An engine runs at \( 11\,000 \ \mathrm{rpm} \) with an inlet temperature ratio of 0.80. Its corrected speed is approximately
Answer: A
Dividing by \( \sqrt{0.80} = 0.894 \) gives about 12298 revolutions per minute. Cold air raises the corrected speed for the same shaft rpm, which is why an engine behaves differently on a cold day at an unchanged physical speed.
Please log in or register to report this question.
Corrected mass flow is \( \dot m \sqrt{\theta}/\delta \). With \( \dot m = 60 \ \mathrm{kg/s} \), \( \theta = 0.85 \) and \( \delta = 0.35 \), the corrected flow is approximately
Answer: B
Correct answer: 158 kg/s
\[ \dot m \frac{\sqrt{\theta}}{\delta} = 60 \times \frac{\sqrt{0.85}}{0.35} = 60 \times \frac{0.922}{0.35} = 158\ \text{kg/s.} \] The corrected value greatly exceeds the physical flow because the pressure ratio in the denominator is well below one - at altitude the air is thin, so a modest physical flow corresponds to a large corrected flow.
Please log in or register to report this question.
A working line sits at a pressure ratio of 24 where the surge line is at 30. The surge margin on pressure ratio is
Answer: C
Correct answer: 25 per cent
Surge margin is the surplus expressed as a fraction of the working line value: \[ \frac{30 - 24}{24} = \frac{6}{24} = 0.25 = 25\ \text{per cent.} \] Dividing by the surge line value instead gives 20 per cent, which is the distractor. The convention matters, because a margin quoted on the wrong base understates the protection available.
Please log in or register to report this question.
Opening the turbine nozzle throat area by 3 per cent increases the corrected flow passed by approximately
Answer: D
Correct answer: 3 per cent
With the nozzle choked the corrected flow it passes is directly proportional to its throat area, so a 3 per cent increase in area passes 3 per cent more corrected flow. This is exactly why nozzle area is such an effective control: a small, easily manufactured change in throat area produces a proportional and predictable shift in the whole engine’s matching.
Please log in or register to report this question.
An engine's exhaust gas temperature margin falls from 80 to 20 kelvin over 6000 cycles. The average rate of loss is
Answer: A
Correct answer: 1 K per 100 cycles
\[ \frac{80 - 20}{6000} = \frac{60}{6000} = 0.01\ \text{K per cycle} = 1\ \text{K per 100 cycles.} \] Rates of this kind are what let an operator forecast a removal: knowing the loss per cycle and the margin remaining gives the number of cycles left before the engine can no longer make its rating.
Please log in or register to report this question.
A turbine must produce \( 152 \ \mathrm{kJ/kg} \) to drive a compressor absorbing \( 150 \) at a mechanical efficiency of 0.99. The check is
Answer: B
The work balance requires the turbine to produce the compressor demand divided by the mechanical efficiency. Bearing and windage losses are small but are met by the turbine, so the figure always exceeds the compressor's requirement.
Please log in or register to report this question.
An engine is flat rated to \( 120 \ \mathrm{kN} \) up to 30 degrees Celsius and could produce \( 138 \) at 15 degrees. The surplus held in reserve at 15 degrees is
Answer: C
Correct answer: 15 per cent
\[ \frac{138 - 120}{120} = \frac{18}{120} = 0.15 = 15\ \text{per cent.} \] Flat rating deliberately holds the thrust down on cool days so that the same thrust is available up to the flat rating temperature. The surplus is not wasted - it is turned into turbine life, because the engine runs cooler than it could for the thrust it is producing.
Please log in or register to report this question.
Compressor efficiency falls from 0.86 to 0.83 through deterioration. The fractional loss is
Answer: D
Correct answer: 3.5 per cent
The loss is expressed as a fraction of the original value: \[ \frac{0.86 - 0.83}{0.86} = \frac{0.03}{0.86} = 0.035 = 3.5\ \text{per cent.} \] Quoting the difference of 0.03 as “3 per cent” confuses percentage points with a percentage change, and the distinction matters when efficiency losses are being accumulated across several components.
Please log in or register to report this question.
A twin spool engine's low pressure shaft runs at 92 per cent of its maximum \( 4200 \ \mathrm{rpm} \) while the high pressure shaft runs at 96 per cent of \( 12\,500 \). Their speeds are
Answer: A
Correct answer: 3864 and 12 000 rpm
\[ 0.92 \times 4200 = 3864\ \text{rpm}, \qquad 0.96 \times 12\,500 = 12\,000\ \text{rpm.} \] The two percentages differ because the shafts are independent: each settles where its own work balance and flow continuity put it, so there is no fixed relationship between the two figures.
Please log in or register to report this question.
Bleeding 5 per cent of a core flow of \( 50 \ \mathrm{kg/s} \) for the airframe leaves the turbine
Answer: B
The bleed removes \( 0.05 \times 50 = 2.5 \), leaving 47.5 kilograms per second. That air was compressed and then removed before doing any work in the turbine, which is why offtakes are always stated with a performance figure.
Please log in or register to report this question.
