22-Mec-B2 Environmental Control in Buildings · December 2016

Question 4 of 8: R-22 vapour-compression heat pump

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

Notes on this paper

Paper format. Professional Engineers of Ontario / Engineers Canada annual examination 07-Mec-B2 Environmental Control in Buildings, December 2016, three hours, open book. Eight problems of 20 points each; candidates are required to solve five, and all questions carry the same value. Psychrometric charts and an R-22 p-h diagram are appended to the paper. All eight problems are solved here.

Reference texts for this subject.

W. P. Jones, Air Conditioning Engineering, 5th ed., Butterworth-Heinemann — the standard reference for this examination code; Ch. 2–3 (psychrometry), Ch. 5–6 (heating and cooling loads), Ch. 15 (fans and duct design).
McQuiston, Parker & Spitler, Heating, Ventilating and Air Conditioning: Analysis and Design, 6th ed., Wiley — Ch. 3 (moist air), Ch. 5 (heat transmission in building structures), Ch. 8 (energy estimating and the degree-day method), Ch. 12–13 (fluid flow, fans and duct design).
ASHRAE Handbook — Fundamentals (2021) — Ch. 1 (psychrometrics), Ch. 14 (climatic design information), Ch. 18 (non-residential cooling and heating load calculations), Ch. 21 (duct design), Ch. 26 (heat, air and moisture control), Ch. 30 (thermophysical properties of refrigerants).
Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 9th ed., Wiley — Ch. 10 (vapour-compression refrigeration and heat pumps).
ANSI/ASHRAE Standard 62.1, Ventilation for Acceptable Indoor Air Quality; ANSI/ASHRAE Standard 55, Thermal Environmental Conditions for Human Occupancy.
Canadian frame: National Building Code of Canada 2020 and its Appendix C design temperatures; National Energy Code of Canada for Buildings 2020; Environment and Climate Change Canada degree-day normals for Toronto and Ottawa; CSA B52 Mechanical Refrigeration Code; Canada Green Building Council LEED Canada and the CaGBC Zero Carbon Building Standard for Problem 5.

Check: assumptions carried through this paper. Cover-page instruction 1 invites a clear statement of any assumption. Standard barometric pressure of 101.325 kPa is used throughout; moist-air properties follow the ASHRAE Handbook — Fundamentals Ch. 1 formulation (Hyland–Wexler saturation pressure, so results agree with the appended chart to chart-reading accuracy rather than being read off it); R-22 properties are on the IIR datum and agree with the appended p-h diagram. Problem-specific assumptions — the coil bypass factor, climate and degree-day data, fuel prices and equipment efficiencies, duct roughness, and the CLTD / SCL / CLF table entries — are stated where they are first used.

Question 4: R-22 vapour-compression 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. An air-source heat pump delivering 16 kW to a house at 20 °C from outdoor air at 4 °C, with R-22 leaving the evaporator as saturated vapour and the condenser as saturated liquid.

Given data and selected saturation states, Problem 4
Quantity	Symbol	Value
Heating duty	ṁ_H	16 kW
Indoor / outdoor temperature	t_i, t_o	20 °C / 4 °C
Evaporating temperature (selected)	t_e	−6 °C
Condensing temperature (selected)	t_c	40 °C
State 1, saturated vapour at t_e	h₁	402.77 kJ/kg
State 2, isentropic discharge at p_c	h₂	436.02 kJ/kg
State 3, saturated liquid at t_c	h₃	249.65 kJ/kg

Find. Suitable evaporator and condenser pressures, then the refrigerant mass flow in kg/min, the compressor power and the coefficient of performance.

Approach. Choose saturation temperatures that give the heat exchangers a workable temperature difference against the two air streams, read the corresponding pressures and the four cycle enthalpies, and then apply steady-flow energy balances to each component.

Select the saturation temperatures, and hence the pressures. The evaporator must draw heat out of 4 °C air and the condenser must reject it into a 20 °C house, so each coil needs a temperature difference to work against. Taking about 10 K on the source side and 20 K on the sink side — ordinary values for finned-tube air coils — gives $t_{e}=-6.0\ ^\circ\text{C}$ and $t_{c}=40.0\ ^\circ\text{C}$, and the R-22 saturation line then fixes$$\boxed{\ p_{e}=407.7\ \text{kPa},\qquad p_{c}=1,533.6\ \text{kPa}\ }$$a pressure ratio of 3.76, comfortably inside the single-stage range for a reciprocating or scroll machine.
Fix the four cycle states. State 1 is saturated vapour at $p_{e}$, so $h_{1}=402.77$ kJ/kg. Compression is taken as isentropic to $p_{c}$, giving $h_{2}=436.02$ kJ/kg at about 61 °C discharge. State 3 is saturated liquid at $p_{c}$, $h_{3}=249.65$ kJ/kg, and the expansion valve is a throttle, so$$h_{4}=h_{3}=249.65\ \text{kJ/kg}$$The specific duties follow immediately: $q_{\text{cond}}=h_{2}-h_{3}=186.37$ kJ/kg, $w=h_{2}-h_{1}=33.25$ kJ/kg and $q_{\text{evap}}=h_{1}-h_{4}=153.12$ kJ/kg, which satisfy $q_{\text{cond}}=q_{\text{evap}}+w$ exactly.
Mass flow, part (a). The condenser is the useful output, so it sets the flow:$$\dot{m}=\frac{\dot{Q}_{H}}{h_{2}-h_{3}}=\frac{16}{186.37}=0.08585\ \text{kg/s}$$$$\boxed{\ \dot{m}=5.1510\ \text{kg/min}\ }$$
Compressor power, part (b).$$\dot{W}=\dot{m}\,(h_{2}-h_{1})=0.08585\times33.25$$$$\boxed{\ \dot{W}=2.855\ \text{kW}\ }$$The evaporator therefore lifts 13.15 kW out of the outdoor air, and 13.15 + 2.855 returns the 16 kW delivered indoors, as it must.
Coefficient of performance, part (c).$$\mathrm{COP}_{\text{hp}}=\frac{\dot{Q}_{H}}{\dot{W}}=\frac{h_{2}-h_{3}}{h_{2}-h_{1}}=\frac{186.37}{33.25}$$$$\boxed{\ \mathrm{COP}=5.605\ }$$Against the reversible limit for the two air temperatures, $\mathrm{COP}_{\text{Carnot}}=T_{i}/(T_{i}-T_{o})=293.15/16=18.32$, the cycle achieves a second-law efficiency of 0.306. Almost all of that shortfall is the 30 K of heat-exchanger temperature difference deliberately chosen in step 1, not compressor irreversibility — which is why oversizing the coils is the cheapest way to raise seasonal performance.

[Figure not reproduced: The cycle on the R-22 p–h diagram appended to the examination paper. The condensing and evaporating pressures are the two horizontal legs; the throttle 3–4 is vertical because it is isenthalpic. See the official exam paper.]

Two design remarks belong with the numbers. First, R-22 is an HCFC and is no longer available for new equipment in Canada — the Ozone-depleting Substances and Halocarbon Alternatives Regulations (SOR/2016-137) ended imports for new systems — so a machine of this kind would today be built on R-410A or R-454B; the analysis is unchanged, only the property table moves. Second, the 16 kW duty at 4 °C says nothing about the design day: at the NBCC Appendix C 2.5 % winter temperature the same house would need substantially more heat while the machine would deliver substantially less, so supplementary heat and a balance-point analysis are part of any real selection.

Problem 4 — results
Quantity	Value
Evaporating temperature / pressure	-6.0 °C / 407.7 kPa
Condensing temperature / pressure	40.0 °C / 1,533.6 kPa
Condenser specific duty	186.37 kJ/kg
Compressor specific work	33.25 kJ/kg
(a) Refrigerant mass flow	5.1510 kg/min (0.08585 kg/s)
(b) Compressor power	2.855 kW
(c) Coefficient of performance	5.605
Evaporator duty	13.15 kW
Carnot COP / second-law efficiency	18.32 / 0.306

Topic. The ideal vapour-compression cycle run as a heat pump, and the selection of saturation temperatures.

Key relations. $h_{4}=h_{3}$ across the throttle; $\dot{Q}_{H}=\dot{m}(h_{2}-h_{3})$; $\dot{W}=\dot{m}(h_{2}-h_{1})$; $\mathrm{COP}_{\text{hp}}=(h_{2}-h_{3})/(h_{2}-h_{1})=\mathrm{COP}_{\text{ref}}+1$; $\mathrm{COP}_{\text{Carnot}}=T_{H}/(T_{H}-T_{C})$.

Why this works. Every component is a steady-flow device with one inlet and one outlet, so the first law reduces to an enthalpy difference in each case; the only modelling decisions are where to put the two saturation temperatures. Those choices are what actually determine the coefficient of performance, because the cycle can never do better than Carnot between the refrigerant temperatures, and every kelvin of heat-exchanger approach widens that lift. Working the problem from the condenser rather than the evaporator matters too: in heat-pump mode the condenser carries the specified duty, and sizing on the evaporator instead would leave the delivered heat short by exactly the compressor power.

Common pitfall. Setting the saturation temperatures equal to the air temperatures, which produces a physically impossible cycle with zero heat-exchanger area and a flattering COP near the Carnot value; and quoting the refrigeration COP where the heat-pump COP is asked for — the two differ by exactly one.

Source. Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 9th ed., Ch. 10; ASHRAE Handbook — Fundamentals (2021), Ch. 2 and Ch. 30; Çengel & Boles, Thermodynamics: An Engineering Approach, 9th ed., Ch. 11.

← Question 3 Next: Question 5 →