PE Power #8

[yaoyaodes]

A 3-phase, 4 wire, neutral-grounded, wye-connected utility line has a phase-to-phase voltage of 13.2 kV. A complex load of (200+j100)KVA is connected between Phase A and neutral, An identical load is connected between B and neutral. The neutral current (amperes) is most nearly.

I do not understand why Ia+Ib+In=0, can anyone explain to me?

 

[akyip]

Some texts/materials have the neutral current In traveling in the same direction as the phase currents, such that the phase currents and neutral current all flow towards the neutral connection point of a wye. I believe the NCEES PE practice exam uses this reference for the neutral point. So if that's the case, then basically since all the phase currents and neutral current are traveling INTO the same node:

Ia + Ib + Ic + In = 0

(In this problem, Ic is just 0)

I prefer referencing the neutral current as flowing out of the neutral connection point, and therefore flowing in the opposite direction than that of the phase currents. So then in my case, the phase currents flow into the neutral connection point of the wye, and the neutral current flows out of the neutral connection point. So that means I prefer:

Ia + Ib + Ic - In = 0

In = Ia + Ib + Ic

In either case, this is just simply KCL at the neutral connection point regardless of how you want to reference the neutral current's flowing direction.

The neutral current's magnitude will be the same regardless of which direction you reference having it flow. Only the angle will change. If you want to flip the neutral current's direction or any phasor's direction, you just simply add +180 degrees to its phase angle (which is equivalent to making the phasor itself negative or multiplying the phasor by -1).

Attached is my work for this problem (this was problem 108 in the previous version of the NCEES practice power exam).

 

[kris7o2]

Hi everyone, this is probably a dumb question, but when reading the question how is someone supposed to know that there is no load on phase C, and neutral? My concern is that I would not identify that the problem didn't state that a load didn't exist on phase C, and that I would struggle to understand that this is an unbalanced load problem. Or is there another trick to identifying an unbalanced load question?

 

[akyip]

Hi everyone, this is probably a dumb question, but when reading the question how is someone supposed to know that there is no load on phase C, and neutral? My concern is that I would not identify that the problem didn't state that a load didn't exist on phase C, and that I would struggle to understand that this is an unbalanced load problem. Or is there another trick to identifying an unbalanced load question?

Yeah, sometimes you really do need to carefully read the problem statement and interpret it correctly. In this case, I guess one way to look at it is that a 3-phase source can be used to power 1-phase and/or 3-phase loads...

 

[Zach Stone P.E.]

Hi everyone, this is probably a dumb question, but when reading the question how is someone supposed to know that there is no load on phase C, and neutral? My concern is that I would not identify that the problem didn't state that a load didn't exist on phase C, and that I would struggle to understand that this is an unbalanced load problem. Or is there another trick to identifying an unbalanced load question?

It's all about the context clues given in the problem.

The problem didn't say there was a load on C-phase. The problem said there was only a load connected A-phase to neutral, and B-phase to neutral:

"A complex load of (200+j100)KVA is connected between Phase A and neutral, An identical load is connected between B and neutral."

 

[kris7o2]

Hi all,

One more question. In the solution they used

7.62 ang (0 deg) and 7.62 ang (+120 deg)

Shouldn't it have been

7.62 ang (0 deg) and 7.62 ang ( -120 deg)

I.e. shouldn't the angle be negative in phase b versus the positive value used?

 

[no_concentrate]

Hi all,

One more question. In the solution they used

7.62 ang (0 deg) and 7.62 ang (+120 deg)

Shouldn't it have been

7.62 ang (0 deg) and 7.62 ang ( -120 deg)

I.e. shouldn't the angle be negative in phase b versus the positive value used?

Yes Angle will be -ve for B phase. Solution of problem is better explain by @akyip above in thread 2

​

 

[akyip]

Yes Angle will be -ve for B phase. Solution of problem is better explain by @akyip above in thread 2

​

To further elaborate on this:

Positive-sequence rotation (ABC rotation) is typically usually assumed if the problem does not state anything about phase sequence.

Positive angular rotation is counterclockwise from the horizontal.

The typical reference is:

Phase A: 0 degrees
Phase B: -120 degrees
Phase C: +120 degrees

So when you draw out the phasors and they rotate counterclockwise, the phase sequence for positive-sequence rotation is:

A B C A B C A B C...

If you draw it out to visualize this, I think you would be able to understand this better.

 

[wesele18]

Some texts/materials have the neutral current In traveling in the same direction as the phase currents, such that the phase currents and neutral current all flow towards the neutral connection point of a wye. I believe the NCEES PE practice exam uses this reference for the neutral point. So if that's the case, then basically since all the phase currents and neutral current are traveling INTO the same node:

Ia + Ib + Ic + In = 0

(In this problem, Ic is just 0)

I prefer referencing the neutral current as flowing out of the neutral connection point, and therefore flowing in the opposite direction than that of the phase currents. So then in my case, the phase currents flow into the neutral connection point of the wye, and the neutral current flows out of the neutral connection point. So that means I prefer:

Ia + Ib + Ic - In = 0

In = Ia + Ib + Ic

In either case, this is just simply KCL at the neutral connection point regardless of how you want to reference the neutral current's flowing direction.

The neutral current's magnitude will be the same regardless of which direction you reference having it flow. Only the angle will change. If you want to flip the neutral current's direction or any phasor's direction, you just simply add +180 degrees to its phase angle (which is equivalent to making the phasor itself negative or multiplying the phasor by -1).

Attached is my work for this problem (this was problem 108 in the previous version of the NCEES practice power exam).

why do you get 29 again when you add the phasors at the end and how do you get that angle when you go to rectangular and back doesnt the magnitude change

 

[Zach Stone P.E.]

why do you get 29 again when you add the phasors at the end and how do you get that angle when you go to rectangular and back doesnt the magnitude change

The calculator will give you the angle if you type it in properly.

See here, there is a much more recent discussion on this problem: https://community.electricalpereview.com/c/third-party-practice-exams/ncees-8