Cooling Capacity Converter
Electricity Consumption
Equivalent Values
These units are often confused because some measure energy, some measure power, and some are specific to cooling. Getting them straight makes AC specifications much easier to read.
| From | To | Multiply by |
|---|---|---|
| BTU per hour | kW (cooling) | divide by 3,412.14 |
| kW (cooling) | BTU per hour | 3,412.14 |
| BTU per hour | Tons | divide by 12,000 |
| Tons | BTU per hour | 12,000 |
| Tons | kW (cooling) | 3.517 |
| kWh (energy) | BTU (energy) | 3,412.14 |
| Tons | BTU per hour | kW cooling | Approx. electricity at COP 3.5 |
|---|---|---|---|
| 0.75 | 9,000 | 2.64 kW | ~0.75 kW |
| 1.0 | 12,000 | 3.52 kW | ~1.0 kW |
| 1.5 | 18,000 | 5.28 kW | ~1.5 kW |
| 2.0 | 24,000 | 7.03 kW | ~2.0 kW |
| 2.5 | 30,000 | 8.79 kW | ~2.5 kW |
A spec sheet lists an AC as 18,000 BTU per hour with a COP of 4.0. Converting:
So a unit advertised as 18,000 BTU is a 1.5 ton AC that moves 5.28 kW of heat while drawing only about 1.32 kW of electricity, the gap being the efficiency benefit of the refrigeration cycle.
How many BTU are in 1 kWh?
1 kWh of energy equals 3,412.14 BTU. This is energy. AC ratings in BTU usually mean BTU per hour, which is power, not energy.
How do I convert BTU per hour to kW?
Divide BTU per hour by 3,412.14. A 12,000 BTU/hr AC is 3.52 kW of cooling; 18,000 BTU/hr is 5.28 kW. This is cooling power, not electricity drawn.
What is the difference between cooling kW and electrical kW?
Cooling kW is heat removed; electrical kW is power drawn. They differ because the AC moves more heat than the electricity it uses. At COP 3.5, it delivers 3.5 kW of cooling per 1 kW of electricity.
How much electricity does a 12,000 BTU AC use?
It delivers 3.52 kW of cooling. At COP 3.5 it draws about 1.0 kW, using about 1 kWh per hour. A less efficient unit at COP 2.8 draws about 1.26 kW.
How many BTU is 1 ton of cooling?
Exactly 12,000 BTU per hour, or 3.517 kW. So 1.5 tons is 18,000 BTU/hr and 2 tons is 24,000 BTU/hr.
Conversions are exact. Electricity consumption estimates depend on the efficiency (COP) you select and assume steady operation; real-world draw varies with conditions.