This set of Aircraft Performance Multiple Choice Questions & Answers (MCQs) focuses on “Cruising Performance – Specific Air Range and Specific Endurance”.

1. In cruise performance the aircraft is considered to be steady, level, straight, symmetric flight with uniform acceleration.

a) True

b) False

View Answer

Explanation: In cruise performance the aircraft is considered to be steady, level, straight, symmetric flight with no acceleration or maneuver.

2. What is a trim in aircraft performance?

a) The state of equilibrium where the forces and moments are balanced

b) The state of equilibrium where the forces and moments are not balanced

c) The state of equilibrium where only the forces are balanced

d) The state of equilibrium where only the moments are balanced

View Answer

Explanation: Trim is a state of equilibrium where the forces and moments are balanced. In this condition

- F
_{N}=D - L = W

Where F_{N} is normal force acting on the aircraft, D is drag, L is lift and W is weight.

3. An aircraft is climbing with a 20° of climb angle and a 5° of angle of attack having a mass of 11,000 kg and aircraft drag is 9,000 N then determine engine thrust.

a) 5257.79 N

b) 1810.97 N

c) 28013.99 N

d) 46082.75 N

View Answer

Explanation: Aircraft weight=mg=11000*9.81=107910 N.

Engine thrust: From equating the horizontal forces we get the following equation: T cos α=D+Wsin γ

T=\(\frac{D+Wsin\gamma}{cos\alpha}\)=\(\frac{9000+107910*sin(20)}{cos(5)}\)

T=46082.75 N.

4. An aircraft is climbing with a 20° of climb angle and a 5° of angle of attack having a mass of 11,000 kg and aircraft drag is 9,000 N then determine aircraft lift.

a) 11453.17 N

b) 9220.07 N

c) 11553.17 N

d) 97385.85 N

View Answer

Explanation: Aircraft weight=mg=11000*9.81=107910 N.

Engine thrust: From equating the forces we get the following equation: T cos α= D+W sin γ

T=\(\frac{D+Wsin\gamma}{cos\alpha}\)=\(\frac{9000+107910*sin(20)}{cos(5)}\)

T=46082.75 N.

From equating the vertical forces we get the following equation: L+Tsinα=Wcosγ

L=107910*cos(20)–46082.75*sin(5)

L=97385.85 N.

5. An aircraft is climbing with a 20° of climb angle and a 5° of angle of attack having a mass of 11,000 kg and aircraft thrust is 46082.75N then determine aircraft drag.

a) 33898.64 N

b) 52708.59 N

c) 82814.78 N

d) 8999.99 N

View Answer

Explanation: Aircraft weight=mg=11000*9.81=107910 N.

Engine thrust: From equating the horizontal forces we get the following equation: Tcosα=D+Wsinγ

D=Tcosα-Wsinγ

D=46082.75*cos(5)–107910*sin(20)

D=8999.99 N.

6. What is specific air range?

a) horizontal distance flown per unit of fuel consumed

b) vertical distance flown per unit of fuel consumed

c) horizontal distance flown per unit time

d) vertical distance flown per unit time

View Answer

Explanation: Specific air range (SAR) is the horizontal distance flown per unit of fuel consumed. It can be expressed as SAR=\(\frac{V}{Q_f}\) where V is true airspeed and Q

_{f}is fuel mass flow. The units of SAR are length/mass i.e. nm/Kg.

7. What is specific endurance?

a) It is instantaneous flight distance per unit of time

b) It is instantaneous flight time per unit of distance

c) It is instantaneous flight distance per unit of fuel consumed

d) It is instantaneous flight time per unit of fuel consumed

View Answer

Explanation: Specific endurance (SE) is the instantaneous flight time per unit of fuel consumed. Specific endurance (SE) can be expressed as SE=\(\frac{1}{Q_f}\) where Q

_{f}is fuel mass flow. The units of SAR are time/mass i.e. hr/Kg.

8. In cruise performance the fuel mass flow determines the rate of change of mass of the aircraft.

a) True

b) False

View Answer

Explanation: In cruise performance the fuel mass flow determines the rate of change of mass of the aircraft. It is expressed by the formula Q

_{f}=\(\frac{-dm}{dt}\) where dm is mass rate and dt is time rate. The fuel mass is always negative as mass of the aircraft always decreases.

9. What will be the specific endurance of an aircraft having fuel mass flow 0.75 kg/s?

a) 4.33 s/kg

b) 3.33 s/kg

c) 1.33 s/kg

d) 2.33 s/kg

View Answer

Explanation: The answer is 1.33. Given Q

_{f}=0.75 kg/s. The specific endurance is given by SE=\(\frac{1}{Q_f}\)

On substituting we get SE=\(\frac{1}{0.75}\)

SE=1.33s/kg.

10. What will be the specific endurance of an aircraft having fuel mass flow 0.75 kg/s and true air speed is 300m/s?

a) 400 m/kg

b) 333 m/kg

c) 133 m/kg

d) 250 m/kg

View Answer

Explanation: The answer is 1.33. Given Q

_{f}=0.75 kg/s. The specific endurance is given by SAR=\(\frac{V}{Q_f}\)

On substituting we get SAR=\(\frac{300}{0.75}\)

SAR=400m/kg.

**Sanfoundry Global Education & Learning Series – Aircraft Performance.**

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