Free stream total pressure is \( 180 \ \mathrm{kPa} \) and the compressor face sees \( 171 \ \mathrm{kPa} \). The total pressure recovery is
Answer: A
Recovery is the ratio of the two total pressures, \( 171/180 = 0.95 \). A five per cent loss sounds modest but typically costs rather more than five per cent in thrust, which is why intake design receives so much attention on fast aircraft.
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Air at Mach 0.85 has a total to static pressure ratio given by \( (1 + 0.2M^2)^{3.5} \). This ratio is approximately
Answer: B
Correct answer: 1.60
\[ \left(1 + 0.2 \times 0.85^2\right)^{3.5} = (1.1445)^{3.5} = 1.60. \] Even at high subsonic speed the ram compression is worth 60 per cent in pressure before the compressor does anything - and since the exponent is 3.5, the ratio climbs very steeply as the Mach number rises further.
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At Mach 2.0, the same relation \( (1 + 0.2M^2)^{3.5} \) gives a total to static pressure ratio of approximately
Answer: C
Substituting gives \( 1.8^{3.5} \approx 7.82 \). Comparing this with the 1.6 available at Mach 0.85 shows how completely ram compression dominates at high speed, and why a ramjet needs no compressor at all in that regime.
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Total pressure ratio across a normal shock at Mach 2.0 is about 0.72. Across two oblique shocks followed by a weak normal shock the recovery is about 0.90. The improvement is
Answer: D
The gain is \( (0.90-0.72)/0.72 = 0.25 \). Splitting the deceleration across several weaker shocks recovers most of what a single strong shock wastes, which is exactly why multi-shock intakes appear above about Mach 1.6.
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An intake of area \( 0.8 \ \mathrm{m^2} \) captures a streamtube of \( 0.6 \ \mathrm{m^2} \) at cruise. The capture area ratio is
Answer: A
The ratio is \( 0.6/0.8 = 0.75 \), meaning the streamtube expands towards the intake as it decelerates. External diffusion of this kind is normal at cruise but must be kept modest, since excessive turning invites separation at the lip.
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An engine demands \( 90 \ \mathrm{kg/s} \) while the intake at that condition would capture \( 110 \ \mathrm{kg/s} \). The mass flow spilled is
Answer: B
The difference of 20 kilograms per second must flow around the intake instead of through it. Turning that air produces spillage drag, which is worst when the engine is throttled at high speed and the capture area cannot be reduced.
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Air enters a diffuser at \( 200 \ \mathrm{m/s} \) and leaves at \( 80 \ \mathrm{m/s} \). With \( c_p = 1005 \ \mathrm{J/(kg\,K)} \), the ideal static temperature rise is approximately
Answer: C
Correct answer: 16.7 K
The kinetic energy given up appears as static temperature: \[ \Delta T = \frac{V_1^2 - V_2^2}{2c_p} = \frac{200^2 - 80^2}{2 \times 1005} = \frac{33{,}600}{2010} = 16.7\ \mathrm{K}. \] Note it is the difference of the squares, not the square of the difference - the latter would give 7.2 K.
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An intake supplies air at Mach 0.5 to a compressor face of area \( 1.2 \ \mathrm{m^2} \). Doubling the area at the same mass flow would give an approximate Mach number of
Answer: D
At low Mach numbers the density change is modest, so halving the velocity roughly halves the Mach number. The relation is not exactly linear because density rises as the flow slows, but the estimate is close enough for a first sizing.
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A distortion index is computed as \( (P_{max} - P_{min})/P_{average} \). With values of 98, 86 and 94 kilopascals respectively, the index is
Answer: A
Correct answer: 0.128
\[ \frac{98 - 86}{94} = \frac{12}{94} = 0.128. \] The index measures the spread across the compressor face relative to the mean, and a figure of this size would concern an engine manufacturer - distortion of a few per cent is enough to move the surge line appreciably.
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A supersonic intake achieves recovery of 0.88 where the specification requires 0.92. The shortfall as a fraction of the requirement is
Answer: B
Correct answer: 4.3 per cent
The shortfall is taken as a fraction of what was required: \[ \frac{0.92 - 0.88}{0.92} = 0.043. \] Expressing it against the achieved figure instead gives 4.5 per cent, and the plain difference of 0.04 gives 4.0 - both offered here, so the denominator is what the question turns on.
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