Chapter 9, Problem 9.35P

To determine

(a)

To calculate:

The Mach number at point 1.

Answer to Problem 9.35P

$M{a}_1=0.498$

Explanation of Solution

Given information:

At stagnation point inside the reservoir,

$\begin{array}{l}{p}_0=150kPa\\ T_0=400K\end{array}$

At section 1,

$A_1=0.1m^2$

The static pressure is equal to,

$p=123kPa$

The pressure ratio is defined as,

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

Assume, for helium gas,

$\begin{array}{l}k=1.66\\ R=2077J/kg.K\end{array}$

Calculation:

Calculate the Mach number at point 1,

$\begin{array}{l}\frac{p_0}{p}={\left[1+\frac{1}{2}\left(k-1\right)M{a}_1{}^2\right]}^{k/k-1}\\ \frac{150kPa}{123kPa}={\left[1+\frac{1}{2}\left(1.66-1\right)M{a}_1{}^2\right]}^{1.66/1.66-1}\\ M{a}_1=0.498\end{array}$

Conclusion:

The Mach number at point 1 is equal to $M{a}_1=0.498$.

To determine

(b)

To calculate:

The mass flow.

Answer to Problem 9.35P

$m=9kg/s$

Explanation of Solution

Given information:

At stagnation point inside the reservoir,

$\begin{array}{l}{p}_0=150kPa\\ T_0=400K\end{array}$

At section 1,

$A_1=0.1m^2$

The static pressure is equal to,

$p=123kPa$

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,

Assume, for helium gas,

$\begin{array}{l}k=1.66\\ R=2077J/kg.K\end{array}$

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

$k$ - Specific heat capacity

The Mach number is defined as,

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

Where,

$V$ - Air velocity

Calculation:

Calculate the temperature at point 1,

$\begin{array}{l}\frac{T_0}{T_1}=\left[1+\frac{1}{2}\left(k-1\right)M{a}_1{}^2\right]\\ \frac{400K}{T_1}=\left[1+\frac{1}{2}\left(1.66-1\right){\left(0.498\right)}^2\right]\\ T_1=369.74K\end{array}$

Calculate the speed of sound,

$a=\sqrt{kR{T}_1}=\sqrt{1.66\left(2077J/kg.K\right)\left(369.74K\right)}=1129.07m/s$

Calculate the velocity at point 1,

$V_1=M{a}_{1}a=\left(0.498\right)\left(1129.07m/s\right)=562.28m/s$

Calculate the density at point 1,

${\rho }_1=\frac{p_1}{R{T}_1}=\frac{123000Pa}{\left(2077J/kg.K\right)\left(369.74K\right)}=0.16kg/m^3$

Calculate the mass flow,

$m={\rho }_1{A}_1{V}_1=\left(0.16kg/m^3\right)\left(0.1m^2\right)\left(562.28m/s\right)=9kg/s$

Conclusion:

The mass flow is equal to $9kg/s$.

To determine

(c)

To calculate:

The temperature at point 1.

Answer to Problem 9.35P

$T_1=369.74K$

Explanation of Solution

Given information:

At stagnation point inside the reservoir,

$\begin{array}{l}{p}_0=150kPa\\ T_0=400K\end{array}$

At section 1,

$A_1=0.1m^2$

The static pressure is equal to,

$p=123kPa$

The temperature ratio is defined as,

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

According to sub-part a,

$M{a}_1=0.498$ Calculation:

Calculate the temperature at point 1,

$\begin{array}{l}\frac{T_0}{T_1}=\left[1+\frac{1}{2}\left(k-1\right)M{a}_1{}^2\right]\\ \frac{400K}{T_1}=\left[1+\frac{1}{2}\left(1.66-1\right){\left(0.498\right)}^2\right]\\ T_1=369.74K\end{array}$

Conclusion:

The temperature at point 1 is equal to $T_1=369.74K$.

To determine

(d)

Calculate $A^{\ast }$.

Answer to Problem 9.35P

$A^{\ast }=0.0755m^2$

Explanation of Solution

Given information:

At stagnation point, inside the reservoir,

$\begin{array}{l}{p}_0=150kPa\\ T_0=400K\end{array}$

At section 1,

$A_1=0.1m^2$

The static pressure is equal to,

$p=123kPa$

Assume, for helium gas,

$k=1.66$

Area change is defined as,

$\frac{A_1}{A^{\ast }}=\frac{1}{M{a}_1}{\left[\frac{1+\frac{1}{2}\left(k-1\right)M{a}_1{}^2}{\frac{1}{2}\left(k+1\right)}\right]}^{\left(1/2\right)\left(k+1\right)/\left(k-1\right)}$

Calculation:

Calculate $A^{\ast }$

$\frac{A_1}{A^{\ast }}=\frac{1}{M{a}_1}{\left[\frac{1+\frac{1}{2}\left(k-1\right)M{a}_1{}^2}{\frac{1}{2}\left(k+1\right)}\right]}^{\left(1/2\right)\left(k+1\right)/\left(k-1\right)}$

According to subpart a,

$M{a}_1=0.498$

Therefore,

$\frac{0.1m^2}{A^{\ast }}=\frac{1}{0.498}{\left[\frac{1+\frac{1}{2}\left(1.66-1\right){\left(0.498\right)}^2}{\frac{1}{2}\left(1.66+1\right)}\right]}^{\left(1/2\right)\left(1.66+1\right)/\left(1.66-1\right)}$

Solve to find $A^{\ast }$

$A^{\ast }=0.0755m^2$

Conclusion:

According to the above calculation,

$A^{\ast }=0.0755m^2$.