A cold store's electricity comes from its heat load. The system runs to remove the heat that enters, and the electricity it uses is that heat divided by the refrigeration efficiency (COP). A compressor's kW rating is power at a moment; the bill comes in kWh, power multiplied by running hours.
Daily consumption comes out in four steps. The figures are an example: a 6 × 5 × 4 m room, 80 mm polyurethane panels (U = 0.28 W/m²K), 30 °C outside, 1 °C inside.
- Wall and roof transfer: U × area × temperature difference. 0.28 × 113 m² × 29 K × 24 hours ÷ 1000 ≈ 22 kWh of heat a day, about 24 with the floor.
- Product load: mass × specific heat × temperature drop. Two tonnes a day entering at 10 °C and brought down to 1 °C (specific heat 3.6 kJ/kg K) add about 18 kWh a day.
- Add door openings, people, lighting, forklifts and defrost; these depend on the operation, 8 kWh a day in the example.
- Divide the total heat by the COP: 50 kWh of heat a day at a COP of 2.5 means about 20 kWh of electricity a day.
The calculation shows how much the product load weighs; in the source's typical breakdown, product makes up more than half of the total load. Shortening door-open time, fixing seals and receiving product pre-cooled, close to target temperature, are the first steps and need no investment.
In an existing store, fit a separate meter to the refrigeration plant and log daily kWh with ambient and product entry temperatures. For a new store, get the contractor's heat-load calculation with its assumptions in writing and compare the bills against it.
