68 m/s2.


21)   If stall speed of the aircraft is 10m/s then, find the approximate value of takeoff speed.

a. 11m/s
b. 1m/s
c. 9m/s
d. 7m/s
Answer  Explanation 

ANSWER: 11m/s

Explanation:
Takeoff speed = 1.1*stall speed = 1.1*10 = 11m/s.


22)   If our design has takeoff velocity of 120 knots then, at which value of stall speed we should design our aircraft?

a. 109.08knots
b. 110 knots
c. 150 knots
d. 125 knots
Answer  Explanation 

ANSWER: 109.08knots

Explanation:
Stall speed = 0.909*Takeoff speed = 0.909*120 = 109.08 knots.


23)   Find liftoff distance for an aircraft which has weight of 20kN and thrust of 8kN. Consider thrust is much larger than drag and friction parameters. Consider maximum lift coefficient of 1.1, reference area of 12m2 and sea-level density.

a. 454m
b. 1000m
c. 12m
d. 12.12m
Answer  Explanation 

ANSWER: 454m

Explanation:
No explanation is available for this question!


24)   Find the average velocity at transition if, stall speed is 15 unit.

a. 17.25 unit
b. 19 unit
c. 30 unit
d. 45 unit
Answer  Explanation 

ANSWER: 17.25 unit

Explanation:
Average velocity = 1.15*stall speed = 1.15* 15 = 17.25 unit.


25)   If climb speed during takeoff is 120 unit then, find stall speed.

a. 100knots
b. 240knots
c. 1267knots
d. 215knots
Answer  Explanation 

ANSWER: 100knots

Explanation:
Stall speed = climb speed / 1.2 = 120/1.2 = 100knots.


26)   If Stall speed is 140fps then, find approximate radius of transition arc.

a. 4018 feet
b. 5015 km
c. 1m
d. 10cm
Answer  Explanation 

ANSWER: 4018 feet

Explanation:
Approximate value of radius of transition arc in feet = 0.205*Stall speed2
= 0.205*140*140
= 4018 feet.


27)   Find the horizontal distance travelled during transition segment. Consider radius of transition arc is 2km and thrust loading is 0.6 and L/D is 5.

a. 800m
b. 8000m
c. 80m
d. 0.80m
Answer  Explanation 

ANSWER: 800m

Explanation:
Given, transition arc r = 2km, Thrust loading t = 0.6, and L/D = 5
Now, the horizontal distance H = r*(t – D/L)
= 2*(0.6-0.2) = 0.8 km = 800m.


28)   Find the approach speed if, stall speed is 12 unit.

a. 15.6 unit
b. 12 unit
c. 10 unit
d. 1 unit
Answer  Explanation 

ANSWER: 15.6 unit

Explanation:
Approach speed = 1.3* stall speed = 1.3*12 = 15.6 unit.


29)   Find the approach speed for military aircraft. Considered stall speed is 22unit.

a. 26.4 unit
b. 20 m
c. 12 ft
d. 45 unit
Answer  Explanation 

ANSWER: 26.4 unit

Explanation:
For military aircraft, Approach speed = 1.2*stall speed = 1.2*22 = 26.4 unit.


30)   Find the required amount of touchdown speed if stall speed is 100knots.

a. 115 knot
b. 120 knot
c. 100 knot
d. 99.2 knot
Answer  Explanation 

ANSWER: 115 knot

Explanation:
Touchdown speed = 1.15*stall speed = 1.15*100 = 115 knot.


31)   If stall speed is 120 knots then, find average velocity at flare.

a. 147.6
b. 120
c. 110
d. 567
Answer  Explanation 

ANSWER: 147.6

Explanation:
Average velocity at Flare = 1.23*stall speed = 1.23*120 = 147.6 knots.


32)   For military aircraft touchdown speed is 100 knots then, approximate value of stall speed.

a. 90.90
b. 100
c. 200
d. 223
Answer  Explanation 

ANSWER: 90.90

Explanation:
Stall speed = touchdown speed/1.1 = 100/1.1 = 90.90.


33)   Find the required average velocity during flare for military aircraft if stall speed is 45 unit.

a. 51.75 unit
b. 60 unit
c. 45 unit
d. 100 unit
Answer  Explanation 

ANSWER: 51.75 unit

Explanation:
Average velocity during flare = 1.15*stall speed = 1.15*45 = 51.75 unit.


34)   Determine the value of FAR field length. Consider approach of 100 ft., flare of 120 feet and ground roll of 600ft.

a. 1366.12
b. 109
c. 600
d. 700
Answer  Explanation 

ANSWER: 1366.12

Explanation:
FAR field length = 1.666*(approach + flare + ground roll)
= 1.666*(100 + 120 + 600) = 1366.12.


35)   Final velocity at the end of the landing phase will be ____________

a. zero
b. same as Approach speed
c. touchdown speed
d. same as climb rate
Answer  Explanation 

ANSWER: zero

Explanation:
Final velocity at the end of the aircraft landing phase will be zero. It is not equal to approach speed. Touchdown speed is achieved during flare segment. Rate of climb is vertical velocity of the aircraft.


36)   Which of the following is correct?

a. Aircraft decelerates from Approach to Touchdown speed during flare
b. Lift is always same as weight
c. Drag is always same as Thrust
d. Thrust loading is defined as the ratio of lift to drag
Answer  Explanation 

ANSWER: Aircraft decelerates from Approach to Touchdown speed during flare

Explanation:
During flare aircraft will decelerates from Approach speed to touchdown speed. Lift is not always same as weight. Drag is not always same as Thrust. During cruise, lift is equal to weight and Thrust is equal to drag.