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What is the difference between 100V 30A and 300A 10V ?

  1. Apr 29, 2013 #1
    What is the difference between 100V 30A and 300A 10V ???

    Hello, I am very new at this so please guide me on the following issue;

    Lets say I have an electrical resistor with a resistivity of 1.5*10^-5 ohm m (graphite)

    and lets say the total resistance of this particular graphite heating element is 0.15ohm and I need it to heat up to 2000 kelvin.

    Considering the power required to radiate 2000 K is 3000 watt, how should I determine the required voltage and current...???? I do know that graphite heating elements are only used with dc step down transformers, but how low should the voltage go?!

    if I hook this up to a 100V 30Amp or 10V 300A transformers the result is 3000 Watt in either case, so what is the difference?! I think I am missing something very fundamental here... PLEASE help...

    The way I got 3000 watt as the required power is from the black-body radiation's formula :
    Power=(surface area of blackbody)*(boltzman constant)*emissivity*(Temperature)^4

    is this a good method for a rough approximation (+/- 500 Kelvin) ?
    or correct at all?

    Thanks in advance.
     
  2. jcsd
  3. Apr 29, 2013 #2

    Integral

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    You have the resistance. So compute :

    P=I2 R
     
  4. Apr 29, 2013 #3
    The difference in the two powers is this:

    Start with Ohm's Law.

    If your heating element has a total resistance of 0.15 ohm then if you hook it up to a 10V source you will consume 66.6A and generate about 665W of heat.

    If you hook it up to a 100V, 30A source then the one of those two values is no longer valid: at 100V and 0.15ohm the circuit wants 6666.66A and at 0.15ohm and 30A the circuit wants 4.5V.

    So if you tried the 100V, 30A source and you had a current limiting power supply the power supply voltage would drop to 3.5V with the current pegged at 30A which only gives you 105W. If you have a completely unregulated system something will probably start to smoke.


    How to find what you need:

    You have to balance this with P = I^2*R = V^2/R.

    So with R = 0.15ohms you need 141.42A and 21.213V
     
    Last edited: Apr 29, 2013
  5. Apr 29, 2013 #4

    Thaaaaaaaaaaaaank you very much Floid, this clarifies everything...

    What is your take on my assumption that I will need 3000Watt to reach 2000K ?
    Is this formula applicable in this context?
    Power=(surface area of blackbody)*(boltzman constant)*emissivity*(Temperature)^4

    I got this from the following wikipedia page:

    http://en.wikipedia.org/wiki/Stefan–Boltzmann_law


    Thanks again. :D
     
  6. Apr 29, 2013 #5

    sophiecentaur

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    Now you are adding the thermal performance into the equation, things are much harder. Did you want a particular temperature? 2000K is quite high and would require a specialist heating element, I think. Imo, you would be best to find a manufacturer and ask their advice. The temperature achieved would depend as much on the enclosure you keep it in as the characteristics of the resistor.
     
  7. May 3, 2013 #6
    Yes. A specialist Heating element must be used. Fortunately such high grade graphites are available and already used in Atomic Absorption Spectroscopy (Graphite AAS) by Buck and other manufacturers.

    The unfortunate part is they don't provide any information regarding the heating behavior of the graphite, other than their resistivity (1.5*10^-5)

    But what what if we don't consider effect of temperature due to enclosure, and just consider the sole temperature of the graphite rod... How to calculate it???
     
  8. May 3, 2013 #7

    sophiecentaur

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    That would give you a value for the notional temperature you could reach - just applying the formula in your first post. But the element is not in outer space and it is connected by conducting wires, held in a support, surrounded by air and, presumably, it is being used to 'heat' something up. All those factors will significantly affect the actual temperature reached by the element (in both directions). You would also, of course, need to consider the thermal coefficient of the resistivity, which is significant and negative (leading to possible thermal runaway in carbon filament lamps, for instance). If this has a serious application then you would need to consider some thermostatic control. Also, I can't imagine you would actually want to be operating the element out in the open - even if only from the running cost consideration.
     
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