Calculating load and no load losses of cables 4

[LSpark]

How do you calculate the load and no load losses of a cable in kW please? I'm used to dealing with these for transformers but it doesn't seem to be as straight forward for cables. Thanks!

 

[rbulsara]

W=I^2*R ( I squared * R)

Rafiq Bulsara

http://www.srengineersct.com

 

[LSpark]

What about the no load losses, that's the one I'm really struggling with.

 

[dpc]

The no-load cable losses are generally insignificant unless you are looking at very long runs.

Cable manufacturer can provide the information on dielectric losses.

David Castor

http://www.cvoes.com

 

[waross]

What voltage and length?
What is the conductor configuration? eg Single conductor, three conductor bundle, free air, in conduit.
You have dielectric losses, (resistive and capacitive) capacitive charging currents, and possibly corona losses.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[LSpark]

Thanks for that, so even for fiscal metering I can probably ignore the no load active loss? I'm also struggling to calculate the no load reactive loss (or change) for a cable as the formula I've got uses susceptance and that isn't quoted in any of the data sheets. I've assumed that the reactive load loss is Isq X.

 

[LSpark]

Bill - it's about 500m of 400kV cable with the 3 phases in parallel run in a cable trough.

I'm trying to work out out how to adjust the fiscal metering as the commercial boundary is at the 'wrong' end of the cable.

Another possibly stupid question, I've got values quoted for resistane and inductance, are they likely to be per phase or a 3 phase value?

 

[waross]

"it's about 500m of 400,000V cable with the 3 phases in parallel run in a cable trough."?
But at 400V, no load losses will be negligible. Most of the very small no-load current will be due to cable capacity to ground and will be quadrature current. This will not affect kW metering. A second order effect will be the I²R loss caused by the charging current. My gut feeling is that it may take several years for the value of the no-load losses to equal the cost of your time to calculate them.
But please continue and let us know what you find out (in dollars). I think that this is the type of problem that should be solved ONCE. It will give you a feel for a real world application that you often don't get in class and goes to developing judgment or feel for your work.

Your resistance values will be per foot or per 100 Feet. (Or the SI equivalent). The values will be applied to each cable and the current in the cable. The total losses will be the sum of the individual cable losses.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Marmite]

400 k V Bill.
Dielectric losses are proportional to frequency, system voltage, conductor screen radius, relative permittivity, conductor screen stress and dielectric loss angle. For cables upto about 50kV the dielectric losses are small in comparison to the conductor losses. However, the dielectric losses increase with voltage and are very significant at higher voltages.
Regards
Marmite

 

[dpc]

It really helps to have a little information. 400 feet is short, but 400 kV is high. Any supplier making 400 kV cable can definitely give you the dielectric losses.



David Castor

http://www.cvoes.com

 

[LSpark]

Thanks everyone. It is a short cable run but I need to be able to justify not correcting for a known error, so hence I at least need to be able to approximate it. I normally only deal with CT error VT errors and power transformer losses but I've been asked about cable losses a couple of times recently and now have one of my sites to calculate the correction factor for.

Is the dielectric loss what I am calling the no load loss? I've seen it quote in kW/km (I'm UK based) but wasn't sure if it was the no load loss or the load loss at cable rating.

I need to correct the kVAr metering too hence the interest in the no load reactive loss. I've seen the capacitance quoted, can I calculate the susceptance from that?

Although it probably isn't really worth calculating for just this one site I'll undoubtedly have to do it again sometime. Definately worth a spreadsheet! This is actually a rare chance for me to do classwork (it's been too long since I was at university), I'm normally dealing with faults and commissioning new systems.

 

[dpc]

Dielectric loss is basically no-load loss. There will be a very small amount of current flowing through the insulation (dielectric). The resistive element of this creates kW losses.

A value of kW/km is probably dielectric loss. Load loss is dependent on the load current. Dielectric shunt losses are not.



David Castor

http://www.cvoes.com

 

[LSpark]

Thank you David, so I can use the dielectric loss for the no load resistive loss (which is small anyway).

 

[jghrist]

See page 18 of

http://www.us.prysmian.com/export/s...ineering_Guide/General_Calculations_Rev_4.pdf

 

[dpc]

Yes, that would be my opinion.

That's a nice link, jghrist. Never heard of the company.

David Castor

http://www.cvoes.com

 

[jghrist]

They bought Pirelli.

 

[ScottyUK]

Who bought BICC. Who were a once a great company.


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If we learn from our mistakes I'm getting a great education!

 

[LSpark]

I think General Cable Corporation bought the cable side of BICC then sold parts of that to Pirielli (now Prysmian). BICC remaned themselves Balfour Beatty who you've probably heard of but they're contstruction based rather than manufacturing.

 

[Marmite]

LSpark, the formula for those of us more used to SI units rather than Foot Pounds per Bushel Inch is given in the attached paper.
Regards
Marmite

 

[LSpark]

Thanks Marmite, saves me converting to SI!