Question 2 of 9: R-134a Heat Pump Driven by a Power Cycle

Question 2: R-134a Heat Pump Driven by a Power Cycle (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Heat pump heating capacity $\dot Q_H=500$ kJ/min $=8.333$ kJ/s. Evaporator: saturated vapor at $-12\ ^\circ$C ($P_{evap}=185.24$ kPa). Condenser pressure $1$ MPa, compressor $\eta_c=0.80$. Expansion-valve inlet (condenser exit, subcooled liquid): $0.96$ MPa, $34\ ^\circ$C. Driving power cycle: $\eta_{cycle}=0.25$; 80% of its rejected heat reaches the heated space. $T_0=15\ ^\circ$C.

Find. (a) $\dot W_c$ [kW]; (b) $\text{COP}_{HP}$; (c) ratio of total heat delivered to $\dot Q_{in,cycle}$; (d) $\dot S_{gen}$ [kJ/K·s]; (e) $\dot X_{dest}$ [kJ/s].

Approach

Fix the compressor-inlet (saturated vapor) and condenser-exit (given subcooled liquid) states, run the isentropic compressor leg and apply its efficiency to get the actual compressor-exit state, then use the condenser energy balance (duty $=\dot Q_H$) to solve for the refrigerant mass flow rate. Compressor power and COP follow directly; the power-cycle side is then sized from the compressor power via its efficiency, and the entropy/exergy analysis closes out the compressor's irreversibility.

1. Compressor inlet and isentropic exit. Saturated vapor at $-12\ ^\circ$C: $h_1=391.46$ kJ/kg, $s_1=1.7348$ kJ/kg·K ($P_{evap}=185.24$ kPa). Isentropic exit at 1 MPa: $h_{2s}=426.59$ kJ/kg, so $w_{c,s}=35.14$ kJ/kg. Actual: $$w_{c,a}=\frac{w_{c,s}}{\eta_c}=\frac{35.14}{0.80}=\boxed{43.92\ \text{kJ/kg}},\qquad h_2=435.38\ \text{kJ/kg},\ \ s_2=1.7620\ \text{kJ/kg}\cdot\text{K}.$$
2. Condenser exit / expansion-valve inlet. Subcooled liquid at 0.96 MPa, 34°C: $h_3=247.53$ kJ/kg (throttled isenthalpically to the evaporator pressure, $h_4=h_3$).
3. Mass flow rate from the condenser energy balance. $$\dot m=\frac{\dot Q_H}{h_2-h_3}=\frac{8.333}{435.38-247.53}=\boxed{0.04436\ \text{kg/s}}.$$
4. Compressor power (part a) and COP (part b). $$\dot W_c=\dot m\,w_{c,a}=0.04436\times43.92=\boxed{1.948\ \text{kW}}.$$ $$\text{COP}_{HP}=\frac{\dot Q_H}{\dot W_c}=\frac{8.333}{1.948}=\boxed{4.277}.$$
5. Power-cycle sizing. The compressor is driven entirely by the power cycle's work output, so $\dot W_c=\eta_{cycle}\dot Q_{in,cycle}$: $$\dot Q_{in,cycle}=\frac{\dot W_c}{\eta_{cycle}}=\frac{1.948}{0.25}=7.794\ \text{kW},\qquad \dot Q_{out,cycle}=\dot Q_{in,cycle}-\dot W_c=5.845\ \text{kW}.$$
6. Ratio of total delivered heat to power-cycle heat input (part c). The heated space receives the heat pump's output plus 80% of the power cycle's rejected heat: $$\dot Q_{total}=\dot Q_H+0.80\,\dot Q_{out,cycle}=8.333+0.80\times5.845=13.010\ \text{kW}.$$ $$\text{Ratio}=\frac{\dot Q_{total}}{\dot Q_{in,cycle}}=\frac{13.010}{7.794}=\boxed{1.6692}.$$
7. Entropy generation rate in the compressor (part d). The compressor is adiabatic, so all entropy production shows up as $\Delta s$ across it: $$\dot S_{gen}=\dot m(s_2-s_1)=0.04436\times(1.7620-1.7348)=\boxed{0.001206\ \text{kJ/K}\cdot\text{s}}.$$
8. Exergy destruction rate in the compressor (part e). $$\dot X_{dest}=T_0\,\dot S_{gen}=288.15\times0.001206=\boxed{0.3474\ \text{kJ/s}}.$$