Worked example
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Cooling COP, instantaneous
- Electrical input per ton of cooling
- kW/ton
- Equivalent EER
- Btu/Wh
- Cooling capacity
- kW
- Electrical input within the boundary
- 200 kW
Instantaneous, time-aligned sanity check only; this is not IPLV, NPLV or certified rating-point performance.
Underlined figures are rounded. Select one to copy its full value.
Method and assumptions
Formula
COP = cooling kW ÷ electrical input kW; kW/ton = electrical input ÷ refrigeration tonsOne refrigeration ton is 3.5168525 kW of cooling. EER = COP × 3.412141633 Btu/(Wh).
Limits of this result
- This is an instantaneous ratio, not IPLV, NPLV or a certified rating-point result.
- Cooling capacity and electrical input must cover the same interval and stated equipment or plant boundary.
- Compare performance only after accounting for load, temperatures, flow, fouling, controls and measurement uncertainty.
Technical sources
Put the value to work
Read the related engineering guidance, or carry this task into System Builder to identify the signal path and EpiSensor products.