Unbalanced Wye

[ptonsparky]

Ref: Recent thread on utilizing only a single phase panel feed by a three phase transformer.

How does that affect the primary delta currents?

 

[charlie b]

I am not certain, but I believe that one leg of the delta primary will have zero current.

 

[Jpflex]

Ref: Recent thread on utilizing only a single phase panel feed by a three phase transformer.

How does that affect the primary delta currents?

I believe I was the one who made several post on this setup.

At work I established power to a 208 volt charge station for battery powered mine carts.

Source is 50KVA? 3 phase 480 volts (actually 509 v) from ungrounded delta delta system transformer 4,800 volts to 480 (509v) 3 phase

This feeds another step down 3 phase transformer 30KVA, 480 volt delta to 208Y120. Three legs of primary delta 30KVA transformer are fed BUT only two legs of secondary are utilized to feed single phase panel for 208/120 volts

I would also like to know specifics about unused third secondary winding that is intentionally left without a connection and transformer imbalance?

Would it be better to use a 3 phase panel to supply single phase loads by using all 3 sets of windings and using a single phase A,B and C along with derived neutral and is this code legal?

 

[Jpflex]

Step down with Taps adjusted for higher input 509 voltage resolved in target 208/120 volt output

Unused secondary X 3 terminal not connected for single phase panel

 

[ramsy]

BUT only two legs of secondary are utilized to feed single phase panel for 208/120 volts

Isolation transformers don't have extra legs, with no wiring between Pri & Secondary.

 

[wwhitney]

I am not certain, but I believe that one leg of the delta primary will have zero current.

That would be true for a wye-wye transformer, but not a delta-wye transformer.

If we call the 3 primary legs L1, L2, and L3, and the 3 secondary legs L1', L2', and L3', let's say that the primary coil L1-L2 provides the voltage L1'-N, etc. Then with only secondary loads on L1', L2', and N, the delta primary will only see loads L1-L2 and L2-L3, with no load L1-L3.

Cheers, Wayne

 

[Jpflex]

Isolation transformers don't have extra legs, with no wiring between Pri & Secondary.

I think you misunderstand. This is a separately derived system transformer with no connection between primary and secondary besides grounding and bonding for grounded systems. The extra leg I was referring to was secondary lug not being used

 

[Jpflex]

That would be true for a wye-wye transformer, but not a delta-wye transformer.

If we call the 3 primary legs L1, L2, and L3, and the 3 secondary legs L1', L2', and L3', let's say that the primary coil L1-L2 provides the voltage L1'-N, etc. Then with only secondary loads on L1', L2', and N, the delta primary will only see loads L1-L2 and L2-L3, with no load L1-L3.

Cheers, Wayne

I was wandering if there would be any drawbacks terminating on x1 and x3 rather than x1 and x2 for secondary output to single phase panel.

All phases are 120 degrees opposed from any set of phases so why would it matter which pair of phases are used in a Y?

 

[ptonsparky]

I was wandering if there would be any drawbacks terminating on x1 and x3 rather than x1 and x2 for secondary output to single phase panel.

All phases are 120 degrees opposed from any set of phases so why would it matter which pair of phases are used in a Y?

...

Would it be better to use a 3 phase panel to supply single phase loads by using all 3 sets of windings and using a single phase A,B and C along with derived neutral and is this code legal?

There will be no difference using any two of the legs for your 1 phase panel, but use a three phase panel. The transformer is giving you a neutral point for all three phases. It is done all the time.

 

[ptonsparky]

That would be true for a wye-wye transformer, but not a delta-wye transformer.

If we call the 3 primary legs L1, L2, and L3, and the 3 secondary legs L1', L2', and L3', let's say that the primary coil L1-L2 provides the voltage L1'-N, etc. Then with only secondary loads on L1', L2', and N, the delta primary will only see loads L1-L2 and L2-L3, with no load L1-L3.

Cheers, Wayne

Ok, my transformer is 480 to 208/120.
L1' to N =23.7 L2 to N=47.4 L3' to N =0
L1 to L2 =10 L2 to L3 = 20
L1=10, L2=26.46 L3=20

 

[wwhitney]

Ok, my transformer is 480 to 208/120.
L1' to N =23.7 L2 to N=47.4 L3' to N =0
L1 to L2 =10 L2 to L3 = 20
L1=10, L2=26.46 L3=20

The turns ratio on the transformer would be 4:1, as each primary coil sees 480V, and each secondary coil generates 120V L'-N.

So your example would need to have secondary L1' to N = 40A, L2' to N = 80A, L3' to N = 0 to get L1 to L2 load = 10A, L2 to L3 load = 20A.

If the L1' to N and L2' to N currents have identical phase shifts (relative to their supply voltages), so that the secondary currents are 120 degrees out of phase, then your primary L2 computation is correct. If for some reason the secondary phase shifts differed, then you'd get a different answer.

For a new example, say the only secondary load is a 40A, 208V resistive load supplied L1' to L2'. The L1' to N and L2' to N currents are both 40A, but they are in phase with each other. The primary L1 to L2 and L2 to L3 currents are both 10A, and also in phase with each other. So the delta currents would be L1 = 10A, L2 = 20A, and L3 = 10A.

To check that the power accounting works in this new example (same power on the primary and the secondary), the secondary side is just 208V * 40A = 480 * 10 * sqrt(3) VA. On the primary side, the current is at 30 degrees to the voltage through the L1-L2 and L2-L3 coils, with no current through the L1-L3 coil. So the power in each of the first two coils would be 480V * 10A * cos(30 deg) = 480V * 10A * sqrt(3) / 2. Add those two together and you also get 480 * 10 * sqrt(3) VA.

Cheers, Wayne

 

[ptonsparky]

The turns ratio on the transformer would be 4:1, as each primary coil sees 480V, and each secondary coil generates 120V L'-N.

So your example would need to have secondary L1' to N = 40A, L2' to N = 80A, L3' to N = 0 to get L1 to L2 load = 10A, L2 to L3 load = 20A.

If the L1' to N and L2' to N currents have identical phase shifts (relative to their supply voltages), so that the secondary currents are 120 degrees out of phase, then your primary L2 computation is correct. If for some reason the secondary phase shifts differed, then you'd get a different answer.

For a new example, say the only secondary load is a 40A, 208V resistive load supplied L1' to L2'. The L1' to N and L2' to N currents are both 40A, but they are in phase with each other. The primary L1 to L2 and L2 to L3 currents are both 10A, and also in phase with each other. So the delta currents would be L1 = 10A, L2 = 20A, and L3 = 10A.

To check that the power accounting works in this new example (same power on the primary and the secondary), the secondary side is just 208V * 40A = 480 * 10 * sqrt(3) VA. On the primary side, the current is at 30 degrees to the voltage through the L1-L2 and L2-L3 coils, with no current through the L1-L3 coil. So the power in each of the first two coils would be 480V * 10A * cos(30 deg) = 480V * 10A * sqrt(3) / 2. Add those two together and you also get 480 * 10 * sqrt(3) VA.

Cheers, Wayne

I'll study it.
I don't know why I'm doing this now that retired has come about. I can only think of a time or two that I may have measured both primary and secondary currents on three phase transformers.

 

[ptonsparky]

Well that endeavor made me want to find a sharp stick for my eye.
I learned enough to know I don't know enough..

Thank you,

 

[ptonsparky]

Ok not done yet.
I have line to neutral loads on the secondary plus line to line. Do I calculate for L-Ns then add the L-Ls?
Too simple? Resistive loads only.

 

[wwhitney]

I have line to neutral loads on the secondary plus line to line. Do I calculate for L-Ns then add the L-Ls?
Too simple? Resistive loads only.

With only linear loads (which includes resistive loads), superposition applies, so you can calculate the currents in any order and just add them together. The only tricky thing is that the currents add as vectors, so for an exact answer you need to keep track of direction as well as magnitude. If you just want an upper bound on current magnitude for a load calculation, though, you could just add the magnitudes to get an overestimate.

In the standard way we draw transformers, the coils are drawn on the page effectively as vectors already (showing both correct direction and relative magnitude). And for resistive loads, the current will be in phase with the voltage, so for L-N secondary loads, the currents are in the direction of the coil to which they are connected.

For an L-L resistive load, the current will flow through two coils, so it needs to get added to both coil currents. And the phase offset of the current will be 30 degrees relative to each of those coils. If you just want the magnitude (not the direction) of the sum of two vectors X and Y at an angle A to each other, you can use the law of cosines. Here |X| means the magnitude of the vector X:

|X + Y|² = |X|² + |Y|² + 2 |X| |Y| cos A.

In the case we're discussing, A = 30 degrees, and cos A = sqrt(3)/2.

So for example if you had resistive loads of 10A L1-N, 20A L2-N, and 30A L1-L2, the current on the L1-N coil would be sqrt(10² + 30² + sqrt(3) * 10 * 30) = 39.0 A. You can see that for this example at least, the deviation from the approximation of just summing the magnitudes 10 + 30 = 40A is not very large. Worst case of the error would occur when |X| = |Y|, in which case the sum would be sqrt(2 + sqrt(3)) = 1.93 |X|, rather than the approximation 2 |X|.

Worst case, that is, for the case that all loads are L1-N, L2-N, or L1-L2. If you have L1-L2, L1-L3, and L1-N loads, and you calculate the L1-N coil current just by summing up the magnitudes, the error is going to be much larger, as you are missing the significant cancellation you would get from adding the L1-L2 and L1-L3 loads. Which, if their currents are equal and resistive, will sum to a factor of sqrt(3) rather than 2, as should be familiar.

Cheers, Wayne

P.S. My background is in math, so I am working out the physics as I go for educational purposes, which means there is some chance of error.