# Figuring out how hot a wire will get

Discussion in 'Electronic Basics' started by Harry Muscle, Dec 11, 2003.

1. ### Harry MuscleGuest

I'm not sure if this is the right p to ask, but since it's dealing with
electricity, I hope it is.

Is there a way to estimate how hot a length of wire will get ...

I'm gonna be running 7V 14.3A through 24 feet of 22 Gauge teflon coated wire
submersed in water. I would like to maintain 55 gallons of water at 25.5C
or 78F. I'm obviously going to need to know the thermal resistance of
teflon and some sort of formula to figure out how quickly heat is going to
pass from the copper, thru the teflon, and into the water. But I don't no
either

The main reason why I'm trying to figure this out is cause I would like to
know how much the resistance of the copper wire is going to increase, since
if it increases too much I might need to use one gauge higher.

Any help is highly appreciated,
Thanks,
Harry

2. ### Ian StirlingGuest

It may measurably increase.
It will not change the wattage much though.

3. ### John PopelishGuest

Besides the dimensions of the insulation, and the correct formula for
resistivity of teflon. I am assuming that your insulation is PTFE or
something close. Here are some links with thermal resistivity:
http://www.bayplastics.co.uk/data sheets/dsm/Product data sheet - (PTFE100).htm
http://www.pragatiplastics.com/pt_tubing.htm
If you google "thermal resistivity" and teflon or PTFE, or whatever,
you might come up with other references.

You may be able to pull the formula for the resistance out of:
http://www.delphi.com/pdf/techpapers/2000-01-0301.pdf
page 9 of:
http://baen.tamu.edu/users/rmoreira/bsen366/pdf/LecChap04r.pdf

The bigger mystery is how you arrived at 100 watts as the correct
amount (or at least enough) of heat to keep 55 gallons of water
exposed to an unknown environment to 78 degrees F.

4. ### Rich GriseGuest

How do you plan to regulate it? Just do it empirically. That is,
open your control loop and put in various control signals, and
see where the temp. of the tank settles out. Draw a graph.
some information that nobody's bothered to try to get before.
(Using copper for a heating element? Why not some calibrated,
known, waterproof heating element wire?)

Good Luck!
Rich

5. ### Harry MuscleGuest

Thanks for those links, I'll check them out. But to answer you question
about how I figured out to use 100W ... that was the easy part. I looked at
formulas used in determining heater sizes for aquariums, like those on the
following sites:

http://www.kernsanalysis.com/HeaterCalculator.cgi
http://www.geocities.com/ResearchTriangle/Lab/2024/thermo1.html

Harry

6. ### Harry MuscleGuest

I was planning on using a temp controller to either turn it on or off.
However, I'm trying to figure out the highest temp of the actual wire not
the water since I would like to know roughly the change in resistance that I
can expect. As for the reason for using copper is that I need a heating
element about 24 feet long, and running small gauge copper seems to fit the
bill very nicely.

Thanks,
Harry

8. ### Dan AkersGuest

Harry wrote;
"Is there a way to estimate how hot a length of wire will get ...
I'm gonna be running 7V 14.3A through 24 feet of 22 Gauge teflon coated
wire submersed in water. I would like to maintain 55 gallons of water at
25.5C or 78F. I'm obviously going to need to know the thermal resistance
of teflon and some sort of formula to figure out how quickly heat is
going to pass from the copper, thru the teflon, and into the water. But
I don't no either
The main reason why I'm trying to figure this out is cause I would like
to know how much the resistance of the copper wire is going to increase,
since if it increases too much I might need to use one gauge higher."
_____________________________________
Re;
For the wire proper, the temperature rise from the copper surface to the
centerline is (for all practical purposes) dependent only upon the
"uniform volumetric heat generation rate", the dimensions of the wire
and the thermal conductivity of copper (386W/mC). In this case, the
volumetric heat generation rate is about 4.196E7W/m^3. However, the
relatively high length to radius ratio and the high thermal conductivity
of copper, yields for this case, a temperature rise of only 0.0028 C;
for all practical purposes; ZERO; you can assume isothermal conditions
through it's diameter.

The heat flux at the surface of the copper is going to be about 6760
W/m^2. Disregarding the teflon insulation for the moment, if you take a
look at what is called the "boiling curve" (usually found in heat
transfer and fluid flow literature), this heat flux will result in a
Tsurface-Tsat of only about 6C; implying a copper surface temp. of about
106C. The heater would be operating in the "nucleate boiling regime"; a
very efficient natural convection heat flux. However, I suspect that
the insulation is going to require the copper temp. to be significantly
higher. You need to post the thickness of the teflon insulation.
As for wire size, if you DECREASE the wire diameter (and not increase
it's length), as I think you suggest, then you will proportionally
increase the heat flux (assuming wattage is the same). This will worsen
the heat flux situation, resulting in even higher heater surface temps.
Thus, if you decrease the diameter with the same total power, then
increase the length to maintain the heat flux below 10,000W/m^2.
The teflon insulation is the key unknown; you need to know it's
thickness and thermal conductivity to accurately predict your copper
temperature.
As for the temperature effect on the resistance of your heater; an
increase to 150C would result in the heater resistance increasing to
remains the same).

-Dan Akers

9. ### John PopelishGuest

Dan Akers wrote:
(snip)
(snip)

Are you saying that if the copper were in contact with the water, and
the heat flux were 6760 W/m^2, the interface would be at boiling
temperature even though the bulk water temperature is only 25 degrees
C? I don't think this heat flux is anywhere near that high. What am
I missing?

10. ### Dan AkersGuest

Harry wrote;
<snip>
"The main reason why I'm trying to figure this out is cause I would like
to know how much the resistance of the copper wire is going to increase,
since if it increases too much I might need to use one gauge higher."
_____________________________________
Re;
Hi again. I did a search for the thermal conductivity of TFE insulation
and came up with 8.2W/mC. Thus, for 10mil thick insulation on your
wire, that would correspond to only a 0.5 C temp rise or so across the
insulation proper. Given the stated geometry of your problem; 24ft long,
about 1.15mm (assuming 10 mil insulation) total diameter; for 100W
dissipated at a heat flux of 3800W/m^2 on the insulation's surface, in
still water and 10 mil TFE insulation, the copper temperature would be
about 107C. I based this on looking up the given heat flux, which puts
you well on the thermodynamic boiling curve for surface heat transfer in
water. The required surface temp. is about 6C above the prevailing
saturation temperature (which would be about 100C assuming the heater
isn't to be submerged to any great depth or that the aquarium is not
under some external overpressure). So adding all of this up, gives an
insulation surface temp. of about 106C and a copper temp. of just
slightly higher, about 107C or so. I hope that helps...

-Dan Akers

11. ### Dan AkersGuest

John Popelish wrote;
"Are you saying that if the copper were in contact with the water, and
the heat flux were 6760 W/m^2, the interface would be at boiling
temperature even though the bulk water temperature is only 25 degrees C?
I don't think this heat flux is anywhere near that high. What am I
missing?"
______________________________________
Re;
Yes. The length of the wire is 7.31m and the bare circumference is
2.02E-3m (22AWG; 0.644mm dia.), making the heat transfer area only
1.48E-2m^2. Divide this into 100W and you get 6752W/m^2. This heat
flux in still water wil result in some boiling at the surface of the
heater even though the bulk liquid remains substantially subcooled. It
is known as nucleate boiling, where minute vapor bubbles form at the
heat transfer surface, detach themselves through buoyancy and the force
of the prevailing convective currents, and then subsequently collapse as
they condense in the bulk fluid, a short distance from the heat transfer
surface. This mode of heat transfer is very efficient since the small
vapor bubbles rapidly carry the relatively large amounts of latent heat
into the bulk fluid. This phenomena begins in earnest at a heat flux of
about 1000 W/m^2; below that is simple convection.

-Dan Akers

12. ### Harry MuscleGuest

Wow, thank you so much. I hope you don't mind if I ask about one thing.
I'm trying to get the temp of the water to reach around 25.5C, not boiling
(100C). So if I understand the above explanation correctly, the wire will
reach 106-107C however, it will be just a matter of turning off the heat
once I reach my temp. I just wanted to verify that cause I'm a little
confused by the "6C above prevailing saturation temperature" ... I just want
to make sure that it's not supposed to be 6C above my temp of 25.5C.

Thanks,
Harry

13. ### Kent RegalGuest

I'm gonna be running 7V 14.3A through 24 feet of 22 Gauge teflon coated wire
After 13 messages and how many days of trying to re-invent the wheel ??? or at
least reverse engineer a readily available hobbiest item ..... Aquarium
heater......... they are available everywhere ! ! at reasonable prices AND
they have the control circuit already builtin ...... of course they run off the
house power supply of 120VAC and also available in 220 volt models.

Why the exercise and possible expense of cooked fish???

14. ### Harry MuscleGuest

Actually I'm not trying to reinvent a standard aquarium heater, but
substrate heating cables, that run around \$150 no matter where you look. So
given the fact they that are ridiculously overpriced and the fact that I
already have a large enough transformer lying around and the only cost
involved for me is going to be the control circuit and teflon wire, I'm
looking at saving at least \$100. Which makes it quite attractive to me.