Chapter 6.11, Problem 101P

A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at −35°C by rejecting 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 50°C and leaves at the same pressure subcooled by 5°C. If the compressor consumes 3.3 kW of power, determine (a) the mass flow rate of the refrigerant, (b) the refrigeration load, (c) the COP, and (d) the minimum power input to the compressor for the same refrigeration load.

FIGURE P6–107

To determine

The mass flow rate of the refrigerant.

Answer to Problem 101P

The mass flow rate of the refrigerant is $\underset{\_}{0.0498\text{kg}/\text{s}}$.

Explanation of Solution

Determine the rate of heat transferred to the water.

${\dot{Q}}_H={\dot{m}}_w{\left(h_2-h_1\right)}_w$ (I)

Here, the mass flow rate of the water is ${\dot{m}}_w$, the enthalpy of saturated liquid that is entering the inlet of the condenser is $h_{w,1}$ and the enthalpy of saturated liquid which is leaving the condenser is $h_{w,2}$.

Determine the mass flow rate of a refrigerant.

${\dot{m}}_{\text{R}}=\frac{{\dot{Q}}_H}{h_1-h_2}$ (II)

Conclusion:

From the Table A-13, “Superheated refrigerant R-134a” obtain the value of enthalpy of the refrigerant at the inlet of the condenser at the 1.2 MPa of pressure and $50°\text{C}$ of temperature as,

$h_1=278.28\text{kJ}/\text{kg}$.

From the Table A-13, “Superheated refrigerant R-134a” obtain the value of temperature of the refrigerant at the inlet of the condenser at the 1.2 MPa of pressure as,

$T_{\text{sat@1}\text{.2MPa}}=46.29°\text{C}$

Calculate the exit temperature of the refrigerant in the condenser.

$T_2=T_{\text{sat@1}\text{.2MPa}}+\Delta T_{\text{subcool}}$ (III)

Here, the temperature leave from the condenser is $\Delta T_{\text{subcool}}$.

Substitute $46.23°\text{C}$ for $T_{\text{sat@1}\text{.2MPa}}$ and $-5°\text{C}$ for $\Delta T_{\text{subcool}}$ in Equation (III).

$\begin{array}{c}{T}_2=46.23°\text{C+}\left(-5°\text{C}\right)\\ =41.29°\text{C}\end{array}$

Refer to Table A-11, “Saturated refrigerant R-134a”, obtain the below exit enthalpy of the condenser at compressed liquid state on the basis of exit temperature of $41.29°\text{C}$ using interpolation method of two variables.

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)

Here, the variables denote by x and y are temperature and enthalpy of vaporization.

Show the temperature at $40°\text{C}$ and $42°\text{C}$ as in Table (1).

S. No	Temperature, $°\text{C}$ $\left(x\right)$	enthalpy of vaporization $\text{kJ}/\text{kg}$ $\left(y\right)$
1	$40°\text{C}$	$108.28\text{kJ}/\text{kg}$
2	$41.29°\text{C}$	$y_2=?$
3	$42°\text{C}$	$111.28\text{kJ}/\text{kg}$

Calculate exit enthalpy of the condenser at compressed liquid state on the basis of exit temperature of $41.29°\text{C}$ for liquid phase using interpolation method.

Substitute $40°\text{C}$ for $x_1$, $41.29°\text{C}$ for $x_2$, $42°\text{C}$ for $x_3$, $108.28\text{kJ}/\text{kg}$ for $y_1$, and $111.28\text{kJ}/\text{kg}$ for $y_3$ in Equation (IV).

$\begin{array}{c}{y}_2=\frac{\left(41.29°\text{C}-40°\text{C}\right)\left(111.28\text{kJ}/\text{kg}-108.28\text{kJ}/\text{kg}\right)}{\left(42°\text{C}-40°\text{C}\right)}+108.28\text{kJ}/\text{kg}\\ =110.19\text{kJ}/\text{kg}\end{array}$

From above calculation the exit enthalpy of the condenser at compressed liquid state on the basis of exit temperature of $41.29°\text{C}$ is $110.19\text{kJ}/\text{kg}$.

Repeat the above Equation (IV) to obtain the value of enthalpy of saturated liquid that entering the inlet of the condenser at the $18°\text{C}$ of temperature from the Table A-4, “Saturated water-Temperature “as,

$h_{w,1}=75.54\text{kJ}/\text{kg}$.

Repeat the above Equation (IV) to obtain the value of enthalpy of saturated liquid which is leaving the condenser at the $26°\text{C}$ of temperature from the Table A-4, “Saturated water-Temperature “as,

$h_{w,2}=109.01\text{kJ}/\text{kg}$.

Substitute $0.25\text{kg}/\text{s}$ for ${\dot{m}}_w$, $75.54\text{kJ}/\text{kg}$ for $h_{w,1}$, and $109.01\text{kJ}/\text{kg}$ for $h_{w,2}$ in Equation (I).

$\begin{array}{c}{\dot{Q}}_H=\left(0.25\text{kg}/\text{s}\right)\left(109.01\text{kJ}/\text{kg}-75.54\text{kJ}/\text{kg}\right)\\ =\left(0.25\text{kg}/\text{s}\right)\left(33.47\text{kJ}/\text{kg}\right)\\ =8.367\text{kJ}/\text{s}\\ =8.367\text{kJ}/\text{s}\times \left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\end{array}$

      $=8.367\text{kW}$

Substitute $8.367\text{kW}$ for ${\dot{Q}}_H$, $278.28\text{kJ}/\text{kg}$ for $h_1$, and $110.2\text{kJ}/\text{kg}$  for $h_2$ in Equation (II).

$\begin{array}{c}{\dot{m}}_{\text{R}}=\frac{8.367\text{kW}}{\left(278.28-110.2\right)\text{kJ}/\text{kg}}\\ =\frac{8.367\text{kW}\times \left(\frac{1\text{kg}/\text{s}}{1\text{kW}}\right)}{\text{168}\text{.08}\text{kJ}/\text{kg}}\\ =0.04977\text{kg}/\text{s}\\ \cong 0.0498\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of the refrigerant is $\underset{\_}{0.0498\text{kg}/\text{s}}$.

To determine

The refrigeration load of the refrigerator.

Answer to Problem 101P

The refrigeration load of the refrigerator is $\underset{\_}{5.07\text{kW}}$.

Explanation of Solution

Determine the refrigeration load of the refrigerator.

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

Here, the power input consumed by compressor is ${\dot{W}}_{\text{in}}$.

Conclusion:

Substitute $8.367\text{kW}$ for ${\dot{Q}}_H$ and $3.3\text{kW}$ for ${\dot{W}}_{\text{in}}$ in Equation (IV).

$\begin{array}{c}{\dot{Q}}_L=\left(8.367\text{kW}\right)-\left(3.3\text{kW}\right)\\ =5.067\text{kW}\\ \cong 5.07\text{kW}\end{array}$

Thus, the refrigeration load of the refrigerator is $\underset{\_}{5.07\text{kW}}$.

To determine

The COP of a reversible refrigerator operating between the same temperature limits.

Answer to Problem 101P

The COP of a reversible refrigerator operating between the same temperature limits is $\underset{\_}{1.54}$.

Explanation of Solution

Determine the coefficient of performance of the refrigerator.

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

Conclusion:

Substitute $5.07\text{kW}$ for ${\dot{Q}}_L$ and $3.3\text{kW}$ for ${\dot{W}}_{\text{in}}$ in Equation (IV).

$\begin{array}{c}{\text{COP}}_{\text{R}}=\frac{5.07\text{kW}}{3.3\text{kW}}\\ =1.536\\ \cong 1.54\end{array}$

Thus, the COP of a reversible refrigerator operating between the same temperature limits is $\underset{\_}{1.54}$.

To determine

The minimum power input to the compressor.

Answer to Problem 101P

The minimum power input to the compressor is $\underset{\_}{1.13\text{kW}}$.

Explanation of Solution

Determine the maximum coefficient of performance of the reversible refrigerator operating between the same temperature limits.

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

Here, the temperature of higher temperature body is $T_H$ and the temperature of lower temperature body is $T_L$.

Determine the minimum power input to the condenser for the same refrigerator load.

${\dot{W}}_{\text{in,min}}=\frac{{\dot{Q}}_L}{{\text{COP}}_{\text{max}}}$ (VII)

Conclusion:

Substitute $18°\text{C}$ for $T_H$ and $-35°\text{C}$ for $T_L$ in Equation (VI).

$\begin{array}{c}{\text{COP}}_{\text{max}}=\frac{1}{\frac{\left(18°\text{C}\right)}{\left(-35°\text{C}\right)}-1}\\ =\frac{1}{\frac{\left(18°\text{C+273}\right)}{\left(-35°\text{C+273}\right)}-1}\\ =\frac{1}{\left(\frac{291\text{K}}{238\text{K}}\right)-1}\\ =4.49\end{array}$

Substitute $5.07\text{kW}$ for ${\dot{Q}}_L$ and $4.49$ for ${\text{COP}}_{\max }$ in Equation (VII).

$\begin{array}{c}{\dot{W}}_{\text{in,min}}=\frac{5.07\text{kW}}{4.49}\\ =1.129\text{kW}\\ \cong 1.13\text{kW}\end{array}$

Thus, the minimum power input to the compressor is $\underset{\_}{1.13\text{kW}}$.