# NCEES Practice Exam, No. 135



## BigWheel (Apr 16, 2017)

The problem states:

An electric generation facility uses a turbine-driven synchronous generator rated 3PH, 150MVA, 13.8-kV.

Per-unit reactances are: X"d=0.15; X'd=0.25; Xd=Xq=1.20

Assume terminal voltage (Et)=1.0 pu.

For a transient stability study at rated MVA, rated voltage, and unity power factor, the internal voltage (pu) and reactance (pu) for the generator should be, respectively:

A) 1.01 and 0.15

B) 1.01 and 0.25

C) 1.03 and 0.25

D) 1.10 and 0.25

I know the per-unit reactance for the generator should be 0.25, because we're told that we're doing a "transient stability study," but I'm having a hard time with the internal voltage part. The solution manual says I should be using (1.0+j0) pu for my current in the calculation of E'0, but doesn't elaborate any further, nor is it obvious to me why I would.

Does it have anything to do with the fact that I'm not given information about the circuit the generator is connected to, and therefore can't calculate the load current, and so must assume the base current is the same as the short-circuit current of the generator? This is the only thing I can think of, but I'm confident in this assessment.

Any help or insight would be very much appreciated. 

Thanks.


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## BigWheel (Apr 16, 2017)

Sorry, I meant say that I'm *not* confident in my assessment.


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## Limamike (Apr 16, 2017)

@BigWheel

So first look at X'd which is the transient component.  So eliminate A.  Now, Vt PU is 1 .  So Ea (internal voltage) = Vt + It*(Xs)  +R (there is none)  Formula from chapman sync gen chapter.......So this turns to 1+.25(1+j), which they round to 1.03.


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## rg1 (Apr 16, 2017)

BigWheel said:


> The problem states:
> 
> An electric generation facility uses a turbine-driven synchronous generator rated 3PH, 150MVA, 13.8-kV.
> 
> ...


I think for transient study we take transient reactance = .25. current pu =1&lt;0 so voltage drop in transient reactance = 1&lt;0X.25j=.25j. So the internal voltage = E+IX= 1+j.25= 1.0308&lt;14.04. So Answer- C


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## rg1 (Apr 17, 2017)

BigWheel said:


> The problem states:
> 
> An electric generation facility uses a turbine-driven synchronous generator rated 3PH, 150MVA, 13.8-kV.
> 
> ...


The solution manual says I should be using (1.0+j0) pu for my current in the calculation of E'0, but doesn't elaborate any further, nor is it obvious to me why I would. ------

Even if resolve the problem in actual values of V, I and Z the answer is to be brought back into pu only so it is better the problem is resolved in pu only. The problem says unity pf at rated load so it is clear from here that the current is 1pu at zero angle from the voltage which is also 1pu with zero angle. Hope this answers your doubt


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## BigWheel (Apr 18, 2017)

rg1 said:


> The solution manual says I should be using (1.0+j0) pu for my current in the calculation of E'0, but doesn't elaborate any further, nor is it obvious to me why I would. ------
> 
> Even if resolve the problem in actual values of V, I and Z the answer is to be brought back into pu only so it is better the problem is resolved in pu only. The problem says unity pf at rated load so it is clear from here that the current is 1pu at zero angle from the voltage which is also 1pu with zero angle. Hope this answers your doubt


Thank you for the replies. Very helpful.


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## rg1 (Apr 18, 2017)

BigWheel said:


> Thank you for the replies. Very helpful.


 My Pleasure. Please let me know if something else is there? Discussion for learning


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## Omer (Sep 23, 2017)

Bringing back this question.

I have some confusion to be cleared.

This is a 3 phase problem and to solve for the internal voltage usually we use the single phase circuit.

1 pu voltage is 13.8 kv (since it is the base).So for the single phase, terminal pu voltage is 1/(sqrt3). 

Doing the calculations, the answer would be nearly 1.1 pu Eg. answer D.

I know I am wrong somewhere. Just tell me where exactly.


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## rg1 (Sep 23, 2017)

Omer said:


> Bringing back this question.
> 
> I have some confusion to be cleared.
> 
> ...


Omer. Very nice doubt!!!. I will tell you where you are wrong when dealing with pu and 3/1 phases. In pu if you have taken three phase voltage as 1pu, then single phase voltage is also 1pu and not to be divided by sqrt3 while dealing with pu.

Now I solve a simple question for you.

Say the three phase Voltage is 13.8kV and I have to find 1phase Voltage.

I take 3 phase Voltage pu as 1.

By actual quantity method- the 1-phase Voltage from actual Quantity 13.8/sqrt3=7.96kV

FRom pu method- As I said the pu single phase will also be 1pu, but Base Voltage single phase is 7.69kV

So Actual single phase Voltage = Vpu*BaseV=1*7.69Kv=7.69kV. 

Did it make sense?? I can explain it in many ways so do not hesitate to raise it again.

When in pu you do not worry about 1 or 3 phase. When you come put of pu in world of actual quantities you will be on right track if you choose and use  your base appropriately.  You will find this true for MVA, I, or V anything.


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## Omer (Sep 24, 2017)

Thanks @rg1.

That was right to the point.

Interesting.


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## wolfpacknc (Oct 15, 2017)

Below is another way to work the problem. This method helped me understand the answer a little bit better.

1. Solve for the current base = S/(SQRT(3)*Vll) = 6275.5 (the base current is the same as the actual current in this example)

2. Solve for Z base = V/(SQRT(3) * Ibase) = 1.2696

3. Solve for Z actual = Z base * Z p.u. = 0.25j * 1.2696

4. Draw the single phase diagram , with the terminal voltage of 13.8/(SQRT(3)) KV (phase to ground voltage)

5. Perform a KVL : Vgen = Vterminal + Ibase * Z actual 

6. Vgen magnitude (phase to ground) = 8212.7V

7. Vgen phase to phase magnitude = 8212.7* SQRT(3) = 14208V

8. Vgen (phase to phase) p.u. = 14208/13800 = 1.03

9. The answer is thus, 1.03 V and 0.25 reactance

Hopefully this helps someone.

-Alex


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## FPar (Oct 15, 2017)

I like @wolfpacknc way of double checking the p.u. with the actual values. You can skip step #8 and do 8212/7967=1.03

It was really good Alex, Thanks!


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## Stephen2awesome (Oct 16, 2017)

wolfpacknc said:


> Below is another way to work the problem. This method helped me understand the answer a little bit better.
> 
> 1. Solve for the current base = S/(SQRT(3)*Vll) = 6275.5 (the base current is the same as the actual current in this example)
> 
> ...


Go Pack.


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## wolfpacknc (Oct 16, 2017)

Stephen2awesome said:


> Go Pack.


Yessir!


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