Chapter 7, Problem 97P

(a)

To determine

The mass flow rate of the refrigerant.

Explanation of Solution

Given:

The refrigerated temperature is $-35°\text{C}$.

The cooling water inlet temperature $\left(T_{w1}\right)$ is $18°\text{C}$.

The cooling water outlet temperature $\left(T_{w2}\right)$ is $26°\text{C}$.

The mass flow rate of the water $\left({\dot{m}}_w\right)$ is $0.25\text{kg}/\text{s}$.

The inlet and outlet pressure of the condenser is 1.2 MPa.

The power input consumed by compressor $\left({\dot{W}}_{\text{in}}\right)$ is $3.3\text{kW}$.

Calculation:

First find out the state properties of the system as shown below:

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,

  $h_1=278.28\text{kJ}/\text{kg}$.

At state 2, the refrigerant is subcooled by $5°\text{C}$. Thus the exit temperature of the refrigerant from the condenser is expressed as follows:

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

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

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 $\left(T_2\right)$ of the condenser using the equation (I).

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

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

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.

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 using interpolation method.

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

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 (II) 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,

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

Calculate the rate of heat transferred to the water.

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

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

        $=8.367\text{kW}$

Calculate the mass flow rate of a refrigerant.

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

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

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

(b)

To determine

The refrigeration load of the refrigerator.

Explanation of Solution

Calculate the refrigeration load of the refrigerator.

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

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

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

(c)

To determine

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

Explanation of Solution

Determine the coefficient of performance of the refrigerator.

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

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

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

(d)

To determine

The minimum power input to the compressor.

Explanation of Solution

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

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

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

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

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

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

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

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