Chapter 9, Problem 9.51P

To determine

(a)

If the flow will still be supersonic in the throat.

Answer to Problem 9.51P

The Mach number at throat is equal to $M{a}_{throat}=3.967$

The flow is supersonic.

Explanation of Solution

Given information:

Mach number at entrance is equal to,

$Ma=7$

The altitude is $10000m$

Inlet area is $1m^2$

The minimum area is $0.1m^2$

Exit area is $0.8m^2$

For perfect gas, where $k=1.4$

Area change is defined as,

$\frac{A}{A^{\ast }}=\frac{1}{Ma}\frac{{\left(1+0.2M{a}^2\right)}^3}{1.728}$

Calculation:

Area change is defined as,

$\frac{A}{A^{\ast }}=\frac{1}{Ma}\frac{{\left(1+0.2M{a}^2\right)}^3}{1.728}$

Therefore,

Therefore,

$\frac{A_1}{A^{\ast }}=\frac{1}{M{a}_1}\frac{{\left(1+0.2M{a}_1{}^2\right)}^3}{1.728}=\frac{1}{7}\frac{{\left(1+0.2{\left(7\right)}^2\right)}^3}{1.728}=104.14$

Throat area will be,

$A^{\ast }=\frac{A_1}{104.14}$

For section 2,

$\frac{A_2}{A^{\ast }}=\frac{A_2}{\left(\frac{A_1}{104.14}\right)}=\frac{0.1m^2}{\left(\frac{1m^2}{104.14}\right)}=10.414$

According to the table which represents the isentropic flow of a perfect gas,

$\begin{array}{l}\frac{A}{A^{\ast }}=9.7990\to Ma=3.9\\ \frac{A}{A^{\ast }}=10.7188\to Ma=4.0\end{array}$

Therefore by interpolation,

$M{a}_{throat}=\frac{\left(10.414-9.7990\right)}{\left(10.7188-9.7990\right)}\left(4.0-3.9\right)+3.9=3.967$

The flow is supersonic.

Conclusion:

The Mach number at throat is equal to $M{a}_{throat}=3.967$

The flow is supersonic.

To determine

(b)

To find: the exit Mach number.

Answer to Problem 9.51P

$M{a}_4=6.655$

Explanation of Solution

Given information:

Mach number at entrance is equal to,

$Ma=7$

The altitude is $10000m$

Inlet area is $1m^2$

The minimum area is $0.1m^2$

Exit area is $0.8m^2$

For perfect gas, where $k=1.4$

Area change is defined as,

$\frac{A}{A^{\ast }}=\frac{1}{Ma}\frac{{\left(1+0.2M{a}^2\right)}^3}{1.728}$

Calculation:

Area change is defined as,

$\frac{A}{A^{\ast }}=\frac{1}{Ma}\frac{{\left(1+0.2M{a}^2\right)}^3}{1.728}$

Therefore,

$\frac{A_1}{A^{\ast }}=\frac{1}{M{a}_1}\frac{{\left(1+0.2M{a}_1{}^2\right)}^3}{1.728}=\frac{1}{7}\frac{{\left(1+0.2{\left(7\right)}^2\right)}^3}{1.728}=104.14$

Throat area will be,

$A^{\ast }=\frac{A_1}{104.14}$

At exit,

$\frac{A_4}{A^{\ast }}=\frac{A_4}{\left(\frac{A_1}{104.14}\right)}=\frac{0.8m^2}{\left(\frac{1m^2}{104.14}\right)}=83.312$

According to the table which represents the isentropic flow of a perfect gas,

$\begin{array}{l}\frac{A}{A^{\ast }}=80.3\to Ma=6.6\\ \frac{A}{A^{\ast }}=85.8\to Ma=6.7\end{array}$

Therefore by interpolation,

$M{a}_4=\frac{\left(83.312-80.3\right)}{\left(85.8-80.3\right)}\left(6.7-6.6\right)+6.6=6.655$

Conclusion:

The Mach number at exit is equal to $M{a}_4=6.655$.

To determine

(c)

To calculate:

The exit velocity.

Answer to Problem 9.51P

$V_4=2085.9m/s$

Explanation of Solution

Given information:

Mach number at entrance is equal to,

$Ma=7$

The altitude is $10000m$

Inlet area is $1m^2$

The minimum area is $0.1m^2$

Exit area is $0.8m^2$

The temperature ratio is defined as,

$\frac{T_0}{T}=\left[1+\frac{1}{2}\left(k-1\right)M{a}^2\right]$

Speed of sound is defined as,

$a=\sqrt{kRT}$

Where,

$R$ - Gas constant

$R=287m^2/s^2.K$

$k$ - Specific heat capacity

The Mach number is defined as,

$Ma=\frac{V}{a}$

Where,

$V$ - Air velocity

Calculation:

At altitude of $10000m$,

According to table A.6 which represents the properties of standard atmosphere,

$\begin{array}{l}{p}_1=26416Pa\\ T=223.16K\\ \rho =0.4125kg/m^3\end{array}$

Calculate the stagnation temperature,

$\begin{array}{l}\frac{T_0}{T}=\left[1+\frac{1}{2}\left(k-1\right)M{a}^2\right]\\ \frac{T_0}{223.16K}=\left[1+\frac{1}{2}\left(1.4-1\right)7^2\right]\\ T_0=2410.13K\\ \end{array}$

Calculate the exit temperature,

$\begin{array}{l}\frac{T_0}{T_4}=\left[1+\frac{1}{2}\left(k-1\right)M{a}_4{}^2\right]\\ \frac{2410.13K}{T_4}=\left[1+\frac{1}{2}\left(1.4-1\right){\left(6.655\right)}^2\right]\\ T_4=244.5K\end{array}$

Calculate the exit velocity,

$V_4=M{a}_4\sqrt{kR{T}_4}=\left(6.655\right)\sqrt{1.4\left(287m^2/s^2.K\right)\left(244.5K\right)}=2085.9m/s$

Conclusion:

The exit velocity is equal to $V_4=2085.9m/s$.

To determine

(d)

To calculate:

The exit pressure.

Answer to Problem 9.51P

$p_4=36357Pa$

Explanation of Solution

Given information:

Mach number at entrance is equal to,

$Ma=7$

The altitude is $10000m$

Inlet area is $1m^2$

The minimum area is $0.1m^2$

Exit area is $0.8m^2$

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:

At altitude of $10000m$,

According to table A.6 which represents the properties of standard atmosphere,

$\begin{array}{l}{p}_1=26416Pa\\ T=223.16K\\ \rho =0.4125kg/m^3\end{array}$

Calculate the stagnation pressure,

$\begin{array}{l}\frac{p_0}{p_1}={\left[1+\frac{1}{2}\left(k-1\right)M{a}_1{}^2\right]}^{k/k-1}\\ \frac{p_0}{26416Pa}={\left[1+\frac{1}{2}\left(1.4-1\right){\left(7\right)}^2\right]}^{1.4/1.4-1}\\ p_0=1.0935\times {10}^{8}Pa\end{array}$

Calculate the exit pressure,

$\begin{array}{l}\frac{p_0}{p_4}={\left[1+\frac{1}{2}\left(k-1\right)M{a}_4{}^2\right]}^{k/k-1}\\ \frac{1.0935\times {10}^{8}Pa}{p_4}={\left[1+\frac{1}{2}\left(1.4-1\right){\left(6.655\right)}^2\right]}^{1.4/1.4-1}\\ p_4=36357Pa\end{array}$

Conclusion:

The exit pressure is equal to $p_4=36357Pa$.