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.

Question

Accepted Answer

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 [latex] M _1 [/latex]. We have

[latex] M _1=\frac{V_1}{\sqrt{k R T_1}}=\frac{550 m / s }{\sqrt{1.4(286.9 J / kg \cdot K )(273+20) K }}=1.6032 [/latex]

Since [latex] M _1>1 ext {, we expect } M _2<1 [/latex] and for both the pressure and temperature to increase in front of the wave.
Using [latex] M _1=1.6032 ext {, the value of } M _2 [/latex] 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

[latex] M _2^2=\frac{ M _1^2+\frac{2}{k-1}}{\frac{2 k}{k-1} M _1^2-1} \quad M _1> M _2 [/latex] (13-73)

[latex] \frac{p_2}{p_1}=\frac{1+k M _1{ }^2}{1+k M _2{ }^2} [/latex] (13-68)

[latex] \frac{T_2}{T_1}=\frac{1+\frac{k-1}{2} M _1^2}{1+\frac{k-1}{2} M _2^2} [/latex] (13-69)

[latex] \begin{aligned} & M _2=0.66746 \ & \frac{p_2}{p_1}=2.8321 \ & \frac{T_2}{T_1}=1.3902 \end{aligned} [/latex]

Thus,

[latex] \begin{aligned} & p_2=2.8321(30 kPa )=85.0 kPa \ & T_2=1.3902(273+20) K =407.3 K \end{aligned} [/latex]

The speed of the air in front of the shock wave can be determined from Eq. 13–71, or, since [latex] T_2 [/latex] is known, we can use Eq. 13–22.

[latex] \frac{V_2}{V_1}=\frac{ M _2}{ M _1}\left[\frac{1+\frac{k-1}{2} M _1^2}{1+\frac{k-1}{2} M _2^2} ight]^{1 / 2} [/latex] (13-71)

[latex] V= M \sqrt{k R T} [/latex] (13.22)

[latex] V_2= M _2 \sqrt{k R T_2}=0.66746 \sqrt{1.4(286.9 J / kg \cdot K )(407.3 K )}=270 m / s [/latex]

Question 13.16: The pipe in Fig. 13–37 transports air at a temperature of 20...

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