Water-source R-134a heat pump: 22-Mec-B2 Environmental Control in Buildings, December 2014, Q8

Question 8: Water-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. A water-source heat pump on the R-134a cycle, with the compressor inlet and outlet states and the condenser exit temperature all specified, serving a house whose heat loss is known.

Given data — Problem 8
State / quantity	Symbol	Value
House heat loss	$q_H$	70,000 Btu/hr
Source water temperature	$t_{\mathrm{well}}$	$45^\circ\text{F}$
Compressor inlet (1)	$p_1$ / $t_1$	30 psia / $20^\circ\text{F}$
Compressor outlet (2)	$p_2$ / $t_2$	120 psia / $140^\circ\text{F}$
Condenser exit (3)	$p_3$ / $t_3$	120 psia / $90^\circ\text{F}$
House temperature (for Carnot)	$t_H$	$70^\circ\text{F}$

Find. The compressor power, the heat absorbed from the well water, the actual and Carnot coefficients of performance, a comparison with resistance heating, and a discussion of ground-source heat pumps.

Problem 8 — the cycle on the HFC-134a pressure-enthalpy diagram attached to the paper. Compression 1→2 carries the vapour into the superheat region; the condenser 2→3 rejects heat to the house; the expansion valve 3→4 is isenthalpic; and the evaporator 4→1 takes heat from the well water.

Approach. Read the three enthalpies off the attached pressure-enthalpy diagram, close the mass flow from the condenser duty (which must equal the house loss), then obtain the compressor work, the evaporator duty and the coefficients of performance in turn.

Read the enthalpies from the attached chart. At 30 psia the saturation temperature is 15.4°F, so state 1 at 20°F carries 4.6°F of superheat and $h_1 = 106.3\ \text{Btu}/\text{lb}$. At 120 psia the saturation temperature is 90.5°F; state 2 at 140°F is well into the superheat region, $h_2 = 127.4\ \text{Btu}/\text{lb}$, and state 3 at 90°F is barely subcooled, $h_3 = 41.6\ \text{Btu}/\text{lb}$. The expansion valve is isenthalpic, so $h_4 = h_3$.
Close the mass flow on the condenser. The condenser must supply the whole house load, so $$\dot m = \frac{q_H}{h_2-h_3} = \frac{70{,}000}{127.4-41.6} = 816\ \text{lb}/\text{hr}$$ rejecting 85.8 Btu per pound of refrigerant circulated.
(a) Compressor power. The work of compression is $w = h_2-h_1 = 21.1\ \text{Btu}/\text{lb}$, so $$\dot W = \dot m\,(h_2-h_1) = 816(21.1) = 17{,}250\ \text{Btu}/\text{hr} = \boxed{5.06\ \text{kW}}$$
(b) Heat absorbed from the well water. The evaporator takes the refrigerant from state 4 to state 1, $$q_L = \dot m\,(h_1-h_4) = 816(106.3-41.6) = \boxed{52{,}750\ \text{Btu}/\text{hr}}$$ and the cycle closes: $52{,}750 + 17{,}250 = 70{,}000\ \text{Btu}/\text{hr}$, exactly the house load, which confirms the chart readings.
(c) Actual coefficient of performance. For a heat pump the useful output is the heat delivered, so $$\mathrm{COP} = \frac{q_H}{\dot W} = \frac{70{,}000}{17{,}250} = \boxed{4.06}$$
(c) Carnot coefficient of performance. Between the reservoirs the question names — well water at 45°F (504.7 R) and the house at 70°F (529.7 R) — $$\mathrm{COP}_{\mathrm{Carnot}} = \frac{T_H}{T_H-T_L} = \frac{529.7}{529.7-504.7} = 21.2$$ The actual cycle reaches only 19 % of this, which looks poor until one notices that the machine is not working between 45 and 70°F at all: it evaporates at 15.4°F and condenses at 90.5°F, and the Carnot value between those temperatures is 7.32. The real cycle achieves 55 % of that, a normal second-law efficiency. The gap between 21.2 and 7.32 is the cost of the heat-exchanger temperature differences, and it is where a designer would look first for improvement.
(d) Comparison with electric resistance heating. A resistance heater must supply the whole 70,000 Btu/hr as electricity, $$\dot W_{\mathrm{res}} = 70{,}000\ \text{Btu}/\text{hr} = \boxed{20.5\ \text{kW}}$$ against 5.06 kW for the heat pump — a factor of 4.06, the COP itself. In an average Canadian heating season the difference is on the order of 20,000 kWh for a house of this load, and it is the whole economic argument for the heat pump. Resistance heating remains attractive only for its negligible capital cost and as a supplementary source during the coldest hours.
(e) Ground-source heat pumps. This machine is a water-source heat pump using an open-loop groundwater source, and the reason it performs so much better than an air-source machine is visible in the numbers: it evaporates at 15.4°F against a 45°F source, whereas an air-source unit on the same January night would have to evaporate below a −20°C outdoor temperature, roughly halving the COP exactly when the load is greatest. Ground temperatures below about 6 m are stable at close to the local mean annual air temperature — 6 to 10°C across most of southern Canada — so the source is both warmer and steadier than the air, and the same loop provides efficient cooling in summer by rejecting heat to the ground. The costs are the ground loop itself, which typically doubles installed cost and dominates the payback; the need for adequate land or borehole depth; and, for an open-loop system like this one, groundwater quantity, quality and the provincial regulatory approvals for withdrawal and reinjection, along with fouling and corrosion risk in the evaporator. Closed vertical boreholes with a water-glycol loop avoid the water-quality problems at higher drilling cost. Long-term thermal balance also matters: a loop that is only ever used for heating gradually cools the ground around it, so in a heating-dominated Canadian climate the loop must be sized for the seasonal energy extracted, not merely for the peak load.

The cycle analysed here is a competent but unremarkable one, and the two Carnot comparisons are worth keeping side by side because they answer different questions. The reservoir-based figure of 21.2 states what a perfect machine could do with this source and this sink, and its distance from 4.06 measures everything lost in the whole installation, heat exchangers included. The cycle-temperature figure of 7.32 states what a perfect machine could do with the evaporating and condensing temperatures this equipment actually establishes, and its distance from 4.06 measures only the compressor's own irreversibility. Improving the first without improving the second means bigger heat exchangers and more pumping; improving the second means a better compressor.

Problem 8 — answers
Quantity	Symbol	Result
Enthalpies read from the chart	$h_1$ / $h_2$ / $h_3=h_4$	106.3 / 127.4 / 41.6 Btu/lb
Evaporating / condensing temperature	—	$15.4^\circ\text{F}$ / $90.5^\circ\text{F}$
Refrigerant mass flow	$\dot m$	816 lb/hr
(a) Compressor power	$\dot W$	17,250 Btu/hr = 5.06 kW
(b) Heat absorbed from the water	$q_L$	52,750 Btu/hr
(c) Heat-pump COP	COP	4.06
(c) Carnot COP (45 to 70°F reservoirs)	$\mathrm{COP}_C$	21.2
(c) Carnot COP (cycle temperatures)	—	7.32
(d) Resistance heater equivalent	$\dot W_{\mathrm{res}}$	20.5 kW — 4.06× the heat pump

Topic. Vapour-compression heat-pump cycle analysis on the pressure-enthalpy diagram, and second-law benchmarking against Carnot.

Key relations. $q_H = \dot m(h_2-h_3)$; $\dot W = \dot m(h_2-h_1)$; $q_L = \dot m(h_1-h_4)$ with $h_4 = h_3$; energy balance $q_H = q_L + \dot W$; $\mathrm{COP}_{\mathrm{hp}} = q_H/\dot W$; $\mathrm{COP}_{\mathrm{Carnot}} = T_H/(T_H-T_L)$ with absolute temperatures; $1\ \text{kW} = 3{,}412\ \text{Btu}/\text{hr}$.

Why this works. The pressure-enthalpy diagram turns the cycle into four line segments whose horizontal projections are the energy transfers per pound of refrigerant, so the entire analysis reduces to reading three enthalpies and closing one mass balance. Which duty fixes the mass flow depends on what the machine is for: in a heat pump the condenser serves the load, so $\dot m$ comes from $q_H/(h_2-h_3)$, whereas in a chiller it would come from the evaporator. The first law then supplies a free check, since the compressor work and the evaporator duty must sum to the condenser duty — if they do not, one of the chart readings is wrong.

Common pitfall. Computing the Carnot COP from the reservoir temperatures and then treating the ratio to the actual COP as the machine's efficiency. The reservoir figure includes all the heat-exchanger temperature differences and is always far above what any real cycle can reach; the honest comparison for compressor quality is against the Carnot value at the actual evaporating and condensing temperatures. A second common slip is using the evaporator duty as the useful output for a heat pump — that is the refrigeration COP, one less than the heating COP.

Source. Stoecker & Jones, Refrigeration and Air Conditioning, Ch. 10 (vapour-compression cycle) and Ch. 15 (heat pumps); ASHRAE Handbook — Fundamentals Ch. 2 and the HFC-134a property tables; Cengel & Boles, Thermodynamics: An Engineering Approach, Ch. 11.

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

Question 8: Water-source R-134a heat pump (20 points)