# Complex Imaginary volume 4, prob 14



## cabby (Apr 8, 2012)

Has anyone worked this problem yet? 34.5kv, 3phase, 4 wire, Y-connected, distrobution panel is experiencing unbalanced load conditions. Phase A load=(120+65j)kva @ (19.9 angle 0) kv. Phase C load=(65+20j)kva @ (19.9 angle 0)kv. Calculate the neutral current.

for the life of me, I do not understand the solution nor do I think the problem is set up correctly. I thought the voltage for phase C would be 19.9 at angle 240 or something to that affect. If anyone has some examples of unbalanced load problems it would be great.


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## nmh0408 (Apr 9, 2012)

Since it doesn't specify where the load is connected and it is asking for In you would assume the loads are connected between each leg and the neutral and therefore In=load A/(V/root3)angel 0 + load B/(V/root3)angel 120+ load B/(V/root3)angel 240


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## stinkycheese (Apr 9, 2012)

I was just looking at this one! I'm taking the week off as final prep. The Phase C voltage should be &lt;240. However, I don't think their answer is right. I get that IA = 6.86&lt;-28.44 and IC = 3.42&lt;137.1 (I think they got confused with conjugates). IN = -(IA+IC) gives me IN = 3.65&lt;165 for an answer of 3.65A.


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## nmh0408 (Apr 9, 2012)

nmh0408 said:


> Since it doesn't specify where the load is connected and it is asking for In you would assume the loads are connected between each leg and the neutral and therefore In=load A/(V/root3)angel 0 + load B/(V/root3)angel 120+ load C/(V/root3)angel 240


Note the correction.


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## nmh0408 (Apr 9, 2012)

stinkycheese said:


> I was just looking at this one! I'm taking the week off as final prep. The Phase C voltage should be &lt;240. However, I don't think their answer is right. I get that IA = 6.86&lt;-28.44 and IC = 3.42&lt;137.1 (I think they got confused with conjugates). IN = -(IA+IC) gives me IN = 3.65&lt;165 for an answer of 3.65A.


If you have three unbalanced loads, why you are not adding up Ib to the neutral current?


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## stinkycheese (Apr 9, 2012)

nmh0408 said:


> If you have three unbalanced loads, why you are not adding up Ib to the neutral current?


There is no load on phase B.


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## nmh0408 (Apr 9, 2012)

stinkycheese said:


> nmh0408 said:
> 
> 
> > If you have three unbalanced loads, why you are not adding up Ib to the neutral current?
> ...


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## stinkycheese (Apr 9, 2012)

nmh0408 said:


> Since I do t have there book, what was the answer?
> 
> Hence, the question is giving you [email protected] degrees, therefore you start your sequence with C, I solved it based on CBA sequence and I got 5.23angel 119?


The CI answer is 6.6. As I wrote above, I got 3.65A.


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## stbtigerr (Apr 10, 2012)

What is their math? I don't have that volume.


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## stbtigerr (Apr 10, 2012)

This is what I did using 240 for the Phase C voltage:

Ia=(120+65i)/(19.9 @ 0) = [email protected] =&gt; [email protected]

Ic=(65+20i)/([email protected]) = [email protected] =&gt; [email protected]

In= [email protected] + [email protected] = [email protected]


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## stinkycheese (Apr 10, 2012)

It looks like I was the one confused about conjugates. Argh. Thanks!


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## cabby (Apr 10, 2012)

Thank you everyone. The answer to the problem was 6.6 @ -58 degrees. The magnitude therefore is 6.6A for the neutral current.


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## wvengineer (Feb 27, 2018)

I apologize for bringing up an old thread but I have a question about this problem.  Why would I assume phase C voltage would be at an angle of 240 when the problem specifically states "On phase A (19.9 angle 0 kV) there is a 120+j65 load, on phase C (19.9 angle 0 kV) there is a 65+j20 kVA load"?  I took it to mean that phase A and C voltages were in phase.  If they wouldn't of had that voltage in parenthesis for each phase I would have assumed 19.9 kV at 0 and 240 for A and C.  Am I missing something?  Is this some notation that I'm unaware of?  Thanks.


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## Zach Stone P.E. (Feb 28, 2018)

wvengineer said:


> I apologize for bringing up an old thread but I have a question about this problem.  Why would I assume phase C voltage would be at an angle of 240 when the problem specifically states "On phase A (19.9 angle 0 kV) there is a 120+j65 load, on phase C (19.9 angle 0 kV) there is a 65+j20 kVA load"?  I took it to mean that phase A and C voltages were in phase.  If they wouldn't of had that voltage in parenthesis for each phase I would have assumed 19.9 kV at 0 and 240 for A and C.  Am I missing something?  Is this some notation that I'm unaware of?  Thanks.


I think there may be an error in your edition of the book. 

The voltages in my copy are 19.9kV&lt;0º for phase A and 19.9kV&lt;240º for phase C, the loads are the same as yours.


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## wvengineer (Mar 2, 2018)

Zach Stone said:


> I think there may be an error in your edition of the book.
> 
> The voltages in my copy are 19.9kV&lt;0º for phase A and 19.9kV&lt;240º for phase C, the loads are the same as yours.


It's an older set.  Thanks!


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