Cold Room Refrigeration Load Calculation: Worked Example

To size a cold room refrigeration system, add up four heat loads over 24 hours—transmission through the panels, product cooling, air infiltration, and internal heat from fans, lights and people—add a safety margin of about 10%, then divide by the planned compressor running time, typically 16–20 hours per day. The result in kW is the capacity your condensing unit must deliver at the design evaporating temperature and ambient temperature.

This guide explains each load, works through a complete example for a +2 °C vegetable chiller, and then shows how the same method changes for a −18 °C frozen store.

What heat loads make up a cold room refrigeration load?

Every cold room calculation, from a 10 m³ restaurant walk-in to a 2,000 m³ distribution store, uses the same four components. Their proportions change significantly with the application, which is why a single "watts per cubic meter" figure is only suitable for rough budgeting.

These ranges apply to well-built prefabricated rooms. A room with frequent door openings or warm incoming product can fall well outside them.

What design data do you need before calculating?

• Internal dimensions (length × width × height) and panel thickness.
• Room temperature and humidity, for example +2 °C / 90% RH for vegetables or −18 °C for frozen food.
• Design ambient temperature around the room and at the condenser. Use the local summer design value, not the annual average: 35 °C is common, while 43–46 °C is realistic for many Gulf and Sahel sites.
• Daily product intake in kg/day, entry temperature and required pull-down time.
• Product type, which determines specific heat, latent heat and respiration heat.
• Door usage: number of openings, door size, and whether a strip curtain or air curtain is fitted.
• Power supply: 220 V single-phase, 380 V or 440 V three-phase, 50 or 60 Hz.

Worked example: how much refrigeration does a +2 °C vegetable chiller need?

Design basis for the example:

• Internal size 6.0 × 5.0 × 3.0 m, giving a volume of 90 m³
• Room temperature +2 °C at 90% RH; ambient 35 °C
• 75 mm polyurethane (PU) sandwich panels on walls, ceiling and floor, thermal conductivity λ = 0.022 W/m·K
• Product intake of 2,000 kg/day of mixed vegetables at 25 °C, cooled to +2 °C within 24 hours; stored stock of 8,000 kg
• Moderate door use with a PVC strip curtain

Step 1: Transmission load

Transmission heat is Q = U × A × ΔT, where the panel U-value is λ divided by the insulation thickness.

• U = 0.022 ÷ 0.075 = 0.293 W/m²·K
• Area: walls 2 × (6 × 3) + 2 × (5 × 3) = 66 m²; ceiling 30 m²; floor 30 m²; total 126 m²
• ΔT = 35 − 2 = 33 K
• Q = 0.293 × 126 × 33 ≈ 1.22 kW

If the room stands in direct sun, add 3–5 K to ΔT for exposed walls and roof. If the floor is uninsulated concrete on ground, calculate it separately using the ground temperature.

Step 2: Product load

Sensible heat removed from the product is Q = m × c_p × ΔT. Fresh vegetables have a specific heat of roughly 3.8–4.0 kJ/kg·K above freezing; this example uses 3.9.

• 2,000 kg × 3.9 kJ/kg·K × (25 − 2) K = 179,400 kJ/day
• Spread over the 24-hour pull-down: 179,400 ÷ 86,400 s ≈ 2.08 kW
• Respiration heat of the stored stock, because fresh produce keeps generating heat: 8,000 kg × about 0.03 W/kg ≈ 0.24 kW

Respiration rates vary widely—leafy greens and broccoli respire several times faster than potatoes or onions—so use published product data for single-crop stores.

Step 3: Infiltration load

The air-change method estimates how often door openings replace the room's air. Published reference tables give roughly 10–12 air changes per 24 hours for a 90 m³ room above 0 °C under average use; heavy traffic can double that.

• Outdoor air at 35 °C / 50% RH has an enthalpy of about 81 kJ/kg; room air at +2 °C / 90% RH about 12 kJ/kg, so Δh ≈ 69 kJ/kg
• 90 m³ × 11 changes × 1.2 kg/m³ × 69 kJ/kg ≈ 82,000 kJ/day
• 82,000 ÷ 86,400 ≈ 0.95 kW

Step 4: Internal loads

• Evaporator fans: 2 × 150 W running continuously ≈ 0.30 kW
• LED lighting: 6 W/m² × 30 m² for 8 h/day ≈ 0.06 kW averaged
• One person working 2 h/day at about 260 W ≈ 0.02 kW averaged
• Total ≈ 0.38 kW

Step 5: Add up, apply the safety factor and running time

The 18-hour running time leaves room for off-cycle defrost and for days hotter than design. Select a condensing unit that delivers at least 7.1 kW at an evaporating temperature of about −6 to −8 °C—a temperature difference of 8–10 K below the room—and 35 °C ambient. That works out to roughly 80 W per m³ for this room, a useful sanity check but not a substitute for the calculation.

How does the calculation change for a −18 °C freezer?

The method stays the same, but several inputs shift:

• Thicker panels. 120 mm PU (U ≈ 0.18 W/m²·K) is typical for −18 to −25 °C. Even so, the larger ΔT—53 K at 35 °C ambient—makes transmission the dominant load.
• Product usually enters frozen in storage applications, so product load is small. If unfrozen goods must be frozen in the room, include the latent heat of freezing (around 230–280 kJ/kg for fish and meat) and consider a dedicated blast freezer instead.
• Higher infiltration enthalpy difference, typically 90–100 kJ/kg, and the moisture that enters ends up as frost on the coil.
• Defrost heat. Electric defrost heaters add load; include it in internal loads or use a running time of 16–18 h/day.
• Lower evaporating temperature, about −25 to −28 °C, where compressor capacity is far lower than at −8 °C.

For the same 6 × 5 × 3 m room at −18 °C with 120 mm panels, transmission is about 0.183 × 126 × 53 ≈ 1.22 kW. Adding a modest frozen-product intake, infiltration, defrost and fans typically brings the requirement to around 4–6 kW at −25 °C evaporating.

What rules of thumb are safe for quick budgeting?

Rule-of-thumb figures help when a buyer needs a ballpark quotation before the full design data is ready. Replace them with a proper calculation before ordering equipment.

Per-m³ figures fall as rooms get larger because the surface-to-volume ratio drops. For rooms above about 500 m³, a per-m³ rule can overstate the required capacity substantially.

Which common mistakes lead to undersized or oversized systems?

1. Comparing capacity at the wrong conditions. Catalog capacity at −10 °C evaporating is much higher than at −30 °C. Ask for the figure at your evaporating and ambient temperatures.
2. Ignoring hot ambient. Air-cooled condensers lose capacity as ambient rises. In 45 °C climates, size the condenser and compressor for that temperature.
3. Underestimating product load. Loading warm product into a store designed for pre-cooled goods is a frequent reason rooms never reach set point.
4. Stacking safety factors. Several 20% margins on top of each other cause short cycling, poor humidity control and higher capital cost.
5. Forgetting the evaporator match. The unit cooler must deliver the same capacity at the chosen temperature difference. An undersized coil forces a lower evaporating temperature and heavier frosting.

How do you turn the load into an equipment list?

Once the required capacity and evaporating temperature are known, the equipment follows:

• A condensing unit rated for the capacity at design evaporating and ambient temperatures and matched to your voltage and frequency.
• One or more evaporators sized at the selected temperature difference, with fin spacing suited to the room temperature and an air throw that covers the room length.
• Expansion valve, solenoid valve, controller with defrost timer, and correctly sized copper piping.
• Insulated cold room panels and doors at the thickness assumed in the calculation.

Send us the room dimensions, temperature, product, daily intake and site conditions, and our engineers will run the full calculation and propose a matched package. Request a load calculation with your project details.

Frequently asked questions