NCEES #135

Can anyone assist me in starting NCEES problem #135? I've been going through my references but haven't found anything that really explains it. I may have missed it or maybe it's different notation. I have two versions of the Grainger/Stevenson book but they go into way more detail than what I think the intent of #135 was. Any help would be appreciated as always. Thanks.

I don't have any good references for transient stability studies. However, I don't think one is really needed here. The key to this problem is knowing which impedance to apply:

X''d = Sub-transient Reactance

X'd = Transient Reactance

Xd = Direct Reactance

If you are doing a transient stability study, you aren't interested in either the sub-transient or direct reactances.

The internal voltage required to generate 1.0 pu voltage at the terminals with a 0.25 pu reactance is found by:

EInternal = 1.0 puVolts + 1.0puAmps*j0.25puohms = 1.0 + j0.25 = 1.03/14o

Flyer_PE said:

I don't have any good references for transient stability studies. However, I don't think one is really needed here. The key to this problem is knowing which impedance to apply:
X''d = Sub-transient Reactance

X'd = Transient Reactance

Xd = Direct Reactance

If you are doing a transient stability study, you aren't interested in either the sub-transient or direct reactances.

The internal voltage required to generate 1.0 pu voltage at the terminals with a 0.25 pu reactance is found by:

EInternal = 1.0 puVolts + 1.0puAmps*j0.25puohms = 1.0 + j0.25 = 1.03/14o

Click to expand...

Thanks for the help Flyer_PE. But how does one know to use 1.0pu for the current? That is where I was grid locked. I must be missing something fundamental here but that certainly did NOT seem intuitive to me.

knight1fox3 said:

Thanks for the help Flyer_PE. But how does one know to use 1.0pu for the current? That is where I was grid locked. I must be missing something fundamental here but that certainly did NOT seem intuitive to me.

Click to expand...

The problem states that you are doing the study at rated MVA and voltage. If both voltage and MVA are 1.0, current is 1.0 by definition.

Flyer_PE said:

The problem states that you are doing the study at rated MVA and voltage. If both voltage and MVA are 1.0, current is 1.0 by definition.

Click to expand...

So the key here is to realize that one must select 150MVA as Sbase and 13.8kV as Vbase. Which yields 1.0pu for both V and S. And by definition, S/V or 1/1 = 1.0pu for I. Practice, practice, practice. I need to be quicker at making the realization for problems of this type. Unfortuntately, PU is not one of my strong areas.

Flyer_PE said:

I don't have any good references for transient stability studies. However, I don't think one is really needed here. The key to this problem is knowing which impedance to apply:
X''d = Sub-transient Reactance

X'd = Transient Reactance

Xd = Direct Reactance

If you are doing a transient stability study, you aren't interested in either the sub-transient or direct reactances.

The internal voltage required to generate 1.0 pu voltage at the terminals with a 0.25 pu reactance is found by:

EInternal = 1.0 puVolts + 1.0puAmps*j0.25puohms = 1.0 + j0.25 = 1.03/14o

Click to expand...

I am not really familiar with Power System Stability or questions related to it. Can we expect to see any power system stability equations involving intertia constants, and swing equations of a generator etc...?

^I doubt you'll see any of that. My own feeling is that it's outside the scope of the exam.

So basically NCEES problem 135 is more of a per-unit problem than a transient stability problem?

this problem is actually a lot easier than it seems; you should only know what it means by "transient". Generators have three types of impedances: Xd, X'd and X"d. Xd is used for steady-state analysis, X'd for transient and X"d for sub-transient analysis.

So you are going to have to use X'd.

The rest of it pretty easy; generator internal voltage is its terminal voltage plus output current times the internal impedance! |1+1xj0.25|=1.03

The problem also gives a Xq parameter in addition to the others. I know it is extra unneeded information in this problem, but if it was needed I wouldn't know what to do with it. I think it means "quadrature axis synchronous reactance" and has something to do with the small amount of flux perpendicular to the excitation winding. Anyone have any words of advice pertaining to the Xq parameter?

I have another problem with this problem. Is it really possible to close the turbine governor fast enough to keep the current stable during the transient event? I would expect the current to spike and come down before governor could react. I think a more reasonable question would ask what the current would be while keeping the internal voltage constant.

knight1fox3 said:

Flyer_PE said:

The problem states that you are doing the study at rated MVA and voltage. If both voltage and MVA are 1.0, current is 1.0 by definition.

Click to expand...

So the key here is to realize that one must select 150MVA as Sbase and 13.8kV as Vbase. Which yields 1.0pu for both V and S. And by definition, S/V or 1/1 = 1.0pu for I. Practice, practice, practice. I need to be quicker at making the realization for problems of this type. Unfortuntately, PU is not one of my strong areas.

Click to expand...

Should the definition S/V stated above for I , actually be S/(sqrt3 *V), Since this is 3 phase and not single phase.

^You have to be very careful in keeping it sorted out as to whether you are determining an actual numerical value in amps or a per-unit value. I don't have the NCEES practice exam with me here but I'll try an example.

There are four quantities in any pu analysis: Power, Voltage, Current, Impedance

In setting up a pu analysis you get to pick two of these, the other two are determined based on the values you choose (typically voltage and power).

Given a 150MVA 13.8kV system you have the following for Per Unit:

1. Select 150MVA as a base which is defined as 1.0 pu

2. Select 13.8kV as the base voltage, also defined as 1.0pu

3. Now that you have selected the first two, the base line current is then determined by IBase=SBase/(sqrt(3)*VLine)

So IBase=150MVA/(sqrt3*13.8kV) = 6276 Amps = 1.0 pu

Good thread here, thanks all for the responses. I too was struggling why the current I = 1 pu. Now I understand, wow this question is a lot simpler than it seems.

Camara's Power Reference Manual has a couple of good pages on transient conditions and fault analysis. Look starting on pg 36-6. He also has a diagram of the subtransient and transient period models and briefly describes the values (X''d, X'd, etc) in each.

PS- I just found Camara's book online, it looks like the whole thing. Try this link to the page I mentioned above: http://books.google.com/books?id=r_rDg_E97...p;q&f=false

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