# Help needed-Intro to 3-phase AC motors

1. Nov 8, 2013

1. The problem statement, all variables and given/known data
hello everyone
I'm a little confused with the whole 3- phase system.
for example:
I tried solving using differenet methods we were taught, but every way i try i get a different result.

For example: 1) I thought that the wattmeter will measure the Real Power of a single load, which should be a third of the total Real Power of the motor. so:

on the other hand:

I'm starting to feel lost in this, so any help would be appreciated.

P.S. Maybe I should have done a Y-DELTA transformation? Although I'm not sure what good would that do...

Thank you and have a great weekend

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2. Nov 9, 2013

### Staff: Mentor

A wattmeter connected as described will sense the voltage of each phase of the motor, but it will sense more than the current of a single phase of the motor, it will sense the phase current x √3.

With this correction, does that confirm your analysis?

3. Nov 9, 2013

when I chekced the reading of the wattmeter (the second solution) I did multply the phase current by sqrt(3).
Or am I not getting your point?

4. Nov 9, 2013

### Staff: Mentor

What you calculated in the first step is the power in one phase of the motor. You will want to multiply this by √3 to compare it with your second calculation.

Where you write VRS / Z I think there is a missing √3.

and a warm greeting ..... http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

Last edited by a moderator: May 6, 2017
5. Nov 9, 2013

Thanks for the nice greeting

Sadly, Im still unsure of what I am doing wrong.
This is my understaing of the setup:

To my understanding, in a delta configuration (such as shown),
VLINE=VPHASE , so wouldn't that mean that IPHASE=$\frac{V}{Z}$? where does the √3 come from (if I want to use the power formula I mentioned in the first solution)?

BTW it has come to my attention that this question was originally a multiple-choise question.
The possible answers were: A. (-44.03) B. 22.07 C. (-14.68) D. 66.21 (all in [kW])
bit I have no Idea which is the correct one.

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6. Nov 9, 2013

### Staff: Mentor

A wattmeter has 2 pairs of sensing coils, one for voltage and another for current. I think it is unreasonable to believe you can routinely break open a delta motor and connect something in series with one of its phases to sense phase current!

You performed 2 calculations. Basically, the first is for 1/3 of the motor's power. The second is for 1/3 of (√3 of the motor's power).

My contention is that the best you can do is to have the wattmeter measuring the latter (whether you like it or not), sensing VRS and IR. You can't sense the motor's phase current, even if you wish you could. Well, that's my thinking.

7. Nov 9, 2013

### Staff: Mentor

Is that the wording of the original question? It sounds a sloppy way of expressing the arrangement.

It seems motor phase power is 22kW and total power is 66kW. So they must want the answer of 22kW.

8. Nov 9, 2013

Yes, that is the original statement, and in addition there is a drawing like the one I have posted.

I agree with you, I also think the right answer is 22. That means I shouldn't have multiplied by √3 in second solution, but the reason escapes me.

Last edited: Nov 9, 2013
9. Nov 10, 2013

### Staff: Mentor

I'd like to see the original drawing. Can you attach it here?

10. Nov 11, 2013

here you go, though its not very informative.
The switches are for a different, unrelated, part of the question- the setup is changed to Wye.

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11. Nov 11, 2013

### Staff: Mentor

So it looks like the construction brings out the ends of each motor phase so that you can connect a meter in series with a single phase of the motor to meter the phase current! (Or at least it is intended to represent that.)

Thanks.