§ 18

Cold store: heat ingress through panels and from goods

How much heat should the evaporator remove from the panels and the new goods?

Answer: A cold room's cooling requirement consists of several items. The two largest in a small cold room are the heat leaking in through the panels and the heat from new goods that must be cooled down. The calculator works out the panel's U-value from thickness and insulation's λ, transmission across walls, ceiling and floor, and the average power needed to cool the day's goods down to 24 hours.

Formula

U = 1 ÷ (Rsi + d ÷ λ + Rse), Rsi + Rse ≈ 0,13 + 0,04 = 0,17 m²K/W (DS 418)

Transmission = U × A × (t_ude − t_rum)

Vareafkøling = m × c × (t_vare − t_rum) ÷ 86.400 s, når døgnets varer køles ned på 24 timer

Panel's steel plates have so little resistance they're not counted

Symbols and units in the formula
Symbol	Meaning	Unit
areal	Total area of walls, ceiling and floor	m²
tykkelse	Panel insulation thickness	mm
lambda	Insulation's λ (manufacturer's declared value)	W/(m·K)
t_rum	Temperature in cooler room	°C
t_ude	Temperature outside room	°C
varer	New goods per day	kg/døgn
cp_vare	Specific heat capacity of items	kJ/(kg·K)
t_vare	Temperature of items upon entry	°C
=	Total heat gain (transmission + goods)	W
=	Panel's U-value	W/(m²·K)
=	Transmission through the panels	W
=	Cooling of goods (spread over 24 hours)	W
=	Energy per day	kWh/døgn

Validity and source

Source: DS 418 (U = 1 ÷ (Rsi + d/λ + Rse), Rsi + Rse = 0,17 m²K/W) + Q = m·c·ΔT

Does not apply when

Only two of the items. Air changes at door opening, people, lighting, fans, defrosting and product respiration come on top and the evaporator is sized for operating time (typically 16-20 hours per day) and safety allowance. Floor to earth has a different temperature than outside air.

Worked example

The calculator's own default values inserted. The figures are calculated with exactly the formula the tool uses.

Substituted

Total area of walls, ceiling and floor60 m²
Panel insulation thickness100 mm
Insulation's λ (manufacturer's declared value)0.02 W/(m·K)
Temperature in cooler room2 °C
Temperature outside room25 °C
New goods per day500 kg/døgn
Specific heat capacity of items3.60 kJ/(kg·K)
Temperature of items upon entry12 °C

Giver

Total heat gain (transmission + goods)501 W
Panel's U-value0.212 W/(m²·K)
Transmission through the panels293 W
Cooling of goods (spread over 24 hours)208 W
Energy per day12.0 kWh/døgn

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Sådan bruger du den

Et kølerums kølebehov består af flere poster. Beregneren regner de to største i et lille kølerum: transmissionen gennem panelerne og afkølingen af nye varer.

Step by step

Mål arealet af alle panelflader.
Find panelets tykkelse og isoleringens λ.
Indtast temperaturer.
Indtast varemængden pr. døgn.
Aflæs varmeindfald og energi pr. døgn, og læg de andre poster til.

Typical errors

At glemme døråbning.
At bruge λ for ældre isolering.
At regne gulv mod jord som mod luft.

Examples from everyday life

Kølerum hos en slagter

Et kølerum på 2 × 3 × 2,4 m (ca. 60 m² flader) med 100 mm PUR-paneler. Rummet holder 2 °C, og omgivelserne er 25 °C. Der kommer 500 kg kød ind pr. døgn ved 12 °C.

Total area of walls, ceiling and floor: 60 m²
Panel insulation thickness: 100 mm
Insulation's λ (manufacturer's declared value): 0.022 W/(m·K)
Temperature in cooler room: 2 °C
Temperature outside room: 25 °C
New goods per day: 500 kg/døgn
Specific heat capacity of items: 3.6 kJ/(kg·K)
Temperature of items upon entry: 12 °C

It means: Det samlede indfald er 501 W. Panelerne har U = 0.212 W/(m²·K), og transmissionen er 293 W. Varerne giver 208 W. Det bliver 12 kWh/døgn.

Frostrum

Et frostrum med samme areal og 150 mm paneler holder −20 °C. Der kommer ingen nye varer ind i dette døgn.

Total area of walls, ceiling and floor: 60 m²
Panel insulation thickness: 150 mm
Insulation's λ (manufacturer's declared value): 0.022 W/(m·K)
Temperature in cooler room: -20 °C
Temperature outside room: 25 °C
New goods per day: 0 kg/døgn
Specific heat capacity of items: 3.6 kJ/(kg·K)
Temperature of items upon entry: 12 °C

It means: Transmissionen er 386 W med U = 0.143 W/(m²·K). Selv med tykkere paneler er den større end i kølerummet, fordi temperaturforskellen nu er 45 K.

Rule of thumb

100 mm PUR giver ca. 0,2 W/(m²·K).

How to check the answer

Transmission + varer = samlet.