An engine at sea level static passes \( 200 \ \mathrm{kg/s} \). At a pressure ratio \( \delta = 0.30 \) and temperature ratio \( \theta = 0.85 \), the physical flow at the same corrected condition is approximately
Answer: C
Correct answer: 65 kg/s
Invert the correction to recover the physical flow: \[ \dot m = \dot m_{corr} \frac{\delta}{\sqrt{\theta}} = 200 \times \frac{0.30}{0.922} = 65\ \text{kg/s.} \] The engine is doing the same aerodynamic work as at sea level static, but the air is thin, so it swallows less than a third of the mass. This is precisely why thrust falls so steeply with altitude.
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