# Question on problem #518 NCEES Sample Problems



## Platinum (Mar 28, 2007)

Hey guys,

I have a stupid question on problem #518 power afternoon question. Its asking for the load current in the generator.

The transformer turns ratio is a= 13.2/132 = 0.1

So the I primary is 759.3A, which is I phase of the delta?

So to get I Line shouldn't you multiply by sqrt3, to get the answer of 1320A?

I know I'm missing something stupid...

BTW, the correct answer is (A) 760A


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## Dark Knight (Mar 28, 2007)

Platinum said:


> Hey guys,I have a stupid question on problem #518 power afternoon question. Its asking for the load current in the generator.
> 
> The transformer turns ratio is a= 13.2/132 = 0.1
> 
> ...


Plat,

They are giving you the current at the transmission line. I don't understand your question. I checked my sample test and this is one of the problems I don't have any comments on red.Meaning? Got it right without a hassle.

There is nothing tricky on it. I used the relationship

Vp/Vs=a=Is/Ip and worked from there.

I understand, or I think I understand now your confusion. You are talking about the SQRT(3) change between the ILine and I phase in a delta. That is referred to the current flowing inside the delta my friend. make a drawing of three lines, A,B,&amp;C conected to a load configured in delta. The current flowing into the delta branches go thru that sqrt(3) conversion in respect to the phase currents.

My friend, I don't know if I am making any sense to you since I am dumber than I was when I took the PE but the bottom line is that using the relationship for the transformer turns ratio solves the problem.

If that helped OK. If not let me know how can I help.


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## grover (Mar 28, 2007)

You're given delta voltage on both sides. As (balanced) 3-phase line current is the same regardless of whether you're measuring it delta or wye, the actual wiring of the transformer is as irrelevant to the problem as the given impedances. Don't let them trick you with a simple problem!

Ratio of V1:V2 is 10, so the current ratio is the same.


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## Platinum (Mar 28, 2007)

Thanks for the responses guys. I was just used to solving a similar problem, but instead they give you a voltage down at one end and ask for the line voltage at the generator. For that problem you had to adjust your voltage levels with the turns ratio and multiply or divide by sqrt(3) depending on what type of transformer. I just thought you would have to do the same with the current.


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## grover (Mar 30, 2007)

Your in-box is full! I'll post the reply here, it might help others, too:



> *539.* An overcurrent relay is used to protect a circuit with a maximum 3-phase fault current of 8,000A promary. A 400:5 current transformer ratio and a %-A relay tap setting are selected. If the total secondary burder on the current transformer for the maximum fault condition is 1.10 Ohms, the excitation voltage (V) that must be developed by the current transformer is most nearly:
> ( A ) 100
> 
> ( B ) 110
> ...





> Hey grover,
> Could you check your notes on question #539? I tried drawing out the problem but im having a hard time visualizing it. I dont know what the primary or secondary currents are....or where the current transformer is applied on what side of the main transformer?
> 
> I worked thru the NCEES problems and i have some questions on various problems.
> ...


Are you familiar with current transformers? If not, they're quite literally loops of small gage wire that the main cable passes through. Same thing you see on clamp-type ammeters- the power conductor going through the coil is 1 loop, and considered the primary. I don't normally think of the current transformer as a "secondary", but the coil of small gage wire really is the secondary, if you think about it. Current passing through the main conductor excites current through the current transorfmer loop.






OK, the current transformer is 400:5 (80:1), which means 8000A through the cable it's wrapped around will push 100A through the current transformer "secondary". If the resistance of the current transformer is 1.1 Ohms, V=IR &amp; voltage is 110V. And, incidently, the relay exceeds 5A and will trip :Locolaugh:

Hope that helps!


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## Dark Knight (Mar 30, 2007)

Same thing happened to me with your box when I tried to PM you:

About the problem 531:

Platinum,

I am sorry I can't help you with this one. I wrote my notes in red on that one, with an asterisk. That means that I got it wrong when I practiced it (red writing) and the asterisk means that I had problems understanding the explanation at that moment.

I had a footnote in blue with the following: *GE*

That means that I found a formula on a GE reference manual I borrowed from a former co-worker/friend.

kVAR new/kVAR rated=sqr(V new/ Vold)

Don't have the manual with me anymore but it was a power distribution reference manual an old friend from Puerto Rico let me use my first try. It was valuable for him because he warned me not to loose it, or else. I did not use the manual during the test on my first two tries so after October 2005 I sent it back to him. For what it worth, it was a very old manual from the 60's

Again, I am sorry I can't help you with this one. About your other questions *Bring Them On*. I may or may not be able to help but I will always try.

Luis


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## grover (Mar 30, 2007)

Luis said:


> That means that I found a formula on a GE reference manual I borrowed from a former co-worker/friend.kVAR new/kVAR rated=sqr(V new/ Vold)


Yeah, I had trouble with that one, too- none of my references had that formula. I would have missed it if it had come up on the exam.


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## Dark Knight (Mar 30, 2007)

grover said:


> Yeah, I had trouble with that one, too- none of my references had that formula. I would have missed it if it had come up on the exam.


Yeah. 539 was something I was doing on my job on a regular basis on my third try. I think I saw a couple of problems on the test where I had to use that experience to solve them.

Anyways, what I wanted to say now: You are a great asset to this Board Grover. Your explanation on 539 was super. Nice illustrations. Keep it up my friend.


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## tobeeepe (Mar 30, 2007)

Luis said:


> Yeah. 539 was something I was doing on my job on a regular basis on my third try. I think I saw a couple of problems on the test where I had to use that experience to solve them.
> Anyways, what I wanted to say now: You are a great asset to this Board Grover. Your explanation on 539 was super. Nice illustrations. Keep it up my friend.



I am a new member of this forum and will be taking the PE exam this April. I apologize if I am posting my first question at the wrong place:

My question is on the answer to #536. The answer given in the book is (D), i.e. voltage drop is largest at 0.707 pf, based on the approximate equation. But this approximate equation is not mentioned in the question. Looking at the problem in a simpler way, the voltage drop on the feeder is I*Z. The magnitude is the product of the two magnitudes which is = |I|*sqrt(R**2+X**2), which is independent of the power factor. (The problem says I is CONSTANT). So I think the correct answer is (A).

Any comments? Thanks in advance.


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## Platinum (Mar 31, 2007)

Sorry guys, i cleaned up my message inbox.

Awesome illustration grover......thats exactly what i needed. I kept trying to visualize a primary and secondary, but i see what you mean now. I was trying to apply the equation Ip/Is = 1/a with Ip being 8000A. But then Is=80*8000, which is wrong. It seems like you can't treat this problem as an actual transformer problem(as you stated), so i just treat it as a ratios problem: 80/1 = 8000/x.

Luis,

That's okay about problem #531.

I knew it was one of those problems that required some magical equation. I hate those!

Tobeeepe-

Maybe you should create a new topic for that problem. Good to see someone else posting questions rather than just me 

I tried working that problem and i don't fully understand it either.


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## tobeeepe (Mar 31, 2007)

Platinum said:


> Sorry guys, i cleaned up my message inbox.
> Awesome illustration grover......thats exactly what i needed. I kept trying to visualize a primary and secondary, but i see what you mean now. I was trying to apply the equation Ip/Is = 1/a with Ip being 8000A. But then Is=80*8000, which is wrong. It seems like you can't treat this problem as an actual transformer problem(as you stated), so i just treat it as a ratios problem: 80/1 = 8000/x.
> 
> Luis,
> ...


I tried a new topic before I 'replied' here. But I am getting an error message and its not allowing me to start a new topic. I will try contacting the administrator.


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## Dark Knight (Mar 31, 2007)

tobeeepe said:


> I tried a new topic before I 'replied' here. But I am getting an error message and its not allowing me to start a new topic. I will try contacting the administrator.


I think you can start a new topic after 5 posts. Not sure about that. Anyways, welcome to EB. Glad to see you made the jump here. This is a nice place to get help and after you pass you will have the chance to help others. We are a big family with all kind of characters(all good).

Don't hesitate if you have questions. We will do our best to help.


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## grover (Mar 31, 2007)

Don't give me too much credit on that illustration, I just grabbed it off a quick google image search


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## tobeeepe (Apr 2, 2007)

That's okay about problem #531.

I knew it was one of those problems that required some magical equation. I hate those!

Platinum,

Here is how the 'magical equation' is derived for #531:

3 phase power in to the capacitor kVar = sqrt(3)*Vline*Ic (equation 1)

Ic is the capacitor current, Ic = Vline/Zc (equation 2)

Zc is the impedance of the capacitor

When you substitute equation 2, in 1, you get kVar = sqrt (3) *(Vline**2)/Zc

Write this eaqauion twice, one for each voltage, new and old. Divide one by the other to eliminate the unknown Zc. You get this formula for the answer:

kVar new = kVar old *[(new voltage)**2]/[(old voltage)**2]


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## Platinum (Apr 2, 2007)

Thanks for that derivation tobeeepe....was wondering where that came from.

Great to have you aboard


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