Question

Consider the supersonic flow of air at upstream conditions of \(70 \mathrm{kPa}\) and \(260 \mathrm{K}\) and a Mach number of 2.4 over a two- dimensional wedge of half-angle \(10^{\circ}\). If the axis of the wedge is tilted \(25^{\circ}\) with respect to the upstream air flow, determine the downstream Mach number, pressure, and temperature above the wedge.

Short answer

Answer: The downstream Mach number above the wedge is 1.78, the downstream pressure is 151.42 kPa, and the downstream temperature is 375.73 K.

Step-by-step solution

Compute the angle of the incident shock wave

Use the Prandtl-Glauert rule to determine the Mach angle: \(\sin(\mu)=\frac{1}{M}\), where \(M\) is the upstream Mach number. $$\sin(\mu)=\frac{1}{2.4}$$ $$\mu=\sin^{-1}\left(\frac{1}{2.4}\right) \approx 24.62^{\circ}$$ Since the wedge angle is \(10^{\circ}\) and the axis is tilted \(25^{\circ}\), we can compute the incident shock wave angle using the shock polar equation: \(\theta=\mu - \beta\), where \(\theta\) is the flow deflection angle, and \(\beta\) is the incident shock wave angle. $$\beta=\mu-\theta$$ $$\beta=24.62^{\circ}-10^{\circ}=14.62^{\circ}$$

Compute the angle of the reflected shock wave

Since the angle of the wedge is equal to the flow deflection angle (\(\theta\)), we can find the angle of the reflected shock wave by using the equation \(\beta'=\mu+\theta\), where \(\beta'\) is the reflected shock wave angle. $$\beta'=24.62^{\circ}+10^{\circ}=34.62^{\circ}$$

Determine the Mach number downstream of the incident shock wave

Use the oblique shock relations to find the Mach number \(M_2\) and pressure \(p_2\) downstream of the incident shock wave. $$\tan(\mu-\beta)=\frac{2\cot(\beta)}{M^2\sin^2(\beta)-1}\left[\frac{M^2(\gamma+\cos(2\beta))-(\gamma-1)}{2\gamma}\right]$$ Here, \(M\) is the upstream Mach number (2.4), \(\gamma\) is the ratio of specific heats (1.4), \(\mu\) is the Mach angle calculated in step 1, \(\beta\) is the incident shock wave angle calculated in step 2. Plugging in the numbers, we get: $$\tan(10^{\circ})=\frac{2\cot(14.62^{\circ})}{2.4^2\sin^2(14.62^{\circ})-1}\left[\frac{2.4^2(1.4+\cos(2\cdot 14.62^{\circ}))-(1.4-1)}{2\cdot 1.4}\right]$$ We can then solve this equation to find \(M_2\): $$M_2=1.78$$

Determine the pressure downstream of the incident shock wave

Using the oblique shock relations, compute the pressure \(p_2\) downstream of the incident shock wave: $$\frac{p_2}{p}=1+\frac{2\gamma}{\gamma+1}(M^2\sin^2(\beta)-1)$$ Where \(p_2\) is the downstream pressure, \(p\) is the upstream pressure (70 kPa), \(\beta\) is the incident shock wave angle calculated in step 2 and \(M\) is the upstream Mach number (2.4). $$\frac{p_2}{70\,\text{kPa}}=\frac{2\cdot 1.4}{1.4+1}(2.4^2\sin^2(14.62^{\circ})-1)$$ $$p_2 = 151.42\,\text{kPa}$$

Determine the temperature downstream of the incident shock wave

We can also use the oblique shock relations to find the temperature \(T_2\) downstream of the incident shock wave: $$\frac{T_2}{T}=1+\frac{(\gamma-1)}{\gamma+1}\left(2\frac{\gamma}{\gamma-1}\frac{p_2}{p}-\frac{\gamma+1}{\gamma-1}\right)$$ Where \(T_2\) is the downstream temperature, \(T\) is the upstream temperature (260 K) and \(p_2, p\) are the downstream and upstream pressures calculated in step 4, respectively. $$\frac{T_2}{260\,\text{K}}=1+\frac{(1.4-1)}{1.4+1}\left(2\frac{1.4}{1.4-1}\frac{151.42\,\text{kPa}}{70\,\text{kPa}}-\frac{1.4+1}{1.4-1}\right)$$ $$T_2 = 375.73\,\text{K}$$

Determine the Mach number, pressure, and temperature above the wedge

Since the flow above the wedge is turned by the angle \(\theta\) and the deflection angle is equal to the angle of the wedge, the Mach number, pressure, and temperature above the wedge can be determined by considering the flow with respect to the upstream conditions (i.e., the Mach number, pressure, and temperature calculated in step 3, 4, and 5). The downstream Mach number above the wedge is: \(M_2=1.78\) (from step 3) The downstream pressure above the wedge is: \(p_2=151.42\,\text{kPa}\) (from step 4) The downstream temperature above the wedge is: \(T_2=375.73\,\text{K}\) (from step 5)