Unbalanced transformer loads

[Pierre C Belarge]

That is why I asked what the load is, A load that large is somewhat unusual to be limited to a single element or other component and if multiple components often are capable of being reconfigured for single or three phase, multiple voltages, etc.

That may be generally true, but the OP has 2 large single phase loads supplied by a three-phase transformer.

 

[david luchini]

Don't confuse line current and phase current; they're not the same in Delta (which I presumed the OP to intend to use, and which wouldn't change the power anyway.)

I haven't confused line and phase current. I presumed a wye connected transformer secondary, but it doesn't make a difference if its wye or delta.

A 150kVA xfmr has a secondary rated current of 416A at 208V (delta or wye.) Connecting a single phase 208V, 320A load between A-B and A-C will cause 554A to flow in phase A in either a delta or wye connected transformer.

 

[kwired]

That may be generally true, but the OP has 2 large single phase loads supplied by a three-phase transformer.

I'm not trying to be disagreeable, I'm just trying to find out what is single phase only, not able to be separated into multiple circuits, and draws 320 amps @ 208 volts?

Something could possibly be overlooked.

Heating equipment has been suggested, that kind of wattage is certainly not going to be a single element is it?

 

[LarryFine]

A 150kVA xfmr has a secondary rated current of 416A at 208V (delta or wye.) Connecting a single phase 208V, 320A load between A-B and A-C will cause 554A to flow in phase A in either a delta or wye connected transformer.

Are you talking about an A-ph conductor or an A-ph transformer secondary winding?

The currents are the same as the lines in a Wye system, but the voltages are different, whereas in a Delta system, the voltages are the same, but the currents are different.

If you have a Delta secondary supplying a Delta load, each secondary will see the same current as the portion of the load connected between the same two conductors, but the line currents will be different.

That's why I felt you were confusing the two.

 

[david luchini]

Are you talking about an A-ph conductor or an A-ph transformer secondary winding?

The currents are the same as the lines in a Wye system, but the voltages are different, whereas in a Delta system, the voltages are the same, but the currents are different.

If you have a Delta secondary supplying a Delta load, each secondary will see the same current as the portion of the load connected between the same two conductors, but the line currents will be different.

That's why I felt you were confusing the two.

I think you're overcomplicating things. In a delta secondary transformer, which is the A-ph secondary winding, the one from A-B or the one from A-C? In a 150kVA 208V delta secondary is the 416A rated secondary current the rating of the phase conductor or the rating of the winding?

It doesn't matter if the transformer secondary is a 208V Wye or a 208V Delta connection, the 320A, 208V single phase loads connected from A-B and A-C will create the same current flow in the A Phase (Conductor)

 

[LarryFine]

In a delta secondary transformer, which is the A-ph secondary winding, the one from A-B or the one from A-C? In a 150kVA 208V delta secondary is the 416A rated secondary current the rating of the phase conductor or the rating of the winding?

That was my point. Obviously, the 416a is line current. If there was a third 320a load, balancing the Delta, all three lines would see the 416a, and all three secondaries would see 320a.

It doesn't matter if the transformer secondary is a 208V Wye or a 208V Delta connection, the 320A, 208V single phase loads connected from A-B and A-C will create the same current flow in the A Phase (Conductor)

Agreed. However, the shared Wye secondary would also see the 416a, whereas each 208v Delta secondary would only see the 320a across its two lines.

In a Wye, the 416a would be each shared phase's line current and winding current, but that's at 120v, not at 208v. That's why the required power rating of the transformer remains constant throughout this discussion.

Added: I will add that, now that I look back, during this discussion, I've been envisioning the source as a Delta, but, at 208v, it's almost definitely a Wye.

 

[rattus]

Missed something!

Missed something!

How did we arrive at 416A line current?

 

[LarryFine]

How did we arrive at 416A line current?

I dunno. With two 320a loads sharing a 3ph conductor, it should see 553.6a. I was just goin' with the flow.

 

[kwired]

How did we arrive at 416A line current?

Post 25, 416 amps would be full load current per phase of mentioned 150kVA, 208 volt three phase transformer.

 

[hurk27]

How did we arrive at 416A line current?

150kva transformer@208 volts
150,000/208/1.732=416.37?

Whoopes
416.37/3=138.79 amps per leg:roll:

 

[hurk27]

Post 25, 416 amps would be full load current per phase of mentioned 150kVA, 208 volt three phase transformer.

should have stopped while I was ahead:grin:

 

[scott thompson]

Maybe some clarification is needed...

Maybe some clarification is needed...

I see where my mistake was made - I figured an Open Delta connected Transformer, simply because the Loads are L-L. As so, the total KVA was added up in an "Open Delta" fashion - hence the 150 KVA Rating.

Per the Delta Secondary, the qualifying "Phase" description would be at the connection points for Two individual Windings' leads... i.e.: Terminal X1 being "Phase A" is made at the Left-Hand side Lead of Secondary Winding "A", along with the Right-Hand side Lead of Winding "C".
This Open Delta would be setup as a Right Triangle, with Phase A derived from the lower Right end, Phase B at the top of the Triangle, and Phase C from the lower left end of the Triangle.

With each separate Winding being capable of supplying 75 KVA, the combination of both will allow for Two 75 KVA Single Phase L-L Loads to be connected as:
Load #1 connected between Phase "A" and Phase "B",
Load #2 connected between Phase "C" and Phase "A".

Phase "A" will be common to both Loads as far as Termination and Feeder (individual Feeders to each Load, or a single Feeder connected to both Loads), but the Windings will have the same Apparent Power overall drawn by the Loads - in this case, each Winding will have 66.5 KVA drawn by the Load, with each Winding having a Maximum Apparent Power Capacity of 75 KVA.

If we apply this to a Wye Connected Secondary, the common Load on Phase "A" will be 66.5 KVA (66,560 VA), whereas the Load on Phase "B" or Phase "C" will only be 33.28 KVA (33,280 VA).

In this case, the "Per Phase" Load is also the "Per Winding" Load, so the Apparent Power drawn per Phase Winding "A" is 66.5 KVA.
A 150 KVA Wye connected Secondary will have a Maximum "Per Phase" (per Winding) Apparent Power Capacity of 50 KVA, which is 75% of the required Apparent Power needed on the "Common Phase A", but is 227% of the required Apparent Power for Phase "B" and Phase "C".

This Design _MIGHT_ perform without issues, since we might have Non-Continuous Loads at hand (some factor of Load Diversity where the Phase A Load would be at Maximum KVA draw for less than 180 Minutes), but for me I would rather compile the Design per the highest Phase Load of 66.5 KVA, so a 225 KVA Wye Connected Transformer would be applied to the Project.
The 225 KVA Transformer will have a Per-Winding Apparent Power Rating of 75 KVA, resulting in 112% of the highest Load KVA draw per Phase (Ph. A).

The 225 KVA Wye Transformer is more "Overkill" for these Loads, as compared to the 150 KVA Open Delta Transformer.
Availability would be more of a key decision than anything else.

Question to david luchini

If you were to compile a Panel Schedule for this scenario, how would you apply the Load VA per Circuit, then the Panelboard's Rating?

Examples:
Load #1: Ckt 1/3,
Load #2: Ckt. 5/7
System is derived from a 208V 3 Phase Wye Connected SDS.

Method #1:
Ckt #1 (Phase A): 33,280 VA
Ckt #3 (Phase B): 33,280 VA
----------------------------
----------------------------
Ckt #5 (Phase C): 33,280 VA
Ckt #7 (Phase A): 33,280 VA
----------------------------
----------------------------
Total Loads: Phase A: 66,560 VA; Phase B: 33,280 VA; Phase C: 33,280 VA.
Highest Phase: 66,560 VA (554.67 Amps)

...............................................................................................................
...............................................................................................................


Method #2:
Ckt #1 (Phase A): 66,560 VA
Ckt #3 (Phase B): 66,560 VA
----------------------------
----------------------------
Ckt #5 (Phase C): 66,560 VA
Ckt #7 (Phase A): 66,560 VA
----------------------------
----------------------------
Total Loads: Phase A: 133,120 VA; Phase B: 66,560 VA; Phase C: 66,560 VA.
Highest Phase: 133,120 VA (1109.34 Amps)

...................................................................................................
...................................................................................................

Method #3:
Ckt #1 (Phase A): 33,280 VA
Ckt #3 (Phase B): 33,280 VA
----------------------------
----------------------------
Ckt #5 (Phase C): 33,280 VA
Ckt #7 (Phase A): 33,280 VA
----------------------------
----------------------------
Total Loads: Phase A: 66,560 VA; Phase B: 33,280 VA; Phase C: 33,280 VA.
Highest Phase: 66,560 VA (320.00 Amps)

------------------------------------------------------------------------


Just curious...

Scott

 

[david luchini]

That was my point. Obviously, the 416a is line current. If there was a third 320a load, balancing the Delta, all three lines would see the 416a, and all three secondaries would see 320a.

I think we're on the same page, but obviously, if we added a third 320A load, all three lines would see 554A, and all three lines would exceed their rating.

Agreed. However, the shared Wye secondary would also see the 416a, whereas each 208v Delta secondary would only see the 320a across its two lines.

Yes, but the secondary winding is "rated" for 240A in the delta transformer. So we'd be exceeding the rating of the A line, and of two of the windings.

In a Wye, the 416a would be each shared phase's line current and winding current, but that's at 120v, not at 208v. That's why the required power rating of the transformer remains constant throughout this discussion.

Yes, in the wye configuration, the windings would be "rated" for 416A. With the 320A loads connected A-B & A-C, the A line and one of the secondary windings would exceed the rating (by carrying 554A.) The other two windings would only see 320A.

So in the 150kVA transformer, the two 320A loads would exceed the rating of one of the lines, and exceed the ratings of either one (wye) or two (delta) of the secondary windings.

 

[harbormaster]

Why not use 2 single phase 100 KVA dry type (One for each load) and let the utility deal with the horrific imbalance?

Locate the transformers near the loads and save feeder money too!

 

[kwired]

Why not use 2 single phase 100 KVA dry type (One for each load) and let the utility deal with the horrific imbalance?

Locate the transformers near the loads and save feeder money too!

What if the POCO way of dealing with horrific imbalance is charging customer for improvements in POCO equipment? Not that that wouldn't be a problem anyway with a three phase transformer.

To balance the load one would need to either generate single phase from three phase source, not necessarily the most efficient way of doing it, or derive single phase from three phase through solid state equipment similar to how VFD's work.

We still haven't found out what kind of load uses 320 amps @ 208Volts that can't be reconfigured to operate on three phases.

 

[topgone]

I come up with a much different Apparent Power Value than posted by others.

Looks like a maximum Load of 133.12 KVA, as the OP defined (2) L-L 208V Heaters, drawing 320 Amps each.

If both Loads are to be driven Coincidentally, Minimum 3 Phase Transformer Capacity would be 150 KVA.
If Loads are Non-Coincidental, the Minimum Capacity would be 75 KVA.

Connect Loads as follows:
Heater #1: Connect between Line "A" and Line "B"
Heater #2: Connect between Line "C" and Line "A"

Line "A" would be common to both Heaters and their corresponding 2 Wire Branch Circuits. The Heaters would be connected in an Open Delta fashion.

Scott

You nailed it! I got the same numbers as yours:

• Ia = 320 A.
• Ib = 524.26 A
• Ic = 320 A
• Total kva = 133.12 kVA

I guess 150 kVA would suffice.

 

[david luchini]

You nailed it! I got the same numbers as yours:

• Ia = 320 A.
• Ib = 554.26 A
• Ic = 320 A
• Total kva = 133.12 kVA

I guess 150 kVA would suffice.

As mentioned earlier, a 150kVA, 208V, 3ph transformer secondary would be rated for 416.4A. Your Ib listed above exceeds the rated secondary current by 33%.

 

[topgone]

As mentioned earlier, a 150kVA, 208V, 3ph transformer secondary would be rated for 416.4A. Your Ib listed above exceeds the rated secondary current by 33%.

OOPSSS! Missed that one! Thank you for correcting me. On 554A, then 200kVA it is!