Transformer Question

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blue302

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I have the utility company feeding a service with 277/480v wye. I need to get 120/240v or 120/208v for some regular recepts and lights. I'm not too familiar with transformer setups. What would be the appropriate transformer to get the lower voltage from the 277/480v wye utility power? Thanks in advance.
 

augie47

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Tennessee
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State Electrical Inspector (Retired)
Normally you would feed a 480/208Y-120 transformer with the phases of the 480 system. You do not connect the neutral on the 480/277 side to the transformer.
 

templdl

Senior Member
Location
Wisconsin
I have the utility company feeding a service with 277/480v wye. I need to get 120/240v or 120/208v for some regular recepts and lights. I'm not too familiar with transformer setups. What would be the appropriate transformer to get the lower voltage from the 277/480v wye utility power? Thanks in advance.

It is strange that you did not state the kva required. But, guessing that your outlets and lighting loads are minimal simply get a common 1ph transformer of sufficient KVA,, 480-120/240 1ph3w. Select any two of the the 480v 3ph supply and connect the transfortmer applyig a proper pimary OCPD and grounding the X0 of the secondary. The X0 also becomes your neutral. Protect the transformer with a properly sized 2p OCPD.
Because your question is somewhat vague as it does not appears that you need a three phase transformer with a 208y/120 secondary. Just because your supply is a 480,y/277 disregard the 277L-N voltage as it is the L--L voltages that are important. And pair of the 480v 3ph primary voltages provides 480v 1ph.
 
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Carultch

Senior Member
Location
Massachusetts
I have the utility company feeding a service with 277/480v wye. I need to get 120/240v or 120/208v for some regular recepts and lights. I'm not too familiar with transformer setups. What would be the appropriate transformer to get the lower voltage from the 277/480v wye utility power? Thanks in advance.

The 120/208 WYE system is a lot easier to balance the phases, than the 120/240 high leg delta.
Dry-type transformers are common for this application. They are 480V delta on the primary (with no neutral), and 120/208WYE on the secondary.

On WYE systems, simply distribute your 120V loads onto each phase, and try to get the amperes on each to add up to the same value on each phase.

On 120/240 Delta systems, your 120V loads CANNOT connect to the B-phase, since it is 208V to neutral. Instead, your A-to-N-to-C winding gets all the loads, and the windings that connect to the B-phase carry no load. The main application for using the high leg delta secondary, is if 240V loads dominate the system with only a few 120V loads.
 

JoeStillman

Senior Member
Location
West Chester, PA
Transformer primaries connected to grounded-wye systems (as in 480Y/277) usually ought to be connected in delta. The secondary neutral does not reflect into the primary. That's why it's called "Separately Derived".

My rule of thumb for 120/208 or 120/240 secondaries is 3Ø for 30kVA and larger, 1Ø for below 30kVA. The capacity of each individual phase leg is larger for the 1Ø than for the 3Ø of the same kVA rating. But over 30 kVA, it's too much imbalance for my taste.
 

GoldDigger

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Other than simply because of what transformers are readily available, is there any particular reason why the primary is usually delta?

To avoid high (even damaging) circulating current involving the neutral IF the secondary is delta and the primary voltages are unbalanced for any reason.
 

GoldDigger

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Location
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Retired PV System Designer
The presence of a closed delta secondary requires that the voltages on the primary wye coils obey the same vector equation that secondary delta coils obey.
If the incoming wye voltages do not obey this relationship, enough current must flow in the primary neutral to offset its voltage by IR drop so that the relation is satisfied with the new center point voltage.
All of this happens with no load on the secondary.
Some might not think this situation is exactly circulating current, but it is close enough for me.
The currents in the wye primary correspond to currents in the delta secondary.
The nature of the wye source and how its output became unbalanced does not really matter.
But the stiffer the source (low source impedance), the greater the transformer currents.
Leaving the primary wye point unconnected allows its voltage to float to a value that allows the equation to be satisfied.
 
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Bugman1400

Senior Member
Location
Charlotte, NC
The presence of a closed delta secondary requires that the voltages on the primary wye coils obey the same vector equation that secondary delta coils obey.
If the incoming wye voltages do not obey this relationship, enough current must flow in the primary neutral to offset its voltage by IR drop so that the relation is satisfied with the new center point voltage.
All of this happens with no load on the secondary.
Some might not think this situation is exactly circulating current, but it is close enough for me.
The currents in the wye primary correspond to currents in the delta secondary.
The nature of the wye source and how its output became unbalanced does not really matter.
But the stiffer the source (low source impedance), the greater the transformer currents.
Leaving the primary wye point unconnected allows its voltage to float to a value that allows the equation to be satisfied.

Are you referring to circulating current in the delta or between one g-wye xfmr to another? I'm still having a tough time understanding what you're saying. You also lost me on why a low source impedance causes greater xfmr currents. Are you referring to a fault?
 

GoldDigger

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Location
Placerville, CA, USA
Occupation
Retired PV System Designer
Are you referring to circulating current in the delta or between one g-wye xfmr to another? I'm still having a tough time understanding what you're saying. You also lost me on why a low source impedance causes greater xfmr currents. Are you referring to a fault?
Look at it this way:

The three windings in the delta secondary (call them a, b and c) form a triangle. The vector equation a = b +c has to be satisfied to keep current from circulating. The bold face letters indicate vector (or phasors if you prefer). So far so good?

Now a, b and c are each directly related by the turns ratio to the primary winding currents A, B and C. In the wye the lettered currents are in wye windings and each corresponds to a service conductor, in the secondary the lettered currents are in the delta windings and the currents out to the load will each come from both windings at that (un lettered) terminal.
Still with me?

Since the wye service conductors may be derived from a wye bank and may have unbalanced loads on them, we cannot assume that the magnitude of the line to neutral voltages are all equal.
In fact we cannot be sure that A = B + C.
Since a = b +c and A is not equal to B + C, something has to give. (Given that the two sets of voltages are each related by the same turns ratio.)

The slack will be taken up in two ways by the same current flows:
Currents will run in the delta windings until the IR losses have made the resulting voltages satisfy the equation in conjunction with the corresponding currents in the primary causing IR drops to help make the second equation be satisfied. Neither mechanism will be responsible by itself for the balance.
On the primary side one contribution toward balance will be an IR based shift in the neutral point, the result of really high neutral currents, since the currents in the three windings will be far from balanced.

If you instead just open the neutral of the wye primary and let the center point float, it does not take any significant current flow to make the primary voltages obey the vector equation. In fact, at idle there will be no circulating currents and no primary currents except for the magnetizing current of the transformer.


Without drawing a picture and waving my hands, that is the best I can do.

In answer to your second question, to get the necessary IR drops on the primary side will take more current if the service contribution to the IR drop on each phase line and the neutral is is reduced.

The IR contributions on the secondary side remain unaffected by source impedance, but will require more current the lower the transformer winding resistances are.
 

Bugman1400

Senior Member
Location
Charlotte, NC
Look at it this way:

The three windings in the delta secondary (call them a, b and c) form a triangle. The vector equation a = b +c has to be satisfied to keep current from circulating. The bold face letters indicate vector (or phasors if you prefer). So far so good? The delta triangle cannot have a current eqn of a = b + c. Perhaps you are referring to each of the leads where, phase A lead (outside the delta) is equal to the sum of the two connected legs of the delta (a + b inside the delta) times 1.73. What type of circulating current in the delta are you talking about....positive, negative, or zero?

Now a, b and c are each directly related by the turns ratio to the primary winding currents A, B and C. In the wye the lettered currents are in wye windings and each corresponds to a service conductor, in the secondary the lettered currents are in the delta windings and the currents out to the load will each come from both windings at that (un lettered) terminal.
Still with me?

Since the wye service conductors may be derived from a wye bank and may have unbalanced loads on them, we cannot assume that the magnitude of the line to neutral voltages are all equal.
In fact we cannot be sure that A = B + C.
Since a = b +c and A is not equal to B + C, something has to give. (Given that the two sets of voltages are each related by the same turns ratio.)

The slack will be taken up in two ways by the same current flows:
Currents will run in the delta windings until the IR losses have made the resulting voltages satisfy the equation in conjunction with the corresponding currents in the primary causing IR drops to help make the second equation be satisfied. Neither mechanism will be responsible by itself for the balance.
On the primary side one contribution toward balance will be an IR based shift in the neutral point, the result of really high neutral currents, since the currents in the three windings will be far from balanced.

If you instead just open the neutral of the wye primary and let the center point float, it does not take any significant current flow to make the primary voltages obey the vector equation. In fact, at idle there will be no circulating currents and no primary currents except for the magnetizing current of the transformer.


Without drawing a picture and waving my hands, that is the best I can do.

In answer to your second question, to get the necessary IR drops on the primary side will take more current if the service contribution to the IR drop on each phase line and the neutral is is reduced.

The IR contributions on the secondary side remain unaffected by source impedance, but will require more current the lower the transformer winding resistances are. I thought lower wdg resistances just meant less losses.

I think I need pictures and some hand waving.
 

kwired

Electron manager
Location
NE Nebraska
I have the utility company feeding a service with 277/480v wye. I need to get 120/240v or 120/208v for some regular recepts and lights. I'm not too familiar with transformer setups. What would be the appropriate transformer to get the lower voltage from the 277/480v wye utility power? Thanks in advance.
Some of this has been mentioned.
First thing is to determine (in kVA) how much load you need to supply. If it is mimimal you may possible settle for less cost and a smaller single phase transformer. If load is heavier, you may want a three phase secondary so you can better balance the load as it applies to the 480 volt system.

Wye to wye issues have nothing to with the fact the 480/277 is from a wye system it is because the transformer primary windings are wye configured. Most typical single core dry type transformers you would use for this application use a delta connected primary and don't have those issues. Delta connected primary doesn't mean the source must be from a delta secondary - they are just connected in a delta fashion. Input voltage just needs to be 480 volts and phase displacement needs to be 120 degrees - which you get with either a wye or delta secondary on the supply system.
 

GoldDigger

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Location
Placerville, CA, USA
Occupation
Retired PV System Designer
I think I need pictures and some hand waving.
I sympathize, but have no way to do that.
Try thinking of what would happen when you take three separate transformers with unequal inputs and isolated windings. Measure the secondary voltages and draw your phase diagrams, and then look at what will happen when you try to connect those three windings into a delta. The voltages do not add up, so something has to give. It is harder to figure out just what does give, but it involves circulating currents.
 
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