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Conversion of Dissipated Heat in Watts to °F or °C

  1. Mar 25, 2007 #1
    Greetings All,
    I'm restarting a personal project (after having to put it on hold) involving magnetically induced electricity and I'm trying to stay within the wire manufacturer temperature tolerance specs. I know the formula (P = I^2*R) for figuring out how much heat will be dissipated by a wire coil based on the outcome of my coil design specs and I can figure out all of the other output parameters of a coil; resistance, volts, amps etc., however I'm obviously missing something (which is probably right in front of me) as I can't find/figure out how the dissipated wattage is converted into °F or °C, which is what I need to know so I don't create a coil that will burn the insulation off of itself or do other nasty things.

    I'd appreciate any helpful information that could get me going in the right direction.
    Last edited: Mar 25, 2007
  2. jcsd
  3. Mar 26, 2007 #2


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  4. Mar 26, 2007 #3
    Thanks for the link to the "Specific Heat Capacity" wiki page. I guess I've got a whole new lexicon of symbology to learn. Any hints?
  5. Mar 27, 2007 #4
    Specific heat is a measure of how much energy is required to raise the temperature of a certain amount of a given material. Physically, it can be measured with the material at a constant volume or at a constant pressure. For measurements at a constant volume, the material does no work on its environment as its temperature is raised. This is ideal, but impractical, as it requires an awful lot of energy to keep almost any substance a constant volume (i.e. to prevent thermal expansion). As such, when engineers/scientists measure specific heat they do so at a constant pressure, and that is what is given on the Wiki article.

    Let's assume that you're using copper wire. Practically, there are going to be three parameters you are going to want to know before you proceed: the specific heat of your material (which you have thanks to Wiki, though don't hesitate to look for a more reliable source), its resistivity, and its density. It seems as if you know more or less know what's up with the energy being dissipated into the wire as heat. Resistivity can be related to resistance for a given length of a gauge, from there, you can use your handy power equation to figure out how much heat is being dumped into your wire. Once you know that, you can use the density and volume of your wire to get a rough estimate of its mass, and from there, you can use the specific heat to figure out the change in temperature/unit time (remember, power is energy/time). This is where things get hairy, and you have to figure out the steady state of the system. The temperature of the wire does not increase indefinitely at this rate, and you have to take into account the wire's interaction with its environment (i.e. the insulator and the room it is in). All of a sudden we're spending way too much time modelling for what should be a simple design.

    Fortunately for both of us, engineers have a wonderful tendency of keeping track of results like this, that are more rigorously tested. The steps I've outlined above are very "back of the envelope", but I'm guessing your design probably doesn't need to be all that accurate. If you're looking for a real, practical solution to your problem, look up the "ampacity" of your wire. Here's a link: http://www.opamp-electronics.com/tutorials/conductor_ampacity_1_12_03.htm [Broken]

    I hope you find my post helpful!
    Last edited by a moderator: May 2, 2017
  6. Mar 27, 2007 #5


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    Mufusisrad, I got a chuckle out of your post towards the end. I don't recall if it was this forum or another one, but I used the term 'ampacity' once upon a time and was accused of making up a new word. Just thought I'd forwarn you. :)
  7. Mar 27, 2007 #6
    Haha, I can definitely see that. Hell, I certainly never learned about ampacity in my coursework at Uni. The only time I ever came across it was back in my garage tinkering days. It really does sound like a made up word though, then again, so do most things in EE. My personal favorites are buckboost and mho.
  8. Mar 27, 2007 #7


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    While we're on the subject of seemingly made up words does anyone know or remember what the word gimmick means in electronics? Be fair about this and no googling or wiki'ing. I can't really prevent this, but if you do at least post your answer in white or keep it to yourself.
  9. Mar 28, 2007 #8
    1)i thought C was the speed of light? 2)i was looking at the wiki page on the boltzmann constant, shouldn't you use that and convert J to K and vice versa, unless your using some non si unit?
    Last edited: Mar 28, 2007
  10. Mar 28, 2007 #9
    Usually the speed of light is indicated by a lower case c (while the origins are a bit murky, it stands for 'celeritas'=latin for swiftness). Sadly, physicists, mathematicians, and engineers have only so many letters (both Greek and Latin alphabets, upper and lower case) to describe the universe. Some of them had to be used twice. For specific heat, we tend to use an upper case C with a subscript V or P to indicate which thermodynamic variable is held constant.

    As far as Boltzmann's Constant is concerned, fundamentally, it is just a conversion constant that turns temperature into energy, much in the same way that the speed of light squared is just a constant for turning mass into energy and Planck's constant is just a constant for turning frequency into energy. However, you have to be careful how you use it (i.e. you have to have some basis for applying it in an equation). The most basic application relating temperature to energy by way of Boltzmann's constant is the equipartition theorem which basically says that the average energy associated with the motion of a single quadratic degree of freedom is 1/2*k*T. You can make a hand wavy argument using the Ideal Gas Law (PV=nRT=NkT) to derive it, but the bottom line is, just because a constant exists to convert between two units, doesn't mean it can be applied so easily. Dimensional analysis is a tremendously useful tool in many cases, but it can only get you so far :)

    To be a bit more specific, if you were working with an ideal gas at a constant volume (with specific heat C_v), you certainly could use Boltzmann's constant. C_v is equal to heat energy / change in temperature. From the First Law of Thermodynamics, you know that the change in the internal energy of a system is equal to heat applied + work performed. At a constant volume, there is no work done, so the heat applied is equal to the change in the system's internal energy. From the equipartition theorem (explained above) we know that the internal energy of the system is equal to 1/2 * the number of degrees of freedom/particle * the number of particles * Boltzmann's Constant * Temperature. Taking the derivative of this w.r.t temperature, we get 1/2 * the number of degrees of freedom/particle * the number of particles * Boltzmann's Constant, which is the specific heat of an ideal gas at constant volume. From there, if you know how much heat energy you are pumping into your gas, you can calculate its change in temperature, and sure enough, it is proportional to Boltzmann's constant, as you predicted. On a fundamental level, you can always have a relationship such that thermal energy is proportional to Boltzmann's constant, it all depends on how you shake the equation of state. For example, a virial expansion is an infinite sum in powers of density, but each term is still proportional to NkT, so once all is said and done, if you really want to have a k in your specific heat, you can if you play with the units just right.

    Averagesupernova, I wish I could say I got your riddle without Googling, but I've never actually heard it referred to as that. I remember in my senior EE design class at Uni the department tech spent a day showing us tricks for when prototypes go bad, and that was one of them, but he never called it a gimmick...I'll have to keep that one in my bag of tricks.
  11. Mar 28, 2007 #10
    i thought the upper case C belonged to coulumb? can't say i've ever heard of ampicity, is that real :confused: i can't belive something just because it's on the internet.
  12. Mar 28, 2007 #11


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    Would we lie to you? I've seen it in books many times long before the internet existed.
  13. Mar 28, 2007 #12
    mufusisrad, thank you very much for the information. Although I think it may be a bit beyond what I need, I'll go through it and see if I can work out some example coils. In the mean time, to excerpt an example I found a while back (http://www.bcae1.com/wire.htm" [Broken]):

    "We will assume that some imaginary piece of wire (we don't want to destroy a real piece of wire) has .01 ohms of resistance (e.g. a 15 foot long piece of 8 gauge wire) and that wire is connected directly to the positive terminal of the battery (without a fuse... that should scare you). Now let's say that the other end of the wire is allowed to touch to the chassis of the vehicle (which, in most vehicles, is connected to the negative terminal of the battery). The two battery terminals are basically shorted together by the wire (through the chassis). In this situation, a very large amount of current will flow through the piece of wire.
    If we wanted to calculate the current flow through the wire, we would use the Ohm's law formula I=E/R. If we use the ideal automotive battery, which is rated at 12 volts, and divide it by the resistance of the wire which is approximately .01 ohms, we get a current of 1200 amps.
    I = E/R
    I = 12/0.01
    I = 1200 amps
    Then plug the current into the formula P=I^2*R. We get:
    P = I2*R
    P = (1200*1200)*0.01
    P = 14,400 Watts
    This shows that the wire would dissipate 14,400 watts of heat which would melt the wire's insulation and more than likely ignite everything that comes in contact with the wire (fuel lines, other wires, carpet, plastic, insulation)."

    I can figure out all the same information on the wire coils I'm dealing with and making; however, because the manufacturers all primarily site the thermal limits of their wire insulation in terms of 105°C etc., that's where I need to be able to figure the conversion from the above example's 14,400 Watts to °C. Being more of the "cut and dried" type (My method of explaining the First Law of Thermodynamics - throw a sealed can of beans on a campfire and wait a couple of minutes, it will explode:surprised), I usually work better with just the formulas involved (and an explanation of the elements of the formulas if they're new to me) as all of the minutia of theory can make my brain trip over itself. :-) Thanks again for the feedback.
    Last edited by a moderator: May 2, 2017
  14. Apr 1, 2007 #13
    I've finally had some time out of my week to go over and into the answers I've received to my OP and I've got some additional questions.

    If I'm understanding all of this correctly, it makes sense to me that; based on power flow through the wire coil being "x" (as once a coil is constructed the speed of the magnetic field is the only changable variable while using that coil); heat produced within the wire coil would be divided by the mass of the wire in the coil. However, what I'm still not sure of is what unit-basis of heat (watts?), divided by what unit-basis of wire volume (cubic meters? cubic centimeters?) will produce what unit-basis of heat (K? °C? °F?).

    Or, am I completely wrong having jumped the track in a confusion of input overload?
  15. Apr 2, 2007 #14
    Could you perhaps elaborate a bit on the application of the coil you're attempting to make (unless, of course, it's a secret)? If I know more about the application, I may be able to help you make some approximations that would make the problem a bit easier.

    To address the second part of your post, you know how much power your coil is going to be absorbing because you know its resistance and the amount of current you will be pumping into it. The equation, I^2*R tells you the rate of energy passing through your wire in Joules/second, assuming you measure current in Amperes and resistance in Ohms. You can assume that the majority of this energy will go to heat, though some of it will inevitably go to the thermal expansion of the wire as well. Even knowing the specific heat and mass of your wire, you still don't have enough information to know how "hot" your wire will become because that is a problem of kinetics and not simple thermodynamics. The heat that is generated by the current passing through the wire will not just stay inside of the wire and constantly contribute to an increase in the temperature of the wire. This heat energy will instead be exchanged with the wire's environment (i.e. its insulation and the ambient air around it). The amount of heat transferred to the insulation and the rate at which that heat is distributed to the air is a function of the material that the insulation is made out of and the temperature of the environment at large (i.e. the air in the room). To use mathematics to get a reliable estimate of the steady state behavior of the thermodynamic behavior of the wire at a constant current is not going to be as simple as plugging a few numbers into a formula. This is where this stops becoming a physics homework problem and becomes an engineering design issue. Engineers designing cabling for power transmission typically use complicated finite-element modeling software to address these sorts of problems. For simpler applications, the engineers who designed your wire have already characterized and tabulated its thermal limits in many different environments. This is where ampacity comes in.

    Ampacity is a metric for judging whether or not a particular type of wire can withstand a certain current at a certain temperature. If you know the ampacity of your wire for a particular temperature, then you know the maximum current that can be safely passed through it according to specs. So long as you stay below that threshold, you should be OK. As far as getting a direct conversion from power dissipated to temperature change in the conductor, that problem is too complicated to address directly knowing just the current, resistance, and specific heat of the conductor. For a good example of another layer of complexity that can be added to the problem, remember that the resistance/resistivity of your wire is a function of temperature as well.

    To address your question regarding units, heat is a form of energy, and as such, in the MKS system it is given in Joules. Power is the rate at which heat (or more fundamentally, energy) flows into/out of something and is given in Watts (Joules per second) in MKS. In my experience, you should be careful, as material densities for conductors are usually given in grams per centimeter cubed (g/cm^3) and specific heats are usually given in Joules per gram Kelvin (J/gK). Keep in mind that when you use a specific heat, you are dealing with a change in temperature. As Celsius degrees and Kelvin "degrees" are the same size (the zero reference is just shifted), your result will give you a meaningful answer in both units. Hopefully, you'll find the points I've addressed useful.
  16. Apr 3, 2007 #15
    Good Evening All,
    I appreciate all of the help, information and feedback I’ve received to my OP. Thank you very much. Along with all of the help here, I’ve been doing a good bit of searching and researching on the web and I’ve found a very useful site called Aus-e-tute at http://www.ausetute.com.au/" [Broken]. They list a formula I believe I can use to figure temperatures from watts, that I would like feedback on. Preliminary calculations work out well for my needs, but I would like opinions on it from more experienced people. The formula is based as: The amount of heat energy (q) gained or lost by a substance is equal to the mass of the substance (m) multiplied by its specific heat capacity (Cg) multiplied by the change in temperature (final temperature - initial temperature) and is stated as: q = m x Cg x (Tf - Ti). Now, using the conversion of 1 lb. = 453.59g (since I can figure out the weight of the wire in lbs.); the specific heat capacity (Cg) for copper they list as being 0.39 (though I prefer 0.385); 30°C (86°F) as a basic start at room temperature for the initial coil temp (Ti), and since the wattage output is known, I can mathematically restate the equation to solve for the final temp (Tf) caused by that wattage; with my restatement of the equation being:

    [q (known watts stated as J/s) / (m)ass * 0.39 (Copper Cg)] + 30°C = (Tf)°C

    Basically, what I need to know is if this calculation will actually work as I think it will; even if it only gets me an accuracy of +/- a couple of °F; or if I’m only wishfully hallucinating. Thanks to All.
    Last edited by a moderator: May 2, 2017
  17. Apr 4, 2007 #16
    i'll attempt to solve.

    wire 18 guage is 6.385 ohms at 20 degrees celcius per 1000 feet, 1000 feet = 4.917 pounds;
    6.385/4.917=1.468 ohms per pound;
    temp coefficent for copper is 0.00426 per ohm at 0 degees celcius;
    starting temp 30 degrees celcius,
    0.426 + 1.468= 1.894 ohms per 1 pound spool

    (1 + (coefficent at 0 degrees * temp1) / 1 + (coefficent at 0 degrees * temp2) x resistance at temp1 = resistance at temp2

    (1 + (.00462 * 0) / 1 + (.1278 * 30) x 1.894 = .389

    thats going from 0 degrees to 30 celcius, i short i don't think you can solve this unless you have another know quanity, wattage by itself is 3.
    Last edited: Apr 4, 2007
  18. Apr 6, 2007 #17
    Hi Light_Bulb,
    Thanks for taking a stab at the equation I presented. I have a couple of questions for you though.

    Okay, I'll buy that,

    but my calculator says that 6.385/4.917=1.29856 ohms per pound or 6.385/1000=0.006385 ohms per foot

    What is the basis for 0.00426 per ohm at 0 degrees C? I thought that a temp. coefficient was a % per degree C?

    Why did the temp. coefficient of 0.00426 get multiplied by 1000 to get the 0.426? A one pound spool has 203.376 feet of 18 guage wire in it.

    What's the purpose of adding 1 where you did?

    I don't at all understand what you're doing in this part. 1+[?] (co-e*start temp.) / [divided?] by 1+[?](.1278*30)[why muiltiply the co-e by 30 and then multiply it by 30 again??] x resistance at temp 1?

    If I re-run the equation I get the following:
    (1 + (coefficent at 0 degrees * temp1) / 1 + (coefficent at 0 degrees * temp2) x resistance at temp1 = resistance at temp2

    (1+(0.00426 * 0) / 1 + (0.00426 * 30) * 1.29856) = resistance at temp2

    (1+0 / 1 + (.1278) * 1.29856) = resistance at temp2

    (1 / 1 + .165955968) = resistance at temp2

    1 / 1.165955968 = .857665321

    This seems like an interesting equation, but I need you to correct and elaborate on it please. How did you come up with 3 watts? And what does this have to do with the specific heat equation I presented?
  19. Apr 6, 2007 #18
    i didn't come up with three watts, the equation you want needs three variables :tongue: the one your missing is the final resistance. the rest is by the book, (basic electricity and electronics, schuler/fowler).
    Last edited: Apr 6, 2007
  20. Apr 11, 2007 #19
    Hi Mufusisrad,
    I apologize for not getting back sooner. It's been a busy couple of weeks. In the specific heat equation that I found; q = m x Cg x (Tf - Ti); which I was restating as [q (known watts stated as J/s) / (m)ass * 0.39 (Copper Cg)] + (Ti)°C = (Tf)°C, I was using a Ti of 30°C (86°F) as a "room-temp" starting point. It is the ending temperature (Tf) equivalence in °C or °F at a power flow of x watts of heat that I'm after. that's why I was asking for verification of the formula I found and was re-stating.
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