# Transformer Ratios



## krummelt1 (Oct 11, 2015)

I feel like I might be missing something fundamental.

Take a look at this problem:

http://www.spinupexams.com/pgm-download_media.php?name=Week_2_QFTW_Question_Solution_Rev2.pdf

The ratio provided is 10:1 yet it is delta to wye. Is the ratio that is provided not the turns ratio? If it where the turns ratio wouldn't the Line to Neutral voltage be 460 and the Line to Line be 460*sqrt(3)

A similar example would be example 118 of the NCEES practice exam. It is a one line diagram with a Delta-Wye Transformer. Transformer is listed as 132-13.2. The answer for this one involves transforming the current across the transformer and is just a straight 10:1.

What am I missing?

Thanks

WEEK_2_QFTW_QUESTION_SOLUTION_REV2 (2).PDF


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## Ship Wreck PE (Oct 12, 2015)

Those are some of the free help me pass problems.


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## krummelt1 (Oct 14, 2015)

So am I missing something or am I right on and the problem is flawed?

I guess I understand the conventions as this: When it is a one line diagram and the voltages are provided on both sides, these are line to line voltages. If you wanted to find a turns ratio you would need to adjust by Sqrt3 accordingly. However in the spin up example above where you have the a primary voltage and the turns ratio the voltage from phase to phase is what you need to consider.

. . . .


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## NJmike PE (Oct 14, 2015)




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## sunguy (Oct 22, 2015)

NCEES 118 problem: The line current in delta transformer is equal to phase current*sqrt(3). So when you apply the transformer turns ratio to calculate current on the 13.2 delta side its the phase current you are calculating (not the line current). So to obtain line current provided by generator you have to multiply the phase current by sqrt(3)

The transformer ratio is 13.2/(132/sqrt(3))= 0.1732. The phase current in 13.2 delta is 75.93/0.1732 = 438.4 A. The line current in the 13.2 delta is 438.2*sqrt(3)=759.3A.


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## sunguy (Oct 22, 2015)

something is not right with that spinup problem. Even the description says it a step-down tranformer. D-Y transformer is typically used as a step up transformer.


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## cupojoe PE PMP (Oct 22, 2015)

sunguy said:


> something is not right with that spinup problem. Even the description says it a step-down tranformer. D-Y transformer is typically used as a step up transformer.




Many dry type transformers for lighting panels in the 15-45kVA range are delta-wye.


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## sunguy (Oct 22, 2015)

480 to 208/120?


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## Ryan P.E. (Oct 22, 2015)

This all hinges on what 10:1 means. I am with the OP that it means turns ratio for a pair of windings on the same core, then calculate what the voltage would be depending on configuration, which would mean the LL voltage is 460*sqrt(3). But given NEMA typical voltage ratings and whats in the NEC, they must mean the voltage ratio of the 3phase transformer is simply 4600V LL on delta side-460V LL on Wye side.

Spinup should give another free practice problem


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## cupojoe PE PMP (Oct 22, 2015)

Ryan P.E. said:


> This all hinges on what 10:1 means. I am with the OP that it means turns ratio for a pair of windings on the same core, then calculate what the voltage would be depending on configuration, which would mean the LL voltage is 460*sqrt(3). But given NEMA typical voltage ratings and whats in the NEC, they must mean the voltage ratio of the 3phase transformer is simply 4600V LL on delta side-460V LL on Wye side.
> 
> Spinup should give another free practice problem




What you described is inconsistent with the NCEES questions. If they give 10:1, that should mean the actual winding ratio, regardless of connection scheme. You would then need to use the connection scheme to determine the voltages and currents seen through the transformer. If they give the voltages on each side, that should be taken as is. The spinup book seems to have an error.


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## cupojoe PE PMP (Oct 22, 2015)

sunguy said:


> 480 to 208/120?




Yes, that would be one example. Delta-Wye is a common connection scheme for step-down and step-up transformers.

Here is an eaton catalog of transformers. The 480-208Y/120 transformers are on page 17, but there is page after page of D-Y transformers. http://www.eaton.com/ecm/idcplg?IdcService=GET_FILE&amp;allowInterrupt=1&amp;RevisionSelectionMethod=LatestReleased&amp;noSaveAs=0&amp;Rendition=Primary&amp;dDocName=VOL02_TAB02


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## ZcoreX29 (Oct 22, 2015)

sunguy said:


> something is not right with that spinup problem. Even the description says it a step-down tranformer. D-Y transformer is typically used as a step up transformer.


Virtually all of our distribution transformers are delta-wye.


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## Kovz (Oct 23, 2015)

I've seen a lot of distribution transformers that are delta-wye. In fact, a three phase one could be 480V-208/120V. Meaning the primary is 480V L-L and the secondary is 208V L-L or 120V L-N.

I don't think Spin-up has an error here. It's 4600V L-L on the primary with a 10:1 step-down ratio giving you 460V L-L on the secondary. The L-N would be 460/1.73=265V.

Besides, when have you ever heard of a 795V motor... that should throw a red flag that something isn't right.


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## Ryan P.E. (Oct 24, 2015)

Kovz said:


> I've seen a lot of distribution transformers that are delta-wye. In fact, a three phase one could be 480V-208/120V. Meaning the primary is 480V L-L and the secondary is 208V L-L or 120V L-N.
> 
> I don't think Spin-up has an error here. It's 4600V L-L on the primary with a 10:1 step-down ratio giving you 460V L-L on the secondary. The L-N would be 460/1.73=265V.
> 
> Besides, when have you ever heard of a 795V motor... that should throw a red flag that something isn't right.




The 10:1 is supposed to be turns ratio, not the voltage ratio. With that being said, the LN voltage would 460V on the secondary or 795V LL, but I'm with you that this isn't a typical voltage for motor. If this were a question from school, I would've said 795V and put a note for the TA/teacher that the question is confusing. It's all about understanding the concept for the NCEES exam anyways.


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## cupojoe PE PMP (Oct 24, 2015)

Ryan P.E. said:


> Kovz said:
> 
> 
> > I've seen a lot of distribution transformers that are delta-wye. In fact, a three phase one could be 480V-208/120V. Meaning the primary is 480V L-L and the secondary is 208V L-L or 120V L-N.
> ...


+1


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## Kovz (Oct 26, 2015)

Ryan P.E. said:


> Kovz said:
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> 
> > I've seen a lot of distribution transformers that are delta-wye. In fact, a three phase one could be 480V-208/120V. Meaning the primary is 480V L-L and the secondary is 208V L-L or 120V L-N.
> ...




Agreed... 10:1 is the turns ratio... so given the transformer equation Vp/Vs = N1/N2 ====&gt; 4600/Vs = 10/1 Solve for Vs = 460V. That's the L-L voltage on the secondary.


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## ZcoreX29 (Oct 26, 2015)

Kovz said:


> Ryan P.E. said:
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Not on a delta-wye transformer. It would be 795V LL


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## Ryan P.E. (Oct 26, 2015)

Kovz said:


> Ryan P.E. said:
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As zcore has said, not for a delta-wye transformer. The voltage across one phase of the high voltage side is 4600, and it is connected in delta already. The voltage across the corresponding low voltage winding is 460V, but 460V is the voltage from LN. The LL voltage would be 795V.


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## cupojoe PE PMP (Oct 29, 2015)

Ryan P.E. said:


> Kovz said:
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This would work with a Wye-Wye(w/o buried delta) and a Delta-Delta. Or a single phase transformer.


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## eksor_PE (Nov 2, 2015)

Kovz said:


> Ryan P.E. said:
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I don't see the error on that Spinup question either. Kovz solution is correct.

The turns/winding ratio of the transformer is equal to the voltage ratio. The connection schemes of the transformer changes/remain the phase of the voltage.


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## ZcoreX29 (Nov 2, 2015)

eksor_PE said:


> Kovz said:
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On a delta-wye or wye-delta transformer, a 10:1 turns ratio does not equal Vp/Vs where Vp and Vs are line-line voltages on the primary and secondary respectively.


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## eksor_PE (Nov 3, 2015)

ZcoreX29 said:


> eksor_PE said:
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Yes, they are line-to-line voltages, V_p and V_s, and that is why the line-to-neutral voltage will be V_ln = 266V.


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## ZcoreX29 (Nov 3, 2015)

eksor_PE said:


> ZcoreX29 said:
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The line to neutral voltage would be 460 V. Below is the closest I can find to a proof at the moment. On a delta-wye transformer, there's always a 1.73 difference in line to line voltages in addition to the turns ratio
http://www.drhenrylouie.com/uploads/17-Three_Phase_Transformers_part2.pdf


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## eksor_PE (Nov 4, 2015)

ZcoreX29 said:


> eksor_PE said:
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> 
> > Yes, they are line-to-line voltages, V_p and V_s, and that is why the line-to-neutral voltage will be V_ln = 266V.
> ...


From this thread, there is a misconception between the turn ratio and the impedance of a transformer. The turn ratio of a transformer does not change regardless of its connection configuration such as delta-delta, delta-wye, or others. The turn ratio of a transformer will only change if the number of turns, physically, increase or decrease, V = 4.44*f*N*Phi.

The connection configuration of a transformer changes the value of the transformer’s impedance; hence, the current across each phase will change either by a multiple or a ratio of 1.732.

With all that said and other users implied, if the turn ratio of a transformer is given, the equation V_p/V_s = N_1/N_2 or I_1/I_2 = N_2/N_1 is used regardless of the transformer’s connection configuration.

For practical application, you can build a three-phase transformer from a three single-phase transformers bank with same characteristics. I attached an actual nameplate of a transformer, which is in delta-wye configuration. At full load, you can evaluate the turn ratio, currents, and voltages of a delta-wye configuration.

Like what others have said, this kind of question is a free pass question.


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