22-Mec-B2 Environmental Control in Buildings · December 2013
Question 3 of 8: R-134a heat pump — COP, displacement and running cost
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 2013 — 3 hours, open book, non-communicating calculator permitted.
Eight problems of 20 points each; candidates answer any five.
ASHRAE psychrometric charts (SI and I-P) and an R-134a pressure–enthalpy diagram are
appended to the paper. All eight problems are solved below, because the set is
intended as a study resource.
Reference texts.
ASHRAE Handbook — Fundamentals (Ch. 1 Psychrometrics, Ch. 15 Fenestration,
Ch. 16 Ventilation and Infiltration, Ch. 18 Nonresidential Cooling and Heating Load Calculations,
Ch. 21 Duct Design);
McQuiston, Parker & Spitler, Heating, Ventilating, and Air Conditioning: Analysis and
Design, 6th ed.;
Çengel & Boles, Thermodynamics: An Engineering Approach, 9th ed. (R-134a
property tables);
ANSI/ASHRAE Standard 55 Thermal Environmental Conditions for Human Occupancy;
ANSI/ASHRAE Standard 62.1 Ventilation for Acceptable Indoor Air Quality.
Canadian practice: National Energy Code of Canada for Buildings (NECB) and CSA F280 where a
Canadian code reference is required.
Check — table-derived coefficients. This is an
open-book paper whose cover page instructs candidates to “submit a clear statement of the
assumption(s)” made. Where a solution needs a value that is read from a handbook table or
chart rather than given in the question — shading coefficients, air-space thermal
resistances, curtain-wall leakage rates, surface-averaged wind pressure coefficients, window and
door U-factors, CO2 generation rates — the value used is stated explicitly at the
point of use with its source. The method, not the table entry, is what the marks follow; a
different edition of the table shifts the arithmetic but not the answer structure.
Find. The cycle drawn on the p–h chart; the heating COP; the refrigerant mass flow;
the swept volume per revolution of the compressor; and the hourly running cost, compared against direct
electric resistance heating.
Figure 3.1 — System schematic (part a). Ground water at 5 °C feeds the evaporator at −10 °C; outside air at 10 °C is heated to 32 °C over the condenser at 40 °C.
Figure 3.2 — The cycle on the R-134a p–h chart (part a). 1–2 isentropic compression, 2–3 condensation to saturated liquid (no undercooling), 3–4 throttling, 4–1 evaporation to dry saturated vapour.
Approach. The two temperature constraints fix the cycle completely: the evaporator sits
15 °C below the 5 °C ground water, and the 1.0164 MPa delivery pressure is the saturation pressure
at 40 °C. Size the condenser from the air-side load, then work the cycle per kilogram to get COP, mass
flow, suction volume flow and compressor power.
Fix the cycle state points. The evaporator must be 15 °C below the 5 °C
ground water, so $t_1 = -10\ \text{°C}$ ($p_1 = 200.6$ kPa); the delivery pressure 1.0164 MPa is the
saturation pressure of R-134a at 40 °C, so $t_3 = 40\ \text{°C}$. Reading the attached p–h
chart (values from the standard R-134a tables on the same IIR datum):
$$h_1 = 392.7\ \text{kJ/kg},\quad s_1 = 1.7334\ \text{kJ/(kg}\cdot\text{K)},\quad
v_1 = 0.0996\ \text{m}^3/\text{kg}$$
$$h_2 = 426.5\ \text{kJ/kg}\ (t_2 = 46.3\ \text{°C, superheated}),\qquad
h_3 = h_4 = 256.4\ \text{kJ/kg}$$
State 2 is found by following the constant-entropy line from 1 up to 1.0164 MPa; state 3 is saturated
liquid because the question says there is no undercooling.
Size the condenser from the air side. The air is delivered at 32 °C and
101.325 kPa, so
$$\rho_a = \frac{p}{R_a T} = \frac{101\,325}{287 \times 305.15} = 1.157\ \text{kg/m}^3,
\qquad \dot m_a = 0.8 \times 1.157 = 0.926\ \text{kg/s}$$
$$\dot Q_{\text{cond}} = \dot m_a c_{pa}(t_{a2} - t_{a1}) = 0.926 \times 1.005 \times (32 - 10)
= \boxed{20.5\ \text{kW}}$$
Coefficient of performance (part b). Per kilogram of refrigerant the condenser
rejects $h_2 - h_3$ and the compressor absorbs $h_2 - h_1$:
$$\text{COP}_{hp} = \frac{h_2 - h_3}{h_2 - h_1} = \frac{426.5 - 256.4}{426.5 - 392.7}
= \frac{170.1}{33.8} = \boxed{5.03}$$
For comparison the Carnot limit between −10 °C and 40 °C is
$T_c/(T_c - T_e) = 313.15/50 = 6.26$, so the cycle achieves about 80% of the ideal — reasonable, the shortfall being
the throttling loss and the superheat horn.
Refrigerant mass flow (part c).
$$\dot m_r = \frac{\dot Q_{\text{cond}}}{h_2 - h_3} = \frac{20.5}{170.1}
= \boxed{0.120\ \text{kg/s}}$$
The corresponding evaporator duty is $\dot m_r(h_1 - h_4) = 0.120(392.7 - 256.4) = 16.4$ kW, which is the
heat the ground water must give up — the difference from 20.5 kW is exactly the 4.1 kW of compressor
work added to the refrigerant.
Swept volume (part d). The compressor inducts saturated vapour at state 1, so the
actual volume flow entering is $\dot m_r v_1$, and the swept volume rate is that divided by the volumetric
efficiency:
$$\dot V_{sw} = \frac{\dot m_r v_1}{\eta_v} = \frac{0.120 \times 0.0996}{0.85}
= 0.01410\ \text{m}^3/\text{s}$$
Single acting at 240 rpm means one induction stroke per revolution, i.e. 4 strokes per second, so
$$V_{sw} = \frac{0.01410}{240/60} = 3.524 \times 10^{-3}\ \text{m}^3
= \boxed{3524\ \text{cm}^3\ \text{per revolution}}$$
Running cost (part e). The indicated compressor power is
$\dot W = \dot m_r(h_2 - h_1) = 0.120 \times 33.8 = 4.07$ kW; with the combined compressor/motor
efficiency of 87% the electrical input is
$$\dot W_{\text{elec}} = \frac{4.07}{0.87} = 4.68\ \text{kW}
\;\Rightarrow\; \text{cost} = 4.68 \times \$0.11 = \boxed{\$0.51\ \text{per hour}}$$
Comparison with electric radiators, and comment (part e). Resistance heaters would
have to supply the same 20.5 kW of useful heat, at unity efficiency:
$$\text{cost}_{\text{resistance}} = 20.5 \times \$0.11 = \boxed{\$2.25\ \text{per hour}}$$
The heat pump costs about 4.4 times less to run, saving roughly $1.74 per hour of
operation. The physical reason is that the radiator converts one unit of high-grade electrical energy into
one unit of low-grade heat, whereas the heat pump uses that unit of work only to lift 4 units of
free ground heat from −10 °C to 40 °C. The practical caveats worth stating: the saving must
pay back a much higher capital cost plus the ground-water well and circulating pump (whose power is
neglected here); the seasonal average COP is lower than this design-point value; and the comparison
assumes the ground-water source stays at 5 °C, which requires an adequate aquifer or a properly sized
ground loop.
Check — property datum and stated assumptions. R-134a enthalpies are on the IIR datum used by the attached p–h chart ($h_f = 200$ kJ/kg for saturated liquid at 0 °C); COP, mass flow, swept volume and power are all differences of enthalpy and so are datum-independent. Isentropic compression, no pressure drop in the heat exchangers or lines, no undercooling and no superheat at compressor suction are assumed, as the question states. Fan and ground-water pump power are excluded from the cost comparison because the question specifies only the compressor/motor efficiency.
