Chapter 8, Problem 8.7P

The flow just upstream of a normal shock wave is given by $p_1=1\text{atm},T_1=288\text{K}$, and $M_1=2.6$. Calculate the following properties just downstream of the shock: $p_2,T_2,{\rho }_0,M_2,p_0,T_{0,2},$, and the change in entropy across the shock.

To determine

The downstream pressure.

The downstream temperature.

The downstream density.

The Mach number of downstream shock.

The pressure $p_0{}_2$ .

The temperature $T_0{}_2$ .

The entropy change.

Answer to Problem 8.7P

The downstream pressure is $7.72\text{atm}$ .

The downstream temperature is $658.66\text{K}$ .

The downstream density is $\text{4}\text{.138}\text{kg}/{\text{m}}^3$ .

The Mach number of downstream shock is $\text{0}\text{.503}$ .

The pressure $p_0{}_2$ is $\text{9}\text{.1759}\text{atm}$ .

The temperature $T_0{}_2$ is $692\text{K}$ .

The entropy change is $\text{22}\text{.898}\text{J}/\text{kg}\cdot \text{K}$ .

Explanation of Solution

Given:

The upstream Mach number is $M_1=2.6$ .

The upstream static pressure is $p_1=1\text{atm}$ .

The upstream temperature is $T_1=288\text{K}$ .

Formula used:

The expression for the downstream pressure is given as,

  $\frac{p_2}{p_1}=1+\frac{2.8}{\gamma +1}\left(M_1{}^2-1\right)$

Here, $\gamma$ is the adiabatic index of air and its value is $1.4$ .

The expression for the downstream temperature is given as,

  $T_2=\frac{p_2}{R{\rho }_2}$

The expression for the downstream Mach number is given as,

  $M_2{}^2=\frac{1+\left(\frac{\gamma -1}{2}\right)M_1{}^2}{\gamma M_1{}^2-\frac{\left(\gamma -1\right)}{2}}$

The expression for the downstream density is given as,

  $\frac{{\rho }_2}{{\rho }_1}=\frac{\left(\gamma +1\right)M_1{}^2}{2+\left(\gamma -1\right)M_1{}^2}$

The expression for the pressure $p_0{}_2$ is given as

  $\frac{p_{o2}}{p_2}={\left(1+\frac{\left(\gamma -1\right)}{2}{M}_2{}^2\right)}^{\frac{\gamma }{\gamma -1}}$

The expression for the temperature $T_0{}_2$ is given as,

  $\frac{T_{o2}}{T_2}={\left(\frac{p_{o2}}{p_2}\right)}^{\frac{\gamma -1}{\gamma }}$

The expression for the entropy change is given as,

  $\Delta s=-R\ln \frac{p_{o2}}{p_o{}_1}$

Here, $R$ is the ideal gas constant and its value is $287\text{J}/\text{kg}\cdot \text{K}$ .

The expression for the pressure $p_o{}_1$ is given as,

  $\frac{p_{o1}}{p_1}={\left(1+\frac{\left(\gamma -1\right)}{2}{M}_1{}^2\right)}^{\frac{\gamma }{\gamma -1}}$

Calculation:

The downstream pressure can be calculated as,

  $\begin{array}{l}\frac{p_2}{p_1}=1+\frac{2.8}{\gamma +1}\left(M_1{}^2-1\right)\\ p_2=1\text{atm}\left(1+\frac{2.8}{1.4+1}\left({2.6}^2-1\right)\right)\\ p_2=7.72\text{atm}\end{array}$

The downstream density can be calculated as,

  $\begin{array}{l}\frac{{\rho }_2}{{\rho }_1}=\frac{\left(\gamma +1\right)M_1{}^2}{2+\left(\gamma -1\right)M_1{}^2}\\ {\rho }_2=1.2\text{kg}/{\text{m}}^3\left(\frac{\left(1.4+1\right){\left(2.6\right)}^2}{2+\left(1.4-1\right){\left(2.6\right)}^2}\right)\\ {\rho }_2=4.138\text{kg}/{\text{m}}^3\end{array}$

Here, at pressure $1\text{atm}$ and temperature $\text{288}\text{K}$ the standard density is ${\rho }_1=1.2\text{kg}/{\text{m}}^3$ .

The temperature of the downstream can be calculated as,

  $\begin{array}{l}{T}_2=\frac{p_2}{R{\rho }_2}\\ T_2=\frac{\left(7.72\text{atm}\times \text{101325}\right)\text{Pa}}{287\text{J}/\text{kg}\cdot \text{K}\times 4.138\text{kg}/{\text{m}}^3}\\ T_2=658.66\text{K}\end{array}$

The Mach number at the downstream can be calculated as,

  $\begin{array}{l}{M}_2{}^2=\frac{1+\left(\frac{\gamma -1}{2}\right)M_1{}^2}{\gamma M_1{}^2-\frac{\left(\gamma -1\right)}{2}}\\ M_2{}^2=\frac{1+\left(\frac{1.4-1}{2}\right){\left(2.6\right)}^2}{1.4\times {\left(2.6\right)}^2-\frac{\left(1.4-1\right)}{2}}\\ M_2=\sqrt{\frac{2.352}{9.264}}\\ M_2=0.503\end{array}$

The pressure $p_{o2}$ can be calculated as,

  $\begin{array}{l}\frac{p_{o2}}{p_2}={\left(1+\frac{\left(\gamma -1\right)}{2}{M}_2{}^2\right)}^{\frac{\gamma }{\gamma -1}}\\ p_{o2}=7.72\text{atm}{\left(1+\frac{\left(1.4-1\right)}{2}{\left(0.503\right)}^2\right)}^{\frac{1.4}{1.4-1}}\\ p_{o2}=9.1759\text{atm}\end{array}$

The pressure $p_{o1}$ can be calculated as,

  $\begin{array}{l}\frac{p_{o1}}{p_1}={\left(1+\frac{\left(\gamma -1\right)}{2}{M}_1{}^2\right)}^{\frac{\gamma }{\gamma -1}}\\ p_{o1}=1\text{atm}{\left(1+\frac{\left(1.4-1\right)}{2}{\left(2.6\right)}^2\right)}^{\frac{1.4}{1.4-1}}\\ p_{o1}=19.95\text{atm}\end{array}$

The temperature $T_0{}_2$ can be calculated as,

  $\begin{array}{l}\frac{T_{o2}}{T_2}={\left(\frac{p_{o2}}{p_2}\right)}^{\frac{\gamma -1}{\gamma }}\\ \frac{T_{o2}}{658.66\text{K}}={\left(\frac{9.1759\text{atm}}{7.72\text{atm}}\right)}^{\frac{1.4-1}{1.4}}\\ T_{o2}=692\text{K}\end{array}$

The change in entropy can be calculated as,

  $\begin{array}{l}\Delta s=-R\ln \frac{p_{o2}}{p_o{}_1}\\ \Delta s=-287\text{J}/\text{kg}\cdot \text{K }\ln \frac{9.176\text{atm}}{19.95\text{atm}}\\ \Delta s=222.898\text{J}/\text{kg}\cdot \text{K }\end{array}$

Conclusion:

Therefore, the downstream pressure is $7.72\text{atm}$ .

Therefore, the downstream temperature is $658.66\text{K}$ .

Therefore, the downstream density is $\text{4}\text{.138}\text{kg}/{\text{m}}^3$ .

Therefore, the Mach number of downstream shock is $\text{0}\text{.503}$ .

Therefore, the pressure $p_0{}_2$ is $\text{9}\text{.1759}\text{atm}$ .

Therefore, the temperature $T_0{}_2$ is $692\text{K}$ .

Therefore, the entropy change is $\text{22}\text{.898}\text{J}/\text{kg}\cdot \text{K}$ .