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Formula for determining the heat dissipation of a radiator

  1. Jun 11, 2016 #1
    Are there any equasions out there which can calculate the heat dissipation of a water-cooled radiator with a fan on it? I would assume some of the variables would include:
    -Volume of the radiator (240 x 199 x 45mm)=2149200mm^3
    -Specific heat of water (4.187 kJ/kgK)
    -Specific heat of the copper radiator fins (0.385 kJ/kgK)
    -ΔT between the temperature of the coolant (variable we are solving for), and the ambient air temperature (lets assume 20 degrees C)
    -Rate at which air moves over the radiator fins (3.115 cubic meters/minute)
    -Rate at which water moves through the fins (3.785 Liters/minute)
    -Energy being dumped into the system (500 watt heat source)

    Ultimately what I want to accomplish, is find out if said radiator can adequately radiate 500W of heat (ΔT of less than 40 degrees C) in an ideal environment. Is there some equasion into which I can plug these variables? Did I leave anything out?
     
  2. jcsd
  3. Jun 12, 2016 #2

    Simon Bridge

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    Welcome to PF;
    The short answer is "yes".

    Forced convection: a good model would be "Newtonian cooling" carrying off heat on the outside. The rate of heat transfer is directly proportional to the temperature difference. The trick is determining the constant of proportionality ... which you can do experimentally, or there may be tables you can look up.
    ie. http://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html

    The water flow is more about keeping the radiator surface at a constant temperature by moving heat from some source to the inside of the heater. So long as the flow can deliver heat at the same rate it is being dissipated you are fine.
    You will be more interested in stead-state situation ... the situation as the heater warms up will be much more difficult to model and not tell you much.
    For strong convection, there is a point where increasing the fan speed makes no difference to the rate of heat transfer ... which is what your description sounds like. What the fan does is blow away the warmer air close to the radiator... ie maintaining the temperature difference.
     
  4. Jun 12, 2016 #3

    jack action

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