- 2068 Bhadra · 8 marks
Heat pump at 50% of ideal: minimum surrounding temperature
Short answer
Using actual COP = 0.5 × 293.15/ΔT and heat load 0.8ΔT kW with power 1.5 kW, solve 0.8ΔT = (146.6/ΔT)×1.5 → ΔT ≈ 16.6 K. Minimum surrounding temperature is about 3.4°C.
A heat pump having a coefficient of 50% of the theoretical maximum maintains a house at a temperature of 20°C. The heat leakage from the house occurs at a rate of 0.8kW per degree temperature difference. For a maximum power input of 1.5kW, determine the minimum surrounding temperature for which the heat pump will be sufficient?
Given: house at T_H = 20 °C = 293.15 K, COP = 0.5 × COP_max, heat leakage 0.8 kW per °C temperature difference, maximum power W = 1.5 kW. Let the outside temperature be T_L = 293.15 − ΔT.
Step 1: COP of the heat pump
COP_max = T_H/(T_H − T_L) = 293.15/ΔT COP_actual = 0.5 × 293.15/ΔT = 146.6/ΔT
Step 2: Heat that must be supplied
Heat leaks out at Q_H = 0.8 ΔT kW, so the heat pump must supply Q_H = 0.8 ΔT kW.
Step 3: Use the power limit
Q_H = COP × W
0.8 ΔT = (146.6/ΔT) × 1.5 ΔT² = 146.6 × 1.5/0.8 = 274.8 ΔT = 16.58 K
Step 4: Minimum surrounding temperature
T_L = 20 − 16.58 = 3.4 °C
Answer: the heat pump is sufficient only if the outside temperature is at least about 3.4 °C. Check: Q_H = 0.8 × 16.58 = 13.26 kW, COP = 8.842, W = 1.50 kW ✓.
✓ Answer checked by Claude Opus 5.5 (high effort)
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