Question 4 of 8: Ground-water heat pump with R-134a (20 marks)

Question 4: Ground-water heat pump with R-134a (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. A simple vapour-compression cycle whose two saturation temperatures are fixed by the source approach and the stated delivery pressure.

Find. (a) the coefficient of performance, (b) the refrigerant mass flow, (c) the compressor swept volume in cm³, and (d) the hourly running cost against direct electric resistance heating.

Approach. The two saturation temperatures fix all four cycle states; the air-side duty then fixes the mass flow, the suction specific volume fixes the swept volume, and the compressor work divided by the drive efficiency gives the electrical demand to be priced.

1. Fix the two saturation temperatures. The evaporator must sit 14 K below the ground water for heat to cross: $$t_{1} = t_{w} - \Delta t = 4 - 14 = -10\ ^\circ\text{C} \quad\Longrightarrow\quad p_{1} = 200.6\ \text{kPa}$$ and the delivery pressure is the condensing pressure, which for R-134a corresponds to $$p_{2} = 1.0164\ \text{MPa} \quad\Longrightarrow\quad t_{\text{cond}} = 40.0\ ^\circ\text{C}$$ That the stated pressure lands exactly on a round 40 °C is the paper's way of confirming the reading; a condensing temperature 5 K above the 35 °C air leaving the coil is also physically necessary, so the data are self-consistent.
2. Read the four cycle states. State 1 is dry saturated vapour at $-10\ ^\circ\text{C}$; 1–2 is isentropic compression to 1.0164 MPa; state 3 is saturated liquid at 40 °C because there is no undercooling; and 3–4 is a throttle at constant enthalpy.

   Cycle state points (IIR datum: h = 200 kJ/kg for saturated liquid at 0 °C)

   State	Condition	p, MPa	t, °C	h, kJ/kg	s, kJ/kg·K
   1	dry saturated vapour	0.2006	-10.0	392.7	1.7334
   2	superheated discharge	1.0164	46.3	426.5	1.7334
   3	saturated liquid	1.0164	40.0	256.4	1.1900
   4	wet, after the throttle	0.2006	-10.0	256.4	dryness 0.338

   The suction specific volume, needed in part (c), is $v_{1} = 0.09959\ \text{m}^{3}/\text{kg}$.
3. Part (a) — the coefficient of performance. For a heat pump the useful output is the condenser duty and the input is the compression work: $$\text{COP}_{hp} = \frac{h_{2} - h_{3}}{h_{2} - h_{1}} = \frac{426.5 - 256.4}{426.5 - 392.7} = \frac{170.08}{33.81} = \boxed{5.03}$$ As a sanity check the reversed-Carnot limit between the same two saturation temperatures is $T_{\text{cond}}/(T_{\text{cond}} - T_{\text{evap}}) = 313.15/50 = 6.26$, so the cycle achieves 80 % of the ideal — high, but that is what a throttle-only, no-superheat cycle over a 50 K lift should give.
4. Get the air-side duty the condenser has to meet. The volume flow is quoted at the delivery state, so the density is evaluated there: $$\rho = \frac{p}{R\,T} = \frac{101\,325}{287.0 \times 308.15} = 1.1455\ \text{kg/m}^{3},\qquad \dot{m}_{a} = 0.8 \times 1.1455 = 0.9164\ \text{kg/s}$$ $$\dot{Q}_{\text{cond}} = \dot{m}_{a}\,c_{p}\,(t_{\text{out}} - t_{\text{in}}) = 0.9164 \times 1.005 \times (35 - 6) = \boxed{26.71\ \text{kW}}$$ If instead the 0.8 m³/s were read at the 6 °C inlet the duty would be 29.5 kW, 10 % higher; the delivery reading is taken because the question states the air is delivered at 35 °C at that rate.
5. Part (b) — refrigerant mass flow. Every kilogram of refrigerant rejects $h_{2} - h_{3}$ in the condenser, so $$\dot{m}_{r} = \frac{\dot{Q}_{\text{cond}}}{h_{2} - h_{3}} = \frac{26.71}{170.08} = \boxed{0.157\ \text{kg/s}} \;(9.42\ \text{kg/min})$$
6. Part (c) — swept volume of the compressor. The volumetric efficiency relates the induced volume to the volume swept per revolution; single acting means one induction stroke per revolution: $$\dot{V}_{1} = \dot{m}_{r}\,v_{1} = 0.157 \times 0.09959 = 0.01564\ \text{m}^{3}/\text{s}$$ $$V_{\text{swept}} = \frac{\dot{V}_{1}}{\eta_{\text{vol}}\,(N/60)} = \frac{0.01564}{0.85 \times 5.0} = 3.680 \times 10^{-3}\ \text{m}^{3} = \boxed{3{,}680\ \text{cm}^{3}}$$ A 3.7 litre displacement is large, but at only 300 rev/min it is what a slow-speed reciprocating machine of this duty needs — for reference, a bore and stroke of about 165 mm on a single cylinder.
7. Part (d) — the running cost. The shaft work follows from the same mass flow, and the drive efficiency converts it to electrical demand: $$\dot{W}_{\text{shaft}} = \dot{m}_{r}\,(h_{2} - h_{1}) = 0.157 \times 33.81 = 5.31\ \text{kW}, \qquad \dot{W}_{\text{elec}} = \frac{5.31}{0.87} = 6.10\ \text{kW}$$ $$\text{cost}_{hp} = 6.10\ \text{kW} \times 0.10\ \frac{\$}{\text{kWh}} \times 1\ \text{h} = \boxed{\$0.61\ \text{per hour}}$$
8. Part (d) — compare with electric radiators, and comment. Resistance heating delivers the same 26.71 kW of heat for 26.71 kW of electricity: $$\text{cost}_{\text{resistance}} = 26.71 \times 0.10 = \boxed{\$2.67\ \text{per hour}}$$ so the heat pump saves $2.06 per hour, 77 % of the bill, and the comparison is properly made on the overall coefficient of performance $\dot{Q}/\dot{W}_{\text{elec}} = 26.71/6.10 = 4.38$, not the 5.03 of the refrigeration cycle alone — the drive losses are real and have to be carried. Three qualifications belong with that number. First, it is an instantaneous figure at one operating point; a seasonal average is lower, because the load falls as the source stays fixed and the machine cycles. Second, the advantage here rests on a ground-water source at a steady 4 °C, which is exactly why ground-coupled machines are specified in Canadian climates: an air-source unit at a Canadian heating design temperature would be operating near its low-ambient cut-out with resistance back-up carrying the coldest hours at a COP of 1. Third, the capital cost of the well, the ground loop and a 3.7 litre compressor is very much larger than that of radiators, so the economic case is a life-cycle one; at a saving of $2.06 per hour of full-load operation and, say, 2,000 equivalent full-load hours a year, the annual saving is about $4,100 and a simple payback follows directly from the installed-cost difference. On the environmental side the answer depends on the grid: on the largely hydro-electric grids of British Columbia, Manitoba and Quebec the heat pump is unambiguously better on both energy and carbon, whereas on a fossil-heavy grid the primary energy comparison has to include the generating and transmission efficiency before the same conclusion can be drawn.

[Figure not reproduced: Problem 4: the cycle on the R-134a p-h diagram appended to the examination paper. 1-2 is isentropic compression, 2-3 de-superheating and condensation with no undercooling, 3-4 the throttle, 4-1 evaporation to dry saturated vapour. See the official exam paper.]