Transformer Ratios

[krummelt1]

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

 

[Ship Wreck PE]

Those are some of the free help me pass problems.

 

[krummelt1]

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.

. . . .

 

[NJmike PE]

 

[sunguy]

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.

 

[sunguy]

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.

 

[cupojoe PE PMP]

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.

 

[sunguy]

480 to 208/120?

 

[Ryan P.E.]

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

 

[cupojoe PE PMP]

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.

 

[cupojoe PE PMP]

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&allowInterrupt=1&RevisionSelectionMethod=LatestReleased&noSaveAs=0&Rendition=Primary&dDocName=VOL02_TAB02

 

[ZcoreX29]

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.

 

[Kovz]

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.

 

[Ryan P.E.]

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.

 

[cupojoe PE PMP]

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.

+1

 

[Kovz]

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.

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

 

[ZcoreX29]

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.

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

Not on a delta-wye transformer. It would be 795V LL

 

[Ryan P.E.]

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.

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

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.

 

[cupojoe PE PMP]

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.

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

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.

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

 

[eksor_PE]

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.

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

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.