Aircraft Performance Questions and Answers – Range and Endurance for Aircraft with Power Producing Engines

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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

Answer: c
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.
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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

Answer: d
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

Answer: d
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.
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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) 0.135 kg/kW-hr
b) 7.43 kg/W-hr
c) 1.546 kg/kW-hr
d) 7.43 kg/kW-hr
View Answer

Answer: d
Explanation: The answer is 7.43 kg/kW-hr. Given Qf=42,800 kJ/kg and P=5760 kW-hr. From the formula C=\(\frac{Q_f}{P}\) we will find C. On substituting we get C=\(\frac{42800}{5760}\).
On solving above equation we get the value of C=7.43 kg/kW-hr.

5. Specific air range is maximum when the aircraft is flying at minimum drag speed.
a) True
b) False
View Answer

Answer: a
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.
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6. Specific endurance is maximum when the aircraft is flying at minimum drag speed.
a) True
b) False
View Answer

Answer: a
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

Answer: a
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.
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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

Answer: a
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

Answer: d
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.
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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

Answer: b
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

Answer: b
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.

Sanfoundry Global Education & Learning Series – Aircraft Performance.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter