How do I calculate the temperature rise in a copper conductor?

Question

If I pass a current through a copper conductor, how can I calculate how hot the conductor will get?

For example, if I have a 7.2kW load powered by 240VAC, the current will be 30A. If I transmit this power to the load via a \$2.5mm^2\$ copper conductor, how do I calculate how hot this conductor will get?

UPDATE:

From the comments and answer from Olin and Jason, I've created the following graph showing Watts per foot of \$2.5mm^2\$ copper wire:

But how do I translate this into the the actual temperature rise. I understand that the missing variable is the rate of cooling, but I just need to get an idea of what the maximum safe current is that can be passed through copper cable of a given thickness.

Assuming a constant current, and that there is no cooling at all, how do I calculate the degrees of temperature rise per hour per Watt for the foot length of copper cable in question?

Answer 1

In your edit, what's missing is that the rate of cooling will depend on the temperature. In general, the cooling rate will increase as the temperature increases. When the temperature rises enough that the cooling rate matches the heating rate, the temperature will stabilize.

But the actual cooling rate is very difficult to calculate. It depends on what other materials the copper is in contact with (conductive cooling), the airflow around the conductor, etc.

As an added complication, the heating rate will also depend on temperature, because the resistance of the copper will increase at higher temperatures.

So without much more detailed information about your conductor and its environment, its not really possible to give a precise answer to your initial question, how hot will it get?.

As for the second question, how fast will it heat up if there's no cooling, you can calculate that from the heat capacity of copper, which Wikipedia gives as 0.385 J / (g K), or 3.45 J / (cm^3 K).

Answer 2

Olin's comment has a good start on the quantitative analysis, but keep in mind that the effect of a watt or two per foot in an 18ga AWG wire (approx 1mm diam) is quite different from a 38ga wire (approx 0.1mm diam). 2.5mm^2 = approx 0.89mm radius 1.78mm diam = approx 13ga AWG wire which is pretty large and a watt per foot is probably fine, but let's see:

The wikipedia page for AWG = American wire gauge shows the National Electric Code copper wire "ampacity" (current capacity) at several temperatures for insulated wire, and 13AWG (not a standard product) is midway between the 12AWG rating of 25A at 60C-rated insulation, and the 14AWG rating of 20A at 60C-rated insulation, so my guess is that at 30A it would get pretty hot (probably >= 100C at 25C ambient) without convective cooling.

The wikipedia page also lists copper resistance of 13AWG as 2 milliohms per foot, so P = 2milliohms * 30A^2 = 1.8W/foot; the 22.5A "rating" at 60C rated insulation (average of neighboring ratings) has dissipation of very nearly 1W/foot.

Answer 3

Purely theoretically with no cooling at all:
\$ P=I^2*R(T) \$
\$ E(t)=\int{P dt}\$
\$ T=T0+dT \$
\$ dT=\frac{E(t)}{m*C} \$
\$ m=V*density \$
\$ V=l*A \$
\$ R(T)=l/A*r(T) \$

The above can be condensed into a linear approximation:
R(T)~=l/A*(r+T*\alpha) -> R(dT)~=l/A*(r0+dT*\alpha)

combining all this: dT ~= int{I^2*l/A*(r0+dT*alpha) dt}/(l*A*density*C) = I^2/(A^2*density*C)*int{r0+dT*alpha dt}

if dT*alpha << r0 then dT ~= I^2*r0*dt/(A^2*density*C)

unless I messed up something :) and it would melt eventually

I: current, R:resistance, P: power, T: temperature, t:time, E:energy, m:mass, V:volume, l:length, A: cross section area of wire, C:heat capacity of copper

Of course some kind of heat transfer always exists: conduction, convection, radiation. A good rule of thumb is to allow 2.5A/mm^2 on a copper wire in a coil with multiple layers, 4..5 A/mm^2 for single layer (without heat insulation) and 8..9 A/mm^2 will require active cooling.