Question 13.16: The pipe in Fig. 13–37 transports air at a temperature of 20......
The pipe in Fig. 13–37 transports air at a temperature of 20°C having an absolute pressure of 30 kPa and a speed of 550 m/s, measured just behind a standing shock wave. Determine the temperature, pressure, and speed of the air just in front of the wave.
Fluid Description. Shock wave formation is an adiabatic process. Steady flow occurs in back and in front of the wave.
Analysis. The control volume contains the shock, as shown in Fig. 13–37. To use the necessary equations or Table B–4, we must first determine M_1. We have
Since M_1 > 1, we expect M_2 < 1 and for both the pressure and temperature to increase in front of the wave.
Using M_1 = 1.6032, the value of M_2 in front of the shock (subsonic) and the ratios of pressure and temperature are found from Eqs. 13–73,
13–68, and 13–69. They are
M_2^2 = \frac{M_1^2 + \frac{2}{k – 1}}{\frac{2k}{k – 1} M_1^2 – 1} M_1 > M_2 (13-73)
\frac{T_2}{T_1} = \frac{1 + \frac{k – 1}{2} M_1^2}{1 + \frac{k – 1}{2} M_2^2} (13-69)
\frac{p_2}{p_1} = \frac{1 + kM_1^2}{1 + kM_2^2} (13-68)
M_2 = 0.66746
\frac{p_2}{p_1} = 2.8321\frac{T_2}{T_1} = 1.3902
Thus,
p_2 = 2.8321 (30 kPa) = 85.0 kPa
T_2 = 1.3902 (273 + 20) K = 407.3 K
The speed of the air in front of the shock wave can be determined from Eq. 13–71\frac{V_2}{V_1}=\frac{M_2}{M_1}[\frac{1+\frac{k-1}{2}M_1^2}{1+\frac{k-1}{2}M_2^2}]^{1/2}, or, since T_2 is known, we can use Eq. 13–22.
V_2 = M_2\sqrt{kRT_2} = 0.66746 \sqrt{1.4 (286.9 J/kg · K) (407.3 K)} = 270 m/s