Calculating Temperature Increase in Copper Wires

[josefbrandt]

I'm wondering if there's a formula that shows how much the temperature is going to increase in a copper wire with a specific resistance. The only formula I can find is the one that shows how to calculate te final resistance in the wire after the increase of temperature, but I don't know how much the temperarure is going to increase so that one doesn't help me too much. I don't want to buy a too thin wire so it gets too hot...I must find this formula but no one seem to know it :/

 

[lalbatros]

energy balance solves this kind of problem

Hi josefbrandt,

You need to evaluate the heat lost by you wire, and include this in the power balance to find out the stationary temperature reached by the wire.

The heat can be lost by different ways:

heat conduction throught the surrounding air
forced heat convection if there is a cooling air flow
natural convection, because hot air goes away
radiation​

For the details, you need to find an handbook on Heat Transfer.
Personally, I use "Heat Transfer, a basic approach" by Özisik, Mac Graw Hill.

If your wire is getting hot enough and if there is no forced convection, radiation will likely be the dominant heat transfer channel. The heat lost in this way will be calculated with a formula like:

Q = s S (T^4-To^4)​

where, s is the Stephan-Boltzman constant, S is the apparent surface of the wire, T is the temperature of the wire (in Kelvin), To is the ambiant temperature.
If your wire is inside a box or close to a reflector, things will be more complicated.

The power absorbed by the wire is P = R(T) I², where R(T) is the temperature-dependant resistance of the wire and I the current.

The power balance gives:

R(T) I² = s S (T^4-To^4)​

Solving this equation for T will give you the temperature of the wire.
Remember that I considered radiation-dominated heat transfer. Things are similar in other situation but the formulas will be different. See the handbook.

Michel

Postscriptum
=========
The equation above can be solved by successive approximation.
It can be written as:

R(T) I² = s S (T-To) P​

where

P = T^3 + T^2 To + T To^2 + To^3​

Assuming P does not change too fast with the temperature, and assuming R(T)=a+bT, you get a linear equation to solve:

(a + b T) I² = s S (T-To) P​

So, assume first T=1000K, calculate P and solve the linear equation.
Repeat with the new value of T.
This converges very fast usually.

 

[Andrew Mason]

I'm wondering if there's a formula that shows how much the temperature is going to increase in a copper wire with a specific resistance. The only formula I can find is the one that shows how to calculate te final resistance in the wire after the increase of temperature, but I don't know how much the temperarure is going to increase so that one doesn't help me too much. I don't want to buy a too thin wire so it gets too hot...I must find this formula but no one seem to know it :/

Lalbatross is quite correct. The temperature is going to rise due to the conversion of electrical energy into heat. You can determine how much heat is being produced in the wire but that will not give you the temperature. But the temperature depends on all sorts of factors relating to how the heat is dissipated. The wire will keep getting hotter until the rate at which heat is being added (I^2R) equals the rate at which heat is being lost.

So the reason no one has the 'formula' is that no formula for the temperature exists. It depends on how the wire is set up to dissipate heat.

AM

 

[LURCH]

From the OP, it is not clear wether you know how to calculate the amount fo heat the wire will generate in the first place. If that is what you're asking, then there is a formula. It is the formula for power. Any power not going into usefull work is given off as radiation.

So, if you know the resistence of your wire, and you know how much voltage you're going to put through it, you can multiply the voltage by the current (P=VI), and determine how many Watts your wire will produce. Since the wire itself (or more specifically, the resistence within the wire) is not doing any usefull work, all of these Watts will be given off as heat.

This of course neglects any EM that might be given off as visible light or other frequencies, but I assume you want to calculate for a wire that won't get hot enough to glow!

 

[lalbatros]

Dear LURCH,
Dear All,

This of course neglects any EM that might be given off as visible light or other frequencies, but I assume you want to calculate for a wire that won' get hot enough to glow!

Very approximately, the glow of the wire will be also given by the black-body radiation law: Q = s S (T^4-To^4) . This will not be a bad approximation for a solid wire because the optical depth will be very small, of course. For a gas lamp this will not be so good because the optical depth is likely much larger that the bulb itself, making the emissivity much lower that 1.

In general the emissivity of the emmiter much be taken into account:

Q = emissivity s S (T^4-To^4)​

For solids the emissivity is most often between 0.5 and 1.0.

If the temperature does not rise much, then things maybe a little bit more complicated. The radiation may not dominate the heat loss, and empirical formulas for free convection of heat must be used.

It is funny also to consider possible effects of geometry.
For example wiring the wire like a spring may decrease the radiation losses by at least 50% and this may increase the (absolute) temperature by more than 10%. (for a wire at 3000 K, you can see that it push the temperature by more than 500 K)

Therefore, a precise answer needs more information on the expected temperature, the geometry, ...

 

[LURCH]

Quite right. I meant to mention this at the end of my previous post, but forgot. The equation I mentioned above was only for determining the amount of heat the wire will produce. Environmental conditions (as mentioned in earlier posts) will determine how much of that heat is retained and how much is shed, and this will give you some predictions as to the temeratures you can expect to see.