# Kaplan Power Afternoon Test Questions



## chicago (Sep 24, 2007)

For those who have access or are familiar with the Kaplan Power PM Exam Solutions, I have a few questions:

1) Problem 4.2, how did they convert the delta--&gt;wye impedances to be 4Z/3?

2) Problem 4.4, why is the cos 22.62 degrees lagging? Is it because it is positive so that means inductive?

3) Problem 4.16, why is I_1=(Y/2)*V®?

4) Problem 4.17, how exactly do you calculate cosh (0.3560/_81.748 degrees)?

5) Problem 4.25, why is R_sc assumed to be split equally between the primary and secondary windings?

6) Problem 4.30, can anyone solve this per unit problem in a less complex method than they way Kaplan proposes?

7) Problem 4.38, how does I_l have cos(PF) = cos(0.4) as its angle? Is it due to 0.4 = R/X = 5/12?

I know that's a lot of questions, but I'm getting desperate and wound-up as the test date is inching closer by the day!

Thanks a bunch!


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## chicago (Oct 2, 2007)

Already went once. Going twice now for any help for the questions??


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## Dark Knight (Oct 2, 2007)

chicago said:


> Already went once. Going twice now for any help for the questions??


Chicago, I just saw this. I will get back to you as soon as I have the chance. It is late and I am on my way to my bed. Tomorrow I will try to help.For starters I think there is a formula for the conversion of the impedances. Don't remember from the top of my head but will check.Stay tuned.


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## jdd18vm (Oct 3, 2007)

Chi

Im at work and need to do a few problems myself. Will answer what little I know. Some of these threw me too.

1) Look at the diagram on page 127. They converted the Delta Z's to Wye (1/3 the Delta Z) then added the additional series Z for a total of 4Z then divided again by the 3 lines. They converted and did the math on the 3 phase total.WTF?

I kept the analysis on a per phase 1/3 Z (4.333) plus 1Z (13) 17.333, so that 13.6/Sqrt 3=7852 Volt/17.3333 Ohm got the 453 Amps.

Didn't do the transmission Line one yet.

Regarding 5, look at an earlier post, I think Art explained that best, I calculated the Total R as i recall and then solved Rp or Rs forget which.

Regarding 30, I know what you mean, dont get overwhelmed by it. I think it important to realize thats 5 questions, and a lot going on. Understand whats going on in each section, you need to perform the same calcs regardless of the method.

Sorry I cant be more help.

John


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## jdd18vm (Oct 3, 2007)

disregard my comment about the 5, you KNOW about that post you posted it. Did you see Arts answer?


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## Ilan (Oct 3, 2007)

Let me try to answer these questions. Gurus please correct me if I am wrong..

4.2. jdd18vm has already given the answer to this. delta is converted to wye (Z/3) and added to the series resistance (Z) to give 4Z/3. Line voltage AB is considered as the reference vector, so when converted to phase voltage, it is divided by sqrt(3) and lags the line voltage by 30 deg. Iaa1 = [(13600(-30deg)/sqrt(3)]/(4Z/3) = 453(-52.62 deg)

4.4. The key here is the power factor as seen by the source. When the complex power delivered from the source has +VAR --&gt; lagging, -VAR --&gt; leading.

4.6. I guess its Kaplan's way of doing it. If you look at the figure in page 135, they show the medium line model. Current through the admittance = Voltage across the admittance * admittance. Anyway, I used the OTHER book's equation for the medium line model and got the same solution. Vs = (1+YZ/2)*Vr + (Z(1+YZ/4))*Ir

4.17. cosh(A+iB) = coshA*cosB + i sinhA*sinB

sinh(A+iB) = sinhA*cosB + i coshA*sinB

I hope this is what you were looking for..

4.25. It is usually assumed that the leakage reactance (Xs) is divided equally between primary and secondary. I guess Kaplan used the same assumption for resistance too. In the solution it should say, R1 + a^2*R2. I am not sure whether we can theoritically derive separate impedances for primary and secondary. :dunno:

4.30. :dunno:

4.38. It is a typo. it is actually arc cos (.9 lag) = -25.84

No worries Mate! We will get through this....

Good Luck.

Ilan.

:thumbs:


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## Ilan (Oct 3, 2007)

4.25. I referred EERM as per jdd18vm's comments on the other post. It states that transformers are normally designed with R1 = a^2*R2 and X1 = a^2*X2. EERM derived that equation R1 = Psc/2*Isc^2 with this assumption.


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## Dark Knight (Oct 4, 2007)

Went home late last night and did not have time to check on this. I am glad Ilan took care of you. Let us know if need extra help.


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## chicago (Oct 6, 2007)

Ilan, jdd18vm, and BringItOn thanks a lot for your answers.

You guys were very helpful in making me understand the questions I had.

I'm still working through Problem 4.30 (the generator/trans. line/load per unit one). It's like weeding through weeds with a weedwacker! I sure hope there's a much simpler per unit problem in the Power PM with only two regions

As far as Problem 4.17, Ilan, how do you solve for cosh(A)? Wouldn't take like 10 minutes to solve on an actual test? I guess I need to brush up on using trig identities on phasors. I sure hope there's no long transmission line problems on the actual test. The short and medium line models are mathematically less complex.

Ok well now I'm gonna attempt the Kaplan morning exam. With all these errors in the Kaplan test, I might have more questions on the morning exam as well.


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## benbo (Oct 6, 2007)

Your actual exam problems will likely be much simpler than these Kaplan problems. And they will certainly have less significant figures in the answers. But if you can basically understand most of the solutions here (notwithstanding the errors) it will be very helpful along with the NCEES exam. I just thought the Camara practice exam, at least for the PM ECC module, though somewhat good for practice, was a bit too easy.


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## Ilan (Oct 7, 2007)

chicago said:


> As far as Problem 4.17, Ilan, how do you solve for cosh(A)? Wouldn't take like 10 minutes to solve on an actual test? I guess I need to brush up on using trig identities on phasors. I sure hope there's no long transmission line problems on the actual test. The short and medium line models are mathematically less complex.


Did you mean how to calculate cosh of a real number? In Casio, press "hyp" and then the trig. functions for the hyperbolic trig functions. With my Casio FX 115MS, I have no choice but to use this long route to calculate the cosh of a complex number. Lets hope that they don't ask long transmission line in the actual exam. I was frustrated in the beginning too to go through this cumbersome calculations, I am used to it now. One more thing on the equation, the calculator needs to be in Radian mode when you calculate coshA, cosB, sinhA and sinB. After calculating cosh(A+iB) and sinh(A+iB) values, don't forget to change the mode back to degree.... :2cents:

Ilan.


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