Total Motor Currents


Closed Thread
Results 1 to 30 of 30

Hybrid View

  1. #1
    Join Date
    Jul 2003
    Location
    Colorado Springs
    Posts
    4,959

    Default 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.
    There's gotta to be a reason somewhere.
    DT

  2. #2
    skimask's Avatar
    skimask Guest


    Did you find this post helpful? Yes | No

    Default

    Quote Originally Posted by Darrel Taylor View Post
    There's gotta to be a reason somewhere.
    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?

  3. #3
    Join Date
    Nov 2003
    Location
    Wellton, U.S.A.
    Posts
    5,924


    Did you find this post helpful? Yes | No

    Default

    the rotors generate a back-current out of phase with the incoming current
    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.
    Dave
    Always wear safety glasses while programming.

  4. #4
    Join Date
    Aug 2005
    Posts
    57


    Did you find this post helpful? Yes | No

    Default

    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.

  5. #5
    Join Date
    Jul 2003
    Location
    Colorado Springs
    Posts
    4,959


    Did you find this post helpful? Yes | No

    Default

    Quote Originally Posted by arniepj View Post
    Ouch,this is starting to hurt thinking about this stuff.
    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.

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

    Thanks again,
    DT

  6. #6
    Join Date
    Nov 2003
    Location
    Wellton, U.S.A.
    Posts
    5,924


    Did you find this post helpful? Yes | No

    Default

    Quote Originally Posted by Darrel Taylor View Post
    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.
    Dave
    Always wear safety glasses while programming.

  7. #7
    Join Date
    Jul 2003
    Location
    Colorado Springs
    Posts
    4,959


    Did you find this post helpful? Yes | No

    Default

    Quote Originally Posted by mackrackit
    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!
    DT

  8. #8
    Join Date
    Jul 2003
    Posts
    2,358


    Did you find this post helpful? Yes | No

    Default

    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.

  9. #9
    Join Date
    Nov 2003
    Location
    Wellton, U.S.A.
    Posts
    5,924


    Did you find this post helpful? Yes | No

    Default

    Quote Originally Posted by Darrel Taylor View Post
    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.
    Dave
    Always wear safety glasses while programming.

  10. #10
    Join Date
    Dec 2005
    Posts
    1,073


    Did you find this post helpful? Yes | No

    Default

    Quote Originally Posted by Darrel Taylor View Post
    Can you just multiply by the "Power Factor" of the motors?
    Naahhh, that would be too easy.
    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.

  11. #11
    Join Date
    Jul 2003
    Location
    Colorado Springs
    Posts
    4,959


    Did you find this post helpful? Yes | No

    Default

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

    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. $$$
    DT

  12. #12
    Join Date
    Jul 2003
    Location
    Colorado Springs
    Posts
    4,959


    Did you find this post helpful? Yes | No

    Default

    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.
    DT

  13. #13
    Join Date
    Feb 2005
    Location
    Kolkata-India
    Posts
    563


    Did you find this post helpful? Yes | No

    Default GOD saved your barber

    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...

    Quote Originally Posted by Darrel Taylor View Post
    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.
    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.
    Regards

    Sougata

  14. #14
    Join Date
    Aug 2006
    Location
    Look, behind you.
    Posts
    2,818


    Did you find this post helpful? Yes | No

    Default

    Quote Originally Posted by dhouston View Post
    . . . Most residential meters measure watts but most commercial and industrial sites have demand meters which measure peak current. . . .
    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.
    If you do not believe in MAGIC, Consider how currency has value simply by printing it, and is then traded for real assets.
    .
    Gold is the money of kings, silver is the money of gentlemen, barter is the money of peasants - but debt is the money of slaves
    .
    There simply is no "Happy Spam" If you do it you will disappear from this forum.

  15. #15
    Join Date
    Dec 2005
    Posts
    1,073


    Did you find this post helpful? Yes | No

    Default

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

Similar Threads

  1. Controlsystem for compact sporting (clay shooting)
    By Fredrick in forum mel PIC BASIC Pro
    Replies: 11
    Last Post: - 30th July 2009, 16:48
  2. Total Newbie...
    By Frozen001 in forum mel PIC BASIC Pro
    Replies: 17
    Last Post: - 9th August 2008, 07:28
  3. Replies: 1
    Last Post: - 18th April 2006, 19:11
  4. Realtime video
    By RUBiksCUbe in forum mel PIC BASIC Pro
    Replies: 37
    Last Post: - 28th November 2005, 07:46

Members who have read this thread : 0

You do not have permission to view the list of names.

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts