Chapter 9, Problem 9.43P

To determine

(a)

Compute $V_2$.

Answer to Problem 9.43P

$V_2=806.43m/s$

Explanation of Solution

Given information:

At section 1,

$\begin{array}{l}{A}_1=20c{m}^2\\ p_1=300kpa\\ {\rho }_1=1.75kg/m^3\\ M{a}_1=0.25\end{array}$

At section 2,

The area is same but flow is faster than at section 1

The temperature ratio is defined as,

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

For perfect gas, where $k=1.4$

Area change is defined as,

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

In the above equation,

$A^{\ast }$ - Throat diameter

The velocity at point 2 is defined as,

$V_2=M{a}_2\sqrt{kR{T}_2}$

The density at section 1 is defined as,

${\rho }_1=\frac{p_1}{R{T}_1}$

For ideal gas,

$\begin{array}{l}R=287m^2/s^2.K\\ k=1.4\end{array}$

Calculation:

Calculate the temperature at point 1,

${\rho }_1=\frac{p_1}{R{T}_1}$

Rearrange,

$T_1=\frac{p_1}{R{\rho }_1}$

Substitute,

$T_1=\frac{p_1}{R{\rho }_1}=\frac{300000Pa}{287m^2/s^2.K\left(1.75kg/m^3\right)}=597.31K$

Calculate the stagnation temperature,

$\frac{T_0}{T_1}=\left[1+\frac{1}{2}\left(k-1\right)M{a}_1{}^2\right]$

Substitute for known values,

$\frac{T_0}{597.31K}=\left[1+\frac{1}{2}\left(1.4-1\right){\left(0.25\right)}^2\right]$

Solve to find $T_0$

$T_0=604.78K$

Calculate the area change at point 1,

$\begin{array}{l}\frac{A_1}{A^{\ast }}=\frac{1}{M{a}_1}\frac{{\left(1+0.2M{a}_1{}^2\right)}^3}{1.728}\\ =\frac{1}{0.25}\frac{{\left(1+0.2{\left(0.25\right)}^2\right)}^3}{1.728}\\ =2.4027\end{array}$

Since the area is same at point 2, the area change will also be the same

Therefore,

$\frac{A_2}{A^{\ast }}=2.4027$

Therefore, according to the table B.1 which represents the isentropic flow of perfect gas

The Mach number at point 2 is equal to,

$M{a}_2\approx 2.4$

Now, find the temperature at point 2,

$\frac{T_0}{T_2}=\left[1+\frac{1}{2}\left(k-1\right)M{a}_2{}^2\right]$

Substitute for known values,

$\frac{604.78K}{T_2}=\left[1+\frac{1}{2}\left(1.4-1\right){\left(0.25\right)}^2\right]$

Solve to find $T_2$,

$T_2=281K$

Calculate the velocity at point 2,

$V_2=M{a}_2\sqrt{kR{T}_2}$

Substitute for known values,

$V_2=\left(2.4\right)\sqrt{\left(1.4\right)\left(287m^2/s^2.K\right)\left(281K\right)}$

Solve to find $V_2$

$V_2=806.43m/s$

Conclusion:

The velocity at point 2 is equal to $V_2=806.43m/s$.

To determine

(b)

Compute $M{a}_2$.

Answer to Problem 9.43P

$M{a}_2\approx 2.4$

Explanation of Solution

Given information:

At section 1,

$\begin{array}{l}{A}_1=20c{m}^2\\ p_1=300kpa\\ {\rho }_1=1.75kg/m^3\\ M{a}_1=0.25\end{array}$

At section 2,

The area is same but flow is faster than at section 1

For perfect gas, where $k=1.4$

Area change is defined as,

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

In above equation,

$A^{\ast }$ - Throat diameter

Calculation:

Calculate the area change at point 1,

$\begin{array}{l}\frac{A_1}{A^{\ast }}=\frac{1}{M{a}_1}\frac{{\left(1+0.2M{a}_1{}^2\right)}^3}{1.728}\\ =\frac{1}{0.25}\frac{{\left(1+0.2{\left(0.25\right)}^2\right)}^3}{1.728}\\ =2.4027\end{array}$

Since the area is same at point 2, the area change will also be the same

Therefore,

$\frac{A_2}{A^{\ast }}=2.4027$

Therefore, according to the table B.1 which represents the isentropic flow of perfect gas

The Mach number at point 2 is equal to,

$M{a}_2\approx 2.4$

Conclusion:

The Mach number at point 2 is equal to $M{a}_2\approx 2.4$.

To determine

(c)

Compute $T_2$.

Answer to Problem 9.43P

$T_2=281K$

Explanation of Solution

Given information:

At section 1,

$\begin{array}{l}{A}_1=20c{m}^2\\ p_1=300kpa\\ {\rho }_1=1.75kg/m^3\\ M{a}_1=0.25\end{array}$

At section 2,

The area is same but flow is faster than at section 1

The temperature ratio is defined as,

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

Calculation:

Calculate the temperature at point 1,

${\rho }_1=\frac{p_1}{R{T}_1}$

Rearrange,

$T_1=\frac{p_1}{R{\rho }_1}$

Substitute,

$T_1=\frac{p_1}{R{\rho }_1}=\frac{300000Pa}{287m^2/s^2.K\left(1.75kg/m^3\right)}=597.31K$

Calculate the stagnation temperature,

$\frac{T_0}{T_1}=\left[1+\frac{1}{2}\left(k-1\right)M{a}_1{}^2\right]$

Substitute for known values,

$\frac{T_0}{597.31K}=\left[1+\frac{1}{2}\left(1.4-1\right){\left(0.25\right)}^2\right]$

Solve to find $T_0$

$T_0=604.78K$

Calculate the area change at point 1,

$\begin{array}{l}\frac{A_1}{A^{\ast }}=\frac{1}{M{a}_1}\frac{{\left(1+0.2M{a}_1{}^2\right)}^3}{1.728}\\ =\frac{1}{0.25}\frac{{\left(1+0.2{\left(0.25\right)}^2\right)}^3}{1.728}\\ =2.4027\end{array}$

Since the area is same at point 2, the area change will also be the same

Therefore,

$\frac{A_2}{A^{\ast }}=2.4027$

Therefore, according to the table B.1 which represents the isentropic flow of perfect gas

The Mach number at point 2 is equal to,

$M{a}_2\approx 2.4$

Now, find the temperature at point 2,

$\frac{T_0}{T_2}=\left[1+\frac{1}{2}\left(k-1\right)M{a}_2{}^2\right]$

Substitute for known values,

$\frac{604.78K}{T_2}=\left[1+\frac{1}{2}\left(1.4-1\right){\left(0.25\right)}^2\right]$

Solve to find $T_2$,

$T_2=281K$

Conclusion:

The temperature at point 2 is equal to $T_2=281K$.

To determine

(d)

To compute: mass flow.

Answer to Problem 9.43P

$m=0.428kg/s$

Explanation of Solution

Given information:

At section 1,

$\begin{array}{l}{A}_1=20c{m}^2\\ p_1=300kpa\\ {\rho }_1=1.75kg/m^3\\ M{a}_1=0.25\end{array}$

At section 2,

The area is same but flow is faster than at section 1

The mass flow is defined as,

$m={\rho }_1{A}_1{V}_1$

Where,

$A_1$ - Area at section 1,

For ideal gas,

$\begin{array}{l}R=287m^2/s^2.K\\ k=1.4\end{array}$

Calculation:

Calculate the velocity at point 1,

$\begin{array}{l}{V}_1=M{a}_1\sqrt{kR{T}_1}\\ V_1=\left(0.25\right)\sqrt{\left(1.4\right)\left(287m^2/s^2.K\right)\left(597.31K\right)}\\ V_1=122.47m/s\end{array}$

Calculate the mass flow,

$m={\rho }_1{A}_1{V}_1$

Substitute,

$m=\left(1.75kg/m^3\right)\left(20\times {10}^{-4}{m}^2\right)\left(122.47m/s\right)$

Solve to find mass flow,

$m=0.428kg/s$

Conclusion:

The mass flow is equal to $m=0.428kg/s$.

To determine

(e)

If there is a sonic throat exit and the value.

Answer to Problem 9.43P

$A^{\ast }=8.32c{m}^2$

Explanation of Solution

Given information:

At section 1,

$\begin{array}{l}{A}_1=20c{m}^2\\ p_1=300kpa\\ {\rho }_1=1.75kg/m^3\\ M{a}_1=0.25\end{array}$

At section 2,

The area is same but flow is faster than at section 1

Area change is defined as,

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

In above equation,

$A^{\ast }$ - Throat diameter

Calculation:

According to sub-part a,

We have found that,

The area change is equal to,

$\frac{A_1}{A^{\ast }}=2.4027$

But we know,

The area of both sections is equal, but the velocity of the flow differs. Therefore, a throat exist in between section 1 and 2.

Therefore, to find the throat area,

$\begin{array}{l}\frac{A_1}{A^{\ast }}=2.4027\\ \frac{20c{m}^2}{A^{\ast }}=2.4027\\ A^{\ast }=8.32c{m}^2\end{array}$

Conclusion:

The throat diameter is equal to $A^{\ast }=8.32c{m}^2$.