How much volt power needed to heat copper rod.?

Discussion in 'General Electronics Chat' started by Anzar Anz, Mar 13, 2016.

  1. Anzar Anz

    Thread Starter New Member

    Mar 13, 2016
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    How much volt power needed to be supplied to heat a copper rod upto 70°c and length=3feet..? What's is the calculation.
    Am using a temperature controller to heat the coil.
    Am not much involved in electricals. . . TIA
     
  2. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    You need the cross section area of the copper rod AND the length to figure resistance.
    Then power lost to heat can be calculated. Current(squared) times resistance.

    THEN you need to know the RATE of heat lost to the environment around the copper rod, before you can calculate the power needed to raise temp by 70C
     
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  3. Pinkamena

    New Member

    Apr 20, 2012
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    Hi Anzar! What heats the rod will be current, not voltage. Your coil will function as a resistor, and the current passing through it will deposit its energy in it. The equation for power deposited in a resistor is
    P = R*I^2 , where P: Power in watts, R: Resistance of your coil in Ohm, and I: Current in Ampere.​

    To solve this, you will need to first find the resistance of your coil. You can either measure it directly if you own a precision ohmmeter, or you can approximate it by the following equation:
    R = ρ*L/Ac, where ρ:Resistivity of copper, L: length of your rod in meters, Ac: cross sectional area of the rod​
    The resistivity of copper is about 1.68*10^-8 ohm-meter.

    We can now look at the current required to heat it to 70 degrees. This will depend on how much energy will be lost from the rod to its surroundings. This is not easy to calculate, but we can make some assumptions. If we assume that most of the energy lost is due to convection and not radiative, we only need one equation:
    q = hc*A*dT, where q is the energy loss in watts per second, hc is the convective heat transfer coefficient, A is the total area of the rod, and dT is the temperature difference between the rod and its surroundings.​

    hc varies strongly. For natural convection in air, it's between 10-100. The area of your rod is easy to calculate with the equation for the surface area of a cylinder. dT will simply be 70 minus your ambient temperature.
    So lets put all this together now to find the current needed to reach, and maintain, 70 degrees. The rod will stop increasing its temperature when the energy deposited in it equals the energy it loses to its surroundings. Therefore, by combining the three equations so far:

    P = R*I^2
    P = (ρ*L/Ac)*I*2
    This is the power deposited in the coil from the current. This power must equal the power lost to the surroundings. We substitute q for P:
    hc*A*dT = (ρ*L/Ac)*I^2
    And solve for I^2:
    I^2 = {hc*A*dT}/{ρ*L/Ac}) = {hc*A*dT*Ac}/{ρ*L}
    Since the total area A=L*2*pi*r, and Ac = pi*r^2, this equation simplifies to
    I^2 = hc*2*pi^2*r^3/ρ
    Take the square root of I^2 to get the final result
    I = pi*r*sqrt(hc*2*r/ρ)
    With this equaton, you only need to know the radius of your rod (r), the convective heat transfer coefficient (hc), and the resistivity of copper (ρ). Naturally this is only a coarse approximation. We have ignored other energy loss mechanisms, and you cannot know which value exactly to use for hc. But it will give you a ballpark idea of what current is required for thermal equilibrium at 70 degrees. You will have to do some trial and error to find the correct value, by measuring the temperature of the rod directly. When it reaches 70 degrees, and doesn't change, you know the current you require.
     
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  4. hp1729

    Well-Known Member

    Nov 23, 2015
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    What diameter? About 12 mm? How many meters long?
    How much power do you have available?
     
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  5. Anzar Anz

    Thread Starter New Member

    Mar 13, 2016
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    6mm rod. 1m length am using a 240V AC temperature controller for heating and surrounding of rod is insulated
     
  6. Pinkamena

    New Member

    Apr 20, 2012
    22
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    If the rod itself is insulated then the energy loss will be much smaller than if it was just standing free in air. It's impossible for us to give you an exact figure, so you should really just find it by trial and error. Try with a low current first, say 2A, and see where that gets you, then increase until you reach a stable 70 degrees.

    EDIT: From the picture it looks like you will be using the rod to heat something up, like a container or something... If the rod is meant to be a heating element, then naturally it will lose most of its energy into the medium you are attempting to heat up. If this is the case, you would perhaps want to use something else than copper. It has quite a low resistivity, so you will need very large currents to heat it. This current will also have to pass through your power supply, which would probably heat up substantially as well unless you got a supply with very high current capabilities. I would ditch the copper rod and get a heating element. You can buy these on ebay, for example.
     
    Last edited: Mar 13, 2016
  7. hp1729

    Well-Known Member

    Nov 23, 2015
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    Ouch! 240 V through an exposed coil? Did the original system use a step down transformer?
     
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  8. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    5,813
    1,105
    Copper is about the worst metal you could choose for a heating element, in view of its low resistivity. Do you have access to nichrome wire?
     
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  9. KMoffett

    AAC Fanatic!

    Dec 19, 2007
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  10. Anzar Anz

    Thread Starter New Member

    Mar 13, 2016
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  11. hp1729

    Well-Known Member

    Nov 23, 2015
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    Could this be an induction heating system? 10 KHz, or such. Not exposed 240 V AC coil?.
     
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  12. Pinkamena

    New Member

    Apr 20, 2012
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    Uh, buddy, you merely quoted my post and added no text yourself.
     
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  13. Pinkamena

    New Member

    Apr 20, 2012
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    Really doesn't look like it from the picture, and no mentions of any induction heating. No, seems like OP wants to just heat a copper rod to 70 degrees with DC. Good luck to him.
    Anzar: You're gonna get nowhere with copper as a heating element.
     
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  14. TheButtonThief

    Active Member

    Feb 26, 2011
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    That rod in your picture looks a lot thicker than 6mm. Are you hoping to heat the rod up to 70 or are you using the rod to heat something up to 70?
     
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  15. Anzar Anz

    Thread Starter New Member

    Mar 13, 2016
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    Hey thank all buddies for. Actually it's an induction heater it must heat refrigerants that are use in refrigeration system. The vapour point of the refrigerant is very less. I must increase the pressure by heating the refrigerant gas in a closed chamber. That is setup I made.

    As the copper is high thermal conductivity the pressure of refrigeration increases very fast.
     
  16. TheButtonThief

    Active Member

    Feb 26, 2011
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    Copper is indeed a very good conductor of heat but because it's a very good conductor of electricity, it doesn't make a very good heating element. For a heating element, the poorest conductors do the best job.
     
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  17. hp1729

    Well-Known Member

    Nov 23, 2015
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    Do you have any more info on what the rest of the induction heater was like? Frequency could be anything between 10 K W to 100 K W.
    What is the biggest circuit breaker you have available? Maybe 400 Amps?
    My guess is a lower frequency, 10 K Hz to 50 K Hz, but I wouldn't estimate power requirements. I have never seen a coil shaped like that before.

    (edited to add ...)
    Dodgydave and Kermit2 have responded to these questions in the past.
     
    Last edited: Mar 14, 2016
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