Chapter 9, Problem 9.55P

To determine

(a)

To compute: the pressure just downstream of this shock.

Answer to Problem 9.55P

$p_2=260.91kPa$

Explanation of Solution

Given information:

Stagnation pressure is equal to $450kPa$

Throat area is $12c{m}^2$

Shock is at $A_1=20c{m}^2$

The pressure ratio is defined as,

$\frac{p_0}{p}={\left[1+\frac{1}{2}\left(k-1\right)M{a}^2\right]}^{k/k-1}$

Where,

$k=1.4$

Calculation:

Calculate the area ratio,

$\frac{A_1}{A^{\ast }}=\frac{20c{m}^2}{12c{m}^2}=1.667$

According to the table B.1 which represents the isentropic flow of perfect gas

$\begin{array}{l}\frac{A}{A^{\ast }}=1.5553\to Ma=1.90\\ \frac{A}{A^{\ast }}=1.6875\to Ma=2.0\end{array}$

By interpolation,

$M{a}_1=\frac{\left(1.667-1.5553\right)}{\left(1.6875-1.5553\right)}\left(2.0-1.9\right)+1.9=1.9845$

Calculate the pressure at point 1,

$\begin{array}{l}\frac{p_0}{p_1}={\left[1+\frac{1}{2}\left(k-1\right)M{a}_1{}^2\right]}^{k/k-1}\\ \frac{450kPa}{p_1}={\left[1+\frac{1}{2}\left(1.4-1\right){\left(1.9845\right)}^2\right]}^{1.4/1.4-1}\\ p_1=58.91kPa\end{array}$

According to the table B.2 which represents the normal shock relations for a perfect gas,

$\begin{array}{l}Ma=1.90\to \frac{p_2}{p_1}=4.0450\\ Ma=2.0\to \frac{p_2}{p_1}=4.5\end{array}$

By interpolation,

$\frac{p_2}{p_1}=\frac{\left(1.9845-1.90\right)}{\left(2.0-1.9\right)}\left(4.5-4.0450\right)+4.0450=4.429$

Therefore, the pressure $p_2$ just downstream of shock,

$\begin{array}{l}\frac{p_2}{p_1}=4.429\\ p_2=4.429\left(58.91kPa\right)\\ =260.91kPa\end{array}$

Conclusion:

The pressure just downstream of shock is equal to $p_2=260.91kPa$.

To determine

(b)

To estimate: $p_3$.

Answer to Problem 9.55P

$p_3=314.22kPa$

Explanation of Solution

Given information:

Stagnation pressure is equal to $450kPa$

Throat area is $12c{m}^2$

Shock is at $A_1=20c{m}^2$

The pressure ratio is defined as,

$\frac{p_0}{p}={\left[1+\frac{1}{2}\left(k-1\right)M{a}^2\right]}^{k/k-1}$

Where,

$k=1.4$

Calculation:

According to sub-part a,

We have found,

$M{a}_1=1.9845$

According to the table B.2 which represents the normal shock relations for a perfect gas,

$\begin{array}{l}Ma=1.90\to \frac{A_2{}^{\ast }}{A_1{}^{\ast }}=1.3032\\ Ma=2.0\to \frac{A_2{}^{\ast }}{A_1{}^{\ast }}=1.3872\end{array}$

By interpolation,

$\frac{A_2{}^{\ast }}{A_1{}^{\ast }}=\frac{\left(1.9845-1.90\right)}{\left(2.0-1.9\right)}\left(1.3872-1.3032\right)+1.3032=1.3742$

Therefore,

$A_2{}^{\ast }=1.3742\left(12c{m}^2\right)=16.49c{m}^2$

At $A_3=30c{m}^2$

$\frac{A_3}{A_2{}^{\ast }}=\frac{30c{m}^2}{16.49c{m}^2}=1.82$

Therefore, calculate the relevant Mach number $M{a}_3$

According to table B.1

$M{a}_3=\frac{1.82-1.5901}{2.0351-1.5901}\left(0.4-0.3\right)+0.3=0.35$

According to the table B.2 which represents the normal shock relations for a perfect gas,

$\begin{array}{l}Ma=1.9\to \frac{p_{0_2}}{p_{0_1}}=0.7674\\ Ma=2.0\to \frac{p_{0_2}}{p_{0_1}}=0.7209\end{array}$

By interpolation,

$\frac{p_{0_2}}{p_{0_1}}=\frac{\left(1.9845-1.9\right)}{\left(2.0-1.9\right)}\left(0.7674-0.7209\right)+0.7209=0.76$

Therefore,

$p_{0_2}=0.76\left(450kPa\right)=342kPa$

Calculate the pressure at point 3,

$\begin{array}{l}\frac{p_{0_2}}{p_3}={\left[1+\frac{1}{2}\left(k-1\right)M{a}_3{}^2\right]}^{k/k-1}\\ \frac{342kPa}{p_3}={\left[1+\frac{1}{2}\left(1.4-1\right){\left[0.35\right]}^2\right]}^{1.4/1.4-1}\\ p_3=314.22kPa\end{array}$

Conclusion:

The pressure at point 3 is equal to $p_3=314.22kPa$.

To determine

(c)

To estimate: throat area.

Answer to Problem 9.55P

$A_2{}^{\ast }=16.49c{m}^2$

Explanation of Solution

Given information:

Stagnation pressure is equal to $450kPa$

Throat area is $12c{m}^2$

Shock is at $A_1=20c{m}^2$

According to sub-part a,

We have found,

$M{a}_1=1.9845$

Therefore, by using table B.2, we can find the relevant area ratio.

Calculation:

According to the table B.2 which represents the normal shock relations for a perfect gas,

$\begin{array}{l}Ma=1.90\to \frac{A_2{}^{\ast }}{A_1{}^{\ast }}=1.3032\\ Ma=2.0\to \frac{A_2{}^{\ast }}{A_1{}^{\ast }}=1.3872\end{array}$

By interpolation,

$\frac{A_2{}^{\ast }}{A_1{}^{\ast }}=\frac{\left(1.9845-1.90\right)}{\left(2.0-1.9\right)}\left(1.3872-1.3032\right)+1.3032=1.3742$

Therefore,

$A_2{}^{\ast }=1.3742\left(12c{m}^2\right)=16.49c{m}^2$

Conclusion:

The throat area is equal to $A_2{}^{\ast }=16.49c{m}^2$.

To determine

(d)

To estimate: $M{a}_3$.

Answer to Problem 9.55P

$M{a}_3=0.35$

Explanation of Solution

Given information:

Stagnation pressure is equal to $450kPa$

Throat area is $12c{m}^2$

Shock is at $A_1=20c{m}^2$

According to sub-part a,

We have found,

$M{a}_1=1.9845$

Therefore, by using table B.2 and B.1 we can able to estimate the Mach number at point 3

Calculation:

According to the table B.2 which represents the normal shock relations for a perfect gas,

$\begin{array}{l}Ma=1.90\to \frac{A_2{}^{\ast }}{A_1{}^{\ast }}=1.3032\\ Ma=2.0\to \frac{A_2{}^{\ast }}{A_1{}^{\ast }}=1.3872\end{array}$

By interpolation,

$\frac{A_2{}^{\ast }}{A_1{}^{\ast }}=\frac{\left(1.9845-1.90\right)}{\left(2.0-1.9\right)}\left(1.3872-1.3032\right)+1.3032=1.3742$

Therefore,

$A_2{}^{\ast }=1.3742\left(12c{m}^2\right)=16.49c{m}^2$

At $A_3=30c{m}^2$

$\frac{A_3}{A_2{}^{\ast }}=\frac{30c{m}^2}{16.49c{m}^2}=1.82$

Therefore, calculate the relevant Mach number $M{a}_3$

According to table B.1

$\begin{array}{l}\frac{A}{A^{\ast }}=1.5901\to Ma=0.4\\ \frac{A}{A^{\ast }}=2.0351\to Ma=0.3\end{array}$

$M{a}_3=\frac{1.82-1.5901}{2.0351-1.5901}\left(0.4-0.3\right)+0.3=0.35$

Conclusion:

The Mach number at point 3 is equal to $M{a}_3=0.35$.