A turboprop delivers \( 2500 \ \mathrm{N\,m} \) at \( 1600 \ \mathrm{rpm} \) at the propeller shaft. The shaft power is approximately
Answer: A
Angular speed is \( 1600 \times 2\pi/60 = 167.6 \) radians per second, so the power is \( 2500 \times 167.6 \approx 419 \) kilowatts. With rpm governed constant, torque alone then tracks power, which is why it serves as the power setting parameter.
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A gas generator turns at \( 30\,000 \ \mathrm{rpm} \) and the propeller at \( 1500 \ \mathrm{rpm} \). The gearbox reduction ratio is
Answer: B
Correct answer: 20 to 1
The reduction ratio is the input speed over the output speed:
\[ \frac{30\,000}{1500} = 20 \]Reductions of this order are unavoidable: the gas generator must turn fast for its small-diameter turbomachinery to work efficiently, while the propeller is limited to a low speed by its tip Mach number. The gearbox that bridges them is heavy, expensive, and often the limiting component in the whole installation.
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A turboprop produces \( 1200 \ \mathrm{kW} \) of shaft power and \( 1.5 \ \mathrm{kN} \) of residual jet thrust at \( 100 \ \mathrm{m/s} \). Taking propeller efficiency as 0.85, the equivalent shaft power is approximately
Answer: C
Correct answer: 1376 kW
Equivalent shaft power credits the residual jet thrust by converting it to the shaft power that would have been needed to produce the same thrust through the propeller:
\[ \mathrm{ESHP} = \mathrm{SHP} + \frac{F_j V}{\eta_p} = 1200 + \frac{1500 \times 100}{0.85 \times 1000} \]\[ = 1200 + \frac{150}{0.85} = 1200 + 176 = 1376 \text{ kW} \]Adding the jet power of 150 kW without dividing by the propeller efficiency gives 1350 kW, which under-credits the jet: the propeller would have needed more than 150 kW of shaft power to deliver that thrust.
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A propeller of \( 3.4 \ \mathrm{m} \) diameter turns at \( 1200 \ \mathrm{rpm} \). Its tip speed due to rotation is approximately
Answer: D
Correct answer: 214 m/s
The rotational tip speed is
\[ V_{tip} = \pi D N = \pi \times 3.4 \times \frac{1200}{60} = \pi \times 3.4 \times 20 \approx 214 \text{ m/s} \]The blade also meets the forward flight speed, so the resultant relative velocity at the tip is the vector sum of the two, and it is that resultant which must stay clear of compressibility effects. This is why propeller speed is limited and why a large diameter propeller must be geared down so far.
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At a true airspeed of \( 130 \ \mathrm{m/s} \), the propeller of the previous question has a resultant tip speed of approximately
Answer: A
The two components are perpendicular, so the resultant is \( \sqrt{214^2 + 130^2} \approx 250 \) metres per second. Compressibility losses begin well below the speed of sound, which is why the margin matters even at this speed.
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A turboprop develops \( 1500 \ \mathrm{kW} \) at a propeller efficiency of 0.86 and a flight speed of \( 120 \ \mathrm{m/s} \). The propulsive thrust is approximately
Answer: B
The thrust power is \( 1500 \times 0.86 = 1290 \) kilowatts, and dividing by the flight speed gives \( 1290\,000/120 \). The relation shows why a propeller gives such large thrust at low speed and why that advantage falls away with speed.
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An engine is flat rated to \( 1000 \ \mathrm{kW} \) and could produce \( 1150 \ \mathrm{kW} \) on a cold day. The surplus available is
Answer: C
Correct answer: 15 per cent
The surplus is taken on the rated figure:
\[ \frac{1150 - 1000}{1000} = 0.15 = 15\% \]That surplus is not wasted - it is the whole point of flat rating. Because the engine reaches its rated 1000 kW without being pushed to its temperature limit, it runs cooler, lasts longer, and delivers full rated power on a hot day when an unrated engine would be falling short.
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A torque limit of \( 3000 \ \mathrm{N\,m} \) applies at \( 1500 \ \mathrm{rpm} \). The corresponding power limit is approximately
Answer: D
Angular speed is \( 1500 \times 2\pi/60 = 157 \) radians per second, giving \( 3000 \times 157 \approx 471 \) kilowatts. The gearbox is often the limiting component at low altitude, well before any temperature limit is approached.
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A helicopter rotor turns at \( 258 \ \mathrm{rpm} \) driven through a main gearbox from a power turbine at \( 21\,000 \ \mathrm{rpm} \). The overall reduction is approximately
Answer: A
Correct answer: 81 to 1
The overall reduction is
\[ \frac{21\,000}{258} \approx 81 \]A helicopter main gearbox is therefore an even more demanding component than a turboprop’s, reducing a very high turbine speed to a rotor speed of a few hundred rpm in two or three stages, while carrying the whole weight of the aircraft through its rotor mast. It is among the most safety-critical and maintenance-intensive items in the machine.
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An auxiliary power unit supplies \( 90 \ \mathrm{kW} \) of electrical power at a generator efficiency of 0.92. The shaft power required is approximately
Answer: B
Correct answer: 98 kW
The shaft must supply more than the electrical output because the generator is not perfect:
\[ P_{shaft} = \frac{90}{0.92} \approx 97.8 \text{ kW} \]Multiplying rather than dividing by the efficiency gives 83 kW, which would have the generator producing more power than it received. The missing 8 kW appears as heat in the windings and iron, and must be carried away by the generator’s cooling.
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