Star Delta Transformer Confusion help please

  • #1

Main Question or Discussion Point

Hi,

From the basics of 3 phase power, I believed that if you had a delta source which had Line Voltages and Phase Voltages of 100V, if you converted this into a star source, the Equivalent Star Line voltage would still be 100V and the equivalent star phase voltage would then be 100V/sqrt(3). Could someone confirm that this is correct?

The reason I ask is because I am now learning about transformers and have been looking at a star delta transformer, and I cannot quite get my head around the primary and secondary voltages.

The book I have been looking at gives an example where there is a transformer with a star primary winding and delta secondary winding, with a turns ratio of 10/1. It says the delta line voltage is 240V. It then explains that the star phase voltage is simply 10*240V = 2400V and the star line voltage is sqrt(3) * 2400V = 4157V. Surely, the star line voltage should be 2400V and the star phase voltage be 2400V/sqrt(3), as I always believed that when converting voltages from star to delta and vice versa, the line voltage magnitudes stayed the same, as explained above. If someone could shed some light on this I would be very grateful. Thanks.
 

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  • #2
dlgoff
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From the basics of 3 phase power, I believed that if you had a delta source which had Line Voltages and Phase Voltages of 100V, ...
For a Delta source like this, there's no neutral, hence no Line Voltage:

http://www.federalpacific.com/university/T-Basics/Charts/section-3-scan-6.gif [Broken]


For a secondary Wye like this, there is a neutral, hence a Line Voltage:

http://www.federalpacific.com/university/T-Basics/Charts/section-3-scan-10.gif [Broken]

Image source: http://www.federalpacific.com/university/university.html
 
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  • #3
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I believed that if you had a delta source which had Line Voltages and Phase Voltages of 100V, if you converted this into a star source, the Equivalent Star Line voltage would still be 100V and the equivalent star phase voltage would then be 100V/sqrt(3). Could someone confirm that this is correct?

...

The book I have been looking at gives an example where there is a transformer with a star primary winding and delta secondary winding, with a turns ratio of 10/1. It says the delta line voltage is 240V. It then explains that the star phase voltage is simply 10*240V = 2400V and the star line voltage is sqrt(3) * 2400V = 4157V.
The book you have been looking at is correct. You must remember that in a delta configuration, both the line voltages and the phase voltages are measured across only one transformer winding, whereas line voltages (line-to-line) in star systems are measured across 2 transformer windings - phase voltages are measured across individual windings. Also, realize that each individual star winding induces voltage onto only one individual delta winding. Hence, in the case of a 10:1 step-up star-delta transformer (as described in your book), each individual star winding will induce a voltage that is 10 times greater onto each individual delta winding. So, if you have 240 V across each delta winding, there MUST be 10 times that across each star winding, or 2400V. Since the phase voltages (individual windings) are 120 degrees out of phase, the line voltage (which is measured across two windings) is then 1.73 times larger than the phase voltage, or approximately 4150 V.


For a Delta source like this, there's no neutral, hence no Line Voltage:

http://www.federalpacific.com/university/T-Basics/Charts/section-3-scan-6.gif [Broken]
Not true ... line voltage is measured from line-to-line (leg-to-leg). In the case of a delta configuration, that would be from one "corner" to another.

For a secondary Wye like this, there is a neutral, hence a Line Voltage:

http://www.federalpacific.com/university/T-Basics/Charts/section-3-scan-10.gif [Broken]

Image source: http://www.federalpacific.com/university/university.html
Again, not (completely) true. Line voltage in a wye (or star) system is also measured from leg-to-leg. The neutral is typically taken from the center connection point and has no bearing on the line voltage. Technically, it has no bearing on the phase voltage either, in that the phase voltage exists across each winding whether there is a neutral, or not. The existence of a neutral gives the user access to that connection point.
 
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  • #4
dlgoff
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The book you have been looking at is correct. You must remember that in a delta configuration, both the line voltages and the phase voltages are measured across only one transformer winding, whereas line voltages (line-to-line) in star systems are measured across 2 transformer windings - phase voltages are measured across individual windings. Also, realize that each individual star winding induces voltage onto only one individual delta winding. Hence, in the case of a 10:1 step-up star-delta transformer (as described in your book), each individual star winding will induce a voltage that is 10 times greater onto each individual delta winding. So, if you have 240 V across each delta winding, there MUST be 10 times that across each star winding, or 2400V. Since the phase voltages (individual windings) are 120 degrees out of phase, the line voltage (which is measured across two windings) is then 1.73 times larger than the phase voltage, or approximately 4150 V.



Not true ... line voltage is measured from line-to-line (leg-to-leg). In the case of a delta configuration, that would be from one "corner" to another.


Again, not (completely) true. Line voltage in a wye (or star) system is also measured from leg-to-leg. The neutral is typically taken from the center connection point and has no bearing on the line voltage. Technically, it has no bearing on the phase voltage either, in that the phase voltage exists across each winding whether there is a neutral, or not. The existence of a neutral gives the user access to that connection point.
The images that I was referring too have changed link locations. Here they are from their new locations.

http://federalpacific.com/training/transformer-basics/images/section-3-scan-6.gif [Broken]

http://federalpacific.com/training/transformer-basics/images/section-3-scan-10.gif [Broken]

The point I was making; if there is no neutral wired, there's no way to measure a line-to-neutral voltage.

Images from http://federalpacific.com/training/transformer-basics/chapter-3.htm
 
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  • #5
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The point I was making; if there is no neutral wired, there's no way to measure a line-to-neutral voltage.
I understand what you are trying to say, but typically, you have access to the terminals of a transformer whether the neutral was connected or not. So you could still physically measure the phase voltage (relatively easily).

My point was that regardless of your ability to access the star connection point (it's only a neutral if it is grounded), the phase voltage still exists.
 
  • #6
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To clarify...
In the lower diagram (the wye connection), there is 120V across each individual winding whether the neutral wire is connected or not. It just becomes more difficult to measure.
 
  • #7
jim hardy
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It's a matter of being rigorous in your terminology.

"Line" voltage is usually understood to mean "line to line voltage"


Sadly, "Phase" is used interchangeably to mean one line of a transmission scheme or one winding of a transformer. So - In Don's excellent figures above, "Phase A" might refer to either transformer winding P1 or wire H1.


Upper figure: It should be clear that Line(to line) voltage in the Delta connection is same as the transformer's Phase voltage .
Lower figure: It should be clear that Line(to line) voltage in the Wye connection is sqrt(3) X transformer's phase voltage.

Myself i prefer to use terminology that removes all doubt, by subscripting:

Vab is line A to line B voltage
Vba is line B to line A voltage
Van is line A to neutral
Vna is neutral to line A
That subscripting really helps out when drawing phasor diagrams.

When it's not stated, as on a one line diagram, i assume Vline means line to line, Vphase means across one transformer winding. Usually one can figure out if that's what the writer had in mind from his accompanying text.
 
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  • #8
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jim hardy:
I fully agree with your statements. However, dlgoff orginally stated
"For a Delta source like this, there's no neutral, hence no Line Voltage"
which simply isn't true, regardless of what terminology you use OR whether or not a neutral exists.
 
  • #9
dlgoff
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jim hardy:
I fully agree with your statements. However, dlgoff orginally stated
"For a Delta source like this, there's no neutral, hence no Line Voltage"
which simply isn't true, regardless of what terminology you use OR whether or not a neutral exists.
touché. You are correct of course.
 
  • #10
jim hardy
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SSWheels:
I too agreed with everything in your post #3.

We old guys often have to dredge back through forty or fifty years of life's detritus to remember our basics

and our ram isn't completely nonvolatile

so an occasional mismatch of word to concept and the accompanying embarrassment is part of everyday life for us. We take no offense when others correct us, au contraire it's appreciated.


The purpose of my post #7 was to demonstrate the utility of carefully defining our terms before using them because not all authors use the same conventions, or at least didn't in the '60s when i went through school.
I thought the best way to do so was to reiterate your post 3 but with just a little more rigor in terminology, and succinctness. I hope you're okay with that . I shoulda given credit.

When i got out of school people studying for the PE used to come to me with 3 phase problems that had them stumped. Usually their troubles stemmed from lack of rigor in labeling their phasors. It made me appreciate my teacher Dr Gross for having pounded that rigor into us boys.

old jim
 
  • #11
davenn
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We old guys often have to dredge back through forty or fifty years of life's detritus to remember our basics and our ram isn't completely nonvolatile

That's classic .... in my signature it goes :smile:

Dave
 
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