Chapter 11.10, Problem 18P

A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at −30°C by rejecting its waste heat to cooling water that enters the condenser at 18°C at a rate of 0.25 kg/s and leaves at 26°C. The refrigerant enters the condenser at 1.2 MPa and 65°C and leaves at 42°C. The inlet state of the compressor is 60 kPa and −34°C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet, (b) the refrigeration load, (c) the COP of the refrigerator, and (d) the theoretical maximum refrigeration load for the same power input to the compressor.

FIGURE P11–22

To determine

The quality of the refrigerant at the evaporator inlet.

Answer to Problem 18P

The quality of the refrigerant at the evaporator inlet is $0.4796$.

Explanation of Solution

Show the T-s diagram for the refrigeration cycle as in Figure (1).

Express specific enthalpy at state 3.

$h_3=h_{f@42\text{°C}}+v_{f@42\text{°C}}\left[P_3-P_{\text{sat@42°C}}\right]$ (I)

Here, specific enthalpy at saturated liquid and temperature of $\text{42°C}$ is $h_{f@42\text{°C}}$, specific volume at saturated liquid and temperature of $\text{42°C}$ is $v_{f@42\text{°C}}$, pressure at state 3 is $P_3$ and saturated pressure at temperature of $\text{42°C}$ is $P_{\text{sat@42°C}}$.

Express the quality of the refrigerant at the evaporator inlet.

$x_4=\frac{h_3-h_{f@60\text{kPa}}}{h_{g@60\text{kPa}}-h_{f@60\text{kPa}}}$ (II)

Here, specific enthalpy at saturated liquid and pressure of $\text{60}\text{kPa}$ is $h_{f@60\text{kPa}}$ and specific enthalpy at saturated vapor and pressure of $\text{60}\text{kPa}$ is $h_{g@60\text{kPa}}$.

Conclusion:

Perform unit conversion of pressure at state 1 from $\text{kPa}\text{to}\text{MPa}$.

$\begin{array}{c}{P}_1=60\text{kPa}\left[\frac{\text{MPa}}{\text{1000}\text{kPa}}\right]\\ =0.06\text{MPa}\end{array}$

Refer Table A-13, “superheated refrigerant-134a”, and write the value of specific enthalpy at state 1 $\left(h_1\right)$ corresponding to initial pressure of $\text{0}\text{.06}\text{MPa}$ and initial temperature $\left(T_1\right)$ of $-34\text{°C}$ using interpolation method.

Write the formula of interpolation method of two variables.

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

Here, the variables denote by x and y is initial temperature and specific enthalpy at state 1 respectively.

Show the specific enthalpy at state 2 corresponding to specific entropy as in Table (1).

Initial temperature $T\left(\text{°C}\right)$	Specific enthalpy at state 1 $h_1\left(\text{kJ}/\text{kg}\right)$
$-36.95$ $\left(x_1\right)$	227.80 $\left(y_1\right)$
$-34$ $\left(x_2\right)$	$\left(y_2=?\right)$
$-20$ $\left(x_3\right)$	240.78 $\left(y_3\right)$

Substitute $-\text{36}\text{.95°C,}-\text{34°C}\text{and}-\text{20°C}$ for $x_1,x_2\text{and}{x}_3$ respectively, $227.80\text{kJ}/\text{kg}$ for $y_1$ and $240.78\text{kJ}/\text{kg}$ for $y_3$ in Equation (III).

$\begin{array}{c}{y}_2=\frac{\left[-\text{34°C}-\left(-\text{36}\text{.95°C}\right)\right]\left[\left(240.78-227.80\right)\text{kJ}/\text{kg}\right]}{\left[-\text{20°C}-\left(-\text{36}\text{.95°C}\right)\right]}+227.80\text{kJ}/\text{kg}\\ =230.04\text{kJ}/\text{kg}\\ =h_1\end{array}$

Perform unit conversion of pressure at state 2 from $\text{kPa}\text{to}\text{MPa}$.

$\begin{array}{c}{P}_2=1200\text{kPa}\left[\frac{\text{MPa}}{\text{1000}\text{kPa}}\right]\\ =1.20\text{MPa}\end{array}$

Refer Table A-13, “superheated refrigerant-134a”, and write the value of specific enthalpy at state 2 $\left(h_2\right)$ corresponding to pressure at state 2 of $\text{1}\text{.20}\text{MPa}$ and temperature at state 2 $\left(T_2\right)$ of $65\text{°C}$ using interpolation method.

Show the specific enthalpy at state 2 corresponding to temperature as in Table (2).

Temperature $T_2\left(\text{°C}\right)$	Specific enthalpy at state 2 $h_2\left(\text{kJ}/\text{kg}\right)$
$60$ $\left(x_1\right)$	289.66 $\left(y_1\right)$
$65$ $\left(x_2\right)$	$\left(y_2=?\right)$
$70$ $\left(x_3\right)$	300.63 $\left(y_3\right)$

Use Excel by taking the values from Table (2), and using Equation (III) to get specific enthalpy at state 2.

$h_2=295.18\text{kJ}/\text{kg}$

Refer Table A-11, “saturated refrigerant 134a-temperature table”, and write the properties corresponding to temperature at state 3 of $\text{42°C}$.

$\begin{array}{l}{h}_{f@42\text{°C}}=111.28\text{kJ}/\text{kg}\\ v_{f@42\text{°C}}=0.0008786{\text{m}}^{\text{3}}/\text{kg}\\ P_{\text{sat@42°C}}=1072.8\text{kPa}\end{array}$

Substitute $111.28\text{kJ}/\text{kg}$ for $h_{f@42\text{°C}}$, $0.0008786{\text{m}}^{\text{3}}/\text{kg}$ for $v_{f@42\text{°C}}$, $1072.8\text{kPa}$ for $P_{\text{sat@42°C}}$ and $\text{1200}\text{kPa}$ for $P_3$ in Equation (I).

$\begin{array}{c}{h}_3=111.28\text{kJ}/\text{kg}+0.0008786{\text{m}}^{\text{3}}/\text{kg}\left[1200\text{kPa}-1072.8\text{kPa}\right]\\ =111.25\text{kJ}/\text{kg}\end{array}$

From Figure (1), write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

$\begin{array}{c}{h}_3\cong h_4\\ =111.25\text{kJ}/\text{kg}\end{array}$

Here, specific enthalpy at state 4 is $h_4$.

Refer Table A-12, “saturated refrigerant 134a-pressure table”, and write the properties corresponding to pressure at state 4 of $\text{60}\text{kPa}$.

$\begin{array}{l}{h}_{f@60\text{kPa}}=3.837\text{kJ}/\text{kg}\\ h_{g@60\text{kPa}}=227.80\text{kJ}/\text{kg}\end{array}$

Substitute $111.25\text{kJ}/\text{kg}$ for $h_3$, $3.837\text{kJ}/\text{kg}$ for $h_{f@60\text{kPa}}$ and $227.80\text{kJ}/\text{kg}$ for $h_{g@60\text{kPa}}$ in Equation (II).

$\begin{array}{c}{x}_4=\frac{111.25\text{kJ}/\text{kg}-3.837\text{kJ}/\text{kg}}{227.80\text{kJ}/\text{kg}-3.837\text{kJ}/\text{kg}}\\ =0.4796\end{array}$

Hence, the quality of the refrigerant at the evaporator inlet is $0.4796$.

To determine

The refrigeration load.

Answer to Problem 18P

The refrigeration load is $\text{5}\text{.404}\text{kW}$.

Explanation of Solution

Express the mass flow rate of the refrigerant from an energy balance on the compressor.

${\dot{m}}_R\left(h_2-h_3\right)={\dot{m}}_w\left(h_{w2}-h_{w1}\right)$ (IV)

Here, mass flow rate of the water is ${\dot{m}}_w$, initial and final specific enthalpy of water is $h_{w1}\text{and}{h}_{w2}$ respectively.

Express the rate of heat supplied from the refrigerant.

${\dot{Q}}_H={\dot{m}}_R\left(h_2-h_3\right)$ (V)

Express compressor power input.

${\dot{W}}_{\text{in}}={\dot{m}}_R\left(h_2-h_1\right)-{\dot{Q}}_{\text{in}}$ (VI)

Here, rate of heat gained by compressor is ${\dot{Q}}_{\text{in}}$.

Express the refrigeration load.

${\dot{Q}}_L={\dot{Q}}_H-{\dot{W}}_{\text{in}}-{\dot{Q}}_{\text{in}}$ (VII)

Conclusion:

Refer Table A-4, “saturated water-temperature table”, and write the initial specific enthalpy of water corresponding to temperature of $\text{18°C}$ using interpolation method.

Show the initial specific enthalpy of water corresponding to temperature as in Table (3).

Temperature $T\left(\text{°C}\right)$	Initial specific enthalpy of water $h_{w1@h_f}\left(\text{kJ}/\text{kg}\right)$
15 $\left(x_1\right)$	62.982 $\left(y_1\right)$
18 $\left(x_2\right)$	$\left(y_2=?\right)$
20$\left(x_3\right)$	83.915 $\left(y_3\right)$

Use Excel by taking the values from Table (3), and using Equation (III) to get initial specific enthalpy of water.

$h_{w1}=75.47\text{kJ}/\text{kg}$

Refer Table A-4, “saturated water-temperature table”, and write the initial specific enthalpy of water corresponding to temperature of $\text{26°C}$ using interpolation method.

Show the initial specific enthalpy of water corresponding to temperature as in Table (3).

Temperature $T\left(\text{°C}\right)$	Final specific enthalpy of water $h_{w2@h_f}\left(\text{kJ}/\text{kg}\right)$
25 $\left(x_1\right)$	104.83 $\left(y_1\right)$
26 $\left(x_2\right)$	$\left(y_2=?\right)$
30$\left(x_3\right)$	125.74 $\left(y_3\right)$

Use Excel by taking the values from Table (3), and using Equation (III) to get final specific enthalpy of water.

$h_{w2}=108.94\text{kJ}/\text{kg}$

Substitute $\text{295}\text{.18}\text{kJ}/\text{kg}\text{and}\text{111}\text{.25}\text{kJ}/\text{kg}$ for $h_2\text{and}{h}_3$ respectively, $\text{0}\text{.25}\text{kg}/\text{s}$ for ${\dot{m}}_w$, $\text{108}\text{.94}\text{kJ}/\text{kg}\text{and}\text{75}\text{.47}\text{kJ}/\text{kg}$ for $h_{w2}$ and $h_{w1}$ in Equation (IV).

$\begin{array}{l}{\dot{m}}_R\left(\text{295}\text{.18}-\text{111}\text{.25}\right)\text{kJ}/\text{kg}=\left(\text{0}\text{.25}\text{kg}/\text{s}\right)\left(\text{108}\text{.94}-\text{75}\text{.47}\right)\text{kJ}/\text{kg}\\ {\dot{m}}_R=0.04549\text{kg}/\text{s}\end{array}$

Substitute $0.04549\text{kg}/\text{s}$ for ${\dot{m}}_R$, $\text{295}\text{.18}\text{kJ}/\text{kg}\text{and}\text{111}\text{.25}\text{kJ}/\text{kg}$ for $h_2\text{and}{h}_3$ respectively in Equation (V).

$\begin{array}{c}{\dot{Q}}_H=0.04549\text{kg}/\text{s}\left(\text{295}\text{.18}-\text{111}\text{.25}\right)\text{kJ}/\text{kg}\\ =8.367\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =8.367\text{kW}\end{array}$

Substitute $0.04549\text{kg}/\text{s}$ for ${\dot{m}}_R$, $\text{295}\text{.18}\text{kJ}/\text{kg}\text{and}\text{230}\text{.04}\text{kJ}/\text{kg}$ for $h_2\text{and}{h}_1$ respectively and $\text{450}\text{W}$ for ${\dot{Q}}_{\text{in}}$ in Equation (VI).

$\begin{array}{c}{\dot{W}}_{\text{in}}=0.04549\text{kg}/\text{s}\left(\text{295}\text{.18}\text{kJ}/\text{kg}-\text{230}\text{.04}\text{kJ}/\text{kg}\right)-\text{450}\text{W}\left[\frac{\text{kW}}{\text{1000}\text{W}}\right]\\ =0.04549\text{kg}/\text{s}\left(\text{295}\text{.18}\text{kJ}/\text{kg}-\text{230}\text{.04}\text{kJ}/\text{kg}\right)-0.\text{450}\text{kW}\\ =2.963\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]-0.\text{450}\text{kW}\end{array}$

$\begin{array}{c}=2.963\text{kW}-0.\text{450}\text{kW}\\ =2.513\text{kW}\end{array}$

Substitute $8.367\text{kW}$ for ${\dot{Q}}_H$, $2.513\text{kW}$ for ${\dot{W}}_{\text{in}}$ and $\text{450}\text{W}$ for ${\dot{Q}}_{\text{in}}$ in Equation (VII).

$\begin{array}{c}{\dot{Q}}_L=8.367\text{kW}-2.513\text{kW}-\text{450}\text{W}\left[\frac{\text{kW}}{\text{1000}\text{W}}\right]\\ =8.367\text{kW}-2.513\text{kW}-0.\text{450}\text{kW}\\ =\text{5}\text{.404}\text{kW}\end{array}$

Hence, the refrigeration load is $\text{5}\text{.404}\text{kW}$.

To determine

The COP of the refrigerator.

Answer to Problem 18P

The COP of the refrigerator is $2.15$.

Explanation of Solution

Express the coefficient of performance of the refrigerator.

$\text{COP}=\frac{{\dot{Q}}_L}{{\dot{W}}_{\text{in}}}$ (VIII)

Conclusion:

Substitute $\text{5}\text{.404}\text{kW}$ for ${\dot{Q}}_L$ and $2.513\text{kW}$ for ${\dot{W}}_{\text{in}}$ in Equation (VIII).

$\begin{array}{c}\text{COP}=\frac{\text{5}\text{.404}\text{kW}}{2.513\text{kW}}\\ =2.15\end{array}$

Hence, the coefficient of performance of the refrigerator is $2.15$.

To determine

The theoretical maximum refrigeration load.

Answer to Problem 18P

The theoretical maximum refrigeration load is $\text{12}\text{.72}\text{kW}$.

Explanation of Solution

Express the reversible COP of the refrigerator for the similar temperature limits.

${\text{COP}}_{\text{max}}=\frac{1}{\frac{T_H}{T_L}-1}$ (IX)

Here, high and low source temperature is $T_H\text{and}{T}_L$ respectively.

Express the theoretical maximum refrigeration load.

${\dot{Q}}_{L,\max }={\text{COP}}_{\text{max}}{\dot{W}}_{\text{in}}$ (X)

Conclusion:

Substitute $18\text{°C}\text{and}-30\text{°C}$ for $T_H\text{and}{T}_L$ in Equation (IX).

$\begin{array}{c}{\text{COP}}_{\text{max}}=\frac{1}{\frac{18\text{°C}}{-30\text{°C}}-1}\\ =\frac{1}{\frac{\left(18+273\right)\text{K}}{\left(-30+273\right)\text{K}}-1}\\ =5.063\end{array}$

Substitute $5.063$ for ${\text{COP}}_{\text{max}}$ and $2.513\text{kW}$ for ${\dot{W}}_{\text{in}}$ in Equation (X).

$\begin{array}{c}{\dot{Q}}_{L,\max }=\left(5.063\right)\left(2.513\text{kW}\right)\\ =12.72\text{kW}\end{array}$

Hence, the theoretical maximum refrigeration load is $\text{12}\text{.72}\text{kW}$.