# An oblique shock wave is generated at a compression corner having a deflection angle of $$20^{^{\circ}}$$. The oblique shock wave is reflected from a straight horizontal wall which is present above the compression corner. The upstream flow properties are $$M_{1}=3.4,\,p_{1}=1\,atm,\,T_{1}=300\,K$$. Find the Mach number, pressure and temperature behind the reflected shock wave and angle made by the reflected shock wave with the upper wall.

An oblique shock wave is generated at a compression corner having a deflection angle of $$20^{^{\circ}}$$. The oblique shock wave is reflected from a straight horizontal wall which is present above the compression corner. The upstream flow properties are $$M_{1}=3.4,\,p_{1}=1\,atm,\,T_{1}=300\,K$$. Find the Mach number, pressure and temperature behind the reflected shock wave and angle made by the reflected shock wave with the upper wall.

techAir Asked on 26th March 2021 in

Reflected oblique shock

Wave angle $$\beta _{1}$$  can be obtained from the $$\theta\, -\beta\, – M$$ diagram. For, deflection angle of $$20^{^{\circ}}$$  and Mach number $$M_{1}$$ of $$3.4$$, $$\beta _{1}=35^{^{\circ}}$$

Therefore,

$M_{n1}=M_{1}Sin\beta _{1} = 3.4Sin35^{^{\circ}}=1.95$

For a normal shock wave for $$M_{n1}=1.95$$,

$$\frac{p_{2}}{p_{1}}=4.27$$, $$\frac{T_{2}}{T_{1}}=1.6473$$ and $$M_{n2}=0.586$$

Therefore,

$M_{2}=\frac{M_{n2}}{Sin\left ( \beta _{1}-\theta \right )}=\frac{0.586}{Sin\left ( 35^{^{\circ}}-20^{^{\circ}} \right )}=2.26$

For the reflected shock, from $$\theta -\beta\, – M$$  diagram, for $$M_{2}=2.26$$ and $$\theta = 20^{^{\circ}}$$ , wave angle, $$\beta _{2}=46.75^{^{\circ}}$$

Therefore,

$M_{n2}=M_{2}Sin\beta _{2} = 2.26Sin46.75^{^{\circ}}=1.65$

For a normal shock wave, for $$M_{n2}=1.56$$

$$\frac{p_{3}}{p_{2}}=3.001$$, $$\frac{T_{3}}{T_{2}}=1.423$$ and  $$M_{n3}=0.654$$

Therefore,

$M_{3}=\frac{M_{n3}}{Sin\left ( \beta _{2}-\theta \right )}=\frac{0.654}{Sin\left ( 46.75^{\circ}-20^{\circ} \right )}=1.453$

and $p_{3}=\left ( \frac{p_{3}}{p_{2}} \right )\left ( \frac{p_{2}}{p_{1}} \right )\left ( p_{1} \right )=\left ( 3.001 \right )\left ( 4.27 \right )\left ( 1 \right )=12.814\,atm$

and $T_{3}=\left ( \frac{T_{3}}{T_{2}} \right )\left ( \frac{T_{2}}{T_{1}} \right )\left ( T_{1} \right )=\left ( 1.423 \right )\left ( 1.6473 \right )\left ( 300\,K \right )=703.23\,K$

Angle made by the reflected shock wave with the upper wall = $$\beta _{2}-\theta =46.75^{\circ}-20^{\circ}=26.75^{\circ}$$

Worldtech Answered on 31st March 2021.