Discussion
Total air temperature is 268 K at Mach 0.8 with a recovery factor of 1. Using \( T_t = T_s(1 + 0.2M^2) \), the static air temperature is approximately
Answer: B
Correct answer: 238 K
Rearranging the total temperature relation for the static value: \[ T_s = \frac{T_t}{1 + 0.2M^2} = \frac{268}{1 + 0.2(0.8)^2} = \frac{268}{1.128} = 237.6 \approx 238\ \text{K.} \] The static temperature must be the smaller of the two, since bringing the air to rest at the probe heats it - here by about 30 K, which is why the raw probe reading cannot be used as ambient temperature.
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