2
\$\begingroup\$

I am very new to electrical power math. So please be friendly. :-)

I am getting 28 gauge nichrome wire to make my own heating element. Here are the specs and numbers of what I am attempting to do and what I know. I would appreciate a second pair of eyes confirming my numbers and also help recommending the correct power supply.

A. 10ft of 28 gauge wire, straight = Ohms Resistance 42.52. I plan to coil the wire myself and then loop the coil through ceramic insulators B. I need to reach temperatures of 450 F (I don't need over, but I def need to consistently hit this temperature.


Using the above info, an online calculator has provided me the following:

Volts = 55.276

Required Amps.  1.3
Required Watts: 71.85
Resistance per foot: 4.252  
Total Resistance: 42.52.

Questions:

  1. Is 55.276 volts the amount of volts needed going into my power supply, or the amount of volts coming out of my power supply?

  2. I have a spare (universal) HP laptop power supply with the following specs, is this power supply sufficient for my needs? : Input: 100-240v, 50-60Hz, 2.0A Output: 18.5v --- 6.5A, 120W

  3. When I cut the laptop connector off I assume I am going to see at least 2, probably three wires. black, white and green (maybe red). Do I connection both black/white to one end of the nichrome wire or do I connect black to one end and white to the other end?

A quick diagram:

110      Power Supply (PS)          PS split      Nichrome Wire
     _________________________    
3]---|>>100/240v | 18v/120W>>|____/---black---o---{Wire EndA}--\
     -------------------------    \---white---o---{Wire EndB}--/
  1. I may have to get 26 or 30 gauge wire...knowing the above, will my power supply still be sufficient with those gauge wires for what I am looking to do?
  2. Lastly, what, if anything am I missing or possibly not understanding?

Thank you in advance. This is all to make my own homemade plastic vacuum molding device.

\$\endgroup\$
  • \$\begingroup\$ Are you heating an enclosed chamber or are you just using local radiant heat? Is 450-deg the desired temp of the plastic, or of the heating element? \$\endgroup\$ – bitsmack Jul 24 '15 at 16:20
  • \$\begingroup\$ 450 is the desired temp of the heating element and the heating chamber will be like an enclosed oven with no top. I will be forming ABS plastic and the plastic will be secured to a frame and the frame is lowered onto the enclosure (essentially becoming the 'top' of the enclosure) which will put the plastic at 6" over the heating element. \$\endgroup\$ – rolinger Jul 24 '15 at 20:02
3
\$\begingroup\$

Is 55.276 volts the amount of volts needed going into my power supply, or the amount of volts coming out of my power supply?

That is the voltage you need to put across the nichrome wire, so it will be the output. You're not going to find a 55.276V power supply on the shelf, you're going to have to make your own.

I have a spare (universal) HP laptop power supply with the following specs, is this power supply sufficient for my needs? : Input: 100-240v, 50-60Hz, 2.0A Output: 18.5v --- 6.5A, 120W

18.5V < 55.276V, so no.

When I cut the laptop connector off I assume I am going to see at least 2, probably three wires. black, white and green (maybe red). Do I connection both black/white to one end of the nichrome wire or do I connect black to one end and white to the other end?

It won't work like you want, so don't bother, but even if it would work, there is no standard for the color or number of wires in laptop power supplies.

I may have to get 26 or 30 gauge wire...knowing the above, will my power supply still be sufficient with those gauge wires for what I am looking to do?

Your power supply isn't sufficicent, but 26 or 30 gauge nichrome wire will have slightly different resistivity than 28 gauge. You'll need to redo your calculations for the gauge you're actually going to use.

It seems like you're pulling a lot of numbers out of thin air. Further, I would be very cautious trusting an online calculator's calculation. The 450F is probably the wire temperature itself. Is that what you really want?

\$\endgroup\$
  • \$\begingroup\$ I knew my math was too good to be true. Well, the math might actually be pretty close but it was my understanding of how all this works that was way off. Yes, I am heating ABS plastic thus I need to get the temp to at least 400, thus I figured 450 would need to be the high end. However, now knowing my power supply won't work what suggestions might you make in order to get all this to work? I am almost on the verge of just getting some floor heaters and pulling them apart and using the heating mechanisms from one...probably two...of those. \$\endgroup\$ – rolinger Jul 24 '15 at 20:08
0
\$\begingroup\$

18.5 volt x 6.5 amps is 120 watts this is to much power you need the volt x amps to add to the correct wattage so your looking for something to put out 72 watts of power any combo of amps x volts will do as long as you hit the correct wattage.

\$\endgroup\$
  • \$\begingroup\$ Welcome to EE.SE, Sam. Your statement that "any combo of amps x volts will do as long as you hit the correct wattage" is not correct if the resistance is fixed as it is in the question. If you reduce the voltage, the current will go down too and the power will go down by the square of the ratio. (Half voltage -> half current -> one quarter power.) Note that the question is from 2015. \$\endgroup\$ – Transistor Oct 1 '18 at 20:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.