Carnot Heat Pump — House Heating: 04-BS-10, December 2013, Q9

Question 9: Carnot Heat Pump — House Heating (15 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. Carnot heat pump, source (outdoor) $T_L=2\ ^\circ$C $=275.15$ K, sink (house) $T_H=20\ ^\circ$C $=293.15$ K. House heat-loss rate $\dot Q_{loss}=82{,}000$ kJ/h (constant). Compressor power while operating: $\dot W=8$ kW. Electricity price $0.085\ \$/\text{kWh}$.

Quantity	Value
$T_H,\ T_L$	293.15 K, 275.15 K
$\dot Q_{loss}$	82,000 kJ/h
$\dot W$ (while running)	8 kW
Price	8.5¢/kWh

Find. (a) Run time that day; (b) heat-pump electricity cost; (c) resistance-heat electricity cost.

Approach

The Carnot COP fixes how much heat the pump delivers per kW while running; since that delivery rate exceeds the (lower, average) house loss rate, the pump must cycle off part of the time, so total run time is the total daily heat demand divided by the pump's heat-delivery rate while on. Electricity cost follows from the actual energy consumed (power × run time), and the resistance-heating comparison uses the full daily heat demand at COP = 1.

Carnot COP. $$\text{COP}_{HP}=\frac{T_H}{T_H-T_L}=\frac{293.15}{293.15-275.15}=\frac{293.15}{18.0}=16.286.$$
Heat delivered while running, and required daily run time (part a). While operating, the pump delivers $\dot Q_H=\text{COP}\times\dot W=16.286\times8=130.29$ kW $=469{,}030$ kJ/h. The house needs $\dot Q_{loss}\times24\ \text{h}=82{,}000\times24=1{,}968{,}000$ kJ over the day: $$t_{run}=\frac{1{,}968{,}000\ \text{kJ}}{469{,}030\ \text{kJ/h}}=\boxed{4.196\ \text{h}}\ (4\ \text{h}\ 11.7\ \text{min}).$$
Heat-pump electricity cost (part b). Energy consumed: $$E=\dot W\times t_{run}=8\times4.196=33.57\ \text{kWh}.$$ $$\text{Cost}_{HP}=33.57\times0.085=\boxed{\$2.85}.$$
Resistance-heating cost (part c). A resistance heater converts electricity to heat with COP = 1, so it must supply the full daily demand directly as electrical energy: $$E_{res}=\frac{1{,}968{,}000\ \text{kJ}}{3600\ \text{kJ/kWh}}=546.67\ \text{kWh}.$$ $$\text{Cost}_{res}=546.67\times0.085=\boxed{\$46.47}.$$

Quantity	Result
(a) Run time	4.196 h (4 h 12 min)
(b) Heat-pump cost	$2.85
(c) Resistance-heat cost	$46.47

Check

Assumes the heat pump can cycle on/off freely with negligible start-up transients (standard idealization for this class of problem) and that outdoor/indoor temperatures stay exactly at 2°C/20°C for the full 24-hour period.

Concept & Theory

Topic. Carnot heat pump cycling analysis and electricity-cost comparison against resistance heating.

Key relations. Carnot heat-pump COP $=T_H/(T_H-T_L)$ (absolute temperatures); heat delivered while running $\dot Q_H=\text{COP}\times\dot W$; resistance heating has COP = 1 by definition (all electrical energy becomes heat).

Why this works. Because the Carnot COP (16.3) is fixed by the temperature limits alone and the compressor's rated power (8 kW) is also fixed, the heat-delivery rate while running is fixed too — and since it comfortably exceeds the house's average loss rate, the only way to match supply to demand on average is for the unit to run intermittently, which is exactly why part (a) asks for a run time rather than treating the pump as running continuously.

Common pitfall. Computing $\dot W\times 24\,\text{h}$ for the energy consumed (i.e., assuming the compressor runs the full day) instead of $\dot W\times t_{run}$ — this single error, order of ~5.7× too large, is the most common mistake on this exact problem type. The COP ratio $\text{Cost}_{res}/\text{Cost}_{HP}=46.47/2.85=16.3=\text{COP}_{HP}$ is a useful self-check.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 6 (The Second Law of Thermodynamics, Carnot heat pumps).

Question 9 of 9: Carnot Heat Pump — House Heating

Question 9: Carnot Heat Pump — House Heating (15 marks)

Approach