22-Mec-A1 Applied Thermodynamics and Heat Transfer · December 2019
Question 4 of 8: Freon-12 heat pump: COP and operating saving
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examination 16-Mec-A1 Applied Thermodynamics and Heat Transfer, 3 hours, open book. Eight questions of equal value: Part A — Thermodynamics (Q1–Q4) and Part B — Heat Transfer (Q5–Q8). A complete paper is any five questions (three from one part and two from the other).
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — ideal-gas processes and entropy generation, polytropic compression, rigid-vessel charging, the reciprocating air compressor, throttling/flash separation, the steam turbine, and vapour-compression refrigeration/heat-pump cycles; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — steady radial conduction through a cylindrical wall with convection, conduction with uniform internal generation, transient (lumped) cooling by combined convection and radiation, and the effectiveness–NTU method for shell-and-tube exchangers. Steam and Freon-12 (R-12) properties are evaluated, which reproduces the IAPWS steam tables and the standard R-12 property tables to graphing accuracy; enthalpy differences (the only quantities used) are datum-independent. Air and combustion gases are treated as ideal with constant specific heats ($\gamma=1.4$, $R=0.287\ \text{kJ/kg}\cdot\text{K}$, $c_p=1.005$, $c_v=0.718\ \text{kJ/kg}\cdot\text{K}$).
Question 4 — Freon-12 heat pump: COP and operating saving (Part A, equal value)
Find. Coefficient of performance as a heat pump and the running saving versus direct (resistance) heating.
Ideal R-12 cycle: 1 sat.-vapour compressor inlet (−10 °C) → 2 isentropic discharge → 3 sat.-liquid condenser exit (32.5 °C) → 4 throttle back to the evaporator. The house receives $Q_H=Q_L+W$.
Approach. Fix the four state enthalpies from the two saturation temperatures (sat. vapour in, isentropic compression to the condenser pressure, sat. liquid out, isenthalpic throttle); size the refrigerant flow from the evaporator load; then form $\dot W$, $\dot Q_H$, the heat-pump COP, and the saving against resistance heating.
State enthalpies (R-12 tables). $h_1=h_g(-10\ ^\circ\text{C})=348.3$, $s_1=1.5644\ \text{kJ/kg}\cdot\text{K}$; isentropic to $P_\text{cond}$: $h_2=371.2$; $h_3=h_f(32.5\ ^\circ\text{C})=231.6$; throttle $h_4=h_3=231.6\ \text{kJ/kg}$.
Refrigerant flow from the evaporator load.
$$\dot m=\frac{\dot Q_L}{h_1-h_4}=\frac{23.5}{348.3-231.6}=\boxed{0.201\ \text{kg/s}}$$
Compressor power and heat delivered.
$$\dot W=\dot m(h_2-h_1)=0.201(371.2-348.3)=4.61\ \text{kW},\quad \dot Q_H=\dot m(h_2-h_3)=0.201(371.2-231.6)=28.1\ \text{kW}$$
Heat-pump COP.
$$\text{COP}_\text{HP}=\frac{\dot Q_H}{\dot W}=\frac{28.1}{4.61}=\boxed{6.10}$$
(The reversible ceiling is $\text{COP}_\text{Carnot}=T_H/(T_H-T_L)=298/25=11.9$.)
Running saving versus resistance heating. Resistance heating would draw the full $\dot Q_H$ as electricity; the heat pump draws only $\dot W$. The saved electrical power equals the heat pulled free from ambient, $\dot Q_H-\dot W=\dot Q_L=23.5$ kW:
$$\text{saving}=(\dot Q_H-\dot W)\times\text{rate}=23.5\ \text{kW}\times 0.045\ \tfrac{\text{dollar}}{\text{kWh}}=\boxed{1.06\ \tfrac{\text{dollar}}{\text{h}}}\;(\approx 25.4\ \text{dollar/day})$$
Check — electricity price units.
The question prints "4.5 cents per kJ", which is physically absurd (it would price a kWh at 162 dollars). It is taken as 4.5 cents per kWh, the only sensible reading. Only the absolute saving scales with this figure; the COP and all state values are independent of it. R-12 enthalpies here use a different datum from the exam's appended R-12 table, but every result uses enthalpy differences, so it is unaffected.
