Heat Pump COP
An air-to-air heat pump operates between -10°C on the cold side and 45°C on the hot side. What is the max theoretical efficiency?
Expand Hint
For a Carnot heat pump cycle, the Coefficient of Performance (COP) is:
$$$COP_c=\frac{T_H}{(T_H-T_L)}$$$
where
$$T_H$$
and
$$T_L$$
are the absolute high and low temperatures respectively in either Kelvin or Rankine.
Hint 2
To convert Celsius to Kelvin:
$$$K\:=\:^{\circ}C+273$$$
For a Carnot heat pump cycle, the Coefficient of Performance (COP) is:
$$$COP_c=\frac{T_H}{(T_H-T_L)}$$$
where
$$T_H$$
and
$$T_L$$
are the absolute high and low temperatures respectively in either Kelvin or Rankine. To convert Celsius to Kelvin:
$$$K=\:^{\circ}C+273$$$
Thus, the high and low temperatures are:
$$$K_H=45^{\circ}C+273=318\:K$$$
$$$K_L=-10^{\circ}C+273=263\:K$$$
Finally,
$$$COP_c=\frac{318K}{(318K-263K)}=\frac{318}{55}=5.78$$$
5.78
Time Analysis
See how quickly you looked at the hint, solution, and answer. This is important for making sure you will finish the FE Exam in time.- Hint: Not clicked
- Solution: Not clicked
- Answer: Not clicked
Similar Problems from FE Sub Section: Carnot Cycle
298. Carnot Cycle
454. Heat Pump
602. Carnot Fridge
Similar Problems from FE Section: Basic Cycles
249. Refrigeration Cycle
298. Carnot Cycle
454. Heat Pump
583. Heat Pump Cycle
586. Fridge Cycle
602. Carnot Fridge