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Finding a temperature rise in a resistor. HELP!

  1. Jun 25, 2005 #1
    I'm building an electric furnace using a coiled resistance heating element. The wire is a high temp alloy made for kilns and furnaces (Kanthal A-1 if you care.) Anyway... I can't find a way to calculate how hot the wire will get based on the current load and amount of wire. If the wire is too long then it won't get hot enough and too short and it will melt. The element has to be at least as hot as what I'm melting of course so I need to calculate before I buy the wire. Now I know how to calculate power consumption/dissipation in watts, but not how to translate that into an actual temperature for the wire.

    So, what's the relationship between power dissipation and temperature? If a wire 8 feet long is dissipating 6700 watts how hot would it get? I'm sure it's based on the diameter and properties of the wire but I just need the equation!

    Thanks,
    Alan
     
  2. jcsd
  3. Jun 25, 2005 #2
    The increase in temperature is proportional to the increase in energy:

    [tex] \Delta E = m C \Delta T [/tex]

    E is energy, m is mass, C is specific heat (look it up for copper) and T is temperature. Differentiate this formula wrt to time:

    [tex] P = m C \frac{d T}{dt}[/tex]

    If you want to take in to account heat loss (the wire is constantly dissipating heat through radiation) then we have an extra term proportional, to the area and T^4:

    [tex] P = m C \frac{d T}{dt} + k A T^4[/tex]

    k = 5.67*10^-8 in SI units. So given P = 6700 watts you can solve the differential equation to give you T at any time t.
     
  4. Jun 25, 2005 #3

    pervect

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    I'd recommend taking a look at Kanthal's recommendations

    kanthal handbook

    They have some recommended maximum surface loads (watts/cm) plotted vs furnace temperature in figure 4 or the above document.

    Hopefully the above link will work, if not browse around http://www.kanthal.com/

    So the general procedure - first estimate how much power you need to heat up your furnace to the desired temperature. This will depend on how well insulated your furnace is (the thermal conductivity of the walls). You may want extra power capability if you want it to reach the operaating temperature in a certain amount of time (this will depend on the thermal mass of the furnace). If you have extra power capability to heat up the furnace in a shorter amount of time, you will need a control circuit of course (because if you left the heat on all the time, you'd overheat).

    Next, chose a heating element surface area that can handle this power level, using the curves I mentioned from Kanthall (their recommendations on maximum watts/cm^2 vs temperaturue). Knowing the wire diameter and the length will give you the surface area.
     
    Last edited: Jun 25, 2005
  5. Jun 25, 2005 #4

    Gokul43201

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    In the steady state, [itex]dT/dt=0 [/itex]. So, it's pretty straighforward as long as you know the emissivity [itex]\epsilon [/itex] of the resistor material.
     
  6. Jun 25, 2005 #5

    rbj

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    i always thought that whe reaching steady state, [tex] \frac{d T}{dt} = 0 [/tex], that the power dissipated is inversely proportional to insulation thickness, proportional to the insulation area and the temperature difference between the inside temp and the ambient temperature. the constant of proportionality is a function of the the insulation quality.

    i know this is not heating a home or dealing with power dissapation of semiconductor devices, but what is different here?

    r b-j
     
  7. Jun 26, 2005 #6

    Gokul43201

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    What insulation ?
     
  8. Jun 26, 2005 #7
    Thanks for the responses, they were very helpful. Thanks for pointing me to the kanathal handbook, pervect, I must have missed it on their site.

    I do have a question about that big nasty equation:
    I'm not familiar with differential equations, though it probably justs look more complicated than it is. And what is "k = 5.67*10^-8 in SI units" I don't understand it's purpose. A constant?

    I have an emissivity factor for the wire of .7 so maybe that could be used in a different equation that Gokul43201 mentioned.
    Anyway, thanks
    Alan
     
  9. Jun 26, 2005 #8

    pervect

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    Yes, that's the Stefan-Boltzman constant
    http://scienceworld.wolfram.com/physics/Stefan-BoltzmannConstant.html

    you might also want to look at

    http://scienceworld.wolfram.com/physics/Stefan-BoltzmannLaw.html

    or even the planck radiation formula
    http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html

    The Stefan-Boltzman law gives the total energy flux (energy/unit surface area) emitted from an ideal blackbody.

    It does NOT include

    the emissivity constant (though including it is trivial, just multiply by the energy given by the formula by the emissivity, see the bottom page of the second link).

    Nor does it include conduction or convection. (You can probably ignore conduction, but it's likely that you'll have appreciable convection, IMO.

    This means that the formula will over-estimate the temperature.

    On the whole, I'd favor going with the engineering approach on this one as far as design goes - rather than calculate the theoretical temperature, I'd look for some semi-emperical rating or recommendations given by the manufacturer (which appears to be the case).
     
  10. Jun 26, 2005 #9

    Gokul43201

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    I agree with Pervect - a theoretical calculation will only give you an upper bound...unless you can say something about the conditions affting covective heat transfer (lots of fins, lots of area).

    To incorporate the emissivity :

    [tex] P = \epsilon \sigma A (T-T_{out})^4 + kC(T-T_{out})~;~~\epsilon = 0.7[/tex]

    [itex]\sigma = 5.67*10^{-8}[/itex] in SI units. k is a convective heat transfer constant for the particular geometry and flow characteristics; C is the heat capacity.

    Skipping the second term (for convection) will give you an upper bound. If there's no airflow, you could reasonably neglect the convective component at high temperatures.
     
    Last edited: Jun 26, 2005
  11. Jun 26, 2005 #10

    rbj

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    is the room that this furnace is in being heated to the same temperature? if yes, what about the neighboring rooms? there is something that exists between the core of this furnace and its surroundings. call it what you will.
     
  12. Jun 26, 2005 #11

    Gokul43201

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    To the level of approximation that we are attempting to solve this problem, you chose an appropriate system (namely the container that immediately surrounds the heater) and assume it's perfectly insulated from the surroundings (neighboring rooms and beyond, things outside the furnace/kiln, the room outside the duct ,etc.)
     
  13. Jun 26, 2005 #12

    rbj

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    i'm just a Neanderthal electrical engineer and i know about blackbody radiation, but i do not think that this approach is the correct one. if the furnace element is perfectly insulated from ambiance, then when you turn it on, energy continues to flow into but no energy flows out (unless you mean the E&M radiation itself, but then it is not assumed that it's perfectly insulated).

    the rate of energy being conducted from the furnace core is proportional to the temperature difference of the core to the ambient. as long as the (electrical) power in exceeds the power lost, the core will continue to increase in temperature until there is an equilibrium when the temperature difference is large enough that the net power is zero or the power in is equal to the power lost. this depends on lots of details, like how well is the core isolated from the ambient.

    of course if there is a thermostat or some kind of control system, and the set-point temperature is less than the maximum you could get by just applying current to the heating element, then the temperature can be any value.

    if it were perfectly insulated and you had non-stop power applied, the upper bound of the temperature of the element is unbounded.

    what is it that i'm missing?

    r b-j
     
  14. Jun 26, 2005 #13

    Gokul43201

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    rbj, you are right. There is an important unstated assumption in my calculation.

    The equation I wrote out is for a semi-empirical determination of the temperature of a heater element in a furnace and I assume there may be some reducing/inert/other gas flow in this furnace which keeps the furnace temperature different from the element temperature. I further assume that this is the dominant mode of heat removal (hence the assumption of perfect insulation).

    In the absence of gas flow, one would have to do your neanderthal calculation :wink:, of setting up a boundary value problem and solving the differential equations. And with the assumption that radiation/convection from the outer surface of the furnace is negligible (ie : outer surface at room temp) and the inner surface is isothermal, this simply reduces to the conduction equation you described above :

    [tex]P = \frac {kA(T_{in} - T_{out})} {t} [/tex]

    where :
    k = thermal conductivity of furnace wall insulation
    A = total furnace wall area,
    t = average insulation thickness.
     
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