The structure of a house is such that it loses heat at a rate of 3800 kJ/h per °C difference between the indoors and outdoors. A heat pump that requires a power input of 4 kW is used to maintain this house at 24°C. Determine the lowest outdoor temperature for which the heat pump can meet the heating requirements of this house.
The correct answer and explanation is :
Answer:
The lowest outdoor temperature for which the heat pump can meet the heating requirements is −20°C.
Explanation:
A heat pump extracts heat from a cold environment and delivers it to a warmer space. The efficiency of the heat pump is measured using the Coefficient of Performance (COP) for heating, given by:
[
COP = \frac{Q_{in}}{W_{in}}
]
where:
- ( Q_{in} ) = Heat delivered to the house (kJ/h)
- ( W_{in} ) = Power input to the heat pump (kJ/h)
Step 1: Heat Loss Calculation
The rate of heat loss from the house is given as:
[
Q_{\text{loss}} = 3800 \times (T_{\text{in}} – T_{\text{out}})
]
Given:
- Indoor temperature ( T_{\text{in}} = 24^\circ C )
- Outdoor temperature ( T_{\text{out}} ) (to be determined)
- Heat pump power input ( W_{\text{in}} = 4 \text{ kW} = 14400 \text{ kJ/h} )
At steady-state, the heat delivered by the heat pump must equal the heat loss:
[
Q_{\text{in}} = Q_{\text{loss}}
]
Step 2: COP of a Reversible Heat Pump
For an ideal (Carnot) heat pump, the COP is given by:
[
COP_{\text{max}} = \frac{T_{\text{hot}}}{T_{\text{hot}} – T_{\text{cold}}}
]
where temperatures must be in Kelvin:
[
T_{\text{hot}} = 24 + 273 = 297K, \quad T_{\text{cold}} = T_{\text{out}} + 273
]
Thus,
[
COP_{\text{max}} = \frac{297}{297 – (T_{\text{out}} + 273)}
]
[
Q_{\text{in}} = COP_{\text{max}} \times W_{\text{in}}
]
Step 3: Solving for ( T_{\text{out}} )
Setting ( Q_{\text{in}} = 3800 \times (24 – T_{\text{out}}) ):
[
\frac{297}{297 – (T_{\text{out}} + 273)} \times 14400 = 3800 \times (24 – T_{\text{out}})
]
Solving, ( T_{\text{out}} = -20^\circ C ).