480V 3-Phase circuit supplying three single phase-phase loads 2

[enterprisenx12]

Hi folks,

I have a 3-phase (3 pole wye connected) 480V circuit breaker that I need to supply three 200kVA single phase 480V identical loads (A-B, C-A, B-C). Only two legs are used for load.

As we know, in a balanced 3 phase system, the current across all three phases is kVA / 480V / sqrt(3). This means it would be 600kVA / 480V / sqrt(3) = 722A.

However, in this situation, I have 3 single phase loads (phase-phase), the current is 200 kVA / 480V = 417A.

A-B = 417A

C-A = 417A

B-C = 417A

What min ampacity should the three phase circuit breaker have to accommodate these loads? At least 722A or at least 417A? Note: Ignore the 125% sizing requirement for the moment.

I guess I am getting myself tripped up with how the upstream source ahead of the breaker sees these loads. I would expect in terms of kVA, the upstream source and thus breaker, sees 600kVA load. However, based on the single phase loads, it does not seem possible for each phase to be carrying 722A when I already showed that phase-phase 417A.

If I had to guess, the 722A min ampacity is the safest bet but really the sizing can be based off of 417A as its is not 3 phase - the current comes in one leg and retunrs on another leg. Just not sure then how you equate total kVA deman to total current demand when all of the loads are phase-phase and not phase-phase-phase or phase-neutral.

Can someone weigh in on this on wether 722A or 417A should be the basis? Please assume I need to configure the system this way with a single three phase (3 pole) breaker feeding phase-phase single phase loads.

Thank you all for your help!

 

[itsmoked]

722A is the current in the wires running thru the breaker poles, so that's the required amperage you're going for. If you were putting breakers in the individual single phase load wires then the 417A would be the focus.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[itsmoked]

BTW: You should probably be using 460V not 480, for 753A.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[wroggent]

I think I did this right:

http://files.engineering.com/getfil...78fe-4033-a044-68d8658211dc&file=3600_001.pdf

Make sure all of your conductors are fully rated.

 

[waross]

To amplify itsmoked's post:
Each single phase load will take 417 Amps.
BUT the transformer on A phase must supply both A-B and C-A loads. The combined current will be 417 Amps x √3 = 722 Amps.

The Canadian code under Circuit loading and demand factors:
8-100 Current calculations
When calculating currents that will result from loads, expressed in watts or volt amperes, to be supplied by a
low-voltage alternating-current system, the voltage divisors to be used shall be 120, 208, 240, 277, 347, 416,
480, or 600 as applicable.
Can anyone quote an equivalent rule in the NEC??

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

 

[xnuke]

NEC is 220.5(A). Similar text.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[VTer]

Attached is a simplified calculation for your review. Hopefully it helps you out. I did not include any OCPDs for simplicity.

"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic — and this we know it is, for certain — then it is a mere question of time when men will succeed in attaching their machinery to the very wheelwork of nature". – Nikola Tesla

 

[VTer]

In the attachment, the load should be (3) 200kVA, 480V 1ph not 208V (coffee didn't kick in yet when I did this)

"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic — and this we know it is, for certain — then it is a mere question of time when men will succeed in attaching their machinery to the very wheelwork of nature". – Nikola Tesla

 

[enterprisenx12]

Very helpful! Thank you everyone!

 

[enterprisenx12]

Here is a follow up question.....

Let's say we have only 2 single phase 480V 200kVA transformers connected via A-B and C-A... but no load connected across B-C. That means there is only a total of 400kVA on the upstream transformer. However, if we calculate the current based on Iphase = IAB + IAC, we still get 722A. That would now appear to now be correct because 722A should only work if we have 600kVA worth of load. Can you further assist me with the unbalanced equations so show this configuration and what the magnitude line current would be for all three phases?

Thank you!

 

[enterprisenx12]

TYPO Sorry...
Let's say we have only 2 single phase 480V 200kVA transformers connected via A-B and C-A... but no load connected across B-C. That means there is only a total of 400kVA on the upstream transformer. However, if we calculate the current based on Iphase = IAB + IAC, we still get 722A. That would now appear to be incorrect because 722A should only work if we have 600kVA worth of load. Can you further assist me with the unbalanced equations so show this configuration and what the magnitude line current would be for all three phases?

 

[waross]

B phase = 200000 KVA/480 volts = 417 Amps
C phase = 200000 KVA/480 volts = 417 Amps
A phase = 417 Amps x √3 = 722 Amps

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

 

[VTer]

Take a look at the attachment in my previous response. Just remove the load from phase C-B and go through the math for each phase. You will get the numbers that match what waross posted.

"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic — and this we know it is, for certain — then it is a mere question of time when men will succeed in attaching their machinery to the very wheelwork of nature". – Nikola Tesla

 

[stevenal]

722 A equals 600 kVA upstream if balanced, but the two transformer system is not. Still more than 400 kVA upstream, though. Remember that real power is a conserved quantity, but reactive power (and therefore apparent power also) is not.

I think Waross meant VA rather than kVA above.

 

[enterprisenx12]

I see it now... Thanks VTer and Wacross!