Chapter 5.5, Problem 82P

(a)

To determine

The temperature of the air at the exit.

Answer to Problem 82P

The temperature of the air at the exit is $16.2°\text{F}$.

Explanation of Solution

Draw the schematic diagram for the evaporator section.

Write the formula to calculate the specific enthalpy of steam $\left(h\right)$ from tables.

$h=h_f+x\left(h_{fg}\right)$ (I)

Here, specific enthalpy of saturated liquid is $h_f$, specific enthalpy of saturated liquid-vapor mixture is $h_{fg}$ and dryness fraction of water is $x$.

Write the ideal gas equation for specific volume of air $\left(v\right)$.

$v=\frac{RT}{P}$ (II)

Here, gas constant for air is $R$, temperature of air is $T$, and the pressure of air is $P$.

Calculate the volume flow rate $\left({\dot{V}}_1\right)$ of the air at the inlet.

${\dot{V}}_1={\dot{m}}_a{v}_1$ (III)

Here, mass flow rate of air is ${\dot{m}}_a$ and specific volume of air is $v_1$.

Write the expression for the mass balance.

$m_{in}-m_{out}=\Delta m$

$m_{in}-m_{out}=0\left(\text{For steady state}\right)$ (IV)

Here, mass of the fluid entering into the system is $m_{in}$, mass of fluid leaving out of the system is $m_{out}$ and change in mass is $\Delta m$.

Write the energy rate balance equation for a control volume.

${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=d{\dot{E}}_{\text{system}}/dt$

${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\left(\text{For steady state}\right)$ (V)

Here, total energy rate at inlet is ${\dot{E}}_{\text{in}}$, total energy rate at outlet is ${\dot{E}}_{\text{out}}$ and change in net energy rate is $d{\dot{E}}_{\text{system}}/dt$.

Conclusion:

Refer Table A-1E, “Gas constant of common gases”, obtain the gas constant of air as $0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$.

Refer Table A-2E, “Ideal – gas specific heats of common gases”, obtain the constant pressure specific heat of air as $0.240\text{Btu}/\text{lbm}\text{.}°\text{F}$.

Refer Table A-12E, “Saturated R-134a-Pressure table”, obtain the properties of saturated R-134a at pressure $\left(P_3\right)$ of $20\text{ psia}$.

$\begin{array}{l}{h}_f=\text{11}\text{.436}\text{Btu}/\text{lbm}\\ h_{fg}=\text{91}\text{.302}\text{Btu}/\text{lbm}\end{array}$

Substitute $\text{11}\text{.436}\text{Btu}/\text{lbm}$ for $h_f$, 0.3 for $x_3$ and $91.302\text{Btu}/\text{lbm}$ for $h_{fg}$ in Equation (I).

$\begin{array}{c}{h}_3=\text{11}\text{.436}\text{Btu}/\text{lbm}+\left(0.3\right)91.302\text{Btu}/\text{lbm}\\ =38.83\text{Btu}/\text{lbm}\end{array}$

Refer Table A-12E, “Saturated R-134a-Pressure table”, obtain the specific enthalpy $\left(h_4\right)$ of saturated vapor of R-134aat pressure $\left(P_4\right)$ of $20\text{ psia}$ as $102.74\text{Btu}/\text{lbm}$.

Substitute $\text{0}\text{.3704}\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$ for $R$, $\text{90}°\text{F}$ for $T_1$, and $\text{14}\text{.7}\text{psia}$ for $P_1$ in Equation (II).

$\begin{array}{c}{v}_1=\frac{R{T}_1}{P_1}\\ =\frac{\text{0}\text{.3704}\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\left(\text{90}°\text{F}\right)}{\text{14}\text{.7}\text{psia}}\\ =\frac{\text{0}\text{.3704}\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\left(\text{90}+460\right)\text{R}}{\text{14}\text{.7}\text{psia}}\end{array}$

$v_1=13.86{\text{ft}}^{\text{3}}/\text{lbm}$

Substitute $\text{200}{\text{ft}}^{\text{3}}/\text{min}$ for ${\dot{V}}_1$, and $13.86{\text{ft}}^{\text{3}}/\text{lbm}$ for $v_1$ in Equation (III).

$\begin{array}{c}{\dot{m}}_a=\frac{\text{200}{\text{ft}}^{\text{3}}/\text{min}}{13.86{\text{ft}}^{\text{3}}/\text{lbm}}\\ =14.43\text{lbm}/\text{min}\end{array}$

Rewrite Equation (I) for the mass flow rate of air $\left({\dot{m}}_a\right)$.

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

Here, mass flow rate of air at inlet is ${\dot{m}}_1$ and mass flow rate of air at outlet is ${\dot{m}}_2$.

Rewrite the Equation (I) for the mass flow rate of R-134a $\left({\dot{m}}_R\right)$.

${\dot{m}}_3={\dot{m}}_4={\dot{m}}_R$

Here, mass flow rate of R-134a at inlet is ${\dot{m}}_3$, and mass flow rate of R-134a at outlet is ${\dot{m}}_4$.

Rewrite Equation (II) for the energy balance in the steam heating system.

${\dot{m}}_1{h}_1+{\dot{m}}_3{h}_3={\dot{m}}_2{h}_2+{\dot{m}}_4{h}_4$ (V)

Substitute ${\dot{m}}_a$ for ${\dot{m}}_1$, ${\dot{m}}_a$ for ${\dot{m}}_2$, ${\dot{m}}_R$ for ${\dot{m}}_3$, and ${\dot{m}}_R$ for ${\dot{m}}_4$ in Equation (V).

$\begin{array}{l}{\dot{m}}_a{h}_1+{\dot{m}}_R{h}_3={\dot{m}}_a{h}_2+{\dot{m}}_R{h}_4\\ {\dot{m}}_R\left(h_3-h_4\right)={\dot{m}}_a\left(h_2-h_1\right)\\ {\dot{m}}_R\left(h_3-h_4\right)={\dot{m}}_a{c}_p\left(T_2-T_1\right)\end{array}$

$T_2=T_1+\frac{{\dot{m}}_R\left(h_3-h_4\right)}{{\dot{m}}_a{c}_p}$ (VI)

Here, specific enthalpy of air at the inlet is $h_1$, specific enthalpy of air at the outlet is $h_2$, constant pressure specific heat of air is $c_p$, temperature of air at inlet is $T_1$ and temperature of air at the outlet is $T_2$.

Substitute $\text{38}\text{.83 }\text{Btu}/\text{lbm}$ for $h_3$, $\text{102}\text{.74 }\text{Btu}/\text{lbm}$ for $h_4$, $0.240\text{Btu}/\text{lbm}\text{.}°\text{F}$ for $c_p$, $4\text{lbm}/\text{min}$ for ${\dot{m}}_R$, $14.43\text{lbm}/\text{min}$ for ${\dot{m}}_a$, and $90°\text{F}$ for $T_1$ in Equation (VI).

$\begin{array}{c}{T}_2=90°\text{F}+\frac{4\text{lbm}/\text{min}\times \left(\text{38}\text{.83 }\text{Btu}/\text{lbm}-\text{102}\text{.74 }\text{Btu}/\text{lbm}\right)}{14.43\text{lbm}/\text{min}\times 0.240\text{Btu}/\text{lbm}\text{.}°\text{F}}\\ =16.2°\text{F}\end{array}$

Thus, the temperature of the air at the exit is $16.2°\text{F}$.

(b)

To determine

The rate of heat transfer from the air to the refrigerant.

Answer to Problem 82P

The rate of heat transfer from the air to the refrigerant is $255.6\text{Btu}/\text{min}$.

Explanation of Solution

Write the formula to calculate the rate of heat transfer from the air to the refrigerant $\left(Q_{\text{air,out}}\right)$.

$Q_{\text{air,out}}=-{\dot{m}}_a{c}_p\left(T_2-T_1\right)$ (VII)

Conclusion:

Substitute $14.43\text{lbm}/\text{min}$ for ${\dot{m}}_a$, $\text{0}\text{.240}\text{Btu}/\text{lbm}\cdot \text{°F}$ for $c_p$, $16.24\text{°F}$ for $T_2$ and $\text{90°F}$ for $T_1$ in Equation (VII).

$\begin{array}{c}{Q}_{\text{air,out}}=-\left(14.43\text{lbm}/\text{min}\right)\left(\text{0}\text{.240}\text{Btu}/\text{lbm}\cdot \text{°F}\right)\left(16.24\text{°F}-\text{90°F}\right)\\ =255.6\text{Btu}/\text{min}\end{array}$

Thus, the rate of heat transfer from the air to the refrigerant is $255.6\text{Btu}/\text{min}$.