Register to reply

Conversion of Dissipated Heat in Watts to F or C (Part 2)

by GarageTinker
Tags: conversion, dissipated, heat, watts
Share this thread:
GarageTinker
#1
Apr13-07, 08:21 PM
GarageTinker's Avatar
P: 38
Hello All,
This is an edited re-posting. I am looking for verification on a formula. As part of a personal design project (not for school or homework) I’m designing coils of magnet wire (i.e. coils in a generator) and I have all of the computational resources I need, with the exception of a formula to calculate the conversion of a power flow in watts to a temperature (°C or °F). I’ve been doing a good bit of searching and researching on the web and I’ve found a formula that I would like feedback on.

Now, to utilize a very useful example I found on the web; let’s say that silly me is goofing around with a 15 foot piece of 8 gauge (AWG) copper wire, which has .01 ohms of resistance and that the wire is connected directly to the positive terminal of a 12V automotive battery (no fuse, silly me) and I clumsily allow the other end to contact the negative terminal of the same battery. Dummy me is now holding a wire with 1200 amps running through it. (ouch)

I = E/R
I = 12/0.01
I = 1200 amps

I take this current and plug it into the formula P=I^2*R (or more simply E*I=P) and we get:

P = I2*R
P = (1200*1200)*0.01
P = 14,400 Watts
Or
E*I=P
12V * 1200A = 14,400W

This shows that the wire would dissipate 14,400 watts of heat which would melt the wire's insulation and more than likely ignite everything that comes into contact with the wire (fuel lines, other wires, carpet, plastic, insulation, my skin…).

I can figure out all the same information on the wire coils I'm dealing with and making for my project; however, because the manufacturers all primarily site the thermal limits of their wire insulation in terms of °C or °F etc., that's where I need to be able to figure the conversion from the above example's 14,400 Watts to °C or °F.

On a very educational site called Aus-e-tute at http://www.ausetute.com.au/. They list a specific heat formula for calculating the gain or loss of heat energy in a substance. The formula is based as: The amount of heat energy (q) in Joules gained or lost by a substance is equal to the mass of the substance (m) in grams multiplied by its specific heat capacity (Cg) multiplied by the change in temperature (final temperature - initial temperature) and is stated as: q = m x Cg x (Tf - Ti).

Now, knowing that 1 watt = 1 Joule/second; and using the conversion of 1 lb. = 453.59g, 15 feet of 8 AWG copper wire is 338.83 grams; the specific heat capacity (Cg) for copper is listed (on average) as being 0.385 and lets say the wire was laying in the shade during one of our balmy Florida summer days and it’s starting temperature (Ti) is a cool 30°C (86°F) and finally, we know from my silliness above that the wire has 14,400 watts running through it. So, from the basic laws of mathematics, I believe I can restate the formula as:
q / [(m x Cg) + Ti] = Tf

which, when the numbers are plugged in would give me:
14,400 (J/s) / {[338.83173g * 0.385 (Copper’s Cg)] + 30(°C)} = 89.75(°C) or 193.55(°F)

Preliminary calculations seem to work out well for my needs, but I would like opinions on my restatement of the formula. Basically, what I need to know is, will this calculation actually work as I think it will; even if it only gets me an accuracy of +/- a couple of °F; or am I only wishfully hallucinating.
Phys.Org News Partner Engineering news on Phys.org
A spray-on light show on four wheels: Darkside Scientific
Research project on accident-avoiding vehicle concluded
Smaller artificial magnetic conductors allow for more compact antenna hardware
sunday
#2
Apr14-07, 06:36 AM
P: 31
I've seen a copper cable 0.5" diameter melting with 2kA in few seconds.

The formula should be:

Tf = Ti + [q / (m x Cg)]

But that q is energy (J), not power (J/s). So Tf will be the temperature of the wire after 1 second of having those 1200A passing through it.

With the data you provided, you could expect a temperature increase in 1 second of 110.39C. So, for instance, in 5 seconds the wire will be at 582C , and so on. But depending of the battery will die before the wire melts and, probably, the battery acid inside will boil, with some funny consequences, like explosions, fumes, and acid spilling all over. How many Ah do your battery have?

Supposing a battery with infinite energy, the final wire temperature will depend on a thermal balance between the heat generated in the wire by resistive dissipation and the heat emitted to the surrounding air-convection- and the supporting structure -conduction. That emission of heat will depend on the convective coefficient, that is very dependent on how the wire is laid, i.e., if the wire is tightly coiled, the coefficient will be small, if the wire is laid uncoiled on the floor, the coefficient will be higher.

With higher coefficients you'll have a lesser wire temperature.

Regards.

Jose
GarageTinker
#3
Apr14-07, 01:55 PM
GarageTinker's Avatar
P: 38
Sunday, thank you very much for your clarification on the equation. I knew something was wonky and your re-ordering of my re-stated equation makes a lot more sense now that I look at it. I used to be really good at re-stating equations long ago back in my school days; I guess I've gotten kind of rusty at it.

No worries about a battery exploding, as I'm not using any batteries in the project, a battery was only being used in the hypothetical example I was citing. The coils I'm designing are to be used in the production of magnetically induced electricity, so your re-ordering of my re-statement of the equation will work perfectly since it is to be used for calculating the heat energy running through a coil at any given time. As I can adjust the speed of the movement of the magnetic fields I'm using, thus adjusting the Voltage and Amparage, the end result is the capability of adjusting the wattage in the coil and thus the temperature of the wire coil.

The coils are air coils (at least at this point) of varying sizes and shape, and they are going to be layed in the manner of coils in a motor or generator. I'm planning on using the equation to predetermine minimum wire gauges for the coils I'm designing under variable conditions. As for the co-efficient for the heat dissipation, I don't have any way to calculate for it, so I'll figure for worse-case (insulation melting temperature) senario; unless someone out there has a method they're willing to helpfully impart along with an explanation in layman's terms.

sunday
#4
Apr14-07, 07:28 PM
P: 31
Conversion of Dissipated Heat in Watts to F or C (Part 2)

Well, the mathematical model of the heating of the coils is based on differential equations, and seems difficult to know all the constants involved, so I'd recommend to take an experimental approach:

Take the coils one by one. Apply to each coil an increasing amperage, measuring the temperature at the same time. Then you'll arrive at the maximum intensity every coil could withstand. The experiment could be also made with reducen lengths of cable, in order to save money.
GarageTinker
#5
Apr15-07, 02:36 PM
GarageTinker's Avatar
P: 38
Thanks Sunday, I appreciate the suggestion. While I'm doing as much as I can mathematically using a programmed Excel sheet to get me as close as possible, I know there's nothing as valuable as Empirical / Practical data derived from live experimenting. Thanks again for your help and feedback.


Register to reply

Related Discussions
Entropy and heat: part 2 Advanced Physics Homework 1
Conversion of Dissipated Heat in Watts to F or C Electrical Engineering 18
X Watts to run = x Watts of heat? Classical Physics 27
Can you measure watts through heat? General Physics 14
Amount of watts needed to heat water Introductory Physics Homework 5