# Current-Temperature equation?

1. Dec 19, 2009

### Jake110

Since running a current through a wire will increase the temperature of the wire, is there an equation for calculating the increase in temperature?

I’m thinking it would vary depending on the cross sectional area of the wire, the temperature of the surrounding material, the starting temperature of the wire, the specific heat capacity of the wire.

I’m planning on making an electromagnet and I started thinking about this when I was trying to decide what grade wire to use.

2. Dec 19, 2009

### f95toli

It can be calculated, but it is somewhat tricky (unless you have a VERY simple case with a loing straight wire with no insulation and only conductive heating you probably need numerical simulations) and for real applications it is usually better to just look in the datasheet.

Also, an electromagnet is not a wire(!); the fact that you many layers of wire will have a significant effect on the temperature. Your best bet is to simply use a wire/curent where the increase in temperature is so small that it won't have any practical effect.

3. Dec 19, 2009

### Bob S

It is best to look up the resistance of the wire in a table. See awg table in
http://en.wikipedia.org/wiki/American_wire_gauge

Decide on what wattage the coil can dissipate; 1 W, 5 W, etc.

Select wire gauge, number of turns, calculate W = I2R (where I is current and R is the coil resistance). Iterate as necessary.

Bob S

4. Dec 19, 2009

### Jake110

f95toli: that’s why I’m having trouble figuring out what grade of wire to use, sure there are amp limits for non bundled wire on the AWG wiki page but I can’t find any amp limit for the wires when they are bundled together like in an electromagnet. I was thinking I could use a chain equation where the surrounding temperature for the outer most layer of wire would use the air temperature and then each lower layer would use the temperature of the layer above it.

It’s really annoying because I don't want to get the wire and then find that the layers of wire at the core have melted.

Bob S: at the moment I don’t know what wattage I need because I’m putting together an excel document that calculates different values for an electromagnet. My aim is to end up with something where you just input some basic values and it will give you the harder to find details you need for actually building them.

The number of turns depends on the thickness of the wire and that would depend on the number of layers and the voltage of the power supply. If I could find some kind of max amps for 1 layer of this type of wire but decrease by a ratio of Y for each additional layer. That would be perfect.

The idea I had was that it’s ok if I have a really hard time figuring everything out now so that I can make it easier for if I do it again later.

I guess I could try to put an equation together myself if I could get a load of different grades of wire, although that would be pretty hard since I’m a student with no money lol.

5. Dec 19, 2009

### whome9

Heating above ambient temp. is proportional to I^2 and the surface area available to dissipate that heat.

6. Dec 20, 2009

### Jake110

i dont know, it seems to simple. wouldn't it use more variables than just current, surface area?

f95toli what way were you thinking of? i can deal with somewhat tricky.

7. Dec 22, 2009

### Yeti08

This would be my approach to your problem: For a given wire guage determine how long the wire will need to be (i.e. number of windings/layers). Find the total resistance from the resistivity (rho=1.72E-8 ohm-m for copper, rho = R*L/A, L=length, A=cross sectional area). For a given current this will give you a dissipated power; for a 30 cm long 30AWG wire carrying 0.5 amps it comes out to 25 mW (Q).

Assume an exposed surface area (can estimate from geometry/#layers), I'll just make a guess of 1/4 of the overall wire surface area (A_s), and guess at a convection coefficient (h) of 10 W/m2-C (in the range of natural convection). Plugging this in to Q=h*A_s*(T-T_0) while using 25C as ambient temperature (T_0) gives me a temperature of 67C. Then play with the variables, make parametric tables, etc.

Then a few questions you need to ask yourself - how accurate does this need to be - e.g. is a 20% error acceptable? Will conduction be a large factor (check thermal resistances)? Are there enough layers that there will be a substantial temperature gradient between layers? Is the wire insulated? And probably some others that I'm missing.

8. Dec 23, 2009

### Jake110

Your using the equation W = I2R that Bob S posted to find Q right? Q being power.

So T_0 is the temperate it’s at to begin with and T is the final temperature? It can be rearranged to form

$$T = \frac{Q}{h*A_0} + T_0$$

And using the melting point of copper (1357.77 K) I can determine if it’s an acceptable level of current and Gauge of wire. Well that’s just the basic version; the wire I’ll use will be insulated so ill need to look into what type of insulation it is and make sure it doesn't go above its melting point.

How many layers would be needed to cause a substantial temperature gradient? I could use anywhere between 5-20 layers of wire depending on what size power source I use.

The convection coefficient (h) needed in the above equation, how would you calculate it?

9. Dec 23, 2009

### Yeti08

Well, first I should mention that if I were doing this as a project I would probably use the rated current capacity of the wire.

The T_0 is actually the ambient temperature, but you could assume that is the starting temperature of the wire as well if everything were at ambient conditions. As far as the temperature gradient, you could approximate the windings as being a plane wall with an adiabatic inner surface, and estimate the overall thermal conductivity. The inner (i.e. maximum) temperature (T_inner) would be given by

$$T_{inner}=\frac{\dot{g}\cdot th^{2}}{2 k}+T_{s}$$

where g_dot is the volumetric internal heat generation (from Joule heating, i.e. I^2*R), k is the thermal conductivity, th is the thickness, and T_s is the surface temperature. I plugged in some numbers and I don't think this should be an issue. For the convection coefficient, h, it can get pretty complex, and there are more equations than I have time to write. If you want to get into that take a look at a heat transfer textbook and look at the section(s) on free convection. Otherwise, I've seen typical free convection coefficients listed as being 2-25 W/m2-K for gases.

10. Dec 23, 2009

### Bob S

When I have built electromagnets in the past, I usually start by calculating the required amp turns. This is easy to calculate if it is either air core, or iron dominated (with a small air gap) electromagnet. Then I would use Excel or similar program to calculate the heat output of various coil configurations. If possible, use higher amperage and lower number of turns, because heat conduction is better. For convection air-cooled coils, my guess is that 5W for a fist-size coil is about the limit. Use nylon, enamel or Formvar coated wire for best packing density. My gauge of heat limit is whether the coil (and/or the iron) is too hot to touch after 1 hour of operation. If you can smell it, it is too hot.
Bob S

11. Dec 23, 2009

### Jake110

thats a pretty good rule "If you can smell it, it is too hot".

the only recommended current limits i've found are just for single wires, not windings.

i'll aim to keep it using the least number of layers so i dont need to worry about temperature gradients.

as for the convection coefficient, would just using 25 be reasonable?

12. Dec 23, 2009

### Bob S

I think 25 is too high. I would go for 10 W/m2degC or less. You haven't said whether the coil is on CW or pulsed on and off.
Bob S

13. Dec 23, 2009

### whome9

Onderdonk and Preece

16 gauge copper wire: Tmelt = 1083°C, Area = 2581 circ mil, Time = 5 sec,diam = .0524 inches, Tamb = 25°C

E= Area in CM
B = Tmelt - Tamb in deg. C
D = 234-Tambient in deg. C
T= time in seconds.
So, E = 2581, B= 1058, D=209, T=5
Then
Ifuse = E* SQRT {<LOG[(B/D)+1]>/(T*33)}
Ifuse = 2581* SQRT {<LOG[(1058/210)+1]>/165}
Ifuse = 2581* SQRT {<LOG(6.04)>/165}
Ifuse = 2581* SQRT {0.781/165}
Ifuse = 2581* SQRT {.00473}
Ifuse = 2581* 0.0688
Ifuse = 178A

Solve Onderdonk for T
(Ifuse/E)^2 = <LOG[(B/D)+1]>/(T*33)
(T*33)= <LOG[(B/D)+1]>/{(Ifuse/E)^2}
T= <LOG[(B/D)+1]>/[33*{(Ifuse/E)^2}]

So, Ifuse = 178A, E = 2581, B= 1058, D=209

T= <LOG[6.06]>/[33*{.00476}]
T= <0.782>/[0.157]
T= 4.98 sec.

For #14 copper, CM = 4110.
B=1058
D=210
E=4110
I = 400A
B/D + 1 = 6.0381
(400/4110)^2= 9.4719E-3
<>=0.78090
[]=0.31257
T= <LOG[(B/D)+1]>/[33*{(Ifuse/E)^2}]
T=~2 sec

Using Preece equation: with 10244 for copper
and .0524 diameter
= 10244*.0524"^1.5 = 123 Amps

And

Discovered by Joseph Fourier in the 1807, Equation No. 1 is sometimes called the Fourier heat transfer equation. The equation in section 310-15(c) of the NEC, as seen below, called the Neher-McGrath equation, is a more complex version of the Fourier heat transfer equation. The Neher-McGrath equation was discovered by two cable engineers in 1957. In the Neher-McGrath (NM) equation, Delta TD, is a term added to the ambient temperature, TA, to compensate for heat generated in the jacket and insulation for higher voltages. Delta TD is called the dielectric loss temperature rise and is insignificant for voltages below 2000. Another term in the NM equation, (1+YC), is a multiplier used to convert direct current resistance (RDC) to alternating current resistance or impedance. For wire sizes smaller than No. 2 this term becomes insignificant.

14. Dec 23, 2009

### Jake110

It’s going to be pulsed. I’ll just use 10 for h then. Thanks for all the help guys, really appreciate it.