The upstream Mach number, pressure and temperature of a supersonic flow over an expansion corner having deflection angle $\theta =25^{^{\circ}}$ is $M_{1}=2.2,\,p_{1}=0.8\,atm$ and $T_{1}=360\,K$. Find the downstream Mach number,  pressure, temperature, density , total pressure, total temperature, and the angles made by the forward and rearward Mach lines with local flow direction.

Prandtl-Meyer expansion waves

Downstream Mach number can be calculated from Prandtl-Meyer function
$$\nu (M)=\sqrt{\frac{\gamma +1}{\gamma -1}}tan^{-1}\sqrt{\frac{\gamma -1}{\gamma +1}\left ( M^{2}-1 \right )}-tan^{-1}\sqrt{M^{2}-1}$$

For a Mach number of $2.2$, $\nu \left ( M_{1} \right )=31.73^{^{\circ}}$

$$\theta =\nu \left ( M_{2} \right )-\nu \left ( M_{1} \right )$$

$$\Rightarrow \nu \left ( M_{2} \right )=\theta + \nu \left ( M_{1} \right ) = 25^{\circ}+31.73^{\circ}=56.73^{\circ}$$

for, $\nu \left ( M_{2} \right )=56.73^{^{\circ}}$, Mach number $M_{2}=3.389$

From Isentropic flow properties for, $M_{1}=2.2$

$$\frac{p_{01}}{p_{1}}=10.69$$ and
$$\frac{T_{01}}{T_{1}}=1.968$$

For, $M_{2}=3.389$
$$\frac{p_{02}}{p_{2}}=65.0798$$ and
$$\frac{T_{02}}{T_{2}}=3.297$$

Total pressure and total temperature is constant through the expansion wave, so

$p_{01}=p_{02}$ and $T_{01}=T_{02}$

Therefore,
$$p_{2}=\left ( \frac{p_{2}}{p_{02}} \right )\left ( \frac{p_{01}}{p_{1}} \right )p_{1}=\left ( \frac{1}{65.0798} \right )\left ( 10.69 \right )0.8=0.1314\,atm$$

$$T_{2}=\left ( \frac{T_{2}}{T_{02}} \right )\left ( \frac{T_{01}}{T_{1}} \right )T_{1}=\left ( \frac{1}{3.297} \right )\left ( 1.968 \right )360=214.886\,K$$

$$p_{2}=\rho _{2}RT_{2}$$

$$\Rightarrow 0.1314=\rho _{2}\left ( 287 \right )\left ( 214.886\,K \right )$$

$$\Rightarrow \rho _{2}=\frac{0.1314}{\left ( 287 \right )\left ( 214.886 \right )}=2.13\times 10^{-6}\,kg/m^{3}$$

Since,
$$p_{01}=p_{02}\Rightarrow p_{02}=\left ( \frac{p_{01}}{p_{1}} \right )\left ( p_{1} \right )=\left ( 10.69 \right )\left ( 0.8 \right )=8.552\,atm$$

Also,

$$T_{01}=T_{02}\Rightarrow T_{02}=\left ( \frac{T_{01}}{T_{1}} \right )\left ( T_{1} \right )=\left ( 1.968 \right )\left ( 360 \right )=708.48\,K$$

Mach angle is given by $\mu = Sin^{-1}\left ( \frac{1}{M} \right )$

Therefore Mach angle $\mu _{1}$ made by upstream Mach wave for $M_{1}=2.2$ is $27.04^{\circ}$

Mach angle $\mu _{2}$ made by downstream Mach wave for$M_{2}=3.389$ is $17.16^{\circ}$

and, angle made by forward Machline is $27.04^{\circ}$ and angle made by the rearward Machline is $17.16^{\circ}-25^{\circ}=-7.84^{\circ}$