what is the temperature equilibrium for a copper conductor conducting 4000 Amps

I have a copper conductor with a total resistance of 1 miliohm and a passing current of 4000 Amps.

I want to know what the temperature equilibrium is for a conductor in the electrical grid (conductor is located inside medium voltage switchgear). At what point is the temperature it creates equal to the temperature it loses to the environment?

The cable is inside a container full of oil (used for cooling). as said before it has a total resistance of 1 miliohm and a current of 4000 Amps passes through the conductor. the cable has a diameter of 2.5cm the temperature of the environment is 20 degrees. time can be considered infinite for the cable cannot be turned off.

when will the temperature stabilize? has my conductor melted before it could reach that point?

• You know the voltage across it and hence the power dissipation. So the question comes down to, at what temperature rise does your chosen cooling system remove 16kW? Look at some 16kw heaters for examples. 7kW and water cooling gives a pretty feeble hot shower, 10kw is OK, so liquid cooling (pumped, with a decent way of dumping heat) looks doable. – Brian Drummond Mar 23 '18 at 12:54
• Is this DC current? – Carl Witthoft Mar 23 '18 at 13:00
• @CarlWitthoft "the electrical grid" and "medium voltage" implies no. – Brian Drummond Mar 23 '18 at 13:01
• That 20C. Is that the temperature of the bulk oil, or the temperature the oil maintains at the copper conductor surface? – Neil_UK Mar 23 '18 at 15:58
• @Neil_UK 20C is the start temperature of the copper, the oil. and the atmosphere outside the installation (i suppose some heat will be dissipated to the air once the casing of the installation starts warming up). – Doonz Apr 19 '18 at 13:01

If you know the viscocity, thermal conductivity and thermal expansivity of the oil, it should be possible to solve the free convection equations from first principles, and so work out at what temperature the power generated in the copper from $I^2R$ heating (the easy bit) equals the power lost to the convecting oil (the very hard bit).

However, fun though it may be to set up those flow equations, this is normally done empirically in practice, by measuring temperature rise against current of scale models, and extrapolating from those.

If you do want help solving those flow equations for oil, then you'd be better off on the physics site, it's in no way an electrical problem.

• The problem states "the temperature of the environment is 20 degrees", i.e. convection flow etc. doesn't have to be solved just thermal conduction inside the Cu conductor. So basically the problem is solving the radial diffusion equation (a PDE that can be reduced to 1 dimension in space) with constaint T=20 @ r=2.5cm. – Curd Mar 23 '18 at 10:40
• @Curd good question, is the 20C the temperature of the bulk oil, or the temperature the oil maintains at the copper surface? – Neil_UK Mar 23 '18 at 15:58

This does not answer your question in total, but takes a stab at the second part:

has my conductor melted before it could reach that point?

A copper "wire" with 2.5 cm diameter and 1 m$\Omega$ is around 29 m long. The power dissipation is 16.000 W.

With this we can calculate the temperature rise per second without cooling:

$$16000 \text{W}/(\pi*(2,5\text{cm}/2)^2*29\text{m}*(3.45\text{J}/(1\text{cm}^3*1\text{K}))) = 0.32 \frac{\text{K}}{\text{s}}$$

So even without cooling the copper will heat rather slowly and will reach melting temperature only after about an hour or so. I'm neglecting here, that the temperature rise will change the resistivity, but for a first guess to see if something will go wrong very fast it should be good enough.

Basically you have to solve the inhomogeneous Heat equation, in general a PDE in 3-dimensional space and time:

$\frac{\partial}{\partial t}u(\boldsymbol{x},t) - a\Delta u(\boldsymbol{x}, t) = p(\boldsymbol{x}, t)$

where
$u(\boldsymbol{x},t)$ is temperature at place $\boldsymbol{x}$ at time $t$,
$p(\boldsymbol{x}, t)$ is power dissipation density at place $\boldsymbol{x}$ at time $t$,
and $a$ is thermal coefficient of conduction.

In your case the general PDE can be simplified much because of following facts:

• $\frac{\partial}{\partial t}... = 0$ because of equilibrium condition
• $p(\boldsymbol{x}, t)$ is a constant over space and time
• the constraints $u(\rho)=20°C$ @ $\rho$=2.5cm are radially symmetric, i.e. space dependency can be reduced to one dimension, the radius $\rho$ (see Laplacian $\Delta$ for cylindrical coordinates)

So what will be left is solving a simple 1-dimensional ordinary differential equation.

As a solution you will get a temperature vs. radius function. Maximum will be at $\rho=0$. This will be the temperature you are interested in.

• And take into account the change in Cu resistivity with temperature (admittedly a small factor) – Carl Witthoft Mar 23 '18 at 12:59
• @Carl Witthoft: that will make the problem much more difficult, because then p(x, t) will also be depending on x. Also if current is AC and skin effect has to be considered.... – Curd Mar 23 '18 at 13:06
• This answer assumes that heat conductance is the dominant transport mechanism, which seems unlikely seeing as oil is a very poor heat conductor. Convection, as discussed by Neil_UK, would be much more important – which is the reason oil is used for cooling. – leftaroundabout Mar 23 '18 at 14:50
• @leftaroundabout: That's not unlikely at all if the oil is pumped around the conductor; nobody says that there is only convection. If 4000A are going through a conductor of 2.5cm diameter it is more likely that you don't rely on convection-only cooling. Also: otherwise this would be rather a flow simulation question and a lot of information would be missing... also there wouldn't be any convection if temerature stays as low as 20°C! – Curd Mar 23 '18 at 15:00
• @Curd what you're describing there is convection too – if it's driven by a pump or by thermal expansion, at any rate it's a fundamentally different mechanism from heat conductance. – leftaroundabout Mar 23 '18 at 15:22

1 miliohm and a passing current of 4000 Amps.

P = I^2 * R ==> P = (4000)^2 * 0.001 ==> P = 16,000,000 * 0.001 ==> P = 16,000W

So you have a 16kW heater, inside an oil cooled switchgear unit, with an ambient temperature of 20C.

The missing part of the problem is the heat carrying and dissipation capacity of the cooling solution. A 16kW heater is not that big a deal, a consumer wall heater is 10% of that, and a garage heater is 25% of that, so as long as enough oil is moved by convection or pumps I doubt you'll reach the melting point of the copper - that would require insulation, as a copper bar of that style would probably dissipate that much heat into air just fine without melting.

If you can describe the heat carrying and dissipation of the switchgear then a better solution can be calculated, but with the information provided we can only calculate this far, and make guesses about the rest.