Total Motor Currents

[mackrackit]

Can you just multiply by the "Power Factor" of the motors.
Naahhh, that would be too easy.

If you knew:

The motor's efficiency
How different loads effect the motor
Monitor the line voltage
Monitor the ground (earth) potential
and maybe some other stuff, you might be able to run a calc. The formula... I have not a clue.

[Darrel Taylor]

If you knew: ...

Well, the system also collects the Pressure and Flow readings from each pump.

I suppose I could calc a "Load" value from that, and try to match it to some curve I've created from observed data.

But, I'd rather not go there.

And just to add fuel to the fire ....

Since these systems are part of a larger system that has it's own compressors and motors, and whatever. Isn't any "True" reading I get, going to combine with whatever bigger motors that might be running. Making that "True" reading useless?

Ugh!

[Melanie]

In large industrial installations, you are supposed to switch-in banks of Capacitors in the plant room to keep the Power Factor within limits set by the electric company if it is distorted by large amounts of (inductive load) machinery. If you don't, you are liable to fines and/or surcharges by the electric company on your energy usage. If you know how, and throw your current and voltage sufficiently out of phase, you can actually run your entire house for 'free'. If you are really clever, by the correct phase difference, you can not only slow-down your electricity meter, you can stop it completely - and with some older mechanical types have the meter go BACKWARDS, so the electric company ends up paying you for the electricity you use. You may find the 'small print' on your electricity supply contract threatens you with dire consequences if you extend your experiments in self-sufficiency in this direction.

[skimask]

In large industrial installations, you are supposed to switch-in banks of Capacitors in the plant room to keep the Power Factor within limits set by the electric company if it is distorted by large amounts of (inductive load) machinery.

What kind of cap's are we talking about here? Are we talking about those huge oil filled jobs (not the transformers, I know what those are), switched in and out with a huge set of contactors at will by a 'Power Factor Controller' or something to that effect?

[Acetronics2]

Hi, Darrel

When adding the current for the 2 motors, you should calculate a VECTORIAL addition ...

but the sum modulus of two vectors is not compulsory equal to the sum of the modulii ... generally, It's lower ...


Your results mean current for both motors have a different de-phasing ( ? ) ,called Cos Phi in Europe, from voltage.

Alain

[sougata]

Hi, Darrel
Your results mean current for both motors have a different de-phasing ( ? ) ,called Cos Phi in Europe, from voltage.
Alain

While that explains the phenomena, you should be busy working up the remedy.

Possibly you would need to sample the voltage as well and find the phase difference between the current and voltage to calculate active/reactive/average power. While zero-cross detection can be used on the current and voltage sampled approach can also work.
Consult AN939 from the mchip website. And here is something that may interest you http://www.circuit-magic.com/acpower.htm

[dhouston]

Darrel, you need to read up on Power Factor. It is an issue with all non-linear (i.e. reactive) loads, even very small loads. Voltage and current are out of phase so measurements can be misleading. Wikipedia is as good a place to start as any - it explains the fundamentals.

• http://en.wikipedia.org/wiki/Power_factor

You wrote that the PIC measures currents. How are you doing this?

[Melanie]

What kind of cap's are we talking about here? Are we talking about those huge oil filled jobs (not the transformers, I know what those are), switched in and out with a huge set of contactors at will by a 'Power Factor Controller' or something to that effect?

Yes, absolutely correct.

Don't you all recall having to remember the word CIVIL at school? (referring to the fact that in a Capacitive C load, the current I leads the voltage V. In an inductive L load the voltage V leads the current I). Measure time difference between both (zero cross is easiest), simple math determines the rest.

[skimask]

Don't you all recall having to remember the word CIVIL at school?

I just saw that 'CIVIL' acronym at Wikipedia...never seen that one before, only 'ELI the ICE man'.
I've never had any sort of 'formal' electronic training, unless you count my USAF tech school (didn't learn anything there) or that 7 week Micro-Mini-Circuit-Card-Repair school I took about 18 months ago (learned almost too much there! and lost of a lot of bad habits!).

The Cap Banks - I just never put 2 and 2 together as far as those banks go. I guess I always thought they were transformer banks of some sort...

[Darrel Taylor]

"CIVIL"?

Never heard of it.

After this thread ... that was probably obvious.

[mackrackit]

Since these systems are part of a larger system that has it's own compressors and motors, and whatever. Isn't any "True" reading I get, going to combine with whatever bigger motors that might be running. Making that "True" reading useless?

I think your best bet would be to pick a time when the load on the entire system is running normal or average. Then play with moving things around on different branch circuits if possible until the reading near the utility meter is close to what you know it should be. Then just leave it at that. From there it should average out.

Like Melanie said, you can make the utility meter slow or stop or run backwards. In some rare occasions the utility meter will run faster than it should. That is why the system needs to be "balanced". Good electrical contractors on new construction or revamped systems try to do this.
That is why you see them running around with the clamp meters at the end of the job doing pretty much what you are doing.