Question 4 of 8: Ground-source R-134a heat pump

Problem 4: Ground-source R-134a heat pump (20 points)

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. An ideal single-stage vapour-compression heat pump on R-134a, delivering its condenser heat to an air stream.

Find. The COP, refrigerant mass flow, compressor swept volume, and the hourly running cost against direct electric radiators.

Approach. The two stated temperatures fix the cycle completely: the ground temperature less the 15 K approach fixes the evaporating temperature, and the delivery pressure fixes the condensing temperature. Then run the air-side energy balance to scale the cycle up to a real mass flow.

1. Fix the cycle temperatures. The evaporator must sit 15 K below the ground to draw heat from it: $$t_e = 5-15 = -10\,{}^{\circ}\text{C}\quad\Rightarrow\quad p_e = 200.6\ \text{kPa}$$ and the stated delivery pressure of 1.0164 MPa is the R-134a saturation pressure at $t_c = 40.0\,{}^{\circ}\text{C}$ — a deliberate choice by the examiner, not a coincidence.
2. State the four cycle points. With dry-saturated suction, isentropic compression and no sub-cooling (properties on the IIR reference, $h_f = 200\ \text{kJ/kg}$ at 0 °C): $$1:\ h_1 = 392.7\ \text{kJ/kg},\quad s_1 = 1.7334\ \text{kJ/kg}\cdot\text{K},\quad v_1 = 0.09959\ \text{m}^3/\text{kg}$$ $$2:\ p_2 = 1.0164\ \text{MPa},\ s_2 = s_1 \Rightarrow t_2 = 46.3\,{}^{\circ}\text{C},\ h_2 = 426.5\ \text{kJ/kg}$$ $$3:\ \text{saturated liquid at }40\,{}^{\circ}\text{C},\ h_3 = 256.4\ \text{kJ/kg};\qquad 4:\ h_4 = h_3$$ Only differences in $h$ enter the answer, so a table on any other reference state gives the same result.
3. Part (b) — coefficient of performance. Specific duties: $$q_c = h_2-h_3 = 170.1\ \text{kJ/kg},\qquad w = h_2-h_1 = 33.81\ \text{kJ/kg},\qquad q_e = h_1-h_4 = 136.3\ \text{kJ/kg}$$ $$\text{COP}_{\text{heating}}=\frac{q_c}{w}=\frac{170.1}{33.81}=\boxed{5.03}$$ The cycle balance $q_c = q_e + w$ closes exactly. For orientation, the Carnot heat-pump COP between 263.15 K and 313.15 K is 6.26, so this cycle achieves 80 % of the ideal — a reasonable figure for a dry-saturated, no-subcool cycle with throttling losses.
4. Air-side duty. At 32 °C and atmospheric pressure, $\rho = 101{,}325/(287.0\times305.15)=1.157\ \text{kg/m}^3$, so $$\dot m_a = 1.5(1.157)=1.735\ \text{kg/s},\qquad Q_c = \dot m_a c_p \Delta t = 1.735(1.005)(32-8)=41.9\ \text{kW}$$ The density is taken at the delivery state, which is where the 1.5 m³/s is specified.
5. Part (c) — refrigerant mass flow. The condenser must transfer that 41.9 kW: $$\dot m_r=\frac{Q_c}{q_c}=\frac{41.85}{170.1}=\boxed{0.246\ \text{kg/s}}$$
6. Part (d) — compressor swept volume. The volume the compressor must actually induct at suction is $$\dot V_1=\dot m_r v_1 = 0.246(0.09959)=0.02451\ \text{m}^3/\text{s} = 24.5\ \text{L/s}$$ A single-acting machine sweeps once per revolution, so with $\eta_v = 0.85$ at $N = 350/60$ rev/s, $$V_{sw}=\frac{\dot V_1}{\eta_v N}=\frac{0.02451}{0.85(5.833)}=4.943\times10^{-3}\ \text{m}^3 =\boxed{4{,}940\ \text{cm}^3\ \text{per revolution}}$$ Nearly five litres of displacement at 350 rpm is a large, slow, industrial reciprocating machine — a modern packaged unit would run at 1,450 or 2,900 rpm on roughly a fifth of the displacement.
7. Part (e) — running cost of the heat pump. Isentropic shaft power and the electrical input it needs: $$W_s=\dot m_r w = 0.246(33.81)=8.32\ \text{kW},\qquad W_e=\frac{8.32}{0.87}=9.56\ \text{kW}$$ $$\text{cost}=9.56\times0.10=\boxed{\$0.96\ \text{per hour}}$$ The effective system COP, condenser heat over electricity bought, is $41.85/9.56 = 4.38$.
8. Compare with electric radiators, and comment. Resistance heating delivers the same 41.85 kW from 41.85 kW of electricity: $$\text{cost}_{\text{res}}=41.85\times0.10=\boxed{\$4.19\ \text{per hour}}$$ so the heat pump costs 4.4 times less to run, saving $3.23 an hour, or about $9,700 over a 3,000-hour Canadian heating season. Three qualifications belong with that number. First, the saving is bought with capital: a compressor, a ground loop and its boreholes or horizontal field, against a coil of wire. Second, the advantage is stable precisely because the source is the ground — at 5 °C year-round the evaporating temperature never collapses the way an air-source machine's does at −25 °C, which is the main argument for a ground loop in a Canadian climate. Third, the cycle here is ideal; a real machine with an isentropic efficiency near 0.70, superheat at the suction and pressure drops through the loop would land nearer a system COP of 3.0 to 3.5, still three times better than resistance. On carbon, the comparison depends entirely on the grid: in Quebec, Manitoba or British Columbia both options are near-zero-carbon and the argument is purely economic, while on a fossil-fired grid the heat pump's factor-of-four advantage in electricity is also a factor-of-four advantage in emissions.