Question 5 of 8: Ground-source heat pump on R-134a — cycle, COP, refrigerant flow and running cost

Question 5: Ground-source heat pump on R-134a — cycle, COP, refrigerant flow and running cost (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.

Find. A system diagram with the cycle plotted on the p–h chart, the heating coefficient of performance, the refrigerant mass flow, the power input, and the hourly running cost compared with electric resistance radiators.

1. Part (a) — confirm that the stated data are self-consistent, then draw the system. Ground at 45 °F with a 30 °F approach puts the evaporating refrigerant at 45 − 30 = 15 °F, and the saturation temperature of R-134a at the stated 30 psia is 15.4 °F. The two independent statements agree to half a degree. The refrigerant therefore leaves the evaporator as saturated vapour at 15.4 °F and picks up 4.6 °F of suction-line superheat on its way to the compressor, reaching the stated 20 °F. Condensing at 200 psia corresponds to 125.3 °F, comfortably above the 90 °F air off the coil.
2. Fix the four cycle states. Properties are on the datum of the attached chart (h_f = s_f = 0 at −40 °F), the compression is taken as isentropic, and the condenser is taken to leave saturated liquid:

   State	Description	p (psia)	t (°F)	h (Btu/lb)
   1	compressor suction (superheated 4.6 °F)	30	20.0	106.27
   2	compressor discharge, isentropic	200	142.1	123.53
   3	condenser exit, saturated liquid	200	125.3	54.30
   4	after throttling, x = 0.422	30	15.4	54.30

   with s₁ = 0.2258 Btu/lb·°R and v₁ = 1.569 ft³/lb.
3. Part (b) — the heating coefficient of performance. Heat delivered per unit of compressor work is $$\text{COP}_{h} = \frac{q_{cond}}{w_{comp}} = \frac{h_2 - h_3}{h_2 - h_1} = \frac{123.53 - 54.30}{123.53 - 106.27} = \frac{69.23}{17.27}$$ $$\text{COP}_{h} = \boxed{4.01}$$ The reversed-Carnot limit between the same two saturation temperatures is 585.0/(125.3 − 15.0) = 5.30, so this cycle achieves 76 % of the ideal — a sound figure, the shortfall being the throttling loss and the superheat horn.
4. Size the condenser from the air side. The 3 200 CFM is stated at the delivery condition, so evaluate the density there: $$\rho_{90} = \frac{p}{R\,T} = \frac{14.696 \times 144}{53.35 \times 549.67} = 0.07216\ \text{lb/ft}^3$$ $$\dot m_{air} = 3\,200 \times 60 \times 0.07216 = 13\,856\ \text{lb/h}$$ $$\dot Q_{cond} = \dot m_{air}\,c_p\,\Delta T = 13\,856 \times 0.240 \times (90 - 45) = \boxed{149\,600\ \text{Btu/h}}\;=\;43.9\ \text{kW}$$ (Had the flow been metered at the 45 °F inlet instead, the duty would be 163 000 Btu/h; the wording ties the CFM to the delivery state, and the assumption is stated here.)
5. Part (c) — the refrigerant mass flow. The refrigerant must carry the condenser duty: $$\dot m_{ref} = \frac{\dot Q_{cond}}{h_2 - h_3} = \frac{149\,641}{69.23} = \boxed{2\,161\ \text{lb/h}} \;=\; 36.0\ \text{lb/min}$$ The compressor must therefore swallow 2 161 × 1.569/60 = 56.5 cfm of suction vapour, which at a volumetric efficiency near 0.80 calls for about 70 cfm of swept volume.
6. Part (d) — the power input. The compression work is $$\dot W_{comp} = \dot m_{ref}\,(h_2 - h_1) = 2\,161 \times 17.27 = 37\,320\ \text{Btu/h} = \boxed{10.94\ \text{kW}}$$ The evaporator then absorbs ṁ_ref(h₁ − h₄) = 112 300 Btu/h from the ground loop — 9.4 tons of refrigeration, which sets the borehole or trench length — and the first law closes on the machine: 112 300 + 37 320 = 149 600 Btu/h, exactly the condenser duty.
7. Part (e) — hourly cost, and the comparison with resistance radiators. Applying the 87 % combined compressor-and-motor efficiency to the shaft work gives the electrical draw: $$P_{elec} = \frac{\dot W_{comp}}{\eta} = \frac{10.937}{0.87} = 12.57\ \text{kW} \quad\Longrightarrow\quad \text{cost} = 12.57 \times 0.10 = \boxed{1.26\ \text{\$/h}}$$ Electric radiators would have to supply the whole 149 600 Btu/h as electricity, at 149 641/3 412 = 43.86 kW: $$\text{cost}_{resistance} = 43.86 \times 0.10 = \boxed{4.39\ \text{\$/h}}$$ The heat pump therefore costs 3.49 times less per hour to run — a saving of $3.13 per hour, or 71 % — because its effective coefficient of performance including motor losses is 4.01 × 0.87 = 3.49 against the radiator's 1.00.
8. Comment. Three things qualify that result and belong in the answer. First, the comparison is a site-energy comparison: on a hydro-dominated grid such as British Columbia's, Manitoba's or Quebec's, both options are close to carbon-free and the heat pump simply wins on cost, whereas on a fossil-fired grid the heat pump wins on emissions too, by roughly its coefficient of performance. Second, this is a ground-source machine, and that is what makes the number durable: the 45 °F source persists all winter, so the rated coefficient of performance is close to the seasonal one. An air-source unit at a Canadian design temperature would be below its low-ambient cut-out, and the resistance backup carrying the coldest hours would drag the seasonal figure far below 4.0. Third, the capital cost of the ground loop — the 9.4 tons of source-side capacity from step 6 — is what has to be paid against the $3.13/h operating saving; at 2 000 heating hours a year that is about $6 300 saved annually, which pays back a typical residential ground-loop premium in well under a decade.