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10 GoodAbout Zach Stone, P.E.

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Lead Instructor for Electrical PE Review INC (EPR)
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http://www.electricalpereview.com
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Male

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USA

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Teaching Electrical Engineering concepts
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Electrical Power, Industrial Controls, Field Troubleshooting, Construction Management, Turbo Generators and Power Generation, 24hr Production Facilities

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PE

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TI

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Electrical
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Preparation for October 2018 Power PE Exam
Zach Stone, P.E. replied to chener16's topic in Power Exam Sub Forum
Glad you enjoyed it. If you need anything feel free to reach me directly via email at zach@electricalpereview.com 
Zach Stone, P.E. started following Complex Imaginary Volume 1, Problem 22, NCEES Question 530, Complex Power Current Conjugate S=VI* and and 6 others

Correct. And since Vpu = 1, the per unit short circuit current is equal to the inverse of the equivalent per unit impedance, or: Ipu = 1/Zeqpu. In my opinion, these type of simple fault problems are easier to solve with the MVA method, including this one. We can use it to verify the solution. For example, the total apparent power at the fault is 15.38MVA: Sf = 40MVA//(1,000kVA/4%) Sf = 15.38MVA The magnitude of the shortcircuit current at the faulted 480V bus is 18.5kA: Isc = 15.38MVA/(√3·480V) Isc = 18.5kA

Complex Power Current Conjugate S=VI*
Zach Stone, P.E. replied to ellen3720's topic in Power Exam Sub Forum
Hi @ellen3720, The first formula uses complex numbers and solves for single phase complex power (both magnitude and angle) as the product of the complex phase voltage and the conjugate of the complex phase current: S1Φ =VpIp* S1Φ = S1Φ<θ To solve for three phase power, you can multiply the same formula by three, which also uses complex numbers and the current conjugate: S3Φ = 3•S1Φ S3Φ = 3•VpIp* S3Φ = S3Φ<θ The third formula uses magnitudes only, and just solves for three phase apparent power, which by definition is the magnitude of three phase complex power without the angle: S3Φ =√3•VL•IL Typically when using this formula, you'll obtain the power angle from the inverse cosine of power factor: θ = cos1(PF) S3Φ = √3•VL•IL<cos1(PF) The following cheat sheet gives examples to help to avoid similar mistakes with power formulas. You can print it out and take it with you to the PE exam: http://www.electricalpereview.com/biggestmistakecommonlymadethreephasepowerformulas/ 
Hi @rmsg, My answer would be the same as the text that you quoted for the same reasons. While I've never personally been asked this before nor read anything specifically stating magnifying glasses I would consult official representation from either NCEES via email or from their documentation. I'd consider contacting them since I do not believe the examine guide mentions magnifying glasses. My best guess is that it would be okay, however, I do not represent nor am I affiliated with NCEES.

I was going over problem 537 of the Pe sample exam . To get a good understanding of this question would this be like the example you had used to describe Icharging current . Which is the line length act as shunt capacitor to ground therefore the line length , the voltage to ground and the radius of the conductor would play a part in Icharging current. If this is the case shouldn't the radius of the conductor be included in this type of problem?

Index for "Electrical Machines, Drives, and Power Systems" by Theodore Wildi
Zach Stone, P.E. replied to jnspark's topic in Power Exam Sub Forum
I read a lot of electrical engineering books. Wildi's is by far one of my favorites, and in my opinion, one of the best reference books you can take with you to the exam. The best way to get the most out of any reference book specifically for the PE exam, is to get familiar with it ahead of time. The order that material appears in each chapter is very intuitive and builds on each other. The best thing to do is skim every chapter that is a subject on the exam and note the headings, subheadings, formulas, and diagrams so that you'll have a general sense of the information laid out in each the chapter. When you come across pain points like areas you've struggled with, or particular points of interest as it applies to the PE exam, stop and tab the page and read a little more indepth. You could probably skim the entire book in about two hours and have a general sense of what's in it so that you know where to look. "This is a tough question but I remember a similar one in Wildi's book. This is a sync. generator question, let me start in the sync generator chapter and flip quickly until I find the heading that fits the problem." 
Index for "Electrical Machines, Drives, and Power Systems" by Theodore Wildi
Zach Stone, P.E. replied to jnspark's topic in Power Exam Sub Forum
These are my sentiments exactly. The book is essentially his Magnus opus and life's work. It's everything he figured out along the way when all of this stuff was new. Imagine that. Think about how old induction motors are. In his career, it was all groundbreaking and state of the art technology that was being figured out by trial and error. Think mad scientists and crazy laboratories. It's pretty crazy when you stop and think about it. Now, all this is just part of our modern everyday life that no one ever even gives a second thought to except for engineers who apply it towards practical problems. 
Max Power Transfer Question  Updated NCEES Power #531
Zach Stone, P.E. replied to jnspark's topic in Power Exam Sub Forum
Hi @jnspark, the equation for reactive power that you're trying to use (quoted above) calculates the reactive sending power. NCEES is asking for the reactive power dissipated (lost) in the line reactance when the sending active power is at it's maximum. Slight but important difference. "When the sending power is at its maximum" tells you what angle to use for your complex voltages. NCEES was even (surprisingly) kind enough to tip you in the right direction of what angle to use for delta if you are unfamiliar with max power conditions (sin(90º) = 1). To calculate the power lost in the line reactance just treat it like a two terminal device. If you know the voltage across a two terminal device and the impedance of the two terminal device, then you can calculate the complex power consumed by the impedance (or supplied by it if the device is a source): S = V^2/Z (O+jQ) = V^2/(0+jX) Q = (VAVB)^2/X Where VA and VB are the linetoline complex voltages with VA leading VB by 90º, X is the per phase line reactance, and Q is the three phase reactive power consumed by the per phase line reactance. Alternatively, you could plug in the linetoneutral complex voltages VA_1ø and VB_1ø with the same angular displacement of VA_1ø leading VB_1ø by 90 degrees to solve for the per phase reactive power loss, then multiply by three to calculate the total three phase reactive power loss since it is a three phase transmission system: Q_1ø = (VA_1øVB_1ø)^2/X Q_3ø =3·Q_1ø Q_3ø = 3·(VA_1øVB_1ø)^2/X Or you could solve by calculating the current first using Ohm's law then solve for reactive power like others have pointed out in this thread similar to the NCEES solution. All roads lead to Rome. 
Hi @rmsg, NCEES tests on codebooks the year after they have been updated. That means that the 2018 NFPA 70E will not be tested on until 2019. For the Oct 2018 PE exam the 2015 edition of the NFPA 70E is the correct book.

Glad it helps. It's fun to speculate, but it's always nice to know what the official rules are.

This is a question that pops up often. Your best resource to answer what is/is not allowed will always be the official examinee guide by NCEES. Otherwise, rumors and hearsay may confuse or mislead you. Here is an official link to the examine guide: https://ncees.org/wpcontent/uploads/ExamineeGuide_June2018.pdf

Hi @kduff70, The trick is realizing that it is actually complex number (or vector) addition and not just adding magnitudes. The surprising thing when you do this is that the longest leg of the center tapped delta is not the connection with the greatest voltage magnitude. Here is a cheat sheet PDF of the center tapped delta directly from our online course from our chapter on Transformer Connections. It has the diagram, proper labeling, formulas, phasor diagram, and an example of a 240V/208V/120V center tapped delta. Feel free to print it out and take it with you to the exam: Center Tapped Delta  Electrical PE Review.pdf It's a large PDF. When you open it don't forget to zoom in to see the details. Also, if you need help with the open delta connection which is even more confusing, here is a detailed free article that will be a HUGE help (you can also print this out and take it with you to the exam): Electrical PE Review  Open Delta Transformer Connection Hope this helps!

Question For Those Who Have Taken Exam in TN
Zach Stone, P.E. replied to MEtoEE's topic in Power Exam Sub Forum
That is my understanding. Post it notes already used for book tabs are okay, blank "new" or "unused" postit notes are a no go. I'd be in trouble. I NEVER use standard book tabs. Too expensive and I can never keep enough around (I read voraciously and tab A LOT!). I have a custom way of making book tabs using postit notes. I've considered making a video demonstration of it since it is such a money saver and you never run out of book tabs. 
Preparation for October 2018 Power PE Exam
Zach Stone, P.E. replied to chener16's topic in Power Exam Sub Forum
Thanks for the kind words and the mention. Glad you enjoyed your time with us! 
Complex Imaginary Volume 1, Problem 22
Zach Stone, P.E. replied to ellen3720's topic in Power Exam Sub Forum
Hi @ellen3720, When two transformers are connected in parallel, the problem is very similar to a standard current division circuit from circuit analysis. I find that it helps to see why Z2 ends up in the numerator to really understand this problem. Let's derive it. In the example you posted above, the total load current will equal the sum of T1's and T2's current contribution since they are in parallel (KCL relationship): I_L = I_1 + I_2 Assuming both transformers have the same ratio and voltage (V1 = V2 = V), we can substitute Ohm's law for I_1 and I_2 using the impedance of each transformer (Z1 and Z2) and the voltage (V): I_L = I_1 + I_2 I_L = V/Z1 + V/Z2 Next, we can add both fractions: I_L = V/Z1 + V/Z2 I_L = (V/Z1)•(Z2/Z2) + (V/Z2)•(Z1/Z1) I_L = (V•Z2)/(Z1•Z2) + (V•Z1)/(Z1•Z2) I_L = V(Z1+Z2) / (Z1•Z2) Then, plug in I_1 = V/Z1 since we are solving for I_1: I_L = V(Z1+Z2) / (Z1•Z2) I_L = I_1 • (Z1+Z2) / Z2 Now, solve for I_1 (This is where Z2 ends up in the numerator): I_1 = I_L • Z2/(Z1+Z2) Finally, we can plug in the values from the problem and check our answer with your provided solution: I_1 = I_L • Z2/(Z1+Z2) I_1 = 11,500A • 4.75%/(5.25%+4.75%) I_1 = 5462.5A Hope this helps.