Chapter 6, Problem 75P

(a)

To determine

The temperature of the air at the exit.

Explanation of Solution

Given:

The inlet pressure of Refrigerant $\left(P_3\right)$ is 20 psia.

The inlet quality of Refrigerant $\left(x_3\right)$ is $30\%$.

The outlet pressure of Refrigerant $\left(P_3\right)$ is 20 psia.

The outlet quality of steam $\left(x_4\right)$ is 1 (saturated vapor).

The inlet pressure of air $\left(P_1\right)$ is 14.7 psia.

The inlet temperature of air $\left(T_1\right)$ is $90°\text{F}$.

The volume flow rate of air $\left({\dot{V}}_1\right)$ is $200{\text{ ft}}^3/\min$.

The mass flow rate of refrigerant $\left({\dot{m}}_R\right)$ is $4\text{ lbm/min}$.

Calculation:

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 $d{\dot{E}}_{\text{system}}/dt$.

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 $h_f=\text{11}\text{.436}\text{Btu}/\text{lbm}$, $x_3=0.3$, and $h_{fg}=91.302\text{Btu}/\text{lbm}$ 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 $R=\text{0}\text{.3704}\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$, $T_1=\text{90}°\text{F}$, and $P_1=\text{14}\text{.7}\text{psia}$ 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 ${\dot{V}}_1=\text{200}{\text{ft}}^{\text{3}}/\text{min}$ and $v_1=13.86{\text{ft}}^{\text{3}}/\text{lbm}$ 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 ${\dot{m}}_a={\dot{m}}_1$, ${\dot{m}}_2={\dot{m}}_a$, ${\dot{m}}_3={\dot{m}}_R$, and ${\dot{m}}_R$ for ${\dot{m}}_4$ in Equation (V).

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

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

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

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

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.

Explanation of Solution

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

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

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

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

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