# Find the pressure, temperature , Mach number,  total pressure and total temperature behind the oblique shock wave.

An oblique shock wave is generated with a wave angle of $${30^ \circ }$$ in a Mach $$4$$ flow, at an altitude of $$9000\,m$$. Find the pressure, temperature , Mach number,  total pressure and total temperature behind the oblique shock wave.

Asked on 11th March 2021 in

Oblique shock wave

Normal component of the upstream Mach number to the oblique shock wave,

$M_{n1}=M_{1}Sin\beta =4\left ( Sin30^{\circ} \right )=2$

From Normal shock wave properties table, for $$M_{n1}=2$$

$$\frac{p_{2}}{p_{1}}=4.5$$, $$\frac{T_{2}}{T_{1}}=1.687$$,  $$\frac{p_{02}}{p_{01}}=0.7209$$, $$M_{n2}=0.5774$$

At an altitude of $$9000\, m$$ , $$p_{1}=3.08\times 10^{4} N/m^{2}$$, $$T_{1}=229.74K$$

Therefore,

$P_{2}=\left ( \frac{P_{2}}{P_{1}} \right )P_{1}=\left ( 4.5 \right )\left ( 3.08\times 10^{4} \right)= 13.86\times 10^{4} N/m^{2}$

$T_{2}=\left ( \frac{T_{2}}{T_{1}} \right )T_{1}=\left ( 1.687 \right )229.74=387.571K$

From the $$\theta -\beta -M$$ diagram , for, wave angle of $$\beta =30^{\circ}$$ and Mach number of $$4$$, deflection angle, $$\theta =17.7^{\circ}$$.

Therefore,

$M_{2}=\frac{M_{n2}}{Sin\left ( \beta -\theta \right )}=\frac{0.5774}{Sin\left ( 30^{\circ} -17.7^{\circ}\right )}=2.710$

From the table of isentropic flow properties , for

$M_{1}=4,\frac{p_{01}}{p_{1}}=151.8,\frac{T_{01}}{T_{1}}=4.2$

Total pressure,

$p_{02}=\left ( \frac{p_{02}}{p_{01}} \right )\left ( \frac{p_{01}}{p_{1}} \right )p_{1}=\left ( 0.0709 \right )\left ( 151.8 \right )\left ( 3.08\times 10^{4} \right )=331488.696N/m^{2}$

Total temperature,

$T_{02}=T_{01}=\left ( \frac{T_{01}}{T_{1}} \right )T_{1}=4.2\times 229.74K=964.908K$