g wrote:

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.

1) The actual voltage increase will relate to the ratio of grid
impedance vs local impedance, i.e. your local power consumers
(fridges, heaters etc) has a much higher impedance relatively,
thus the grid will "take" the majority of the generated power.

I'm not arguing that the grid can't or won't take any, the majority, or
all of the generated power.

The question here is - what exactly must the invertors do in order to
get as much current as the PV system can supply into the grid.

If our analogy is pipes, water, and water pressure, then we have some
pipes sitting at 120 PSI and we have a pump that must generate at least
121 PSI in order to push water into the already pressurized pipes. So
the local pipe system now has a pressure of 121 PSI. If you measure the
pressure far away from your pump, it will be 120 psi.

The _only_ increase in voltage you will see results from the
voltage drop in the grid components.

Not sure I understand what you're trying to say there.

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?

2) Pretty complex calculation, but yes, _somewhere_ one or
more generating pieces of machinery will reduce its output.
Makes sense intuitively, does it not?

No, I don't agree.

Hypothetically speaking, let's assume the local grid load is just a
bunch of incandecent lights. A typical residential PV system might be,
say, 5 kw. At 120 volts, that's about 42 amps. How are you going to
push out 42 amps out to the grid? You're not going to do it by matching
the grid voltage. You have to raise the grid voltage (at least as
measured at your service connection) by lets say 1 volt. So all those
incandescent bulbs being powered by the local grid will now see 121
volts instead of 120 volts. They're going to burn a little brighter -
they're going to use all of the current that the local grid was already
supplying to them, plus they're going to use your current as well.

Doesn't matter if we're talking about incandescent bulbs or AC motors.
Switching power supplies - different story - but they're not a big part
of the load anyways.

Not if your typical load device in homes surround the PV
system will simply operate at a higher wattage.

3) You just set your PV system to operate at max power, the
grid system will balance out automatically. See 1) above

I don't see how - not at the level of the neighborhood step-down
transformer. I don't see any mechanism for "balancing" to happen there.

The only way that a neighborhood PV system can actually
supplement municipal utility power is when the PV system
is wired up as a dedicated sole supply source for a few
select branch circuits.

5) That will be a very inefficient way to utilize your PV
system.

If you're getting paid for every kwh of juice you're feeding into some
revenue load, then the concept of "efficiency" doesn't apply. What does
apply is ergonomics and practicality. I agree that a small-scale PV
system can't be counted on to supply a reliable amount of power 24/7 to
a revenue load customer (or even a dedicated branch circuit of a revenue
load customer) to make such an effort workable - but I still stand by my
assertion that the extra current a small PV system injects into the
local low-voltage grid will not result in a current reduction from the
utility's sub station to the local step-down transformer.

The extra current injected by the PV system will result in a small
increase in the local grid voltage which in turn will be 100% consumed
by local grid loads (motors, lights) and converted into waste heat with
no additional useful work done by those load devices.