Delta versus Wye Voltages

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jbellino

jbellino

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Hello gentlemen,

There's a question that I have about the nature of voltage between delta and wye configurations. Why is it that voltage between phases is additive and current is consistent in a wye configuration, and voltage is consistent but current is additive in a delta configuration? In delta the phases are connected in series with each other, and in wye they are connected in parallel with each other (once phase-to-phase is established). So it seems that the voltage and current rules for delta/wye should be vice versa, and I'm not sure why it isn't? I'd appreciate input from anyone who can help me figure this out, or point out if I have an improper understanding of something here.

Thank you,

-Jesse
 
Try not to use just the word 'phase' when describing conductors, instead use line or hot. Phase voltage is measured between two points either line-line or line-neutral.
In engineering school we studied 3-phase systems for some 6 months before we spent about 2weeks on 1-phase ones. My suggestion is to just accept that 120/240V 1-phase 3wire systems, like you are used to, are really the exception and shouldn't be the starting point of learning about poly phase systems.

The windings/coils of delta sources are not connected in series, likewise those of wye sources are not connected in parallel. Delta voltages are line-line while wye voltages are line-neutral.
 
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jbellino said:
... Why is it that voltage between phases is additive and current is consistent in a wye configuration, and voltage is consistent but current is additive in a delta configuration? In delta the phases are connected in series with each other, and in wye they are connected in parallel with each other (once phase-to-phase is established).
Click to expand...
To be blunt, none of that is an accurate or useful way to describe things. In particular, nothing is connected in parallel in a wye.

Maybe try first understanding the vector math of why voltages add up the way they do, then go back to Ohm's Law to understand the current flow.
 
I think I see where the OP question is coming from.

In a wye source, the L-L voltage is 1.732 * coil voltage, but terminal current is equal to coil current.

In a delta source, the L-L voltage is equal to coil voltage, but terminal current is 1.732 * coil current.

As @jaggedben notes, the answer to these questions comes from understanding how vector math is used for AC systems.

The basis of using vector math is that we use simplified numbers to describe what is happening in an AC system.

Alternating current is supplied by a source with a continuously varying output voltage. The voltage starts at zero, climbs to a positive maximum, goes back to zero, then to a negative maximum, and back to zero. A true description of what is present at the wall outlet would be a continuous signal, not a single number.

But usually we don't need this complete description. We can have a single value which represents a sort of average of the voltage signal, and a single number which represents its timing. These are voltage and phase angle. A quantity described by both magnitude and direction is a 'vector quantity', and there are rules for how vectors add. These rules are simple once you connect them to the graphic of 'adding line segments up'.

https://www.allaboutcircuits.com/te...rrent/chpt-2/introduction-to-complex-numbers/

is a pretty good introduction to AC waveforms represented with vectors and how they add up.

-Jon
 
winnie said:
I think I see where the OP question is coming from.

In a wye source, the L-L voltage is 1.732 * coil voltage, but terminal current is equal to coil current.

In a delta source, the L-L voltage is equal to coil voltage, but terminal current is 1.732 * coil current.

As @jaggedben notes, the answer to these questions comes from understanding how vector math is used for AC systems.

The basis of using vector math is that we use simplified numbers to describe what is happening in an AC system.

Alternating current is supplied by a source with a continuously varying output voltage. The voltage starts at zero, climbs to a positive maximum, goes back to zero, then to a negative maximum, and back to zero. A true description of what is present at the wall outlet would be a continuous signal, not a single number.

But usually we don't need this complete description. We can have a single value which represents a sort of average of the voltage signal, and a single number which represents its timing. These are voltage and phase angle. A quantity described by both magnitude and direction is a 'vector quantity', and there are rules for how vectors add. These rules are simple once you connect them to the graphic of 'adding line segments up'.

https://www.allaboutcircuits.com/te...rrent/chpt-2/introduction-to-complex-numbers/

is a pretty good introduction to AC waveforms represented with vectors and how they add up.

-Jon
Click to expand...
Thank you, Jon. You were by far the most helpful.
 
I think @winnie's answer is spot on, but thought I would try to simplify it some more because I think this is a great way to quickly remember how voltage and current changes from a three-phase wye system to a delta wound system, without being perplexed by the vector mathematics of it all.

When the line (aka terminal) current travels from the wye winding and enters a delta wound machine, the line current gets divided between two of the windings in the delta system. Since we are dealing with three-phase systems and the way that vectors mathematically add, it so happens that the current in a delta winding is equal to the line current divided by 1.732 (which is the square root of 3).

Now, looking at the voltages between delta and wye systems... The line-to-line voltage in a delta system is equal to the delta winding voltage, since each winding is connected line-to-line. In the wye system, two of the three windings are connected between line-to-line. The line-to-line voltage is divided between two of the three windings in the wye system. Again, since it is a three-phase system and the way the vectors add, the wye winding voltage is equal to the line-to-line voltage divided by 1.732.

This all given... there are many assumptions made and this all holds true only when everything is balanced and ideal (i.e. not the real world).

Hope this helps and expands or simplifies @winnies's answer.
 
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