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 = \(\beta _{2}-\theta =46.75^{\circ}-20^{\circ}=26.75^{\circ}\)