Insulated power cable watts loss

[Exhibitionist]

I am in need of advice on how to calculate watts losses from insulated power cables. It is easy to calculate the watts loss of the actual conductor based on current and resistance (I squared R), but when the cable is insulated, what are the actual losses through the insulation.

We are trying to calculate the amount of cooling required in a large switchroom and using the straight I2R calculation the amount of cooling require seems too high, therefore expensive.

The type of cable we are using is a flexible fire rated power cable (110 deg rating). Each circuit is aroung 2500A and we are using 5 x 240mm sq cable per phase & neutral (415V 3ph 50hz system).

We have sought help from the cable manufacturer but they had never been asked the question before and could not give us the answer. The engineer there wanted to put a length of cable in an oven to see how hot it got (basically had no idea).

Look forward to input on this issue.

 

[Zogzog]

PF Test

 

[Exhibitionist]

what is PF test. Power factor??

 

[Skogsgurra]

I see no reason why you shouldn't use the result you arrive at. The heat from a 240 sqmm cable is accurately calculated using I^2R. Skin effect will probably make the losses a little higher than what you calculate, but that is only marginal at 50/60 Hz and 240 sqmm.

The insulation has no effect in steady state. It does not stop heat from getting out and it does not cause any extra losses (yes, some very minor dielectric losses, but nothing you need to calculate).

Why do you think that it costs too much to cool the switch room? If you have to cool it, you have to. Or use more copper to reduce current density. Half the density will reduce heat to 25 percent of what you have now. But that will be a lot of copper to have less cooling.

Gunnar Englund

http://www.gke.org

 

[Warpspeed]

Gunnar is right. The temperature of the conductors within the cable will continue to rise until the heat generated equals the heat dissipated.

Work out your total copper losses, and that is the power that will be dissipated once steady thermal equilibrium is reached.

Skin effect may start to become significant with those larger cables, also resistance will increase with temperature rise. Most accurate assessment would be to measure actual voltage drops because that will take everything into account.

 

[jraef]

For future reference Exhibitionist, please try to refrain from cross posting. Most of us read several fora. If you feel compelled to do so in the future, post the question in just one forum, then post just a link to that thread in the others. You can highlight and copy the "thread number" at the top of the page just under the title like this thread237-155605

Eng-Tips: Help for your job, not for your homework Read faq731-376

 

[Exhibitionist]

jraef, I have already had someone else point that out to me in one of the other threads and I do apologise. Being a first timer I was unsure where the best spot to post my question was.

I will certainly not do it again.

 

[jraef]

Welcome.

Just trying to help and keep you from getting your posts deleted. Cross posting will usually draw a red flag, but sometimes the admins can accidentally delete both postings and you lose all of your answers, or at the least lose good answers in one thread or the other.


No need to appologize. Trust me, mistakes are tolerated here! Well, most of them anyway...

 

[itsmoked]

The power forum allows no mistakes... That's what my power prof always shouted at us. (Miroslav Markovic)

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[debodine]

Skogsgurra and Warpspeed have hit the nail on the head. Heat generated is heat generated. It has to go somewhere.

Once the insulation has absorbed and trapped all the heat it can contain after current begins to flow in the conductor, it reaches steady state and then begins to allow any excess heat to pass through into the room. If the insulation can pass the excess heat through fast enough, then the room temperature rises. If the insulation cannot pass the excess heat through fast enough, the insulation overheats and is damaged. Once steady state is reached, effectively all heat generated in the conductor is transferred into the space around the cables.

If you had some means of heat transfer embedded in the insulation, such as cooling coils in the insulation that carry the heat outside the room before it can transfer all the way through the insulation, then the insulation would help lower your room cooling requirements. Otherwise the insulation is no help in reducing room temperature.

Assuming a given amount of power (that you are not permitted to reduce) must be transferred through the switchroom, the two primary options you have are to reduce heat buildup by stopping heat creation at it's source (larger diameter cables or more conductive material having less resistance) or carrying the heat away after generation (some method of removing heat from the room).

debodine

 

[logbook]

Curious posting. You didn't like the results of using I squared R for the power so looked for an alternative? Dielectric loss will also occur but is additional to the I squared R loss, and at 415V probably isn't that significant.

I also can't understand why putting cable in an oven tells you anything about how hot it gets when you put current through it.

The key thing of interest is how hot the inside of the insulation gets for a given internal power dissipation. It's easy to measure the outside of the insulation with a thermocouple, but the inside is more difficult. A thermocouple drilled into the insulation would become live, which is pretty nasty, but with care you could run a battery powered thermometer live and get away with it.