high leg load

[Do You Like Ham]

Hi! I'm trying to understand how one would determine the primary line currents in a 480 delta to 240/120 split phase 3ph 4w delta system. E.g., for example if there is an 80A (208v) load connected on high-leg to neutral on the secondary, how does one determine the currents on the primary line conductors?

 

[LarryFine]

You shouldn't load a delta like that. A high-leg delta IS a 1ph 120/240v source and a 3ph 240v source superimposed on one another. Use it as one or the other, but not high leg to neutral.

 

[jim dungar]

Hi! I'm trying to understand how one would determine the primary line currents in a 480 delta to 240/120 split phase 3ph 4w delta system. E.g., for example if there is an 80A (208v) load connected on high-leg to neutral on the secondary, how does one determine the currents on the primary line conductors?

All of your 208V 80A load will flow through only 1/2 of the center tapped winding, in addition to any other single phase loading. Has your transformer bank been sized to accommodate this?

 

[Do You Like Ham]

@LarryFine thanks for the professional guidance, however, this is not a question on application, it is a question on the theory of such a calculation.

 

[Do You Like Ham]

All of your 208V 80A load will flow through only 1/2 of the center tapped winding, in addition to any other single phase loading. Has your transformer bank been sized to accommodate this?

@jim dungar Can you elaborate on why that is? Why would it be different other other parallel feeds?

re: "Has your transformer bank been sized to accommodate this?" hence the question.. if one wanted to to size for such a condition, what is the computation to do so?

 

[jim dungar]

@jim dungar Can you elaborate on why that is? Why would it be different other other parallel feeds?

re: "Has your transformer bank been sized to accommodate this?" hence the question.. if one wanted to to size for such a condition, what is the computation to do so?

I assumed, you were dealing with a typical open-delta transformer bank.
For example, if you were drawing the current path, your 80A 'leaves' the B, high leg terminal, and enters the N terminal. The current then 'flows' from the N terminal to the A terminal, which is 1/2 of the center tapped coil "A", and finally back to B through transformer 'B').

With a typical open delta transformer sizing formulas of:
KVA(A) = .58T+S
KVA(B) = .58T
where T = the total 3-phase load, and S = the total balanced 1-phase load.

If you were creating you own transformer bank of 3 units, connected in a wye primary - delta secondary arrangement the overall transformer would be sized using:
KVA = 5Su+T
where T = the total 3-phase load, and Su = the maximum unbalanced 1-phase (120V) load.
or
KVA=2.5S+T
where T = the total 3-phase load, and S = the total balanced 1-phase load.

 

[Do You Like Ham]

@jim dungar - ah, ok.. yes, I was meaning a wye-delta configuration... but ultimately, what I'm after is the current in the primary feeders.

 

[wwhitney]

I'm teaching myself about topics like this, so the following bears further confirmation by others:

Using the nomenclature in the diagram below, say you have an 80A secondary load connected L2 - N. Then if the impedance of coil A + coil C1 exactly matches the impedance of coil B and coil C2, the current would split with 40A flowing through A + C1 and 40A flowing through B + C2.

If the positive current direction is clockwise through A, C1/C2, and B, and away from the transformer for L1, L2, L3, and N, and if the 80A load is considered +80A on L2 and -80A on N, then the currents on the coils (A, B, C1, C2) are (-40, 40, -40, 40).

On the primary side, the currents are just half as much, and I believe the current in the C coil is just the transformation of the average current on C1 and C2 (not 100% sure on this, but what else could it be?) That would make the primary currents (A, B, C) to be (-20, 20, 0). Then the primary line currents are given by L1 = A - C, L3 = C - B, and L2 = B - A. That gives (L1, L2, L3) = (-20, 40, -20). The phase angle of those currents is in phase with the original secondary load current L2 - N.

Now contrast this with another way to get secondary currents of 40A in magnitude in each coil, namely just apply a 40A 2-wire load across each coil A, B, and C1/C2. That gives you 3 * 240 * 40A = 28.8 kVA of apparent power, while the 80A 208V L2-N load is just 16.6 kVA of apparent power, a factor of 1/sqrt(3) less.

Cheers, Wayne

 

[jim dungar]

Wayne, If there is current flowing through C1 and C2 on the secondary, you will have some current flowing in Cprimary. Have you considered the circulating currents in the delta?

This is one reason these types of transformer banks are usually Wye primary or open-delta.

 

[wwhitney]

Wayne, If there is current flowing through C1 and C2 on the secondary, you will have some current flowing in Cprimary.

Thanks for responding, this may be a limit of my physics understanding. But what happens when the current in secondary C1 is the negative of the current in secondary C2? (For a consistent choice of positive for primary C and secondary C1 and C2.)

Superposition is telling me that the primary current attributable to secondary C1 cancels the primary current attributable to secondary C2, and you'd get zero current in the primary C. And if the current L2-N splits evenly between coils A and B, it looks like that implies the currents in C1 and C2 are opposites. Unless I've made a subtle sign error.

Have you considered the circulating currents in the delta?

If the arrangement is symmetric (coils A and B have the same impedance, and coils C1 and C2 do as well), I don't see how you could get a circulating current in one direction rather than the other?

I'm still a bit puzzled about all this.

Thanks, Wayne

 

[jim dungar]

Thanks for responding, this may be a limit of my physics understanding. But what happens when the current in secondary C1 is the negative of the current in secondary C2? (For a consistent choice of positive for primary C and secondary C1 and C2.)

Superposition is telling me that the primary current attributable to secondary C1 cancels the primary current attributable to secondary C2, and you'd get zero current in the primary C. And if the current L2-N splits evenly between coils A and B, it looks like that implies the currents in C1 and C2 are opposites. Unless I've made a subtle sign error.



If the arrangement is symmetric (coils A and B have the same impedance, and coils C1 and C2 do as well), I don't see how you could get a circulating current in one direction rather than the other?

I'm still a bit puzzled about all this.

Thanks, Wayne

So, why is the C primary coil even needed?
Were you thinking of (3) cores or a common core?
How would this work if the primary was connected in a wye, as per the OP in post #7?

 

[wwhitney]

So, why is the C primary coil even needed?
Were you thinking of (3) cores or a common core?

Good questions, and I'm not really sure. But I believe the math is saying that you could delete the C primary coil. You'd be left with an open delta transformer, and the secondary C1-C2 acting as an autotransformer to provide the neutral point. Which seems entirely plausible to me.

How would this work if the primary was connected in a wye, as per the OP in post #7?

The secondary side analysis is unchanged. On the primary side, now the voltage ratio between coils is 277V : 240V = 2/sqrt(3). So the primary coil currents would be sqrt(3)/2 times the average secondary coil currents, or (A, B, C) = (-20 sqrt(3), 20 sqrt(3), 0). The positive convention I chose for the secondary coil currents gives a positive convention on the primary of current moving from the neutral to the line terminals. So if the positive convention of the neutral/line conductors is away from the transformer, the conductor currents (A, B, C, N) would be (-20 sqrt(3), 20 sqrt(3), 0, 0). Again the primary C coil could be deleted, giving an open wye-open delta configuration, with C1/C2 as an autotransformer across the open leg of the delta.

I think. : -)

Cheers, Wayne

 

[jim dungar]

Have you considered the angles between the A, B, C1, and C2 coils that would exist in your 'open autotransformer'.

Your output cannot be an open delta, as you need the common N terminal.

 

[wwhitney]

Have you considered the angles between the A, B, C1, and C2 coils that would exist in your 'open autotransformer'.

I can't quite tell if you're saying my answer in post #8 is wrong (which it may be, although I don't see it yet), or if you're just surprised by the answer (as I was) and so raising concerns that you'd like me to address. Anyway:

If you're referring to voltage phase angles, I don't see how deleting the C primary coil in the diagram in post #8 changes any of them.

As to the current phase angle, with a single 2-wire load L2-N, the load will determine the phase angle. If the load is resistive, it will be in phase with L2-N and thus purely reactive in secondary coil C1/C2. Either way, all the currents in the system are in phase with the load current, which is why my computations are just with scalars, rather than phasors.

Your output cannot be an open delta, as you need the common N terminal.

The equivalency I'm making for the diagram in post #8 (at least if it is made with individual single phase transformers) with the primary C coil deleted is to a 3-wire open delta, plus an autotransformer to derive a (single phase) neutral.

I.e. ignoring usual practice and naming conventions (so as to match the figure in post #8), say the POCO provides you with an ungrounded 240V 3-wire delta service. It's supplied by an open delta with single phase transformers L1 - L2 and L2 - L3. You supply your own center-tapped autotransformer and connect it L1 - L3. The midpoint is now your N terminal, which you earth.

Cheers, Wayne

 

[jim dungar]

I can't quite tell if you're saying my answer in post #8 is wrong (which it may be, although I don't see it yet), or if you're just surprised by the answer (as I was) and so raising concerns that you'd like me to address. Anyway:

If you're referring to voltage phase angles, I don't see how deleting the C primary coil in the diagram in post #8 changes any of them.

As to the current phase angle, with a single 2-wire load L2-N, the load will determine the phase angle. If the load is resistive, it will be in phase with L2-N and thus purely reactive in secondary coil C1/C2. Either way, all the currents in the system are in phase with the load current, which is why my computations are just with scalars, rather than phasors.


The equivalency I'm making for the diagram in post #8 (at least if it is made with individual single phase transformers) with the primary C coil deleted is to a 3-wire open delta, plus an autotransformer to derive a (single phase) neutral.

I.e. ignoring usual practice and naming conventions (so as to match the figure in post #8), say the POCO provides you with an ungrounded 240V 3-wire delta service. It's supplied by an open delta with single phase transformers L1 - L2 and L2 - L3. You supply your own center-tapped autotransformer and connect it L1 - L3. The midpoint is now your N terminal, which you earth.

Cheers, Wayne

Would the end points of your two 'open-delta' legs be at the same voltage potential, if not they could not be combined into a neutral point.
Assuming VA = 240@0° and VB= 240@120° then V_C = 240@240° with V_C1 = 120@240° and V_C2 =120@240°

In a typical open delta autotransformer does V_A+V_C1 = V_B+V_C2?

 

[wwhitney]

Would the end points of your two 'open-delta' legs be at the same voltage potential, if not they could not be combined into a neutral point.
Assuming VA = 240@0° and VB= 240@120° then V_C = 240@240° with V_C1 = 120@240° and V_C2 =120@240°

I'm not sure I'm following your question. If you give me any 3 wire 240V delta, I can take one voltage, say V_C and use a single coil center tapped autotransformer to get a voltage point N at the midpoint of C. So if I do that, and call the voltages on the two sides of N (in opposite directions, one towards N, one away from N) V_C1 and V_C2, yes the voltage vectors would match what you've listed, for the appropriate conventions.

The above is true whether the delta you give me is supplied from an open delta or a closed delta. Of course, if it is supplied by a closed delta, it would be simpler just to center tap one of the secondary windings rather than use a separate autotransformer on the secondary side. [In the typical diagrams I've seen for a 4-wire high leg open delta supply, one of the two windings is center-tapped; in the arrangement I have been describing, the "phantom winding" is center tapped via this autotransformer.]

In a typical open delta autotransformer does V_A+V_C1 = V_B+V_C2?

I'm taking "open delta autotransformer" to mean "open delta plus an autotransformer," which is what I have been describing as equivalent to "closed delta minus the primary C coil."

For the sign convention implied by your question, sure. I'm used to the sign convention where V_A + V_B +V_C = 0 (as in the vectors spelled out above). Then whenever V_C1 + V_C2 = V_C (again as in the vectors above), you have V_A + V_B + V_C1 + V_C2 = 0, or V_A + V_C1 = - V_B - V_C2

Cheers, Wayne

 

[herding_cats]

There is no "high-leg" in a 208 system. High-leg hot to neutral single-phase loads are not usable. The voltage will fluctuate between 160v-250v.

 

[jim dungar]

There is no "high-leg" in a 208 system. High-leg hot to neutral single-phase loads are not usable. The voltage will fluctuate between 160v-250v.

Yes, there is a high leg. It is not usable because its loading is not usually taken into account during the transformer sizing, so the voltage regulation is poor. If voltage fluctuates between 160-250V, then your 240V across the open leg also is fluctuating, impacting the use of 3-phase loads.
The high leg voltage is fixed as 1.732X the Line-Neutral voltage.

 

[herding_cats]

Yes, there is a high leg. It is not usable because its loading is not usually taken into account during the transformer sizing, so the voltage regulation is poor. If voltage fluctuates between 160-250V, then your 240V across the open leg also is fluctuating, impacting the use of 3-phase loads.
The high leg voltage is fixed as 1.732X the Line-Neutral voltage.

You are saying that you have a 208v delta system with a high (wild) leg?

 

[herding_cats]

None of what you are asking makes much sense.

I have NEVER seen a 208v delta system in any building. Not saying you couldn’t build a transformer to do this, but why would you?

Additionally, I have never heard of any 480/460v system with a wild leg, off of a delta type transformer.

Either way, there is no usable function for any load hot to neutral off of a delta transformer from the wild leg.

If you are asking about single or three phase loads from a delta transformer, you calculate them the same whether they are connected to the wild leg or not. Every delta transformer I have ever seen has secondary voltages of 236-240 vac.