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How can i figure how hot this wire is getting?

  1. May 1, 2007 #1
    To start things off i am using DSC (Dipsersion Strenghtened Copper)
    99.58% copper
    0.19% Aluminum Oxide
    0.16% Oxygen
    Less than .01% Iron
    Less than .01% Lead
    max tensile is about 10 breaking pounds or 88ksi
    i think for most properties i will assume it is pure copper

    the diameter of the wire is 0.012". they are putting 30volts and 124 amps through it. its resistance is 0.2414 ohms.

    i did come across this equation:
    Cg-Specific Heat (i am going to assume it is pure copper)
    Tf-final temperature
    Ti-initial temperature

    can i use this formula? or do i need to get the energy (J/s) going through the wire to the heat(degrees)
  2. jcsd
  3. May 1, 2007 #2
    i am not going to use pure copper properties because our material starts off as a copper powder with aluminum oxide, along with a oxygen free copper cladding.

    But can i use that formula?
  4. May 1, 2007 #3


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    Staff: Mentor

    Power dissipation is 3720W. If you knew the "thermal resistance coefficient" of the length of wire (in degrees C per Watt), you could calculate the temperature rise. That's how we do it for heat sinks.

    As for wire, you would need to know how well the wire dissipates heat. Like, if you blow a fan on it, it will have a lower effective thermal resistance coefficient.

    I would think that a full calculation would be fairly involved, so it might just be best to do a couple medium-power experiments to figure out what the coefficient is for your wire and environment (moving or still air, wire vertical or horizontal, spread out or spooled, etc.), and then use that coefficient to extrapolate to the full power that you mentioned.
  5. May 2, 2007 #4
  6. May 2, 2007 #5


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    Staff: Mentor

    Put a known power into the wire, and measure the temperature rise after it stabilizes. The temperature rise will be linear with power input over some range (at least that's the simplification used in heat sink design in electronics).
  7. May 3, 2007 #6
    i don't have the equipment to do that. can i just use specific heat? and put it in
    [energy in wire(Joules)]=mass*specificHeat*(change in temperature)
  8. May 3, 2007 #7


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    Staff: Mentor

    No, I don't think so. The heat dissipation capacity enters into it, right? If the heat is carried away from the wire more easily (due to airflow or whatever), then the quiescent temperature rise above ambient will be less.

    BTW, you can buy a thermocouple-based thermometer attachment for you DVM for something like $100 USD or less, I think. Is that still too expensive?

    EDIT -- Wow, here's one for $12 on e-bay!

  9. May 3, 2007 #8
    wow thanks for new equipment. just curious what is the difference between "heat dissipation capacity" and "specific heat coefficient"?
    is one for how well it will release temperature and other how well it will storage energy or temperature?
  10. May 3, 2007 #9


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    Staff: Mentor

    Yes, something like that. BTW, I googled wire table maximum current, and got some good hits that may help you. Check out the hit list to see if these tables talk about power dissipation and temperature rise:

  11. May 4, 2007 #10

    In steady-state conditions, the specific heat or the mass of the wire has no effect on its temperature.
    In steady-state, the power absorbed by the wire is totally dissipated in the environment.
    Only during the warm-up transient will the specific heat and density of the wire play some role.
    Highed heat capacity will slow-down the heat-up process, for example.

    If the temperature is high, the dominant mode of cooling of the wire will be radiation.
    If the wire is straigth and not isolated, then in first approximation, the "black-body" radiation law can be applied.
    Usually an additional factor is used called "emissivity" that takes the quality of the radiating surface into account.
    The best radiatiors have an emissivity = 1 .
    The "black-body" radiation law reads:

    (emitted power) = Pemit = eps sig (T^4-To^4) Surf


    eps < 1 is the emissivity (close to 1 for good 'radiatiors')
    sig = 5.67E-9 W/m²/K^4 is the Stefan-Boltzmann Constant
    T is the temperature of the emitting body in K
    To is the temperature of the surrounding in K
    Surf is the (apparent) surface of the emitting body (wire)​

    As an example, let us assume that your wire had a length of 1 meter.
    Then you have these likely values:
    eps = 1 (assumption with a slight effect when in the usual range)
    To = 300 K (assumption with negligible impact)
    P = 3720 W
    Surf = 0.001 m²​

    Solving the balance equation Pemit = P for the the wire temperature gives T = 2842 K.

    These calculations will not be correct for another geometry (I assumed a straight wire far fom any object).
    For example, when the wire is torsaded, then its "apparent surface" is decreased, and its temperature will increase.
    Similarly, if the wire is enclosed within a glass cylinder, its temperature will increase a little bit. For such a geometry, the calculations are similar, but the (radiative) properties of the glass have to be taken into account, and the temperature of the glass must be solved simultaneously to the temperature of the wire.

    As you can guess from the high temperatures above, the radiation cooling process will always dominate. However, if you introduce a strong forced convective cooling, you could still calculate the wire temperature in a similar way. All you need to do, is to include an additional term for the additional cooling mode (and solve numerically the equation). Usually the corresponding heat tranfer coefficient will depend on the speed of the cooling fluid and its temperature. These are given by empirical correlations. The same applies for natural convection, but this will be even more negligible. Usually the heat transfered is calculated by a formula like
    powerTransfered [W] = h S (T-Tfluid)


    h is the heat exchange coefficient [W/m²/K], given by empirical correlations
    S is the exposed surface
    T is the temperature of the wire
    Tfluid is the temperature of the fluid

    There are many variants and all the details can be found in thick books on heat transfer.

    Additional note:
    Keeping in mind how the radiative heat transfer works and increases with temperature, it should be clear that measuring the temperature by contact would modifiy the temperature quite a lot. Therefore, a non-contact method should be used. Google about hot-wire techniques for measuring flame temperatures, this might suggest you some way of measuring your wire temperature. There are many other techniques and strategies available.
    Last edited: May 4, 2007
  12. May 4, 2007 #11
    our machine has a solution of alcohol and water falling onto the wire. other guys are working on measuring the percentage of each in the solution. if i can get the heat exchange coefficient for the wire i am using can i use that formula above? i have power going into the wire, the exposed surface, and can get the water/alcohol solution temperature.

    do i need the heat exchange coefficient for my water/alcohol solution?
  13. May 4, 2007 #12
    By definition, when only radiation is involved, the exchange coefficient will be given by:

    k = eps sig (T^4-To^4)/(T-To)​

    In this way the heat exchanged with the surrounding is written in the standard form with the standard definition of the heat exchange coefficient:

    (emitted power) = Pemit = k (T-To) Surf​

    UNFORTUNATELY, if there is an alcohol/water cooling system, then the heat exchange will not be determined by radiation, and there is little hope to find a correlation in the litterature to find the heat exchane coefficient.

    FORTUNATELY, there is still another possibility.
    Let's assume first your wire is in pure copper.
    You can find some data on wiki (http://en.wikipedia.org/wiki/Electrical_resistivity" [Broken]) for the copper resistivity.
    It changes with the temperature as follows:

    rho(copper)[nOhm.m] = 17.2 + 0.039*(T[°C]-20)​

    This could help you to determine the wire temperature with some precision.
    You measure the resistance of the wire at 20°C, then you measure it under operation, and by comparing you can evaluate the temperature.

    But there are still problems.
    First, you don't have pure copper, maybe this introduces an uncertainty.
    Second, even with copper, the resisticity might depend on its crystaline state.
    Third, for a short temperature range the effect will be rather small and therefore, you should use a precision resistance measurement, if possible. Using the http://en.wikipedia.org/wiki/Wheatstone_bridge" [Broken] with two wires (one hot and one cold) could offer you a very high precision. You could in this way consider measuring a calibration curve of your wire resistance, in an oven. Applying much power on one wire within a Wheatstone bridge might be an additional difficulty, but could be overcome by measuring the resistance at some frequency. The electrical measurement method to be used depends on the precision required.

    Now, since your temperatures are probably rather low (I don't know), then some sort of thermometer, thermistor, thermocouple, ..., might also be an easy solution. If the cooling flow is rather large, then the pertubation introduced by the measuring device could be negligible.

    Additional note:
    Have a look at this web page: http://www.minco.com/tools/sensorcalc/rtd/default.aspx
    It can generate tables of resistance versus temperature for different alloys and models of resistances.
    You could easily generate the same data by yourself from other source of information, but it nevertheless illustrates the possibility to use a resistance measurement to find the temperature of your wire.
    Last edited by a moderator: May 2, 2017
  14. May 4, 2007 #13
    the temperature is proportional to the resistance :) you might be able ratio the lead/copper coeffecients. someone more experienced than me should be able to do this.

    crystalline state should change only at melting point.
  15. May 4, 2007 #14
    Annealed copper has a slightly higher resistivity than crystalline copper.
    (see http://en.wikipedia.org/wiki/Electrical_conductivity)
    This should be no surprise.
    But this should be a negligible effect in the context discussed here, and maybe also as compared to the effect of some Al2O3 addition.
  16. May 4, 2007 #15

    From the above, it seems likely that a direct measurement is probably more suited than an indirect one based on heat balance.

    However, I could investigate that a bit further if you could give us a more detailled description of your experiment. A description of the cooling system is necessary.
  17. May 4, 2007 #16
    what you were saying before along with Light_bulb about resistivity. i think this is what you guys are talking about:

    α - temperature coefficient

    i have values for all the variables except for temperature coefficient and t2(which i am solving for) seems like this would be accurate if we don't account for the water/alcohol flowing over the wire.
  18. May 4, 2007 #17
    the cooling system is very basic. it pumps water/alcohol to the top and it falls over the wire which are on rollers. the rollers are applying the current to the wire. hopefully soon enough i will know the percent alcohol content.

    R1/t1=R2/t2 R1 = .2414 ohms t1 = 23.8 Celsius R2 = 0.8769 ohms t2=?

    t2 = 86.45 celsius >>> about 187 Fahrenheit.

    sounds reasonable considering we havn't acoounted for any cooling from the water/alcohol
  19. May 4, 2007 #18
    what kind of heat exchange is there occuring between the wire and the water/alcohol? is it conduction?
  20. May 4, 2007 #19
    or do i mean convection?
  21. May 4, 2007 #20
    convection, if you have a tank of water and you treat the water like it's in a vacuum except for the ambient temperature you should be able to work out based on how much time it takes to reach an equilibrium temp using a set volume of water.

    same thing different properties.
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