# NCEES 518



## jdd18vm (Sep 30, 2007)

Man I give up this should be a no brainer. Why isn't this current of 75.93 Amps (multiplied by the Turns Ratio of 10) on the WYE side then multiplied by Sqrt 3 on the Delta side?

Is it normal to feel like you'll just never make it 26 days out?

John


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## Flyer_PE (Sep 30, 2007)

John,

I'm just making a quick guess at this:

There are two tricks here that you need to be careful of:

1. The wye side of transformer 1 is connected to the delta side of transformer 2.

2. The generator voltage is 13.2kV while the load is 13.8kV.

Since all of the devices here are on a 50MVA base, I set this one up as a quick pu problem.

I can give more detail later today if you still need it.

Jim


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## Dark Knight (Sep 30, 2007)

They are talking about the current on both sides of the transformer. I don't think you have to make the delta-wye conversion there because on a transformer you use the turns ratio for that so you don't worry about delta-wye.

John,

You make very good questions and I think you are very well prepared for the test. Just relax and don't be hard on yourself. If I have to tell by the kind of questions you ask and the comments you have made here I would bet my house you are going to pass.

Another thing, take a break every now and then. Don't over do it. You don't want to look like a CT curve(after saturation) for the day of the test.

I will be here today since it is raining like crazy so if you have questions let me know.

Good Luck!!!!


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## Art (Sep 30, 2007)

this may help:

Pin = Pout (neglecting losses):

1.73 Vip Ipl = 1.73 Vsp Isl (V ph-ph, I line)

and:

Vip = N Vsp

substitution and reduction yeilds:

N Ipl = Isl

you seem well prepared...do not over think this stuff, it's basic, minimum compentency...you'll do just fine


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## jdd18vm (Sep 30, 2007)

Thanks for the votes of confidence BOI and Art. It means a lot. Dont bet that house Luis...lol

Yes I think that helps Art, I read something in Grainger that I need to revisit. I'm not entirely convinced yet, because I swear I've done similar problems where I had to apply the SqRt3 whats the diff..... Spent the day on DC Machines, done.

What about this sanity check (couldn't resist looking at it as I was typing this.) as another approach. Power doesn't change, using Line current 75.93ampsX132KV=10.02 MVA. 10.2MVA/13.2 KV= 759 Amps. Or am I going INsane

John


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## Flyer_PE (Oct 1, 2007)

jdd18vm said:


> What about this sanity check (couldn't resist looking at it as I was typing this.) as another approach. Power doesn't change, using Line current 75.93ampsX132KV=10.02 MVA. 10.2MVA/13.2 KV= 759 Amps. Or am I going INsane
> John


John,

The only trick you are missing with your sanity check is that the power on each side is sqrt3*VLine*ILine. In this particular problem, the sqrt 3 divides out.

Power=sqrt3*132kV*79.53A = sqrt3*13.2kV*IGen

The thing I missed in my initial response is that they give us the transmission line current. I was a little distracted when I wrote the response.

You should do fine on the test for the simple reason that you're taking it seriously and putting in the time. Having said that, I'm not sure which is worse, watching the countdown clock to the exam or the clock waiting for results.

Jim


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## Art (Oct 1, 2007)

jdd18vm said:


> Thanks for the votes of confidence BOI and Art. It means a lot. Dont bet that house Luis...lol
> Yes I think that helps Art, I read something in Grainger that I need to revisit. I'm not entirely convinced yet, because I swear I've done similar problems where I had to apply the SqRt3 whats the diff..... Spent the day on DC Machines, done.
> 
> What about this sanity check (couldn't resist looking at it as I was typing this.) as another approach. Power doesn't change, using Line current 75.93ampsX132KV=10.02 MVA. 10.2MVA/13.2 KV= 759 Amps. Or am I going INsane
> ...


the sqrt 3 factors in when converting delta to wye when making a single ph equivalent circuit transformation...

75.93 A X 132KV=10.02 MVA

10.2 MVA/13.2 KV= 759 A

that makes perfect sense and as would be expected

N = 10 = 132/13.2 = 759/75.9


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## shellbell500 (Oct 11, 2007)

jdd18vm said:


> Man I give up this should be a no brainer. Why isn't this current of 75.93 Amps (multiplied by the Turns Ratio of 10) on the WYE side then multiplied by Sqrt 3 on the Delta side?
> Is it normal to feel like you'll just never make it 26 days out?
> 
> John


i think you may be making it hard on yourself. i looked @ the fact that it looked like a per unit (PU) problem, and found the current of 75.93 amps in PU based on 50MVA and 132KV, then took that PU and multiplied it by the Ibase on the generator side to find the actual current... using the PU eliminates all of the turns ratios, square roots of three, etc.

btw, i feel the same way - can't figure out if i'm well prepared or about to face destruction!


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## jdd18vm (Oct 12, 2007)

shellbell500 said:


> i think you may be making it hard on yourself. i looked @ the fact that it looked like a per unit (PU) problem, and found the current of 75.93 amps in PU based on 50MVA and 132KV, then took that PU and multiplied it by the Ibase on the generator side to find the actual current... using the PU eliminates all of the turns ratios, square roots of three, etc.
> btw, i feel the same way - can't figure out if i'm well prepared or about to face destruction!


You're right sometimes I feel like I over analyze. I've been in consulting doing power for over 20 years. Just never had to it like all this..lol

Hang in there Shell, these guys are way helpful.

John


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## chicago (Oct 21, 2007)

shellbell500 said:


> i looked @ the fact that it looked like a per unit (PU) problem, and found the current of 75.93 amps in PU based on 50MVA and 132KV, then took that PU and multiplied it by the Ibase on the generator side to find the actual current... using the PU eliminates all of the turns ratios, square roots of three, etc.


OK I agree this transformer problem is a great example of utilizing per unit. So, I'm trying to analyze it the way shellbell has outlined above. But I can't seem to get the current _his_ way.

Can someone give me the two or three equations needed to set up the per unit base currents?

I'm getting base current = 50MVA/(sqrt3)(132kV) on the line side = 218A but don't know what to do from there.


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## mudpuppy (Oct 21, 2007)

chicago said:


> OK I agree this transformer problem is a great example of utilizing per unit. So, I'm trying to analyze it the way shellbell has outlined above. But I can't seem to get the current _his_ way.
> Can someone give me the two or three equations needed to set up the per unit base currents?
> 
> I'm getting base current = 50MVA/(sqrt3)(132kV) on the line side = 218A but don't know what to do from there.


You've calculated I-base for the line side of the transformer. Now calculate I-base on the generator side using the same equation (Sbase will be the same--50 MVA, Vbase will be the generator-side line-line voltage).

The only other equation you need is I-per-unit = I-Amps/I-base, and you need to know that I-per-unit is the same on both sides of the transformer.

I don't have the problem handy, but it sounds like the line-side current is 75.93 A and the generator line-line voltage is 13.2 kV. If this is the case then Ibase-generator = 50000 kVA/(sqrt(3)*13.2 kV) = 2186.9 A. Using the line-side values, I-per-unit = 75.93 A/218 A = 0.347. I-generator then is I-per-unit*Ibase-generator or 0.347*2186.9 A = 759 A.

Let me know if the numbers work out right (and if this makes sense).


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## Flyer_PE (Oct 21, 2007)

chicago said:


> Can someone give me the two or three equations needed to set up the per unit base currents?
> I'm getting base current = 50MVA/(sqrt3)(132kV) on the line side = 218A but don't know what to do from there.


From there, the pu current on the transmission line side is IActual/IBase = 75.93/218.7 = 0.347

The current in pu will be equal on both sides of the transformer. The big thing to keep in mind here is that the pu value will change if the power base changes (i.e. .35pu on a 30MVA base is not the same as 0.35pu on a 50MVA base). Since all of the devices here are on a 50MVA base, there are no conversions necessary.

The next thing you need is the base current on the generator side.

This can be found the same way as for the transmission line:

IBaseGen=50MVA/(sqrt3*13.2kV)=2187 Amps

Then IActualGen=Ipu*IBase = 0.347 * 2187 = 759 Amps

As can be seen in the posts above, there are several ways to skin this particular cat. All of which will give you the correct answer. It's just a matter of which one works best for you.

Jim


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## mudpuppy (Oct 21, 2007)

Jim's solution looks prettier than mine.



IFR_Pilot said:


> The big thing to keep in mind here is that the pu value will change if the power base changes (i.e. .35pu on a 30MVA base is not the same as 0.35pu on a 50MVA base).


I was just about to add a note saying the same thing. An important part of the assumptions for per-unit to work is that all elements are on the same Sbase. This would be fairly easy for them to test for on the exam as well.


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## jdd18vm (Oct 21, 2007)

chicago said:


> OK I agree this transformer problem is a great example of utilizing per unit. So, I'm trying to analyze it the way shellbell has outlined above. But I can't seem to get the current _his_ way.
> Can someone give me the two or three equations needed to set up the per unit base currents?
> 
> I'm getting base current = 50MVA/(sqrt3)(132kV) on the line side = 218A but don't know what to do from there.


I'll give it a shot, this worked out for me, someone correct me if I am wrong here.

your off to the right start you found the Ibase=218 amps

next put into per unit=ACT/BASE 75.93/218.7=.347 pu amps

Ibase gen=50MVA/(13200xsqrt3)=2187 Amps

pu=act/base therefore ACT new=(pu)(BASE new) yields (.347)(2187)=759 amps

There are some good formulas in Schaums and I like the ones in Grainger

Let me know if you need those.

John

guess I shouldnt have stopped for lunch...


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## chicago (Oct 21, 2007)

Ok just came back from making a visit out to the test site prior to this Friday's exam. I'm so glad I did because apparently the hotel name on the admission notice and the actual hotel name on the side of the expressway was not the same. It turns out that the former Best Western is now under new ownership but nobody bothered to change the name on the building to the new hotel name! So, advice to the others taking it this Friday, check out the test site before then, especially for anyone taking it in Chicago.

mudpuppy, Jim &amp; John,

Thanks so much for the awsome feedback. Not only did you all show me how to solve it, but also reiterated the most important thing about per unit bases being equal.

Here comes another one of my hypothetical questions again (I feel I will make a good test maker one of these days!).

Suppose the line side MVA rating was actually 30 MVA and the generator side rating remains at 50 MVA.

Obviously they are different bases, so after find I_pu on the line side, I need to have the following equation before continuing to the generator side, correct?

I_pu_new = I_pu_old * (30 MVA / 50 MVA) = 0.35 * (30 MVA / 50 MVA) = 0.21 pu

Also, John, did you use equations 1.50 and 1.51 in the Grainger book (p.26-27) to solve this?


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## Flyer_PE (Oct 21, 2007)

chicago said:


> Here comes another one of my hypothetical questions again (I feel I will make a good test maker one of these days!).
> Suppose the line side MVA rating was actually 30 MVA and the generator side rating remains at 50 MVA.
> 
> Obviously they are different bases, so after find I_pu on the line side, I need to have the following equation before continuing to the generator side, correct?
> ...


That is the correct method for converting from one MVA to another. The thing to pay attention to when doing pu analysis is the bases for the system as well as the bases for the components in the system. The power base for the analysis is your choice. You just have to make sure to convert any values for the given devices to the base you have chosen. The idea is to choose your base wisely.

The big one where this can get you in the real world as well as the exam is transformer nameplate data. Example: If you hvae a transformer rated at 13.8kV and you're using in a 12.5kV system and you're base voltage is 12.5kV, you will have to adjust the transformer impedance accordingly.

Hope I haven't thrown a monkey wrench in the works here.

Jim


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## jdd18vm (Oct 21, 2007)

chicago said:


> Ok just came back from making a visit out to the test site prior to this Friday's exam. I'm so glad I did because apparently the hotel name on the admission notice and the actual hotel name on the side of the expressway was not the same. It turns out that the former Best Western is now under new ownership but nobody bothered to change the name on the building to the new hotel name! So, advice to the others taking it this Friday, check out the test site before then, especially for anyone taking it in Chicago.
> mudpuppy, Jim &amp; John,
> 
> Thanks so much for the awsome feedback. Not only did you all show me how to solve it, but also reiterated the most important thing about per unit bases being equal.
> ...



Yes those are the formulas from Grainger.

What would you do with the 0.21? Right approach but I'm lost where you're going. (but dont let that bother you, Im senile). I have to think about that. Like Jim said be careful, here we are looking for current using the power as a base.

I went back and worked on your example and it worked equally well, the thing to remember is apply the same base throughout.

here is how I approached your example on a 30MVA base

Ibase is now 30000/Sqrt3x132=131 Amps

pu=75.93/131=.579

Ibase gen= 30000/(sqrtx13.2)=1312

act=1312x.579=759.7

Another example on 100 MVA Base

Ibase=100000/(sqrt 3x132)=437Amps

pu=75.93/437.4=.1736

Ibase gen 100000/(sqrt3x13.2)=4373

act=.1736x4373=759

Thank you Shell for mentioning this

again make sure this is blessed by one of our mentors...lol I'm fried i may have typoed as well.

John


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## cabby (Oct 21, 2007)

Could you guys take a look at this and see if I am on the right track? After reviewing all the comments, I believe I was able to get the hang of this one.

You guys are great,


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## chicago (Oct 21, 2007)

chicago said:


> I_pu_new = I_pu_old * (30 MVA / 50 MVA) = 0.35 * (30 MVA / 50 MVA) = 0.21 pu


Sorry John, I wrote the values in the equation wrong. I didn't mean to confuse you, I just didn't proofread. Let me show you how I came up with the equations using 30 MVA on the line side.

Line:

Ibase = 30MVA/sqrt3*132kV = 131A

Ipu = 75.92/131 = 0.58A

Looking from the generator side now, apply EERM equation 34.38:

I_pu_new = 0.58(30MVA/50MVA) = 0.35

Generator:

Ibase = 50MVA/sqrt3*13.2kV = 2187A

Ipu = Ipu_new*Ibase = 0.35*2187 = 759A

So I essentially _adjusted_ the base from 50MVA to 30MVA on the generator side which then ended up with the same answer.


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## jdd18vm (Oct 21, 2007)

I think we've got it. arty-smiley-048:

Cabby you can see pu works on a per phase basis as well wtg.

Chi got it now, I did that as well. Important thing as Jim said be consistent throughout.

John


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## mudpuppy (Oct 21, 2007)

Chicago and jdd: Both of your methods look like they work. I personally use the method jdd is using because I find I make fewer errors that way (in chicago's method I always forget if it's Sold/Snew or Snew/Sold). But as always, you should use whatever makes the most sense to you.

Cabby: looks like you've hit the nail on the head. I've always calculated per-unit problems on a phase basis too (it takes more calcs, but I have to remember fewer equations that way).


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## grownupsara (Oct 22, 2007)

Maybe I'm over-simplifying this, but I just used eqn 28.11 from EERM (basic equations for xfmr turns ratio): IpVp=IsVs, then solved for Ip=Is*Vs/Vp=75.93A*132kV/13.2kV=759.3A


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## Flyer_PE (Oct 22, 2007)

grownupsara said:


> Maybe I'm over-simplifying this, but I just used eqn 28.11 from EERM (basic equations for xfmr turns ratio): IpVp=IsVs, then solved for Ip=Is*Vs/Vp=75.93A*132kV/13.2kV=759.3A


You're not oversimplifying it at all for the solution to the problem in the NCEES book. This thread just took a right turn into PU analysis since that particular problem/circuit is a good example for use in playing with the concept.

Jim


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## grownupsara (Oct 22, 2007)

IFR_Pilot said:


> You're not oversimplifying it at all for the solution to the problem in the NCEES book. This thread just took a right turn into PU analysis since that particular problem/circuit is a good example for use in playing with the concept.
> Jim


Oops, okay. Sorry for just jumping in on the discussion!


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## EngrinSF (Mar 11, 2018)

Hello,

I wanted to revisit this problem since I am a bit confused on the terms used. I calculated the load current in the generator as the full load current: 

IGen=50MVA/(sqrt3*13.2kV)=2187 Amps

and chose this answer which turns out is wrong. Why is this  answer not right?


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## knight1fox3 (Mar 11, 2018)

Whoa! @ptatohed, this is quite possibly the largest time bump I've seen to date. Almost 11 years! Well done!  :appl:


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## ptatohed (Mar 11, 2018)

LOL.

Grave Digger!


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## EngrinSF (Mar 12, 2018)




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