Total Motor Currents

I have 2 pumps being controlled by a PIC, 1 has a 1/2 hp motor, the other has a 2.5hp motor. They can be 110v or 220v 50 or 60hz, it depends on the installation but the voltage or frequency does not change the results.

Using a Fluke meter with an Amp Clamp, I read the 1/2HP as drawing 3.2 amps, and the 2.5hp motor is drawing 18.5 amps. These readings always closely match the FLA number on the motors faceplate.

So it seems that if I add the 2 numbers together, the total current draw should be 3.2 + 18.5 = 21.7 amps total current.

But when I put the amp clamp on the wire feeding both pumps, I always get a reading of around 20.7. 1 amp less than the 2 numbers added together.

Doesn't matter what voltage, frequency, single phase or 3-phase, the total is always less than the 2 combined.

I am also reading the currents with the PIC, and it comes up with the exact same numbers.
Of course the customer says that it's WRONG.

Please help keep my barber in business. :eek:
There's gotta to be a reason somewhere.

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Originally Posted by Darrel Taylor

There's gotta to be a reason somewhere.

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Is this one of those things where the 'power factor' comes into play?
I've read that in large factories, plants, whatever, the use large electric induction motors, the rotors generate a back-current out of phase with the incoming current, and while that back-current doesn't try to 'spin the meter backwards', it is enough to trick the KWH meters into showing energy usage that is far below what the factory actually uses, even though the power company has still made a lot of heat in their generators and used a lot of fuel to spin them.
I don't remember what that whole theory is called, and I don't know if it applies with smaller motors. Seems to me that this sort of thing only applied to large, heavily loaded motors...
http://en.wikipedia.org/wiki/Power_factor <-there is it...
Can you load and unload the motors at the same time and see if the meter changes readings?

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the rotors generate a back-current out of phase with the incoming current

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and being the two motors are not balanced to each other, the back EMF from one motor is helping to feed the other. The larger the motors the greater the effect. Most people do not notice it on the small motors or do not get to excited about an amp or so.

We can have a 30 and a 40 hp running and then run a 10 almost for free. Well we could if the power company did not have a "special" meter that is supposed to correct for this.

Lentz's law kinda sorta helps to explain it too.

Taking a quick look in the old basic electric circuits book seems to verify what you are seeing.On the subject of parallel ac circuits involving impedance,the calculated branch currents are higher than the total calculated generated current.The basic reason is the phase angle of each branch.Ouch,this is starting to hurt thinking about this stuff.

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Originally Posted by arniepj

Ouch,this is starting to hurt thinking about this stuff.

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Yeah, tell me about it. OUCH!

What you guys are saying does seem to make a lot of sense.
And I do remember seeing different waveforms from the current sensors on each motor, so perhaps the larger motor, which is driving a Positive Displacement pump at 1000psi, is putting enough force back into the circuit to partially drive the 1/2hp motor (some of the time), effectively reducing the overall current reading by an Amp.

I guess the problem now is ... Is the difference something that's calculatable.

I only installed current sensors on each motor. I didn't put one on the main feed. If I put one there, then it reads the number the customer wants. But I didn't. :o

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

Thanks again,

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Originally Posted by Darrel Taylor

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

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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.:o

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Originally Posted by mackrackit

If you knew: ...

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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!

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.

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Originally Posted by Darrel Taylor

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?

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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.

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Originally Posted by 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.

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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?

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

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?

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Originally Posted by Acetronics

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

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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

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Originally Posted by Darrel Taylor

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

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It depends on exactly what you want to know. The measured currents are real (and determine wire size, etc.) - it's just that not all of the current is doing work. Average Volts x Average Current gives VA (VoltAmps) and multiplying that by PF gives WATTS. Most residential meters measure watts but most commercial and industrial sites have demand meters which measure peak current.

For small loads there is an inexpensive meter available that measures all of these things. However, it's only for 120VAC and a maximum of 15A.

• http://www.newegg.com/Product/Produc...82E16882715001

Well, Thanks for all the information.
At least now I know where I screwed up. :o

The problem is that these things are spread all around the world. And many are on ships at sea, or sitting on oil rigs in the Gulf of Mexico. There's no way for me to go retro-fit anything. A software update was the only possibility.

The current is sensed with current transformers, and processed with an RMS converter which by definition, averages the waveforms. The result is measured with a 16F676. The voltage is rectified, filtered, and read with a 12F675. So I can't see anything like phase angles or zero-crossing.

I was hoping there would be an easy way to take the 2 current readings that I have, and figure out the actual total. But apparently, I'm out of luck. At least the individual motor currents are correct.

I guess they'll just have to buy the latest "System Upgrade", a third current sensor.
If they absolutely have to have the exact combined reading. $$$

And just an afterthought.
The Total current reading was never part of the original requirements.

Since I had the current reading for both motors, I thought I'd just add them together and display the total current, just because I could.

Wasn't till some technician decided to verify it, that it became a problem.
Didn't even need to be there, and now it's a problem.

Be careful what you add to your designs. ;)

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Originally Posted by dhouston

. . . Most residential meters measure watts but most commercial and industrial sites have demand meters which measure peak current. . . .

• http://www.newegg.com/Product/Produc...82E16882715001

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And as I understand it, the leg of the circuit which is consuming the greatest power is the one which you pay for, which is to say if you have 3 phase comming into the building and you are using 120 1 phase @ 2000 watts and 3 phase @ 2000 watts you will be billed @ 4000 watts.

Just charge them extra for the 1 amp you're magically saving.

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Originally Posted by dhouston

Just charge them extra for the 1 amp you're magically saving.

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ROFL.

I Like that idea!

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Originally Posted by skimask

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?

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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.

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Originally Posted by Melanie

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

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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...

"CIVIL"?

Never heard of it.

After this thread ... that was probably obvious. :eek:

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Originally Posted by Darrel Taylor

"CIVIL"?
Never heard of it.
After this thread ... that was probably obvious. :eek:

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A little more time down in K-town, and we might've heard of it eventually eh? :D

Hi Darrel,

Great to know that at least your barber wouldn't be out of business. We all try to enhance our designs cause the customer may not know the possibilities of the system, and for me every project has some or other emotional attachment too. More $$$$= more emotions...:D

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Originally Posted by Darrel Taylor

The current is sensed with current transformers, and processed with an RMS converter which by definition, averages the waveforms. The result is measured with a 16F676.

The voltage is rectified, filtered, and read with a 12F675. So I can't see anything like phase angles or zero-crossing.

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BTW if you remove the rectification, filtration and still read the voltage (half cycle) you can determine the zero cross programatically. Also in most cases if your input is not AN3 you already have the option of determining zero-cross in some way or the other. Cin+, Cin-, INT. Not very precise though.

Hi, Darrel

May be you could find a turnaround ...

if you found an info, on motor id. plates ; a number like "cos xx = .85"

Generally it's written there. May be also in the motors datas ...


your intensity will be

SQR ( ( I1_meas. * sin xx1 + I2_meas. * sin xx2 )^2 + ( I1_meas. * cos xx1 + I2_meas. * cos xx2 )^2 )

with I1_meas, xx1 related to motor 1 and I2, xx2 to motor 2.

this could seriously simplify things ... don't you think ???

Alain

Hi,

Distributed Power factor correction (PFC):
http://www.drasa.co.za/news_articles_detail2.html

(From the above link)

If a customer only applies PFC at the utility metering point then a load reduction is only
experienced on the utility’s network and not on the customer’s internal reticulation network.
If however, the PFC is applied at the points where the inductive reactive energy is consumed,
then a loading reduction and consequently also an associated Capex reduction will result on the
customer’s own network. It then follows that the customer will enjoy the same benefit for the
reduction in loading due to the application of PFC as the utility enjoys.

* * *

PFC: (ABB - Asea Brown Boveri )
http://library.abb.com/global/scot/s...with%20PFC.pdf

* * *

Capacitors for Power Factor Correction:
http://www.epcos.com/web/generator/W...locale=en.html

Applications & Cases
http://www.epcos.com/web/generator/W...product=247926


Best regards,

Luciano

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Originally Posted by Acetronics

your intensity will be

SQR ( ( I1_meas. * sin xx1 + I2_meas. * sin xx2 )^2 + ( I1_meas. * cos xx1 + I2_meas. * cos xx2 )^2 )

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Got me all excited with that one Alain.

Made a spreadsheet, was playing with the numbers, kinda almost could see how it might work.
Went to find the cos = .xx ...
Oh no! Don't have that info.

http://www.pbpgroup.com/files/halfHP_Spec.JPG

http://www.pbpgroup.com/files/dejected.gif

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Originally Posted by Darrel Taylor

Went to find the cos = .xx ...
Oh no! Don't have that info.

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For an inductive circuit like this, it's the Power Factor. However, for more complex non-linear circuits it gets more complicated.

• http://findarticles.com/p/articles/m..._/ai_n17364837

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Originally Posted by Darrel Taylor

Got me all excited with that one Alain.

Went to find the cos = .xx ...
Oh no! Don't have that info.

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Hi, Darrel

You've one left chance ... on the motor itself, there's a metal little sheet where it MIGHT be written. ( a summary of the main characteristics ... including this COSxx )

Alain

HEYYYYYYYYYYYYYYYYYYYYYY ..

POWER FACTOR = 66 ...% on 17th line !!! .... means Cos xx = .66 and Sin xx = .75

You had it under the eyes !!!

the only way to make a utility meter to run backwords is to supply power back out to them.
a pure capacative load will draw amps at 90' but not turn meter.
don