Chapter 9, Problem 9.7P

To determine

(a)

The flow.

Answer to Problem 9.7P

The flow is isentropic, and the stagnation properties are:

$T_o=605K$,

$p_o=319kPa$, and

${\rho }_o=1.805kg/m^3$

All the above values are constant in the flow from section 1 to section $2$.

Explanation of Solution

Given Information:

$\begin{array}{l}{A}_1=20c{m}^2\\ p_1=300kPa\\ {\rho }_1=1.75kg/m^3\\ V_1=122.5m/s\\ {\rho }_2=0.266kg/m^3\\ T_2=281K\end{array}$

Calculation:

The flow is isentropic, and the stagnation properties are:

$T_o=605K$,

$p_o=319kPa$, and

${\rho }_o=1.805kg/m^3$

All the above values are constant in the flow from section 1 to section $2$.

To determine

(b)

The mass flow at section 2.

Answer to Problem 9.7P

The required value of the mass flow is $0.0429kg/s$.

Explanation of Solution

Given information:

$\begin{array}{l}{A}_1=20c{m}^2\\ p_1=300kPa\\ {\rho }_1=1.75kg/m^3\\ V_1=122.5m/s\\ {\rho }_2=0.266kg/m^3\\ T_2=281K\end{array}$

Formula used:

$m_1=m_2={\rho }_1{A}_1{V}_1$

Calculation:

We know that the mass flow should be constant:

$\begin{array}{l}{m}_1=m_2={\rho }_1{A}_1{V}_1\\ \Rightarrow \left(1.75kg/m^3\right)\left(0.0020m^2\right)\left(122.5m/s\right)\\ \Rightarrow 0.0429kg/s\end{array}$.

To determine

(c)

The value of V₂.

Answer to Problem 9.7P

The required value of the V₂is $806m/s$

Explanation of Solution

Given Information:

$\begin{array}{l}{A}_1=20c{m}^2\\ p_1=300kPa\\ {\rho }_1=1.75kg/m^3\\ V_1=122.5m/s\\ {\rho }_2=0.266kg/m^3\\ T_2=281K\end{array}$

Formula used:

$V_2=\frac{m}{{\rho }_2{A}_2}$

Calculation:

We know that; $V_2=\frac{m}{{\rho }_2{A}_2}$ ;

Put the values in the above equation;

$\begin{array}{l}{V}_2=\frac{m}{{\rho }_2{A}_2}=\frac{0.0429kg/s}{\left(0.266kg/m^3\right)\left(0.002m^2\right)}\\ \Rightarrow 806m/s\end{array}$

We know that, $a_2={\left(kR{T}_2\right)}^{1/2}$ ;

Put the values in the above equation;

$\begin{array}{l}{a}_2={\left(kR{T}_2\right)}^{1/2}={\left[1.4\left(287\right)\left(281\right)\right]}^{1/2}\\ \Rightarrow 366m/s\end{array}$

Therefore, the Mach number at section $2$ is;

$\begin{array}{l}M{a}_2=V_2/a_2\\ \Rightarrow 806/336\\ \Rightarrow 2.40\end{array}$

So, the flow at section $2$ is supersonic.

To determine

(d)

The value of p₂.

Answer to Problem 9.7P

The required value of the p₂ is $21450Pa$.

Explanation of Solution

Given Information:

$\begin{array}{l}{A}_1=20c{m}^2\\ p_1=300kPa\\ {\rho }_1=1.75kg/m^3\\ V_1=122.5m/s\\ {\rho }_2=0.266kg/m^3\\ T_2=281K\end{array}$

Formula used:

$p_2={\rho }_{2}R{T}_2$

Calculation:

We know that, $p_2={\rho }_{2}R{T}_2$ ;

Put the values in the above equation;

$\begin{array}{l}{p}_2={\rho }_{2}R{T}_2=\left(0.266kg/m^3\right)\left(287m^2/s^2-K\right)\left(281K\right)\\ \Rightarrow 21450Pa\end{array}$

Again,

$\begin{array}{l}{T}_1=\frac{p_1}{R{\rho }_1}\\ \Rightarrow \frac{\left(300000Pa\right)}{\left(287m^2/s_2-K\right)\left(1.75kg/m^3\right)}\\ \Rightarrow 597K\end{array}$.

To determine

(e)

The difference of s₂-s₁.

Answer to Problem 9.7P

The required value is $0J/kg-K$

Explanation of Solution

Given Information:

$\begin{array}{l}{A}_1=20c{m}^2\\ p_1=300kPa\\ {\rho }_1=1.75kg/m^3\\ V_1=122.5m/s\\ {\rho }_2=0.266kg/m^3\\ T_2=281K\end{array}$

Formula used:

$s_2-s_1=c_p\ln \left(\frac{T_2}{T_1}\right)-R\ln \left(\frac{p_2}{p_1}\right)$

Calculation:

We know the equation, $\left(9.8\right)$ ;

$s_2-s_1=c_p\ln \left(\frac{T_2}{T_1}\right)-R\ln \left(\frac{p_2}{p_1}\right)$

Put the values in the above equation;

$\begin{array}{l}{s}_2-s_1=c_p\ln \left(\frac{T_2}{T_1}\right)-R\ln \left(\frac{p_2}{p_1}\right)=1005\ln \left(\frac{281}{597}\right)-287\ln \left(\frac{21450}{300000}\right)\\ \Rightarrow -757+757\\ \Rightarrow 0J/kg-K\end{array}$.