Answer:
The lowest outdoor temperature for which the heat pump can meet the requirement of the house is -9.6°C
Explanation:
The coefficient of performance of a heat pump, COP, is given by the relation;
[tex]COP_{hp} = \dfrac{T_2}{T_2 - T_1} = \dfrac{Q_1\times (T_2 - T_1)}{\Sigma W}[/tex]
The rate of heat loss [tex]\dot Q_1[/tex] = 3800 kJ/h = 3800 ×1/60 × 1/60 kJ/s = 1.06 kJ/s
The power input = 4 kW = 4 kJ/s
The temperature indoors = 24 + 273.15 = 297.15 K
Therefore, we have;
[tex]COP_{hp} = \dfrac{297.15}{297.15- T_1} = \dfrac{1.06 \times (297.15- T_1)}{4}[/tex]
4×297.15= (297.15- T₁) × 1.06×(297.15- T₁)
Which gives;
18/19·T₁²-627.32·T₁+92014.97=0
(T₁ -263.59)·(T₁ - 330.71) = 0
The solutions are;
T₁ = 263.59 K or T₁ = 330.71 K
The lowest temperature = 263.59 K which is -9.6°C
The lowest temperature = -9.6°C.
