Chapter 9, Problem 45P

To determine

Find the average density of the gas exiting the engine.

Answer to Problem 45P

The average density of the gas exiting the engine in $\frac{\text{kg}}{{\text{m}}^{\text{3}}}$, $\frac{\text{slugs}}{{\text{m}}^{\text{3}}}$, and $\frac{\text{lbm}}{{\text{ft}}^{\text{3}}}$ are $0.155\frac{\text{kg}}{{\text{m}}^{\text{3}}}$, $3\times {10}^{-4}\frac{\text{slugs}}{{\text{m}}^{\text{3}}}$ and $0.0097\frac{\text{lbm}}{{\text{ft}}^{\text{3}}}$ respectively.

Explanation of Solution

Given data:

Consider that the inlet area of the engine as $5{\text{ft}}^2$.

$A_{\text{in}}=5{\text{ft}}^2$ (1)

Consider that the outlet area of the engine as $4{\text{ft}}^2$.

$A_{\text{out}}=4{\text{ft}}^2$ (2)

Formula used:

Consider the following expression.

${\left(\text{air}\text{mass flow rate}\right)}_{\text{in}}+{\left(\text{fuel}\text{mass flow rate}\right)}_{\text{in}}={\left(\text{combustion mass flow rate}\right)}_{\text{out}}$

In this problem, the fuel mass flow rate is neglected since it is negligible value when compared air mass flow rate.

Consider that the mass flow rate of air enters is equal to the mass flow rate of exists.

${\left(\text{mass flow rate}\right)}_{\text{in}}={\left(\text{mass flow rate}\right)}_{\text{out}}$ (3)

Write the expression for the mass flow rate.

$\text{mass flow rate}=\left(\text{density}\right)\left(\text{volume flow rate}\right)$ (4)

Calculation:

Convert the unit of the value in equation (1) from ${\text{ft}}^2$ to ${\text{m}}^2$.

$\begin{array}{c}{A}_{\text{in}}=5\times \left(\frac{1}{{\left(\text{3}\text{.28084}\right)}^2}\right){\text{m}}^2\left[\therefore \text{1}\text{m}=\text{3}\text{.28084}\text{ft}\right]\\ =0.464{\text{m}}^2\end{array}$

Convert the unit of the value in equation (2) from ${\text{ft}}^2$ to ${\text{m}}^2$.

$\begin{array}{c}{A}_{\text{out}}=4\times \left(\frac{1}{{\left(\text{3}\text{.28084}\right)}^2}\right){\text{m}}^2\left[\therefore \text{1}\text{m}=\text{3}\text{.28084}\text{ft}\right]\\ =0.372{\text{m}}^2\end{array}$

Substitute equation (4) in (3),

$\begin{array}{l}{\left[\left(\text{density}\right)\left(\text{volume flow rate}\right)\right]}_{\text{in}}={\left[\left(\text{density}\right)\left(\text{volume flow rate}\right)\right]}_{\text{out}}\\ {\rho }_{\text{in}}{\left(\text{volume flow rate}\right)}_{\text{in}}={\rho }_{\text{out}}{\left(\text{volume flow rate}\right)}_{\text{out}}\\ \left({\rho }_{\text{in}}\right)\left(V_{\text{in}}\right)\left(A_{\text{in}}\right)=\left({\rho }_{\text{out}}\right)\left(V_{\text{out}}\right)\left(A_{\text{out}}\right)\left[\therefore \text{volume flow rate}=\left(\text{velocity}\right)\left(\text{area}\right)\right]\end{array}$

Substitute $0.45\frac{\text{kg}}{{\text{m}}^{\text{3}}}$ for ${\rho }_{\text{in}}$, and $180\frac{\text{m}}{\text{s}}$ for $V_{\text{in}}$, $650\frac{\text{m}}{\text{s}}$ for $V_{\text{out}}$, $0.464{\text{m}}^2$ for $A_{\text{in}}$, and $0.372{\text{m}}^2$ for $A_{\text{out}}$,

$\left(0.45\frac{\text{kg}}{{\text{m}}^{\text{3}}}\right)\left(180\frac{\text{m}}{\text{s}}\right)\left(0.464{\text{m}}^2\right)=\left({\rho }_{\text{out}}\right)\left(650\frac{\text{m}}{\text{s}}\right)\left(0.372{\text{m}}^2\right)$

${\rho }_{\text{out}}=0.155\frac{\text{kg}}{{\text{m}}^{\text{3}}}$ (5)

Convert the unit of the value in equation (5) from $\frac{\text{kg}}{{\text{m}}^{\text{3}}}$ to $\frac{\text{slugs}}{{\text{m}}^{\text{3}}}$.

$\begin{array}{c}{\rho }_{\text{out}}=0.155\left(\frac{1}{\text{14}\text{.5939}}\right)\frac{\text{slugs}}{{\text{m}}^{\text{3}}}\left[\therefore 1\text{slug}=\text{14}\text{.5939}\text{kg}\right]\\ =3\times {10}^{-4}\frac{\text{slugs}}{{\text{m}}^{\text{3}}}\end{array}$

Convert the unit of the value in equation (5) from $\frac{\text{kg}}{{\text{m}}^{\text{3}}}$ to $\frac{\text{lbm}}{{\text{ft}}^{\text{3}}}$.

$\begin{array}{c}{\rho }_{\text{out}}=0.155\frac{\text{kg}}{{\text{m}}^{\text{3}}}\left(\frac{2.20462\text{lbm}}{1\text{kg}}\right)\frac{1{\text{m}}^{\text{3}}}{{\left(\text{3}\text{.28084}\right)}^3{\text{ft}}^3}\left[\begin{array}{l}\therefore 1\text{kg}=2.20462\text{lbm,}\text{and}\\ \text{1}\text{m}=\text{3}\text{.28084}\text{ft}\end{array}\right]\\ =0.0097\frac{\text{lbm}}{{\text{ft}}^{\text{3}}}\end{array}$

Conclusion:

Thus, the average density of the gas exiting the engine in $\frac{\text{kg}}{{\text{m}}^{\text{3}}}$, $\frac{\text{slugs}}{{\text{m}}^{\text{3}}}$, and $\frac{\text{lbm}}{{\text{ft}}^{\text{3}}}$ are $0.155\frac{\text{kg}}{{\text{m}}^{\text{3}}}$, $3\times {10}^{-4}\frac{\text{slugs}}{{\text{m}}^{\text{3}}}$ and $0.0097\frac{\text{lbm}}{{\text{ft}}^{\text{3}}}$ respectively.