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. $\dot Q_H=500$ kJ/min $=8.333$ kW. Evaporator: sat. vapour at $-12\ ^\circ$C. Condenser pressure $=1$ MPa, $\eta_{comp}=0.80$. Expansion-valve inlet: $0.96$ MPa, $34\ ^\circ$C. $T_0=15\ ^\circ$C.

Find. (a) $\dot W_c$ [kW]; (b) $\text{COP}_{HP}$; (c) $\dot S_{gen}$ [kJ/K·s]; (d) $\dot X_{dest}$ [kJ/s].

Approach

Fix state 1 (saturated vapour), find the isentropic compressor exit at 1 MPa, apply the 80% compressor efficiency to get the actual exit enthalpy and hence entropy, and fix state 3 directly from its given pressure and temperature (checking first whether it is sub-cooled). The condenser energy balance $\dot Q_H=\dot m(h_2-h_3)$ then sizes the mass flow rate, and everything else follows.

1. Evaporator exit and isentropic compression. Saturated vapour at $-12\ ^\circ$C: $h_1=391.45$ kJ/kg, $s_1=1.7348$ kJ/kg·K. Isentropic compression to 1 MPa: $h_{2s}=426.59$ kJ/kg, so $w_{s}=35.14$ kJ/kg. Actual: $$w_{c,a}=\frac{w_s}{\eta_{comp}}=\frac{35.14}{0.80}=43.92\ \text{kJ/kg},\qquad h_2=h_1+w_{c,a}=435.38\ \text{kJ/kg}\ (T_2=54.19\ ^\circ\text{C},\ s_2=1.7620\ \text{kJ/kg}\cdot\text{K}).$$
2. Valve-inlet state. At 0.96 MPa, $T_{sat}=37.88\ ^\circ$C, and the stated $34\ ^\circ$C is below this, so state 3 is a sub-cooled liquid (about $3.9\ ^\circ$C of sub-cooling), not saturated liquid: $h_3=247.53$ kJ/kg.
3. Mass flow rate and compressor power (part a). Condenser energy balance, $\dot Q_H=\dot m(h_2-h_3)$, with $\dot Q_H=500/60=8.333$ kW: $$\dot m=\frac{8.333}{435.38-247.53}=0.04436\ \text{kg/s}.$$ $$\dot W_c=\dot m\,w_{c,a}=0.04436\times43.92=\boxed{1.948\ \text{kW}}\quad\textbf{(a)}.$$
4. Coefficient of performance (part b). $$\text{COP}_{HP}=\frac{\dot Q_H}{\dot W_c}=\frac{8.333}{1.948}=\boxed{4.277}\quad\textbf{(b)}.$$
5. Entropy generation rate (part c). The compressor is adiabatic, so all its entropy generation is the state-1-to-2 rise: $$\dot S_{gen}=\dot m(s_2-s_1)=0.04436\times(1.7620-1.7348)=\boxed{0.001206\ \text{kJ/K}\cdot\text{s}}\quad\textbf{(c)}.$$
6. Exergy destruction rate (part d). With $T_0=15+273.15=288.15$ K, $$\dot X_{dest}=T_0\,\dot S_{gen}=288.15\times0.001206=\boxed{0.3474\ \text{kJ/s}}\quad\textbf{(d)}.$$