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How to calculate the volts/amps of a battery which is connected to a simple wire for heat generation?

That is, how to use:

known wire length, diameter/area, resistivity

to calculate the required volts/amps to heat the wire to 30 degrees Celsius?

the surrounding medium is air,

the ambient temperature is 10 degrees Celsius.

There is no movement of air.

The wire is more or less in a straight line.

The wire is not insulated.


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3 Answers 3

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Step 1 is to figure out how much power the wire will dissipate when it is at 30 C. That is not electrical engineering. It is thermal engineering. It depends very critically on the ambient temperature and airflow conditions. So I am not going to explain how to do that. You have to figure it out some other way. But I will caution you that any change in airflow or ambient temperature will lead to a change in the required power. So if you need to maintain the wire at 30 C, you probably need to use temperature sensing and feedback rather than a fixed voltage and current.

Let's assume you have solved for 'Pdis', the power dissipation of the wire.

You can calculate the resistance of the wire as follows:

  1. R = rho * L / A

where R is the resistance of the wire, rho is the resistivity of the wire material (in Ohm-meters) L is the length of the wire (in meters), and A is the cross sectional area of the wire (in square meters).

Please note that for many conductors, rho is very sensitive to temperature (4% increase every 10 degrees). So if you are using copper or aluminum or steel or stainless steel, you may need to allow for this by using the corrected rho.

Now we have Pdis and R. The easiest thing to do is solve for voltage first. Start with the formula for power dissipation:

Pdis = V2 / R

Solve for V.

V = sqrt(Pdis * R)

That is the voltage needed to keep your wire at the desired temperature. If you want to solve for current, you can just use Ohm's law, re-arranged to solve for current:

I = V / R

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  • \$\begingroup\$ thanks; does this Pdis have something to do with thermal conduction? \$\endgroup\$
    – ivanantuns
    Commented Apr 17, 2022 at 19:45
  • \$\begingroup\$ in any case thanks for clarifying the rest of the problem! \$\endgroup\$
    – ivanantuns
    Commented Apr 17, 2022 at 19:46
  • \$\begingroup\$ Yes, the Pdis is going to depend on how the wire is situated. Like is it stretched out in open air? What temperature is the air? Or is the wire wrapped with thermal insulation like a wire running through a wall in a house? Etc. Basically, to me figuring out Pdis is the hard part. For me the electrical part is easy if I know Pdis. \$\endgroup\$
    – user57037
    Commented Apr 17, 2022 at 20:23
  • \$\begingroup\$ cheers. that is much more clear now. And apart from the thermal engineering part , i am also considering how this would be structured as an electrical system. Could there maybe be a voltage regulator between battery and wire, so when the temperature is reached the power can be reduced to maintain temperature? \$\endgroup\$
    – ivanantuns
    Commented Apr 18, 2022 at 7:23
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    \$\begingroup\$ Also yes I now realize that is not very hot for a wire. This is because I need the wire to be continually warming the air over a few hours, while the battery should be as lightweight as possible - but fortunately I don't believe it will require a lot of power to do so (as you said, 30 deg C is not a lot, and also the surrounding air does not lose too much heat through power distribution to the outside environment. The wire is actually inside a container which has air about 10 deg C and there is some good insulation to the outside of the container) \$\endgroup\$
    – ivanantuns
    Commented Apr 21, 2022 at 12:27
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calculate the required volts/amps to heat the wire to 30 degrees Celsius?

Volts and amps produce power dissipation in the wire. That power raises the temperature of the wire (above ambient) until the heat taken away by the wire's surroundings matches the heat produced by the power. At that point, the temperature stabilizes but, that temperature is entirely dependant on the heat taken away by the surroundings.

For instance, if the wire were suspended in a vacuum, the wire would heat up until the temperature rose to such a high level that visible light is being emitted and, if left unchecked, the wire would probably overheat and melt. End of that story.

On the other hand, if bonded in a prescribed way to a heatsink that had a thermal resistance of (say) 10°C per watt then, if dissipating 3 watts of power, the wire would reach a temperature of 30°C above ambient.

In simple terms, power does not directly relate to temperature; you need to consider the thermal resistance of the surroundings and the localized ambient temperature all along the wire.

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  • \$\begingroup\$ thank you for the answer and pointing out what i missed , i will edit the question \$\endgroup\$
    – ivanantuns
    Commented Apr 17, 2022 at 18:23
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    \$\begingroup\$ "if the wire were suspended in a vacuum, the wire would heat up until the temperature rose to such a high level that visible light is being emitted and, if left unchecked, the wire would probably overheat and melt." This misconstrues what happens in a vacuum. Nothing magical about a vacuum prevents heat from leaving the wire. Conduction and convection become negligible, but radiation still works. More correctly, the wire will reach a temperature where radiative power loss equals the input electrical power. This doesn't need to be visible light and it won't necessarily cause melting. \$\endgroup\$
    – J...
    Commented Apr 18, 2022 at 12:24
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    \$\begingroup\$ @Andyaka The point is that checking or not checking doesn't matter - the amount of input power is what matters, and the construction of the wire. These are all computable quantities and systems can be designed so that a current-carrying wire in a vacuum comes to any desired temperature. Nothing about being in a vacuum means it will melt simply by not being observed. \$\endgroup\$
    – J...
    Commented Apr 18, 2022 at 14:46
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    \$\begingroup\$ You don't have to mention vacuums or melting at all - the first paragraph is correct and applies to vacuum as well as non-vacuum. The wire will always stabilize at a temperature where energy in equals energy out. Nothing changes when going into a vacuum other than the mechanisms available for heat rejection. You don't need to go down that path, certainly, but if you are going to go down there (which you did) you should at least go the right way. \$\endgroup\$
    – J...
    Commented Apr 18, 2022 at 15:05
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    \$\begingroup\$ The only thing I can see that we disagree with is whether the wire will melt and how we state that. I say "probably" and you say it won't necessarily melt. I'm not seeing any other difference other than it may be visible light or not. Maybe you should consider making an answer to this question. \$\endgroup\$
    – Andy aka
    Commented Apr 18, 2022 at 16:57
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That is, how to use: known wire length, diameter/area, resistivity to calculate the required volts/amps to heat the wire to 30 degrees Celsius?

Can't be done without a good deal more information.

To start, you must specify the thermal value of the wire insulation. Second, you must specify the surrounding environment - what is it's temperature, what material surrounds the wire, and what is the velocity of the surrounding material.

You also need to specify exactly how the wire is configured.

So, what is the surrounding medium? Air? Water? Oil?

How fast is it flowing?

What is its temperature? Just as an illustration, if the wire is suspended in air at 30 C, it will require 0 volts and amps for the wire to reach 30 C. Or are you asking for a 30 C temperature rise above ambient?

Is the wire insulated? If it is, it will heat up more for a given power.

Configuration - Is the wire suspended in more-or-less a straight line, or is it formed into a tight coil? If it is a coil, heat lost from any part of the wire will tend to heat another part.

Unless you can provide all of these details, your question is unanswerable.

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  • \$\begingroup\$ thanks for claryfing! i will edit the question \$\endgroup\$
    – ivanantuns
    Commented Apr 17, 2022 at 18:22
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    \$\begingroup\$ Once I was measuring the steady state temperature of a flex pcb resistor under constant voltage condition. I attached a temperature logger and then went away for a while. When I came back, it looked like it had reached steady state so I bumbled around for a while and disconnected everything then looked over the log data. When I zoomed in I realized that just me walking up to the resistor and looking over the test setup caused very noticeable perturbation to the steady state temperature. This was just the airflow of me moving around near the resistor. \$\endgroup\$
    – user57037
    Commented Apr 18, 2022 at 16:25
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    \$\begingroup\$ @mkeith I cannot say how many times I was presented by data showing that my temperature instrumentation had "a problem" because of odd "cycles" and their "shapes." Of course, it turned out that it was on a 24 hour cycle and that an air conditioner was at fault. Or something quite similar. A lot of things can be detected with good equipment. \$\endgroup\$
    – jonk
    Commented Apr 19, 2022 at 16:57
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    \$\begingroup\$ @jonk sometimes I joke that everything is a thermometer. It is not hard to design something that varies with temperature. The hard thing is to design something that DOESN'T vary with temperature. I remember looking at something, maybe it was RF power data vs time, and we saw this clear sawtooth waveform. Eventually we figured out that it was going down rapidly when the AC was on, and slowly creeping up when the AC was off. Yeah, kind of temperature sensitive. \$\endgroup\$
    – user57037
    Commented Apr 19, 2022 at 17:01
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    \$\begingroup\$ @mkeith - and don't get me started on electrical noise. Quite a few years ago a guy at MIT was having trouble with noise on his instrumentation. Occurred regularly at 10 minute intervals. Turned out to be pickup from the subway line - drive motor arcing. \$\endgroup\$ Commented Apr 19, 2022 at 19:46

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