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Heat transfer using heat sinks

  1. Jan 8, 2012 #1
    hello
    I need help regarding a small project on thermoelectric refrigerator. I need to cool a 5.4 ltr capacity refrigerating space to maintain it at 10-12 degrees. I have the cooling module in place.I needed to dissipate 98 W of power from one module via a heat sink.

    I have a heat sink that is of pin type measuring 96*96 mm with 98 pins of rectangular cross section.I need to find:

    a) The thermal resistance of this heat sink when coupled with a cooling fan of 12.5W

    b)how to choose a heat sink so that the temperature rise can be restricted to 20 degrees from the ambient temperature.

    anyone having any idea in this topic please reply!!!
     
  2. jcsd
  3. Jan 8, 2012 #2

    russ_watters

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    Staff: Mentor

    Welcome to PF!

    Temperature rise is a function of the airflow of the fan on the heatsink, so just multiply the heat you are dissipating by the specific heat of air and desired delta-T.

    If you already have all your parts, I'd start finding some of the answers by experimentation: calculating the performance of a heatsink from scratch is near impossible.
     
  4. Jan 8, 2012 #3

    berkeman

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    Staff: Mentor

    Generally the datasheet for the heat sink will give you numbers for its "thermal resistance" versus airflow and orientation. Can you find a datasheet for it? If not, you could try some Google Images searching to find similar heat sinks, and look for their datasheets.
     
  5. Jan 8, 2012 #4
    thank you for both the replies!!

    1) yes i did figure out id have to experiment and find out the temperature rise of the heat sink( coupled with the fan )corresponding to the 90 W dissipation and see if it rises above the 20degrees from ambient.But isnt there any basic formulas behind it??(i did google it....not much help) Can you suggest me where it?

    2) Yeah they do have a datasheet, but it has a thermal resistance (without the air flow) of 1.1 K/W which is a lot (because 1.1 *90 =99 degrees rise!...i do believe that is what it means right?) whereas by using the formula:

    Th = Tamb + (Qh)*Rt
    Th=50
    Qh=90
    Tamb=30

    Rt comes out to be 0.22k/w ,ie to maintain temperature rise within the 20 degrees mark,i need something that is of this thermal resistance.

    So i do want to know if the thermal resistance can fall this much,ie from 1.1 to .22 when coupled with a cooling fan.If so if theres any method to evaluate it theoritically?

    (though my pin fin heat sink and fan should be sufficient.... practically because it was provided by the thermoelectric suppliers corresponding to the particular module i had selected for cooling..so......)

    yeah Ill be starting with the experimenting part !
     
  6. Jan 8, 2012 #5

    AlephZero

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    Science Advisor
    Homework Helper

    Inproving the cooling by a factor of 5 by using forced convection doesn't seem unreasonable to me.

    Almost all the formulas you will find on the web or in heat transfer handbooks are based on experimental correlations. The big problem with using them correctly is knowing what physical situation they apply to. Outside of their proper range they can sometimes be ridiculously wrong - even giving negative heat transfer (from cold to hot).

    It may seem counter-intuitive, but it's actually easier to calculate heat transfer "from first principles" for much more extreme situations, like hypersonic flow (e.g. space shuttle re-entry) than for air wafting around at low speed where the flow pattern is strongly affected by the change in air temperature and the resulting changes in density and bouyancy.
     
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