Chapter 9, Problem 9.2P

Consider an oblique shock wave with a wave angle of $30°$ in a Mach 4 flow. The upstream pressure and temperature are $2.65\times {10}^4{\text{N/m}}^{\text{2}}$ and 223.3 K, respectively (corresponding to a standard altitude of 10,000 m). Calculate the pressure, temperature, Mach number, total pressure, and total temperature behind the wave and the entropy increase across the wave.

To determine

The pressure.

The temperature.

The Mach number.

The total pressure.

The total temperature behind the wave.

The entropy increase.

Answer to Problem 9.2P

The pressure is $P_2=11.925\times {10}^4\text{N}/{\text{m}}^{\text{2}}$ .

The temperature is $T_2=376.81\text{K}$ .

The Mach number is $M_2=2.728$ .

The total pressure is $P_{o2}=28.96\times {10}^5\text{N}/{\text{m}}^{\text{2}}$ .

The total temperature behind the wave is $T_{o2}=937.86\text{K}$ .

The entropy increase is $\Delta s=94.33\text{J}/\text{kgK}$ .

Explanation of Solution

Given:

The upstream temperature is $T_1=223.3\text{K}$ .

The upstream pressure is $P_1=2.65\times {10}^4\text{N}/{\text{m}}^{\text{2}}$ .

The wave angle of the shock wave is $\beta =30\text{°}$ .

The Mach number is $M_1=4$ .

Formula Used:

The expression for the normal component of upstream Mach number is given as,

  $M_{n1}=M_1\sin \beta$

Here, $\beta$ is the Mach angle and $M_1$ is the Mach number.

The expression for total upstream pressure is given as,

  $\frac{P_{o1}}{P_1}={\left(1+\frac{\gamma -1}{2}{M}_1^2\right)}^{\frac{\gamma }{\gamma -1}}$

Here, $\gamma$ is the heat capacity ratio.

The expression for the upstream temperature is given as,

  $T_{o1}=T_1{\left(1+\frac{\gamma -1}{2}{M}_1^2\right)}^{\frac{\gamma }{\gamma -1}}$

The expression for the downstream static pressure is given as,

  $P_2=P_1\left(1+\frac{2\gamma }{\gamma -1}\right)\left(M_{n1}^2-1\right)$

The expression for the downstream static temperature is given as,

  $T_2=T_1\frac{P_2}{P_1}\left(\frac{2+\left(\gamma -1\right)M_{n1}^2}{\left(\gamma +1\right)M_{n1}^2}\right)$

The expression for the downstream normal Mach number is given as,

  $M_{n1}=\sqrt{\frac{1+\left(\frac{\gamma -1}{2}\right)M_{n1}^2}{\gamma M_{n1}^2-\frac{\gamma -1}{2}}}$

The expression for the entropy increase across oplique shock wave is given as,

  $\Delta s=c_p\ln \left(\frac{T_{o2}}{T_{o1}}\right)-R\ln \left(\frac{P_{o2}}{P_{o1}}\right)$

Here, $R$ is the ideal gas constant.

Calculation:

Thenormal component of upstream Mach number can be calculated as,

  $\begin{array}{c}{M}_{n1}=M_1\sin \beta \\ M_{n1}=4\times \sin 30\text{°}\\ M_{n1}=2\end{array}$

The total upstream pressure can be calculated as,

  $\begin{array}{c}\frac{P_{o1}}{P_1}={\left(1+\frac{\gamma -1}{2}{M}_1^2\right)}^{\frac{\gamma }{\gamma -1}}\\ \frac{P_{o1}}{2.65\times {10}^4\text{N}/{\text{m}}^{\text{2}}}={\left(1+\frac{1.4-1}{2}\times 4^2\right)}^{\frac{1.4-1}{2}}\\ P_{o1}=40.23\times {10}^5\text{Pa}\end{array}$

The upstream temperature can be calculated as,

  $\begin{array}{l}{T}_{o1}=T_1{\left(1+\frac{\gamma -1}{2}{M}_1^2\right)}^{\frac{\gamma }{\gamma -1}}\\ T_{o1}=223.2\text{K}{\left(1+\frac{1.4-1}{2}\times 4^2\right)}^{\frac{1.4}{1.4-1}}\\ T_{o1}=937.86\text{K}\end{array}$

The downstream static pressure can be calculated as,

  $\begin{array}{l}{P}_2=P_1\left(1+\frac{2\gamma }{\gamma -1}\right)\left(M_{n1}^2-1\right)\\ P_2=\left(2.65\times {10}^4\text{N}/{\text{m}}^{\text{2}}\right)\times \left(1+\frac{2\times 1.4}{1.4-1}\right)\left(2^2-1\right)\\ P_2=11.925\times {10}^4\text{Pa}\end{array}$

The downstream static temperature can be calculated as,

  $\begin{array}{l}{T}_2=T_1\frac{P_2}{P_1}\left(\frac{2+\left(\gamma -1\right)M_{n1}^2}{\left(\gamma +1\right)M_{n1}^2}\right)\\ T_2=\left(223.3\text{K}\right)\times \left(\frac{11.925\times {10}^5\text{N}/{\text{m}}^{\text{2}}}{2.65\times {10}^4\text{N}/{\text{m}}^{\text{2}}}\right)\left(\frac{2+\left(1.5-1\right)\times 2^2}{\left(1.4+1\right)\times 2^2}\right)\\ T_2=376.81\text{K}\end{array}$

The downstream normal Mach number can be calculated as,

  $\begin{array}{l}{M}_{n1}=\sqrt{\frac{1+\left(\frac{\gamma -1}{2}\right)M_{n1}^2}{\gamma M_{n1}^2-\frac{\gamma -1}{2}}}\\ M_{n1}=\sqrt{\frac{1+\left(\frac{1.4-1}{2}\right)\times 2^2}{\left(1.4\times 2^2\right)-\frac{1.4-1}{2}}}\\ M_{n1}=0.5774\end{array}$

The deflection angle can be calculated as,

  $\begin{array}{c}\theta ={\tan }^{-1}\left(2\cot \beta \times \frac{M_1^2{\sin }^2\beta -1}{M_1^2\left(\gamma +\cos 2\beta \right)+2}\right)\\ \theta ={\tan }^{-1}\left(2\cot 30\text{°}\times \frac{4^2{\sin }^{2}30\text{°}-1}{4^2\left(1.4+\cos 2\left(30\text{°}\right)\right)+2}\right)\\ \theta =17.78\text{°}\end{array}$

The angle between downstream flow and oblique shock wave can be calculated as,

  $\begin{array}{c}\sin \left(\beta -\theta \right)=\frac{M_{n2}}{M_2}\\ \sin \left(30\text{°}-17.78\text{°}\right)=\frac{0.5774}{M_2}\\ M_2=2.728\end{array}$

The total downstream pressure can be calculated as,

  $\begin{array}{l}{P}_{o2}=P_2{\left(1+\frac{\gamma -1}{2}{M}_2^2\right)}^{\frac{\gamma }{\gamma +1}}\\ P_{o2}=\left(11.925\times {10}^4\text{N}/{\text{m}}^{\text{2}}\right){\left(1+\frac{1.4-1}{2}\times {\left(2.728\right)}^2\right)}^{\frac{1.4}{1.4+1}}\\ P_{o2}=28.96\times {10}^5\text{N}/{\text{m}}^{\text{2}}\end{array}$

The temperature does not change across shock wave and will be,

  $T_{o2}=T_{o1}=937.86\text{K}$

The expression for the entropy increase across oplique shock wave is given as,

  $\begin{array}{l}\Delta s=c_p\ln \left(\frac{T_{o2}}{T_{o1}}\right)-R\ln \left(\frac{P_{o2}}{P_{o1}}\right)\\ \Delta s=0-\left(287\text{J}/\text{kg}\text{K}\right)\ln \left(\frac{28.96\text{N}/{\text{m}}^{\text{2}}}{40.23\text{N}/{\text{m}}^{\text{2}}}\right)\\ \Delta s=94.33\text{J}/\text{kg}\text{K}\end{array}$

Conclusion:

Therefore, the pressure is $P_2=11.925\times {10}^4\text{N}/{\text{m}}^{\text{2}}$ .

Therefore, the temperature is $T_2=376.81\text{K}$ .

Therefore, the Mach number is $M_2=2.728$ .

Therefore, the total pressure is $P_{o2}=28.96\times {10}^5\text{N}/{\text{m}}^{\text{2}}$ .

Therefore, the total temperature behind the wave is $T_{o2}=937.86\text{K}$ .

Therefore, the entropy increase is $\Delta s=94.33\text{J}/\text{kgK}$ .