Auto Transformer NCEES #535 Solution Clarification

[wfg42438]

For the Question below im having trouble understanding the simplifications made.




So far i understand we are given 3P quantities and we are after the ratings of each 1P XFRM that make up the 3P auto XFMR where:

3Ph MVA= 180

Therefore, this means the 1P MVA= 60
Assuming this is an ideal XFRM this means MVA in =MVA out= 60MVA
for which the output is taken across the common winding so why wouldn't MVA_Common=60MVA?

I find it confusing that we assume 60MVA at the output yet we derive a new MVA on the common winding.

If anyone can help clarify it would be much appreciated.

 

[main197]

Autotransformers are different from a two-winding transformer. In a two-winding transformer, power is transferred inductively. But, in an autotransformer, power is transferred both inductively (at common winding) and conductively (series winding). There is no electrical isolation between primary and secondary.

Output VA of above autotransformer, Sout= V_2 * I_2

Or, Sout= S_condctive + S_inductive = V_2*I_1 + V_2*I_c

I recommend you to read and understand theory first before trying to solve the problems. There will be more qualitative problems than you see in that practice test ( about 30% or more), so you'll need to know the theory as well.

 

[wfg42438]

Autotransformers are different from a two-winding transformer. In a two-winding transformer, power is transferred inductively. But, in an autotransformer, power is transferred both inductively (at common winding) and conductively (series winding). There is no electrical isolation between primary and secondary.

Output VA of above autotransformer, Sout= V_2 * I_2

Or, Sout= S_condctive + S_inductive = V_2*I_1 + V_2*I_c

I recommend you to read and understand theory first before trying to solve the problems. There will be more qualitative problems than you see in that practice test ( about 30% or more), so you'll need to know the theory as well.

Based on further review of the auto transformer theory i believe the following is true in this question:
1. Each 1P Auto Transformer is rated for 60MVA this means the power at the input and output is 60MVA
2. The question specifically calls for the power across the common winding therefore this is simply I_Common *Vcommon
3. Now we find Scommon=16.5 MVA given the relationship that Scommon=Sseries this means S_Series=16.5MVA as well

So based on the items above how would we find Srated which is 60MVA, is there a relationship im missing here?
Im assuming there must be a third relationship that lets us derive the rated power given the common and series coil power

 

[main197]

1. Correct
2. Correct
3. Sc= V2*Ic, Sseries= V2*I1 or you if you have Sc, Sseries= Srated-Sc= 60MVA-16.5MVA= 43.5 MVA

Srated= V1*I1 or V2*I2

 

[main197]

For step up auto transformer, Scond=V1*I2 and Sind= V1*Ic

 

[wfg42438]

1. Correct
2. Correct
3. Sc= V2*Ic, Sseries= V2*I1 or you if you have Sc, Sseries= Srated-Sc= 60MVA-16.5MVA= 43.5 MVA

Srated= V1*I1 or V2*I2

Based on the references i saw S_Series=S_Common why is that we are saying this is not true?

Based on Zachs video below thats what i gathered am i missing something here?

Step DOWN Autotransformers Explained for the PE CBT Exam (including circuit analysis formulas) - YouTube

 

[main197]

Based on the references i saw S_Series=S_Common why is that we are saying this is not true?

Based on Zachs video below thats what i gathered am i missing something here?

Step DOWN Autotransformers Explained for the PE CBT Exam (including circuit analysis formulas) - YouTube

View attachment 27269

Interesting. If we use Zach's equation, Sc=Sse= 16.5 MVA. This gives Srated= Sc+Sse= 16.5+16.5= 33MVA. They should add up to 60MVA, no? Srated is 60 MVA.
I am not sure if I am missing something. Maybe @Zach Stone P.E. would like to chime in.

 

[wfg42438]

Interesting. If we use Zach's equation, Sc=Sse= 16.5 MVA. This gives Srated= Sc+Sse= 16.5+16.5= 33MVA. They should add up to 60MVA, no? Srated is 60 MVA.
I am not sure if I am missing something. Maybe @Zach Stone P.E. would like to chime in.

Agreed this is why i was confused it seems there something missing here

@Zach Stone P.E. your feedback would be much appreciated

 

[Zach Stone P.E.]

Interesting. If we use Zach's equation, Sc=Sse= 16.5 MVA. This gives Srated= Sc+Sse= 16.5+16.5= 33MVA. They should add up to 60MVA, no? Srated is 60 MVA.
I am not sure if I am missing something. Maybe @Zach Stone P.E. would like to chime in.

Agreed this is why i was confused it seems there something missing here

@Zach Stone P.E. your feedback would be much appreciated

Unfortunately, summing the series and common winding power does not add up to the power rating of the auto-transformer.

However, you can use the series or common winding power along with the turns ratio of the auto-transformer to calculate the power rating of the auto-transformer using the following formula that is available in section 4.3.1.7 Autotransformers of the Reference Handbook:



Where SIO is the power rating of the auto-transformer ("input-output power"), and SW is the winding power of either the series or common winding, since they are equal.

Or, you can also calculate the power rating of the auto-transformer by taking the product of either the primary voltage and primary current, or the product of the secondary voltage and secondary current.

 

[Zach Stone P.E.]

Based on the references i saw S_Series=S_Common why is that we are saying this is not true?

Based on Zachs video below thats what i gathered am i missing something here?

Step DOWN Autotransformers Explained for the PE CBT Exam (including circuit analysis formulas) - YouTube

View attachment 27269

Thanks for sharing this video.

There is an updated version of this video that uses the same variables used in the Reference Handbook for added clarity:

​

 

[wfg42438]

Unfortunately, summing the series and common winding power does not add up to the power rating of the auto-transformer.

However, you can use the series or common winding power along with the turns ratio of the auto-transformer to calculate the power rating of the auto-transformer using the following formula that is available in section 4.3.1.7 Autotransformers of the Reference Handbook:

View attachment 27333

Where SIO is the power rating of the auto-transformer ("input-output power"), and SW is the winding power of either the series or common winding, since they are equal.

Or, you can also calculate the power rating of the auto-transformer by taking the product of either the primary voltage and primary current, or the product of the secondary voltage and secondary current.

Thanks for the input Zach!
So in such cases it seems its not straightforward to try to derive the rated power of the auto XFMR. It seems knowing the formula may be a good idea in this case.

I'm curious if there any other alternative since in all NCEES and other practice auto XFMR materials i have seen the turn ratios in your equations are not typically provided.

 

[Zach Stone P.E.]

Thanks for the input Zach!
So in such cases it seems its not straightforward to try to derive the rated power of the auto XFMR. It seems knowing the formula may be a good idea in this case.

I'm curious if there any other alternative since in all NCEES and other practice auto XFMR materials i have seen the turn ratios in your equations are not typically provided.

Determining the rated power of an auto-transformer It's just about as straightforward as determining the rated power of a regular transformer.

The same formula applies:

|Sin| = |Sout|
|V1|·|I1| = |V2|·|I2|

It's just the product of the primary voltage and current or the product of the secondary voltage and current.

The same video above goes over it at at the 12:30 mark:

​

 

[wfg42438]

Determining the rated power of an auto-transformer It's just about as straightforward as determining the rated power of a regular transformer.

The same formula applies:

|Sin| = |Sout|
|V1|·|I1| = |V2|·|I2|

It's just the product of the primary voltage and current or the product of the secondary voltage and current.

The same video above goes over it at at the 12:30 mark:

​

You know what i see it now.

In the solution we find Scommon= Ic*Vout
If we wanted Srated then Srated=Sout=I2*Vout

or we use the relationship of the turns ratio provided.

On a unrelated note just for clarity
Nse-Series Winding turn ratio
Nc=Common winding turn ratio

How would you find the turn ratios above?
Are these simply related to the 1P XFMRs Nprim and Nsecondary turn ratios?

 

[Zach Stone P.E.]

On a unrelated note just for clarity
Nse-Series Winding turn ratio
Nc=Common winding turn ratio

Yep!

How would you find the turn ratios above?
Are these simply related to the 1P XFMRs Nprim and Nsecondary turn ratios?

Turns ratio is a = N1/N2.

For a step up autotransformer, N1 = Nc and N2 = Nse + Nc.

For a step down autotransformer, N1 = Nse + Nc and N2 = Nc.

See the video above at 21:30 for an explanation of both.