Chapter 9, Problem 9.49P

To determine

(a)

To calculate:

The pressure at point 1.

Answer to Problem 9.49P

$p_1=188.45kPa$

Explanation of Solution

Given information:

The diameter at section 1 is $D=5cm$

The diameter at section 2 is $d=3cm$

The stagnation temperature is equal to $T_0=300K$

Velocity at point 1 is $V_1=72m/s$

Throat pressure is $124kPa$

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

The stagnation temperature is defined as,

$T_0=T+\frac{V^2}{2C_p}$

Where,

$C_p=1005m^2/s^2.K$

Area change is defined as,

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

The pressure ratio is defined as,

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

Calculation:

Calculation the temperature at point1,

$\begin{array}{l}{T}_0=T_1+\frac{V^2}{2C_p}\\ 300K=T_1+\frac{{\left(72m/s\right)}^2}{2\left(1005m^2/s^2.K\right)}\\ T_1=297.42K\end{array}$

Calculate the Mach number at point 1,

$M{a}_1=\frac{V}{a}=\frac{V}{\sqrt{kR{T}_1}}=\frac{72m/s}{\sqrt{1.4\left(287m^2/s^2.K\right)\left(297.42K\right)}}=0.208$

Find the throat diameter,

$\begin{array}{l}\frac{A_1}{A^{\ast }}=\frac{1}{M{a}_1}\frac{{\left(1+0.2M{a}_1{}^2\right)}^3}{1.728}\\ \frac{\pi {\left(0.025m\right)}^2}{A^{\ast }}=\frac{1}{0.208}\frac{{\left(1+0.2{\left(0.208\right)}^2\right)}^3}{1.728}\\ A^{\ast }=6.877\times {10}^{-4}{m}^2\end{array}$

Calculate the Mach number at point 2,

$\begin{array}{l}\frac{A_2}{A^{\ast }}=\frac{1}{M{a}_2}\frac{{\left(1+0.2M{a}_2{}^2\right)}^3}{1.728}\\ \frac{\pi {\left(0.015m\right)}^2}{6.877\times {10}^{-4}{m}^2}=\frac{1}{M{a}_2}\frac{{\left(1+0.2{\left(M{a}_2\right)}^2\right)}^3}{1.728}\\ M{a}_2=0.827\end{array}$

Calculate the stagnation pressure,

$\begin{array}{l}\frac{p_0}{p_2}={\left[1+\frac{1}{2}\left(k-1\right)M{a}_2{}^2\right]}^{k/k-1}\\ \frac{p_0}{124kPa}={\left[1+\frac{1}{2}\left(1.4-1\right){\left(0.827\right)}^2\right]}^{1.4/1.4-1}\\ p_0=194.22kPa\end{array}$

Calculate the pressure at point 1,

$\begin{array}{l}\frac{p_0}{p_1}={\left[1+\frac{1}{2}\left(k-1\right)M{a}_1{}^2\right]}^{k/k-1}\\ \frac{194.22kPa}{p_1}={\left[1+\frac{1}{2}\left(1.4-1\right){\left(0.208\right)}^2\right]}^{1.4/1.4-1}\\ p_1=188.45kPa\end{array}$

Conclusion:

The pressure at point 1 is equal to $p_1=188.45kPa$.

To determine

(b)

To calculate:

The Mach number at point 2.

Answer to Problem 9.49P

$M{a}_2=0.827$

Explanation of Solution

Given information:

The diameter at section 1 is $D=5cm$

The diameter at section 2 is $d=3cm$

The stagnation temperature is equal to $T_0=300K$

Velocity at point 1 is $V_1=72m/s$

Throat pressure is $124kPa$

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

The stagnation temperature is defined as,

$T_0=T+\frac{V^2}{2C_p}$

Where,

$C_p=1005m^2/s^2.K$

Area change is defined as,

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

Calculation:

Calculation the temperature at point1,

$\begin{array}{l}{T}_0=T_1+\frac{V^2}{2C_p}\\ 300K=T_1+\frac{{\left(72m/s\right)}^2}{2\left(1005m^2/s^2.K\right)}\\ T_1=297.42K\end{array}$

Calculate the Mach number at point 1,

$M{a}_1=\frac{V}{a}=\frac{V}{\sqrt{kR{T}_1}}=\frac{72m/s}{\sqrt{1.4\left(287m^2/s^2.K\right)\left(297.42K\right)}}=0.208$

Find the throat diameter,

$\begin{array}{l}\frac{A_1}{A^{\ast }}=\frac{1}{M{a}_1}\frac{{\left(1+0.2M{a}_1{}^2\right)}^3}{1.728}\\ \frac{\pi {\left(0.025m\right)}^2}{A^{\ast }}=\frac{1}{0.208}\frac{{\left(1+0.2{\left(0.208\right)}^2\right)}^3}{1.728}\\ A^{\ast }=6.877\times {10}^{-4}{m}^2\end{array}$

Calculate the Mach number at point 2,

$\begin{array}{l}\frac{A_2}{A^{\ast }}=\frac{1}{M{a}_2}\frac{{\left(1+0.2M{a}_2{}^2\right)}^3}{1.728}\\ \frac{\pi {\left(0.015m\right)}^2}{6.877\times {10}^{-4}{m}^2}=\frac{1}{M{a}_2}\frac{{\left(1+0.2{\left(M{a}_2\right)}^2\right)}^3}{1.728}\\ M{a}_2=0.827\end{array}$

Conclusion:

The Mach number at point 2 is equal to $M{a}_2=0.827$.

To determine

(c)

To calculate:

The mass flow.

Answer to Problem 9.49P

$m=0.312kg/s$

Explanation of Solution

Given information:

The diameter at section 1 is $D=5cm$

The diameter at section 2 is $d=3cm$

The stagnation temperature is equal to $T_0=300K$

Velocity at point 1 is $V_1=72m/s$

Throat pressure is $124kPa$

The density at section 1 is defined as,

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

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:

According to sub-part a,

$p_1=188.45kPa$

$T_1=297.42K$

Calculate the density at point 1,

${\rho }_1=\frac{p_1}{R{T}_1}=\frac{188.45kPa}{\left(287m^2/s^2.K\right)\left(297.42K\right)}=2.207\times {10}^{-3}kg/m^3$

Calculate the mass flow,

$m={\rho }_1{A}_1{V}_1=\left(2.207kg/m^3\right)\left(\pi {\left(0.025m\right)}^2\right)\left(72m/s\right)=0.312kg/s$

Conclusion:

The mass flow is equal to $m=0.312kg/s$.