Is there any math to tell what the average temperature would be if you had for example a constant dc current of 60 VDC (20 AMPS MAX current from the source) running through a coil of wire with a resistance of 30 ohms?

 

Twinkle twinkle
Little star
Power's equal
I squared R

So,

\( (20\text{ Amperes})^2\times 30\text{ Ω}\;=\;12,000\text{ watts} \)

That will give you the power in the wire. Now there are three basic mechanisms for heat transfer:

1. Radiation
2. Conduction
3. Convection

https://phys.libretexts.org/Bookshelves/University_Physics/Book:_Physics_(Boundless)/13:_Heat_and_Heat_Transfer/13.4:_Methods_of_Heat_Transfer

The environment that the coil of wire exists in will determine how much heat is removed from the vicinity of the coil of wire and that will determine the average temperature. There are too many variables for there to be a simple formula, and you need to be more specific about the conditions surrounding the coil.

ETA; The above power is equivalent to 120, 100 watt, light bulbs each of which would be too hot to touch, which would certainly be enough to keep a large room quite toasty.

ETA 2: (From Wikipedia)

A 100-watt incandescent light bulb has a filament temperature of approximately 4,600 degrees Fahrenheit. The surface temperature of incandescent light bulbs varies from 150 to more than 250 degrees, whereas compact fluorescent light bulbs have a surface temperature of 100 degrees Fahrenheit.​

 

In addition what is the wire gauge? As Papabravo points out:

The environment that the coil of wire exists in will determine how much heat is removed from the vicinity of the coil of wire and that will determine the average temperature. There are too many variables for there to be a simple formula, and you need to be more specific about the conditions surrounding the coil.

Plenty of variables.

Ron

 

Milage may vary from cable to cable, and depending on whether the wire is coated or not, but there are a few ways to approximate this. As @Papabravo just mentioned, you have to calculate power. Then, from that, you can get a temperature approximation from thermal conductivity of the wire. This will only be an approximation, however, as, to his same point, there are a LOT of things that go into temperatures.

If you have a straight copper wire of a specific length \(L\) and a cross-sectional area of \(A = \pi r^2\) (for some given wire radius \(r\), which depends on the gauge of the wire), then the thermal resistance \(R_\theta\) is approximated by:
\[R_\theta = \frac{L}{\left(398\,\mathrm{W\cdot m^{-1}\cdot K^{-1}}\right)A}\]
Your \(\Delta T\) from starting temperature (in C or K) will then be, for a given power \(P = I^2R\) (for DC):

\[\Delta T = PR_\theta\]

Now, if it's a very tight coil with insulated wire (with some thin insulation like acrylic or enamel) with some cross-sectional area of \(A_{coil}\) (or, rather, a coil with a radius of \(r_{coil}\)) and \(N\) turns, then you can VERY roughly approximate the coil temperature a number of ways, the simplest being (ignoring wire-to-wire thermal conduction effects and other effects) the following crude method:
First, approximate each loop of the coil to be basically circular, giving each loop a length of \(l \approx 2\pi r_{coil} = 2\frac{A_{coil}}{r_{coil}}\). Then, if you have \(N\) loops of that, the total length of the wire is roughly approximated as \(L_T \approx Nl\). You can then use that in the above equation for \(R_\theta\).

As a note: these are only VERY rough/crude approximation that ignores a lot of things. They're starting points, but don't rely on the numbers you get from them too much.

 

In an simple answer: No.
You cannot tell the temperature of the wire because you do not know the thermal dynamics equations.

You know the energy input into the coil.
You don't know how much energy is lost from the coil.

Edit:
Well actually, you do know how much energy is lost from the coil. You just don't know at what temperature it will reach thermal equilibrium.

 

@Papabravo:

60V, 30Ω

I = V / R = 2A

P = VI = 120W

The correct formula for V and R is V^2 / R = 60^2 / 30 = 3600 / 30 = 120W

 

@Papabravo:

60V, 30Ω

I = V / R = 2A

P = VI = 120W

The correct formula for V and R is V^2 / R = 60^2 / 30 = 3600 / 30 = 120W

You are of course correct, but the TS did not posit a consistent set of numbers. I believed he was interested in pushing 20 amps through 30 Ω with whatever voltage was required.

 

one can tell with certainty that the temperature will be simply some value between absolute zero and infinity.

 

one can tell with certainty that the temperature will be simply some value between absolute zero and infinity.

I think you can be pretty confident that it will be below 1085°C - the melting point of copper (although he didn't actually say it was copper wire) but it will definitely be below 3422°C (melting point of tungsten).

Seriously, assuming most countries' electrical codes are similar, there are different current ratings for different types of cable in different situations (clipped to the wall, in fibreglass insulation, buried underground, grouped with other cables) so it is pretty clear that the situation in which the wire is used is almost as important as its resistance when it comes to calculating its temperature rise.

 

(20 Volts)2×30 Ω=12,000 watts

(2A)^2 x 30Ω = 120W

 

Is there any math to tell what the average temperature would be if you had for example a constant dc current of 60 VDC (20 AMPS MAX current from the source) running through a coil of wire with a resistance of 30 ohms?

Sure, you can model the heat loss as a function of temperature, and then the equilibrium temperature is where the heat loss equals the heat input. You generally need to iterate to solve for heat loss as a function of a guessed temperature, and then keep adjusting the guess until it all converges.

You don't need to model the heat input because it's "known".

A simple model of heat loss would use just one factor to summarize the rate constant of heat loss due to radiation, convention and conduction, eg. dH/dt = k∆T. You could get more elaborate by breaking apart the three modes of heat loss and model each one: dH/dt = k∆T for conduction, = ƒ(∆T) for both convention and radiation. And how do you figure out what k is? There's no substitute for data, but with enough research you can probably find reported values of k for various coil configurations.

Just don't forget the first rule of models, "All models are wrong, some are useful." - Box

If the temperature of a coil is really important, you're far better off measuring it - collecting actual data - than trying to model it and hope your model is accurate.