Caculating heat loss from cooling fins?

[rsalmon]

Hello,

I am currently studying transformer design, in particular cooling methods.
I am tyring to calculate the cooling effect (increased heat transfer) of adding cooling fins to a transformer.

Does anyone know of some general equations/methods for calculating the increased heat radiation achieved by adding cooling fins.

Thanks,

Rob

 

[edgepflow]

Basically, fins increase the heat transfer rate by adding surface area. But there is an efficiency associated with the fins. The heat transfer rate is:

q_Total = Nfin * eff * hfin * Af * (Tsurface - Tsurr) + hbase * Ab * (Tsurface - Tsurr)


q_Total = heat transfer rate
Nfin = number of fins
eff = fin efficiency
hfin = convection heat transfer coefficient of fin
Af = surface area of one fin
Tsurface = surface temperature
Tsurr = surroundings temperature
hbase = convection heat transfer coefficient of base
Af = surface area of base

Notice the first term in the above equation is the heat transfer from the fin and the second term is from the base.

These calculations are a bit of an art. The trick is to make simplyfying assumptions that do not introduce too much error.

Let me know if you need help getting started.

 

[Low-Q]

Just have in mind that adding metals close to a transformer will generally reduce the transformers efficiency. When a heat sink is placed upon a toroide transformer for example, the transformer is "seeing" a short circuit secondary winding because the magnetic field wants to induce electric current through the heat sink. As the heat sink (usually made by copper or aluminum) is a good conductor, it will for sure reduce the efficiency of the transformer - which in turn makes it even hotter than without heat sink at all.

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