IV.A.5. Fault Current Analysis Example - NCEES Electrical Power PE Exam

[@pqwwhuongnguyen4490]

Compared to all the other PE reference books and videos I have watched, your method is the best. It pretty much cleared up any confusion I had.

[@electricalpereview]

Thanks for the kind words, glad it helped!

[@randyw84832]

23.35kA

[@DamianWSower]

One of my favorite topics covered by Zach! His analysis cuts to the heart of what to expect on the PE exam and his patient explanation will guide you carefully through the process of understanding. I highly recommend his course! If you aren't familiar with fault contributions of transformers, generators and lines to short circuit current (MVA or PU method) this is a must have video.

[@electricalpereview]

Thanks Damian!

[@KarthikVijapurapu]

I had the greatest confusion about the pu method. Zach's class and his examples just declutter every doubt you have on these methods.

[@mikemuchesi9895]

This video demystifies fault current analysis. The step by step approach and understanding fault contributions was key for me. Thank you.

[@electricalpereview]

Thanks Mike, glad you enjoyed.

[@caseywhitcher4576]

Zach's explination for Fault Analysis just makes sence. I took other courses before, such has the Test Master's Power Review Course, and the material covered here just wasn't as straight forward. Thanks Zach, your course has really helped these concepts to make sence to me.

[@isaacmatute5325]

This video helped me clarify how to solve Fault Currents and find a way that makes it understandable. I especially like the simplicity of Zach's explanation to help decide which method suits me better.

[@electricalpereview]

Thanks Isaac

[@danbeesmer8554]

Another great video from Zach. Fault Current Analysis isn't so bad after watching and listening to the explanations followed by this great example. Thank you Zach!

[@electricalpereview]

Thanks for the kind words Dan, glad you enjoyed it

[@katelynnjourdan4593]

Showing both methods side by side helped me clear up confusion for addressing both types of problems.

[@yanasseri]

Wow. I would not have thought to use the PU method with MVA bases. I'm used to using that for impedance values to find fault current. Thanks Zach!

[@justinobenchain93]

This is an extremely helpful video, and really helped make this topic clear for me. Fault current analysis is a big part of the PE exam, and Zach does a great job making it intuitive.

[@electricalpereview]

Thanks for the kind words Justin, glad you enjoyed.

[@DB-iv6ks]

Excellent side-by-side comparison of PU vs MVA, I had not used either in a long time and only had notes on the Point-to-Point Method

[@jennifernolan7400]

I've watched this video a few times...its helped me understand per unit method. Its not as scary to approach this kind of problem. Thank you

[@TuberKMT]

Holy cow, that is so much easier than per unit! Thank you!

[@electricalpereview]

Yes, much easier!

[@AlbertaDMode]

Ahead of the solution portion, I paused and solved using both methods selecting transformer values as a base. Got the same results. Another thing to remember is MVA method is only good for a three-phase short circuit.

[@rockerkid1006]

This is also one of the topics Zach does well. I went from being iffy on these types of problems, to pretty much being able to understand and doing it both PU method and MVA method.

[@electricalpereview]

Thanks!

[@alisondecleene6940]

So great to see PU and MVA solved side by side.

[@electricalpereview]

Absolutely. While the per unit method is a little tougher at first, seeing them side by side really helps to understand both.

[@JuanpysGoProVideos]

this video definitely helped me better understand fault current analysis, this plus along side with the course review module. thanks Zach.

[@ralieghcowboy]

This was a good video. I have trouble using the PU method. This video helps me understand it better even though i perfer the MVA method.

[@fredericosimonflores8998]

Very detailed Fault Current Analysis explanation recommended to try the premium course.

[@electricalpereview]

Thanks Frederico, we enjoyed having you as a student in our online program this semester. Best of luck on the PE exam.

[@derekedwards2680]

Totally agree, MVA method makes more sense than the PU method. But its good to see how both methods can be used.

[@issybella7]

Enjoyed learning this topic from you!

[@hardefarmy7532]

Is it 23.35?

[@komalpatel712]

getting the same answer as 23.348KA....

[@SF-fb6lv]

So no one got the part where the fault current is wrong. 19.41MVA / (SQRT(3) x 480V) = 23,347kA, not 22,350.

[@amitbhatia5681]

Sir what i meant to ask is that while calculating fault current we need base kva and maximum fault at utility. How do we get those

[@electricalpereview]

Hi Amit, in this problem the fault duty of the utility is given. This value is the fault power in volt-amps available at the utility connection feeding the system. It would be equal to the MVA available at the utility divided by the equivalent percent impedance at the utility if given. When given fault duty, this calculation has already been calculated.

[@eEnergyVA17]

actually Ifault = 23.35kA, please check solution to guide people. Thank you.

[@electricalpereview]

Hi Reyred, that is the correct answer same as shown in the video for both the MVA and Per Unit Method.

[@midoemara867]

The answer shown in your video is 22.35kA, not 23.35kA. The methodology is correct, the final answer in the video is not. I spent 15 minutes trying to figure out why I arrived at a different answer than you did in your video. It is a good and beneficial video and thank you for spending the time to make it. I would maybe add a comment at end video with the correct answer so that others do not spend time like me trying to figure out where the issue is.

[@rafihaq3485]

@@midoemara867 soo whats the solution

[@berniesayson6612]

@@rafihaq3485 Mido is right it seems typo error only, the actual answer is 23.348 kA

[@deeds1463]

why are the MVA values calculated as parallel equipment even though they are in series?

[@electricalpereview]

Hi Collin, good question. The short answer is that since the MVA method uses fault duty values, which are calculated by dividing rated power by percent impedance, we are really working with admittance values of each device/machine. Since admittance is the inverse of impedance (Y = Z^-1) the rules for summing the values are opposite of the rules we typically follow for impedance when working with series or parallel components.

[@deeds1463]

@@electricalpereview ahhh i see, thanks!

[@amitbhatia5681]

sir how do we get to know the fault level at utiity ??

[@electricalpereview]

Hi Amit, draw the fault located between the utility and the transformer instead of showing it after the transformer. This will change how you calculate your values in either the per unit method or MVA method since a fault between the utility and transformer will put them in parallel with each other, instead of in series.

[@sohailjanjua123]

Hi Zach, Your lecture is excellent. Happy new year . Thanks

[@electricalpereview]

Thank you, glad you enjoyed it

[@sungmopark0405]

Can you explain why symmetrical fault L-L-L-G and L-L-L is same?

[@sohailjanjua123]

Hi Zach, I like your lecture

[@electricalpereview]

Glad you enjoyed

[@yuechen2421]

why xfmr duty pu= (1.5mva/0.05)*(1/55mva)?

[@adambodom]

any video on harmonic analysis?

[@electricalpereview]

Not yet but soon. It will be published inside the course.

[@tinaquayum8910]

How did you got 30mva

[@electricalpereview]

30 MVA is the fault duty of the transformer. You can calculate the fault duty of a transformer by dividing it's rated power by it's percent impedance. 1.5 MVA / 5% = 30 MVA.

[@khanhlankhanhlan9731]

Character In the video It's great, I like it a lot $$

[@danyjoshua6434]

Thank you.

[@majeds2445]

U r the best

[@electricalpereview]

Thank you! Glad you enjoyed.

[@MrDands]

Wasn't clear why you add apparent power in parallel, good video though!