Calculators
Calculate control cabinet cooling
A control cabinet gives off part of its heat through the enclosure surface. How much depends on the effective area, and that depends on how the cabinet is installed. Only what exceeds this has to be removed actively.
Result
- Required cooling capacityIn the design case
- —W
- Cooling capacity in the normal case
- —W
- Air flow of a filter fanFree-blowing, in the design case
- —m³/h
- Effective surface areaTo IEC 60890
- —m²
- Self-dissipation of the enclosure
- —W
- Cooling assessment
- —
- Dew point of the ambient airAt maximum temperature and assumed humidity
- —°C
- Heating power against condensationTo hold the internal temperature at the dew point at standstill
- —W
- Governing criterion for the heating power
- —
- Condensation assessment
- —
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A guide, not a design to standard. We give no warranty for the correctness of the results. The applicable standards and case-by-case verification govern.
The grey values are an example. Simply enter your own.
Matching services
Calculating is the easy part.
A formula gives you a number. Designing a plant also demands installation method, grouping, discrimination, the standards in force and a look at the installed base. That is what we take on.
Method and questions
Enclosure cooling
Self-dissipation of the enclosure and required cooling capacity.
Discuss a projectWorked example
The default case the calculator starts with, worked through once. Change the values above and it recalculates immediately.
Inputs
- 2 × Variable-frequency drive 7.5 kW
- 390 W
- 2 × Variable-frequency drive 4 kW
- 220 W
- 1 × Power supply 24 V / 20 A
- 55 W
- 1 × Controller with modules
- 45 W
- 5 × Contactors and motor protection
- 40 W
- Heat dissipation inside the enclosure
- 750 W
- Width
- 1,200 mm
- Height
- 2,000 mm
- Depth
- 600 mm
- Installation
- Free-standing, all sides free
- Top surface
- Free
- Maximum internal temperature
- 35 °C
- Minimum internal temperature
- 5 °C
- Maximum ambient temperature
- 30 °C
- Expected ambient temperature
- 25 °C
- Minimum ambient temperature
- 5 °C
- Assumed relative humidity
- 60 %
Result
- Required cooling capacity
- 544 W
- Cooling capacity in the normal case
- 338 W
- Air flow of a filter fan
- 337 m³/h
- Effective surface area
- 7.49 m²
- Self-dissipation of the enclosure
- 206 W
- Cooling assessment
- The requirement is in the range of a cooling unit. Filter ventilation is no longer sufficient once ambient temperature approaches the permissible internal temperature.
- Dew point of the ambient air
- 21.4 °C
- Heating power against condensation
- 675 W
- Governing criterion for the heating power
- The dew point governs the heating power. More heating is needed against condensation than the minimum internal temperature demands.
- Condensation assessment
- Uncritical in operation, but not at standstill. If the enclosure cools to the minimum ambient temperature, it drops below the dew point. An enclosure heater with a hygrostat prevents the condensation.
Formula and standards
Formula
- Heat output Q = k · A · ΔT with k = 5.5 W/(m²·K) for painted sheet steel
- ΔT = internal temperature − ambient temperature
- Cooling requirement = heat dissipation − heat output
- Free-standing cabinet A = 1.8 · H · (W + D) + 1.4 · W · D
- With a covered top, the top area 1.4 · W · D drops out
- Dew point by the Magnus formula with α = ln(φ/100) + 17.62 · t / (243.12 + t), τ = 243.12 · α / (17.62 − α)
- Heating power against condensation P = k · A · (τ − minimum ambient temperature)
Assumptions and standards
- Effective surface to IEC 60890 depending on the installation type.
- k = 5.5 W/(m²·K) applies to painted sheet steel. Stainless steel is lower, aluminium higher.
- If the ambient is warmer than the permissible internal temperature, filter fans are ineffective, only a cooling unit will do.
- The air flow follows V = 3.1 · Q / ΔT. The factor 3.1 m³·K/(W·h) is the reciprocal of air density times specific heat capacity, converted to the hour, and applies at sea level. Density falls with altitude, so the factor rises.
- The figure is the free-blowing air flow. Filter mat, outlet grille and internal components create back pressure, so a fan delivers less than its free-blowing rating once installed. Select the unit by its air flow under actual back pressure.
- The design case governs, that is the warmest ambient assumed. At the expected ambient temperature the difference is larger and the required air flow considerably smaller.
- The covered top surface is simplified here, and the top area drops out entirely. IEC 60890 lists separate rows with their own factors for covered surfaces. The simplification errs on the safe side. It reduces the area and therefore increases the calculated cooling requirement.
- The heating power is the value needed to hold the enclosure at the dew point at standstill. An upper bound. In operation a hygrostat switches, and the heat from the installed components contributes.
- Solar gain, installation in recesses and dirty filters worsen the result in operation.
Where do I get the heat dissipation figure?
From the data sheets of the installed components. Drives, power supplies and transformers dominate. Contactors and terminals barely matter. As a rough figure, drives dissipate about 3 % of rated power.
How do I calculate enclosure heating?
Heating power follows from the effective enclosure surface and the temperature difference the heater has to hold, P = k · A · ΔT, counted from the minimum ambient temperature. The higher of two limits applies. One is the required minimum internal temperature, for displays or electronics for example. The other is the dew point, so that no condensation forms at standstill. The calculator deliberately does not offset the heat dissipation of the components, because it is gone at standstill. The result states which of the two limits determines the heating power.
Is a filter fan enough?
Only if the ambient temperature is below the permissible internal temperature. A fan cannot cool below ambient. Otherwise a cooling unit or an air-to-water heat exchanger is needed.
Why does the required air flow rise so steeply?
Because the temperature difference sits in the denominator. At 10 K it takes half the air of 5 K. At 2 K it takes two and a half times as much. As the ambient approaches the permissible internal temperature, the air flow tends to infinity. That is exactly where filter ventilation stops being an option.
Why does the installation matter so much?
A side placed against a wall or built into a row gives off hardly any heat. Between free-standing and enclosed on both sides there is easily a 40 % difference in effective area. The same applies to a covered top. A platform or cable floor resting on it takes the top area out of the calculation.
Why three ambient temperatures?
Because three different questions sit behind them. The cooling unit is selected for the maximum temperature, otherwise it falls short on the hottest day. The expected temperature says how much the unit actually runs. And the minimum temperature decides whether a heater is needed, the case most often overlooked.
What is the humidity for?
Temperature and relative humidity give the dew point. If the enclosure cools below that value at standstill, water condenses on the components. This is why corrosion appears in an unheated plant over winter even though cooling was adequate in summer.
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