3000 KVA transformer

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Ilvaro

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Need help with this and understand it better, we have two 2500 kva transformer,( source A and B), Primary Full-Load Current: 115.751 A
Secondary Full-Load Current: 3007.121 A
Our drawing calls out 12 sets of 500 kcmil, from the transformers feeding our switch gear down stream with A section 3000A and B section 3000A, and a tie in between for backup, let ‘em explain the tie, if B source loses power the tie in between bus A and B closes, meaning both gears are fed from source A, and vise versa, if A looses power B will kick in, my question is should the transformers at full load current be rated for 6000A for this scenario and also what is the reason why the 2500 kva Transformers is calling out for 12 sets of 500 kcmil on the secondary side feeding A or B, but doing the calcs on it, that’s little over 5000 amps.
 
1) What are the loads on the A section and B section? If the total load is less than 3000A then perhaps the system was designed so that either A or B can supply the total load. If the total load is more than 3000A, then is there load shedding so that a single section can continue to power the facility?

2) How did you calculate that 12 sets of 500 kcmil is a little over 5000A? What is the conductor material, the insulation temperature rating, the termination temperature rating, and the number of parallel sets in each conduit (if you have conduit)?

3) Read up on 'significant figures'. 2500 KVA at 480V is not 3007.121A (You used the correct equation, but left more digits in the answer than is appropriate, implying an exactitude that really isn't present in the data given.)

-Jon
 
What you have is often called a double-ended lineup.
It is rare to find these where either side can carry the combined loading. Most customers plan on reducing/shedding their loads if they ever 'lose' a transformer.
 
The UG duct bank which have been cal base on either Neher–McGrath method or NEC annex B.
 
winnie said:
1) What are the loads on the A section and B section? If the total load is less than 3000A then perhaps the system was designed so that either A or B can supply the total load. If the total load is more than 3000A, then is there load shedding so that a single section can continue to power the facility?

2) How did you calculate that 12 sets of 500 kcmil is a little over 5000A? What is the conductor material, the insulation temperature rating, the termination temperature rating, and the number of parallel sets in each conduit (if you have conduit)?

3) Read up on 'significant figures'. 2500 KVA at 480V is not 3007.121A (You used the correct equation, but left more digits in the answer than is appropriate, implying an exactitude that really isn't present in the data given.)

-Jon
Click to expand...
Thanks for that reply

Section A 3000A and Section B 3000A

Calculated the 12 sets little over 5000 A 90c(194f) xhhw cable nec table 310.16

On Kva calculation I used

Single Phase Transformer Full-Load Current (Amps)= kVA × 1000 / V

Three Phase Transformer Full-Load Current (Amps) = kVA × 1000 / (1.732 × V)
 
Ilvaro said:
Thanks for that reply

Section A 3000A and Section B 3000A
Click to expand...

Section A and Section B are both 3000A switchgear each carrying 3000A calculated load, or you don't know the calculated load connected?

Ilvaro said:
Calculated the 12 sets little over 5000 A 90c(194f) xhhw cable nec table 310.16
Click to expand...

90C cable is generally limited to lower temperature use because of terminations. There is additional derating if that cable shares a raceway with other conductors. If the raceway is underground, then you have the thermal insulation of the soil to consider, leading to a different derating calculation.

Only if your cables are Copper (not Aluminum), connected to 90C terminals in 90C equipment, running in above ground raceways with no more than 3 current carrying conductor in each raceway can you calculate the full 12 * 430A for the cable set. My _guess_ is that there is some detail which you left out which results in lower total ampacity.

Ilvaro said:
Three Phase Transformer Full-Load Current (Amps) = kVA × 1000 / (1.732 × V)
Click to expand...

As I said, you used the correct equation, but left too many significant figures in your result. By saying 3007.121 you are implying that you actually know the number is correct down to that final 1. You can't really know that. The noise on the various nominal values (2500 KVA, 480V, etc) is larger than the implied accuracy of your result. Let this point go for now; I apologize for being pedantic. The important point in my reply is about the ampacity calculation.

-Jon
 
winnie said:
As I said, you used the correct equation, but left too many significant figures in your result. By saying 3007.121 you are implying that you actually know the number is correct down to that final 1. You can't really know that. The noise on the various nominal values (2500 KVA, 480V, etc) is larger than the implied accuracy of your result. Let this point go for now; I apologize for being pedantic. The important point in my reply is about the ampacity calculation.
Click to expand...
It's like when PVWatts predicts the yearly output of a PV system to be 45713kWh per year with a 10% margin of error. :D
 
Ilvaro said:
Need help with this and understand it better, we have two 2500 kva transformer,( source A and B), Primary Full-Load Current: 115.751 A
Secondary Full-Load Current: 3007.121 A
Our drawing calls out 12 sets of 500 kcmil, from the transformers feeding our switch gear down stream with A section 3000A and B section 3000A, and a tie in between for backup, let ‘em explain the tie, if B source loses power the tie in between bus A and B closes, meaning both gears are fed from source A, and vise versa, if A looses power B will kick in, my question is should the transformers at full load current be rated for 6000A for this scenario and also what is the reason why the 2500 kva Transformers is calling out for 12 sets of 500 kcmil on the secondary side feeding A or B, but doing the calcs on it, that’s little over 5000 amps.
Click to expand...
12 sets of 500 kcmil is a lot of wire to install and test. Why not use a bus duct?
 
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