# Internal Voltage generated



## rg1 (May 14, 2017)

What is the internal generated phase voltage in a 3 phase Synchronous generator. Given base values 250 KV, 100 MVA and Xd= 1.1 pu,  steady state current =.97 pu


----------



## Don 3.0 (May 26, 2017)

For a sync. generator E = V + X*I where E is the internal generated voltage, V is the terminal voltage per phase, X is the sync. reactance, and I is the output phase armature current of the generator. Key things given to you that you need to catch right away and understand: "phase voltage", "generator", "base value", and per unit values. These determine how to approach the problem.


----------



## rg1 (May 27, 2017)

Don 3.0 said:


> For a sync. generator E = V + X*I where E is the internal generated voltage, V is the terminal voltage per phase, X is the sync. reactance, and I is the output phase armature current of the generator. Key things given to you that you need to catch right away and understand: "phase voltage", "generator", "base value", and per unit values. These determine how to approach the problem.


Don 3.0 I was trying to do it like that - E= 1+j 1.1 X 0.97; that gives you E= 1.46 pu ; the base phase voltage is 250/ sqrt3= 144 kV so the answer should be E=1.46 X 144=211 kV. But CI gives answer as 268KV. Am I wrong some where.


----------



## rg1 (May 30, 2017)

I got it now. The steady state condition given at I=.97 pu is a condition of fault and not a condition of a healthy Generator. CI is right it is E=Id *Xd and not V+IX.


----------



## cos90 (Jul 2, 2017)

rg1 said:


> I got it now. The steady state condition given at I=.97 pu is a condition of fault and not a condition of a healthy Generator. CI is right it is E=Id *Xd and not V+IX.


I have the same problem with this CI Exam 3 Problem 34. What is the clue that the generator has been short circuited? Is it because the current is called "steady state?"


----------



## rg1 (Jul 2, 2017)

cos90 said:


> I have the same problem with this CI Exam 3 Problem 34. What is the clue that the generator has been short circuited? Is it because the current is called "steady state?"


Cos90, the question C-3(34) does not mention explicitly about short ckt. but yes as you say transient and steady state terms are used for fault conditions. How ever steady state condition can also a healthy condition but I think it is less frequently used. More confusing here was .97pu of current. It is near rated current and so it looks more like a healthy condition but I was happy to get the answer by multiplying something and I expressed it, specially in absence of contribution from anyone else on the forum.

Having assumed that it is a steady state of the fault condition (ie it is a short ckt condition) the answer should have been E=0+Id*Xd= .97*1.1=1.067pu and answer should be 1.067*144(144 kV being phase base Voltage; given 250KV, 100MVA  base value must be for 3 phases)=154KV. But you find answer given by CI is (b) 268KV which is Line voltage. So there are many wrongs in this question. Hope you agree with me?


----------



## cos90 (Jul 2, 2017)

I think the convention for these equivalent circuit of an AC generator problems is that Eg is the line to neutral voltage of one terminal. 

Please see this image from Elements of Power System Analysis by Stevenson

I think CI is right for this problem as long as you agree the generator terminals are shorted.


----------



## cos90 (Jul 2, 2017)

Image:


----------



## rg1 (Jul 3, 2017)

cos90 said:


> I think the convention for these equivalent circuit of an AC generator problems is that Eg is the line to neutral voltage of one terminal.
> 
> Please see this image from Elements of Power System Analysis by Stevenson
> 
> I think CI is right for this problem as long as you agree the generator terminals are shorted.


Yes Eg is line to neutral Voltage and in the question also, it is asking phase voltage. Now the figures 250KV and 100MVA which are called rated values for the generator must be meant for 3phase as nothing otherwise is mentioned there. If that be the case the answer should be 154 KV and not 268KV.


----------



## Zach Stone P.E. (Jul 11, 2017)

I find that these types of problems (equivalent generator and motor circuits) are always MUCH easier when we draw the diagram out.

This problem is essentially asking:

"what is the internal generated (stator) voltage (per phase) that would result in a stator current of 0.97pu if the stator reactance (per phase) is j1.1pu"

Then, they give you the base value of the voltage to multply your per unit answer with to get the actual value in volts.

Like this:




*Make sense?*

*The stator terminals are not necessarily shorted.  *

Motor and generator equivalent circuits can be really difficult at first, but you'll find they are quite easy.

If you'd like more practice, we have a tremendous amount of worked out motor and generator equivalent circuits just like this one available inside our free electrical power review course

Electrical PE Review - Free Electrical Power PE Exam Review Course


----------



## rg1 (Jul 11, 2017)

Electrical PE Review said:


> I find that these types of problems (equivalent generator and motor circuits) are always MUCH easier when we draw the diagram out.
> 
> This problem is essentially asking:
> 
> ...


----------



## Zach Stone P.E. (Jul 11, 2017)

Hi @rg1

It is easy to over complicate the per unit system but it helps to remember that Base values are just scalar multipliers.

Three phase ratings of a machine are commonly used as base values. 

Since they gave you the base value, AND gave you the answers in volts instead of pu, you have to work with what you've got and assume.   

Making the correct assumptions comes with practice but is pretty straight forward after awhile.


----------



## rg1 (Jul 11, 2017)

Electrical PE Review said:


> Hi @rg1
> 
> It is easy to over complicate the per unit system but it helps to remember that Base values are just scalar multipliers.- I do not have any confusion in pu system
> 
> ...


----------



## cos90 (Jul 12, 2017)

I think rg1 is correct. There is an identical question in CI that solves it by dividing the base voltage by sqrt 3. I will point out the discrepant questions when I get home.


----------



## Zach Stone P.E. (Jul 12, 2017)

rg1 said:


> So the 250 KV given here is Line to Line Voltage


Correct. 

For any type of three phase or machine, the ratings and base values can almost always be assumed to be the three phase values unless explicitly stated otherwise. 

For example, if you are sizing a three phase transformer, then you are most concerned with the total three phase power it can deliver - because it is going in a three phase system, and the three phase line to line voltage rating - because it is going to be connected across all three phases.

In this case, it is a three phase generator: 



> "What is the internal generated phase voltage in a 3 phase Synchronous generator. Given base values 250 KV, 100 MVA and Xd= 1.1 pu,  steady state current =.97 pu"


So the 250kV is assumed to be the line to line voltage, and the 100MVA is assumed to be the total three phase apparent power it will deliver to the three phase system.

The 1.1pu Xd is assumed to be a *single phase*, or *per phase* reactance used in the single phase equivalent circuit since impedance is always dealt with on a per phase basis. 

The Steady State Current of 0.97pu is assumed to be the stator current, as a result of the internally generated voltage (what the question is asking for) and the per phase reactance, Xd.



rg1 said:


> My doubt is not in pu but whether the answer 268 KV is Line Voltage or Phase Voltage and did the question ask line Voltage or Phase Voltage? I suppose the Phase to Neutral Voltage in this case, as asked by the Question is 154 KV and options does not contain that option.
> 
> 
> 
> ...




I understand your confusion. Sometimes either the line voltage or phase voltage will be selected as the base. However, if the problem points it out directly, it is best to use the given the base value at face value:

*"Given base values 250 KV"*

Hope this helps.


----------



## Zach Stone P.E. (Jul 12, 2017)

cos90 said:


> I think rg1 is correct. There is an identical question in CI that solves it by dividing the base voltage by sqrt 3. I will point out the discrepant questions when I get home.




I would be interested in seeing the other problem, it would make for good discussion.  

Please post the CI number when you get a chance so I can look it up and we can continue the discussion.


----------



## rg1 (Jul 12, 2017)

Electrical PE Review said:


> Correct.
> 
> For any type of three phase or machine, the ratings and base values can almost always be assumed to be the three phase values unless explicitly stated otherwise.
> 
> ...


----------



## TNPE (Jul 12, 2017)

Phase voltage (Vt) for a wye connected system must be divided by sqrt3 if working from a line quantity.  I can check tomorrow in some texts and corroborate, but I'm confident this must be considered, thus, a connection should be given.  If not, I don't know that you can fully answer the question without making gross assumptions that would be atypical for this exam.

Does the problem statement give any indication about the connection?


----------



## Zach Stone P.E. (Jul 13, 2017)

rg1 said:


> It does not help sir. It creates more confusion. I am clear. The issue here is only of convention and even after agreeing that 250kV is L-L voltage we are getting different answers.??? The student in me is not accepting your logic.


I'd be happy to discuss it further with you.

Can you explain where you are confused?


----------



## rg1 (Jul 13, 2017)

Electrical PE Review said:


> I'd be happy to discuss it further with you.
> 
> Can you explain where you are confused?


Lol I am not confused, you are!!!. Anyways.

Given base values 250 KV, 100 MVA and Xd= 1.1 pu,  steady state current =.97 pu

To find per phase Internal Generated Voltage?

pu internal generated Volatge= I*Xd=1.1*.97=1.067pu

Base Line to line Voltage=250kV (Given)

Internal generated Line to Line Volatge (6.7% more than base) = 1.067*250=266.75 kV

Internal generated Phase Voltage=266.75/sqrt3= 154kV

Where is the confusion, let us discuss step by step, let me know where you find the issue, instead of whole problem at one go.


----------



## Zach Stone P.E. (Jul 13, 2017)

rg1 said:


> Lol I am not confused, you are!!!. Anyways.


Just here to help 



rg1 said:


> Given base values 250 KV, 100 MVA and Xd= 1.1 pu,  steady state current =.97 pu
> 
> To find per phase Internal Generated Voltage?
> 
> pu internal generated Volatge= I*Xd=1.1*.97=1.067pu


Correct. I am in agreement with you here.



rg1 said:


> Internal generated *Line to Line *Volatge (6.7% more than base) = 1.067*250=266.75 kV


Internal generated voltage tends to be a per phase quantity, I think the issue is assuming that it is line to line.



rg1 said:


> Internal generated Phase Voltage=266.75/sqrt3= 154kV


This is not necessary since we've already calculated the internal generated phase voltage in the previous step.

To arrive at the correct answer according to the author of this problem, than you need to use the given base value along with the internal generated phase voltage per unit quantity that we calculated in the previous step.

Make sense?

A great resource for these problems is Wildi's Electrical Machines Drives and Power Systems if you are looking for additional examples.


----------



## TNPE (Jul 13, 2017)

No one has given credence to my points raised previously.  Right or wrong, a confirmation of the winding configuration is necessary to determine what the "phase" voltage generated actually is?  I didn't get a chance to review today, but as previously stated, the winding configuration will dictate how this problem should be treated.  I don't feel the need to drudge up the basic 3-phase analysis scenarios for wye/delta, and the associated differences with phase and line quantities in voltage and current with each respective configuration.  That said, does the problem statement indicate the configuration?  If not, you can't answer this problem with certainty!!  Unless someone can convince me otherwise, I'll stay put where I'm at.


----------



## rg1 (Jul 13, 2017)

TNPE said:


> No one has given credence to my points raised previously.  Right or wrong, a confirmation of the winding configuration is necessary to determine what the "phase" voltage generated actually is?  I didn't get a chance to review today, but as previously stated, the winding configuration will dictate how this problem should be treated.  I don't feel the need to drudge up the basic 3-phase analysis scenarios for wye/delta, and the associated differences with phase and line quantities in voltage and current with each respective configuration.  That said, does the problem statement indicate the configuration? No it does not.  If not, you can't answer this problem with certainty!!  Still we can Unless someone can convince me otherwise, I'll stay put where I'm at. It may take time but I will do that.


TNPE, Line configuration in this question is derived from the fact that it is steady state faulty condition of 3 phase bolted fault at the terminals of the generator. The terminal voltage is therefore taken as zero and whole of generated voltage is used to circulate the fault current. The reactance given, Xd, is for one line, as generally given in 3 phase questions(in this case actually it is one phase of the armature of the generator).  The convention  for line configuration is always to take the ckt as star (Y) if nothing is mentioned, for purposes of ease of understanding only, else  you can take it will not effect the results. This I learned when I was in college and I think it is mentioned somewhere in Wildi too.  I will try to give you exact reference.

You do this question by assuming delta connection, the answer to internal phase voltage generated will be same, because in that case the phase current will be reduced by a factor of sqrt3. ( The given reactance is for one phase).


----------



## rg1 (Jul 14, 2017)

Electrical PE Review said:


> Just here to help
> 
> Correct. I am in agreement with you here.
> 
> ...


----------



## TNPE (Jul 14, 2017)

Then if you use that assumption across the board, the generated voltage per phase in this circumstance should be divided by sqrt3.  I don't agree that this assumption is "safe."  It's not like we're working with a motor/generator/XFMR that looks entirely to be an inductive load at start-up, whereby, it is a safe assumption to neglect R and any associated I^2R losses at this given condition.  Furthermore, you have to consider what happens to impedance when you "create" an equivalent wye circuit from a delta, or vice versa.  This problem lacks explicit data IMHO.  This exam is clear and concise.  You will know what they're asking for without having to guess (not to say you can dodge having to use your noggin', cause you can't, but you will be given the appropriate info to solve the problem...along with red herrings and info that's completely unnecessary, just for the sake of tripping candidates up and testing their intuition and subject knowledge).


----------



## rg1 (Jul 14, 2017)

In an earlier post I mentioned

I found many mistakes in Wildi book too.

I correct myself - It is not in the book text, but in some of the Answers to unsolved Questions.


----------



## TNPE (Jul 14, 2017)

Well, if I were you and I wasn't given the proper info, I'd be questioning it; but as previously stated, you will be given all necessary info/ratings on this exam.  You will have to make assumptions, but you will know -if you're prepared, well-versed and understand what's being asked for- what assumption should be made by some trigger in the problem statement.  Guessing what winding configuration you're working with is one I can guarantee you will not see.  Now, it's entirely possible a question can be asked in which the winding configuration is completely irrelevant and unnecessary, but that doesn't hold for this given scenario.  At least it doesn't for me, and I have yet to be convinced otherwise.

Put this another way, what is the secondary phase current of a 3-phase, 15MVA, 69kV - 13.2kV XFMR?  Wouldn't you need to know the winding configuration to accurately tell me?  Same applies to this problem, and gross assumptions are simply that, gross!  You want to talk gross and sick, let these assumptions be the difference between you passing or failing the PE..... 

My advice to you and anyone else, don't put too much stock in the veracity and accuracy of practice problems.  If you find a discrepancy, determine why it is a discrepancy and convince yourself of the "right" way to do a problem.  I see erroneous stuff all over this board, and more frequently than should be normal for mass produced texts/materials!!!


----------



## cos90 (Jul 14, 2017)

CI Test 4 Problem 9 is an unfaulted condition, but its base value is taken to mean line to line voltage and is divided by sqrt 3. This is the problem I mentioned earlier.


----------



## rg1 (Jul 14, 2017)

Let me solve the question at hand by assuming connection of the armature winding of the Generator as delta as well as Y connected.

What is the internal generated phase voltage in a 3 phase Synchronous generator. Given base values 250 KV, 100 MVA and Xd= 1.1 pu,  steady state current =.97 pu

 Description- 1. The Xd is always reactance of one of the phase windings of the Generator whether it is delta or star connected. This also appears in the single phase equivalent circuit of the Generator as one phase reactance and I do not think there should be any doubt in this.

                      2. The currents and Voltages are given as Line quantities if not specified otherwise.

A.  Assuming the Generator as Delta connected. 

The phase current will be 0.97/sqrt3= .56 pu giving Eo phase = .56*1.1=0.616pu; resulting generated phase or Line Voltage (Being Delta) =250* 0.616=154kV.

B. Assuming the Generator to be Star connected.

The phase current will be line current and so Eo=1.1*.97=1.067pu; resulting Generated phase voltage to be 1.067*250/sqrt3=154kV.

Does it not make sense. Is there anything wrong with the description 1 and 2.

The example of transformer given by TNPE seems to me out of place mention here?


----------



## TNPE (Jul 14, 2017)

Is phase voltage in a delta configuration not the same as taking a volt meter and placing the leads across any 2 phases?  If not, then any 3-phase academia I've ever learned has been a farce!!  I've still failed to check this due to being consumed with other things, but Vt (terminal voltage) in a wye is divided by root3 to determine phase voltage (if given line quantities).  You can't "parallel" the familiar power equations to say (i.e. P=sqrt3*V*I*cos(theta) and call it the same thing...not that you explicitly did that, but that's what dots you're practically trying to connect), "see, it's the same."  Still not convinced.....


----------



## TNPE (Jul 14, 2017)

Nvm, maybe we're making the same argument.  Mea culpa.


----------



## TNPE (Jul 14, 2017)

I think we're making the same argument, just that I was mistaken due to not fully reading the problem statement.  Since they've provided a base, I would use your approach.  Sorry for the confusion.

Good debate.  Keep the problems coming.  

Since passing the exam, I seem to have a knack for "wanting more."  Blame it on the student in me, but I enjoy tackling problems on this board, and you do a good job of posing some thought provoking scenarios.


----------



## rg1 (Jul 15, 2017)

TNPE said:


> I think we're making the same argument, just that I was mistaken due to not fully reading the problem statement.  Since they've provided a base, I would use your approach.  Sorry for the confusion.
> 
> Good debate.  Keep the problems coming.
> 
> Since passing the exam, I seem to have a knack for "wanting more."  Blame it on the student in me, but I enjoy tackling problems on this board, and you do a good job of posing some thought provoking scenarios.


Thanks, I appreciate it. I too suffer with same syndrome since I do not know when. I enjoy discussions and learn  and refine a lot of concepts during discussions. Problem solving has been my passion and that is why I poke my nose everywhere. I simply, can't stop it. Locked horn is my normal status. I read only concept from the text book and then make formulae satisfying that concept. I generally work the problem from basics and derive formula every time, I do it. I was able to gauge the gap in your concept, there is very very very small gap in full understanding and your understanding; I could feel it, but  there is always a limitation while communicating by messages . I was not able to phrase my replies which satisfies the student in you, of course that was my fault and limitation. Anyways I loved it, I thank you for the positive engagement. Expect a few more sessions like this one.


----------

