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Feeding solar power back into municipal grid: Issues andfinger-pointing

Discussion in 'Home Power and Microgeneration' started by Home Guy, Apr 4, 2011.

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  1. Guest

    That's good for a first order approximation. The fact is that water doesn't
    run down hill and electricity does *not* flow from low potential to high. The
    inverter's instantaneous voltage *must* be higher than the grid for current to
    flow into the grid.

    Think about turning a bicycle crank, one without a clutch makes the point
    better. You're not doing any work unless you're applying pressure to the
    pedals. If you do nothing it drives you.
    No, you can't get on the "freeway" unless you're going faster than 60. If
    you're going slower, they're actively pushing you off.
    It's OK for a first order, but not for discussing the details people are
    trying to get into here.
     
  2. g

    g Guest

    Well, in a simplified scenario, (Grid, load, PV array) there will be
    "load sharing" between the grid and the PV array. With the voltage
    pretty much equal. I say pretty much equal, because there are line
    resistance losses to take into account.

    I think there is some misunderstanding about the concepts here.

    trader4 talks about 2 batteries supplying a load. The example is good
    for calculating load sharing between 2 voltage sources with load
    resistor and line resistance.

    The way I see it is that the LOAD with this battery analogy is the house
    load. One battery models the PV array, the other the grid.

    One thing should be clear: To get power into the grid at all, the house
    load must be lower than what the PV array can supply. If you remove the
    house load, then all the available PV array current will flow into the
    grid, with the inverter at a higher voltage.

    To get back to that Kirchoff's Law example, if we remove the 40 ohms
    resistor (the load), there are basically 2 voltage sources opposing each
    other. When these voltages are equal, no current flows.

    To allow current to flow into the grid, the PV array voltage has to be
    higher, whether there is a house load or not.
     
  3. Home Guy

    Home Guy Guest

    That's exactly what I've been saying - that you "turn it up" (the
    inverter's voltage output) to maximize the PV's current (I) supply into
    the grid.

    But everyone else (or most everyone else) is saying no - that simply
    matching the grid voltage (as measured at your service connection) is
    all that happens (and is all that needs to happen) for the entire PV
    current (I) capacity of the PV system to be "injected" into the grid.

    So now that we agree that PV systems need to raise the grid voltage if
    they're going to "force" their maximal available supply capacity into
    the grid, it's a moot or academic question as to what exactly their
    supply situtation would be (how much current they'd supply into the
    grid) if the invertors simply matched the grid voltage.
     
  4. Guest

    Motors aren't the only things that need PF correction. Capacitors aren't the
    only, or often the, way of doing it.
     
  5. Mho

    Mho Guest

    You guys need to get over your basic understanding of electricity.

    When you connect to a grid the voltage is always **EXACTLY*** the same as
    the connect point. They are in parallel.

    If your impedance source is lower than the grid's at that point you will
    supply the majority of the current. If your impedance source is higher (to
    the load) then you will supply less than the grid.

    One very simple rule. Two supplies in parallel output the same voltage. When
    you measure each of them you will measure across the same two points for
    both measurements.

    --------------------------

    "g" wrote in message
    Well, in a simplified scenario, (Grid, load, PV array) there will be
    "load sharing" between the grid and the PV array. With the voltage
    pretty much equal. I say pretty much equal, because there are line
    resistance losses to take into account.

    I think there is some misunderstanding about the concepts here.

    trader4 talks about 2 batteries supplying a load. The example is good
    for calculating load sharing between 2 voltage sources with load
    resistor and line resistance.

    The way I see it is that the LOAD with this battery analogy is the house
    load. One battery models the PV array, the other the grid.

    One thing should be clear: To get power into the grid at all, the house
    load must be lower than what the PV array can supply. If you remove the
    house load, then all the available PV array current will flow into the
    grid, with the inverter at a higher voltage.

    To get back to that Kirchoff's Law example, if we remove the 40 ohms
    resistor (the load), there are basically 2 voltage sources opposing each
    other. When these voltages are equal, no current flows.

    To allow current to flow into the grid, the PV array voltage has to be
    higher, whether there is a house load or not.
     
  6. Mho

    Mho Guest

    All that "load increases" is a bunch of baloney!

    A fixed load is just that....***FIXED***

    Sources share the load between themselves according to the impedance between
    and including the source and the load. If ht e grid were a perfect conductor
    and had zero impedance no co-gen source could work.

    --------------------------------

    wrote in message

    That was where I was coming from too. However, having gone back
    and revisited that basic circuit diagram of two voltage source driving
    a load, I have come around to where I see Homeguy's point that if a
    new
    source wants to deliver power onto the grid, it can raise the voltage
    at the load. Let's say I have a solar array that is covered up and
    its sunny outside. It's connected via distribution lines to a load
    that
    is a block away. Another power source is located a similar
    distance away from the load on the other side. In other words,
    a case like the circuit example Jim Wilkens provided.


    When I uncover the PV array, for it's additional X KW of
    power to make it to the load down the block, at least one
    of 3 things needs to happen:

    1 - The PV raises the voltage on it's end of the distribution system
    slightly.

    2 - the load increases

    3 - the power source on the other end of the distribution system
    lowers it's voltage.

    Here's the circuit example again that Jim provided:

    http://www.electronics-tutorials.ws/dccircuits/dcp_4.html

    It's example one.

    Voltage source V1 and R1 represent a simple battery,
    with R1 being the internal resistance of the battery.
    Same with V2 and R2. For our purposes a suitable
    model for a PV array and another power source on
    the other end of the distribution lines.


    Let's leave V2 as is at 20V, supplying all
    the current to the load, with no current coming from V1.
    You then have a simple series circuit
    consisting of resistors R2 and R3 connected to voltage
    source V2. A current of 20/(40+20) = .33A is flowing,
    which is I2 in the drawing. The only way for no power to be
    flowing through the other half of the circuit encompassing V1 is
    if V1 is at the exact same potential as the load resistor. With
    ..33A flowing through the load, R3, you have R3 at 13.2 V.
    That means V1 must be 13.2 volts. With V1 at 13.2 volts
    the voltage across R1 is zero and no current flows.
    So, everything is in balance. V1=13.2V, I1=0, the voltage
    on the load is 13.2 volts, and I2= .33A is flowing from V2
    through R2, R3.

    Now, if we want V1 to start supplying part of the power, what
    has to happen? V1 has to increase ABOVE 13.2 volts. And
    when it does, the voltage across the load resistor R3 will
    also increase. As that happens, current will start to flow now
    from V1 through the load resistor and at the same time the current
    from V2 will decrease slightly, as the potential drop across
    R2 is decreased slightly.

    The net result of this is that the voltage across the load
    has increased. Current I1 is now flowing from V1, I2 from V2
    is now slightly less and the combined currents of I1 and I2
    which together are I3 has increased slightly.

    The other ways to get V1 to supply power
    would be for either the load resistor R3 to decrease in
    value or for voltage source V2 to decrease.

    If you want to more closely model the situation, we could
    add two more resistors to model the distribution line
    resistance. A resistor could be added after R1 and after
    R2. But if you go through the analysis, it doesn't change
    the basics of the above analysis. For source V1 to supply
    power, the voltage on it's portion of the distribution system
    and across the load must increase.

    I think this is what Homeguy has been saying all along. So,
    I've come full circle here and now agree with him on that
    issue. I still disagree that the slight increase in voltage
    in a distribution system means that the power is wasted.
    The issue there is how the loads respond to being given
    121V instead of 120V. HG claims that except for resistance
    heaters, that energy goes to waste. And I say there he
    is wrong, but that topic is being covered in another part
    of this thread.
     
  7. g

    g Guest

    Define a fixed load.

    As a teaser, say I buy a 2 kW heater to heat my office in my home on
    those cold winter nights when I am reading posts in
    alt.energy.homepower. Is that 2 kW heater a fixed load?

    I disagree. But I have been known to be wrong :)

    Could you state why it does not work?

    Is it possible to hook up any power source to such a grid?
    If the answer is yes, why is co-gen not possible?
     
  8. Whoa. Where do you get that (using Smitty's analogy)? Sure, you can't be
    going slower; that's self-evident. You say you have to be going faster.
    But you say nothing about going (approximately) *the same speed*, which
    is what Smitty's example was saying. (And is what ever article I've ever
    read about inverters, grid interties, etc., has said. (*None* of them
    say "the voltage of the contributing system has to be slightly higher
    than the grid in order to feed current into it". None of 'em.)

    (Which, by the way, is eggs-ackley the same thing I've been saying here ...)


    --
    The current state of literacy in our advanced civilization:

    yo
    wassup
    nuttin
    wan2 hang
    k
    where
    here
    k
    l8tr
    by

    - from Usenet (what's *that*?)
     
  9. Guest

    "Approximately" means a little faster, or slower. Slower does *not* work.

    (And is what ever article I've ever
    Of course they don't say that. Physics does.
    Whatever you've been saying doesn't change physics.
     
  10. On 4/17/2011 2:20 PM spake thus:

    [cut to the chase, i.e., trader4's example:]
    All I can say is, good example, but wrong conclusion.

    Let's change the example just a little. We'll have a large AC source
    (call it "the grid"), and a smaller AC source (the solar system's
    inverter), connected *in parallel*, and then some resistance (the total
    load of "the grid") after all that, completing the circuit.

    In this example, "the grid" and the PV inverter are operating at
    *exactly* the same voltage.

    Let me argue this negatively, and see what you say to it:
    If you're saying that this will not work (i.e., that the PV inverter
    cannot contribute any power to the circuit because it isn't at a higher
    voltage), then *no* circuit where you have two power sources in parallel
    could ever work under the same circumstances.

    No electric vehicle would ever work, because the battery cells in them
    are (pretty close to) exactly the same voltage, so how would each tiny
    cell (tiny in comparison to the total power of its siblings) ever be
    able to "push" electricity into the circuit?

    That elementary electronics tutorial (Kirchoff's Law, etc.) explains
    everything you need to know here. And it doesn't require any higher voltage.

    I look forward to your reply.


    --
    The current state of literacy in our advanced civilization:

    yo
    wassup
    nuttin
    wan2 hang
    k
    where
    here
    k
    l8tr
    by

    - from Usenet (what's *that*?)
     
  11. Guest

    Simple: The motor is at a lower potential.
    Elementary, sure, but you're still not getting it.
     
  12. MarkK

    MarkK Guest

    test
     
  13. Smitty Two

    Smitty Two Guest

    I've chewed this over a bit, and I still don't like it, and here are my
    reasons:

    1: Voltage sources in parallel do not push *against* one another.

    2: If no voltage source can join the grid without being at a higher than
    grid potential, then every contributing power station would have to be
    at a higher potential than every other one, and that's impossible.

    3: While voltage might *push*, it's the load that it said to *pull* the
    current. If there's a demand, current will flow whether the supply
    voltage is 119, 120 or 121.


    Where is my thinking flawed?
     
  14. Guest

    No, Roy, you are *not* my type. Now go *away*!
    Isn't it cute when Roy Queerjano accuses others of being what he *is*.
    Sorry, Pat Eats Cum, I'm already married - longer than you've been alive,
    moron.
    No it's *not*, Roy.
     
  15. Smitty Two

    Smitty Two Guest

    None of the things you just said mean anything. Saying it's "due to
    physics" is meaningless. I know wire has resistance, so what? And I'm
    sure you know that when I said "I don't like it" I meant "I don't agree
    with it." And, uh, rivers don't fight against tributaries, last I
    checked, but you shouldn't be using analogies if you don't think they
    hold water. So to speak.

    You are billed by how much current you draw. Let's say your normal
    consumption is a steady 1 kw. Now you start generating 100 watts. So now
    you're only drawing 900 watts from the rest of the grid, and that,
    multiplied by hours, is what you get charged for.

    OTOH, let's say you install some bigger panels, and you can generate
    1500 watts. Now your net consumption is *minus* 500 watts, so you're
    pumping 500 watts into the grid, and, hopefully, being compensated for
    that by the power company. You're using 1 kw locally, and delivering the
    rest where it's needed elsewhere.

    I seriously fail to see where the resistance of the wire has anything to
    do with it. Please don't tell me "that's just the way it is." If I'm
    wrong, give me a reasonable and logical argument.
     
  16. Guest

    Let me say it again, perhaps you'll catch on. Physics doesn't care what you
    like. It is what it is.
    Because, like electricity, water always flows "down hill" - high to low.
    Voltage is dropped across a resistance. Not all points in the grid go up
    because your solar cells output more voltage because there is non-zero
    resistance between all points. If you're generating electricity, your house
    will be at a *higher* voltage than the pole. If the resistance of the wire
    were zero this couldn't happen.
     
  17. But... but... but... EVERY car insurance company is cheaper than its
    competitors and every alkaline cell is the longest lasting... LOL

    Seriously, the power stations are not higher than each other, but every one
    is higher than the grid, imagine three power stations, or 3 generators in
    the same station all putting out exactly 48KV, the substation/transformer
    they are connected to is getting slightly less, say 47,990 volts. If they
    then need to bring a fourth generator online they use a syncroscope to
    adjust the prime mover to sync up the frequency and phase, adjust the field
    to match the voltage IE 47,990, close the breakers, and slowly increase the
    field and prime mover so that the load is shared and now all 4 are at 48KV.
     
  18. It *sounds*--and I'm sure you'll correct me if I'm wrong--as if you're
    agreeing with me, and with Smitty, and others when we say that it is
    *not* required that the photovoltaic inverter supply a higher voltage in
    order to transfer current to the grid. (I take this from the last
    sentence in the next-to-last paragraph, where you say " ... will adapt
    itself to the line voltage, whatever it may be".)

    The arguments against this, with all the pseudo-science being thrown
    around (most of it by the ones who are also slinging insults) are
    getting quite tiresome here.


    --
    The current state of literacy in our advanced civilization:

    yo
    wassup
    nuttin
    wan2 hang
    k
    where
    here
    k
    l8tr
    by

    - from Usenet (what's *that*?)
     
  19. Guest

    No, the voltage will STILL be higher if you're supplying current to the grid.
    Wires have resistance. Current sources don't go against physics and really
    are the same thing as voltage sources (Norton/Thevenin duality). Physics
    doesn't lie.
    You're throwing around the pseudo-science. If you don't like the treatment,
    you can easily leave.
     
  20. g

    g Guest

    Well, I was just teasing him, did not really expect any answer.
     
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