Electrical components and enclosures generate heat—due to I²R losses, iron losses, switching losses, and more—that must be dissipated to keep temperatures within safe limits. Heat dissipation calculations determine the temperature rise in equipment or the required cooling needed to maintain acceptable temperatures. These calculations are based on fundamental thermal principles: conduction, convection, and radiation. In steady state, the heat balance can be expressed as:
Heat generated = Heat dissipated
If heat generation exceeds dissipation, the temperature will rise until increased dissipation (or active cooling) balances the generated heat—or, in the worst case, thermal runaway may occur.
P = I²R = 2000² × 50e-6 = 200 W
Q = U × A × ΔT
Q is the total heat (W), A is the surface area (m²), and U is the overall heat transfer coefficient (W/m²K) accounting for convection and radiation.
Q = ṁ × cₚ × ΔT
cₚ is the specific heat capacity of air, and ΔT is the temperature difference—this helps in selecting a fan or air conditioning unit.
Consider an electrical cabinet with dimensions 2 m (H) × 1 m (W) × 0.5 m (D). Assume only five surfaces dissipate heat (with the bottom on the floor). The total effective surface area may be roughly 7 m². If the internal components dissipate Q = 300 W and the enclosure is sealed (no fan), and if the overall heat transfer coefficient is estimated as U ≈ 5 W/m²K (typical for natural convection plus radiation), then the temperature rise is:
ΔT = Q / (U × A) = 300 / (5 × 7) ≈ 8.6 K
With an ambient temperature of 25°C, the internal temperature would stabilize around 33.6°C. For a higher heat load (say 1000 W), the temperature rise would be approximately 28.6 K, bringing the internal temperature to about 53°C, which might be borderline—indicating that additional cooling (vents or a fan) could be needed.
For a power semiconductor with power loss P attached to a heat sink, the thermal resistance chain is analyzed as:
Rtotal = Rjc + Rcs + Rsa
where Rjc is the junction-to-case resistance, Rcs is the case-to-heatsink resistance, and Rsa is the heatsink-to-ambient resistance. The junction temperature rise is then given by:
ΔTj = P × Rtotal
For example, a transistor dissipating 10 W with Rjc=1 K/W, Rcs=0.5 K/W, and Rsa=4 K/W will have a total resistance of 5.5 K/W, yielding a junction temperature rise of 55 K. If the ambient is 40°C, the junction temperature would be about 95°C, which is safely below typical maximum ratings (e.g., 150°C).
When multiple cables run together in a tray, the inner cables may run hotter due to reduced convection. Standards account for this by providing derating factors derived from thermal modeling.
Transformer and motor heating is usually addressed by standard tests or empirical formulas. A motor’s temperature rise under load is often determined by testing or by creating a thermal circuit model. Designers ensure that adequate cooling (such as fans) is provided so that the operating temperature stays within safe limits (e.g., around 105°C for Class F insulation).
Thermal failures are a leading cause of electrical equipment damage—loose connections may overheat and cause fires, and electronic components may be overstressed by high temperatures. Proper heat dissipation design prolongs equipment life and ensures reliability. For example, IEC 61439 for low-voltage switchgear sets maximum temperature rise limits for busbars and terminals, and manufacturers test their assemblies accordingly. In hazardous areas, calculations also ensure that surface temperatures remain below specified limits to prevent ignition.
Heat dissipation calculations blend fundamental thermal principles with empirical data and simulation tools. By following industry standards and using available software, engineers can design enclosures and cooling systems that maintain safe operating temperatures, enhance reliability, and extend equipment life.