Chapter 6.11, Problem 58P

Refrigerant-134a enters the evaporator coils placed at the back of the freezer section of a household refrigerator at 100 kPa with a quality of 20 percent and leaves at 100 kPa and −26°C. If the compressor consumes 600 W of power and the COP of the refrigerator is 1.2, determine (a) the mass flow rate of the refrigerant and (b) the rate of heat rejected to the kitchen air.

FIGURE P6–58

To determine

The mass flow rate of the refrigerant.

Answer to Problem 58P

The mass flow rate of the refrigerant is $\underset{\_}{\text{0}\text{.00414}\text{kg}/\text{s}}$.

Explanation of Solution

Write the expression for the energy balance equation.

$E_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}$ (I)

Here, the total energy entering the system is $E_{\text{in}}$, the total energy leaving the system is $E_{\text{out}}$, and the change in the total energy of the system is $\Delta E_{\text{system}}$.

Simplify Equation (II) and write energy balance relation of refrigrent-134a.

${\dot{W}}_{\text{in}}+\dot{m}{h}_1={\dot{Q}}_{\text{R}}+\dot{m}{h}_2$ (II)

Here, the rate of work to be done into the system is ${\dot{W}}_{\text{in}}$, the mass flow rate of refrigerant is $\dot{m}$, the rate of heat rejected in the refrigerant is ${\dot{Q}}_{\text{R}}$, the initial specific enthalpy of the refrigerant is $h_1$ and the final specific enthalpy of the refrigerant is $h_2$.

Substitute $0$ for ${\dot{W}}_{\text{in}}$ in Equation (II), write the expression for the energy balance of mass flow rate of the refrigerant.

$\begin{array}{l}{\dot{Q}}_{\text{L}}=\dot{m}\left(h_1-h_2\right)\\ \dot{m}=\frac{{\dot{Q}}_{\text{L}}}{h_1-h_2}\end{array}$ (III)

Here, the rate of heat transfer with low-temperature body is ${\dot{Q}}_L$.

Write the expression for the rate of coefficient performance of a refrigerant.

$\begin{array}{l}\text{COP}=\frac{{\dot{Q}}_L}{{\dot{W}}_{\text{in}}}\\ {\dot{Q}}_L=\left(\text{COP}\right){\dot{W}}_{\text{in}}\end{array}$ (IV)

Conclusion:

Determine the initial specific enthalpy of refrigerant.

$h_1=h_f+x{h}_{fg}$ (V)

Here, the specific enthalpy of saturated liquid is $h_f$ , the dryness fraction is $x$, and the specific enthalpy change upon vaporization is $h_{fg}$.

Refer to Table A-13, “Saturated refrigerant-134a-Pressure table”, obtain the value of specific enthalpy of saturated liquid and specific enthalpy change upon vaporization at 100 kPa pressure.

$\begin{array}{l}{h}_f=17.27\text{kJ}/\text{kg}\\ h_{fg}=217.19\text{kJ}/\text{kg}\end{array}$

Substitute $17.27\text{kJ}/\text{kg}$ for $h_f$, 0.2 for $x$, and $217.19\text{kJ}/\text{kg}$ for $h_{fg}$ in Equation (V).

$\begin{array}{c}{h}_1=\left(17.27\text{kJ}/\text{kg}\right)+\left(0.2\right)\left(217.19\text{kJ}/\text{kg}\right)\\ =\left(17.27\text{kJ}/\text{kg}\right)+\left(43.438\text{kJ}/\text{kg}\right)\\ =60.708\text{kJ}/\text{kg}\\ \cong 60.71\text{kJ}/\text{kg}\end{array}$

Refer to Table A-13, “Saturated refrigerant-134a”, obtain the below properties at the final pressure and saturated temperature of 100 kPa and (-26 C) using interpolation method of two variables.

Write the formula of interpolation method of two variables.

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

Here, the variables denote by x and y are saturated temperature and specific enthalpy.

Show the temperature at 31.31 C and 40 C as in Table (1).

Temperature,  C	Specific enthalpy, $\text{kJ}/\text{kg}$
-26.65 C	234.46
-26 C	?
-22.32 C	236.99

Calculate final pressure and saturated temperature of 100 kPa and (-26 C) for liquid phase using interpolation method.

Substitute $-26.65°\text{C}$ for $x_1$, $-26°\text{C}$ for $x_2$, $-22.32°\text{C}$ for $x_3$, $234.46\text{kJ}/\text{kg}$ for $y_1$, and $236.99\text{kJ}/\text{kg}$ for $y_3$ in Equation (VI).

$\begin{array}{c}{y}_2=\frac{\left(\left(-26°\text{C}\right)-\left(-26.65°\text{C}\right)\right)\left(236.99\text{kJ}/\text{kg}-234.46\text{kJ}/\text{kg}\right)}{\left(\left(-22.32°\text{C}\right)-\left(-26.65°\text{C}\right)\right)}+234.46\text{kJ}/\text{kg}\\ =234.84\text{kJ}/\text{kg}\end{array}$

From above calculation the final enthalpy of refrigerant is $234.84\text{kJ}/\text{kg}$.

Substitute $1.2$ for $\text{COP}$ and $0.600\text{kW}$ for ${\dot{W}}_{\text{in}}$ in Equation (IV).

$\begin{array}{c}{\dot{Q}}_L=\left(1.2\right)\left(0.600\text{kW}\right)\\ =0.72\text{kW}\end{array}$

Substitute 0.72 kW for ${\dot{Q}}_L$, $234.84\text{kJ}/\text{kg}$ for $h_2$, and $60.71\text{kJ}/\text{kg}$ for $h_1$ in Equation (III).

$\begin{array}{c}\dot{m}=\frac{0.72\text{kW}}{\left(234.84-60.71\right)\text{kJ}/\text{kg}}\\ =\frac{0.72\text{kW}}{\left(174.13\right)\text{kJ}/\text{kg}}\\ =0.004135\text{kg}/\text{s}\\ \cong 0.00414\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of the refrigerant is $\underset{\_}{\text{0}\text{.00414}\text{kg}/\text{s}}$.

To determine

The rate of heat rejected from refrigerant.

Answer to Problem 58P

The rate of heat rejected from refrigerant is $\underset{\_}{1.32\text{kW}}$.

Explanation of Solution

Write the expression for the rate of conversation of energy principle for refrigerant 134a.

$\begin{array}{l}{\dot{W}}_{\text{in}}={\dot{Q}}_H-{\dot{Q}}_L\\ {\dot{Q}}_H={\dot{Q}}_L+{\dot{W}}_{\text{in}}\end{array}$ (V)

Conclusion:

Substitute $0.72\text{kW}$ for ${\dot{Q}}_L$ and $\text{600}\text{W}$ for ${\dot{W}}_{\text{in}}$ in Equation (V).

$\begin{array}{c}{\dot{Q}}_L=0.72\text{kW}-600\text{W}\\ =0.72\text{kW}-600\text{W}\times \left(\frac{{10}^{-3}\text{kW}}{1\text{W}}\right)\\ \text{=1}\text{.32}\text{kW}\end{array}$

Thus, the rate of heat rejected from refrigerant is $\underset{\_}{1.32\text{kW}}$.