Cable Selection

........and what warranty does an installation of this size have in the event of an anonymous user of CR4 making a mistake?

The correct way to address this task is to consult a qualified person locally to carry out the calculations.

Your question is a little difficult to understand. It sounds like you are saying the following:

1. You have 4 - 50 mva 11 kv transformes feeding your plant

2. There are 25 - 3600 kva transformers connected to a 60 mva substation.

3. You are looking for the cable size for 3150 amps

I am totally confused oner the 60 mva and 50 mva. Can you please be more clear as to how this will be done.

Hi Friend,

Design of such large substations are the job of competent licenced Electrical Engineer and a matter of safety and normally needs the approval from the electrical supplying agency. How ever I can provide general guidance for selection of cable. you will have to make your own calculations depending on the data available at site.

1. Short circuit rating

It will depend on its size ,insulation,initial temperature,finial permissible rise of temperature of the cable ( The data will normally be provided by the cable manufacturer)It will also depend on the type of cable PILC/PVC/ XLPE , Earthed or un earthed system etc. Duration of short circuit depending on the protective system and back up protection which are other important point to be considered.

2.. Voltage drop

You must check the likely voltage drop for which standard chart is available which should be referred.

3. Power loss.

The power transmission loss in cable should be calculated as per the data provided by the the cable manufacturer.

4. Type of cable laying / depth of cable laying soil conditions etc. and other de rating factors to be considered for the calculation.

please let me knoe ampere load u required to feed for your proposed plant. then after cables has to be selected.

When you are bidding on a contract, you should have an intimate knowledge of both the code and the prices of standard size cable and equipment. Because anything that ever goes wrong always goes wrong when you try to save a few dollars! You MUST follow the code book, it exists for only one reason and that is safety.

That being said, the electrical code book will only tell you minimum size. Since "normally" smaller sizes of wire are cheaper, this is useful if you are bidding on a job. However, a few things can mess up the best inended calculations. You should have a good knowledge of what cable sizes are readily available, and their prices. You never know...sometimes a larger cable than necessary according to calculations is cheaper and more readily available because it is a standard size. Same goes for electrical boxes. The only reason why a larger cable might not be suitable is if you are dragging it through a conduit. But if you are laying it on a tray in the open, that helps. Your engineering firm may have in-house standards that are even higher than the minimums laid down in the code book. You should check. Don't let yourself be rushed.

first, you must calculate the Full Load Current (FLC).

in this case 60MVA, FLC = 60MVA/√3*11kV = 3149.1832A≈3150 A

second, you may find in the catalog, how many cables size wiil be used.

for example : if you see cable 1 core, 500 mm^2, having Current Carrying Capacity 650A (direct burried), its mean you need 5 cables to service your 3150 A load.(5x650 =3250A).R needs 5 cables,S needs 5 cables,T need 5 cables.

third,you can calculate voltage drop by :

Vd = I.R.cosΦ + I.XL.sinΦ

Vd 3ph = √3(I.R.cosΦ + I.XL.sinΦ)

R & XL can be found in vendor catalogue(dont forget to multiply with lenght cable will you lay). if you find, Vdrop still not reach your specification(ex:Vdrop=2%), may be you need to add cable become 6 cables.

R = 6 cables, S= 6 cables, T = 6 Cables

thank you...maybe helpfull

first, you must calculate the Full Load Current (FLC).

in this case 60MVA, FLC = 60MVA/√3*11kV = 3149.1832A≈3150 A

second, you may find in the catalog, how many cables size wiil be used.

for example : if you see cable 1 core, 500 mm^2, having Current Carrying Capacity 650A (direct burried), its mean you need 5 cables to service your 3150 A load.(5x650 =3250A).R needs 5 cables,S needs 5 cables,T need 5 cables.

third,you can calculate voltage drop by :

============================================================

value of R & XL can be found in vendor catalogue.

let say, R = 0.0241 Ω/km , reactance = XL = 0.052 Ω/km

if our cable length is 500 m ,so that R = 0.0241 Ω/km * 0.5 km,

XL = 0.052 Ω/km * 0.5 km.

if we use 5 cables : R = 1/5*(0.0241 Ω/km * 0.5 km)

XL =1/5*(0.052 Ω/km * 0.5 km)

============================================================

Vd = I.R.cosΦ + I.XL.sinΦ

Vd 3ph = √3(I.R.cosΦ + I.XL.sinΦ)

R & XL can be found in vendor catalogue(dont forget to multiply with lenght cable will you lay). if you find, Vdrop still not reach your specification(ex:Vdrop=2%), may be you need to add cable become 6 cables.

R = 6 cables, S= 6 cables, T = 6 Cables

thank you...maybe helpfull