Wye-Delta High Leg Calculations

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cdcengineer

cdcengineer

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I've been reading old posts on teh forum for the last hour. I did not see any specific discussion regarding calculations for wye-delta high elg systems.

I have a project (only 2nd time I've seen this) where the building has a 2nd service fed from two 25kW 1-phase xfmr's. The system is 120/240V wye-delta (closed) with the high leg on B phase reading roughly 219V to neutral or ground.

My question: Would we calculate the total current on the system with a 3-phase 30kVA load as:

#1 - 30000/240(SQRT(3)) = ~72A

Or #2 30000/208*(SQRT(3)) = ~83A

I believe it is option #1 above, but I see some say it is #2. Another thing to consider, if we add single phase loads into the mix (served only from A & C phases), how does that impact the calculation?

Any feedback is greatly appreciate.
 
cdcengineer said:
I've been reading old posts on teh forum for the last hour. I did not see any specific discussion regarding calculations for wye-delta high elg systems.
Click to expand...
I pretty certain there are several discussions... but perhaps not specific. You may get better search results if you use search terms "lighter" and "stinger".


I have a project (only 2nd time I've seen this) where the building has a 2nd service fed from two 25kW 1-phase xfmr's. The system is 120/240V wye-delta (closed) with the high leg on B phase reading roughly 219V to neutral or ground.
Click to expand...
I don't see how the delta secondary can be closed if supplied with just two 1? xfmrs... :?


My question: Would we calculate the total current on the system with a 3-phase 30kVA load as:

#1 - 30000/240(SQRT(3)) = ~72A

Or #2 30000/208*(SQRT(3)) = ~83A

I believe it is option #1 above, but I see some say it is #2.
Click to expand...
Option #1. However, there's an additional calc' if open delta. The 3? calculated line current is carried by each secondary winding. 72 ? 240 = 17280VA

Another thing to consider, if we add single phase loads into the mix (served only from A & C phases), how does that impact the calculation?
Click to expand...
Depends. If you have any 1? 240V loads, you can put them on A-B and B-C to balance the load. If substantially unbalanced, the easiest way to calculate is to just compensate by including "not connected" or "future" load current values which balance the system.
 
I don't see how the delta secondary can be closed if supplied with just two 1? xfmrs... :?


I may be wrong in my understanding of the closed secondary. I was under the impression that the neutral / 4-wire connection made it a closed system.

Thx
 
For a 2 transformer bank, with equal sized units, you have two choices for calculating effective 3-phase kVA

1) [kVA(1) + kVA(2)]*.866
or
2) [kVA(1)*3]*.577

Then use the standard formula to figure the amps (line-line voltage =240).
If the transformers are of unequal size then let kVA(1) = the smallest kVA and use formula #2 above.

For a combination of single phase and three phase loading; subtract the single phase load from the transformer and then use a formula from above.

For example.
given:
kVA(1) = 20kVA
kVA(2) = 30kVA center-tapped

What is the effective maximum 3-phase kVA and how much single phase loading is possible?
Effective 3-phase kVA = [20*3]*.577 = 34.6kVA. 20kVA of one unit is being used for the 3-phase load, so 30-20=10kVA is left for single phase loading.

or
given:
kVA(1) = 20kVA
kVA(2) = 30kVA center tapped
Single phase loading =15kVA

What is the effective remaining maximum 3-phase kVA?
15kVA is being used for the single-phase load, so 30-15 = 15kVA left for 3-phase loading) Available effective 3-phase kVA = [15+15]*.866 = 25.98kVA, due to the ratio of single phase and three phase loading, the full capacity of the bank is not being used.
 
jim dungar said:
For a 2 transformer bank, with equal sized units, you have two choices for calculating effective 3-phase kVA
...
Click to expand...
That's pretty much the standard for determining capacity... but makes it unecessarily complicated.

First, it appeared he was working from the load value so I gave a transformer loading formula rather than system load capacity. I chose to use his amp cal value... but it's actually easier working with the kVA value (fewer numbers). For the sake of demonstration let's say the 3? portion of the load is the same as your effective 3? kVA capacity example...
.
34.6kVA load/sqrt(3)=minimum 20kVA/xfmr​
.
1? load just gets added to the per transformer 3? value.

Works the other way around for capacity, too....
.
20kVA smallest xfmr*sqrt(3)=34.6kVA maximum 3? capacity​
.

given:
kVA(1) = 20kVA
kVA(2) = 30kVA center tapped
Single phase loading =15kVA

What is the effective remaining maximum 3-phase kVA?
15kVA is being used for the single-phase load, so 30-15 = 15kVA left for 3-phase loading) Available effective 3-phase kVA = [15+15]*.866 = 25.98kVA, due to the ratio of single phase and three phase loading, the full capacity of the bank is not being used.
Click to expand...
15kVA*sqrt(3)=25.98kVA​
.
... provided other transformer is 15kVA or greater.

But let's say it's not... and only 10kVA...
.
10kVA*sqrt(3)=17.32kVA 3? available.​
 
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