Chapter 9, Problem 9.31P

To determine

(a)

The Mach number at point 2.

Answer to Problem 9.31P

$M{a}_2=0.938$

Explanation of Solution

Given information:

At section 1,

$\begin{array}{l}{p}_1=35lbf/i{n}^2\\ T_1=200°F\\ V_1=400ft/s\end{array}$

At section 2,

$\begin{array}{l}{p}_2=18lbf/i{n}^2\\ V_2=1100ft/s\end{array}$

Speed of sound is defined as,

$a=\sqrt{kRT}$

Where,

$R$ - Gas constant

$R=1716f{t}^2/s^2.°R$

$k$ - Specific heat capacity

The Mach number is defined as,

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

Where,

$V$ - Air velocity

For adiabatic flow, the stagnation temperature is defined as,

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

Where,

$C_p=6009f{t}^2/s^2.°R$

Calculation:

Convert,

$T_1=200°F=659.67°R$

Calculate the stagnation temperature,

$T_0=T_1+\frac{V_1{}^2}{2C_p}=659.67°R+\frac{{\left(400ft/s\right)}^2}{2\left(6009f{t}^2/s^2.°R\right)}=672.98°R$

Calculate the temperature at point 2,

$T_0=T_2+\frac{V_2{}^2}{2C_p}$

Substitute for known values,

$\begin{array}{l}672.98°R=T_2+\frac{{\left(1100ft/s\right)}^2}{2\left(6009f{t}^2/s^2.°R\right)}\\ T_2=572.29°R\end{array}$

Calculate the speed of sound,

$a=\sqrt{kR{T}_2}=\sqrt{1.4\left(1716f{t}^2/s^2.°R\right)\left(572.29°R\right)}=1172.54ft/s$

Calculate the Mach number at section 2,

$M{a}_2=\frac{V_2}{a}=\frac{1100ft/s}{1172.54ft/s}=0.938$

Conclusion:

The Mach number at section 2 is equal to $M{a}_2=0.938$.

To determine

(b)

Calculate $U_{\max }$.

Answer to Problem 9.31P

$U_{\max }=2843.92ft/s$

Explanation of Solution

Given information:

At section 1,

$\begin{array}{l}{p}_1=35lbf/i{n}^2\\ T_1=200°F\\ V_1=400ft/s\end{array}$

At section 2,

$\begin{array}{l}{p}_2=18lbf/i{n}^2\\ V_2=1100ft/s\end{array}$

The maximum value of velocity is defined as,

$U_{\max }=\sqrt{2C_p{T}_0}$

Where,

$T_0$ - Stagnation temperature

$C_p=6009f{t}^2/s^2.°R$ Calculation:

According to sub-part a,

The stagnation temperature is equal to,

$T_0=672.98°R$

Calculate the maximum value of velocity,

$U_{\max }=\sqrt{2C_p{T}_0}=\sqrt{2\left(6009f{t}^2/s^2.°R\right)\left(672.98°R\right)}=2843.92ft/s$

Conclusion:

The maximum value of velocity is equal to $U_{\max }=2843.92ft/s$.

To determine

(c)

Calculate $\frac{p_{0_2}}{p_{0_1}}$.

Answer to Problem 9.31P

$\frac{p_{0_2}}{p_{0_1}}=0.845$

Explanation of Solution

Given information:

At section 1,

$\begin{array}{l}{p}_1=35lbf/i{n}^2\\ T_1=200°F\\ V_1=400ft/s\end{array}$

At section 2,

$\begin{array}{l}{p}_2=18lbf/i{n}^2\\ V_2=1100ft/s\end{array}$

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,

$k=1.4$

Speed of sound is defined as,

$a=\sqrt{kRT}$

Where,

$R$ - Gas constant

$R=1716f{t}^2/s^2.°R$

$k$ - Specific heat capacity

The Mach number is defined as,

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

Where,

$V$ - Air velocity

Calculation:

Calculate the Mach number at section 1,

$M{a}_1=\frac{V_1}{a}=\frac{V_1}{\sqrt{kR{T}_1}}=\frac{400ft/s}{\sqrt{1.4\left(1716f{t}^2/s^2.°R\right)\left(659.67°R\right)}}=0.318$

Convert,

$p_1=35lbf/i{n}^2=5040lbf/f{t}^2$

Calculate the stagnation pressure at section 1,

$\frac{p_{0_1}}{p_1}={\left[1+\frac{1}{2}\left(k-1\right)M{a}_1{}^2\right]}^{k/k-1}$

Substitute for known values,

$\frac{p_{0_1}}{5040lbf/f{t}^2}={\left[1+\frac{1}{2}\left(1.4-1\right){\left(0.318\right)}^2\right]}^{1.4/1.4-1}$

Solve for stagnation pressure at section 1,

$p_{0_1}=5405.87lbf/f{t}^2$

Convert,

$p_2=18lbf/i{n}^2=2592lbf/f{t}^2$

Calculate the stagnation pressure at section 1,

$\frac{p_{0_2}}{p_2}={\left[1+\frac{1}{2}\left(k-1\right)M{a}_2{}^2\right]}^{k/k-1}$

Substitute for known values,

$\frac{p_{0_2}}{2592lbf/f{t}^2}={\left[1+\frac{1}{2}\left(1.4-1\right){\left(0.938\right)}^2\right]}^{1.4/1.4-1}$

Solve for stagnation pressure at section 2,

$p_{0_2}=4571.10lbf/f{t}^2$

Therefore,

$\frac{p_{0_2}}{p_{0_1}}=\frac{4571.10lbf/f{t}^2}{5405.87lbf/f{t}^2}=0.845$

Conclusion:

The stagnation pressure ration is equal to $\frac{p_{0_2}}{p_{0_1}}=0.845$.