Chapter 7, Problem 53P

(a)

To determine

The COP of the heat pump.

Explanation of Solution

Given:

The initial pressure of the heat pump $\left(P_1\right)$ is 800 kPa.

The initial temperature of the heat pump $\left(T_1\right)$ is $35°\text{C}$.

The mass flow rate of heat pump is $\left(\dot{m}\right)$ is $0.018\text{kg}/\text{s}$.

The final pressure of the heat pump $\left(P_2\right)$ is $800\text{ kPa}$.

The quality of the refrigerant at the exit is 0 (saturated liquid).

The rate of required input of the heat pump $\left({\dot{W}}_{\text{net,in}}\right)$ is 1.2 kW.

Calculation:

Convert the unit of pressure from kPa to MPa.

  $\begin{array}{c}P=800\text{kPa}\\ =800\text{kPa}\times {10}^{-3}\frac{\text{MPa}}{\text{kPa}}\\ =0.80\text{MPa}\end{array}$

Refer to Table A-13, “Superheated refrigerant-134a”, obtain the below properties at the superheated pressure and temperature of 800 kPa (0.80 MPa) and 35 C using interpolation method of two variables.

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

Temperature,  C	Specific enthalpy, $\text{kJ}/\text{kg}$
	Saturated liquid, $v_f$
31.31 C	267.34
35 C	?
40 C	276.46

Calculate superheated pressure and temperature of 800 kPa (0.80 MPa) and 35 C for liquid phase using interpolation method.

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

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

Substitute $x_1= 31.31 °\text{C}$, $x_2=35 °\text{C}$, $x_3=40 °\text{C}$, $y_1=267.34\text{kJ}/\text{kg}$, and $y_3= 276.46\text{ kJ/kg}$ in the above Equation.

  $\begin{array}{c}{y}_2=\frac{\left(35°\text{C}-31.31\text{°C}\right)\left(276.46\text{kJ}/\text{kg}-267.34\text{kJ}/\text{kg}\right)}{\left(40\text{°C}-31.31\text{°C}\right)}+267.34\text{kJ}/\text{kg}\\ =271.21\text{kJ}/\text{kg}\end{array}$

From above calculation the initial enthalpy $\left(h_1\right)$ of condenser is $271.21\text{kJ}/\text{kg}$.

Refer to Table A-12, “Saturated pressure table”, at final state of 800 kPa and 0.

  $h_f=h_2=95.48\text{kJ}/\text{kg}$

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.

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

Here, The heat rejected in the condenser is ${\dot{Q}}_{\text{H}}$,

The initial specific enthalpy of the condenser is $h_1$ and the final specific enthalpy of the condenser is $h_2$.

Substitute ${\dot{W}}_{\text{in}}=0$ in Equation (II) and write the expression for the energy balance on the heat rejected in the condenser.

  ${\dot{Q}}_{\text{H}}=\dot{m}\left(h_1-h_2\right)$        (III)

  $\begin{array}{c}{\dot{Q}}_{\text{H}}=\left(0.018\text{kg}/\text{s}\right)\left(271.21\text{kJ}/\text{kg}-95.48\text{kJ}/\text{kg}\right)\\ =\left(0.018\text{kg}/\text{s}\right)\left(175.73\text{kJ}/\text{kg}\right)\\ =3.163\text{kJ}/\text{s}\times \left(\frac{\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =3.163\text{kW}\end{array}$

Write the expression for the rate of coefficient performance of a heat pump.

  ${\text{COP}}_{\text{HP}}=\frac{{\dot{Q}}_H}{{\dot{W}}_{\text{net,in}}}$

  $\begin{array}{c}\text{COP}=\frac{3.163\text{kW}}{1.2\text{kW}}\\ =2.6359\\ \cong 2.64\end{array}$

Thus, the COP of the heat pump is $\underset{\_}{\text{2}\text{.64}}$.

(b)

To determine

The rate of heat absorbed from the outside air.

Explanation of Solution

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

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

  $\begin{array}{c}{\dot{Q}}_L=\left(3.163-1.2\right)\text{kW}\\ \text{=1}\text{.96}\text{kW}\end{array}$

Thus, the rate of heat absorbed from the outside air is $\underset{\_}{1.96\text{kW}}$.