Power cable heat rise calculations

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Discussion Overview

The discussion revolves around calculating the temperature rise of a 1x185mm² PEX copper cable rated for 90°C, used intermittently in a 45°C environment while carrying a load of 600 kW for 40 seconds followed by a 40-second rest period. The focus is on understanding the thermal behavior of the cable under these conditions, including heat dissipation and temperature changes over time.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant notes the need for the cable length to calculate resistance and dissipation, estimating a resistance of about 0.1Ω/km and calculating a current of approximately 570A for the given power and voltage.
  • Another participant emphasizes the importance of thermal resistance between the cable and the environment, suggesting a value of about 5°C/W for thick insulation, leading to a potential temperature rise of 160°C over ambient under continuous load.
  • A participant points out that the cable's large cross-section will affect the heating time, requesting formulas used for calculations.
  • One participant provides a formula for heat capacity and suggests calculating average power dissipation based on the intermittent operation, proposing an average of 320W for the cable.
  • Another participant introduces an adiabatic heating perspective, providing a formula for current and temperature rise, referencing IEEE standards for conductor sizing factors.
  • A later reply mentions that the cable is rated for 485 amps continuous, suggesting that the intermittent loading of 570 amps for 40 seconds on and off would be acceptable.

Areas of Agreement / Disagreement

Participants express various viewpoints on the calculations and assumptions regarding temperature rise and thermal resistance. There is no consensus on the exact temperature rise or the appropriateness of the proposed methods, indicating multiple competing views remain.

Contextual Notes

Limitations include the unspecified length of the cable in some calculations, assumptions about thermal resistance, and the potential inapplicability of certain standards to short-duration heating scenarios.

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

Now you haven't 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.
 
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.
 
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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.
 
Thanks a lot guys
 

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