sizing an open delta transformer bank

[mobo]

Hi, curious as how to know how to size the large pot and the kicker pot for an open delta transformer bank. Looks like one customer will require a 3ph 200A load, another a 3ph 400A load, and another a 1ph 200A load. Voltage will be 240/120/208.


Would I take the 3 phase load as a starting point in order to know the size of the kicker pot and for the base of the larger pot, such as:


400A + 200A=600A therefore, 600A X (240VX1.732)=249.41KVA 249.41/2=124.7KVA


so the kicker pot would be a 125 KVA and also 125 KVA would be the starting point for the larger pot's KVA rating?


As for the single phase load to add to the larger pot's KVA:


200A X 240V=48KVA


so 125 KVA 3ph load + 48 KVA 1ph load = 173KVA for the larger pot.

So a 175 KVA for the larger pot and a 125 KVA for the kicker pot? Would that be correct?

Any input would be appreciated.

 

[Smart $]

The way I always looked at it was this...

Loads determine line current, so calculate the line current on each line (aka phase, but incorrect) the same as you would a full delta. The difference comes after that in that, when the line current gets to the open delta bank, all the B line current will travel through the kicker, and all the C line current will travel through the lighter... assuming the open phase is BC. So all you have to do is multiply the calculated B and C line currents by the line-line voltage to get the minimum rating of the respectively connected transformer.

 

[coop3339]

Hi, curious as how to know how to size the large pot and the kicker pot for an open delta transformer bank. Looks like one customer will require a 3ph 200A load, another a 3ph 400A load, and another a 1ph 200A load. Voltage will be 240/120/208.


Would I take the 3 phase load as a starting point in order to know the size of the kicker pot and for the base of the larger pot, such as:


400A + 200A=600A therefore, 600A X (240VX1.732)=249.41KVA 249.41/2=124.7KVA


so the kicker pot would be a 125 KVA and also 125 KVA would be the starting point for the larger pot's KVA rating?


As for the single phase load to add to the larger pot's KVA:


200A X 240V=48KVA


so 125 KVA 3ph load + 48 KVA 1ph load = 173KVA for the larger pot.

So a 175 KVA for the larger pot and a 125 KVA for the kicker pot? Would that be correct?

Any input would be appreciated.

Multiply this answer 400A + 200A=600A therefore, 600A X (240VX1.732)=249.41KVA 249.41/2=124.7KVA by 1.732 (the 3 phase portion is only good for 57.7% of the 3 phase calculation because it is open delta) this would be about 216kva. This is the minimum size for the small pot. Now add the single phase load to the other 216kva + 48kva = 264kva

Small pot 216kva
Large pot 264kva

 

[Smart $]

Multiply this answer 400A + 200A=600A therefore, 600A X (240VX1.732)=249.41KVA 249.41/2=124.7KVA by 1.732 (the 3 phase portion is only good for 57.7% of the 3 phase calculation because it is open delta) this would be about 216kva. This is the minimum size for the small pot. Now add the single phase load to the other 216kva + 48kva = 264kva

Small pot 216kva
Large pot 264kva

This is not that difficult. Think about it. With 600A of 3Ø current you'll have 600A line current on B line (the high leg), which is the open end of the kicker pot. So all 600A and only the 600A will go through the kicker pot.

600A × 240V = 144,000VA = 144kVA

The phase angle of the current does not matter because the transformer rating is in kVA.

The same would be true of the lighter pot if you did not have the extra 200A single phase load. So a simple conclusion for it is:

144kVA + (200A × 240V) =

144kVA + 48kVA = 192kVA

 

[mobo]

This is not that difficult. Think about it. With 600A of 3Ø current you'll have 600A line current on B line (the high leg), which is the open end of the kicker pot. So all 600A and only the 600A will go through the kicker pot.

600A × 240V = 144,000VA = 144kVA

The phase angle of the current does not matter because the transformer rating is in kVA.

The same would be true of the lighter pot if you did not have the extra 200A single phase load. So a simple conclusion for it is:

144kVA + (200A × 240V) =

144kVA + 48kVA = 192kVA

We're dealing with 3 phase there though, so you're forgetting to multiply your 240V by 1.732.

I believe i understand how to size now just based off the load (i.e., the size of panels each customer would report installing), but is there a derating or diversity percentage to apply to that. For example, we know even though you may have a 400A panel you won't see it maxed out, usually not even 50%.

 

[Smart $]

We're dealing with 3 phase there though, so you're forgetting to multiply your 240V by 1.732.

No I am not... and you do not. When you have full delta, the line current gets distributed to two windings, thus the square root 3 factor. With open delta, all the "open end" line current goes through the connected winding, so no root 3 factor is involved.

Say you have a 173.2A 240V 3Ø load.

On a full delta, 173.2A comes/leaves on each line. 100A (173.2A ÷ √3) is handled by each winding.

240V × 100A × 3 = 72kVA

173.2A × 240V × √3 = 72kVA

However, with open delta, all the "open end" line current goes through the connected winding...

173.2A × 240V × 2 windings = 83,138VA

...to power 72kVA of load

I believe i understand how to size now just based off the load (i.e., the size of panels each customer would report installing), but is there a derating or diversity percentage to apply to that. For example, we know even though you may have a 400A panel you won't see it maxed out, usually not even 50%.

There can be. Only way to know for certain is to do a proper Article 220 load calcualtion.

However, if this is a tenant scenario, where the tenant, as well as the calculated load, could change in the future, one should generally oversize the supply.

 

[coop3339]

This works also but is a lot more difficult and confusing than smarts way

Sizing an open delta transformer for a 600A 240V 3 phase load
this for closed delta (600x240x1.73)/1000=249.12kva
this would be (3) 83.04kva transformers connected in delta

next we need to supply the same 3 phase load with 2 transformers in open delta instead of the 3 for closed delta
When one transformer is removed the remaining 2 can only supply 57.7% of what the 3 could handle

for an open delta bank to supply 249.12kva we would need increase the size of the individual 83kva trans by (1/.57.7) or 1.73
83x1.73=143.89

this can be checked by knowing that 2 trans in open delta can supply 87% of their combined rating
144+144=288 288*.87=250.4 allowing for some rounding errors this is about what we want

so we know that we need (2) 144kva trans for the 3 phase load portion (600A 240V)

now we add the single phase portion of the load to one of the transformers
144+48=192

so the large pot is 192kva or 800x240
and the small is 144kva or 600x240

thanks for the explanation smart I never looked at it that way