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I am using a nichrome wire for my project.
Q.1. I want to know which source should I use as my voltage AC or DC? I know that the average power dissipated in a resistor in one cycle(in AC source) is equal to the average power dissipated in DC source. But I want to know which would be efficient for me ... I want to use this to cut fabric.
Q.2 First of all the power dissipated in the circuit will be in the wire(which will act as a resistor). And also the wire will radiate heat energy(Stefans law depending upon emmisivity). So the temperature of the wire will increase for some time and then become constant.
I want to know if there is energy dissipating in any other form?

Edit 1: I figured out a little bit. Now please tell me if there is something to be added in place of something or not! But the problem is that e i.e. emissivity is varying. Then how to use the below formula. We are trying to find temperature but a term in the formula of temperature is itself dependent on temperature :(

$$ V^2/R=e\sigma AT^4 + mS + Something$$

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Most of the cutting wire's heat loss will be through conduction: into the air, into the mounting hardware, and especially into the material being cut. Melting the material is an endothermic process, meaning it will lower the temperature of your wire more than heat diffusion alone.

All of these heat losses (plus the radiated heat) will increase with the temperature of the wire, though it won't be a simple linear relationship. The wire's temperature will rise to an equilibrium value where the rate of heat loss is equal to the power supplied.

Since the wire is a resistor, the power $P$ will be determined by $P=V^2/R$, where $V$ is the DC voltage or RMS AC voltage, and $R$ is the resistance of the circuit. Assuming the cutting wire has a much higher resistance than the wires supplying it-- which it should-- then $R$ is just the resistance of the cutting wire, and is inversely proportional to its cross-sectional area. Putting that all together, loosely speaking:

$T_{eq}\propto{P}$

$V=IR$

$P=VI=\frac{V^2}{R}$

$R\propto \frac{1}{d^2}$

so

$T_{eq}\propto {V^2}{d^2}$

Where $T_{eq}$ is the equilibrium temperature, $d$ is the wire diameter, and there is some constant of proportionality that would be too hard to work out. In practice you just pick a thin wire and experiment to find the lowest voltage that gets it hot enough to cut your material.

It doesn't make any difference whether you use AC or DC, except that of course the specific choice of power supply affects the overall efficiency. But just to make this explicit, do not run mains AC through your cutting wire! You don't need anything like that much voltage or current, and having a bare wire connected to a live wall outlet is extremely dangerous. And it'd melt your cutting wire anyway.

As I understand it, the reason some hot wire cutters use AC is mainly that something like a 30V AC power supply is simpler and cheaper than a comparable DC supply, though I suspect this was more an issue in the olden days.

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  • $\begingroup$ I figured out a little bit. Now please tell me if there is something to be added in place of something or not! But the problem is that e i.e. emissivity is varying. Then how to use the below formula. We are trying to find temperature but a term in the formula of temperature is itself dependent on temperature :( $$ V^2/R=e\sigma AT^4 + mS + Something$$ $\endgroup$ – brainst Feb 4 '16 at 12:17
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AC or DC is fine, however do not exceed 48 volts. This is the known safe limit to ensure that you will not be injured even in wet conditions. This is most easily done by purchasing a sealed dc power supply.

The heat dissipation by the wire changes primarily based on the energy input and the speed at which it is pulled through the material (phase change conduction). Find your nichrome wire resistance per foot in this table and do some V=IR Ohms Law calculations for getting in the ball park, but I recommend getting a PWM LED dimmer so you can easily adjust the power going to the wire as needed. Note that the linked dimmer is only rated for 12volts 8Amps, but that should be more than enough. This will be very handy especially if you frequently change the wire gauge, length of wire, or material being cut.

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  • $\begingroup$ Thanks for motivating me! I'll definitely keep in mind the precautions. Actually I'll be changing the wire guage and length of the wire(You read my mind!). But what about the temperature? I want to know the temperature at steady state. $\endgroup$ – brainst Feb 4 '16 at 11:43
  • $\begingroup$ It would be a difficult set of calculations and your results would be dependent on a lot of assumptions. Especially where you want to change the temperature for different situations, just design a setup to have variable power and determine your operating parameters empirically (by testing). I would say the 12V 8A supply i recommended could easily power a 4foot hot wire cutter. To deliver max power we want V=IR 12=8*R, R=1.5 So we want our resistance to be 1.5 Ohms or greater. So per the table 19gauge would give you 3ft at max power. $\endgroup$ – ericnutsch Feb 4 '16 at 16:27
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The power a resistor will dissipate is the same for a particular DC voltage applied to it, or AC of the same RMS voltage. The only difference between these AC and DC cases is that the power comes in clumps over time with AC, while continuous with DC.

If the AC is in the form of a sine, then the power is also a sine but at twice the frequency. For example, if you drive your nichrome wire from a transformer output running from 60 Hz power, then the power into the wire will vary from 0 to maximum at 120 Hz.

AC at the same RMS will have the same average power, so the question is at what frequency does the non-continuous power of AC matter. To answer that, you have to look at the thermal time constant of the wire. AT 120 Hz, you get a pulse of heat every 8.3 ms. That is almost certainly a lot shorter than the time the wire can meaningfully change temperature.

Consider that old LEBs (light emitting bulbs) used filaments considerably finer than your cutting wire. The light output is a non-linear function of temperature, such that a little more temperature makes disproportionately more light. Even so, LEBs weren't generally considered to flicker at the line frequency. For example, you could swing a bulb around at the end of a cord and not notice flicker.

Since LEBs exhibited little flicker at 50 and 60 Hz, your much thicker wire with longer thermal time constant isn't going to have meaningful temperature ripple either. For your purposes, 50 and 60 Hz AC is continuous in heating your wire, and you only see the average over a longer period than a AC cycle or two.

For your purposes, 12 V DC and 12 V AC at 50 or 60 Hz are equivalent.

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  • $\begingroup$ If you could please add something about finding the steady state temperature then that would be awesome! $\endgroup$ – brainst Feb 4 '16 at 11:44

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