LOAD CALCULATIONS FOR DELTA/ WYE

[smur]

Hello
I have ben doing power calcs for some years off and on. Recently I joined a new company and I came across the way these folks do the calculations. And they are convincing me to do it their way. Much against my intuition.
Here is the story:
case 1
transformer secondary DELTA 208V
Feeds following loads
3 phase 208V 15Amps VA = 5404
3 phase 208V 20 Amps va=7205
3 phase 208V 18 Amps va=6484
1 phase 208V 10 amps va= 2080
1 phase 208v 15 amps va= 3120
1 phase 208v 20 amps va=4160
My calcs:
Total VA = 28453
Line current = 80 amps
CURRENT DISTRIBUTION BETWEEN PHASES:
15 15 15
20 20 20
18 18 18
10 10
15 15
20 20
TOT 98 78 68
MAIN CIRCUIT BREAKER RATING 100 AMPS
PANEL RATING 125 AMPS
My companys method: Multiply 98 x 1.732 = 170
78 x 1.732 = 135
68 x 1.732 = 118
SIZE CIRCUIT BREAKER FOR 170 AMPS. (His Line current)
This must be wrong ?? My managers solution
What if the transformer secondary is Y instead of delta.?
he says that is because the phase current x 1.73 = line current
My response: Wether it is y or Delta secondary, the following is still true:
MAIN CIRCUIT BREAKER RATING 100 AMPS
PANEL RATING 125 AMPS
He likes 170 Amps. Who is correct?
Thanks.

 

[rbulsara]

A few issues:

You are on the right track but not completely correct. Your 'company' is completely wrong.

Line currents remain the same, whether a system is star or delta. The breaker or panel only see line currents not phase currents.

Your method of adding current individually for lines is correct. Currents you added up are all line currents. So 98A is maximum. Assuming this is a continuous load, you need to multiply it by, at least, 1.25 to meet 80% loading requirement by NEC, unless entire assembly is rated for 100%.

In that case, at minimum you need 125A breaker. You may need larger breaker if these are motor loads and need to allow for starting. For that refer to NEC 430. So in the end you may require a larger breaker/panel.

Your initial answer for 80A based on total kva is only valid if loads on all phases are completely balanced (equal line currents). In reality you need to account for unbalance, so your other method is more appropriate.

 

[waross]

You are correct for three phase loads.
The single phase currents are out of phase with the three phase currents. The vector sum of the three phase currents and the single phase currents will be less, not more, than the simple sum.
With a 10 amp single phase load on "A"-"B", and a 10 amp single phase load on "B"-"C", the current on "B" will be 1.73 x 10 amps = 17.3 amps, not 2 x 10 amps = 20 amps.
Simple addition of the single phase loads will yeild conservative results.
It may be time to get out of the office and poke around an actual panel with an ammeter for awhile.

 

[smur]

You say single phase addition is not correct.
You say
"With a 10 amp single phase load on "A"-"B", and a 10 amp single phase load on "B"-"C", the current on "B" will be 1.73 x 10 amps = 17.3 amps, not 2 x 10 amps = 20 amps. "
PL Clarify this.
thank you

 

[rbulsara]

Waross:

With all due repsect, your statement needs revision. A equipment rated 10A, 208V will draw 10A in two 'lines' connected to the equipment. It will add 10 amps to those two line loads.

No where is such set up, there is a question of calculating or measuring the currents in each winding (phase) of a source or load. The 1.732 factor comes in play only then.

 

[smur]

Thanks everyone for the replies. Appreciate.

 

[jghrist]

I agree with Waross. The phase angle of 1Ø load current going from ØA to ØB is not the same as that of load current going from ØB to ØC. The currents do not add arithmetically. The phase angle of single phase 208V load current is also different than that of the current in balanced 3Ø loads connected to the same phase.

 

[rbulsara]

Yes, it is correct that, if the single phase loads are equally distributed on all three phases, then (3)10A,208V loads will show up as 17.3A on each line, which is 0.866 times the arithmatic sum of 20A. However, for all practical purposes, what smur is doing to add "unbalanced" load is just fine.

It is not that 10A shows up as 17.3A, its the total of two 20A that shows up 20*0.866=17.3 amps. If you had only one single phase 10A,208V load it does not show up as 17.3A. Or addition of a 10A and 15A load will result in something different.

If there are lots of single phase loads and are generally balanced between three phases, the correct method would be to add all KVAs and calculate line amps using a 3 phase power formula of KVA=1.732*V*A

For example: (3)10A, 208V loads will add to (3)*2080=6240VA.
6240/1.732*208 = 17.3A.

 

[smur]

One more question on the 3 phases;
When I made the table in my first post, I have calculated the va of the 3 phase va as:
1.732 x 208 x current of Circuit breaker
Later I divided this va between the 3 phases
example ; 3 phase 208V 15Amps VA = 5404 va
If I do it like this then ?
3 x 208 x 15 = 9369va
I suspect that 208 is not phase voltage, so the 2nd answer is wrong?
5404va is correct.
Pl correct me if I am wrong.
thanks

 

[rbulsara]

smur:

Your first attempt is correct:
3 phase VA=1.732*V*A where V is line to line voltage or 208V in this case, and A is the line current. Agian this for a balanced system so 3-ph 15A means there is 15A in each line and you use 15 in your formula.

Alternate way of seeing the same thing by phase is:
Use per phase current and per Phase voltage to calculate VA per phase and add them.

So 3ph, 15A, 208V: Assuming a star connection

V per phase is 208/1.732= 120V.
A per phase is same as line = 15A
VA per phase is 120*15
Total for 3 phase is 3*120*15

If you assume a delta system:
V per phase is same as V line=208
A per phase is 15/1.732 = 8.66A

You do rest of the math and you should get the same result. (5400).