How is direction of AC power measured?

Discussion in 'General Electronics Chat' started by strantor, Feb 16, 2013.

  1. strantor

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    Oct 3, 2010
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    People with solar power and grid tie inverters can sell power back to the power company in some areas. I've read that to do this, they need a special bidirectional power meter that can differentiate between power delivered to or from the house. I'm curious how these work. The only things I know to measure current are a torroidal current transformer or a hall sensor, and as far as I know, both of these would give you the same output no matter which direction the power was going. Anybody know how it's done?
     
  2. gerty

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    The one that I saw was connected to a barn and had two meters.
    One meter was used to measure the power used by the barn, the second was connected to the incoming line of the first meter. In other words the solar panels are just tied to the grid.
    The inverters are of the 'grid tie' type and must see incoming power, and sync with it before they'll work. If the power goes off during a storm, the owners will have a field full of panels and no power.
    I think I still have some pics in my work 'puter, I'll post them if you are interested.
     
  3. #12

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    Based on no experience whatsoever, When the voltage peaks of the distributed system occur, current to the load has a direction. If you're pouring power into the grid, the current will be going the other way.

    You already know that a current transformer will output a wave that is in matched polarity with the source if you wire it that way. If you're pumping power toward the grid, the output of a current transformer will reverse polarity. Presto. Add magnitude and time, and send me the money.

    (I'm not treating you like a beginner because you aren't.)
     
  4. strantor

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    Is that the typical setup? I'd be interested in pics. Thanks.
    I'm more noob than you know. I have serious gaps in my knowledge, having never been through any (real) education program. I only know things that I've had to learn to do what I want to do, so I'm strong in some areas and void in others.

    What I'm picturing from your reply is sort of like how direction is resolved from a rotary encoder. If wave A leads wave B then, power is flowing to the house, and if wave A lags wave B then, it's from the house. Am I on the right track?

    If so...
    unlike an encoder, there will not be a perfect or consistent phase shift between the two waves depending on the types of loads in the house or on the power line, correct? How much can it vary? If it got all the way out to a 180 degree shift, wouldn't it be impossible to know which leads which? Is it possible to go past 180 degrees? then you'd be paying for the privilege of supplying power.
     
  5. thatoneguy

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    Electric meters measure Volt-Amps rather than Watts (though the display is in Watts). This allows them to accurately meter reactive power, such as inductive loads that return most of the current back to the grid at a different phase.

    For selling electricity, two meters are used, outbound and inbound. Diagram is from the wikipedia article on Net Metering

    [​IMG]
     
  6. debe

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    Sep 21, 2010
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    Hi Strantor. Heres a pic of my import/export meter with my solar system. You press the scroll button for 03 import, 09export. Its a ltron ACE2000 SMB type292 meter. There is a bit on Google on that brand meter. Be aware if you fit solar & feed power out on an ordinary digital meter it will bill you for the exported power. With solar you need an inport/export meter.
     
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  7. #12

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    Excellent. It's been invented.
    I knew, if they could put 3 microprocessors in my kitchen stove, they could design a 2 way power meter.:D
     
  8. gerty

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    I'll see if I can find the pics tomorrow. I don't know about typical, as I have only seen 2 systems up close. They seemed to be the same but I couldn't ask questions to one of the owners. The bottom picture on post #5 is the meter arraingement I was referring to.
     
  9. shazseo

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    Sure. Assume that you're standing in the middle of nowhere with two wires stretching to infinity in both directions. You have any type of measurement equipment you want (but *only* measurement equipment). The fact is, there is no way in which you can determine which direction power if flowing. (And this is true for DC as well as AC)

    However, if you have a low-value resistor, you can now measure with certainty which side goes to a generator and which side goes to a load.
     
  10. gerty

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    Here's a few of the pics. The tracking sensor is what allows the photo cells to follow the brightest sunlight.
    There is also a subpanel located next to the system for field disconnect.
    The inverters are 24 volt in and 120 v out IIRC.
     
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  11. gerty

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    Here's the meter arrangement, bad pic I know, and the disconnect for the cells.
    Notice the warning says that both line and load side may be hot with disconnect open.
     
    Last edited: Feb 18, 2013
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  12. nsaspook

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    Sure you can, by measuring the E/B fields and calculating the Poynting vector power flow.

    http://amasci.com/elect/poynt/poynt.html
     
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  13. strantor

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    Well I didn't understand all that after reading once, but I believe the answer to "how they do it" has been given, so thank you for that. I'll have to sit down and try to wrap my brain around it later.
     
  14. vrainom

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    But would this be practical to implement in a commercial product?

    Now this is simple, because even with a high value resistor in parallel with the load if you measure the current from both sides of the resistor there would be a slight increase in the side providing the current.
     
    Last edited: Feb 18, 2013
  15. nsaspook

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  16. summitville

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    After reading this ...
    http://personal.ee.surrey.ac.uk/Personal/D.Jefferies/poynting.html

    and specifically these paragraphs ...


    "... The solution to this conundrum is to consider the integrated Poynting vector only over an entire closed surface in 3-d space. If the integral over the closed surface is not zero, then we can be sure that there is a flow of power from the inside to the outside of the surface, or from the outside to the inside.
    So we have seen above that when this is not the case, and we persist in our interpretation of the Poynting flux as a local power flow, the difficulty is solved by considering the power to be circulating, such that the flow outwards across a local region of any closed surface is balanced by an equal power flow inwards somewhere else on the same surface.
    In the case of our static charges and our magnet, such an integral over an entire closed surface works out to zero whatever shape the closed surface is.
    Now we consider the case where a battery (d.c.) supplies a pair of wires connected to a resistor. There is power drawn from the battery; it is conveyed by the wires ("waveguide??") to the resistor and dissipated there. There is a magnetic field generated around each of the wires by the current flowing in them, and there is an electric field between the wires because of the voltage drop across the resistor. However, this is a static problem, as was our first example of the magnet and the charged plates. We perform our integral of the Poynting vector in various ways. (case #1) If the surface surrounds the battery only, with the wires passing through the surface, there will be an outward directed total to the integral, this total being equal to the power supplied by the battery. (case # 2) If the surface surrounds the entire circuit, the integral will be zero. (case # 3) If the surface surrounds just the resistor, with the wires coming in through the surface. the integral will give an inward directed power flow just equal to the power dissipated in the resistor. (case #4) If the surface cuts the wires twice, but does not include either the battery or the resistor, the integral is again zero as the power flow in equals the power flow out. ..."

    Regarding a battery at one end of the long pair of wires and a resistor at the other end of the long pair of wires ...
    Case #1 - cannot be performed because you are not near the battery
    Case #2 - cannot be performed because the wires are too long
    Case #3 - cannot be performed because you are not near the resistor
    Case #4 - can be performed but now the Poynting Vector integral is Zero!

    So, how can you use Poynting Vectors with two long wires to determine Source side vs Load side?
     
  17. nsaspook

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    As you see energy flows (outside) the path of both wires from the source to the load so there is no "side", only a direction for energy flow. We measure the EB fields to calculate the Poynting vector S so we also know the +- battery connections.
    http://sites.huji.ac.il/science/stc/staff_h/Igal/Research%20Articles/Pointing-AJP.pdf

    http://electretscientific.com/author/ajp/1962ajpv30n1pp19-21.pdf
    http://www.ifi.unicamp.br/~assis/Found-Phys-V29-p729-753(1999).pdf

    And finally another reason why the 'water' analogy is bad. http://www.matterandinteractions.org/Content/Articles/circuit.pdf
     
    Last edited: Mar 1, 2013
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