Power NCEES, Question 501

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Audienceof1

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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,

 
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,

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

 
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!

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

 
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

 
^ 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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