- 26th December 2002, 11:28 #1

- Join Date
- Feb 2002
- Location
- Pakistan
- Posts
- 839
- Helped
- 57 / 57
- Points
- 5,590
- Level
- 17

## 3 phase power measurement with 1 wattmeter

I want to measure the power KW of a motor that is Delta

connected (3 wires, no neutral). I want to use only 1 watt-hour meter. The motor has balanced load but has harmonics because of an PWM invertor driving it at variable speed. (would it make any difference).

Thanks for any suggestions.

- 26th December 2002, 12:27 #2
## 3 phase...

U can use Atmel sollution (2 chips) for 3 phase powermeter.

Search on atmel site,

best,

//a

- 26th December 2002, 15:45 #3

- Join Date
- Feb 2002
- Location
- Sweden
- Posts
- 120
- Helped
- 0 / 0
- Points
- 3,857
- Level
- 14

Provided the design of the driver is professional and the motor is of standard type I think it is safe to assume a symmetrical load.

Regards

- 26th December 2002, 16:14 #4

- Join Date
- Feb 2002
- Location
- Pakistan
- Posts
- 839
- Helped
- 57 / 57
- Points
- 5,590
- Level
- 17

Thanks. But I want to ask the connections of the wattmeter with the 3 phase wires and the conversion factor to get the 3 phase power from 1 meter.

- 26th December 2002, 16:17 #5

- Join Date
- Jul 2002
- Location
- Middle Earth
- Posts
- 4,632
- Helped
- 477 / 477
- Points
- 37,801
- Level
- 47

## artificial neutral

How about making an artificial neutral point with a Y connected set of resistors which are much lower resistance than the voltage sense winding in your meter? This would solve the connection problem. I woul suspect, as one of the previous posts said, that the three phases are equal in current and power factor and therefore multiply one phase real power by three.

- 26th December 2002, 16:18 #6

- Join Date
- Feb 2002
- Location
- Pakistan
- Posts
- 839
- Helped
- 57 / 57
- Points
- 5,590
- Level
- 17

The @tmel solution looks very high-end. I intend to use An@log devices AD*775O which are very simple to use. Any suggestions on that.

- 26th December 2002, 16:20 #7

- Join Date
- Feb 2002
- Location
- Pakistan
- Posts
- 839
- Helped
- 57 / 57
- Points
- 5,590
- Level
- 17

flatulent, I remember reading a simpler arrangement in one our text books in EE. But that was in the last century and I have forgotten that.

- 26th December 2002, 16:23 #8

- Join Date
- Jul 2002
- Location
- Middle Earth
- Posts
- 4,632
- Helped
- 477 / 477
- Points
- 37,801
- Level
- 47

## how about this

How about conneting the voltage sense between two phases and the current in one line? Then calculate the equivalent line to neutral voltage with the square root of three factor. This should allow multiplying this one phase measurement by the square root of three to get the total power in three phases. Then as a final check, temporarily measure the power with two meters to verify the simpler method.

- 26th December 2002, 16:57 #9

- Join Date
- Feb 2002
- Location
- Pakistan
- Posts
- 839
- Helped
- 57 / 57
- Points
- 5,590
- Level
- 17

Does leaving the 3rd line free make any difference in your suggesion , flatulent

- 26th December 2002, 17:01 #10

- Join Date
- Jul 2002
- Location
- Middle Earth
- Posts
- 4,632
- Helped
- 477 / 477
- Points
- 37,801
- Level
- 47

## I would think not

I would think not. This and all other suggestions above rely on the lines being totally identical in current and power factor. As is usual in all prototyping situations, the comparason with a standard measurement method should be done at several motor loads to make sure some aspect was not over looked.

I overlooked one point on the correction factor. The phase of the adjacent line will be different than the phase of the neutral.

There may be some further correction factor of the reading. I am not in a position now to do the diagrams. But in general, the comparison with the two meter temporary measurement will give you the correction factor.

- 26th December 2002, 17:12 #11

- Join Date
- Feb 2002
- Location
- Pakistan
- Posts
- 839
- Helped
- 57 / 57
- Points
- 5,590
- Level
- 17

Someone has suggested that I just have to multiply the reading of the wattmeter by 2.0 and I will get the correct 3 phase power. No extra power factor corrections needed. I will try it out practically as you suggested. thnx

- 26th December 2002, 17:51 #12sick_manGuest
There is a simple way to measure watts consumed on each of the 3 phases.

You need to multiply the reading from 1 phase {at no drive ie the controller is set to zero rotation}

by pi 3.1457.

To get the power factor of each phase

simply then divide your outcome from above by 3

To get each phases current

do the same over several ranges of the controller

then draw a graph ploting each point

then divide each of these points numerical value by 3

to get the watts on each phase

this can be done over several phases then each can have its own colour

the average plots {add the three phase reading divide by three}

you then have a nice graph shows you how "symetrical" each phase actualy is

not just a guess at it being symertical drive or not {they very seldom are unless you star point earth the whole system}

the simple way to think of phase is a circle divided by three

{if you dont use neutral {basicaly earth or the .1457 left over {this is the differance in potential {energy {reverse current not voltage} left over from three phases that make a circle}{take out of the equassion using earth

same as neurtral}}

neutral isnt a phase

instead neutral is a kind of earth

being just a long run earthed at the substation or power station that supplies your needs....

so .1457 is all the potential {voltage but no foward current} that is can return to the genorator station or substation transformer

lightning comes up from the ground 1.707 X less distance than it travels down.... this is a fact

- 26th December 2002, 21:54 #13

- Join Date
- Jul 2002
- Location
- Middle Earth
- Posts
- 4,632
- Helped
- 477 / 477
- Points
- 37,801
- Level
- 47

## calculatins

I agree with the 2.0 factor for two reasons.

1) I did the calculations at lunch and it shows 2.0

2) A thought experiment with the two meter method shows equal readings for the equal phase loads. Summing the two meter readings will result in twice the reading for one meter.

- 27th December 2002, 03:41 #14

- Join Date
- Feb 2002
- Location
- Pakistan
- Posts
- 839
- Helped
- 57 / 57
- Points
- 5,590
- Level
- 17

Thanks flatulent. I think it solves the problem then