Sizing DC Cables

Short answer - No, the cable is too small for the load current. Always size the cable for both the load current and the volt drop.

Best to use cable and a circuit breaker sized for the load.

But surely all we need to worry about is the power dissipation of the cable. If the cable can disspate 450W @ 150VDC, then surely it can dissipate 450W @ 24VDC.

CR4 doesn't do arguments.

Ah the point is that the cable cannot dissipate 450W, that is why I didn't bother mentioning the supply voltage (it isn't relevant to this particular question) only the current and volt drop across the cable. Also the load is 450W, not the power dissipated across the cable due to its resistance.

Why

I = P/V = 450/24 = 18.75A

Are you planning to short circuit the cable?

I = 3A whatever may be the voltage and the loss is I²R irrespective of the voltage.

The P is the Load + loss power and not the one dissipated in the cable.

Hence the breaker must be rated for the 3A only.

All these calculations are complicating the issue.

The cable is of course not dissipating 450W - it is carrying 450W

Ok, so confusiuon in terminology. To clarify the load is for example 6A @ 24VDC, which the cable must be able to carry: 6 x 24 = 144W. For say 100m of cable, the power dissipated by the cable is I^2R = 36 x 3.84 = 138.24W thus total power the cable must carry is 282.24W. For the maximum ratings the cable can carry 450W. Since the load + loss is less than the maximum, this should be ok? (this is all just hypothetical, just trying to get my thoughts straight). Ignore the voltage drop for now, assume an ideal voltage source.

The capability to dissipate power of the cable is limited by the rated current : I^2*R in W/length. In your case 3^2*3.84/100= 0.3456 W/m!

The maximal voltage it can accept is limited by the insulation it self (max 150V).

So that in fact the cable can transport a power of 450 W for the "user" but can only dissipate ≈0.35 W/m!

AS you see two different aspects NOT related to each other in the way you assume.

What on earth is the point of short-circuiting the supply voltage with a piece of cable? Cables are there to transfer energy from some place to some other place, not to act as room heaters!

So, voltage is unimportant in the current carrying capacity of a conductor, we could essentially put 11kV on this conductor, assuming the insulation was rated for 11kV and it would still carry 3A?

If the rated current is 3A and resitance is 3.84ohms/100m we have 0.35W/m. This is the rated power of the cable. If we were to run 6A through this cable we would have

36 X (3.84/100) = 1.38W/m, approximately 4 times the maximum allowable power, and the reason it is 4 times as it is a squared relationship ie (6A/3A)^2.

So, when the manufacturer says 'the current carrying capacity of this conductor is xxxA' this is the maximum allowable current regardless of the voltage applied (assuming an ideal installation method etc.)

Well done.

"This is the rated power of the cable."

NOT, it is only the power dissipation rate which depends on the heat transfer through the insulation and the way the cable can transfer heat to the ambient.

The dissipation is not the same for a cable in free air, in water or in the wall thus the maximal allowed current will not be the same.

In fact the current is limited by the acceptable losses as well. The resistivity increases with temperature so that if a cable runs hot the voltage drop will increase as well.

Those are the limitations: functional and economical.

For very short times the maximal current can be a lot higher as long as the maximal temperature, considering the cable and insulation accumulating capability and the thermal diffusivity of the insulation, do not go so high that the insulation molecular structure will be harmed.

If one looks at the current limits for different cables it is possible to determine the transfer coefficient of heat for the usual conditions which is between 2 and 4 w/m^2°K. Values based on external cable diameter.

So, when the manufacturer says 'the current carrying capacity of this conductor is xxxA' this is the maximum allowable current regardless of the voltage applied (assuming an ideal installation method etc.)

Since it is entering the theoritical area.

1. remember AC and DC will have different capacity - Due to skin effect.

2. There will be effect of voltage due to the lekage losses. Though not in the practical area where you are going to work with limited voltages, but of course it will be prominent when, as you said at 11KV that is even if the insulation can withstand.