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