# Power NCEES, Question 501



## Audienceof1 (Oct 15, 2011)

Here's the low hanging fruit for the day for somebody:

Question 501 in the latest Electrical: Power NCEES sample question book asks for the voltage magnitude at the secondary terminals of a single-phase transformer, provided resistive load value (kW), voltage at the load, distance from transformer to load, and the load impedance in ohms/1000'.

I worked it out similar to the examples NEC provides after Chapter 9, Table 9, with the main difference being 3-phase vice single-phase as given in this problem. My question is, why is the voltage drop doubled in the solution? What am I missing here? I'm thinking that in the single phase arrangement, the neutral is a current carrying conductor also and must be included in the calculation, thereby doubling the conductor length. Maybe I should be quiet before I expose more ignorance!

Thanks for the help,


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## EEVA PE (Oct 15, 2011)

Audienceof1 said:


> Here's the low hanging fruit for the day for somebody:
> 
> Question 501 in the latest Electrical: Power NCEES sample question book asks for the voltage magnitude at the secondary terminals of a single-phase transformer, provided resistive load value (kW), voltage at the load, distance from transformer to load, and the load impedance in ohms/1000'.
> 
> ...



Yep, your thought process is correct. In a single phase you need to also be concerned with the return. So double the conductor length.


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## Audienceof1 (Oct 15, 2011)

Super. Thanks for the help!


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## ElecPwrPEOct11 (Oct 17, 2011)

For the 3 phase situation do you not worry about the return path because you're assuming balanced phase currents so zero current on the neutral wire? I want to make sure I understand this so I don't miss a 'gimme' on the exam. Thanks!


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## Flyer_PE (Oct 17, 2011)

^Correct. If the system is balanced, the neutral current will be zero and there will be no voltage drop.


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## DK PE (Oct 17, 2011)

I hope to not add confusion but if you use the IEEE-141 method (below) you end up with the line-neutral drop and you have to multiply by 2 for 1Φ and by √3 for 3Φ.

V = IRcosΦ + IX sinΦ


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## ElecPwrPEOct11 (Oct 18, 2011)

^ Thanks for the responses.

DK PE- I've seen those equations before but am used to using the NEC table and Ze. At this point I'm gonna stick with it instead of confusing myself with a new method.


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