Connect with us

Figuring out how hot a wire will get

Discussion in 'Electronic Basics' started by Harry Muscle, Dec 11, 2003.

Scroll to continue with content
  1. Harry Muscle

    Harry Muscle Guest

    I'm not sure if this is the right p to ask, but since it's dealing with
    electricity, I hope it is.

    Is there a way to estimate how hot a length of wire will get ...

    I'm gonna be running 7V 14.3A through 24 feet of 22 Gauge teflon coated wire
    submersed in water. I would like to maintain 55 gallons of water at 25.5C
    or 78F. I'm obviously going to need to know the thermal resistance of
    teflon and some sort of formula to figure out how quickly heat is going to
    pass from the copper, thru the teflon, and into the water. But I don't no
    either :(

    The main reason why I'm trying to figure this out is cause I would like to
    know how much the resistance of the copper wire is going to increase, since
    if it increases too much I might need to use one gauge higher.

    Any help is highly appreciated,
    Thanks,
    Harry
     
  2. Ian Stirling

    Ian Stirling Guest

    It may measurably increase.
    It will not change the wattage much though.
     
  3. Besides the dimensions of the insulation, and the correct formula for
    radial thermal resistance with radial heat flow, you need the thermal
    resistivity of teflon. I am assuming that your insulation is PTFE or
    something close. Here are some links with thermal resistivity:
    http://www.taconic-add.com/de/pdf/tlx.pdf
    http://www.bayplastics.co.uk/data sheets/dsm/Product data sheet - (PTFE100).htm
    http://www.pragatiplastics.com/pt_tubing.htm
    If you google "thermal resistivity" and teflon or PTFE, or whatever,
    you might come up with other references.

    You may be able to pull the formula for the resistance out of:
    http://www.delphi.com/pdf/techpapers/2000-01-0301.pdf
    page 9 of:
    http://baen.tamu.edu/users/rmoreira/bsen366/pdf/LecChap04r.pdf

    The bigger mystery is how you arrived at 100 watts as the correct
    amount (or at least enough) of heat to keep 55 gallons of water
    exposed to an unknown environment to 78 degrees F.
     
  4. Rich Grise

    Rich Grise Guest

    How do you plan to regulate it? Just do it empirically. That is,
    open your control loop and put in various control signals, and
    see where the temp. of the tank settles out. Draw a graph.
    That answers your question, and will probably give the world
    some information that nobody's bothered to try to get before.
    (Using copper for a heating element? Why not some calibrated,
    known, waterproof heating element wire?)

    Good Luck!
    Rich
     
  5. Harry Muscle

    Harry Muscle Guest

    Thanks for those links, I'll check them out. But to answer you question
    about how I figured out to use 100W ... that was the easy part. I looked at
    formulas used in determining heater sizes for aquariums, like those on the
    following sites:

    http://www.kernsanalysis.com/HeaterCalculator.cgi
    http://www.geocities.com/ResearchTriangle/Lab/2024/thermo1.html

    Harry
     
  6. Harry Muscle

    Harry Muscle Guest

    I was planning on using a temp controller to either turn it on or off.
    However, I'm trying to figure out the highest temp of the actual wire not
    the water since I would like to know roughly the change in resistance that I
    can expect. As for the reason for using copper is that I need a heating
    element about 24 feet long, and running small gauge copper seems to fit the
    bill very nicely.

    Thanks,
    Harry
     
  7. John Fields

    John Fields Guest

     
  8. Dan Akers

    Dan Akers Guest

    Harry wrote;
    "Is there a way to estimate how hot a length of wire will get ...
    I'm gonna be running 7V 14.3A through 24 feet of 22 Gauge teflon coated
    wire submersed in water. I would like to maintain 55 gallons of water at
    25.5C or 78F. I'm obviously going to need to know the thermal resistance
    of teflon and some sort of formula to figure out how quickly heat is
    going to pass from the copper, thru the teflon, and into the water. But
    I don't no either :(
    The main reason why I'm trying to figure this out is cause I would like
    to know how much the resistance of the copper wire is going to increase,
    since if it increases too much I might need to use one gauge higher."
    _____________________________________
    Re;
    For the wire proper, the temperature rise from the copper surface to the
    centerline is (for all practical purposes) dependent only upon the
    "uniform volumetric heat generation rate", the dimensions of the wire
    and the thermal conductivity of copper (386W/mC). In this case, the
    volumetric heat generation rate is about 4.196E7W/m^3. However, the
    relatively high length to radius ratio and the high thermal conductivity
    of copper, yields for this case, a temperature rise of only 0.0028 C;
    for all practical purposes; ZERO; you can assume isothermal conditions
    through it's diameter.

    The heat flux at the surface of the copper is going to be about 6760
    W/m^2. Disregarding the teflon insulation for the moment, if you take a
    look at what is called the "boiling curve" (usually found in heat
    transfer and fluid flow literature), this heat flux will result in a
    Tsurface-Tsat of only about 6C; implying a copper surface temp. of about
    106C. The heater would be operating in the "nucleate boiling regime"; a
    very efficient natural convection heat flux. However, I suspect that
    the insulation is going to require the copper temp. to be significantly
    higher. You need to post the thickness of the teflon insulation.
    As for wire size, if you DECREASE the wire diameter (and not increase
    it's length), as I think you suggest, then you will proportionally
    increase the heat flux (assuming wattage is the same). This will worsen
    the heat flux situation, resulting in even higher heater surface temps.
    Thus, if you decrease the diameter with the same total power, then
    increase the length to maintain the heat flux below 10,000W/m^2.
    The teflon insulation is the key unknown; you need to know it's
    thickness and thermal conductivity to accurately predict your copper
    temperature.
    As for the temperature effect on the resistance of your heater; an
    increase to 150C would result in the heater resistance increasing to
    about 0.608 ohm, cutting your power by about 20% (assuming the voltage
    remains the same).

    -Dan Akers
     
  9. Dan Akers wrote:
    (snip)
    (snip)

    Are you saying that if the copper were in contact with the water, and
    the heat flux were 6760 W/m^2, the interface would be at boiling
    temperature even though the bulk water temperature is only 25 degrees
    C? I don't think this heat flux is anywhere near that high. What am
    I missing?
     
  10. Dan Akers

    Dan Akers Guest

    Harry wrote;
    <snip>
    "The main reason why I'm trying to figure this out is cause I would like
    to know how much the resistance of the copper wire is going to increase,
    since if it increases too much I might need to use one gauge higher."
    _____________________________________
    Re;
    Hi again. I did a search for the thermal conductivity of TFE insulation
    and came up with 8.2W/mC. Thus, for 10mil thick insulation on your
    wire, that would correspond to only a 0.5 C temp rise or so across the
    insulation proper. Given the stated geometry of your problem; 24ft long,
    about 1.15mm (assuming 10 mil insulation) total diameter; for 100W
    dissipated at a heat flux of 3800W/m^2 on the insulation's surface, in
    still water and 10 mil TFE insulation, the copper temperature would be
    about 107C. I based this on looking up the given heat flux, which puts
    you well on the thermodynamic boiling curve for surface heat transfer in
    water. The required surface temp. is about 6C above the prevailing
    saturation temperature (which would be about 100C assuming the heater
    isn't to be submerged to any great depth or that the aquarium is not
    under some external overpressure). So adding all of this up, gives an
    insulation surface temp. of about 106C and a copper temp. of just
    slightly higher, about 107C or so. I hope that helps...

    -Dan Akers
     
  11. Dan Akers

    Dan Akers Guest

    John Popelish wrote;
    "Are you saying that if the copper were in contact with the water, and
    the heat flux were 6760 W/m^2, the interface would be at boiling
    temperature even though the bulk water temperature is only 25 degrees C?
    I don't think this heat flux is anywhere near that high. What am I
    missing?"
    ______________________________________
    Re;
    Yes. The length of the wire is 7.31m and the bare circumference is
    2.02E-3m (22AWG; 0.644mm dia.), making the heat transfer area only
    1.48E-2m^2. Divide this into 100W and you get 6752W/m^2. This heat
    flux in still water wil result in some boiling at the surface of the
    heater even though the bulk liquid remains substantially subcooled. It
    is known as nucleate boiling, where minute vapor bubbles form at the
    heat transfer surface, detach themselves through buoyancy and the force
    of the prevailing convective currents, and then subsequently collapse as
    they condense in the bulk fluid, a short distance from the heat transfer
    surface. This mode of heat transfer is very efficient since the small
    vapor bubbles rapidly carry the relatively large amounts of latent heat
    into the bulk fluid. This phenomena begins in earnest at a heat flux of
    about 1000 W/m^2; below that is simple convection.

    -Dan Akers
     
  12. Harry Muscle

    Harry Muscle Guest

    Wow, thank you so much. I hope you don't mind if I ask about one thing.
    I'm trying to get the temp of the water to reach around 25.5C, not boiling
    (100C). So if I understand the above explanation correctly, the wire will
    reach 106-107C however, it will be just a matter of turning off the heat
    once I reach my temp. I just wanted to verify that cause I'm a little
    confused by the "6C above prevailing saturation temperature" ... I just want
    to make sure that it's not supposed to be 6C above my temp of 25.5C.

    Thanks,
    Harry
     
  13. Kent Regal

    Kent Regal Guest

    I'm gonna be running 7V 14.3A through 24 feet of 22 Gauge teflon coated wire
    After 13 messages and how many days of trying to re-invent the wheel ??? or at
    least reverse engineer a readily available hobbiest item ..... Aquarium
    heater......... they are available everywhere ! ! at reasonable prices AND
    they have the control circuit already builtin ...... of course they run off the
    house power supply of 120VAC and also available in 220 volt models.

    Why the exercise and possible expense of cooked fish???
     
  14. Harry Muscle

    Harry Muscle Guest

    Actually I'm not trying to reinvent a standard aquarium heater, but
    substrate heating cables, that run around $150 no matter where you look. So
    given the fact they that are ridiculously overpriced and the fact that I
    already have a large enough transformer lying around and the only cost
    involved for me is going to be the control circuit and teflon wire, I'm
    looking at saving at least $100. Which makes it quite attractive to me.

    Thanks for asking though,
    Harry
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-