Question 13.19: A jet plane is flying horizontally at 845 m/s, at an altitud......
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.
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.
From the geometry shown in Fig. 13–44b, M_1 is resolved into its normal and tangential components relative to the wave. The normal component is M_{1n} = 2.5063 sin 40 = 1.6110. 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 M_{2n} = 0.6651. This is subsonic, which is to be expected so as not to violate the second law of thermodynamics. Also,
\frac{T_2}{T_1} = 1.3956; T_2 = 1.3956(273 + 10) K = 394.95 K = 395 K
\frac{p_2}{p_1} = 2.8619; p_2 = 2.8619(80 kPa) = 228.90 kPa = 229 kPa
The angle θ, shown in Fig. 13–44c, can be obtained directly from Eq. 13–88, since M_1 and β are known.
\tan \theta = \frac{2 \cot \beta (M_1^2 \sin^2 \beta – 1)}{M_1^2 (k + \cos 2\beta) + 2}\tan \theta = \frac{2 \cot 40°[(2.5063)^2 \sin ^2 40° – 1]}{(2.5063)^2 (1.4 + \cos 2(40°)) + 2}
\theta = 17.7°
The result is shown in Fig. 13–44d.
