Discussion
Exhaust gas at 900 K expands isentropically to an exit velocity given by \( V = \sqrt{2 c_p \Delta T} \). With \( c_p = 1150 \) and a drop of 150 K, the velocity is approximately
Answer: B
Substituting gives \( \sqrt{2 \times 1150 \times 150} = \sqrt{345\,000} \approx 587 \) metres per second. Every kelvin of temperature drop that the nozzle fails to convert appears as lost exit velocity and therefore lost thrust.
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