Power cable heat rise calculations

In summary: I suggest 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.
  • #1
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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  • #2
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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  • #3
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.
 
  • #4
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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  • #5
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.
 
  • #7
Thanks a lot guys
 

FAQ: Power cable heat rise calculations

1. What is the purpose of power cable heat rise calculations?

The purpose of power cable heat rise calculations is to determine the maximum temperature that a cable can reach during normal operation. This is important for ensuring the safety and efficiency of the cable, as exceeding the maximum temperature can lead to insulation degradation and potential fire hazards.

2. How are power cable heat rise calculations performed?

Power cable heat rise calculations involve using mathematical equations and data from the cable's manufacturer to determine the expected temperature rise based on factors such as current load, cable size, and ambient temperature. These calculations can be done manually or with the use of specialized software.

3. What factors can affect the heat rise of a power cable?

The heat rise of a power cable can be affected by factors such as the cable's size, insulation material, current load, ambient temperature, and installation conditions. Other factors that can contribute to heat rise include cable bundling, proximity to other heat sources, and the use of cable trays.

4. What is the maximum allowable temperature for a power cable?

The maximum allowable temperature for a power cable varies depending on the type of cable and its intended use. However, as a general rule, most cables should not exceed a temperature of 90°C. Additionally, some industries may have specific regulations or standards for maximum cable temperatures.

5. How can power cable heat rise be controlled?

There are several ways to control power cable heat rise, including proper cable selection, installation in appropriate environments, and managing current loads. Other methods include using heat shields or insulation barriers, using ventilation or cooling systems, and implementing regular cable maintenance and inspections.

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