Power cable heat rise calculations

1. May 13, 2015

Paahei

We have a 90 deg C. rated PEX cable copper sized 1x185mmm2. This cable is to be used in 45 deg C environment (alone in air on a cable tray).
We want to use it in intermittent operation where the cable will be loaded with 600 kW for 40 seconds, then nothing for 40 seconds. Then repeat. The voltage is 1050 V DC.

Lets say the cable is 45 dec C. Then full current 600kW / 1050 V is applied for 40 seconds.

How can we calculate how much the temperature rise will be? What will be the cables temperature after the 40 seconds?

Can we calculate how much the temperature has dropped after the 40 seconds rest period?

2. May 13, 2015

Svein

You haven't specified the length of the cable. But.. 185mm2 cable has a resistance of about 0.1Ω/km. 600kW and 1050VDC gives a current of about 570A. Assuming 1m of cable: The dissipation will be roughly 32W when supplying power.

Now to the next item you haven't specified: The thermal resistance between the cable and the environment! A "standard", thick insulation cable - I would guess at about 5°C/W, meaning that the copper will reach 160°C over ambient temperature under load (not at once, but eventually).
Now we come to the heat capacity of copper. It works the same way as a capacitor does in an electronic circuit - it collects and stores heat. The heat capacity of copper can be found on the Internet, but you need to calculate the mass of the copper (where you need the length of the cable). The energy introduced is 32W*40s = 1280J. The heat leakage to ambient is (Tcopper - Tambient)/Rthermal.

Last edited: May 13, 2015
3. May 14, 2015

Paahei

Thanks. The cable is roughly 20 meters. The insulation is pretty thick as it is fire resistant RFOU offshore/ship cable.

Now you havent mentioned anything about the time aspect of it. Now I understand that the cable will get very hot if you supply it with constant power, but we are only talking 40 seconds here, and a cable with a big cross section will take some time to heat up. Could you please show me the formulas you use? Thanks.

4. May 14, 2015

Svein

The heat capacity of copper you can find at http://en.wikipedia.org/wiki/Heat_capacity. You must multiply by the mass (cable length ⋅ cross section) and you will get the necessary energy to increase the temperature by 1°K. The formula for the heat leakage to ambient: I gave the formula above.

Now you say that you turn the power on for 40 seconds (and then off for 40 seconds). Another way to look at it, is to say that on the average you have half the power all the time - or about 16W/m constant on. For your cable this means an average power dissipation of 320W (the copper heat capacity will act to smooth this out). Pick a period (say 80s), calculate the average energy (320W⋅80s), multiply by the thermal resistance and add the ambient temperature.

I suggest that you do a laboratory experiment - take a piece of the cable, run a known current through it and measure the temperature and the voltage drop. This will give you a better estimate for the thermal resistance and the ohmic resistance.

5. May 15, 2015

If we shall take the heating phenomenon as adiabatic [no heat evacuation from the conductor]

then at 570 A the temperature will reach only 47.1 oC[if the starting temperature it is the ambient] or

87.4 if the conductor was loaded up to 85 [C] .See –for instance- IEEE 80/2000 ch.11.3 conductor sizing factors formula no.37:

I=Amm^2*sqrt(TCAP/tc/ar/ror/10^4*ln((Ko+Tm)/(Ko+Ta))) [KA]

From Table 1 Material constant for soft-drawn annealed copper:

TCAP=3.42;ar=0.00393;ror=1.72;Ko=234

If Tm=47.1,tc=40sec and Ta=45 then I=0.571 [approx.]

There are standards-IEEE 242 for instance allowing for heat evacuation for short-time and the conductor temperature will be only 45.0093.See: IEEE 242/2001 CH.9.5.2.4 Development of intermediate characteristics.

But I think it is suitable for heating time of hours not seconds.

6. May 15, 2015

insightful

7. May 18, 2015

Paahei

Thanks alot guys