shield current capacity

[144x]

if a power cable has a rated short circuit current capacity of 31.5 KA for 3 second does it neccessarily mean that the shield of the cable should have the same rating? if not then what would determine the shield current capacity?
thanks.

 

[gordonl]

The sheild capacity is determined by the maximum ground fault current and duration of ground fault which can be seen by the cable's sheild.

If the system is solidly grounded, the sheild should be capable of caring a bolted ground fault for the time it takes the relaying to clear the fault. The same applies for resistively grounded systems, only now the maximum ground fault current is much reduced.

Try these links to Okonite & Southwire:

http://www.okonite.com/engineering/shielding-currents.html

http://www.okonite.com/engineering/shielding.html

http://www.southwire.com/tech/library/pc/pwrcab04.htm

If I haven't answered the question try and rephrase, and it may sink in.

 

[busbar]

I think possibly the limitations are borne out by the popular habit of furnishing a large [~2/0?] bare-copper conductor paralled with the shield tape.

 

[144x]

Dear friends.
thanks for your response .I read those articles too.what I want to say is that I have ordered a cable with 31.5 KA short circuit capacity.I had assumed that there would be a 3 phase fault somewhere outside cable so the cable would have to resist the current.but now if I assume that there would be an earth fault in one of the cables(single core ) and also assume that cable shield is bonded at both ends to earth then fault current would be shared in two paths ,one in the shield and the other in the earth .the earth fault current would definitely depends on the current capacity of the shield. my situation is that tenderers have offered different shields say:tape ,wire ,tape and wire and with different capacities ,31.5 KA(shared in shield and armor ),7 KA,and even 630 A.I don't know what to do .do I have to force them to give me a larger shield cross section and with special design ?or 630 A is just enough for me assuming that total earth fault current would be large enough for the relays to operate .it's quite possible that I totaly misunderstood the concept .so please help me.
thank you

 

[Bung]

The short circuit capacity of the cable is pretty well directly related to the cross-section of the cores. The rating of the screen / shield should be based on the earth fault protection clearance times, and the the eath fault clearance times should be based on (inter alia) the rating of the shield. (which came fist, the chicken or the egg?) But be careful of earthing both ends of the cable shield - you can end up with currents flowing in the shield and heating the cable due to differences in voltage between the earth mats at the two ends (a ground loop, well known anyone who has ever tried to build a power amp in their misspent youth!) What do your local standards say on the issue? We use cables built according to local standards, and set up the porotection accordingly.

Bung

 

[gordonl]

The worst case would be the sheild grounded at one end only, so the shield would need to be rated for the ground fault capability of the system as previously described.

If you ground at both ends this could still conceivably be the case if the fault is close to the supply end on the cable the shield back to the supply would carry most of the fault. If a cable fault is not the most serious case (ie: line differential relaying) and a through fault could last much longer, then you need to analyze the total copper going back to the source of ground fault current. (Current divider) (ie: shields, sheaths, ground cables) I personally would not include the earth path because of the low resistance of copper.

You may want to read some other threads on shield grounding.

 

[144x]

thanks for all the help.I didn't find any information regarding shield current capacity calculation in any book.
except some formula in the okonite web site but in that formula you have to know the shield current then evaluate the cross section while I don't know what is the fault shield current.with regard to circulating current
actually we intend to perform cross bonding system on our system so I don't think that there would be a problem although again I couldn't find a single formula regarding
circulating shield current calculation and the amount that it would be reduced by this method in any document ,text book or manufactureres web site.any help on this topic?