Sizing of Transformer with unbalanced load.

[reyweesor]

Scenario: In Philippines, we are using 3 single phase Transformer to have a 3 phase supply.
My supply will be Delta - Wye - 230 V 3 phase, 4 wires system. But my load is 4 Single phase loads which is causing me to unbalance load.

Item	Description	Volts	Line Current A	Line Current B	Line Current C
1	18 kVA Single Phase	230	78	78
2	18 kVA Single Phase	230		78	78
3	18 kVA Single Phase	230	78		78
4	18 kVA Single Phase	230	78	78
		Total Feeder Current	234	234	156

My question is,
1.) How to calculate for the size of transformer would it be the total kVA of the loads which is (4*18 kVA) = 72 kVA and use 3 - 25 kVA single phase transformer
or kVA = √3 x Line Voltage x Line Current = √3 x 230 x 234 = 93.2 kVA and use 3 - 37.5 kVA single phase transformer.

Please help. Thanks

 

[bwat]

The second way with the (3) 37.5 kVA.

The proof would be to consider the load on secondary winding on phase A for example. Since it's in a wye configuration, your winding current will equal your line current (234A). The voltage across the winding will be 230/1.73 = 133 volts (i.e. L-N voltage). So the phase A winding power is 133V x 234V = 31.1kVA


(I'm not familiar at all with Philippine voltages. I assume, based on description and table, that 230/130 is what's being talked about here)

 

[reyweesor]

The second way with the (3) 37.5 kVA.

The proof would be to consider the load on secondary winding on phase A for example. Since it's in a wye configuration, your winding current will equal your line current (234A). The voltage across the winding will be 230/1.73 = 133 volts (i.e. L-N voltage). So the phase A winding power is 133V x 234V = 31.1kVA


(I'm not familiar at all with Philippine voltages. I assume, based on description and table, that 230/130 is what's being talked about here)

Thank you so much.

 

[LarryFine]

Using the greatest current as basis for rating of all units allows for any future expansion, which will only reduce the imbalance.

 

[topgone]

Take the highest current and compute the size of the transformer using that value as the line current (234A)! I would suggest for you to use a fudge factor to provide a small margin for future loads expansion of the transformer.
Example:
234 X 1.732 X 230 X 1.4 = 130.5 kVA! Use a three-phase transformer = 112.5 kVA (or 3 X 37.5 kVA)

 

[reyweesor]

Thank you for all your response.

 

[synchro]

The line currents above should be scaled by √3/2 ≅ 0.866 because they are at a 30° angle from the load currents.
So the line currents on phases A and B would be 0.866 x 234A = 203A. And, as bwat described, the kVA on the windings for A and B would each be 133V x 203A = 27.0 kVA (which is 1.5 x 18kVA). This makes sense since half of the 18kVA from the extra load designated "Item 4" above is supplied by phase A, and half by phase B.
But you will still need three 37.5 kVA units because 25 kVA is insufficient.

 

[reyweesor]

Thank you

 

[ussop]

Thanks man