Question 13.20: Air flows over a surface at a speed of 900 m/s, where the ab......
Air flows over a surface at a speed of 900 m/s, where the absolute pressure is 100 kPa and the temperature is 30°C. Expansion waves occur at the sharp transition shown in Fig. 13–47a. Determine the velocity, temperature, and pressure of the flow just to the right of the wave.
Fluid Description. We have steady compressible flow that expands isentropically.
The initial or upstream Mach number for the flow is
Analysis. Since the Prandtl–Meyer expansion function, Eq. 13–92, has been referenced from M = 1, the turning angle ω_1 from this reference is
ω_1 = \sqrt{\frac{k + 1}{k – 1}} \tan^{-1} \sqrt{\frac{k – 1}{k + 1} (M_1^2 – 1)} – \tan^{-1} \sqrt{M_1^2 – 1} (1)
= \sqrt{\frac{1.4 + 1}{1.4 – 1}} \tan^{-1} \sqrt{\frac{1.4 – 1}{1.4 + 1} (2.5798)^2 – 1)} – \tan^{-1} \sqrt{(2.5798)^2 – 1}
= 40.96°
This same value (or one close to it) can also be determined from Table B–5. Since the boundary layer over the inclined surface is very thin, the deflection angle for the streamlines is defined by the same angle as the deflection of the surface, that is, 180° – 170° = 10°, Fig. 13–47b. The downstream wave must therefore have a Mach number M_2 that produces a turning angle of 40.96° + 10° = 50.96° from the M = 1 reference.
Rather than using this angle and trying to solve for M2 using the Prandtl–Meyer function, Eq. 13–92, we can use an Internet website, or use the table with ω_2 = 50.96°. Using the more exact approach, we get
M_2 = 3.0631
In other words, the expansion waves increase the Mach number from M_1 = 2.5798 to M_2 = 3.0631. Realize, however, that this increase does not violate the second law of thermodynamics, because only the normal component of the wave converts from supersonic to subsonic.
Since we have isentropic expansion, we can determine the temperature on the right side of the waves using Eq. 13–69.
T_2 = 0.8104(273 + 30) K = 245.54 K = 246 K
We can also use Table B–1 as follows:
so that again T_2 = 0.8104(273 + 30) K = 246 K.
The pressure is obtained in a similar manner using Eq. 13–17 \frac{p_2}{p_1}=(\frac{T_2}{T_1})^{k/(k-1)}, or Table B–1. Here
so
p_2 = 0.47915(100 kPa) = 47.9 kPa
The velocity of the flow after the expansion is
V_2 = M_2\sqrt{kRT_2} = 3.0631 \sqrt{1.4 (286.9 J/kg · K) (245.54 K)} = 962 m/s
