rg1
Well-known member
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
Internal Voltage generated
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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
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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
Well-known member
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.
rg1
Well-known member
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
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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.
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.
rg1
Well-known member
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.
Well-known member
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
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
Well-known member
Zach Stone P.E.
Well-known member
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.
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.
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rg1
Well-known member
Zach Stone P.E.
Well-known member
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:
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.
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.
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Zach Stone P.E.
Well-known member
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
Well-known member
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?
Does the problem statement give any indication about the connection?
Zach Stone P.E.
Well-known member
rg1
Well-known member
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.
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