Chapter 7.13, Problem 180RP

Air enters the evaporator section of a window air conditioner at 100 kPa and 27°C with a volume flow rate of 6 m³/min. The refrigerant-134a at 120 kPa with a quality of 0.3 enters the evaporator at a rate of 2 kg/min and leaves as saturated vapor at the same pressure. Determine the exit temperature of the air and the rate of entropy generation for this process, assuming (a) the outer surfaces of the air conditioner are insulated and (b) heat is transferred to the evaporator of the air conditioner from the surrounding medium at 32°C at a rate of 30 kJ/min.

FIGURE P7–180

To determine

The exit temperature of air and the entropy generated during the process.

Answer to Problem 180RP

The exit temperature of air is $-15.9°\text{C}$.

The entropy generated during the process is $0.00196\text{kW}/\text{K}$.

Explanation of Solution

Write the expression to calculate the initial enthalpy of the refrigerant.

$h_1=h_f+x_1{h}_{fg}$ (I)

Here, initial enthalpy is $h_1$, saturated liquid enthalpy is $h_f$, initial vapor quality is $x_1$ and evaporated enthalpy is $h_{fg}$.

Write the expression to calculate the initial entropy of the refrigerant.

$s_1=s_f+x_1{s}_{fg}$ (II)

Here, initial entropy is $s_1$, saturated liquid entropy is $s_f$, initial vapor quality is $x_1$ and evaporated entropy is $s_{fg}$.

Write the expression to calculate the mass flow rate of air.

${\dot{m}}_{air}=\frac{P_3{\dot{V}}_3}{R{T}_3}$ (III)

Here, mass flow rate of air is ${\dot{m}}_{air}$, inlet pressure of air is $P_3$, volumetric flow rate is ${\dot{V}}_3$, gas constant is R and inlet temperature of air is $T_3$.

Write the expression for the mass balance of the system.

${\dot{m}}_{in}-{\dot{m}}_{out}=\Delta {\dot{m}}_{system}$ (IV)

Here, mass flow rate into the control system is ${\dot{m}}_{in}$, mass flow rate exit from the control volume is ${\dot{m}}_{out}$ and mass flow rate change in the system is $\Delta {\dot{m}}_{system}$.

Write the expression for the energy balance equation for closed system.

${\dot{E}}_{in}-{\dot{E}}_{out}=\Delta {\dot{E}}_{system}$ (V)

Here, rate of energy transfer into the control volume is ${\dot{E}}_{in}$, rate of energy transfer exit from the control volume is ${\dot{E}}_{out}$ and rate of change in internal energy of system is $\Delta {\dot{E}}_{system}$

Write the expression for the rate of entropy balance for the system.

${\dot{S}}_{in}-{\dot{S}}_{out}+{\dot{S}}_{gen}=\Delta {\dot{S}}_{system}$ (VI)

Here, rate of entropy in the system is ${\dot{S}}_{in}$, rate of entropy exit from the system is ${\dot{S}}_{out}$, rate of entropy generated is ${\dot{S}}_{gen}$ and rate of change of entropy in the system is $\Delta {\dot{S}}_{system}$.

Conclusion:

From Table A-12, “Saturated refrigerant-134a-Pressure table”, Obtain the following properties at saturated pressure of 120 kPa

Saturated liquid enthalpy, $h_f=22.47\text{kJ}/\text{kg}$

Evaporated enthalpy, $h_{fg}=214.52\text{kJ}/\text{kg}$

Saturated vapor enthalpy, $h_2=h_{g@120\text{kPa}}=236.99\text{kJ}/\text{kg}$

Saturated vapor entropy, $s_2=s_{g@120\text{kPa}}=0.9479\text{kJ}/\text{kg}\cdot \text{K}$

Saturated liquid entropy, $s_f=0.09269\text{kJ}/\text{kg}\cdot \text{K}$

Evaporated entropy, $s_{fg}=0.85520\text{kJ}/\text{kg}\cdot \text{K}$

Substitute $22.47\text{kJ}/\text{kg}$ for $h_f$, 0.3 for $x_1$ and $214.52\text{kJ}/\text{kg}$ for $h_{fg}$ in Equation (I).

$\begin{array}{c}{h}_1=22.47\text{kJ}/\text{kg}+\left(0.3\right)\left(214.52\text{kJ}/\text{kg}\right)\\ =86.83\text{kJ}/\text{kg}\end{array}$

Substitute $0.09269\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, 0.3 for $x_1$ and $0.85520\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$ in Equation (II).

$\begin{array}{c}{s}_1=0.09269\text{kJ}/\text{kg}\cdot \text{K}+\left(0.3\right)\left(0.85520\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =0.3492\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Refrigerant –134a enters and leaves at the same pressure. Hence, $P_2=P_1$

From Table A-1E, “the molar mass, gas constant and critical–point properties table”, select the gas constant of air at room temperature as $0.287\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute $\text{100 kPa}$ for $P_3$, $6{\text{m}}^3/\text{min}$ for ${\dot{V}}_3$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for R and 27 C for $T_3$ in Equation (III).

$\begin{array}{c}{\dot{m}}_{air}==\frac{\left(100\text{kPa}\right)\left(6{\text{m}}^3/\text{min}\right)}{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)27°\text{C}}\\ =\frac{\left(100\text{kPa}\right)\left(6{\text{m}}^3/\text{min}\right)}{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\frac{\text{1}\text{kPa}\cdot {\text{m}}^3}{1\text{kJ}}\right)\left(27+273\right)\text{K}}\\ =6.968\text{kg}/\text{min}\end{array}$

Substitute ${\dot{m}}_1$ for ${\dot{m}}_{in}$, ${\dot{m}}_2$ for ${\dot{m}}_{out}$ and 0 for $\Delta {\dot{m}}_{system}$ in Equation (I).

$\begin{array}{l}{\dot{m}}_1-{\dot{m}}_2=0\\ {\dot{m}}_1={\dot{m}}_2={\dot{m}}_R\end{array}$

Here, mass flow rate of refrigerant is ${\dot{m}}_R$.

Substitute ${\dot{m}}_3$ for ${\dot{m}}_{in}$, ${\dot{m}}_4$ for ${\dot{m}}_{out}$ and 0 for $\Delta {\dot{m}}_{system}$ in Equation (IV).

$\begin{array}{l}{\dot{m}}_3-{\dot{m}}_4=0\\ {\dot{m}}_3={\dot{m}}_4={\dot{m}}_{air}\end{array}$

Here, mass flow rate of air is ${\dot{m}}_{air}$.

Substitute ${\dot{m}}_1{h}_1+{\dot{m}}_3{h}_3$ for ${\dot{E}}_{in}$, ${\dot{m}}_2{h}_2+{\dot{m}}_4{h}_4$ for ${\dot{E}}_{out}$ and 0 for $\Delta {\dot{E}}_{system}$ in Equation (V).

$\begin{array}{l}{\dot{m}}_1{h}_1+{\dot{m}}_3{h}_3-\left({\dot{m}}_2{h}_2+{\dot{m}}_4{h}_4\right)=0\\ {\dot{m}}_1{h}_1+{\dot{m}}_3{h}_3={\dot{m}}_2{h}_2+{\dot{m}}_4{h}_4\end{array}$

$\begin{array}{l}{\dot{m}}_R\left(h_2-h_1\right)={\dot{m}}_{air}\left(h_3-h_4\right)\\ {\dot{m}}_R\left(h_2-h_1\right)={\dot{m}}_{air}{c}_p\left(T_3-T_4\right)\\ T_4=T_3-\frac{{\dot{m}}_R\left(h_2-h_1\right)}{{\dot{m}}_{air}{c}_p}\end{array}$ (VII)

From the Table A-2, “Ideal-gas specific heats of various common gases”, select the value of the specific heat at constant pressure value of air as $1.005\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute 27 C for $T_3$, $2\text{kg}/\text{min}$ for ${\dot{m}}_R$, $236.99\text{kJ}/\text{kg}$ for $h_2$, $86.83\text{kJ}/\text{kg}$ for $h_1$, $6.968\text{kg}/\text{min}$ for ${\dot{m}}_{air}$ and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in Equation (VII).

$\begin{array}{c}{T}_4=27°\text{C}-\frac{\left(2\text{kg}/\text{min}\right)\left(236.99\text{kJ}/\text{kg}-86.83\text{kJ}/\text{kg}\right)}{\left(6.968\text{kg}/\text{min}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =-15.9°\text{C}\end{array}$

Hence, the exit temperature of air is $-15.9°\text{C}$.

For the steady flow system, change of entropy in the system is zero.

Substitute ${\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3$ for ${\dot{S}}_{in}$, ${\dot{m}}_2{s}_2+{\dot{m}}_4{s}_4$ for ${\dot{S}}_{out}$, 0 for $\Delta {\dot{S}}_{system}$ in Equation (VI).

$\begin{array}{l}{\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3-\left({\dot{m}}_2{s}_2+{\dot{m}}_4{s}_4\right)+{\dot{S}}_{gen}=0\\ {\dot{S}}_{gen}={\dot{m}}_R\left(s_2-s_1\right)+{\dot{m}}_{air}\left(s_4-s_3\right)\\ ={\dot{m}}_R\left(s_2-s_1\right)+{\dot{m}}_{air}\left(c_p\ln \left(\frac{T_4}{T_3}\right)-R\ln \left(\frac{P_4}{P_3}\right)\right)\end{array}$ (VIII)

Substitute $2\text{kg}/\text{min}$ for ${\dot{m}}_R$, $0.9479\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$, $0.3492\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$, $6.968\text{kg}/\text{min}$ for ${\dot{m}}_{air}$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, –15.9 C for $T_4$, 27 C for $T_3$, and $P_4$ for $P_3$ in Equation (VIII).

${\dot{S}}_{gen}=\left\{\begin{array}{l}2\text{kg}/\text{min}\left(0.9479\text{kJ}/\text{kg}\cdot \text{K}-0.3492\text{kJ}/\text{kg}\cdot \text{K}\right)\\ +6.968\text{kg}/\text{min}\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\ln \left(\frac{-15.9°\text{C}}{27°\text{C}}\right)-R\ln \left(\frac{P_4}{P_4}\right)\right)\end{array}\right\}$

$\begin{array}{c}{\dot{S}}_{gen}=\left\{\begin{array}{l}2\text{kg}/\text{min}\left(0.9479\text{kJ}/\text{kg}\cdot \text{K}-0.3492\text{kJ}/\text{kg}\cdot \text{K}\right)\\ +6.968\text{kg}/\text{min}\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\ln \left(\frac{\left(-15.9+273\right)\text{K}}{\left(27+273\right)\text{K}}\right)-R\ln \left(\frac{P_4}{P_4}\right)\right)\end{array}\right\}\\ =1.1974-\left(6.968\right)0.1551\\ =0.1175\text{kJ}/\text{min}\cdot \text{K}\left(\frac{1\min }{60\sec }\right)\\ =0.00196\text{kW}/\text{K}\end{array}$

Hence, the entropy generated during the process is $0.00196\text{kW}/\text{K}$.

To determine

The exit temperature of air and the entropy generated during the process.

Answer to Problem 180RP

The exit temperature of air is $-11.6°\text{C}$.

The entropy generated during the process is $0.00225\text{kW}/\text{K}$.

Explanation of Solution

Write the expression for the energy balance equation for closed system.

${\dot{E}}_{in}-{\dot{E}}_{out}=\Delta {\dot{E}}_{system}$ (IX)

Here, rate of energy transfer into the control volume is ${\dot{E}}_{in}$, rate of energy transfer exit from the control volume is ${\dot{E}}_{out}$ and rate of change in internal energy of system is $\Delta {\dot{E}}_{system}$

Write the expression for the rate of entropy balance for the system.

${\dot{S}}_{in}-{\dot{S}}_{out}+{\dot{S}}_{gen}=\Delta {\dot{S}}_{system}$ (X)

Here, rate of entropy in the system is ${\dot{S}}_{in}$, rate of entropy exit from the system is ${\dot{S}}_{out}$, rate of entropy generated is ${\dot{S}}_{gen}$ and rate of change of entropy in the system is $\Delta {\dot{S}}_{system}$.

Conclusion:

Substitute ${\dot{m}}_1{h}_1+{\dot{m}}_3{h}_3+{\dot{Q}}_{in}$ for ${\dot{E}}_{in}$, ${\dot{m}}_2{h}_2+{\dot{m}}_4{h}_4$ for ${\dot{E}}_{out}$ and 0 for $\Delta {\dot{E}}_{system}$ in Equation (IX).

$\begin{array}{l}{\dot{m}}_1{h}_1+{\dot{m}}_3{h}_3+{\dot{Q}}_{in}-\left({\dot{m}}_2{h}_2+{\dot{m}}_4{h}_4\right)=0\\ {\dot{m}}_1{h}_1+{\dot{m}}_3{h}_3+{\dot{Q}}_{in}={\dot{m}}_2{h}_2+{\dot{m}}_4{h}_4\\ {\dot{Q}}_{in}={\dot{m}}_R\left(h_2-h_1\right)+{\dot{m}}_{air}{c}_p\left(T_4-T_3\right)\\ T_4=T_3+\frac{{\dot{Q}}_{in}-{\dot{m}}_R\left(h_2-h_1\right)}{{\dot{m}}_{air}{c}_p}\end{array}$ (XI)

Here, rate of heat gain from the surrounding is ${\dot{Q}}_{in}$

Substitute 27 C for $T_3$, $30\text{kJ}/\text{min}$ for ${\dot{Q}}_{in}$, $2\text{kg}/\text{min}$ for ${\dot{m}}_R$, $236.99\text{kJ}/\text{kg}$ for $h_2$, $86.83\text{kJ}/\text{kg}$ for $h_1$, $6.968\text{kg}/\text{min}$ for ${\dot{m}}_{air}$ and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in Equation (XI).

$\begin{array}{c}{T}_4=27°\text{C}+\frac{30\text{kJ}/\text{min}-\left(2\text{kg}/\text{min}\right)\left(236.99\text{kJ}/\text{kg}-86.83\text{kJ}/\text{kg}\right)}{\left(6.968\text{kg}/\text{min}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =-11.6°\text{C}\end{array}$

Hence, the exit temperature of air is $-11.6°\text{C}$.

For the steady flow system, change of entropy in the system is zero.

Substitute ${\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3+\frac{{\dot{Q}}_{in}}{T_{surr}}$ for ${\dot{S}}_{in}$, ${\dot{m}}_2{s}_2+{\dot{m}}_4{s}_4$ for ${\dot{S}}_{out}$, 0 for $\Delta {\dot{S}}_{system}$ in Equation (I).

$\begin{array}{l}{\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3+\frac{{\dot{Q}}_{in}}{T_{surr}}-\left({\dot{m}}_2{s}_2+{\dot{m}}_4{s}_4\right)+{\dot{S}}_{gen}=0\\ {\dot{S}}_{gen}={\dot{m}}_R\left(s_2-s_1\right)+{\dot{m}}_{air}\left(s_4-s_3\right)-\frac{{\dot{Q}}_{in}}{T_{surr}}\\ {\dot{S}}_{gen}={\dot{m}}_R\left(s_2-s_1\right)+{\dot{m}}_{air}\left(c_p\ln \left(\frac{T_4}{T_3}\right)-R\ln \left(\frac{P_4}{P_3}\right)\right)-\frac{{\dot{Q}}_{in}}{T_{surr}}\end{array}$ (XII)

Substitute $2\text{kg}/\text{min}$ for ${\dot{m}}_R$, $0.9479\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$, $0.3492\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$, $6.968\text{kg}/\text{min}$ for ${\dot{m}}_{air}$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, –11.6 C for $T_4$, 27 C for $T_3$,$P_4$ for $P_3$, $30\text{kJ}/\text{min}$ for ${\dot{Q}}_{in}$, and 32 C for $T_{surr}$ in Equation (XII).

${\dot{S}}_{gen}=\left\{\begin{array}{l}2\text{kg}/\text{min}\left(0.9479\text{kJ}/\text{kg}\cdot \text{K}-0.3492\text{kJ}/\text{kg}\cdot \text{K}\right)\\ +6.968\text{kg}/\text{min}\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\ln \left(\frac{-11.6°\text{C}}{27°\text{C}}\right)-R\ln \left(\frac{P_4}{P_4}\right)\right)\\ -\frac{30\text{kJ}/\text{min}}{32°\text{C}}\end{array}\right\}$

$\begin{array}{c}{\dot{S}}_{gen}=\left\{\begin{array}{l}2\text{kg}/\text{min}\left(0.9479\text{kJ}/\text{kg}\cdot \text{K}-0.3492\text{kJ}/\text{kg}\cdot \text{K}\right)\\ +6.968\text{kg}/\text{min}\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\ln \left(\frac{\left(-11.6+273\right)\text{K}}{\left(27+273\right)\text{K}}\right)-R\ln \left(\frac{P_4}{P_4}\right)\right)\\ -\frac{30\text{kJ}/\text{min}}{\left(32+273\right)\text{K}}\end{array}\right\}\\ =1.1974+\left(6.968\right)\left(-0.1384\right)-0.09836\\ =0.1348\text{kJ}/\text{min}\cdot \text{K}\left(\frac{1\min }{60\sec }\right)\\ =0.00225\text{kW}/\text{K}\end{array}$

Hence, the entropy generated during the process is $0.00225\text{kW}/\text{K}$.