Chapter 12, Problem 75EP

To determine

The wave angle, Mach number and temperature after shock.

Answer to Problem 75EP

The wave angle, Mach number and temperature after shock are ${\beta }_{weak}=37.21°\text{and}{\beta }_{strong}=85.05°$, $1.713$ and $567.125°\text{R}$ respectively.

Explanation of Solution

Given:

Air pressure, air temperature and Mach number are $14\text{psia}$, $40°\text{F}$ and $2$ respectively. Flow direction angle is $8°$.

Concept used:

Relation of wave angle is expressed as follows:

  $\tan \theta =\frac{2\left(M{a}_1{}^2{\sin }^2\beta -1\right)}{\tan \beta \left(M{a}_1{}^2\left(k+\cos 2\beta \right)+2\right)}$........ (1)

Here, wave angle is $\beta$, angle of deflection is $\theta$ and Mach number before ramp is $M{a}_1$.

Relation of upstream normal Mach number is expressed as follows:

  ${\left(M{a}_1\right)}_n=M{a}_1\sin {\beta }_{\text{weak}}$........ (2)

Here, wave weak angle is ${\beta }_{\text{weak}}$, upstream normal Mach number is ${\left(M{a}_1\right)}_n$ and Mach number before ramp is $M{a}_1$.

Relation of downstream normal Mach number is expressed as follows:

  ${\left(M{a}_2\right)}_n=\sqrt{\frac{\left(k-1\right){\left({\left(M{a}_1\right)}_n\right)}^2+2}{2k{\left({\left(M{a}_1\right)}_n\right)}^2-k+1}}$........ (3)

Here, upstream normal Mach number is ${\left(M{a}_1\right)}_n$ and downstream normal Mach number is ${\left(M{a}_2\right)}_n$.

Relation of the downstream Mach number is expressed as follows:

  $M{a}_2=\frac{{\left(M{a}_2\right)}_n}{\left(\sin {\beta }_{\text{weak}}-\theta \right)}$........ (4)

Here, wave weak angle is ${\beta }_{\text{weak}}$, angle of deflection is $\theta$ and downstream Mach number is $M{a}_2$.

Relation of downstream pressure is expressed as follows:

  $p_2=p_1\left(\frac{2k{\left({\left(M{a}_1\right)}_n\right)}^2-k+1}{k+1}\right)$........ (5)

Here, downstream pressure is $p_2$, upstream normal Mach number is ${\left(M{a}_1\right)}_n$ and pressure before ramp is $p_1$.

Relation of downstream Temperature is expressed as follows:

  $T_2=T_1\left(\frac{p_2}{p_1}\right)\left(\frac{2+\left(k-1\right){\left({\left(M{a}_1\right)}_n\right)}^2}{\left(k+1\right){\left({\left(M{a}_1\right)}_n\right)}^2}\right)$........ (6)

Here, downstream pressure is $p_2$, downstream temperature is $T_2$, Temperature before ramp is $T_1$ upstream normal Mach number is ${\left(M{a}_1\right)}_n$ and pressure before ramp is $p_1$.

Calculation:

Substitute $8°$ for $\theta$, $2$ for $M{a}_1$ and $1.4$ for $k$ in equation (1).

  $\begin{array}{l}\tan 8°=\frac{2\left({\left(2\right)}^2{\sin }^2\beta -1\right)}{\tan \beta \left({\left(2\right)}^2\left(1.4+\cos 2\beta \right)+2\right)}\\ 0.1405\left(\tan \beta \left(4\left(1.4+\cos 2\beta \right)+2\right)\right)=\left(8{\sin }^2\beta -2\right)\\ 0.787\tan \beta +0.562\tan \beta \cos 2\beta +1.124\tan \beta =\left(8{\sin }^2\beta -2\right)\end{array}$

Solve the above equation by iteration. We get two value of wave angle as ${\beta }_{weak}=37.21°$ and ${\beta }_{strong}=85.05°$.

Substitute $37.21°$ for ${\beta }_{weak}$ and $2$ for $M{a}_1$ in equation (2).

  $\begin{array}{l}{\left(M{a}_1\right)}_n=2\sin 37.21°\\ {\left(M{a}_1\right)}_n=1.21\end{array}$

Substitute $1.21$ for ${\left(M{a}_1\right)}_n$ and $1.4$ for $k$ in equation (3).

  $\begin{array}{l}{\left(M{a}_2\right)}_n=\sqrt{\frac{\left(1.4-1\right){\left(1.21\right)}^2+2}{2\times 1.4{\left(1.21\right)}^2-1.4+1}}\\ {\left(M{a}_2\right)}_n=0.8360\end{array}$

Substitute $0.8360$ for ${\left(M{a}_2\right)}_n$, $37.21°$ for ${\beta }_{weak}$ and $8°$ for $\theta$ in equation (4).

  $\begin{array}{l}M{a}_2=\frac{0.8360}{\left(\sin \left(37.21°-8°\right)\right)}\\ M{a}_2=1.713\end{array}$

Substitute $14\text{psia}$ for $p_1$, $1.4$ for $k$ and $1.21$ for ${\left(M{a}_1\right)}_n$ in equation (5).

  $\begin{array}{l}{p}_2=14\left(\frac{2\times 1.4{\left(1.21\right)}^2-1.4+1}{1.4+1}\right)\\ p_2=21.58\text{psia}\end{array}$

Substitute $14\text{psia}$ for $p_1$, $21.58\text{psia}$ for $p_2$, $500\text{R}$ for $T_1$. $1.4$ for $k$ and $1.21$ for ${\left(M{a}_1\right)}_n$ in equation (5).

  $\begin{array}{l}{T}_2=500\left(\frac{21.58}{14}\right)\left(\frac{2+\left(1.4-1\right){\left(1.21\right)}^2}{\left(1.4+1\right){\left(1.21\right)}^2}\right)\\ T_2=567.125°\text{R}\end{array}$

Thus, the wave angle, Mach number and temperature after shock are ${\beta }_{weak}=37.21°\text{and}{\beta }_{strong}=85.05°$, $1.713$ and $567.125°\text{R}$ respectively.

Conclusion:

The wave angle, Mach number and temperature after shock are ${\beta }_{weak}=37.21°\text{and}{\beta }_{strong}=85.05°$, $1.713$ and $567.125°\text{R}$ respectively.