Engineering Intro to 3-Phase AC Motors: Understanding Wattmeter Measurements | Help Needed

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The discussion revolves around understanding wattmeter measurements in a 3-phase AC motor connected in a Delta configuration. The user is confused about calculating the real power using the wattmeter connected between the RS lines, particularly regarding the role of phase current and the √3 factor. It is clarified that the wattmeter measures the voltage of each phase and the current multiplied by √3, leading to a total power calculation. The correct interpretation suggests that the wattmeter reading corresponds to 1/3 of the total motor power, which is estimated to be 22 kW. Overall, the conversation highlights the complexities of power measurement in 3-phase systems and the importance of correctly applying formulas.
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Homework Statement


hello everyone
I'm a little confused with the whole 3- phase system.
for example:
An AC motor is connected to a 3-phase R-S-T system with 400 volts.
The motor is in Delta configuration and every load is Z=1+2.5j .
A wattmeter is connected between the RS lines.
W=|VRS|*|IR|*cos(∠IR-∠VRS)=?

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:
attachment.php?attachmentid=63770&stc=1&d=1383936769.jpg


on the other hand:
attachment.php?attachmentid=63775&stc=1&d=1383937564.jpg


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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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?
 
Thank you for your reply

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?
 
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 :smile: ... http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif
 
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Thanks for the nice greeting :smile:

Sadly, I am still unsure of what I am doing wrong.
This is my understaing of the setup:
attachment.php?attachmentid=63794&stc=1&d=13839952995.jpg


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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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.
 
A wattmeter is connected between the RS lines.
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.
 
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NascentOxygen said:
Is that the wording of the original question? It sounds a sloppy way of expressing the arrangement.
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.

Thank you for your help
 
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ohadbx said:
Yes, that is the original statement, and in addition there is a drawing like the one I have posted.
I'd like to see the original drawing. Can you attach it here?
 
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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.
attachment.php?attachmentid=63854&stc=1&d=1384200685.jpg
 

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  • #11
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.
 
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