Help settle a debate

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sokoman

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I did some calculations for a some data center equipment and the Engineer that is reviewing it is questioning my calculations: We need to show the KVA per phase.
Here is the back ground, equipment is single phase 208vac, its 5640watts, total of 4 power supplies at 208vac each.
So:

Calculation #1 (myway)
5640 total watts / 208 = 27.11amps total draw
27.11 amps / 4 power supplies = 6.77 amps per power supply (208vac)
6.77amps / 2 = 3.38 amps per phase (120vac)
3.38 amps x 120vac = 405.6 watts / 1000 = .405KVA per phase
Or:
5640watts / 4 power supplies = 1410 watts per power supply
1410watts per power supply /208vac = 6.77 amps at 208vac per power supply
6.77 amps / 2 = 3.38 amps at 120vac per phase
3.38 amps x 120vac = 405.6watts / 1000 = .405kva per phase

Engineers way:
5640 total watts /4 power supplies = 1410 wattts per power supply
1410watts / 1000 = 1.410kva
1.410kva / 2 = .705kva per phase

Which is the correct way to do this? As you can see, my calculation vs his they do not match. Mine is .405kva per phase and his is .705kva per phase. What am I missing here?
 
If you have 4-1Ø, 208 volt power supplys how can the kva per phase be the same across all three phases? For example if you connect them as follows:
A-B
B-C
A-C
A-B
You have more on the A and B phase than on the C phase.
 
infinity said:
If you have 4-1Ø, 208 volt power supplys how can the kva per phase be the same across all three phases? For example if you connect them as follows:
A-B
B-C
A-C
A-B
You have more on the A and B phase than on the C phase.
Click to expand...
I am not calculating the total per phase im totaling the total per phase per power supply. yes if we add up the total connected per phase it would be different, im just calculating the per phase of each power supply. with that said breaker one is AB breaker 2 is CA breaker 3 is BC and breaker 4 is AB so phase A and B are equal and phase C is less. thats not what im asking im asking which is the correct way to figure the per phase of each power supply.
 
Are the voltages (120) separated by 180 degrees,(/2) or are they separated by 120 degrees (/1.732)
 
sokoman said:
I did some calculations for a some data center equipment and the Engineer that is reviewing it is questioning my calculations: We need to show the KVA per phase.
Here is the back ground, equipment is single phase 208vac, its 5640watts, total of 4 power supplies at 208vac each.
Click to expand...

Please clarify:

1) do you have 4 power supplies each rated 5640 watts, or do you have 4 power supplies which together total 5640 watts.
2) are all the power supplies connected on the same 'single phase', meaning say 'A-B' or will the 4 supplies be distributed across the 3 phases?

sokoman said:
So:

Calculation #1 (myway)
5640 total watts / 208 = 27.11amps total draw
27.11 amps / 4 power supplies = 6.77 amps per power supply (208vac)
6.77amps / 2 = 3.38 amps per phase (120vac)
Click to expand...
The above line is a clear error. Assuming that a power supply is drawing 6.77A at 208V, then that same 6.77A is present on _both_ phases. _All_ the current flows through _all_ parts of a circuit.

sokoman said:
3.38 amps x 120vac = 405.6 watts / 1000 = .405KVA per phase
Click to expand...
Thus by your method you should be getting 0.812kVA per phase
This approach (once the factor of 2 is eliminated) does not give the correct kVA for the load, because the load is not seeing the 120V. But it does give the correct kVA that the load imposes on the transformer coils if there are no other loads on the system. It also gives the correct loading applied to say a 'single phase' feeder derived from a three phase wye system.

[....]

sokoman said:
Engineers way:
5640 total watts /4 power supplies = 1410 wattts per power supply
1410watts / 1000 = 1.410kva
1.410kva / 2 = .705kva per phase
Click to expand...

The Engineer's approach gives a more accurate approximation if the load is part of an over-all balanced three phase system. The detailed difference between 0.812kVA and 0.705kVA comes out of the vector math for 3 phase systems, which we can discuss separately, and the _true_ answer will be somewhat between the two numbers depending upon the specific layout of the system. The key thing that you got wrong was that errant factor of 2.

-Jon
 
winnie said:
Please clarify:

1) do you have 4 power supplies each rated 5640 watts, or do you have 4 power supplies which together total 5640 watts.
2) are all the power supplies connected on the same 'single phase', meaning say 'A-B' or will the 4 supplies be distributed across the 3 phases?


The above line is a clear error. Assuming that a power supply is drawing 6.77A at 208V, then that same 6.77A is present on _both_ phases. _All_ the current flows through _all_ parts of a circuit.


Thus by your method you should be getting 0.812kVA per phase
This approach (once the factor of 2 is eliminated) does not give the correct kVA for the load, because the load is not seeing the 120V. But it does give the correct kVA that the load imposes on the transformer coils if there are no other loads on the system. It also gives the correct loading applied to say a 'single phase' feeder derived from a three phase wye system.

[....]



The Engineer's approach gives a more accurate approximation if the load is part of an over-all balanced three phase system. The detailed difference between 0.812kVA and 0.705kVA comes out of the vector math for 3 phase systems, which we can discuss separately, and the _true_ answer will be somewhat between the two numbers depending upon the specific layout of the system. The key thing that you got wrong was that errant factor of 2.

-Jon
Click to expand...
That’s what I was alluding to above.
I got .814 with my calcs
 
sokoman said:
I am not calculating the total per phase im totaling the total per phase per power supply. yes if we add up the total connected per phase it would be different, im just calculating the per phase of each power supply. with that said breaker one is AB breaker 2 is CA breaker 3 is BC and breaker 4 is AB so phase A and B are equal and phase C is less. thats not what im asking im asking which is the correct way to figure the per phase of each power supply.
Click to expand...
I was pointing out that the engineers approach also assumed that the kva is equal in all three phases which is incorrect.
 
winnie said:
Please clarify:

1) do you have 4 power supplies each rated 5640 watts, or do you have 4 power supplies which together total 5640 watts.
2) are all the power supplies connected on the same 'single phase', meaning say 'A-B' or will the 4 supplies be distributed across the 3 phases?


The above line is a clear error. Assuming that a power supply is drawing 6.77A at 208V, then that same 6.77A is present on _both_ phases. _All_ the current flows through _all_ parts of a circuit.


Thus by your method you should be getting 0.812kVA per phase
This approach (once the factor of 2 is eliminated) does not give the correct kVA for the load, because the load is not seeing the 120V. But it does give the correct kVA that the load imposes on the transformer coils if there are no other loads on the system. It also gives the correct loading applied to say a 'single phase' feeder derived from a three phase wye system.

[....]



The Engineer's approach gives a more accurate approximation if the load is part of an over-all balanced three phase system. The detailed difference between 0.812kVA and 0.705kVA comes out of the vector math for 3 phase systems, which we can discuss separately, and the _true_ answer will be somewhat between the two numbers depending upon the specific layout of the system. The key thing that you got wrong was that errant factor of 2.

-Jon
Click to expand...
There is one piece of equipment that piece of equipment pulls 5640 watts total, within is 4 power supplies. so clearly you divide 5640 by 4 to get the wattage of each power supply. So with what you said we would use the engineers way to calculate the kva per phase. I clearly mixed things up when i divided the amps by 2 assuming the amperage would be split between the phases, this is clearly wrong and i acknowledge it. so the KVA per phase is .705. I acknowledge my error in dividing by 2 that makes since then.
 
Engineers way:
5640 total watts /4 power supplies = 1410 wattts per power supply
1410watts / 1000 = 1.410kva
1.410kva / 1.732 = .814 kva per phase
 
sokoman said:
I did some calculations for a some data center equipment and the Engineer that is reviewing it is questioning my calculations: We need to show the KVA per phase.
Here is the back ground, equipment is single phase 208vac, its 5640watts, total of 4 power supplies at 208vac each.
So:

Calculation #1 (myway)
5640 total watts / 208 = 27.11amps total draw
27.11 amps / 4 power supplies = 6.77 amps per power supply (208vac)
6.77amps / 2 = 3.38 amps per phase (120vac)
3.38 amps x 120vac = 405.6 watts / 1000 = .405KVA per phase
Or:
5640watts / 4 power supplies = 1410 watts per power supply
1410watts per power supply /208vac = 6.77 amps at 208vac per power supply
6.77 amps / 2 = 3.38 amps at 120vac per phase
3.38 amps x 120vac = 405.6watts / 1000 = .405kva per phase

Engineers way:
5640 total watts /4 power supplies = 1410 wattts per power supply
1410watts / 1000 = 1.410kva
1.410kva / 2 = .705kva per phase

Which is the correct way to do this? As you can see, my calculation vs his they do not match. Mine is .405kva per phase and his is .705kva per phase. What am I missing here?
Click to expand...
If these are line to line, 208 volt loads, 120 volts never enters into the calculation.
 
sokoman said:
I did some calculations for a some data center equipment and the Engineer that is reviewing it is questioning my calculations: We need to show the KVA per phase.
Here is the back ground, equipment is single phase 208vac, its 5640watts, total of 4 power supplies at 208vac each.
So:

Calculation #1 (myway)
5640 total watts / 208 = 27.11amps total draw
27.11 amps / 4 power supplies = 6.77 amps per power supply (208vac)
6.77amps / 2 = 3.38 amps per phase (120vac)
3.38 amps x 120vac = 405.6 watts / 1000 = .405KVA per phase
Or:
5640watts / 4 power supplies = 1410 watts per power supply
1410watts per power supply /208vac = 6.77 amps at 208vac per power supply
6.77 amps / 2 = 3.38 amps at 120vac per phase
3.38 amps x 120vac = 405.6watts / 1000 = .405kva per phase

Engineers way:
5640 total watts /4 power supplies = 1410 wattts per power supply
1410watts / 1000 = 1.410kva
1.410kva / 2 = .705kva per phase

Which is the correct way to do this? As you can see, my calculation vs his they do not match. Mine is .405kva per phase and his is .705kva per phase. What am I missing here?
Click to expand...
Engineers way is generally correct, but I don't think it really matters what the per-phase kVA per power supply is. As a whole, this is an unbalanced delta arrangement and should be treated as such (load angles aren't pretty). The kVA's on A, B and C phases on a per phase basis (as seen by the main feeder) are:

SA= 2.15-kVA (17.9-A X 120-V)
SB= 2.15-kVA (17.9-A X 120-V)
SC= 1.41-kVA (11.8-A X 120-V)

Note: If this was a perfectly balanced delta (with only three power supplies, then the kVA's on A, B and C phases on a per phase basis (as seen by the main feeder) would be:

SA = SB = SC =1.41-kVA (11.8-A X 120-V)

Adding that 4th power supply changes the kVA distribution.
 
Last edited:
petersonra said:
What difference does it make if it's balanced or not? There's no way to know how well it's balanced upstream anyway.
Click to expand...

Because it changes the line currents and how kVA is distributed. The fourth power supply is between A-B. See post 3.
 
Xptpcrewx said:
Because you can’t have a feeder with different sized conductors on each phase.
Click to expand...
Where does it say that in the code? And anyway what does that have to do with whether or not the phases are balanced? There's nothing in the code that requires phase balancing. Where in the code does it require the overcurrent devices to be the same rating.
 
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