Library / Thermodynamics / THE-15
Thermodynamics
Ideal refrigeration coefficient of performance
Heat removed per unit of work for a perfect refrigerator.
- Ref
- THE-15
- Rev
- v1
- Field
- Thermodynamics
- Result
- COP
T_C is the temperature of the cold space and T_H the temperature of the surroundings it rejects heat to, both in kelvin and with T_H above T_C. The result is the heat removed per unit of work put in for a perfectly reversible refrigerator. No real machine can exceed it.
Heat removed per unit of work for a perfect refrigerator. The smaller the temperature gap, the more heat moved per unit of work.
Calculator
- Work needed per kW of cooling111.1 W
- COP if the temperature gap doubles4.5
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A 30 K gap at 270 K gives an ideal COP of 9: only 111 W of work moves 1 kW of heat. A real machine reaches about 30 to 60 % of that.
- Widening the gap is costly. Doubling it to 60 K halves the COP to 4.5, so keeping a store at −20 °C costs much more than at +5 °C.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing () | Local step | Change in result |
|---|---|---|---|---|---|---|
| TC Cold space temperature | 100 to 500 K | −23.5 | 21.5 | 45 | +10% | +1,000 % |
| TH Hot surroundings temperature | 150 to 600 K | −18 | 18 | 36 | +10% | −50 % |
Sweep of Hot surroundings temperature
| Hot surroundings temperature (K) | Coefficient of performance |
|---|---|
| 150 | −2.25 |
| 206.3 | −4.235 |
| 262.5 | −36 |
| 318.8 | 5.538 |
| 375 | 2.571 |
| 431.3 | 1.674 |
| 487.5 | 1.241 |
| 543.8 | 0.9863 |
| 600 | 0.8182 |
Sweep one variable
Variables and how each one contributes
Coefficient of performance
What the equation gives you. Everything else feeds it.
Cold space temperature
The temperature of the cold space, in kelvin. Raises the COP very strongly at this gap, because it closes the gap: a 10 % rise in T_C (to 297 K) shrinks the gap from 30 K to 3 K and multiplies the COP by eleven.
Hot surroundings temperature
The temperature of the surroundings, in kelvin. Lowers the COP because the gap widens. A 10 % rise in T_H (30 K) cuts the COP by about 50 %.
Worked example with the values above
- Temperature gap: TH − TC = 300 − 270 = 30 K.
- Divide the cold temperature by the gap: 270 / 30.
- COP = 9.
Assumptions and limits
- A reversible cycle, so the result is the upper limit.
- Constant temperatures at the cold and hot reservoirs.
Common mistakes
- Using °C. The formula needs kelvin: 270 K is −3 °C.
- Quoting the Carnot COP as what the machine will achieve.
Statistics and uncertainty
COP depends on the gap TH − TC, a difference of two similar numbers, so errors in either temperature are magnified. A 1 K error on a 30 K gap gives a 3 % error in COP; a 3 K error gives 10 %. Always use temperatures measured at the heat exchanger surfaces, not the room air.
Where it is used
- Judging cooling systems for electronics bays and stores.
- Setting a benchmark for a measured COP.
Maths you need
- Absolute temperature: kelvin
- Second law: reversible machines set the limit