# Graffeo Ex 51 - Using MVA Method



## TWJ PE (Aug 19, 2015)

Something seems to be tripping me up on this problem.

I've tried to solve using the MVA method but I'm missing something.

Ex. 51:

Two generators are connected through two transformers to a bus that is connected to a transmission line. The line is open at the far end, where there is a three phase fault. Prior to the fault, the voltage at the end of the transmission line is 230KV. The ratings are per unit reactances of the equipment based on the equipment ratings are shown below:

G1 = 700 MVA, 12KV, X''d=X2=.11, X0=.06

G2 = 600 MVA, 12 KV, X''d=X2=.14, X0=.08

T1 = 700MVA, 12/230KV, delta/wye, X=.14

T2 = 600MVA, 12/230KV, wye/delta, X=.12

Transmission line = base of 1000MVA, 230KV, X1=X2=.19, X0=.45

All wye winding neutrals are solidly grounded.

Find fault current.


----------



## TWJ PE (Aug 19, 2015)

Thank you in advance!


----------



## sebann (Aug 19, 2015)

I have an issue with that example as well. Everything is pretty straigh forward except I dont understand why the impedances for both generators are not recalculated based on new voltage base but just new 1000 MVA base.

Any advice?

Thank you


----------



## ZcoreX29 (Aug 19, 2015)

Using the MVA method, I came up with 6506A. Since you don't say how you are calculating the current using that method, I can only guess as to where you are going wrong. One possibility is that you are calculating the MVA for the transmission line using the per unit impedance instead of the actual ohmic impedance. What are the MVA values for each item in your circuit?


----------



## TWJ PE (Aug 19, 2015)

I get Isc=(1000*1700))/(sqrt(3)*230)=4267


----------



## ZcoreX29 (Aug 19, 2015)

sebann said:


> I have an issue with that example as well. Everything is pretty straigh forward except I dont understand why the impedances for both generators are not recalculated based on new voltage base but just new 1000 MVA base.
> 
> Any advice?
> 
> Thank you


For the per-unit method, you specify a MVA base for the entire system and a kV base for each section of the system where the transformer acts as the diving line between sections. The kV base for each section is typically the kV rating of the transformer (12 kV and 230 kV). Since the problem told you that the generator impedances were derived from the equipment ratings (12 kV), there is no need for a voltage conversion.


----------



## ZcoreX29 (Aug 19, 2015)

W9TWJ said:


> I get Isc=(1000*1700))/(sqrt(3)*230)=4267


Where do you get the 1000 and 1700?


----------



## TWJ PE (Aug 19, 2015)

1700 from 7243 (from gen and xfmr) || 2222 (transmission)


----------



## TWJ PE (Aug 19, 2015)

Originally thought it was the transmission line tripping me up but now I'm questioning everything.


----------



## ZcoreX29 (Aug 19, 2015)

I'm not sure how you obtained 7243 and 2222. What are your short circuit MVAs for each item in the circuit (G1, G2, T1, T2, Tline)?

You're using the zero sequence impedance for the t-line. For a symmetrical fault (3 phase fault) there is no zero sequence current therefore zero sequence impedance is not used. Try using the positive sequence.


----------



## TWJ PE (Aug 19, 2015)

G1=6363

G2=4285

T1 &amp; T2=5000

Tline=5263


----------



## TWJ PE (Aug 19, 2015)

Nevermind... I was doing something totally off base.

G1||T1= 2799

G2||T2=2307

2307+2799=5106

TLine=1000/.19=5263

5106||5293=2584

Isc=(1000*2584)/(sqrt(3)*1000)=6505

Yeah... I'm not sure what the heck I was doing either. Thanks.


----------



## sebann (Aug 20, 2015)

ZcoreX29 said:


> sebann said:
> 
> 
> > I have an issue with that example as well. Everything is pretty straigh forward except I dont understand why the impedances for both generators are not recalculated based on new voltage base but just new 1000 MVA base.
> ...


Thank you!!


----------



## KatyLied P.E. (Aug 25, 2015)

For what it's worth Graffeo is very good in responding to questions. Nothing wrong with asking here either. Everyone benefits then.


----------



## StinkyTofu (Sep 25, 2015)

W9TWJ said:


> Nevermind... I was doing something totally off base.
> 
> G1||T1= 2799
> 
> ...


 W9TWJ,
Where did you get that 1000 from (in red above)? Shouldn't that be the line to line voltage (230kV)?


----------



## TWJ PE (Sep 29, 2015)

Yes, you are correct - typo. Good catch.


----------



## Limamike (Aug 19, 2016)

You know I have aske graffeo a few times on this one and 49 and no response.  Has anyone one been able to do 49 via MVA method?  Thanks


----------



## TWJ PE (Aug 25, 2016)

I don't have my book any more. Will you post the problem?


----------



## bripgilb (Feb 15, 2018)

@doubleoinfo07 posted the question here 





I'm also interested in the MVA method for this question.

Thanks,

brip


----------



## bripgilb (Feb 15, 2018)

doubleoinfo07 said:


> Can anyone explain how to do Graffeo example 49 using MVA method.
> 
> The problem reads: The following system was operating at no load and rated voltage, when suddenly a tree fell causing a single line to ground fault at bus 2. Find the fault current.
> 
> ...


----------



## bripgilb (Feb 15, 2018)




----------



## bripgilb (Feb 15, 2018)

Thanks!


----------



## supra33202 (Feb 16, 2018)

I am trying to use the MVA method.

Sg = 200MVA/.25 = 800MVA

Stx1 = 200MVA/.12 = 1666.7 MVA

Since they are in series, treat them as resisters in parallel circuit.

S=((1/800MVA) + (1/1666.7))^-1 = 540.5MVA

Isc = S / sqrt(3) * V = 540.5 MVA / sqrt (3) * 230kV = *1356.8 A*

Graffeo's answer is I = *2791.34 A*

What did I do wrong?

Thanks!


----------



## rg1 (Feb 18, 2018)

Guys,  I would like to tell you that derivation of formula for distribution of loads of two sources will be good idea to solve this problem. The starting point of this formula is distribution of loads (currents ) in two impedances in parallel. When two impedances are in parallel the voltage across them is equal. So it gives you. I1*Z1=I2*Z2. Rest is maths. Let me know if you want to understand the basics and if you are not able to get it by yourself. But I am sure, once you understand this, no matter what info the question gives you, will reach the correct answer.


----------



## rg1 (Feb 18, 2018)

One more thing I would like to mention here. If anyone of you need any help to solve a problem, I am here to help. I would like you guys try first to understand among yourself and in case the student in you is not satisfied, just put me in loop and I will chip in the discussion.


----------



## bripgilb (Feb 18, 2018)

@supra33202 @rg1

Thank you both for trying to find the solution to this problem.

RG1 I have attached my work of the problem.  

My result is Ifault= 2281.7A

The books result = 2791.3A

1.) Do you agree with my MVA approach?

2.) Why do you think I'm 500A off?

3.) Could you explain a little more on "formula is the distribution of loads..." I think I was having a little trouble understanding.

Best regards and thanks,

brip


----------



## bripgilb (Feb 18, 2018)

rg1 said:


> Guys,  I would like to tell you that derivation of formula for distribution of loads of two sources will be good idea to solve this problem. The starting point of this formula is distribution of loads (currents ) in two impedances in parallel. When two impedances are in parallel the voltage across them is equal. So it gives you. I1*Z1=I2*Z2. Rest is maths. Let me know if you want to understand the basics and if you are not able to get it by yourself. But I am sure, once you understand this, no matter what info the question gives you, will reach the correct answer.


----------



## Zach Stone P.E. (Feb 19, 2018)

Hi Brip,

If I'm looking at the correct example in the latest edition of Graffeo's book (July 2017 printing), it is a single line to ground fault.  The MVA method you are attempting to use is only for worst case scenario three phase faults. Symmetrical components needs to be used with the single line to ground fault equivalent circuit model.


----------



## bripgilb (Feb 19, 2018)

@Zach Stone, P.E.  Hey Zach!

Thank you for your reply!

I was actually thinking the exact same thing when I started working the problem!  But, we all kept asking for an MVA method on the problem so I figured there had to be one.

I watched your MVA method on youtube to help me get through this problem! Glad that is confirmed.  Symmetrical components takes a bit of muscle.  What are the chances to see a single line to ground fault current analysis question on the PE?

Thanks,

Brip


----------



## Zach Stone P.E. (Feb 19, 2018)

bripgilb said:


> @Zach Stone, P.E.  Hey Zach!
> 
> Thank you for your reply!
> 
> ...


Hi Brip,

Glad you enjoyed our MVA method video on youtube for three phase fault analysis. 

To be honest, symmetrical components only appears complicated at first but once you get the hang of it, it's actually not too bad. 

As long as your circuit analysis is strong, you can usually solve them without too much brain muscle as long as you use the correct equivalent circuit. 

With symmetrical components, each type of fault has it's own unique single-phase equivalent circuit. As long as you have each one in a handy reference, you are will be good to go if you put in the values where they belong and solve accordingly. 

To answer your question, according to the new 2018 NCEES specs, there are a total of 9 questions in II.A. Circuit Analysis and a total of 7 topics, one of which of course is symmetrical components. 

I'd say it is a safe bet to expect at least one or two symmetrical component questions but they will most likely be pretty basic. 

I don't think we've put out any free videos on youtube for symmetrical components, so if it helps, I could probably upload one or two from our online review course for the electrical PE exam as free promo material.


----------



## bripgilb (Feb 19, 2018)

@Zach Stone, P.E.  I just gotta remember what to do with the different types of equipment. And of course the steps.

Once I worked through it a few times I think I started to get the hang of it.

I think we'd all appreciate any practice resources that you would would be willing to share.

Best regards,

Brip


----------



## Zach Stone P.E. (Feb 20, 2018)

bripgilb said:


> @Zach Stone, P.E.  I just gotta remember what to do with the different types of equipment. And of course the steps.
> 
> Once I worked through it a few times I think I started to get the hang of it.
> 
> ...


Hi Brip,

We just added the following video to our YouTube Channel, it's a single line to ground fault symmetrical components example. 

It's not the same as this particular Graffeo example since it's for a transmission line (and not multiple generators/transformer etc). 

In my opinion, this is more comparable to what you can expect on the PE exam. 

This example is comparable to, and the same method you would need to solve the NCEES practice exam question #514. 



The video was just uploaded so the quality max is only at 360p but if you give it a few minutes it will be available in 720p HD once youtube finishes encoding it.


----------



## rg1 (Feb 20, 2018)

bripgilb said:


> @Zach Stone, P.E.  I just gotta remember what to do with the different types of equipment. And of course the steps.
> 
> Once I worked through it a few times I think I started to get the hang of it.
> 
> ...


Hi Brip,

I am sorry, I was a bit occupied. As clarified now, this is a question involving the concept of symmetrical components to resolve non symmetrical faults. I made a mistake in getting the question earlier. The formula I1Z1=I2Z2  should  be  used for distribution of load currents and not for fault currents. It is mainly for sharing of load by  two sources.

Now coming to this question. 

Since the transformers are delta star with star point grounded and it is a single line to ground fault on star side, both the generators will not feed to any fault current. The fault current will only circulate in the delta winding of the transformer compensating the fault ampere turns  on star side. Having said this we conclude that the fault is fed only by transformers.  Generally in dealing with symmetrical components we deal in pu values of V, I and Z; of course no one prevents us from using MVA method. I do not expect a question on both concepts mixed together in PE exam.   Before we discuss solution to this question, I would like you to learn symmetrical components. In a single line to ground fault the fault current in pu is 3 times Voltage (pu) divided by sum of positive , negative and zero sequence impedances, assuming fault impedance is zero. In this case there are two wings feeding the fault, left side only has one transformer so impedances of Xmer only need to be taken and Right side, sum of impedances of TL and Transformer need to be taken. Impedances here means all three Z1, Z2 and Zo. We add these current pu values as they are fed by two sources in parallel and then convert to current to amperes or MVA as the case may be.

Hope it is of some help to readers.


----------



## kshitijs13 (Mar 29, 2018)

My answer comes about 5.879 kA for Three Phase Fault and 5.890 kA for Single Phase to Ground Fault.


----------

