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Selecting a cabinet cooling system without calculating the actual heat load is one of the most common mistakes in electrical enclosure design.
In many projects, cooling capacity is selected according to cabinet dimensions, installed drive power, or previous experience. These methods may provide a rough starting point, but they do not reflect the actual thermal conditions inside the enclosure.
A 30 kW variable frequency drive does not necessarily release 30 kW of heat into the cabinet. Likewise, a large enclosure does not automatically require a large air conditioner. The correct cooling capacity depends on equipment power losses, cabinet surface area, ambient temperature, target internal temperature, installation conditions, and environmental contamination.
This guide explains how engineers calculate electrical cabinet heat load, estimate passive heat dissipation, and determine the cooling capacity required for filter fans, heat exchangers, and enclosure air conditioners.
After understanding the heat load calculation process, engineers can further evaluate the right cabinet cooling method based on environmental conditions and installation requirements.
The first step in cabinet thermal design is understanding that almost all electrical energy consumed inside a cabinet eventually becomes heat.
For example:
A variable frequency drive rated at 2 kW does not transfer all electrical energy into mechanical output. A portion of the energy is lost through semiconductor switching losses and becomes heat inside the enclosure.
Common Sources of Cabinet Heat Load
| Heat Source | Example |
|---|---|
| Power conversion losses | VFD, servo drive, inverter |
| Control electronics | PLC, IPC, communication modules |
| Power supplies | 24VDC supply systems |
| Passive components | Transformers, braking resistors |
| External environment | Solar radiation, high ambient temperature |
The total cabinet heat load usually includes:
1.1 Internal Heat Generation
Internal heat sources are the most important factor.
Typical heat-generating components include:
| Component | Typical Heat Loss |
|---|---|
| PLC power supply | 10–50 W |
| Industrial PC | 50–150 W |
| Network switch | 10–40 W |
| Servo drive | 5–15% of rated power |
| Variable frequency drive | 2–5% of rated power |
| Transformer | Depends on load efficiency |
For example:
A control cabinet contains:
Total internal heat:
Qv = 40 + 100 + 600 + 50 + 30
Qv = 820 W
This means the cabinet must remove approximately 820 watts of heat under normal operating conditions.
The basic cabinet cooling calculation follows a heat balance principle:
Required Cooling Capacity
Qcooling=Qv+Qs
Where:
The internal heat load is usually the dominant factor.
For many industrial cabinets:
Qv≈∑Power Losses
Every component's actual heat dissipation should be considered instead of simply using its rated power.
For example, a 5 kW motor drive does not dissipate 5 kW of heat inside the electrical cabinet. The actual heat generated depends on the drive efficiency and operating conditions.
Assume the drive operates at 97% efficiency:
5,000×(1−0.97)=150W
This means the drive contributes approximately 150 W of heat load to the enclosure under this operating condition.
In real cabinet cooling design, engineers should calculate heat generated by component losses rather than simply adding the rated power of all installed devices. Using equipment nameplate power as the cooling load often leads to unnecessary oversizing, higher energy consumption, and increased equipment cost.
After calculating the total heat load, engineers can select an appropriate cabinet cooling method, such as filter fans, air conditioners, or air-to-air heat exchangers depending on the environmental conditions.
Not all heat generated inside an electrical cabinet requires active cooling.Understanding heat generated inside an electrical cabinets helps engineers determine whether natural heat dissipation is sufficient or additional cooling is required.
A portion of heat naturally escapes through the cabinet walls.
The natural heat dissipation depends on:
For steel cabinets, the heat transfer coefficient is commonly around:
k = 5.5 W/m²K
The heat transferred through the enclosure surface can be estimated as:
Qs=A×k×ΔT
Where:
Project Background
In many industrial automation projects, similar thermal issues are discovered during commissioning when cabinets operate under continuous production conditions.
A machine builder designed a CNC control cabinet for an automated production line.
Cabinet specification:
Installed components:
| Equipment | Heat Loss |
|---|---|
| CNC controller | 120 W |
| Servo drives | 650 W |
| PLC and I/O modules | 80 W |
| Power supply | 60 W |
| Industrial Ethernet devices | 40 W |
Total internal heat:
Qv=120+650+80+60+40=950 W
Step 1: Calculate Natural Heat Dissipation
Cabinet surface area:
Approximately:
A = 6 m²
Temperature difference:
ΔT=45−35=10 K
Natural heat dissipation:
Qs=6×5.5×10=330 W
Step 2: Calculate Required Cooling
The cabinet already releases approximately 330 W naturally.
Therefore:
Qcooling=950−330=620 W
The required cooling capacity is approximately 620 W.
In practice, engineers usually add a safety margin to account for ambient temperature changes, future expansion, and continuous operation.
Recommended cooling capacity:
This ensures reliable cabinet operation without unnecessary oversizing.
Based on the calculated load, the cooling method depends on the installation environment.
For example:
Option 1: Filter Fan
Suitable when:
Option 2: Air-to-Air Heat Exchanger
Suitable when:
Option 3: Enclosure Air Conditioner
Required when:
For outdoor electrical cabinets, internal heat generation is not the only factor that affects the cooling requirement.
Unlike indoor installations, outdoor enclosures are exposed to solar radiation, which can significantly increase the external surface temperature and add additional thermal load to the cabinet.
This additional heat input should be considered during thermal design, especially when:
The solar heat gain can be estimated as:
Qsolar=α×G×Asolar
Where:
In practical cabinet cooling design, solar load is often underestimated because engineers focus mainly on internal components such as drives, PLCs, and power supplies.
However, an outdoor cabinet with a relatively low internal heat load can still experience high internal temperatures if it is exposed to direct sunlight.
An IP rating defines the enclosure’s protection against:
However, it does not indicate protection against solar heating or high ambient temperature.
A cabinet with IP65 protection may still require additional thermal management when installed outdoors.
Common methods to reduce solar heat gain include:
The appropriate solution depends on the calculated heat load, ambient conditions, and required cabinet temperature.
Using connected load instead of heat loss
The total electrical rating of the installed equipment is not equal to the heat released inside the cabinet.
Ignoring passive surface dissipation
Cabinet walls may remove a meaningful portion of the heat when ambient temperature is lower than the internal target.
Counting blocked surfaces as exposed surfaces
Cabinet sides connected to another enclosure or installed against walls do not provide the same heat-transfer performance as exposed surfaces.
Ignoring peak production conditions
A cooling system calculated only from average load may fail during high-speed or full-capacity operation.
Selecting fans from free-air airflow
Filter mats, grilles, cable ducts, and internal restrictions reduce actual airflow.
Ignoring solar radiation
Outdoor cabinets may absorb substantial heat even when internal electrical losses are moderate.
Applying an excessive safety factor
Oversizing does not correct an inaccurate calculation and may reduce the efficiency and service life of active cooling equipment.
Using nominal cooling output without checking the performance curve
Cooling capacity must be evaluated at the expected internal and ambient temperature conditions.
During project design, engineers sometimes select oversized cooling units because they want additional safety margin. However, excessive oversizing does not always improve reliability.
However, oversized cooling can create new problems:
7.1 Higher Energy Consumption
An oversized air conditioner cycles frequently, reducing efficiency.
7.2 Condensation Risk
When cabinet cooling capacity is excessive, internal temperatures may drop below dew point conditions, creating moisture problems.
7.3 Higher Initial Cost
A larger cooling unit increases:
Correct thermal calculation avoids both overheating and unnecessary investment.
Accurate heat load calculation is the foundation of reliable cabinet thermal design. However, selecting the correct cooling solution requires more than knowing the required cooling capacity.
The final cooling method depends on several factors, including ambient temperature, installation environment, enclosure protection requirements, and operating conditions.
After determining how much heat must be removed, the next step is understanding which cooling technology is suitable for the application.
In the next guide, Choosing the Right Cabinet Cooling Method, we will explain how engineers compare filter fans, air-to-air heat exchangers, and enclosure air conditioners, and how to select the most appropriate cooling approach based on real industrial requirements.