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Heat transfer coefficient of bare wire in still air

by xtronics
Tags: heat transfer
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xtronics
#1
Nov7-08, 04:27 PM
P: 5
I'm looking for some number to use for a heat transfer coefficient of bare wire in still air. but not having much luck.

This number just needs to get me into the ball park - I know it would be rather complex to solve with precision.

Think of it this way - I have a length of wire dissipating x number of watts across a known surface area - what would the temperature rise in still air be? Just need to know to about +/- 50% .. I will be under 250C of temperature rise - no incandescence.

I've seen tables - they are based on unknown safety factors etc... not actual rise and for jacketed/insulated wire.

I've got a book ordered that will get me there, but I'm thinking some kind of number must exist. Can anyone point me in the right direction?
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NoTime
#2
Nov8-08, 08:56 AM
Sci Advisor
HW Helper
P: 1,571
Welcome to PF.
You are right that it is a complex problem.
I don't know of any tables for your example, but that doesn't mean much.

Temperature rise will depend on how large the convection air currents generated by the heat from the power dissipated in the wire are (and thermal radiation), so the concept of "still air" is somewhat suspect.
Thus a wire in a small enclosure (or jacketed) will get hotter than a wire suspended in a large room.
xtronics
#3
Nov20-08, 05:38 PM
P: 5
I've found the basic solution. Irradiation dominates to the point you can ignore conduction in still air.

Because Q is about = Ah(T1-T2)

h =about 4(e)(5.67e-8W/(m^2 K^4)(298K)^3

or

h = about 6e W/m^2 K

h is the radiation heat transfer coefficient and e is emissivity.

emissivity is always a fuzzy depending on surface finish and for copper can vary from 0.7 to .88 but for electrical work 0.4 works most of the time.

My new problem is finding emissivity of Manganin.

es1
#4
Nov21-08, 02:32 AM
P: 324
Heat transfer coefficient of bare wire in still air

can you just measure? i.e measure the resistance of the wire. measure initial temp. dump some current into it for x amount of time and then measure the temp, then bump it up and measure again. 3 points make a line.
xtronics
#5
Nov21-08, 04:36 PM
P: 5
Quote Quote by es1 View Post
can you just measure? i.e measure the resistance of the wire. measure initial temp. dump some current into it for x amount of time and then measure the temp, then bump it up and measure again. 3 points make a line.
Sometimes when designing it is nice to start in the correct ballpark. I have it all worked out and wrote it up here:

http://wiki.xtronics.com/index.php/W...ated_in_a_Wire

Now if I could only find believable emissivity numbers for bare Manganin and enamel coated wire I would be happy...
ericgrau
#6
Apr20-09, 11:26 AM
P: 20
Quote Quote by xtronics View Post
I've found the basic solution. Irradiation dominates to the point you can ignore conduction in still air.

Because Q is about = Ah(T1-T2)

h =about 4(e)(5.67e-8W/(m^2 K^4)(298K)^3

or

h = about 6e W/m^2 K

h is the radiation heat transfer coefficient and e is emissivity.

emissivity is always a fuzzy depending on surface finish and for copper can vary from 0.7 to .88 but for electrical work 0.4 works most of the time.

My new problem is finding emissivity of Manganin.

I don't think this is correct. Radiation almost never dominates except in a vacuum. Try this link:

http://en.wikipedia.org/wiki/Convective_heat_transfer

Look for natural convective heat transfer, horizontal cylinder. You can ballpark the emissivity and radiation losses for now and compare it to the convection losses to see if radiation is significant. If it is then you can get more precise.

Conceptually, the heated air around the wire rises and causes convection, even in otherwise still air.
xtronics
#7
Apr20-09, 02:22 PM
P: 5
I did quite a bit of digging - and found the work and formulas of fusing currents etc. The convection effect is quite small if the wire is horizontal - and not that big if vertical. If you can point me to a reference that disagrees I would be interested.

The practical results seem to support the math.

See my page on this at:
http://wiki.xtronics.com/index.php/W...dth_vs_Current
ap_phys_1
#8
May7-09, 02:44 AM
P: 2
What ericgrau posted is accurate. Of the three heat transfer modes from the bare wire, natural convection dominates over conduction and radiation. Any introductory heat transfer book (Incropera and DeWitt, Fundamentals of Heat and Mass Transfer) discusses these heat transfer modes, and their corresponding equations and correlations. Also, as ericgrau pointed out, you can determine for yourself which mode dominates by calculating each and comparing them. The emissivity of metal is usually near zero. Wikipedia has the "Nusselt Number" to calculate the heat transfer coefficient you need.

It is very interesting that you are working this problem, because I just finished today recalculating the fundamental solution for this problem, just as a Professor Preece did over one hundred years ago. If you google "Preece's Law", I think you will find some of this worked out already. Professor Preece's articles can be found in JSTOR. His experiments were simple and elegant, worth reading, and verify his calculations. I performed similar experiments using a high constant current source for further verification of my (re)calculations.
xtronics
#9
May7-09, 01:40 PM
P: 5
Emissivity (sometimes called emittance) of copper can vary from 0.7 to .88 but for electrical work 0.4 is pretty good guess - that is not near zero - emissivity varies from 0 to 1!
ap_phys_1
#10
May12-09, 12:56 AM
P: 2
The emissivity of copper is not from 0.7 to 0.88 unless its surface has been oxidized or significantly modified in some other way. For electrical work, 0.4 does not seem to be a good guess for copper emissivity. Even if you google "copper emissivity", the first hit you get is of a company (Omega) that has been providing this kind of data for years. Here are the numbers from their website: Copper (Etched) 0.09, Copper (Roughly Polished) 0.07. When copper, which is highly pure for electrical work, is drawn into wire, it is polished (burnished) when it passes through the wire die, and is then annealed in an inert gas. Is this not correct? The emissivity of bare copper wire for electrical work cannot be as high as 0.7 unless its surface is significantly degraded, by oxidiation, for example. The emissivity of pure, clean metal is near zero, not near one as you have suggested.

But this debate about the emissivity of copper is not what is important. Until you calculate independently the heat transfer of all three modes, conduction, convection, and radiation, and compare them, as others have suggested, you cannot conclude that radiation dominates. The proof is not by authority nor by debate. The proof is by accurate calculation and repeatable experimentation. Assume a high as well as a low copper emissivity to yield an exhaustive proof. At best, radiation maybe comparable to convection, but this has not been seen as far back as in the work of Professor Preece of over one hundred years ago.

Finally, the experimental data that I have collected from actual bare copper wires used in electrical work matches that of Professor Preece, data that indicates that convection, not radiation dominates in my case as well as his.


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