R-134a Heat Pump — Power Input and Comparison with Resistance Heating: 04-BS-10, May 2013, Q5

Question 5: R-134a Heat Pump — Power Input and Comparison with Resistance Heating (15 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. R-134a heat pump; house heat loss (condenser duty) $\dot Q_H=60{,}000$ kJ/h $=16.667$ kW. Compressor inlet: $P_1=280$ kPa, $T_1=0\ ^\circ\text{C}$ (superheated). Compressor exit: $P_2=1$ MPa, $T_2=60\ ^\circ\text{C}$ (superheated). Condenser exit: $T_3=30\ ^\circ\text{C}$ at 1 MPa — note $T_{sat}(1\ \text{MPa})=39.39\ ^\circ\text{C}$, so the condenser exit is a subcooled liquid, about 9.4°C below saturation, not a saturated-liquid state.

Find. (a) compressor power input; (b) rate of heat absorption from the ground water (evaporator duty); (c) additional electric power needed if a simple resistance heater replaced the heat pump.

Approach

The condenser duty $\dot Q_H$ (given, equal to the house heat loss) fixes the mass flow rate once $h_2$ and $h_3$ are known from the compressor-exit and condenser-exit states; compressor power and evaporator duty then follow directly, and the resistance-heater comparison is a simple energy-balance difference since a resistance heater converts 100% of its electrical input to heat.

Fix the four states. $h_1=h(280\ \text{kPa},0\ ^\circ\text{C})=398.98$ kJ/kg; $h_2=h(1\ \text{MPa},60\ ^\circ\text{C})=441.53$ kJ/kg; $h_3=h(1\ \text{MPa},30\ ^\circ\text{C},\text{subcooled})=241.72$ kJ/kg; throttling to the evaporator is isenthalpic, $h_4=h_3=241.72$ kJ/kg.
Mass flow rate from the condenser energy balance. $$\dot m=\frac{\dot Q_H}{h_2-h_3}=\frac{16.667}{441.53-241.72}=\boxed{0.0834\ \text{kg/s}}.$$
Compressor power input (part a). $$\dot W_{in}=\dot m(h_2-h_1)=0.0834\times(441.53-398.98)=\boxed{3.549\ \text{kW}}.$$
Heat absorbed from the ground water (part b), evaporator. $$\dot Q_L=\dot m(h_1-h_4)=0.0834\times(398.98-241.72)=\boxed{13.117\ \text{kW}}.$$ Check: $\dot Q_L+\dot W_{in}=13.117+3.549=16.667$ kW, matching $\dot Q_H$ to within rounding.
Resistance heater comparison (part c). A resistance heater must supply the full $\dot Q_H$ electrically (COP $=1$), so the increase in electric power over the heat pump is $$\Delta \dot W=\dot Q_H-\dot W_{in}=16.667-3.549=\boxed{13.117\ \text{kW}}.$$

Quantity	Result
(a) $\dot W_{in}$	3.549 kW
(b) $\dot Q_L$	13.117 kW
(c) $\Delta \dot W$ (resistance heater)	13.117 kW

Concept & Theory

Topic. Vapor-compression heat pump cycle sized from a fixed heating-load duty, contrasted with straight resistive heating.

Key relations. Steady-flow energy balances on each component: $\dot Q_H=\dot m(h_2-h_3)$ (condenser), $\dot W_{in}=\dot m(h_2-h_1)$ (compressor), $\dot Q_L=\dot m(h_1-h_4)$ (evaporator), $h_4=h_3$ (isenthalpic throttling); heat-pump COP $=\dot Q_H/\dot W_{in}$, resistance-heater COP $=1$.

Why this works. The condenser duty is the one quantity fixed directly by the application (the house's heat loss), so it — not the compressor — is the natural place to solve for $\dot m$; every other duty then scales off that same mass flow rate. The heat pump's advantage over resistive heating is exactly the evaporator heat it moves "for free" from the ground water, $\dot Q_L$, which here is nearly 4× the electrical work input (COP $\approx 4.7$).

Common pitfall. Reading the condenser exit as saturated liquid at 1 MPa because "condenser exit" is a familiar cue for $x=0$ — here the stated exit temperature (30°C) is below $T_{sat}(1\ \text{MPa})\approx 39.4\ ^\circ\text{C}$, so the state is subcooled and its enthalpy must be looked up (or computed) at the actual $T,P$ pair, not read off the saturation table.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 11 (Refrigeration and Heat Pump Cycles).

Question 5 of 9: R-134a Heat Pump — Power Input and Comparison with Resistance Heating

Question 5: R-134a Heat Pump — Power Input and Comparison with Resistance Heating (15 marks)

Approach