Discussion
A flow has a static temperature of \( 250 \) K and a speed of \( 200 \) m/s. With \( c_p = 1005 \) J/kg K, its stagnation temperature is nearest to
Answer: C
The energy equation gives \( T_0 = T + V^2/(2 c_p) \), so the kinetic term is \( 200^2 / (2 \times 1005) = 40\,000 / 2010 = 19.9 \) K. Adding this to 250 gives \( 269.9 \) K. This form is often quicker than the Mach-number relation when the speed is given directly.
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