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04-BS-10 · December 2019

Question 9 of 9: Heat Pump — Minimum Cost of Operation

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

Notes on this paper

National Examinations — December 2019, 04-BS-10, Thermodynamics. 3-hour closed-book exam (one 8.5×11 in. double-sided note sheet allowed); property tables/charts supplied, no interpolation required. Part A: two of three 20-mark questions; Part B: four of six 15-mark questions. Every question is answered in full.

Reference texts. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed. (Vapor Power Cycles Ch. 10; Gas Power Cycles Ch. 9; Refrigeration Cycles Ch. 11; Gas Mixtures Ch. 13; Psychrometrics Ch. 14; Exergy Ch. 8).

Questions 1, 3, 4, 6, 9 are fully legible and solved as printed. Question 2's specific numbers (pressure ratio, reheat pressure, temperatures, efficiencies) were reconstructed from surviving fragments that match a well-known textbook archetype (see Q2's callout). Questions 5, 7 and 8 have genuinely blank gaps in the source with no fully-confirming fingerprint; clearly-labelled representative values are substituted so every question can still be solved in full (see each question's own check callout).

Question 9: Heat Pump — Minimum Cost of Operation (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. $T_H=20^\circ$C (293.15 K), $T_L=5^\circ$C (278.15 K), $\dot Q_H=30{,}000$ kJ/h, electricity $\$0.08$/kWh.

Find. Minimum operating cost, $/h.

Approach

The Carnot COP between the indoor and outdoor temperatures bounds every real heat pump's performance; the MINIMUM work (hence minimum cost) corresponds to the Carnot (reversible) limit.

  1. Carnot COP and minimum power input. $$\text{COP}_{max}=\frac{T_H}{T_H-T_L}=\frac{293.15}{293.15-278.15}=\boxed{19.54}$$ $$\dot Q_H=\frac{30{,}000\ \text{kJ/h}}{3600\ \text{s/h}}=8.333\ \text{kW},\qquad \dot W_{min}=\frac{\dot Q_H}{\text{COP}_{max}}=\frac{8.333}{19.54}=\boxed{0.4264\ \text{kW}}.$$
  2. Minimum operating cost. $$\text{Cost}=\dot W_{min}\times\$0.08/\text{kWh}=0.4264\times0.08=\boxed{\$0.0341\ \text{per hour}\ (3.41\ \text{cents/h})}.$$
QuantityResult
COP$_{max}$19.54
$\dot W_{min}$0.4264 kW
Minimum cost$0.0341/h (3.41 cents/h)
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