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I Temperature Coefficient of Resistance and Specific Heat

  1. Aug 2, 2016 #1
    I am trying to estimate the amount of electrical power needed to heat up a material from 20C to 1000C in 20 seconds. Assuming it is 18sqin and 3mm thick, I used the specific heat which was given as 0.78 J/gC and weight 11.39g (from the destiny of 3.26).

    0.78 J/gC x 11.39g x (1000C - 20C) / 20sec = 435 J/s = 435 W.

    Now I know I've left out some heat losses.
    I estimate Radiation at 1000C with an emissivity of 0.85 would be around 1470W
    Convective loss around 377W with a 25mph wind
    If heatup is linear, I would expect maybe half of that would be the losses during warmup are around 950W
    so perhaps a total of 1400W needed for the heatup

    Reading data from the manufacturer, they tell me that 700W/sqin (12600W total) are required to heat it from 20C to 1000C in 20sec. I asked them why so much and they say it has a "Temperature Coefficient of Resistance" of 0.0015. I showed them my calculation, they tell me that will give an average power needed, but 12600W is required at the start.

    That makes no sense to me as I would expect the power to be constant and I am unfamiliar with this TCR coefficient. How is TCR related to Specific Heat?
    What am I missing?
     
  2. jcsd
  3. Aug 2, 2016 #2
  4. Aug 2, 2016 #3

    Nidum

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    ' square inches ' .
     
  5. Aug 2, 2016 #4
    I think the mass of your sample is 113 g rather than 11.3 g.
    Isn't it?
     
  6. Aug 2, 2016 #5

    Nidum

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    Exactly how is this plate being heated - it is not at all clear what you are doing from your initial description .
     
  7. Aug 2, 2016 #6
    What is the material that you are trying to heat up? Are you heating it from one side or from both sides? Is the heater in contact with the sample? What is the geometry of the heater?
     
  8. Aug 2, 2016 #7
    the temperature coefficient of resistance tells you how much the resistance of the specimen will change as it warms up (or cools down) this could effect your calculations of power supplied by electrical means.
    The effect is small and I think that you have more to worry about before temperature coefficient of resistance becomes a significant aspect in your calculations.
    By the way, it is very bad to mix up units in calculations...square inches and mm do not work well together.
     
  9. Aug 2, 2016 #8
    It material is being heated electrically, it is a square plate in geometry. I thought that using my calculation using specific heat and then including losses from convection and radiation, I should arrive at an answer fairly close (ie. within 25%). However, the numbers I'm getting from the spec are an order of magnitude different.
     
  10. Aug 2, 2016 #9
    See nasu's post #4.
     
  11. Aug 2, 2016 #10
    AH, I see, I must have missed that post. Now I'm only about 3X off from their estimate. A 3X safety factor seems resonable
     
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