Chapter 6.11, Problem 55P

Refrigerant-134a enters the condenser of a residential heat pump at 800 kPa and 35°C at a rate of 0.018 kg/s and leaves at 800 kPa as a saturated liquid. If the compressor consumes 1.2 kW of power, determine (a) the COP of the heat pump and (b) the rate of heat absorption from the outside air.

FIGURE P6–57

To determine

The COP of the heat pump.

Answer to Problem 55P

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

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.

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

Here, the rate of work to be done into the system is ${\dot{W}}_{\text{in}}$, the mass flow rate of heat pump is $\dot{m}$, 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 $0$ for ${\dot{W}}_{\text{in}}$ in Equation (II), write the expression for the energy balance on the condenser confer heat rejected in the condenser.

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

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

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

Here the rate of required input of the heat pump is ${\dot{W}}_{\text{net,in}}$.

Conclusion:

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.

Write the formula of interpolation method of two variables.

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

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

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

Temperature,  C	Specific enthaply, $\text{kJ}/\text{kg}$
	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.

Substitute 31.31 C for $x_1$, 35 C for $x_2$, 40 C for $x_3$, $267.34\text{kJ}/\text{kg}$ for $y_1$, and $276.46\text{kJ}/\text{kg}$ for $y_3$ in Equation (V).

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

From above calculation the initial enthalpy of condenser is $271.21\text{kJ}/\text{kg}$.

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

$h_f=95.48\text{kJ}/\text{kg}$

$h_g=267.34\text{kJ}/\text{kg}$

Write the expression of final specific enthalpy of a two-phase system for condenser.

$\begin{array}{c}{h}_2=h_f+x{h}_{fg}\\ =h_f+x\left(h_g-h_f\right)\end{array}$ (VI)

Here, the specific enthalpy of condenser is $v_f$, the specific enthalpy of condenser is, and the quality of final state for condenser is $x$.

Substitute $95.48\text{kJ}/\text{kg}$ for $h_f$, $267.34\text{kJ}/\text{kg}$ for $h_g$, and 0 for $x$ in Equation (VI).

$\begin{array}{c}{h}_2=\left(95.48\text{kJ}/\text{kg}\right)+\left(0\right)\times \left(267.34\text{kJ}/\text{kg}-95.48\text{kJ}/\text{kg}\right)\\ =95.48\text{kJ}/\text{kg}\end{array}$

Substitute $0.018\text{kg}/\text{s}$ for $\dot{m}$, $271.21\text{kJ}/\text{kg}$ for $h_2$, and $95.48\text{kJ}/\text{kg}$ for $h_2$ in Equation (III).

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

Substitute $3.163\text{kW}$ for ${\dot{Q}}_{\text{H}}$ and $1.2\text{kW}$ for ${\dot{W}}_{\text{in}}$ in Equation (IV).

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

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

To determine

The rate of heat absorbed from the outside air.

Answer to Problem 55P

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

Explanation of Solution

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

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

Here, the rate of heat rejected in the condenser is ${\dot{Q}}_H$, the rate of consume power by compressor is ${\dot{W}}_{\text{in}}$, and the rate of heat absorbed from the outside air is ${\dot{Q}}_L$.

Conclusion:

Substitute $3.163\text{kW}$ for ${\dot{Q}}_H$ and $1.2\text{kW}$ for ${\dot{W}}_{\text{in}}$ in Equation (V).

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

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