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Cold room heat load basics: transmission, product, infiltration and internal loads at 43 °C

The refrigeration plant for a cold room is chosen from its heat load: the heat that leaks in or is carried in each day and must be pumped out. A proper calculation belongs to the refrigeration engineer, but knowing how it is built lets you check a quote and see which decisions drive cost. This guide explains the four main loads, gives the formulas with units, and works a small chiller at 43 °C.

7 min readUpdated

The four loads in one line each

  • Transmission: heat conducted through walls, ceiling and floor from the warmer surroundings.
  • Product: heat removed to bring incoming goods down to storage temperature, plus freezing if they are frozen, plus respiration for fresh fruit and vegetables.
  • Infiltration: warm, moist air that enters when the door opens and must be cooled and dried.
  • Internal: everything inside that gives off heat, such as lights, people, evaporator fan motors, pallet trucks and defrost heaters.

Transmission: Q = U × A × ΔT

Q is in watts, A is surface area in m², ΔT is the temperature difference in kelvin (one kelvin difference equals one °C difference), and U is the panel's heat transfer coefficient in W/m²·K. For a foam panel U ≈ k ÷ thickness, where k is the insulation's conductivity. India's National Horticulture Board cold storage standard lists PUF at about 0.023 W/m·K, so 100 mm of PUF gives U ≈ 0.023 ÷ 0.100 = 0.23 W/m²·K, before small corrections for surface films and joints.

Use the real temperature on the other side of each surface. A wall facing a kitchen at 43 °C, a roof under direct sun, and a floor on soil each need their own ΔT. Sun on a roof or wall can push the outer surface well above air temperature.

Product load: sensible, latent and respiration

Cooling above freezing: Q (kJ) = m × c × ΔT, with m in kg, c the specific heat in kJ/kg·K and ΔT in K. Foods high in water have c around 3.5 to 3.9 kJ/kg·K above freezing (approximate). Freezing adds latent heat of roughly the water fraction × 334 kJ/kg, and cooling below freezing uses a lower specific heat, roughly half the unfrozen value.

Divide kJ by 3,600 to get kWh. Fresh produce also respires and gives off heat for as long as it is stored, which matters in large fruit and vegetable stores.

Infiltration and internal loads

Infiltration: Q (kJ) = mass of air exchanged (kg) × enthalpy difference (kJ/kg). Enthalpy includes the moisture in the air, which is why humid monsoon air costs more to cool than dry air at the same temperature. Mass of air can be estimated from door openings or from an assumed number of room air changes per day.

Internal: every watt used inside the room ends up as heat. Lights, evaporator fan motors running round the clock and defrost heaters in freezers all count. A person working inside gives off roughly 250 to 400 W depending on how cold the room is and how hard they work (approximate).

Worked example: small chiller at +2 °C in a 43 °C space

Assumptions: internal 3.0 × 2.5 × 2.4 m (18 m³); 100 mm PUF on walls, ceiling and floor, U = 0.23 W/m²·K; walls and ceiling face 43 °C air; floor on ground assumed at 30 °C; 400 kg/day of product arriving at 30 °C, c = 3.7 kJ/kg·K; 20 air changes per day of air at 43 °C, 40% RH (enthalpy about 100 kJ/kg) replacing room air at +2 °C, 90% RH (about 12 kJ/kg), air density 1.25 kg/m³; 20 W of LED light for 8 h; one person for 1 h at 270 W; 80 W of evaporator fans running 24 h. Internal areas are used for simplicity.

Daily heat load for the example chiller (all inputs are assumptions)
LoadWorkingkWh/day
Walls + ceiling33.9 m² × 0.23 × 41 K = 320 W × 24 h7.67
Floor7.5 m² × 0.23 × 28 K = 48 W × 24 h1.16
Product400 kg × 3.7 × 28 K = 41,440 kJ ÷ 3,60011.51
Infiltration18 m³ × 20 × 1.25 kg/m³ × 88 kJ/kg = 39,600 kJ ÷ 3,60011.00
Lights, person, fans0.16 + 0.27 + 1.922.35
Total33.69
With 10% margin33.69 × 1.1037.06

From kWh per day to plant capacity

Plant is not sized to run 24 hours. It needs off-cycle time for defrost and spare capacity for recovery after loading. Designers commonly divide the daily load by about 16 to 20 running hours, depending on room temperature and defrost method; this is design practice, not a fixed rule. Here, 37.06 kWh ÷ 18 h = 2.06 kW of refrigeration, about 0.59 TR, required at the actual operating condition: evaporating temperature to suit +2 °C and condensing in 43 °C air.

Notice where the load comes from. Product and infiltration are each larger than transmission. A strip curtain, faster door habits and pre-cooling warm product can matter as much as thicker panels.

What to ask your installer

  • Show me the load sheet with each component, the ambient and the product intake you assumed.
  • At what evaporating and condensing temperatures is the selected unit's capacity stated, and does it meet the load at 43 °C?
  • How many running hours and what safety margin did you use?
  • How is defrost handled, and has its heat been included for a freezer room?

Frequently asked questions

Why design for 43 °C if summer outside is usually lower?
The condensing unit sits outdoors or in a plant space that can be hotter than the weather report, and the room may sit next to a kitchen. Sizing at a realistic worst case avoids rooms that cannot hold temperature on the hottest afternoons, exactly when they are under most pressure.
Is a 10% safety margin a standard?
No. It is a common allowance; the National Horticulture Board's worked design examples use a 10% factor, for instance. The right margin depends on how confident the designer is in the inputs, so ask what was used and why.

Sources

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