Calculation of heat generation from electrical equipment in the substation

Regarding the heat transfer amount emitted by the control cabinet, if we use the formula for heat transfer, with the average heat transfer rate K (W/m2°C) of the control cabinet, the temperature inside the control cabinet Th (°C), the temperature outside the control cabinet Tc (°C), and the surface area of the control cabinet S (m2), then the heat transfer amount Q emitted by the control cabinet is given by Q = k × (Th – Tc) × S. Therefore, the desired temperature inside the control cabinet is Th, the total heat generation from the air within the control cabinet is P1 (W), and the required cooling capacity is P2 (W). Thus, the necessary cooling capacity is calculated using the following formula. P2 = P1 — k×(Th—Tc)×S. For natural convection in a solid wall exposed to air, the heat transfer coefficient k ranges from 4 to 12 (W/m2℃). For ordinary control cabinets (where there are no cooling fans or anything of the kind), if calculated using a value of 4–6 (W/m2°C), empirical judgment shows that this is basically consistent with the actual values. Using this value, the required cooling capacity for the actual control cabinet is calculated as shown below. Example: · The desired set temperature inside the control cabinet is 40°C. · The temperature outside the control cabinet is 30°C. · The control cabinet is a stand-alone unit with dimensions of 2.5 m in width, 2 m in height, and 0.5 m in depth (the bottom surface should be excluded from the total surface area). · There are 20 SSRs operating continuously at 30A each. · The total heat generation from the control devices other than the SSRs is 500 W. The total heat generation inside the control cabinet, P1, is calculated as follows: P1 = 1.6 V drop in output voltage × load current of 30A × 20 units + total heat generation from devices other than SSRs = 960 W + 500 W = 1460 W. The heat dissipated by the control cabinet, Q2, is calculated as: Q2 = heat transfer rate × (40°C – 30°C) × (2.5 m × 2 m × 2 + 0.5 m × 2 m × 2 + 2.5 m × 0.5 m) = 662.5 W. Therefore, the required cooling capacity, P2, is P2 = 1460 – 662.5 = 797 W. The heat dissipation from just the surface of the control cabinet is not sufficient; it is necessary to find ways to remove more than 797 W of heat from the control cabinet. Normally, fans for adequate ventilation should be installed, but. When the cooling capacity of the fans alone is insufficient, air conditioning for the control cabinet should also be installed. Using air conditioning in control cabinets not only provides cooling but is also effective at preventing moisture and dust, which is very beneficial for the long-term use of such cabinets.

It’s a bit complicated; take some time to process it. But I don’t know the heat generation of the device