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.

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    Oblique shock waveOblique 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,

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

    Answered on 19th March 2021.
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