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Joule Heating

  1. Oct 12, 2009 #1
    I want to find the increase in the temperature of a resistor due to joule heating. I mean that if we apply some electrical power to a resistor then the electrical energy is converted to heat energy due to joule heating. The heat energy will increase the temperature of the resistor. Is it possible to calculate the increase in the temperature of the resistor?
    I will appreciate any help.
     
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  3. Oct 12, 2009 #2

    Mapes

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    Hi charahim94, welcome to PF. The resistor's temperature over time isn't just a function of the electrical power, but also a function of the heat transfer mechanisms to the surroundings. As soon as the resistor's temperature increases above room temperature, conductive, convective, and radiative heat transfer will transfer energy away. You can calculate an upper limit for temperature, if you like, by assuming all these heat transfer mechanisms to be negligible, and dividing the total energy dissipated in the resistor (in Joules or other energy units) by the resistor's heat capacity (in J/K). Remember, though, that the true temperature increase in practice will always be lower. Does this answer your question?
     
  4. Oct 12, 2009 #3
    Mapes, Thanks for the reply. The total energy dissipated in the resistor is given by: E=I^{2}Rt. So , If i assume that there is no heat transfered then the joule heat will increase the temperature of resistor. The Heat energy is given by Q=mc\Delta T. And by comparing the two equations I will get the change in temperature. Is this what you mean? or i have misunderstood!
     
  5. Oct 12, 2009 #4

    Mapes

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    Yes, exactly. But by ignoring heat transfer away from the resistor, your answer after long time periods will tend to be ridiculous (e.g., 500,000°C after one week :bugeye:, when in practice the resistor might reach a steady state temperature increase of 5°C in two seconds, thereafter remaining in thermal equilibrium with its surroundings).
     
  6. Oct 12, 2009 #5
    Hello Mapes, the last thing that i want to know is: In case if i want to take the conductive heat transfer into account, then is it correct to replace Q=mc\Delta T with Q=[mc\Delta T]+[kA/\Delta x],
    where k is the thermal conductivity of the material. I am not sure if Q will be sum of two terms. But i think that due to conductive heat transfer the net temperature of the resistor should be less.
     
  7. Oct 12, 2009 #6

    Mapes

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    The units don't match up, so right away it's clear something's wrong. In general, heat transfer will occur by conduction (not through the resistor material but through the surrounding material), convection, and radiation, so it can be quite complicated to model. One approach is to assume the resistor has reached steady state and to equate the dissipated power with the rate of energy transfer away from the resistor. (And frequently it's easiest just to build a prototype and test it.)
     
  8. Oct 13, 2009 #7
    Hello Mapes, I am a bit confused now because i get some strange numbers by the above equation. I do not know Heat Capacity in J/K of titanium (material of resistor). Therefore, I used \Delta T=(I^{2}Rt)/(mC) and take the mass of resitor into account. The resistor is made of titanium and is 400micron long, 5 micron wide and 100nm thick. I take C=520J/Kg-K and for the above dimensions of resisotr i get m= 9exp-13 kg. Then i get very strange result. Do you have any idea what is wrong? Can i find heat capacity in J/K from some material handbook?
     
  9. Oct 13, 2009 #8

    Mapes

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    If you're getting a very large number, note the caveats I mentioned above.
     
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