Unbalanced line-to-line loads on 3-ph transformer

[ben5243]

I have a 54kVA 3-ph transformer wired for 240V in a Y-config. No neutral wire to loads because all loads are 3-ph or single phase line-to-line.

Loads:
1x 240VAC 3ph with full load 19A (industrial chiller)
1x 240VAC 3ph 15A max input (5000W power supply with 0.95 power factor)
4x 240VAC 1ph 2000W power supplies (0.94 power factor)
4x 240VAC 1ph 650W power supplies (no power factor given - assumed 0.8)


For the 3 phase loads my kVA is:
240*19*1.73 = 7.898 kVA
and
240*15*1.73 = 6.325 kVA

Then I divide the 3 phase loads

For the single phase loads, I use the wattage and power factor to get kVA
2000/0.94 = 2.128 kVA
and
650/0.8 = 0.813 kVA

Because the single phase loads are line-to-line, do I then need to factor in 1.73? I'm pretty sure no, but I wanted to ask here.
From my understanding, I divide the kVA in half between the two phases they are connected to, so here is my load table:

		Voltage	Rated Amps	kVA		U	V	W
Chiller	3Φ	240VAC	19A	7.9		2.6	2.6	2.6
Power Supply	3Φ	240VAC	15A	6.2		2.1	2.1	2.1
Server Power Supply	1Φ	240VAC		2.1		1.1	1.1
Server Power Supply	1Φ	240VAC		2.1			1.1	1.1
Server Power Supply	1Φ	240VAC		2.1		1.1		1.1
Server Power Supply	1Φ	240VAC		2.1		1.1	1.1
Device Power Supply	1Φ	240VAC		0.8			0.4	0.4
Device Power Supply	1Φ	240VAC		0.8		0.4		0.4
Device Power Supply	1Φ	240VAC		0.8		0.4	0.4
Device Power Supply	1Φ	240VAC		0.8			0.4	0.4
					Total kVA:	8.7	9.1	8.0	25.7

Does this appear to be the correct method? Total load is 27.3 kVA?

Thanks

 

[synchro]

Because the single phase loads are line-to-line, do I then need to factor in 1.73? I'm pretty sure no, but I wanted to ask here.

Not if you're keeping your calculations in kVA on both the L-L and L-N sides.
When you add two L-L currents from different phases to get a L-N current you use a factor of 1.73, but the L-N voltage is a factor of 1/1.73 relative to the L-L voltage. Therefore the √3 ≈ 1.73 divides out when multiplying voltage times the current for kVA, and so effectively kVA is conserved when going between the L-L and L-N sides (as you'd expect from a physics point of view).

 

[ben5243]

Thank you, that confirms what I read but I wasn't able to find an explanation of why. Makes perfect sense now!

So for my load table, is it correct practice to divide the 3 phase loads by 3 for each phase, and divide the single phase loads in half for each phase they connect to?

 

[synchro]

So for my load table, is it correct practice to divide the 3 phase loads by 3 for each phase ...?

Yes.

... is it correct practice to ...divide the single phase loads in half for each phase they connect to?

Yes, and it will be sufficiently accurate as long as the L-L loads are reasonably well balanced between phases so that the resulting line currents are in-phase with the corresponding L-N voltage of the wye. Your loads above are sufficiently well balanced.

In the extreme case of unbalance when there's only a single L-L load, the resulting L-N current will be at 30° from the L-N voltage and not in-phase like it will be with balanced L-L loads. That's because the reactive component of this L-L current relative to the L-N voltage is not offset by that from an adjacent L-L load current at a 60 degree relative angle. Therefore with a single L-L load the L-N current will equal the full L-L current instead of being scaled by 1.73/2. And so instead of allocating 1/2 of the L-L load's kVAR to each phase you'd need to allocate 1/2 x 2/1.73 = 0.577 times the L-L kVar to be accurate, but that's not much off from the normally used 1/2 factor even under this extreme case.