Chapter 9, Problem 9.28P

To determine

(a)

The Mach number at point 1.

Answer to Problem 9.28P

$M{a}_1=0.5$

Explanation of Solution

Given information:

At section 1,

$\begin{array}{l}{p}_1=250kPa\\ T_1=125°C\\ V_1=200m/s\end{array}$

At section 2,

$\begin{array}{l}{A}_2=0.25m^2\\ M{a}_2=2.0\end{array}$

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:

Calculate the speed of sound,

For ideal gas,

$k=1.4$

Convert,

$T_1=125°C=398K$

$a=\sqrt{kRT}=\sqrt{\left(1.4\right)\left(287m^2/s^2.K\right)\left(398K\right)}=399.89m/s$

Calculate the Mach number,

$M{a}_1=\frac{V}{a}=\frac{200m/s}{399.89m/s}=0.5$

Conclusion:

The Mach number at section1 is equal to $M{a}_1=0.5$.

To determine

(b)

The temperature at point 2.

Answer to Problem 9.28P

$T_2=232.16K$

Explanation of Solution

Given information:

At section 1,

$\begin{array}{l}{p}_1=250kPa\\ T_1=125°C\\ V_1=200m/s\end{array}$

At section 2,

$\begin{array}{l}{A}_2=0.25m^2\\ M{a}_2=2.0\end{array}$

The temperature ratio can be defined in terms of Mach number as,

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

For ideal gas,

$k=1.4$

Calculation:

Calculate the stagnation temperature,

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

For point 1,

According to sub-part a,

$M{a}_1=0.5$

Substitute for known values,

$\left(\frac{T_0}{398K}\right)=\left[1+\frac{1}{2}\left(1.4-1\right){\left(0.5\right)}^2\right]$

Therefore,

$T_0=417.9K$

Find the temperature at section 2,

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

Substitute for known values,

$\left(\frac{417.9K}{T_2}\right)=\left[1+\frac{1}{2}\left(1.4-1\right){\left(2.0\right)}^2\right]$

Therefore,

$T_2=232.16K$

Conclusion:

The temperature at section 2 is equal to $T_2=232.16K$.

To determine

(c)

To calculate:

The velocity at point 2.

Answer to Problem 9.28P

$V_2=610.84m/s$

Explanation of Solution

Given information:

At section 1,

$\begin{array}{l}{p}_1=250kPa\\ T_1=125°C\\ V_1=200m/s\end{array}$

At section 2,

$\begin{array}{l}{A}_2=0.25m^2\\ M{a}_2=2.0\end{array}$

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:

According to sub-part a,

$T_2=232.16K$

Calculate the speed of sound,

$a=\sqrt{kRT}=\sqrt{\left(1.4\right)\left(287m^2/s^2.K\right)\left(232.16K\right)}=305.42m/s$

Calculate the velocity at point 2,

$V_2=M{a}_{2}a=\left(2.0\right)\left(305.42m/s\right)=610.84m/s$

Conclusion:

The velocity at section 2 is equal to $V_2=610.84m/s$.

To determine

(d)

To calculate:

The mass flow.

Answer to Problem 9.28P

$m=86.74kg/s$

Explanation of Solution

Given information:

At section 1,

$\begin{array}{l}{p}_1=250kPa\\ T_1=125°C\\ V_1=200m/s\end{array}$

At section 2,

$\begin{array}{l}{A}_2=0.25m^2\\ M{a}_2=2.0\end{array}$

The density at section 2 is defined as,

${\rho }_2=\frac{p_2}{R{T}_2}$

The pressure ration is defined as,

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

The mass flow is defined as,

$m={\rho }_2{A}_2{V}_2$

Where,

$A_2$ - Area at section 2,

For ideal gas,

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

Calculation:

Calculate the stagnation pressure,

$\frac{p_0}{250kPa}={\left[1+\frac{1}{2}\left(1.4-1\right){\left(0.5\right)}^2\right]}^{1.4/1.4-1}$

Therefore,

$p_0=296.55kPa$

Calculate the pressure at section 2,

$\frac{296.55kPa}{p_2}={\left[1+\frac{1}{2}\left(1.4-1\right){\left(2.0\right)}^2\right]}^{1.4/1.4-1}$

Therefore,

$p_2=37.9kPa$

Calculate the density at section 2,

${\rho }_2=\frac{p_2}{R{T}_2}=\frac{37900Pa}{\left(287m^2/s^2.K\right)\left(232.16K\right)}=0.568kg/m^3$

Calculate the mass flow,

$m={\rho }_2{A}_2{V}_2=\left(0.568kg/m^3\right)\left(0.25m^2\right)\left(610.84m/s\right)=86.74kg/s$

Conclusion:

The mass flow is equal to $m=86.74kg/s$.