Aircraft Performance Questions and Answers – Cruising Performance – Effect of Alternative Fuel Flow Laws

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This set of Aircraft Performance MCQs focuses on “Cruising Performance – Effect of Alternative Fuel Flow Laws”.

1. The range and endurance are proportional to which of the following factors?
a) fuel flow
b) net thrust
c) bypass ratio
d) airspeed
View Answer

Answer: b
Explanation: The range and endurance are proportional only to net thrust. The fuel flow laws proved to be a constant idealized laws. Though fuel flow laws cannot describe all the characteristics of the engine they are proved to be a constant idealized laws.
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2. Which of the following is a correct form of fuel flow?
a) C=C2θMn
b) C=C2θ\(\frac{1}{2}\)M
c) C=C2θM
d) C=C2θ\(\frac{1}{2}\)Mn
View Answer

Answer: d
Explanation: The correct formula for fuel flow is C=C2θ\(\frac{1}{2}\)Mn where θ is the temperature, M is mach number and n varies from 0.2-0.6 for turbojet to turbofan. It depends on the bypass ratio of the combustion chamber.

3. Thrust produced is the functional form of engine speed and flight Mach number.
a) True
b) False
View Answer

Answer: a
Explanation: Thrust produced is the functional form of engine speed and flight Mach number. The temperature decreases as the altitude increases so that during the cruise-climb situation the rotational speed increases and mach number increases.
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4. The value of ‘n’ in fuel flow formula C=C2\(\theta^{\frac{1}{2}}\)Mn is 0.6 for turbojet.
a) True
b) False
View Answer

Answer: b
Explanation: The correct formula for fuel flow is C=C2\(\theta^{\frac{1}{2}}\)Mn where θ is the temperature, M is mach number and n varies from 0.2-0.6 for turbojet to turbofan. It depends on the bypass ratio of the combustion chamber.

5. The value of ‘n’ in fuel flow formula C=C2\(\theta^{\frac{1}{2}}\)Mn is 0.2 for turbofan.
a) True
b) False
View Answer

Answer: b
Explanation: The correct formula for fuel flow is C=C2\(\theta^{\frac{1}{2}}\)Mn where θ is the temperature, M is mach number and n varies from 0.2-0.6 for turbojet to turbofan. It depends on the bypass ratio of the combustion chamber.
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6. The formula for velocity of maximum specific air range is __________
a) \(\big[\frac{3-n}{1+n}\big]^{\frac{1}{4}}\)
b) \(\big[\frac{3-n}{1+n}\big]^{\frac{1}{3}}\)Vmd
c) \(\big[\frac{3-n}{1+n}\big]^{\frac{1}{4}}\)Vmd
d) \(\big[\frac{3+n}{1+n}\big]^{\frac{1}{4}}\)Vmd
View Answer

Answer: c
Explanation: The formula for velocity of maximum specific air range is \(\big[\frac{3-n}{1+n}\big]^{\frac{1}{4}}\)Vmd where ‘n’ is the exponent and Vmd is maximum velocity. The value of ‘n’ depends on the value of bypass ratio of the engine. This implies that as the value of ‘n’ increases the bypass ratio increases.

7. The formula for velocity of maximum specific endurance is _______________
a) \(\big[\frac{3+n}{1+n}\big]^{\frac{1}{4}}\)Vmd
b) \(\big[\frac{2+n}{1+n}\big]^{\frac{1}{4}}\)Vmd
c) \(\big[\frac{2+n}{2-n}\big]^{\frac{1}{4}}\)Vmd
d) \(\big[\frac{2-n}{2+n}\big]^{\frac{1}{4}}\)Vmd
View Answer

Answer: d
Explanation: The formula forvelocity of maximum specific endurance is \(\big[\frac{2-n}{2+n}\big]^{\frac{1}{4}}\)Vmd where ‘n’ is the exponent and Vmd is maximum velocity. The value of ‘n’ depends on the value of bypass ratio of the engine. This implies that as the value of ‘n’ increases the bypass ratio increases.
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8. What is the velocity of maximum specific air range of the engine when the maximum air speed is 400m/s and the exponential value is 0.2?
a) 300.84
b) 380.43
c) 440.76
d) 494.37
View Answer

Answer: d
Explanation: The answer is 494.37.
Given n=0.2, Vmd=400m/s. From the formula \(\big[\frac{3-n}{1+n}\big]^{\frac{1}{4}}\)Vmd. On substituting the values we get \([\frac{3-0.2}{1+0.2}]^{\frac{1}{4}}\)*400=494.37.

9. What is the velocity of maximum specific endurance of the engine when the maximum air speed is 400m/s and the exponential value is 0.2?
a) 350.43
b) 490.34
c) 494.37
d) 380.43
View Answer

Answer: d
Explanation: The answer is 380.43.
Given n=0.2, Vmd=400m/s. From the formula \(\big[\frac{2-n}{2+n}\big]^{\frac{1}{4}}\)Vmd. On substituting the values we get \(\big[\frac{2-0.2}{2+0.2}]^{\frac{1}{4}}\)*400=380.43.
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10. What is the value of ‘n’ exponential in the fuel flow law for a high bypass ratio turbofan?
a) 0.4
b) 0.5
c) 0.6
d) 0.7
View Answer

Answer: c
Explanation: The value of ‘n’ exponential in the fuel flow law for a high bypass ratio turbofan is 0.6. The value of ‘n’ exponential is directly proportional to the engine bypass ratio i.e. it increases when the bypass ratio increases.

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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