Base Voltage Throught Transformer

[kduff70]

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

 

[rg1]

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. 

 

[TNPE]

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.

 

[rg1]

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.

 

[kduff70]

Here the full Problem  Thank you

View attachment impedanceprob.docx

 

[rg1]

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. 

 

[TNPE]

@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!

 

[kduff70]

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

 

[Omer]

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.

 

[rg1]

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

 

[kduff70]

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]

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.

 

[rg1]

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.

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.

 

[Omer]

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.

 

[rg1]

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.

I will concentrate on your first approach since, I think, it is the one required by the question. The question wanted to ask the option two but it got confused between one and two. No clarity in asking.

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. I have never come across transferring base values across transformers. Is is legislated in some good book, I would like to see it. 

your second approach is interesting and I like it. IMHO this question is of transferring impedance across Xmer and pu system (two concepts mixed). I could get a sense the way it was asked, that is the reason in my first post itself I mentioned there is a probability of this.

 

[Omer]

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.

 

[kduff70]

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. 

 

[rg1]

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.

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. 

 

[kduff70]

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 .

 

[Omer]

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/