Feeding solar power back into municipal grid: Issues and finger-pointing

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On Apr 13, 12:06am, " snipped-for-privacy@att.bizzzzzzzzzzzz"

Do you think they need one more thing they don't understand? ;-)

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wrote:

One? Complex numbers would be a start, but the list is apparently endless.
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David Nebenzahl wrote:

Under the terms of the Ontario microFIT program, you are paid 80 cents per kwh for any electricity your project generates and "makes available to the grid".
For about the first year of the existance of the program, a "behind the meter" connection was allowable, but at some point last year, Measurement Canada (a federal gov't department tasked with regulating commercial scales and other forms of measurement devices) published some sort of guide or position paper stating their disapproval of this method.
"Behind the meter" meant that your house retained the same single hookup to the power mains lines (ie - the grid) and the primary meter be capable of bi-directional current measurement. The meter measuring the power output of your PV system (which also had to be bi-directional) could be connected to the grid through your meter. This is also known as a "series" connection.
Under the new(er) rules, your revenue meter must make a parallel connection to the grid (in parallel with your load meter). One result of this is that you will usually be billed an extra $5 or $10 a month for having a second service connection to the mains grid - even if it's the same physical wires carrying both services to your home.
As a load customer, you are billed based on what your primary meter is reading. As a power generator, you are paid for what your PV meter says you delivered to the grid. This is (and was) the case regardless how the revenue meter was connected.
The whole point of the microFIT program is to encourage home owners to fork out the estimated $35k to $55k to put up 3kw to 10kw PV system on their roof and contractually garantee them a rate of 80 cents per kwh for 20 years. You need approval all up and down several layers of burocracy to get your revenue meter plugged in (the last step of the process) before you start getting paid.
Alternatively, there is nothing stopping you from installing panels on your own home and basically hooking everything up exactly the same way as you would under the microFIT program, except there is no revenue meter. Your load bill would be the net energy you pulled from the grid. The payback you would get from your investment would therefore take much longer.

Unless your invertors were set to operate at a slightly higher output voltage. Even just a few volts differential between the mains voltage and the invertor output would mean that you could push current out into the grid, and by doing that raise the local grid voltage slightly.
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This explains a lot about inverter technology, though not whether they use a higher voltage, a leading phase angle or both to force power into the line: http://www.solarpanelsplus.com/solar-inverters/How-Solar-Inverters-Work-With-Solar-Panels.pdf
jsw
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This explains generator output regulation by varying voltage: http://www.basler.com/downloads/VR_parallel.pdf
and this the effect on current in or out of a synchronous generator caused by varying the leading/lagging phase angle between the internal magnetic field and the line voltage: http://nptel.iitm.ac.in/courses/IIT-MADRAS/Electrical_Machines_II/pdf/2_3.pdf
jsw
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g wrote:

Here's the problem:
Many of the load devices you find in a typical home (primarily electric motors that run cooling systems, air conditioners, fridges and freezers) are not capable of regulating their input voltage.
So when a secondary electricity source comes on-line (like a small PV system) then in order to push it's current into the local grid it will have to *try* to raise it's output voltage in order to see some current flow. It might only be a few volts, maybe less.
But does that mean there will be a measurable net reduction in the current being supplied by the high-voltage substation for that corner of the city?
Not if your typical load device in homes surround the PV system will simply operate at a higher wattage.
The only sort of load that can effectively be regulated by a slight increase in local grid voltage are electric heaters. When you raise their input voltage slightly, they will put out more BTU of heat, and if their heat output set-point doesn't change, then their operational duty cycle will change slightly.
But in the case of an AC compressor, the fact that it might be getting a slightly higher input voltage because a neighboring house is feeding PV power into the local grid won't mean that the AC compressor will reduce it's current consumption from the municipal utility supplier because of the extra current coming from a neighbor's roof-top solar array. It just means the motor will use BOTH sources of current and (I suppose) run a little hotter but in the end not do any extra cooling work in the process (it's rotational speed won't change).
Same theory would hold true for lighting (incandescent especially). If you raise the input voltage, you'll get more light output - the bulb will simply consume all the juice it would normally get from the utility in addition to that being supplied by the neighborhood PV system.
The only way that a neighborhood PV system can actually suppliment municipal utility power is when the PV system is wired up as a dedicated sole supply source for a few select branch circuits. The way I see it, you have to feed certain select loads 100% from a PV system (ie - disconnect them from the municipal energy source) if you're going to make a meaningful contribution to the supply-side of a municipal or city-wide grid.
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On 4/12/2011 9:33 AM, Home Guy wrote:

A devastating analysis.
I am sure when the utilities read it they will stop paralleling generators, since that just causes the amount of electricity used to go up from what would be used by isolated systems.
--
bud--

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"bud--" wrote in message
On 4/12/2011 9:33 AM, Home Guy wrote:

A devastating analysis.
I am sure when the utilities read it they will stop paralleling generators, since that just causes the amount of electricity used to go up from what would be used by isolated systems.
------------------
Careful! Sarcasm does not work well in a text medium, at all!
People cannot see your facial expression and people are never sure unless it is totally ridiculous.
I agree with your point but it is made very poorly in a text only medium.
mike
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bud-- full-quoted:

How many utilities connect the output of new parallel generating sources to the 120/208 connection side of a grid, instead of at the sub-station high-voltage side?
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For once I agree with Harry. We can forget about generators and distributions systems. Just take two 12V batteries and connect them in parallel to a 12ohm resistor. Under Homeguy's theories, I don;t know what he thinks would happen. But clearly he thinks if we put a second AC power source on a distributions system, it has to be at a higher voltage to "push" current out.
So, what happens with the two batteries? Under the laws of physics the rest of us use the voltage would remain at 12 volts and BOTH batteries would be supplying part of the 1 AMP flowing through the resistors.
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On Thu, 14 Apr 2011 08:52:52 -0700 (PDT), " snipped-for-privacy@optonline.net"

Oh, but according to Homeguy, that doesn't work because batteries are VOLTAGE sources. <slap!>

Not on Homeguy's and Vaughn's planet. One of the batteries will be charging the other.
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" snipped-for-privacy@att.bizzzzzzzzzzzz" wrote:

There's your problem right there.
It's usually not a good idea to connect batteries together in parallel, unless they are exactly of the same type, age, condition, etc. If you get a weak cell in one of the batteries it will turn into a load.
And what happens when they are not exactly at the same voltage before being connected together? How do you insure that you always get current flowing out of both of them?

The IEEE paper I posted earlier today shows exactly that - that PV systems raise local grid voltage and the utility company must compensate by reducing primary supply voltage to down-regulate the secondary voltage coming from the distribution transformer.

Take a 12 V car battery and wire it up in parallel with 8 AAA batteries connected in series. Then connect a load and tell me how much current the AAA batteries will supply vs their what their potential current supply could be if they were connected to their own isolated load.

If they are unequal in capacity, then yes that will eventually happen.
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What a dumbass!

They will share in proportion to their capacity. Electricity, water, nor shit flow uphill. ...though you have been pumping enough of the latter here.

Wrong!
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How to properly analyze the problem: http://www.electronics-tutorials.ws/dccircuits/dcp_4.html
Look at Example 1.
jsw
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Thank you Jim. I was sitting here wishing I could draw a simple circuit into the newsgroup that models what we are talking about and shows what really happens. You've gone one better and found a perfect example from an independent and credible source. That circuit is EXACTLY the model for the dual battery example I brought up. Each battery is modeled as an ideal voltage source in series with a resistor. And each supplies part of the current flowing through the load. One never charges the other, nor does one need to have a higher voltage to "push" current.
I suspect we'll be hearing soon from Homeguy about how this isn't a valid way two batteries connected in parallel to a load can be modeled.
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" snipped-for-privacy@optonline.net" wrote:

You dumbass.
Look more closely at the current flow in battery 1. It's NEGATIVE.
==================The negative sign for I1 means that the direction of current flow initially chosen was wrong, but never the less still valid. In fact, the 20v battery is charging the 10v battery. =================
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Home Guy wrote:

Where's trader4?
Where did you go, you coward?
You have no response to what I wrote above?
You disappeared from this tangent thread pretty fast, didn't you?
You absolutely loved that link posted above, showing Example 1 - where you thought it was proving me wrong.
Go ahead and substitute 13.33333 volts for Battery 1 in that example and tell me how much current it's supplying to the load.
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wrote:

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.
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On 4/15/2011 8:53 PM snipped-for-privacy@att.bizzzzzzzzzzzz spake thus:

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 ...)
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wrote:

Yes.
"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.
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