Hot Wire Cutter

[dnb_p]

Hi all,

I'm an architecture student and would be extremely grateful for some advice with a tool/machine that im making. Basically i'm creating a large hot wire cutter and need to pass enough power through to heat a 1.6mm, (or 2mm gauge) 1m long nichrome wire to a temperature suitable for cutting. This gauge increases the stiffness which is necessary when dragging through the foam.

This link below displays a hand tool that works in a similar way. This tool uses 0.8mm nichrome wire and comes with an adjustable transformer that can supply 2 Amps @ 16 volts

https://images-na.ssl-images-amazon.com/images/I/817vqvmvMML._SL1500_.jpg

I've tried using an adjustable power supply at 3 amps, 30 volts with crocodile clips but had no success. I obviously don't have much knowledge in this sector and need to find a way to heat the wire, any help would be great!

 

[Richard Burns]

The thicker the wire the lower the resistance and so more power is needed to generate heat in the wire.
Generally heating elements are fairly low resistance but consume a lot of power.
Your power supply needs to be capable of outputting significant current, which can be expensive.
I am surprised that the tool specified has enough heat output.
However if that one is OK then your wire will need about four times more power to heat to the same level assuming 1.6mm is the diameter of the wire.(resistance is proportional to the square of the radius)
So 2A @ 16V is 32W
you would need 5A at 30V (=150W) or more to generate similar heat I would have thought, off the cuff, 8A would be better.
Generally finding a reasonably price power supply can be expensive and if it is for occasional use then a 12V battery (that can supply a lot of current) can be successful, although I am not sure of the actual calculations involved.

 

[Lucien Nunes]

The current and voltage required depend on the length, gauge of the wire, its resistivity and the heat output needed per unit length. smoke-packet approximation with top of head figures:

Nichrome alloys vary in resistivity, 1.2μΩ.m is average. Resistance of 2mm dia wire will be 1.2/(Π(1/2*2)²)=0.38Ω/m
Heat output required (this is the real guess as there are many thermal factors involved) perhaps the power of an ordinary soldering iron (25W) per 25cm of wire, i.e. 100W/m.
Current = sqrt(100/0.38) = 16A
Voltage = sqrt(100*0.38) = 6.2V
This suggests your previous experiment failed because the supply had inadequate current rating. We don't know the heat input required accurately, but the resistance of around 0.4Ω is fairly certain for 1m of 2mm dia wire. So whatever voltage rating of supply you use, the current rating will need to be around 1/0.4 = 2.5x greater. E.g. a 5V 15A unit might do, but a 15V 5A one won't.

E2A I have used 2mm dia for the example as this is the more demanding situation. Obviously for 1.6mm replace the 2mm in the resistance calculation with 1.6, giving 0.6Ω/m, 13A, 7.7V respectively for the same heat dissipation.

And...

resistance is proportional to the square of the radius

Oops, I think you meant to say inversely proportional

 

[Richard Burns]

And...
Oops, I think you meant to say inversely proportional

You might well be right there, perhaps I should just check with my solicitor... oh, it appears you are entirely right and it is all my fault!

 

[Marvo]

I'd just use a 12v alarm type battery and an adjustable resistor in series with the heater wire to limit the current and give some temperature control. If the voltage is too low to get the wire hot enough then double up on the batteries to get 24v. You'll probably find the current is kinda self limiting as the resistance of the wire will go up with the temperature so ohms law assumptions of current consumption might be inaccurate.

 

[Lucien Nunes]

Trouble is though Marvo, this wire is so thick the resistance is less than half an ohm. With the 2mm version and a 12V battery you would probably need to dissipate more heat in the resistance than in the element itself, and an alarm battery won't last long being discharged at 1-2C.

The way I would do it here as a test, if I didn't have big variable power supplies, would be to get any mains transformer of 200VA or more with some window area available, and wrap a dozen or more turns of 2.5mm² through it as a secondary. A toroidal of that size might have 3-4 turns per volt so by adding or subtracting turns the heater supply could be adjusted in steps of 0.25-0.35V

I deliberately ignored temperature coefficient of resistance as it's small (~0.0004) for typical nichrome alloys, a tenth of that of copper, so the resistance / current won't vary by more than say 10% from cold to full working temp.