# 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.

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.

Asked on 26th March 2021 in

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 $$17.16^{\circ}-25^{\circ}=-7.84^{\circ}$$