A jet plane is flying horizontally at 845 m/s, at an altitude where the air temperature is 10°C and the absolute pressure is 80 kPa. If an oblique shock forms on the nose of the plane, at the angle shown in Fig. 13–44a, determine the pressure and temperature, and the direction, of the air just

Question

A jet plane is flying horizontally at 845 m/s, at an altitude where the air temperature is 10°C and the absolute pressure is 80 kPa. If an oblique shock forms on the nose of the plane, at the angle shown in Fig. 13–44a, determine the pressure and temperature, and the direction, of the air just behind the shock.

Accepted Answer

Fluid Description. The air is considered compressible, and the shock is an adiabatic nonisentropic process. Steady flow occurs as viewed from the plane.

Analysis. The Mach number for the plane must first be determined.

[latex] M _1=\frac{V_1}{c}=\frac{V_1}{\sqrt{k R T_1}}=\frac{845 m / s }{\sqrt{1.4(286.9 J / kg \cdot K )(273+10) K }}=2.5063 [/latex]

From the geometry shown in Fig. 13–44b, [latex] M_1[/latex] is resolved into its normal and tangential components relative to the wave. The normal component is [latex] M _{1 n}=2.5063 \sin 40^{\circ}=1.6110 [/latex]. We can now use Table B–4 or Eqs. 13–83, 13–84, and 13–86 to obtain the speed, temperature, and pressure in front of the shock. We get [latex] M _{2 n}=0.6651 [/latex]. This is subsonic, which is to be expected so as not to violate the second law of thermodynamics. Also,

[latex] M _{2 n}^2=\frac{ M _{1 n}^2+\frac{2}{k-1}}{\frac{2 k}{k-1} M _{1 n}^2-1} [/latex] (13-83)

[latex] \frac{p_2}{p_1}=\frac{2 k}{k+1} M _{1 n}^2-\frac{k-1}{k+1} [/latex] (13-84)

[latex] \frac{T_2}{T_1}=\frac{1+\frac{k-1}{2} M _{1 n}^2}{1+\frac{k-1}{2} M _{2 n}^2} [/latex] (13-86)

[latex]\frac{T_2}{T_1}=1.3956; \quad T_2=1.3956(273+10)K=394.95 K=395 K[/latex]

[latex]\frac{p_2}{p_1}=2.8619: \quad p_2=2.8619(80 kPa)=228.90 kPa=229 kPa[/latex]

The angle θ, shown in Fig. 13–44c, can be obtained directly from Eq. 13–88, since [latex] M _1 ext { and } \beta [/latex] are known.

[latex] an heta=\frac{2 \cot \beta\left( M _1^2 \sin ^2 \beta-1 ight)}{ M _1^2(k+\cos 2 \beta)+2} [/latex] (13-88)

[latex] \begin{gathered} an heta=\frac{2 \cot 40^{\circ}\left[(2.5063)^2 \sin ^2 40^{\circ}-1 ight]}{(2.5063)^2\left(1.4+\cos 2\left(40^{\circ} ight) ight)+2} \ heta=17.7^{\circ} \end{gathered} [/latex]

The result is shown in Fig. 13–44d.

Question 13.19: A jet plane is flying horizontally at 845 m/s, at an altitud...

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