Chapter 6.11, Problem 152RP

A heat pump with refrigerant-134a as the working fluid is used to keep a space at 25°C by absorbing heat from geothermal water that enters the evaporator at 60°C at a rate of 0.065 kg/s and leaves at 40°C. Refrigerant enters the evaporator at 12°C with a quality of 15 percent and leaves at the same pressure as saturated vapor. If the compressor consumes 1.6 kW of power, determine (a) the mass flow rate of the refrigerant, (b) the rate of heat supply, (c) the COP, and (d) the minimum power input to the compressor for the same rate of heat supply.

FIGURE P6–152

To determine

The mass flow rate of the refrigerant.

Answer to Problem 152RP

The mass flow rate of the refrigerant is $\underset{\_}{0.0338\text{kg}/\text{s}}$.

Explanation of Solution

Determine the rate of heat absorbed from 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-11, “Saturated refrigerant R-134a”, obtain the value of saturated pressure of the refrigerant at the inlet temperature of $12°\text{C}$ as below.

$P_1=443.3\text{kPa}$.

Here, the pressure of refrigerant is constant in evaporation.

$P_1=P_2$.

From the Table A-11, “Saturated refrigerant R-134a” to obtain the value of specific enthalpy of the refrigerant at the outlet pressure of $443.3\text{kPa}$ and the dryness faction of 1 as,

$h_2=257.33\text{kJ}/\text{kg}$.

From the Table A-11, “Saturated refrigerant R-134a” to obtain the value of specific enthalpy of saturated liquid and specific enthalpy change upon vaporization of the refrigerant at the inlet temperature of $12°\text{C}$ as,

$\begin{array}{l}{h}_{f,1}=68.18\text{kJ}/\text{kg}\\ h_{fg,1}=189.09\text{kJ}/\text{kg}\end{array}$

Calculate the specific enthalpy of refrigerant at evaporator inlet.

$h_1=h_{f,1}+x_1\times h_{fg,1}$ (III)

Here, the specific enthalpy of saturated liquid is $h_{f,1}$, the dryness fraction is $x_1$, and the specific enthalpy change upon vaporization is $h_{fg,1}$.

Substitute $68.18\text{kJ}/\text{kg}$ for $h_{f,1}$, $189.09\text{kJ}/\text{kg}$ for $h_{fg,1}$, and 0.15 for $x_1$ in Equation (III).

$\begin{array}{c}{h}_1=\left(68.18\text{kJ}/\text{kg}\right)+\left(0.15\right)\times \left(189.09\text{kJ}/\text{kg}\right)\\ =\left(68.18\text{kJ}/\text{kg}\right)+\left(28.3635\text{kJ}/\text{kg}\right)\\ =96.5435\text{kJ}/\text{kg}\end{array}$

From the Table A-4, “Saturated water-temperature” to obtain the value of specific enthalpy of saturated liquid of water at the inlet temperature of $60°\text{C}$ as,

$h_{w,1}=251.18\text{kJ}/\text{kg}$

From the Table A-4, “Saturated water-temperature” to obtain the value of specific enthalpy of saturated liquid of water at the outlet temperature of $40°\text{C}$ as,

$h_{w,2}=167.53\text{kJ}/\text{kg}$

Substitute $0.065\text{kg}/\text{s}$ for ${\dot{m}}_w$, $251.18\text{kJ}/\text{kg}$ for $h_{w,1}$, and $167.53\text{kJ}/\text{kg}$ for $h_{w,2}$ in Equation (I).

$\begin{array}{c}{\dot{Q}}_L=\left(0.065\text{kg}/\text{s}\right)\left(251.18\text{kJ}/\text{kg}-167.53\text{kJ}/\text{kg}\right)\\ =\left(0.065\text{kg}/\text{s}\right)\left(83.65\text{kJ}/\text{kg}\right)\\ =5.437\text{kJ}/\text{s}\\ =5.437\text{kJ}/\text{s}\times \left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\end{array}$

      $=5.437\text{kW}$

Substitute $5.437\text{kW}$ for ${\dot{Q}}_L$, $96.54\text{kJ}/\text{kg}$ for $h_1$, and $257.33\text{kJ}/\text{kg}$  for $h_2$ in Equation (II).

$\begin{array}{c}{\dot{m}}_{\text{R}}=\frac{5.437\text{kW}}{\left(257.33-96.54\right)\text{kJ}/\text{kg}}\\ =\frac{5.437\text{kW}\times \left(\frac{1\text{kg}/\text{s}}{1\text{kW}}\right)}{\text{160}\text{.79}\text{kJ}/\text{kg}}\\ =0.0338\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of the refrigerant is $\underset{\_}{0.0338\text{kg}/\text{s}}$.

To determine

The heating load of the heat pump.

Answer to Problem 152RP

The heating load of the heat pump is $\underset{\_}{7.04\text{kW}}$.

Explanation of Solution

Determine the heating load of the heat pump.

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

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

Conclusion:

Substitute $5.437\text{kW}$ for ${\dot{Q}}_L$ and $1.6\text{kW}$ for ${\dot{W}}_{\text{in}}$ in Equation (IV).

$\begin{array}{c}{\dot{Q}}_L=\left(5.437\text{kW}\right)+\left(1.6\text{kW}\right)\\ =7.037\text{kW}\\ \cong 7.04\text{kW}\end{array}$

Thus, the heating load of the heat pump is $\underset{\_}{7.04\text{kW}}$.

To determine

The COP of a heat pump operating between the same temperature limits.

Answer to Problem 152RP

The COP of a heat pump operating between the same temperature limits is $\underset{\_}{4.40}$.

Explanation of Solution

Determine the coefficient of performance of the heat pump.

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

Conclusion:

Substitute $7.04\text{kW}$ for ${\dot{Q}}_H$ and $1.6\text{kW}$ for ${\dot{W}}_{\text{in}}$ in Equation (IV).

$\begin{array}{c}{\text{COP}}_{\text{HP}}=\frac{7.04\text{kW}}{1.6\text{kW}}\\ =4.40\end{array}$

Thus, the COP of a heat pump operating between the same temperature limits is $\underset{\_}{4.40}$.

To determine

The minimum power input to the compressor.

Answer to Problem 152RP

The minimum power input to the compressor is $\underset{\_}{0.740\text{kW}}$.

Explanation of Solution

Determine the maximum coefficient of performance of the heat pump operating between the same temperature limits.

${\text{COP}}_{\text{max}}=\frac{1}{\frac{T_L}{T_H}-1}$ (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 heat pump load.

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

Conclusion:

Substitute $60°\text{C}$ for $T_H$ and $25°\text{C}$ for $T_L$ in Equation (VI).

$\begin{array}{c}{\text{COP}}_{\text{max}}=\frac{1}{1-\frac{\left(25°\text{C}\right)}{\left(60°\text{C}\right)}}\\ =\frac{1}{1-\frac{\left(25°\text{C+273}\right)}{\left(60°\text{C+273}\right)}}\\ =\frac{1}{1-\left(\frac{298\text{K}}{333\text{K}}\right)}\\ =9.51\end{array}$

Substitute $7.04\text{kW}$ for ${\dot{Q}}_H$ and $9.51$ for ${\text{COP}}_{\max }$ in Equation (VII).

$\begin{array}{c}{\dot{W}}_{\text{in,min}}=\frac{7.04\text{kW}}{9.51}\\ =0.740\text{kW}\end{array}$

Thus, the minimum power input to the compressor is $\underset{\_}{0.740\text{kW}}$.