Chapter 9, Problem 9.42P

To determine

(a)

To calculate:

The Mach number, velocity and temperature at exit plane.

Answer to Problem 9.42P

$M{a}_e=0.9$

$T_e=260.75K$

$V_e=291.31m/s$

Explanation of Solution

Given information:

At stagnation point,

$\begin{array}{l}{p}_0=169.12kPa\\ T_0=30°C\end{array}$

The absolute pressure is equal to $p_a=100kPa$ and $T_a=20°C$

The valve exit diameter is equal to $2mm$

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 air,

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

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 Mach number,

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

Substitute for known values,

$\frac{169.12kPa}{100kPa}={\left[1+\frac{1}{2}\left(1.4-1\right)M{a}_e{}^2\right]}^{1.4/1.4-1}$

Solve to find Mach number,

$M{a}_e=0.9$

Calculate the exit temperature,

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

Convert,

$T_0=30°C=303K$

Substitute for known values,

$\frac{303K}{T_e}=\left[1+\frac{1}{2}\left(1.4-1\right){\left(0.9\right)}^2\right]$

Therefore,

$T_e=260.75K$

Calculate the exit velocity,

$V_e=M{a}_e\left(a\right)=M{a}_e\sqrt{kR{T}_e}$

Substitute for known values,

$V_e=\left(0.9\right)\sqrt{1.4\left(287m^2/s^2.K\right)\left(260.75K\right)}$

Therefore,

$V_e=291.31m/s$

Conclusion:

The exit Mach number is equal to $M{a}_e=0.9$

Exit temperature is equal to $T_e=260.75K$

Exit velocity is equal to $V_e=291.31m/s$.

To determine

(b)

To calculate:

Initial mass flow rate.

Answer to Problem 9.42P

$m=1.223\times {10}^{-3}kg/s$

Explanation of Solution

Given information:

At stagnation point,

$\begin{array}{l}{p}_0=169.12kPa\\ T_0=30°C\end{array}$

The absolute pressure is equal to $p_a=100kPa$ and $T_a=20°C$

The valve exit diameter is equal to $2mm$

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:

Calculate the exit density,

${\rho }_e=\frac{p_e}{R{T}_e}=\frac{100000kPa}{\left(287m^2/s^2.K\right)\left(260.75K\right)}=1.336kg/m^3$

Calculate the mass flow rate,

$\begin{array}{l}m={\rho }_e{A}_e{V}_e\\ =\left(1.336kg/m^3\right)\left(\frac{\pi }{4}{\left(0.002m\right)}^2\right)\left(291.31m/s\right)\\ =1.223\times {10}^{-3}kg/s\end{array}$

Conclusion:

The mass flow is equal to $m=1.223\times {10}^{-3}kg/s$.

To determine

(c)

To calculate:

The exit velocity using incompressible Bernoulli’s equation.

Answer to Problem 9.42P

$V_e=266.59m/s$

The above obtained value of $266.59m/s$ is almost $8\%$ lower than the velocity found in sup-part a $V_e=291.31m/s$

Explanation of Solution

Given information:

At stagnation point,

$\begin{array}{l}{p}_0=169.12kPa\\ T_0=30°C\end{array}$

The absolute pressure is equal to $p_a=100kPa$ and $T_a=20°C$

The valve exit diameter is equal to $2mm$

The density at section 1 is defined as,

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

According to incompressible Bernoulli’s equation,

The exit velocity is defined as,

$V_e=\sqrt{\frac{2\Delta p}{{\rho }_0}}$

Calculation:

Calculate the density,

${\rho }_0=\frac{p_0}{R{T}_0}$

Assume,

$p_0=p_{tire}=169.12kPa$

Therefore,

${\rho }_0=\frac{p_0}{R{T}_0}=\frac{169120Pa}{\left(287m^2/s^2.K\right)\left(303K\right)}=1.945kg/m^3$

Calculate the exit velocity

According to incompressible Bernoulli’s equation,

$V_e=\sqrt{\frac{2\Delta p}{{\rho }_0}}$

Substitute for known values,

$V_e=\sqrt{\frac{2\Delta p}{{\rho }_0}}=\sqrt{\frac{2\left(169120Pa-100000Pa\right)}{1.945kg/m^3}}=266.59m/s$

The above obtained value of $266.59m/s$ is almost $8\%$ lower than the velocity found in sup-part a $V_e=291.31m/s$

Conclusion:

The exit velocity is equal to $266.59m/s$.