Delta to wye transformer

[hhsting]

I have dry type 75kVA transformer primary delta and secondary is wye. Secondary wye side is unbalanced with phase A: 15305VA, phase B: 11316VA, phase C: 13985VA.

No neutral is brought delta side primary. Questions

1. what load does Delta side of the transformer line cable and winding see?

2. Is it ok not to have neutral delta side?



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[jim dungar]

This sounds like an exam question.

What have you come up with so far? Where would you connect a neutral on a delta primary winding configuration?

 

[hhsting]

This sounds like an exam question.

What have you come up with so far? Where would you connect a neutral on a delta primary winding configuration?

It’s actually shown on panel schedule on plans.

No their is no neutral delta side so their would like be circulating currents on delta winding I am assuming.

Primary side of the transformer connects to service switchboard and I was wondering service switchboard schedule phase A, B, C see 121818VA each or each phase service switchboard schedule sees different vA shown post #1?


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[kwired]

The primary will see similar imbalance of loading as is on the secondary.

 

[synchro]

I have dry type 75kVA transformer primary delta and secondary is wye. Secondary wye side is unbalanced with phase A: 15305VA, phase B: 11316VA, phase C: 13985VA.

No neutral is brought delta side primary. Questions

1. what load does Delta side of the transformer line cable and winding see?

2. Is it ok not to have neutral delta side?

Each winding of the delta sees the same kVA load as that on the corresponding winding of the wye. However, the current and kVA in each delta winding is not directly measurable if the windings are internally connected inside a single 3-phase transformer .

The input kVA for each line input of the delta primary is:
kVA line A = 0.577 √ ( kVA² load A + kVA load A x kVA load B + kVA² load B )
kVA line B = 0.577 √ ( kVA² load B + kVA load B x kVA load C + kVA² load C )
kVA line C = 0.577 √ ( kVA² load C + kVA load C x kVA load A + kVA² load A )

So for kVA load A = 15.305, kVA load B = 11.316, kVA load A = 13.985, we have:
kVA line A = 13.36
kVA line B = 12.67
kVA line C = 14.65
This assumes that the power factors on the three phases are near unity or at least the same.
You can see that the delta line input currents are not quite as unbalanced as the load currents on the wye. This is because each line input current to the delta is a vector summation of the currents from two different windings.

 

[kwired]

Each winding of the delta sees the same kVA load as that on the corresponding winding of the wye. However, the current and kVA in each delta winding is not directly measurable if the windings are internally connected inside a single 3-phase transformer .

The input kVA for each line input of the delta primary is:
kVA line A = 0.577 √ ( kVA² load A + kVA load A x kVA load B + kVA² load B )
kVA line B = 0.577 √ ( kVA² load B + kVA load B x kVA load C + kVA² load C )
kVA line C = 0.577 √ ( kVA² load C + kVA load C x kVA load A + kVA² load A )

So for kVA load A = 15.305, kVA load B = 11.316, kVA load A = 13.985, we have:
kVA line A = 13.36
kVA line B = 12.67
kVA line C = 14.65
This assumes that the power factors on the three phases are near unity or at least the same.
You can see that the delta line input currents are not quite as unbalanced as the load currents on the wye. This is because each line input current to the delta is a vector summation of the currents from two different windings.

Yes if you can't easily measure current in a particular delta primary winding you must keep in mind that just clamping onto say the A input leg you will be reading a vector sum of the current in two coils that connect to the A input leg.

 

[jim dungar]

Each winding of the delta sees the same kVA load as that on the corresponding winding of the wye. However, the current and kVA in each delta winding is not directly measurable if the windings are internally connected inside a single 3-phase transformer.

The input kVA for each line input of the delta primary is:

Great answer.
But for those trying this at home, make sure to look at how your transformer is actually constructed. It is not uncommon to find the wye side A winding being connected between the delta line A and line C, in which case you would still use the same same formula but the results for Line A and Line C would be reversed.

 

[synchro]

Great answer.
But for those trying this at home, make sure to look at how your transformer is actually constructed. It is not uncommon to find the wye side A winding being connected between the delta line A and line C, in which case you would still use the same same formula but the results for Line A and Line C would be reversed.

I agree that you need to look carefully how the transformer is connected. One way to confirm which pair of phases on the wye is associated with a particular line current on the delta is to change the load on each phase of the wye, one at a time, while observing which of the wye phases makes a difference on the line current being measured. The two phases on the wye that change that line current are the ones to use in the equations of post #5.
So for example if delta line current A changes with the loads on wye phases A and B then use those loads in the equation for delta line current A. If it instead changes with the loads on wye phases A and C, then use the A and C load kVA values.