# kaplan morning #19



## rick.conner (Oct 17, 2011)

the book definitely calculated the generator pu voltage wrong.

Though I am getting two different answers when working different ways so I might need a little lesson!

Give that the pu current is 0.059-0.01078j and the base is 240MVA. The generator Voltage line-line is 45kv and is the reference. They want to know what the complex power delivered is. the current is across a load several xfmr's down though who cares as it is given in pu!

1) convert I pu to the 45kv base which is 184 at an angle of -10.35, next V*conj(I)*sqrt(3) should give me the total 3-phase mva of 14+2.56j

2) if I convert the gen to pu, multiply sqrt(3)*Vpu*conj(Ipu) then (138/45)^2 so i convert to the 45kV side and then multiply by 240MVA I get 75+13.738j

help please - this is the first pu problem I've problems with and going crazy


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## DK PE (Oct 17, 2011)

I don't have the Kapan reference but it seems like your calculation following 1) is correct assuming you are are given some sort of generator with series reactance diagram that gives the pu current given. The base is 240 MVA and base voltage at generator is 45kV. You calculated base current, actual current and MVA correctly as far I can see with the given values.

I can't follow what you are doing after 2)... while it is true you can choose a any convenient base MVA for the system, you appear to be converting the generator to some other base ... you can't just convert only the generator to another base and leave others alone. Apolgies if I missed what you're doing... maybe I'm just not following.


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## rick.conner (Oct 17, 2011)

thanks for the help.

on 2). i take the generation voltage of 45kV and divide by 138kV to get pu (the problem stated the pu voltage was 138kV. I then take Vpu*conj(Ipu)*sqrt(3). this is where I get stuck. I'm in kva per-unit. and believe Im not converting to the generation kva base and voltage correctly.

if you are interested still i can scan in the prob tomorrow morning.

thanks again


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## DK PE (Oct 18, 2011)

rick.conner said:


> thanks for the help.
> 
> on 2). i take the generation voltage of 45kV and divide by 138kV to get pu (the problem stated the pu voltage was 138kV. I then take Vpu*conj(Ipu)*sqrt(3). this is where I get stuck. I'm in kva per-unit. and believe Im not converting to the generation kva base and voltage correctly.
> 
> ...


When you said in post #1 "_The generator Voltage line-line is 45kv and is the reference"_ ... I interpreted 45kV was base voltage at generator.

Then you said "_generation voltage of 45kV and divide by 138kV to get pu (the problem stated the pu voltage was 138kV)"_ --&gt; I think you meant to say the base voltage was 138kV... doesn't make any sense to say pu voltage was 138kV?

Sorry I can't seem to follow but if the base voltage at the generator is not the same as the rated voltage (i.e the voltage rating where the reactance is specified), then you need to convert using the forumula I'm sure you know.

I believe the best way to solve these problems is to draw out a quick one line noting the ratings on the components. Then assign a base kVA or MVA which applies everywhere. Then pick a place and say that is the base voltage there. Then go through the diagram and write down the base voltages -which are determined by the ratio of line voltage on each side of transformers (don't get confused with delta/wye etc). Then notice that say you have a transformer whose rated MVA doesn't match your chosen base and apply the correct formula to get %Z it the chosen base. Then notice the generator's voltage and MVA rating doesn't match your base and convert that. Then you have a one-line you can solve.

Just remember that the one-line looks simple and even though the same pu current may flow... the base impedance and base current varies at various points in the diagram so you need to think a bit if you are asked for actual current at some point out in middle.

Reading your problem one more time... remember you can have a one line where the base voltage at the generator is say 16kV ... but the machine rating is 20kV --&gt; then you need to convert the reactance using the appropriate forumla... but then they can say the machine is actually operating at 13.2kV. Then in your one-line the pu voltage is actual/base or 13.2/16 = 0.825 pu

Hope this makes sense and helps.


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## rick.conner (Oct 18, 2011)

yea that makes perfect since though still not sure what the heck I'm doing wrong!!! Here is a pic of the problem. thanks again for looking at this


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## knight1fox3 (Oct 18, 2011)

For anyone doing the Kaplan Sample Exam problems, there was a good discussion (not to mention some good humor) in this thread from the Oct. 2010 exam that compiled which problems were wrong for both the morning and afternoon sections. Though a newer sample exam book may have been published by Kaplan since then.


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## rick.conner (Oct 18, 2011)

thanks. these problems are definitely harder than the ncees so i want to keep tackling them even though it makes you second guess your self a lot. Seems like the comlex imaginary are too easy and the camara are focusing on the EERM. if the exam is like the ncees practice I'll do alright. if it is like the kaplan I might be in trouble.


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## knight1fox3 (Oct 18, 2011)

rick.conner said:


> thanks. these problems are definitely harder than the ncees so i want to keep tackling them even though it makes you second guess your self a lot. Seems like the comlex imaginary are too easy and the camara are focusing on the EERM. if the exam is like the ncees practice I'll do alright. if it is like the kaplan I might be in trouble.


Kaplan problems seem to be more in-depth and don't follow the NCEES format. However, they are certainly still good for practicing fundamental concepts. Well, the ones that don't contain errors in the solutions that Kaplan provides.


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## rick.conner (Oct 18, 2011)

so you think i shouldn't be getting freaked out about having problem with some of these? the complex imaginary were pretty easy though not sure if that means anything.


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## knight1fox3 (Oct 18, 2011)

rick.conner said:


> so you think i shouldn't be getting freaked out about having problem with some of these? the complex imaginary were pretty easy though not sure if that means anything.


No. Just compare a Kaplan problem to a problem from the NCEES practice exam and tell me if the amount of information the problem is looking for is similar. Just be aware of the errors that some of the Kaplan problems/solutions contain.


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## DK PE (Oct 18, 2011)

rick, I know it can be rattling when you finish punching the calculator and "none of the above" is not a choice.

In this problem I would recognize that the solution is half finished and you assume they constructed a correct one line reactance diagram (handling the far right transformer that has properties that have to be converted). In this case, they chose 240MVA Sbase and 138kV in region 3 as bases. Therefore, region 2 base voltage is 345kV and region 1 base voltage is 45kV. As you said, the base current in the generator circuit is 3079A and therefore the actual current is ~ 185A. If you notice, you really don't need the complex power since all the real parts of the solution are different. The real power delivered is P = √3 VL IL cosθ = √3 * 45kV * 185 cos10.3̊ =14.1 MW which is none of the above. You could speculate all day what they may have missed like maybe the generator base voltage is 45kV but is is operating at 23kV but who knows.

Just trying to rebuild your confidence... DK


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## rick.conner (Oct 18, 2011)

awesome! thanks for the help on this. I think if i stop working these kaplan problems I'll be good!!!! have the morning done and will hopefully finish the rest here shortly.

thanks for pointing out the real power differences in the solutions - something definitely to look for on the exam.

good luck and thanks again


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## Insaf (Oct 18, 2011)

rick.conner said:


> yea that makes perfect since though still not sure what the heck I'm doing wrong!!! Here is a pic of the problem. thanks again for looking at this


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Which answer was selected by Kaplan?


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