Question 7.57: Refrigerant-134a enters the condenser of a residential heat ......

Refrigerant-134a flows through the condenser of a residential heat pump unit. For a given compressor power consumption the COP of the heat pump and the rate of heat absorbed from the outside air are to be determined.

Assumptions 1 The heat pump operates steadily. 2 The kinetic and potential energy changes are zero.

Properties The enthalpies of R-134a at the condenser inlet and exit are

\begin{aligned} & \left.\begin{array}{l} P_1=800 \mathrm{kPa} \\ T_1=35^{\circ} \mathrm{C} \end{array}\right\} h_1=271.22 \mathrm{~kJ} / \mathrm{kg} \\ & \left.\begin{array}{l} P_2=800 \mathrm{kPa} \\ x_2=0 \end{array}\right\} h_2=95.47 \mathrm{~kJ} / \mathrm{kg} \end{aligned}

Analysis (a) An energy balance on the condenser gives the heat rejected in the condenser

\dot{Q}_{H}=\dot{m}\left(h_{1}-h_{2}\right)=(0.018 \mathrm{~kg} / \mathrm{s})(271.22-95.47) \mathrm{kJ} / \mathrm{kg}=3.164 \mathrm{~kW}

The COP of the heat pump is

\mathrm{COP}=\frac{\dot{Q}_{H}}{\dot{W}_{\text {in }}}=\frac{3.164 \mathrm{~kW}}{1.2 \mathrm{~kW}}=\mathrm{2 . 6 4}

(b) The rate of heat absorbed from the outside air

\dot{Q}_{L}=\dot{Q}_{H}-\dot{W}_{\text {in }}=3.164-1.2=\mathrm{1 . 9 6} \mathrm{k W}