An ideal ramjet has design flight Mach 2.8 and combustor-exit stagnation temperature 2400 K. With γ = 1.4 and R = 287 J/(kg K), find engine-exit air velocity in m/s, to 1 decimal place.
For the ideal ramjet with no total-pressure losses and expansion to ambient pressure, the nozzle-exit Mach number equals flight Mach. Thus $T_e=T_{04}/[1+(\gamma-1)M^2/2]=2400/[1+0.2(2.8)^2]=934.58$ K. Exit speed is $V_e=M\sqrt{\gamma RT_e}\approx1715.8$ m/s. The pressure matching and ideal-loss assumptions are needed to identify the exit Mach.
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