# Base Voltage Throught Transformer



## kduff70 (Jul 27, 2017)

Can any help me lock down the concept of Vbase through a transformer ? I have attach a problem from Kaplan PE Power exam. I'm stuck on how they  solve for the base voltage on the motor side of the problem.

The have the  solution saying Vbase = 22 (220/25) (13/sqrt3*130) give you 11.1775 as the Vbase , Can some show how to go about this .

View attachment 9896


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## rg1 (Jul 27, 2017)

Can u share the complete question including figure. In absence of that; I assume that you have two parts in the question one is generator side and the other is Motor side or say transmission line side and you want to transfer pu values of Zs on motor side of the Xmer. You can assume anything as base for whole of the circuit but when you go from one side of the transformer to the other side you have to take care. Now say there are 3 parts of the circuit 1. Gen and TL up to Xmer 2. Xmer and 3. TL and Motor. and you want all pu Zs transferred to Motor side of the Xmer. For this you can assume a base V and base MVA, convert all Z to this base. Transfer the Gen side pu Z on other side of Xmer as we do normal impedance transfer across a Xmer.


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## TNPE (Jul 27, 2017)

You don't reflect across a XFMR when completing this analysis.  It is taken care of in the system power base and appropriate voltage base to that section of the one line.  Also, %Z of a XFMR is the same looking from either side, so no worry there.

As rg1 said, please post the rest of the problem so that we can help further.

This is a very important concept when completing fault current analysis.  I prefer the pu approach over all others, but that's just me.


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## rg1 (Jul 27, 2017)

TNPE said:


> You don't reflect across a XFMR when completing this analysis.  It is taken care of in the system power base and appropriate voltage base to that section of the one line.  Also, %Z of a XFMR is the same looking from either side, so no worry there.  Very true and very important. I missed it in my explanation.
> 
> As rg1 said, please post the rest of the problem so that we can help further.
> 
> This is a very important concept when completing fault current analysis.  I prefer the pu approach over all others, but that's just me.


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## kduff70 (Jul 27, 2017)

Here the full Problem  Thank you

View attachment impedanceprob.docx


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## rg1 (Jul 27, 2017)

kduff70 said:


> Here the full Problem  Thank you
> 
> View attachment 9897


Prima facie, I have yet to sit with pen and paper; The question is asking to change the base pu value of the motor from its rated base to new base. If that is what is required to be done then we have the formula Zpunew*KVnew**2/MVAnew=Zpuold*KVold**2/MVAold. There is nothing like Base on generator section. It should be - This is the new base at which Zpu of the motor is to be found?? 

There is a probabilty - If Question wants to see Motor impedance at Generator section then we can calculate the real value of Motor Z and take it to Gen end through Xmers. and then devide by Zbase there.


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## TNPE (Jul 28, 2017)

@rg1

No, that is not what's being asked.  Per unit analysis does not involve reflecting any impedances through a XFMR, regardless of side.  You practically develop an equivalent circuit from a system power base and voltage base for each respective zone.  You may be overthinking this.  Moving left to right, find the voltage base on the secondary of T2, use change of base Z formula and wahlaa!


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## kduff70 (Jul 28, 2017)

I'm sorry but can you explain to me  a little more on how the voltage base  for the motor side is calculated. I  don't understand how the voltage base through  transformer 1  is 22(220/25) I thought is should be 22(25/220) if you can explain this part maybe the light bulb in my head will go on , but this part is sticking me .Base on what I think the voltage base should be when it pass through a transformer. Ex. V2= (V2/V1)*V1


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## Omer (Jul 28, 2017)

kduff70 said:


> I'm sorry but can you explain to me  a little more on how the voltage base  for the motor side is calculated. I  don't understand how the voltage base through  transformer 1  is 22(220/25) I thought is should be 22(25/220) if you can explain this part maybe the light bulb in my head will go on , but this part is sticking me .Base on what I think the voltage base should be when it pass through a transformer. Ex. V2= (V2/V1)*V1


I, usually think of it intuitively.

suppose the base voltage of T1 on the primary side is 25 Kv same as its primary rating, then the base on the secondary is 220 Kv same as its secondary rating.

thus, to transfer 25 Kv to secondary you multiply by 220/25 (ratio of transformer).

in your case the base is 22 Kv, you do the same multiply by 220/25 (ratio of transformer) to get base on secondary.

second transformer the same approach.

hope it is clear.


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## rg1 (Jul 28, 2017)

I post two solutions as I was mentioning there are two ways of understanding the Question.1. Change the base of pu. 2. Z of motor seen at Gen end.

 I will put the Question like this- What is pu Z of Motor when seen at Gen end at Gen ratings as Base.  I have never heard of transferring the bases. We know how a impedance on one side of Xmer is seen on the other side. Zprimary seen on secondary=Zprimary*Vs**2/Vp**2. Hope this is okay now. 

View attachment 9901


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## kduff70 (Jul 28, 2017)

Thank you  Omer,

I get what your saying if to take base voltage coming from the primary side of T1 you would use the ratio (220/25) but when you get to T2 the ratio is reversed it is not clear to me.  I'm thinking the  voltage base of T1 would be multiplied  by the T2 Ratio (130Sqrt3/13) but the problem call for the reverse and that is confusing.


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## Omer (Jul 28, 2017)

kduff70 said:


> Thank you  Omer,
> 
> I get what your saying if to take base voltage coming from the primary side of T1 you would use the ratio (220/25) but when you get to T2 the ratio is reversed it is not clear to me.  I'm thinking the  voltage base of T1 would be multiplied  by the T2 Ratio (130Sqrt3/13) but the problem call for the reverse and that is confusing.


As I said, just look at it intuitively and don't concentrate too much on the (ratio or the reverse).

moving from LV side to the HV side then your multiplier (ratio) is bigger than 1, in this case 220/25 for T1.

moving from HV side to the LV side then your multiplier (ratio) is less than 1, in this case 13/130sqrt(3) for T2.

hope it make sense for you.


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## rg1 (Jul 28, 2017)

kduff70 said:


> Thank you  Omer,
> 
> I get what your saying if to take base voltage coming from the primary side of T1 you would use the ratio (220/25) but when you get to T2 the ratio is reversed it is not clear to me.  I'm thinking the  voltage base of T1 would be multiplied  by the T2 Ratio (130Sqrt3/13) but the problem call for the reverse and that is confusing.






Omer said:


> As I said, just look at it intuitively and don't concentrate too much on the (ratio or the reverse).
> 
> moving from LV side to the HV side then your multiplier (ratio) is bigger than 1, in this case 220/25 for T1.
> 
> ...


I think there is nothing like transferring the bases. IMHO it will create more confusion in understanding the pu system and may lead to state of confusion all around. If you are familiar with transferring impedance from one side of Xmer to other side- (if you are not you should do it, may find a question on that ), then question is simple. The question asks only how the motor impedance is seen on Gen side of T1, thats all. It is a question for transferring impedance across a xmer, do not fall into the trap.


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## Omer (Jul 28, 2017)

rg1 said:


> I post two solutions as I was mentioning there are two ways of understanding the Question.1. Change the base of pu. 2. Z of motor seen at Gen end.
> 
> I will put the Question like this- What is pu Z of Motor when seen at Gen end at Gen ratings as Base.  I have never heard of transferring the bases. We know how a impedance on one side of Xmer is seen on the other side. Zprimary seen on secondary=Zprimary*Vs**2/Vp**2. Hope this is okay now.
> 
> View attachment 9901


interesting,

I think you should get both answers same.

I will concentrate on your first approach since, I think, it is the one required by the question.

Base MVA is defined as 200, no problem with this one and it is fixed across the whole system.

base voltage is defined as 22 Kv on the generator section, for the base voltage it will change across different sections when you pass across a transformer with the same transformer ratio. (I think this one you missed on your first approach). you should get the .09 Pu.

your second approach is interesting and I like it.


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## rg1 (Jul 28, 2017)

Omer said:


> interesting,
> 
> I think you should get both answers same. No you will not get both answers same, because one is only change of base and the other is change in actual Z(Za) as well as base.
> 
> ...


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## Omer (Jul 28, 2017)

just to mention,

in the per unit system, voltage across the transformer is 1 Pu, both in primary and secondary, however actual voltages are different , that mean bases are different, right?

this problem is all about getting the new base voltage in the motor side while given base voltage on the generator side.

the new V base will be 22 (220/25) (13/sqrt3*130) give you 11.1775 as the Vbase on the motor side.

plug the value on the equation and you get .09 Pu.


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## kduff70 (Jul 28, 2017)

I Thank all you  for your help on this I understand finding the impedance in pu ,

but it just stuck on the beginning step of establishing the V base as 11.1775V . 

This part has me stuck, For some reason. Is there a way you can show me a step by step approach on just how you were able to get Vbase.


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## rg1 (Jul 28, 2017)

Omer said:


> just to mention,
> 
> in the per unit system, voltage across the transformer is 1 Pu, both in primary and secondary, however actual voltages are different , that mean bases are different, right?
> 
> ...


I know what is happening when your transferring Base Voltage across Xmer. I am able to feel it and understand it. Having said that I have never come across this concept of transferring base values like this in any book. I am really interested in perusing it from a book. Can you share some material on that.


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## kduff70 (Jul 28, 2017)

This was a  practice problem for Kaplan  Sample exam  PE Power book  . I thought I would try some of the problem to get practice  but I never came across a V base problem Like this one  and it just has me stuck . All other Vbase problem I seem to get from other books .  I haven't seen this type of problem any where else to get material on this type of problem .


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## Omer (Jul 28, 2017)

Look at this topic, it does have some good graphs also

http://peguru.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understanding/


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## TNPE (Jul 28, 2017)

@rg1

That is the only approach I've ever used when working with pu.  Not that you're approach is "wrong," but it is convoluted and more difficult than the pu approach.  You can do everything necessary with a given power base and a given voltage base.  From this, move from the generator down line to the motor, changing bases as required across XFMRS (i.e. this could be a voltage and/or power base, depending on the problem statement and the given initial bases and ratings).   

I have never seen your approach in academia (or texts) when working with pu analysis.  Also, it'd be wise to be careful with your approach as you may have to do multiple change of bases.


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## kduff70 (Jul 29, 2017)

Thank you all but I have finally figure it out . I agree with TNPE the base method is the best approach. I attach my hand written calculation to show how I came up with  the answer . When finding the Vbase of the motor  I took the Vbase of T1 over The ratio of T2 and I get the Vbase of the motor and from there I get the Zpu of the motor .Thank you all for the help

View attachment NewDoc2017-07-29.pdf


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## rg1 (Jul 29, 2017)

kduff70 said:


> Thank you all but I have finally figure it out . I agree with TNPE the base method is the best approach. I attach my hand written calculation to show how I came up with  the answer . When finding the Vbase of the motor  I took the Vbase of T1 over The ratio of T2 and I get the Vbase of the motor and from there I get the Zpu of the motor .Thank you all for the help
> 
> View attachment 9903


Yes either you take horse to water or water to horse, the result is same. When you travel through transformers, it will result into transformation of the quantity(V,I,Z not VA) and you can not avoid it in this question. It was a learning for me as I do not remember I did ever take bases across transformers. This too, seems interesting. For this a big thank you to all of you. 

Having said that just for the benefit of the forum transferring impedance across a transformer (the one I used in the solution) is also an important concept, need to be learned. I saw a few questions on that concept in few practice exams and may be in NCEES paper.

Thanks a lot again.


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## cos90 (Jul 29, 2017)

Combining everything together like that confuses me. What helps is remembering the transformer ratio is the relationship between phase voltages.

To answer your original question this way:

T1 ratio is 220/ (1.73*25)=5.086

T2 ratio is 130/13=10

22*5.086 = 111kV ph-n

111/10 = 11.1kV 

That said I missed it first workthrough. This question makes me want to get the Kaplan test. I need to practice with questions harder than the test I am taking to succeed.


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## rg1 (Jul 29, 2017)

cos90 said:


> Combining everything together like that confuses me. What helps is remembering the transformer ratio is the relationship between phase voltages. There is nothing wrong in that provided you can keep track of the ghost sqrt3 and the ratios given but generally in three phase we take Line values. I just want to understand how you do and see if there is any gap or chance of error by doing that way.
> 
> To answer your original question this way:
> 
> ...


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## TNPE (Jul 29, 2017)

rg1 said:


> Yes either you take horse to water or water to horse, the result is same. When you travel through transformers, it will result into transformation of the quantity(V,I,Z not VA) and you can not avoid it in this question. It was a learning for me as I do not remember I did ever take bases across transformers. This too, seems interesting. For this a big thank you to all of you.
> 
> Having said that just for the benefit of the forum transferring impedance across a transformer (the one I used in the solution) is also an important concept, need to be learned. I saw a few questions on that concept in few practice exams and may be in NCEES paper.
> 
> Thanks a lot again.


You're not necessarily taking bases "across" a XFMR, per se. Think of it more along the lines of a zone.  A generator zone up to a XFMR, say.  From the secondary "base" of the XFMR, traveling down the transmission line to the next XFMR... Secondary base of this XFMR down line to the load.  You're merely relating each zone to a starting system-wide base, and adjusting/relating each successive zone as necessary.  But the genius in pu analysis is, everything in the system is related by a scalar and it's easy to spot mistakes.  If you're working with fractional values (ideally, 1pu or less), but you end up with 12.862pu, pretty good indication you've made an error.


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