# New NCEES #108



## kevin t (Mar 14, 2010)

Hi,

I understand the phase rotation doesn't make a difference with the solution, but why is the voltage on Phase B 120 degrees and not -120 degrees. Isn't the normal phase rotation ABC e.g. 0, -120, 120?

Thanks,

Kevin


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## tonyb (Mar 14, 2010)

Hi,

I have a bad habit of trying to justify things.

In this question, the power factor is 200/(sqrt(100^2+200^2) lagging

then use

S=VI*

so with out conjugating this angle, this fact would not hold

so use the conjugate of this load power factor as the current's angle and when you add Ia +Ib it should give the desired magnitude.

...

since [email protected],

...

I*@26.57 and [email protected]

I don't know if this sounds ok but let me know if it is off.


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## BamaBino (Sep 2, 2010)

kevin t said:


> Hi,
> I understand the phase rotation doesn't make a difference with the solution, but why is the voltage on Phase B 120 degrees and not -120 degrees. Isn't the normal phase rotation ABC e.g. 0, -120, 120?
> 
> Thanks,
> ...


That is a good question.

Also, isn't the solution shown in the back IA = ZA / VA ?


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## Flyer_PE (Sep 2, 2010)

BamaBino said:


> Also, isn't the solution shown in the back IA = ZA / VA ?


No. The value 200+j100 is apparent power in kVA, not ohms. It's just odd seeing it represented in rectangular format rather than polar.


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## Kahrlo (Sep 6, 2010)

Actually, the right phase sequence for positive sequence is a-b-c rotated counterclockwise.. Some books have it the other way around..


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## Kahrlo (Sep 7, 2010)

Kahrlo said:


> Actually, the right phase sequence for positive sequence is a-b-c rotated counterclockwise.. Some books have it the other way around..


After actually reading the question from the book i can understand now your question, to calculate for the line current,

I line = S*/V* where * is the conjugate of the phasor

Just like for calculating for power S= V x I* (single-phase)


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## enterprisenx (Mar 8, 2011)

Kahrlo said:


> Kahrlo said:
> 
> 
> > Actually, the right phase sequence for positive sequence is a-b-c rotated counterclockwise.. Some books have it the other way around..
> ...



Hi - I was hoping this solution could be clarified further. I agree that I = S* / V* as we know S = VI* . However, the solution for problem 108 does not take the complex conjugate of the Sload as it shows [ (200+j100) / 7.62&lt;0 ] + [(200+j100) / 7.62&lt;120]. So even if the conjugate was used for the voltage angles ie: theta = - theta such that it was -120 degrees and now it is +120 degrees, there was no conjugate used for the complex power load value.

Between EERM and my Electric Machinery and Power System Fundamentals (Chapman), the angle between phase voltages is 0 , -120, and -240 degrees. If you use -120 degrees, the magntude and angle for your answer is incorrect. Therefore, it seems more than just arbritrary to use 120 rather -120 since it clearly makes a difference.

Thank you all for trying to clear this up as it is appreciated!


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## Dolphin P.E. (Mar 9, 2011)

If you use -120 degrees, the magntude and angle for your answer is incorrect. Therefore, it seems more than just arbritrary to use 120 rather -120 since it clearly makes a difference.


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## enterprisenx (Mar 9, 2011)

[No message]


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## EEVA PE (Sep 14, 2011)

In this problem the following is stated " A complex load of (200 + j100) KVA is ..."

Does using the word "complex" give a different meaning to the problem than if it just said "A load of (200 + j100)KVA is ..."

Should I always assume complex load if I see KVA for both single and 3 phase. And make sure I use the complex conjugate for current?


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## Flyer_PE (Sep 15, 2011)

EEVA said:


> Should I always assume complex load if I see KVA for both single and 3 phase. And make sure I use the complex conjugate for current?


Yes. VA is equivalent to apparent power (S) which is determined by the equation S=VI*.


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## 170B (May 17, 2012)

Maybe i'm missing the forest for the trees, but both my A and B terms are 29.3 (200+j100 in polar [email protected]/7.62=29.3), when added together are giving me the wrong result. ?


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## 170B (May 17, 2012)

Chalk that one up to calculator unfamiliarity and unfortunately coincidental looking numbers...


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## K_Nova (Mar 28, 2013)

I reworded this problem, but I'm having a little difficulty arriving at the correct answer. Working this problem backwards:

If we have the neutral current (calculated from the original problem): 29.3 /_ -33.4 A

Now, we need to solve for the complex load values, as configured identical to the original problem. Noting, both the complex loads between Phase A &amp; Neutral, and Phase B &amp; neutral are identical in value.

I can't seem to arrive back at the original value of (200 + j100) kVA.


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