Caculating heat loss from cooling fins?

  1. 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
     
  2. jcsd
  3. Chegg
    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.
     
  4. Low-Q

    Low-Q 258
    Gold Member

    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.

    Vidar
     
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