? on how to calculate current requirements for resistance wire

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The discussion centers on calculating the current requirements for nichrome wire used to heat plastic laminates to 375 degrees Fahrenheit. The original setup involves 6 feet of wire at 0.8 ohms per foot, drawing 12 amps at 120 volts, but the user seeks to reduce the voltage to 30 volts for safety. Calculations suggest that at 375 degrees, the resistance of 12 feet of wire would be approximately 9.8 ohms, requiring around 4.5 amps at 44.1 volts to achieve the desired temperature. There is also mention of the temperature coefficient of nichrome, which affects resistance as temperature increases. Additionally, a recommendation is made to consult an Acrylite Fabrication Manual for further guidance on heating methods.
tmossberg
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I am attempting to make a set up for bending plastic laminates along a straight line by heating it to 375 degrees using a nichrome hot wire and 120 vac.
What I know so far is 6' of wire @ .8 ohms per foot will reach 1700 degrees with 120v and will draw 12 amp.
My device will need to be 12' long and I want to reduce the voltage to a safer working level, say 30v, is there a way to calculate the amps required to produce 375 degrees at 30 volts?
 
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tmossberg said:
I am attempting to make a set up for bending plastic laminates along a straight line by heating it to 375 degrees using a nichrome hot wire and 120 vac.
What I know so far is 6' of wire @ .8 ohms per foot will reach 1700 degrees with 120v and will draw 12 amp.
My device will need to be 12' long and I want to reduce the voltage to a safer working level, say 30v, is there a way to calculate the amps required to produce 375 degrees at 30 volts?

Is the plastic in contact with the NiChrome when you are measuring the votage?

If not, one might say that your wire is 0.8 Ohms / foot, but that the flow of electrical charge is 25 Amperes. If it is truly 12 Amperes, then the Req of the circuit described is 5.2 Ohms greater than expected.




- Mesmer8
 
Mesmer8, I expect that is a measured actual real-world result. 25 amps is correct if we assume the wire does not increase with temperature. I am sure it is increasing with temperature.
 
Thanks for the reply and please forgive me -- I am really trying to wrap my head around this. The info I gave regarding (6' @ 8 ohms per foot will result in a temp of 1700 degrees was just based on info I pulled from a distributors website. I guess the real question is will a 15 amp variable voltage controller give me the 375 degrees at a safer operating voltage --- say 25 or 30 volts?
 
It's true that metals will increase in resistance as they are heated (just the opposite is true of semiconductors), however, the alpha (temperature coefficient) of NiChrome (the same wire used to heat the cutting stylus of a disc mastering lathe, btw) is extremely small. It is about 0.00017. Therefore, even if the 1700 degree temperature were meant to be in Centigrade, which I gather it was not, the current measured should be no lower than 19 Amperes - more likely 21... This is because the 0.8 Ohms (it was 8/10 Ohm, wasn't it? You might have mistyped it in the second post, which reads simply "8 Ohms") per foot, multiplied by 6 gets heated up from a 20-degree Centigrade reference temperature to resist current at a level of around 6 Ohms - whence the 21 Ampere charge flow. If you ask me, 25 is closer to 21 than 12 is. 4 Amps versus 9. Ergo, again, something is not right in the figures we have been given.

Let's pretend, anyway, that the current drawn will actually be 21 Amperes at 1700 degrees Fahrenheit (i.e., 927 degrees Centigrade). At 375 degrees Fahrenheit (190.5 degrees Centigrade), the original 6 feet of NiChrome wire has risen in resistance from 4.8 Ohms to 4.9 Ohms. At 12 feet of NiChrome wire, the resistance at 375 degrees Fahrenheit should be about 9.8 Ohms. If the Power to dissipate 1700 degrees from NiChrome was 120 Volts squared divided by 6 Ohms = 2400 Watts, then we might be able to divide this figure by the quotient of 1700/375 (= 4.5) and apply that in the calculation of the required current for the lower temperature. Then we can know how much potential energy to apply across the NiChrome wire.

2400 / 4.5 = 533 Watts. The current that should get us to the temperature dissipation we are looking for should be around (533/120 =) 4.5 Amperes. So, the potential energy you need to apply to get this to work with 12 feet of NiChrome wire should be 44.1 V.


I hope this helps you.



- Mesmer8
 
You should try and get an Acrylite Fabrication Manual.
My copy is dated 1967 so it is very likely that manual is not availiable.
They recommend using strip heater for wide bends or a tublar (rod) heater for sharp bends.
If you will send your Email address, will Email the two pages on forming acrylite.
send to carl pugh 1 then the at symbol then att.net leave out all spaces.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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