A jet plane is traveling at M = 1.5, where the absolute air pressure is 50 kPa and the temperature is 8°C. At this speed, a shock forms at the intake of the engine, as shown in Fig. 13–38a. Determine the pressure and velocity of the air just to the right of the shock.

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Fluid Description. The control volume that contains the shock moves with the engine so that steady flow occurs through the open control surfaces, Fig. 13–38b. An adiabatic process occurs within the shock.

Analysis. The velocity [latex] V_2 [/latex] will be obtained from [latex] V_2= M _2 \sqrt{k R T_2} [/latex], and so we must first obtain [latex] M _2 ext { and } T_2 [/latex]. Since k = 1.4, using Eqs. 13–73, 13–76, and 13–69, or Table B–4, for [latex] M _1 [/latex] = 1.5 (supersonic), we have

[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{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] \frac{p_2}{p_1}=\frac{2 k}{k+1} M _1^2-\frac{k-1}{k+1} [/latex] (13-76)

[latex] M _2=0.70109 \quad ext { Subsonic } [/latex]

[latex] \begin{aligned} & \frac{p_2}{p_1}=2.4583 \ & \frac{T_2}{T_1}=1.3202 \end{aligned} [/latex]

Therefore, just to the right of the shock,

[latex] \begin{aligned} & p_2=2.4583(50 kPa )=123 kPa \ & T_2=1.3202(273+8) K =370.98 K \end{aligned} [/latex]

Thus, relative to the engine, the velocity of the air is

[latex] \begin{aligned} & V_2= M _2 \sqrt{k R T_2}=(0.70109) \sqrt{1.4(286.9 J / kg \cdot K )(370.98 K )} \ & V_2=271 m / s \end{aligned} [/latex]

Question 13.17: A jet plane is traveling at M = 1.5, where the absolute air ...

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