Discussion
At Mach 5 the same relation gives a ram pressure ratio of approximately
Answer: B
Correct answer: 529
For isentropic deceleration of air,
\[ \frac{p_0}{p} = \left(1 + \frac{\gamma-1}{2}M^2\right)^{\frac{\gamma}{\gamma-1}} = \left(1 + 0.2 \times 25\right)^{3.5} = 6^{3.5} \approx 529 \]Ram compression alone therefore delivers a pressure ratio far beyond anything a mechanical compressor achieves, which is why a ramjet needs no rotating machinery at all. The corresponding stagnation temperature, however, is 1320 K from an ambient 220 K, and it is that figure rather than the pressure that sets the ramjet’s limit.
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