# NCEES Power Sample Exam #530 and #519



## swooda2

#519:

I'm feeling kind of lost on this one.....can someone explain?

#530:

OK, I think I get most of the given solution to this problem. It looks like they find the p.u. short-ckt current by dividing Vpu by total Zpu. Then they solve for the base short-ckt current by dividing S by sqrt of 3 times the voltage. It's then pretty simple to get the actual short circuit current.

My question is, how did they get 0.025 for the p.u. impedance of the 12.47kV system?

I'm new to the board, and any help would be greatly appreciated. Thanks.


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## Flyer_PE

Question 519:

I can't figure out a way to explain this one without generating graphs.

Question 530:

They are performing the calculation using the transformer base of 1MVA. The utility impedance is given in the form of 40 MVA short circuit. The utility impedance on the transformer base is 1 p.u. * 1MVA/40MVA = 0.025


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## mudpuppy

Flyer_PE, if you want to add graphs, I've found it is not too difficult to make a drawing with pencil &amp; paper, and then take a digital photograph and post it.


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## Flyer_PE

^ I was just too lazy this morning.

The problem is to pick the correct current waveform for A phase on the AC side of a three-phase rectifier with the C phase fuse blown. Basically, the current waveform becomes the same as that of a single phase rectifier.


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## swooda2

Flyer_PE said:


> Question 519:
> I can't figure out a way to explain this one without generating graphs.
> 
> Question 530:
> 
> They are performing the calculation using the transformer base of 1MVA. The utility impedance is given in the form of 40 MVA short circuit. The utility impedance on the transformer base is 1 p.u. * 1MVA/40MVA = 0.025



Flyer, thanks for the help. I get the concept now--you have to convert from the 40MVA base to the 1MVA base, correct?

The equation you solved looks like the equation for converting from one base to another, but where did you get the value of 1.0 for the 40MVA per-unit impedance? Or am I misinterpreting?


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## Flyer_PE

swooda2 said:


> Flyer, thanks for the help. I get the concept now--you have to convert from the 40MVA base to the 1MVA base, correct?
> The equation you solved looks like the equation for converting from one base to another, but where did you get the value of 1.0 for the 40MVA per-unit impedance? Or am I misinterpreting?


Yes you have to convert from the 40 MVA base to a 1 MVA base.

In this problem, they give you the available short circuit current in the form of MVA. This value can be provided directly as an impedance, available short circuit amps, or MVA. Since they give you the value of 40MVA short circuit, the utility impedance is 1 p.u. on a 40 MVA base by definition.


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## swooda2

Flyer_PE said:


> Yes you have to convert from the 40 MVA base to a 1 MVA base.
> In this problem, they give you the available short circuit current in the form of MVA. This value can be provided directly as an impedance, available short circuit amps, or MVA. Since they give you the value of 40MVA short circuit, the utility impedance is 1 p.u. on a 40 MVA base by definition.



got it now. thanks very much for the help.


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## visu212

Flyer_PE said:


> Yes you have to convert from the 40 MVA base to a 1 MVA base.
> In this problem, they give you the available short circuit current in the form of MVA. This value can be provided directly as an impedance, available short circuit amps, or MVA. Since they give you the value of 40MVA short circuit, the utility impedance is 1 p.u. on a 40 MVA base by definition.


Flyer_PE, Thanks a bunch for the explanation. I almost got it but can you please explain me why 1.0 in numerator was taken while calculating Isc pu in the solution.

Isc pu= 1.0/(ZT+ Zsys)= 1/0/(0.04+0.025) What is the significance of 1.0 there? is it vsc p.u value or new 1 MVA base?

*My assumption*

Iscpu = Vscp.u /Zpu but Vsc p.u = 1.0 assumed( or per definition) then Isc pu = 1.0/Zpu

Isc actual = (Isc base X Isc pu)

= 1MVA/(SQRT 3x 480) x 1/(0.04+0.025) = 18507.7 amps Or 18520 amps(Ans)


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## le.boot

Flyer-PE -

Can you explain the following in a little more detail (in reference to the system fault duty)?

"In this problem, they give you the available short circuit current in the form of MVA. This value can be provided directly as an impedance, available short circuit amps, or MVA. Since they give you the value of 40MVA short circuit, the utility impedance is 1 p.u. on a 40 MVA base by definition. "

I get that the per unit voltage and MVA is 1. Is the per unit current also 1 because I=S/V?

So, all these values (Z, Isc, MVA, V) would be 1 pu if we use two knowns as the bases?

(obviously I'm having a little trouble getting my head around this).

Thanks!


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## Flyer_PE

^I've typed this a couple of different ways and I'm not sure if it will help or not. Here goes:

The pu values for all parameters = 1 for the base. You have 4 basic values: S, Z, I, and V. On the base system, you choose the value for any two and the others are calculated from there. For any analysis, the key to getting it done fast and simple is properly choosing the base values to make the math simple.


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## Rei

Thanks Flyer. I think I understand the calculation steps, but still feel like assuming 1pu for the 12.47kV system is not easy to catch on the exam. I think what throw me off is the "assuming the transformer and the 12.47kV system has the same X/R ratio."


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## Flyer_PE

Two things:

1. They _really_ like giving you extraneous information in the problems. In that problem, the X/R ratios are totally irrelevant.

2. You can choose any base voltage you want. In this case it simplifies things if you choose 12.47 kV as the base because it makes the math real easy. Typically, the power and voltage bases are chosen based on either the system or the largest transformer in your analysis. In this problem, if you use something other than 12.47 kV, the 4% impedance value for the transformer will have to be adjusted to the new base and you will no longer be applying a 1 pu voltage to the fault. In the end, you'll still wind up with the same number of amps though.


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## Rei

Flyer_PE said:


> Two things:
> 1. They _really_ like giving you extraneous information in the problems. In that problem, the X/R ratios are totally irrelevant.
> 
> 2. You can choose any base voltage you want. In this case it simplifies things if you choose 12.47 kV as the base because it makes the math real easy. Typically, the power and voltage bases are chosen based on either the system or the largest transformer in your analysis. In this problem, if you use something other than 12.47 kV, the 4% impedance value for the transformer will have to be adjusted to the new base and you will no longer be applying a 1 pu voltage to the fault. In the end, you'll still wind up with the same number of amps though.


I did think about converting the 4%pu from 1MVA to 40MVA base so that the 1pu doesn't need to be converted. But then what is the base current? The 40MVA is from the 12.47kV and the question asked for the 480V.


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## Flyer_PE

Doing the problem on the 40 MVA base rather than the 1 MVA Base:

Base values or voltage are 12.47 kV and 480/277 V.

Transformer impedance on 40 MVA base = 40x4% = 1.6 pu

The utility impedance is now 1 by definition.

Ipu= 1/(1 + 1.6) = 1/2.6 = 0.385

On the 480V level: Ibase= 40MVA/(sqrt3*480V) = 48.11 kA

IActual = Ipu*Ibase=0.385 * 48.11 kA = 18.5 kA.

The math is a little uglier, but wind up with the same answer.


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## Rei

Flyer_PE said:


> Doing the problem on the 40 MVA base rather than the 1 MVA Base:
> Base values or voltage are 12.47 kV and 480/277 V.
> 
> Transformer impedance on 40 MVA base = 40x4% = 1.6 pu
> 
> The utility impedance is now 1 by definition.
> 
> Ipu= 1/(1 + 1.6) = 1/2.6 = 0.385
> 
> On the 480V level: Ibase= 40MVA/(sqrt3*480V) = 48.11 kA
> 
> IActual = Ipu*Ibase=0.385 * 48.11 kA = 18.5 kA.
> 
> The math is a little uglier, but wind up with the same answer.


Thank you!


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## Rei

I do understand about the base conversion, etc. but when it comes to problems, I still have trouble applying the concept. I went through the Kaplan sample exam over the weekend, and could not answer any of the pu problems. Does anyone has any suggestion on how to master this concepts. Any suggestions on good books and practice problems? Thanks.


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## Flyer_PE

Problems were kinda hard to come by for me. What do you have for texts on the subject? The best one I have is Elements of Power System Analysis by Stevenson. For me, I had a really good teacher in college that beat us to death with PU analysis.


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## Rei

Flyer_PE said:


> Problems were kinda hard to come by for me. What do you have for texts on the subject? The best one I have is Elements of Power System Analysis by Stevenson. For me, I had a really good teacher in college that beat us to death with PU analysis.


You are lucky. My professor skipped the pu analysis and now it's biting me.


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## Rei

I tried to work through the NCEES again and missed this question again. After a second thought, I think I still don't get why they assume the system impedance is 1pu.

If the Sbase = 1MVA, then the system impedance should be Z=(12.47k)^2/40MVA and then divided by the Zbase=(12.47k)^2/1MVA to get the pu value. The system pu should not be 1.


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## Rei

Rei said:


> I tried to work through the NCEES again and missed this question again. After a second thought, I think I still don't get why they assume the system impedance is 1pu.
> If the Sbase = 1MVA, then the system impedance should be Z=(12.47k)^2/40MVA and then divided by the Zbase=(12.47k)^2/1MVA to get the pu value. The system pu should not be 1.


Humm...no one comments on this? Why the system impedance is not

Z=(12.47k)^2/40MVA

Zbase=(12.47k)^2/1MVA

there fore Zpu=Z/Zbase which is not 1pu as used in the solution


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## DK PE

Hi Rei, take a look at the reference I listed at the other thread on prob #530.... maybe it will help. You can sort of think of it as trying to find the thevenin equivalent of the utility source upstream.


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## knight1fox3

DK PE said:


> Hi Rei, take a look at the reference I listed at the other thread on prob #530.... maybe it will help. You can sort of think of it as trying to find the thevenin equivalent of the utility source upstream.


I'm having a bit of trouble locating the thread you mentioned. Could you provide a link? Thanks.


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## Flyer_PE

^He's talking about this thread.

The paper he mentions is here.


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## BamaBino

Flyer_PE said:


> The paper he mentions is here.


I worked NCEES # 530 using the same method as the example in that paper. I like the approach and it is the second jpeg below.

When tried working NCEES # 540 using the same general method, I had a problem getting the right answer (close but off 10%). I'm sure the error is in the last step or a step was left out. Would someone point out where it went wrong?

Thanks


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## Flyer_PE

I didn't do all the math. I do see one problem with your solution though. You can't add the pu impedance of the generator (834MVA base) to the pu impedance of the transformer (933MVA Base). It looks like you have chosen the transformer rating (933MVA) as the base for the system. You will need to convert the generator pu impedance from the 834MVA base to the 933MVA base of the transformer.


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## BamaBino

Flyer_PE said:


> I didn't do all the math. I do see one problem with your solution though. You can't add the pu impedance of the generator (834MVA base) to the pu impedance of the transformer (933MVA Base). It looks like you have chosen the transformer rating (933MVA) as the base for the system. You will need to convert the generator pu impedance from the 834MVA base to the 933MVA base of the transformer.


That was the problem. Now I get the correct answer.

Thanks


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## BamaBino

Flyer_PE said:


> I didn't do all the math. I do see one problem with your solution though. You can't add the pu impedance of the generator (834MVA base) to the pu impedance of the transformer (933MVA Base). It looks like you have chosen the transformer rating (933MVA) as the base for the system. You will need to convert the generator pu impedance from the 834MVA base to the 933MVA base of the transformer.


Below is Problem 540 after making this correction to get answer B.


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## jayache80

reading through this, I just wanted to add: by definition, S_sc = S_base / Zpu.

When they give us "fault duty" of 40MVA, that means they gave us S_sc, using 40MVA as S_base:

S_sc = S_base / Zpu

40MVA = 40MVA / Zpu

Zpu should be assumed to be 1pu, as Flyer has said, but it didn't click with me until I realized this definition.

As it's been said, I think the best way to solve these types of problems is using the MVA method:

Ssc_xfmr = S_xfmr_rated / Zpu = 1000kVA / .04 = 25MVA

Ssc_sys = 40MVA (given)

To find the total Ssc, when combining Ssc's in series, it's like resistors in parallel.

Ssc_total = Ssc_xfmr * Ssc_sys / (Ssc_xfmr + Ssc_sys)

Ssc_total = 25MVA * 40MVA / (25MVA + 40MVA)

Ssc_total = 15.3846 MVA (this number looks familiar if you did it with the per unit method first)

Now, to get your Isc, just 'three-phase' divide by the voltage of interest:

Isc = Ssc_total / (rad3 * V)

Isc = 15.3846MVA / (rad3 * 480) = 18,504.8 A


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## saberger_vt

Flyer_PE said:


> Two things:
> 
> 1. They _really_ like giving you extraneous information in the problems. In that problem, the X/R ratios are totally irrelevant.
> 
> 2. You can choose any base voltage you want. In this case it simplifies things if you choose 12.47 kV as the base because it makes the math real easy. Typically, the power and voltage bases are chosen based on either the system or the largest transformer in your analysis. In this problem, if you use something other than 12.47 kV, the 4% impedance value for the transformer will have to be adjusted to the new base and you will no longer be applying a 1 pu voltage to the fault. In the end, you'll still wind up with the same number of amps though.


I would agree with Flyer_PE that in this particular problem, the X/R ratio is irrelevant since it equals 1. From the point of solving short circuits, it is still a viable piece of information. Here is one post on engineersboard covering some calculations using the X/R, and another post of an older GE document covering asymmetrical faults:

X/R Ratio and 3 phase fault
GE Document:

Short-Circuit Current Calculations - GE Industrial Systems


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