Chapter 5.5, Problem 26P

Air enters a nozzle steadily at 50 psia, 140°F, and 150 ft/s and leaves at 14.7 psia and 900 ft/s. The heat loss from the nozzle is estimated to be 6.5 Btu/lbm of air flowing. The inlet area of the nozzle is 0.1 ft². Determine (a) the exit temperature of air and (b) the exit area of the nozzle.

To determine

The exit temperature.

Answer to Problem 26P

The exit temperature is $507.3624\text{R}$.

Explanation of Solution

Write the energy rate balance equation for one inlet and one outlet system.

$\left[{\dot{Q}}_{\text{1}}+{\dot{W}}_{\text{1}}+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+g{z}_1\right)\right]-\left[{\dot{Q}}_{\text{2}}+{\dot{W}}_{\text{2}}+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+g{z}_2\right)\right]=\Delta {\dot{E}}_{\text{system}}$ (I)

Here, the rate of heat transfer is $\dot{Q}$, the rate of work transfer is $\dot{W}$, the enthalpy is $h$ and the velocity is $V$, the gravitational acceleration is $g$, the elevation from the datum is $z$ and the rate of change in net energy of the system is $\Delta {\dot{E}}_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The air flows at steady state through the nozzle. Hence, the rate of change in net energy of the system becomes zero.

$\Delta {\dot{E}}_{\text{system}}=0$

Here, the heat loss from the nozzle is estimated as $6.5\text{Btu}$ per mass flow of air.

$\frac{{\dot{Q}}_{\text{2}}}{\dot{m}}=6.5\text{Btu/lbm}$

There is no heat transfer at inlet i.e. ${\dot{Q}}_1=0$ and no work done in nozzle i.e. $\dot{W}=0$. The inlet, outlet are at same elevation, the potential energy becomes negligible i.e. $gz=0$.

The Equations (I) reduced as follows to obtain the exit enthalpy.

$\begin{array}{l}\left[0+0+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+0\right)\right]-\left[0+{\dot{Q}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+0\right)\right]=0\\ \dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)-\left[{\dot{Q}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)\right]=0\\ \dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)={\dot{Q}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)\\ -{\dot{Q}}_2=\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)-\dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)\end{array}$

$\begin{array}{l}-{\dot{Q}}_2=\dot{m}\left[h_2-h_1+\frac{V_2{}^2-V_1{}^2}{2}\right]\\ \frac{-{\dot{Q}}_2}{\dot{m}}=h_2-h_1+\frac{V_2{}^2-V_1{}^2}{2}\end{array}$ (II)

Here, the heat transfer per mass flow is expressed as,

$q_2=\frac{{\dot{Q}}_2}{\dot{m}}$

Substitute $q_2$ for $\frac{{\dot{Q}}_2}{\dot{m}}$ in Equation (II) and rearrange the equation to obtain exit enthalpy $\left(h_2\right)$.

$\begin{array}{l}-q_2=h_2-h_1+\frac{V_2{}^2-V_1{}^2}{2}\\ h_2=-q_2+h_1-\frac{V_2{}^2-V_1{}^2}{2}\end{array}$ (III)

Refer Table A-17E, “Ideal-gas properties of air”.

The inlet enthalpy $\left(h_1\right)$ corresponding to the temperature of $140\text{}°\text{F }\left(600\text{R}\right)$ is $143.47\text{Btu/lbm}$.

Conclusion:

Substitute $6.5\text{Btu/lbm}$ for $q_2$, $143.47\text{Btu/lbm}$ for $h_1$, $150\text{ft/s}$ for $V_1$ and $900\text{ft/s}$ for $V_2$ in Equation (III).

$\begin{array}{c}{h}_2=-6.5\text{Btu/lbm}+143.47\text{Btu/lbm}-\frac{{\left(900\text{ft/s}\right)}^2-{\left(150\text{ft/s}\right)}^2}{2}\\ =-6.5\text{Btu/lbm}+143.47\text{Btu/lbm}-\left(393750{\text{ft}}^2{\text{/s}}^2\times \frac{1\text{Btu/lbm}}{25037{\text{ft}}^2{\text{/s}}^2}\right)\\ =-6.5\text{Btu/lbm}+143.47\text{Btu/lbm}-15.7267\text{Btu/lbm}\\ =121.2433\text{Btu/lbm}\end{array}$

Refer Table A-17E, “Ideal-gas properties of air”.

The exit temperature $\left(T_2\right)$ corresponding to the enthalpy of $121.2433\text{Btu/lbm}$- using interpolation method.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (IV)

Show the temperature and enthalpy values from the Table A-17E as in below table.

S.No.	x	y
	Enthalpy $\left(h\right)$, in $\text{Btu/lbm}$	Temperature $\left(T\right)$,in $\text{R}$
1	119.48	500
2	121.2433	?
3	124.27	520

Substitute $119.48$ for $x_1$, $121.2433$ for $x_2$, $124.27$ for $x_3$, $500$ for $y_1$, and $520$ for $y_3$ in Equation (IV).

$\begin{array}{c}{y}_2=\frac{\left(121.2433-119.48\right)\left(520-500\right)}{\left(124.27-199.48\right)}+500\\ =507.3624\text{R}\end{array}$

Thus, the temperature corresponding to exit enthalpy of $121.2433\text{Btu/lbm}$ is $507.3624\text{R}$.

Thus, the exit temperature is $507.3624\text{R}$.

To determine

The exit area of the nozzle.

Answer to Problem 26P

The exit area of the nozzle is $0.04795{\text{ft}}^2$.

Explanation of Solution

The carbon dioxide flows through the nozzle at steady state. Hence, the inlet and exit mass flow rates are equal.

${\dot{m}}_1={\dot{m}}_2$

Write equation of conservation of mass flow rate.

$\frac{A_1{V}_1{P}_1}{R{T}_1}=\frac{A_2{V}_2{P}_2}{R{T}_2}$ (V)

Here, the cross-sectional area is $A$ and the velocity is $V$, the gas constant of air is $R$, the temperature is $T$ and the pressure is $P$; the suffix 1 and 2 indicates the inlet and outlet conditions.

Rearrange the Equation (V) to obtain the exit area $\left(A_2\right)$.

$\begin{array}{l}\frac{A_1{V}_1{P}_1}{T_1}=\frac{A_2{V}_2{P}_2}{T_2}\\ A_2=A_1\left(\frac{V_1}{V_2}\right)\left(\frac{P_1}{P_2}\right)\left(\frac{T_2}{T_1}\right)\end{array}$ (VI)

Conclusion:

Substitute $0.1{\text{ft}}^2$ for $A_1$, $150\text{ft/s}$ for $V_1$,$900\text{ft/s}$ for $V_2$, $50\text{psia}$ for $P_1$, $14.7\text{psia}$ for $P_2$, $507.3624\text{R}$ for $T_2$ and $140°\text{F}$ for $T_1$ in Equation (VI).

$\begin{array}{c}{A}_2=\left(0.1{\text{ft}}^2\right)\left(\frac{150\text{ft/s}}{900\text{ft/s}}\right)\left(\frac{50\text{psia}}{14.7\text{psia}}\right)\left(\frac{507.3624\text{R}}{140°\text{F}}\right)\\ =\left(0.1{\text{ft}}^2\right)\left(\frac{150\text{ft/s}}{900\text{ft/s}}\right)\left(\frac{50\text{psia}}{14.7\text{psia}}\right)\left(\frac{507.3624\text{R}}{\left(140+460\right)\text{R}}\right)\\ =0.1{\text{ft}}^2\left(0.1667\right)\left(3.4014\right)\left(0.8456\right)\\ =0.04795{\text{ft}}^2\end{array}$

Thus, the exit area of the nozzle is $0.04795{\text{ft}}^2$.