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Capacitance needed to heat Nichrome wire 34°

  1. Aug 10, 2013 #1
    Hi PF!

    I'm working on a portable heating circuit. I'd like to use an ultracapacitor charged by 2 AA batteries to increase the temperature of 2.25m of 1mm Nichrome wire by 34° in 6 seconds.

    The question is, what capacitance does the ultracapacitor have to have?

    Initial values:
    • Specific Heat of Nichrome: 450 J/(kg ∆K)
    • Density of Nichrome: 8400 kg/m3
    • Resistivity of Nichrome: 1e-6 Ωm

    Calculated values:
    • Mass of Nichrome: 0.015 kg = density*wire length*π*(wire diameter/2)^2
    • Resistance of wire: 2.25e-6 Ω = resistivity/wire length
    • Heat needed to cause temp change: 229.5 J = specific heat*mass*temp change
    • Current needed for Joule heating: 10100 A = √(Heat needed/resistance of wire)
    • Capacitance needed 51 F = heat needed / (0.5 * voltage^2)

    So I'm confused about the relationship between the current needed to cause the appropriate Joule heating in the wire and the energy stored in the capacitor. Can anyone shed some light on this?

    Thanks!

    Michael
     
  2. jcsd
  3. Aug 10, 2013 #2

    Vanadium 50

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    You need to calculate the energy needed to raise the wire temperature, and then match that to the capacitor's stored energy. To get that in 6 seconds means you then need to match the resistance of the wire to the capacitance to get the appropriate time constant.
     
  4. Aug 11, 2013 #3

    Baluncore

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    As you note, the resistivity of nichrome is between 1.0e−6 and 1.5e−6 Ωm. But that actually means the resistance between two opposite faces of a 1m cube of nichrome is about 1.25 millionths of an ohm.

    To calculate your wire's resistance;
    You have 2.25m of 1mm diameter wire. Cross section of 1mm diameter is Pi / 4 = 0.785 mm2
    One square metre is 1,000,000. square mm = 1e6 mm2.
    Your wire will therefore have a resistance of 2.25m * (1.25e-6) * (1e6 / 0.785) = 3.5828 ohms.

    I think that will make a slight difference to your numbers. By 6 orders of magnitude !
     
  5. Aug 11, 2013 #4
    Thanks Baluncore, I'd totally missed that. Recalculating:

    • Wire resistance: 3.6 Ω = resistivity * wire length * cross section area of 1m^3 / wire cross section area
    • Current needed for Joule heating: 8.0 A = √(heat needed / wire resistance
    • Time to discharge 2/3 energy: 180 s = resistance * capacitance

    8 A is a lot more realistic than 10.1 kA!

    So if I short a charged 50F 3V ultracapacitor across 2.25m of nichrome 1mm wire, it should heat 34°C in 3 minutes, right?

    I'm concerned that there isn't a more explicit relationship between the current needed and the capacitance of the ultracap.
     
  6. Aug 11, 2013 #5

    Baluncore

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    Ohm's Law says I = V / R
    Vc = 3V, R = 3.6 ohm therefore I = 0.833 amp maximum.
    You must juggle your resistance if your peak voltage is fixed.

    Are you deliberately neglecting thermal conductive and convective cooling of the wire?
    That cooling will be proportional to temperature above ambient. How is your Nichrome wire mounted in what environment ?
     
  7. Aug 11, 2013 #6
    Hmm, maybe nichrome isn't the right material, or 1mm isn't the best thickness.

    Yes, I am ignoring cooling for now. The application is a steering wheel heater, so it needs to go from 0°F to 60°F very quickly (≤6s) but then the hands will keep it at temperature. The wire will be split between two pads, zig-zagging back and forth to completely fill the pads' area.
     
  8. Aug 11, 2013 #7

    Baluncore

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    Energy in the capacitor is 0.5 * Cap * volt^2 joule
    So; 2 * Energy = Cap * v^2
    Therefore; Cap = (2 * Energy) / v^2
    But Vmax is fixed at 3V.
    You say you need 230J which requires 2 * 230 / 9 = 51.1 farad.
    But you will be unable to get all that energy out of the capacitor in 6 seconds.
    So you will require more capacitance, maybe double.
     
  9. Aug 11, 2013 #8

    Integral

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    Computing temperature form the information you have given is VERY difficult, if not impossible. To get a meaningful temperature you have to know ALL heat exchanges occurring. How much heat will be lost from your wire? For example if it is in water you have a vastly different result then in a vacuum. Even small air currents will effect the final temperature. If your environment is not very stable you will get a different temperature every time you run your heater.
     
  10. Aug 11, 2013 #9
    Good point Integral, but to make this easy I'm just interested in the heat a capacitor can generate via short circuit.

    I'm now investigating lead, nichrome and copper as potential heating elements. So far lead seems to be the best in terms of requiring the lowest voltage. Any suggestions on other materials to consider?

    Here are the equations and values I'm using: https://www.dropbox.com/s/oarnxyyy5r30tsy/heater.py
     
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