Chiller Energy Consumption and COP

COP is cooling capacity divided by electrical input at defined conditions. An illustrative unit delivering 100 kW while drawing 25 kW has a COP of 4; this is not a product performance claim.

Technical assessment

Use equal liquid and ambient conditions and the same measurement boundary. Include pump, fan and tower loads as appropriate to that boundary. Annual energy also requires operating hours and the load profile; nominal COP alone is insufficient.

Information needed for selection

Provide the application, operating hours, cooling load, entering and leaving liquid temperatures, fluid type and concentration, flow, site location and environmental conditions. A nominal capacity alone does not replace a detailed equipment selection.

Frequently asked question

Does this guide replace a product performance sheet?

No. It explains selection inputs and technical limits. Confirm the final model, capacity and options through a manufacturer selection at the actual operating conditions.

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Technical sources

ASHRAE — Liquid-Chilling Systems · US DOE — Electric chillers

How to calculate chiller COP and electricity use

COP = cooling capacity (kW) / electrical power (kW). For example, simultaneous values of 100 kW cooling and 25 kW electrical input give COP 4. This is an arithmetic example, not a performance claim for a Thrace model.

Energy depends on power and time: a constant average input of 25 kW for 8 hours consumes 200 kWh. Actual use varies with load and controls. Include pumps, fans and any cooling tower within the stated system boundary.