This set of Aircraft Performance Multiple Choice Questions & Answers focuses on “Range and Endurance for Aircraft with Power Producing Engines”.
1. Which of the following is the correct performance equation for power-producing engines?
a) ηV=PD
b) ηD=PV
c) ηP=VD
d) V=ηPD
View Answer
Explanation: The correct equation for performance equation is ηP=VD where ŋ is propeller efficiency, P is drag power, D is drag and V is velocity. In a power-producing engine the shaft power is produced as well as propulsive thrust produced by the propeller which is converted into power.
2. What is the unit of specific fuel consumption?
a) kg/kW-sec
b) kg/W-hr
c) kg/hr
d) kg/kW-hr
View Answer
Explanation: The unit of specific fuel consumption is given by kg/kW-hr. The formula for specific fuel consumption is C=\(\frac{Q_f}{P}\) where Qf is fuel flow rate, P is power and C is specific fuel consumption.
3. Which of the following is the correct formula for specific fuel consumption?
a) Qf=\(\frac{P}{C}\)
b) Qf=\(\frac{C}{P}\)
c) C=\(\frac{P}{Q_f}\)
d) C=\(\frac{Q_f}{P}\)
View Answer
Explanation: The unit of specific fuel consumption is given by kg/kW-hr. The formula for specific fuel consumption is C=\(\frac{Q_f}{P}\) where Qf is fuel flow rate, P is power and C is specific fuel consumption.
4. What is the value of specific fuel consumption where the fuel heating is 42,800kJ/kg and power produced is 5760 kW-hr?
a) 2.06 kg/kW-hr
b) 0.206 kg/kW-hr
c) 0.0206 kg/kW-hr
d) 0.00206 kg/kW-hr
View Answer
Explanation: First, lets convert fuel heating value from kJ/kg to kW-hr/kg. Since, 1 kW-hr = 3600 kJ,
(42,800 kJ/kg) / (3600 kJ/kW-hr) = 11.89 kW-hr/kg
Now, calculate SFC (using the converted fuel heating value):
SFC = (Fuel Heating Value) / (Power Produced)
SFC = (11.89 kW-hr/kg) / (5760 kW-hr)
SFC ≈ 0.00206 kg/kW-hr
5. Specific air range is maximum when the aircraft is flying at minimum drag speed.
a) True
b) False
View Answer
Explanation: Specific air range is maximum when the aircraft is flying at minimum drag speed and specific endurance is maximum when the aircraft is flying at minimum power speed. This is the same relation with thrust producing engines.
6. Specific endurance is maximum when the aircraft is flying at minimum drag speed.
a) True
b) False
View Answer
Explanation: Specific air range is maximum when the aircraft is flying at minimum drag speed and specific endurance is maximum when the aircraft is flying at minimum power speed. This is the same relation with thrust producing engines.
7. What is the relation between specific air range and specific enurance?
a) SAR=V*SE
b) SAR=\(\frac{V}{SE}\)
c) SE=V*SAR
d) SE=\(\frac{V}{SAR}\)
View Answer
Explanation: The formula for specific air range is given by: SAR=\(\frac{V}{CP}\) where v is velocity, C is specific fuel consumption and P is power. The formula for specific endurance is given by: SE=\(\frac{1}{CP}\) where C is specific fuel consumption and P is power.
8. Which of the following is the correct formula for specific air range?
a) SAR=\(\frac{\eta}{C}\frac{L}{D}\frac{1}{W}\)
b) SAR=\(\frac{1}{PC}\)
c) SAR=\(\frac{\eta W}{C}\frac{L}{D}\)
d) SAR=\(\frac{1}{C}\frac{L}{D}\frac{1}{W}\)
View Answer
Explanation: The correct formula for specific air range is given by SAR=\(\frac{\eta}{C}\frac{L}{D}\frac{1}{W}\) where η is propeller efficiency, C is specific fuel consumption, L is lift, D is drag, W is weight. SAR can also be written as SAR=\(\frac{V}{CP}\) where v is velocity, C is specific fuel consumption and P is power.
9. Which of the following is the correct formula for specific endurance?
a) SE=\(\frac{V}{C}\frac{L}{D}\frac{1}{W}\)
b) SE=\(\frac{V}{PC}\)
c) SE=\(\frac{\eta}{C}\frac{L}{D}\frac{1}{W}\)
d) SE=\(\frac{\eta}{CV}\frac{L}{D}\frac{1}{W}\)
View Answer
Explanation: The correct formula for specific endurance is given by SE=\(\frac{\eta}{CV}\frac{L}{D}\frac{1}{W}\) where η is propeller efficiency, C is specific fuel consumption, L is lift, D is drag, V is velocity, W is weight. SE can also be written as SE=\(\frac{1}{CP}\) where C is specific fuel consumption and P is power.
10. What is the value of specific air range where the lift to drag ratio is 10, weight is 50000N, efficiency is 85% and specific fuel consumption is 7.43 kg/kW-hr?
a) 3000km
b) 4000 km
c) 1000km
d) 5000km
View Answer
Explanation: The answer is 4000km. Given \(\frac{L}{D}\)=10, W=50,000kg, η=85%, C=4.25×10-5kg/kW-hr. From the equation SAR=\(\frac{\eta}{C}\frac{L}{D}\frac{1}{W}\). On substituting the values we get SAR= \(\frac{0.85}{4.25 \times 10^{-8}}\)*10*\(\frac{1}{50000}\).
On solving above equation we get SAR=4000 km.
11. What is the value of specific endurance where the lift to drag ratio is 10, weight is 50000N, efficiency is 85% , velocity is 250m/s and specific fuel consumption is 7.43 kg/kW-hr?
a) 30hr
b) 57.6hr
c) 10hr
d) 50hr
View Answer
Explanation: The answer is . Given \(\frac{L}{D}\)=10, W=50,000kg, η=85%, C=4.25×10-5kg/kW-hr, V=69.44km/hr. From the equation SE=\(\frac{\eta}{CV}\frac{L}{D}\frac{1}{W}\). On substituting the values we get SE= \(\frac{0.85}{69.44 \times 4.25 \times 10^{-8}}\)*10*\(\frac{1}{50000}\).
On solving above equation we get SE=57.6hr.
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