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22-Mec-A1 Applied Thermodynamics and Heat Transfer · December 2014

Question 4 of 8: Freon-12 Heat Pump versus a Fuel Furnace

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

Notes on this paper

Open-book, 3-hour paper. Part A (Thermodynamics, Q1–Q4) and Part B (Heat Transfer, Q5–Q8); the rubric grades any five (three from one part, two from the other), all of equal value. All eight questions are solved in full. Freon-12 property values are read from the saturated and superheated tables printed in the exam appendix (pages 4–5). Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed.); Çengel & Ghajar, Heat and Mass Transfer (6th ed.); Incropera et al., Fundamentals of Heat and Mass Transfer (8th ed.).

Question 4: Freon-12 Heat Pump versus a Fuel Furnace (equal value)

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. Evaporator 5 °C (sat. vapour $h_1=189.518$, $s_1=0.6937$); condenser 50 °C (sat. liquid $h_3=84.868$, $P_\text{cond}\approx1.22\ \text{MPa}$); $\eta_c=0.85$; $\dot Q_H=30\ \text{kW}$; electricity \$0.077/kW·hr; furnace 70 % efficient, fuel \$5.60 per $10^6$ kJ.

Find. The lower-cost heating option.

Condenser (50°C)Evaporator (5°C)Compη=0.85valve2134Qᴱ = 30 kW (to building)Qₗ from outdoors
Figure 5 — Heat-pump loop. The condenser delivers the 30 kW heating duty; the compressor draws the paid electrical power $\dot W=\dot Q_H/\text{COP}$.

Approach. Build the vapour-compression cycle, get the isentropic then actual compressor work, form the heating COP, convert to electrical demand and hourly cost, and compare with the furnace fuel cost for the same 30 kW.

  1. Isentropic compression. From $s_1=0.6937$ to $P_\text{cond}\approx1.20\ \text{MPa}$, interpolating the superheated table between 50 °C and 60 °C gives $h_{2s}=210.77\ \text{kJ/kg}$.
  2. Actual work. $w=\dfrac{h_{2s}-h_1}{\eta_c}=\dfrac{210.77-189.518}{0.85}=\boxed{25.0\ \text{kJ/kg}}$; actual $h_2=214.5\ \text{kJ/kg}$.
  3. Heating COP. $q_H=h_2-h_3=214.5-84.868=129.6\ \text{kJ/kg}$; $\text{COP}_\text{hp}=\dfrac{q_H}{w}=\dfrac{129.6}{25.0}=\boxed{5.19}$.
  4. Heat-pump cost. Electrical power $\dot W=\dot Q_H/\text{COP}=30/5.19=5.78\ \text{kW}$; cost $=5.78\times0.077=\boxed{\$0.445/\text{hr}}$.
  5. Furnace cost. Fuel energy $=\dfrac{30\times3600}{0.70}=1.543\times10^{5}\ \text{kJ/hr}$; cost $=1.543\times10^{5}\times5.60\times10^{-6}=\boxed{\$0.864/\text{hr}}$.

The heat pump costs about \$0.45/hr against the furnace's \$0.86/hr — roughly half — so the heat pump is the economic choice. Its advantage comes from moving 30 kW of heat while paying for only 5.8 kW of electricity, whereas the furnace must burn more than the delivered heat to cover its 30 % loss.

Check: The comparison assumes the paid electrical input equals the compressor shaft work (motor and drive losses neglected). Including a typical 90 % motor efficiency would raise the heat-pump cost to about \$0.49/hr — still well below the furnace.
QuantityResult
Actual compressor work≈ 25.0 kJ/kg
Heating COP≈ 5.19
Heat-pump electrical demand≈ 5.78 kW
Heat-pump operating cost≈ \$0.45/hr
Furnace operating cost≈ \$0.86/hr
RecommendationHeat pump (≈ 48 % cheaper)