07-Str-B6 · May 2018

Question 3 of 6: R134a Refrigeration Cycles — Ideal COP, Actual COP and COP Efficiency

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Exams, May 2018 — 07-Str-B6 Building Engineering and Services. Three hours, open book, one Casio or Sharp approved calculator. Six questions of equal value (20 marks each); five constitute a complete paper and only the first five appearing in the answer book are marked. All six are solved here. This subject contains no structural analysis at all — the cover page names it Building Engineering and Services, and every question is HVAC, building physics, acoustics or electrical services.

Reference texts (the books an open-book candidate should have on the desk for this subject):

ASHRAE, Handbook — Fundamentals, 2021 — ch. 1 (psychrometrics, and the chart reproduced on page 5 of this paper), ch. 25–27 (thermal resistance of assemblies, surface films, air spaces), ch. 8 (sound and vibration).
F. C. McQuiston, J. D. Parker & J. D. Spitler, Heating, Ventilating, and Air Conditioning: Analysis and Design, 6th ed. — ch. 3 (moist-air processes, coil bypass factor, reheat), ch. 5 (wall U-values), ch. 15 (refrigeration cycles).
Y. A. Çengel & M. A. Boles, Thermodynamics: An Engineering Approach, 9th ed. — ch. 11 (vapour-compression refrigeration, R134a tables), ch. 14 (gas–vapour mixtures).
D. Chadderton, Building Services Engineering, 6th ed. — room acoustics, site electrical distribution, ventilation systems for dwellings.
National Research Council Canada, National Building Code of Canada (NBCC) 2020, Part 6 and Subsection 9.32; National Energy Code of Canada for Buildings (NECB) 2020.
CSA C22.1, Canadian Electrical Code, Part I — Section 76 (temporary wiring on construction sites), Section 10 (grounding and bonding), Rule 26-700 (receptacle ground-fault protection).

Check — conventions adopted across this paper. (1) All psychrometry is worked at the 101.325 kPa sea-level barometric pressure printed on the supplied ASHRAE chart, using the standard moist-air relations rather than by scaling off the printed chart; the chart-read and calculated values agree to within the width of a pencil line, and calculating makes every number auditable. (2) Fan heat and duct gains are neglected, as the question intends: the supply-air state is taken as the state leaving the heating coil, and the return-air state as the room state. (3) In Question 3 the paper writes the second cycle as 4 → 1 → 2a → 3a → 4, reusing the label "4"; the state after throttling from 3a is not the same point as the state after throttling from 3, so it is called 4a here and the difference is exactly what changes the refrigerating effect. (4) Question 3 asks for "ideal COP" — taken as the Carnot COP between the stated evaporating and condensing temperatures, with the plotted vapour-compression cycle giving the "actual" COP and the ratio giving the COP efficiency. (5) Question 4 writes thermal conductivity in W/(m·°K); the degree sign on a kelvin is a typographic slip in the paper, and the units are read as W/(m·K). (6) Questions 2, 5 and 6 are answered in the Canadian frame — NBCC/NECB, CSA C22.1 and CSA/ANSI standards.

Question 3: R134a Refrigeration Cycles — Ideal COP, Actual COP and COP Efficiency (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 single-stage vapour-compression machine charged with R134a, evaporating at −20 °C in both cases and condensing at 40 °C in the first case and 60 °C in the second, with the cycle drawn on the supplied pressure–enthalpy chart as saturated vapour into the compressor, isentropic compression, condensation to saturated liquid, and isenthalpic throttling.

States read from the supplied P–h chart (datum: h = 0 for saturated liquid at −40 °C)
State	Description	P (kPa)	h (kJ/kg)
1	Saturated vapour leaving the evaporator at −20 °C	133	238.4
2	End of isentropic compression to the 40 °C saturation pressure (superheated, ≈ 49 °C)	1017	280.9
3	Saturated liquid leaving the condenser at 40 °C	1017	108.3
4	After throttling from 3 to the evaporator pressure	133	108.3
2a	End of isentropic compression to the 60 °C saturation pressure (superheated, ≈ 70 °C)	1682	291.6
3a	Saturated liquid leaving the condenser at 60 °C	1682	139.4
4a	After throttling from 3a to the evaporator pressure	133	139.4

Find. For each of the two condensing temperatures: the ideal (Carnot) coefficient of performance between the stated temperature limits, the actual coefficient of performance of the plotted vapour-compression cycle, and the ratio of the two expressed as a COP efficiency.

Figure 3.1 — the two cycles on the R134a P–h diagram. Both draw saturated vapour from the same −20 °C evaporator, so the compressor inlet state 1 is shared; raising the condensing temperature from 40 °C (red) to 60 °C (blue) pushes the discharge state further right and, because the liquid leaving the condenser is warmer, moves the throttled state from 4 to 4a — which is where the refrigerating effect is lost.

Approach. Read the four enthalpies of each cycle off the chart, form the refrigerating effect and the compressor work as enthalpy differences, divide to get the actual COP, compute the Carnot COP from the two absolute temperatures alone, and take the ratio.

Write the three cycle quantities as enthalpy differences. For a steady-flow vapour-compression cycle with negligible kinetic and potential energy, each component reduces to a simple enthalpy balance per kilogram of refrigerant circulated: $$q_L = h_1 - h_4 \quad(\text{evaporator}), \qquad w_{in} = h_2 - h_1 \quad(\text{compressor}), \qquad q_H = h_2 - h_3 \quad(\text{condenser})$$ The expansion device is a throttle, so it is isenthalpic and $h_4 = h_3$; that single fact is what makes the refrigerating effect depend on the condenser outlet temperature.
Evaluate cycle 1 (−20 °C to 40 °C). Substituting the chart reads, $$q_L = 238.4 - 108.3 = 130.1\ \text{kJ/kg}, \qquad w_{in} = 280.9 - 238.4 = 42.5\ \text{kJ/kg}$$ $$q_H = 280.9 - 108.3 = 172.6\ \text{kJ/kg} \quad \text{(check: } q_L + w_{in} = 130.1 + 42.5 = 172.6\ \checkmark\text{)}$$ so the actual coefficient of performance for cooling is $$\boxed{\mathrm{COP}_{actual,1} = \frac{q_L}{w_{in}} = \frac{130.1}{42.5} = 3.06}$$
Compute the ideal (Carnot) COP for the same temperature limits. The best any refrigerator can do between a cold reservoir at $T_L$ and a warm reservoir at $T_H$ is the reversed Carnot cycle, and its COP depends on the absolute temperatures alone: $$\mathrm{COP}_{ideal} = \frac{T_L}{T_H - T_L} = \frac{253.15}{313.15 - 253.15} = \frac{253.15}{60}$$ $$\boxed{\mathrm{COP}_{ideal,1} = 4.22}$$
Form the COP efficiency for cycle 1. The COP efficiency (the second-law or exergetic efficiency of the cycle) is the ratio of what the machine achieves to what a reversible machine between the same reservoirs would achieve: $$\eta_{COP,1} = \frac{\mathrm{COP}_{actual}}{\mathrm{COP}_{ideal}} = \frac{3.06}{4.22} = 0.726 \;\Rightarrow\; \boxed{\eta_{COP,1} = 72.6\%}$$
Repeat for cycle 2 (−20 °C to 60 °C), noting that state 4 moves. The compressor now discharges to 1682 kPa, and the liquid leaving the condenser is 20 K warmer, so throttling lands at 4a rather than 4: $$q_L = 238.4 - 139.4 = 99.0\ \text{kJ/kg}, \qquad w_{in} = 291.6 - 238.4 = 53.2\ \text{kJ/kg}$$ $$q_H = 291.6 - 139.4 = 152.2\ \text{kJ/kg} \quad \text{(check: } 99.0 + 53.2 = 152.2\ \checkmark\text{)}$$ $$\boxed{\mathrm{COP}_{actual,2} = \frac{99.0}{53.2} = 1.86}$$
Ideal COP and COP efficiency for cycle 2. With the same evaporating temperature but a 60 °C condenser, $$\mathrm{COP}_{ideal,2} = \frac{253.15}{333.15 - 253.15} = \frac{253.15}{80} = \boxed{3.16}$$ $$\eta_{COP,2} = \frac{1.86}{3.16} = 0.588 \;\Rightarrow\; \boxed{\eta_{COP,2} = 58.8\%}$$

The comparison is the real content of the question. Raising the condensing temperature by 20 K costs the machine 39% of its coefficient of performance, and it does so through two independent mechanisms working in the same direction. The compressor work rises 25%, because the pressure ratio increases from 7.7 to 12.7 and the discharge state moves further into the superheat region. At the same time the refrigerating effect falls 24%, because the liquid arriving at the expansion valve is 20 K warmer and therefore flashes to a higher vapour fraction on throttling, so less of each kilogram is left to evaporate usefully. The second effect is invisible on a temperature–entropy sketch of an idealised Carnot cycle and is the reason the COP efficiency also falls, from 72.6% to 58.8%: the throttling loss is an internal irreversibility that grows with the pressure difference across the valve.

The practical lesson for building services is that condensing temperature is worth money. A fouled air-cooled condenser, a blocked coil, a cooling tower running with insufficient airflow, or simply a head-pressure control set unnecessarily high, all raise the condensing temperature and are paid for continuously in compressor power. The same arithmetic run in reverse is the argument for water-cooled condensing, for free cooling, and for floating head-pressure control — and it is why NECB 2020 and ASHRAE 90.1 both include condenser and head-pressure control requirements rather than only minimum full-load efficiencies. Recovering the throttling loss (a two-stage system with a flash-tank economiser, or an expander in place of the valve) is the other half of the answer, and it is exactly the loss that grows fastest as the lift increases.

Final results — Question 3
Quantity	Cycle 1: −20 °C / 40 °C	Cycle 2: −20 °C / 60 °C
Refrigerating effect q_L	130.1 kJ/kg	99.0 kJ/kg
Compressor work w_in	42.5 kJ/kg	53.2 kJ/kg
Heat rejected q_H	172.6 kJ/kg	152.2 kJ/kg
Pressure ratio P_cond/P_evap	7.7	12.7
Ideal (Carnot) COP	4.22	3.16
Actual COP	3.06	1.86
COP efficiency	72.6%	58.8%

Concept & Theory

Topic. The ideal single-stage vapour-compression refrigeration cycle on a pressure–enthalpy diagram, and its second-law comparison with the reversed Carnot cycle.

Key relations. $q_L = h_1 - h_4$, $w_{in} = h_2 - h_1$, $q_H = h_2 - h_3$, $h_4 = h_3$ (isenthalpic throttling), $s_2 = s_1$ (isentropic compression); $\mathrm{COP}_{R} = q_L/w_{in}$; $\mathrm{COP}_{Carnot} = T_L/(T_H - T_L)$; COP efficiency $= \mathrm{COP}_{actual}/\mathrm{COP}_{ideal}$; energy balance $q_L + w_{in} = q_H$.

Why this works. The pressure–enthalpy diagram is chosen for refrigeration precisely because every process in the cycle becomes a straight line on it: evaporation and condensation are horizontal (constant pressure), throttling is vertical (constant enthalpy), and compression follows a plotted isentrope. The lengths of the horizontal segments read directly as the energy terms per kilogram, so the whole cycle analysis reduces to reading four numbers. The Carnot comparison is meaningful because both machines move heat between the same two temperatures; the gap between them is entirely accounted for by the cycle's internal irreversibilities — superheating the vapour during compression (the compressor discharges hotter than the condensing temperature and that superheat is rejected at a temperature above $T_H$), and the throttling process, which is inherently irreversible and generates entropy in proportion to the pressure drop.

Common pitfall. The dominant error on this question is reusing state 4 for both cycles because the paper's own cycle description writes "3a → 4". Throttling from 3a lands at $h = 139.4$ kJ/kg, not 108.3, and using the wrong value inflates the second COP by nearly a third. The second error is quoting the Carnot COP with Celsius temperatures. The third is computing $q_H$ as $h_2 - h_1$ rather than $h_2 - h_3$, forgetting that the condenser desuperheats as well as condenses — the energy balance check $q_L + w_{in} = q_H$ catches all three.

Source. Çengel & Boles, Thermodynamics: An Engineering Approach, 9th ed., ch. 11 §§11.2–11.4 (the ideal vapour-compression cycle, deviations from it, R134a property tables); ASHRAE Handbook — Fundamentals (2021), ch. 2 (thermodynamics and refrigeration cycles); McQuiston, Parker & Spitler, ch. 15.

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