The stagnation temperature of an isentropic flow of air (k = 1.4) is 360K. If the temperature is 200K at a section, then the Mach number of the flow will be
- A. 1.0
- B. 1.5
- C. 3.0
- D. 2.0
Answer: D. 2.0
Explanation: $T_0/T = 1 + (k-1)/2 M²$. $360/200 = 1 + 0.2 M²$. $1.8 - 1 = 0.2 M²$. $0.8 = 0.2 M² \implies M² = 4 \implies M = 2$.