Question 8 of 8: R-134a air-source heat pump

Question 8: R-134a air-source heat pump (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. An air-source heat pump using R-134a, heating 1000 ft³/min of recirculated house air through a 45 °F rise, with the cycle data tabulated.

Find. The system diagram and the cycle on the p-h chart, the coefficient of performance, the refrigerant mass flow, the hourly heating cost compared with electric radiators, and the limitations of the system in Ottawa.

Approach. Fix the four cycle states on the p-h chart, apply the isentropic efficiency to obtain the real compressor work, form the COP from condenser heat over compressor work, then scale to the air-side heating duty for the mass flow and the running cost.

1. Heating duty required. The maximum load is set by the air side: $$q_H = 1.10\,\dot V\,\Delta t = 1.10\,(1000)\,(45) = \boxed{49{,}500\ \text{Btu/h}}\ (14.5\ \text{kW})$$
2. State 1 — compressor suction. Dry saturated vapour at 30 psia. From the R-134a tables or the attached chart, $$t_1 = 15.4\ ^\circ\text{F}, \qquad h_1 = 105.3\ \text{Btu/lb}$$with enthalpies referred, as the chart is, to zero for saturated liquid at −40 °F.
3. State 2 — compressor discharge. Following the constant-entropy line from 1 up to 160 psia gives the ideal end point $h_{2s} = 120.3$ Btu/lb at 121.0 °F, so the isentropic work is $$w_s = h_{2s} - h_1 = 120.3 - 105.3 = 15.0\ \text{Btu/lb}$$The real compressor delivers the same pressure rise for more work, $$w = \frac{w_s}{\eta_s} = \frac{15.0}{0.90} = 16.67\ \text{Btu/lb} \;\Rightarrow\; h_2 = 105.3 + 16.67 = 122.0\ \text{Btu/lb}$$which on the chart is 127.3 °F, some 6 °F hotter than the isentropic end state — the irreversibility appears as extra superheat.
4. States 3 and 4 — condenser and expansion valve. With no undercooling the refrigerant leaves the condenser as saturated liquid at 160 psia, $$t_3 = 109.5\ ^\circ\text{F}, \qquad h_3 = 48.5\ \text{Btu/lb}$$The expansion valve is a throttle, so the process is isenthalpic and $h_4 = h_3 = 48.5$ Btu/lb, which at 30 psia is a two-phase mixture of quality 0.36 — over a third of the liquid flashes to vapour before it reaches the evaporator, which is why the evaporating effect is much smaller than the latent heat alone would suggest.
5. (a) Coefficient of performance. The useful output of a heat pump is the heat rejected in the condenser: $$q_{cond} = h_2 - h_3 = 122.0 - 48.5 = 73.5\ \text{Btu/lb}$$$$\mathrm{COP}_H = \frac{q_{cond}}{w} = \frac{73.5}{16.67} = \boxed{4.41}$$For comparison, a reversed Carnot cycle between the same saturation temperatures would give $T_3/(T_3 - T_1) = 569.2/(569.2-475.1) = 6.05$, so the real cycle achieves about 73% of the ideal — a reasonable figure given a 90% efficient compressor and a throttling expansion.
6. (b) Refrigerant mass flow. The condenser must deliver the air-side duty, so $$\dot m_r = \frac{q_H}{q_{cond}} = \frac{49{,}500}{73.5} = \boxed{674\ \text{lb/h}}\ (11.2\ \text{lb/min})$$
7. (c) Running cost. The shaft work absorbed is $$P_{shaft} = \dot m_r\,w = 674\,(16.67) = 11{,}230\ \text{Btu/h} = 3.29\ \text{kW}$$and the electricity drawn at the meter, after the 87% compressor-motor losses, is $$P_{elec} = \frac{3.29}{0.87} = 3.78\ \text{kW} \;\Rightarrow\; \text{cost} = 3.78\,(0.10) = \boxed{\$0.378\ \text{per hour}}$$Electric radiators would have to supply the whole 49,500 Btu/h directly, $$P_{res} = \frac{49{,}500}{3412} = 14.5\ \text{kW} \;\Rightarrow\; \text{cost} = \$1.45\ \text{per hour}$$so the heat pump costs 3.8 times less to run at this operating point. Expressed differently, its overall coefficient of performance measured at the meter is $49{,}500/(3.78 \times 3412) = 3.83$, against exactly 1.00 for the radiators.

(d) Limitations, and what to do in Ottawa

Every limitation of an air-source heat pump comes from the same source: the outdoor coil is the evaporator, so as the weather gets colder the evaporating pressure falls, the suction gas becomes less dense, and the compressor — a fixed-displacement machine — pumps less mass. Capacity therefore falls at exactly the moment the building load rises, and the two curves cross at the balance point, typically near −5 to 0 °C for equipment of this vintage. Below that point supplementary heat is required. At the same time the pressure ratio grows, the compressor discharge temperature rises toward the lubricant limit, and the COP falls; the 4.41 calculated above would degrade to below 2 at Ottawa design conditions. Frost is the second limitation: whenever the coil surface is below both freezing and the outdoor dew point, frost accumulates, blocks the air passages and insulates the fins, and periodic defrost cycles — usually by reversing the cycle and robbing heat from the house — cost both energy and comfort. Frosting is worst between about −5 and +5 °C, which in Ottawa is a large part of the season. Third, the supply air leaves the indoor coil at only 35 to 45 °C, cooler than skin temperature, so occupants standing in the draught report it as cold air even though the room is at set point. Fourth, sizing is a genuine conflict: equipment large enough to carry the winter load is grossly oversized for the summer cooling load and will short-cycle and dehumidify badly.

Ottawa has a January design temperature near −25 °C and roughly 4500 °C-days of heating, so the machine described would spend a large part of the winter below its balance point. Four measures are recommended. Select a cold-climate heat pump — an inverter-driven, variable-capacity machine with vapour or economiser injection, which maintains useful capacity and a COP above 2 to about −25 °C — rather than the single-speed unit implied here. Size it to the cooling load or a little above and arrange supplementary heat below the balance point, ideally as a dual-fuel system with a gas furnace locked out above the balance point, since Ontario electricity is expensive at the margin while the grid itself is low-carbon; where gas is unavailable, staged electric resistance with outdoor-temperature lockout is the fallback. Ensure a proper demand-defrost control based on measured coil pressure or fin temperature rather than a timer, and mount the outdoor unit clear of snow accumulation on a raised stand with a free-draining base. Finally, consider a ground-source heat pump: a vertical borefield in the Ottawa area sees an entering water temperature near 0 °C all winter and so holds a COP of 3 to 4 at design conditions with no frost and no defrost penalty at all, at the price of substantially higher capital cost. Whichever route is chosen, the equipment should be selected and rated to CSA/AHRI 210/240 at the Canadian heating standard rating conditions, not on a single mild-day COP.