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

D

#### David Nebenzahl

Jan 1, 1970
0
All rotating machinery is AC.

Really, Harry? Really?

Is the spindle motor in your DVD player AC? How about the spindle motor

You do agree that those are all examples of "rotating machinery", don't you?

--
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*?)

T

#### The Daring Dufas

Jan 1, 1970
0
It's just another form of electricity generation in parallel with the
electricity grid. No different from any other form of generation. When
the PV array is generating the local voltage in the grid will rise a
little.
Or think of it another way. The current coming down the grid power
line will reduce as the PV panel takes up the local load, there will
be less voltage drop.
If the panel generates enough power the current in the grid will
reverse and the voltage close to the panel will be higher than the no

You could set up an interesting classroom experiment using colored water
or oil and clear tubing. A little grid with tiny widgets turning
from the flow of liquid being fed from different directions. It could
be fun. The pressure would represent voltage and flow represent current.

TDD

T

#### The Daring Dufas

Jan 1, 1970
0
How many things are wrong with what you wrote? let's see:

That's not Ohm's Law, not by a long shot. Do you even know what that is?

Yes, electricy flows from a point of higher potential (voltage) to a
lower point. But you're confusing voltage and current here, a common
rookie mistake.
Please enlighten us oh great one. ^_^

TDD

H

#### Home Guy

Jan 1, 1970
0
harry said:
Small scale locally generated power makes no difference to these
problems. In fact it helps.

Our regional and municipal electricity distributors are not pointing the
finger at the capability or specs of residential service panels or
neighborhood distribution / stepdown transformers as the reason why they
won't let small-scale (less than 10kw) roof-top PV systems to connect to
the grid.

They are saying that the local sub-station doesn't have the "capability"
to allow for a handful (or perhaps even a single) small-scale PV systems
to be hooked up and they would need to "upgrade" the sub-station in some
way.

http://www.greenpowertalk.org/archive/index.php/t-13885.html

And less relevant, here:

All the arguments put forward here about why homeowner-operated PV
systems (with nameplate rating under 10kw) are not being allowed to
connect to the grid through their own bi-directional revenue meter have
not addressed the issue as to how the connection of such a PV system can
possibly affect or influence the operation of the regional municipal
sub-station supplying power (at 20 kv?) to the neighborhood in
question. The sub-station is "insulated" from direct exposure to any
individual home by at least 1 step-down distribution transformer (in our
case, a ground-mounted distribution transformer supplying maybe 20 homes
- our electrical service runs underground - not on poles in our
neighborhood).

There may not yet even be a single residential PV system that's been
connected to the grid for the area being served by the sub-station in
question.

H

#### Home Guy

Jan 1, 1970
0
g said:
The grid can be seen as a pretty rigid beast. No small puny inverter
in the sub 1000kW class will much affect the grid voltage as a whole.
When voltage of the converter is attempted to be raised, current will
flow into the grid of course. The voltage increase will hardly be
measurable, as electrical characteristics of the grid will adjust
dynamically.

At any one time, there is a certain load on the grid as a whole. When
Mr. Homeowner adds 10Kw from some solar panels, some other power
generating systems connected to the grid will (have to) reduce their
output. As a result the voltage stays the same overall.

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.

B

#### bud--

Jan 1, 1970
0
This is a fatal flaw in your argument. Transformers are not infinite
sources. A utility transformer might supply a fault current 20x the
rated current (for a "5% impedance" transformer). (While a transformer
will supply a fault current larger than the rated current that is not
likely with PV. PV is basically a constant current source.)

Using a real transformer houses will have far less available fault current.

Cite where 100kA is required.
Only problem with that is that many home service panels use breakers
with an AIR of only 10kA, not 100kA. (my old house, built in 2000 was
10kA, and my new one, built in 2010 is also 10kA, both perfectly correct
by code)

Here's are some modern service panels that come with 10k AIR breakers.

And how many homes in the utilities service area are even up to current
code? I'd bet many homes in many service areas have only 10kA AIR.

I agree that is very likely. One reason is that a higher rating is not
necessary.

(SquareD, if I remember right, has a rating of 20kA downstream from both
the main and branch circuit breaker.)

I doubt many Canadian house panels have fuse protection, or are
different from US panels with circuit breaker protection rated around 10kA.
The utility that is being ultra-conservative may have to consider that
older homes in their service area may not even support this.

Can you just imagine the hue and cry when some homeowners are told they
have to spend a couple hundred bucks to upgrade their service panel
because of changes in the utility's distribution?

daestrom

The interrupt rating required goes up with the service current rating.
For a house, the utility is not likely to have over 10,000kA available
fault current. The transformers become too large, many houses are
supplied with longer wires and higher resistance losses, and the system
is much less safe.

I believe it would take a rather massive amount of PV installations to
cause a problem. The PV installations would all have to be on the
secondary of the same utility transformer. The transformer is then not
likely to support the PV current back to the grid. If the fault current
is 20x the transformer full load current, and the PV current is equal to
the transformer full load current, the PV supply would increase the
fault current by about 5% (assuming the inverter doesn't shut down). If
there were too many PV installations the utility could put fewer houses
on a transformer. Seems like a problem that is not that hard to handle
for the utility, at least until PV generation becomes rather common.

B

#### bud--

Jan 1, 1970
0
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.

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.

V

#### vaughn

Jan 1, 1970
0
David Nebenzahl said:
Is the spindle motor in your DVD player AC? How about the spindle motor on

No magic there! Spindle motors simplyt substitute solid state switching for
mechanical commutation. Actually, what is fed to the windings of a spindle
motor (though you may "nictpick" by calling it pulsating DC) actually more
resembles 3-phase AC. In fact, the windings of a spindle motor are usually
connected in a wye or delta configuration, sound familiar?
You do agree that those are all examples of "rotating machinery", don't you?

I agree that spindle motors are examples of rotating machinery. I don't agree
that proves any particular point for you.

Vaughn

M

#### m II

Jan 1, 1970
0
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Hash: SHA1

On 11-04-12 09:58 AM, m II wrote:

Nothing.

Forgery reported to nntp provider

mike

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G

#### g

Jan 1, 1970
0
Yes it does. Electrical current flows from a point of higher
potential to a lower point. The very first thing you learn. Ohm's Law.

So, you don't increase current by raising the voltage, but you increase
current by having a higher potential.

Now, difference in potential is voltage?

G

#### g

Jan 1, 1970
0
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.

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

4) The grid voltage does actually fluctuate a bit, depending on load.
fluctuations. The average subscriber never knows this.
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.

A simplified way is to look at the grid as a battery. When your PV
system generates more power than your local consumers, the surplus will
flow into the grid. At all other times the grid and the PV will both
supply the needed power to the local consumers.
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.

6) Fairly close to impossible. How do you match local power consumers to
hit the 100% PV capacity?

D

#### David Nebenzahl

Jan 1, 1970
0
No magic there! Spindle motors simplyt substitute solid state switching for
mechanical commutation. Actually, what is fed to the windings of a spindle
motor (though you may "nictpick" by calling it pulsating DC) actually more
resembles 3-phase AC. In fact, the windings of a spindle motor are usually
connected in a wye or delta configuration, sound familiar?

You're referring to steppers. Most spindle motors aren't steppers, but
simple DC motors (2-wire, no feedback sensors or additional windings). I
know; I've got a box full of 'em that I've removed from old CD and DVD
drives. No pulsating DC, no PWM (which in any case isn't AC anyhow).

What you say is true of steppers, which are used for moving the laser
sled on a CD/DVD drive, but not the spindle motor.

--
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*?)

D

#### David Nebenzahl

Jan 1, 1970
0
So, you don't increase current by raising the voltage, but you increase
current by having a higher potential.

No, no, no: increasing the current doesn't increase the potential
(that's voltage). It increases the *flow* of electricity (= current), at
least the maximum possible current. But that's not the same thing as
potential difference.

Example: Let's say you run your house off 12 volt batteries (just for
illustration). The *potential* of your power circuit is 12 volts
(assuming the batteries are fully charged, and they'll actually be
closer to 13.something, but let's call it 12).

Now let's say you add some more stuff to your house and find that your
lights are going dim because the battery can't provide enough *current*
(= amps) to the load. So what you do is add another battery in parallel
with the first one. This doubles the available current (= amps), but it
does *nothing* to change the voltage; it remains at 12 volts (nominal,
as explained above). This is true no matter how many batteries you add
*in parallel* with each other. But each battery increases the
*available* current (= amps) you can draw from your power source.

Notice that adding more batteries does not "push" more current through
the system; it increases the amount of current that can be "pulled"
(drawn from) the batteries.

Which is exactly the situation when you connect your photovoltaic system
to "the grid". It increases the *available current* to the grid. It does
not change the voltage of the grid; there's no need for it to be at a
higher voltage than (but it needs to be at about the *same* voltage as)
the grid.
Now, difference in potential is voltage?

Yes. Please refer to any good basic guide to electricity for more details.

--
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*?)

D

#### David Nebenzahl

Jan 1, 1970
0
Please enlighten us oh great one. ^_^

Read my posting a couple below this one.

--
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*?)

V

#### vaughn

Jan 1, 1970
0
David Nebenzahl said:
You're referring to steppers. Most spindle motors aren't steppers, but simple
DC motors (2-wire, no feedback sensors or additional windings). I know; I've
got a box full of 'em that I've removed from old CD and DVD drives. No
pulsating DC, no PWM (which in any case isn't AC anyhow).

Wrong. Spindle motors either have brushes and commutators or else the solid
state equivalent. Otherwise, they can't work. Just like the commutators &
brush system, that electronic "stuff" may be built directly into the motor where
you can't see it, but it's still there.

http://en.wikipedia.org/wiki/Brushless_DC_electric_motor

Vaughn

H

#### [email protected]

Jan 1, 1970
0
The utility needs to plan & build for all the peak-demand periods.
If every single home in a southern Canadian town installed 5-10
kw solar - the utility would still need to plan & design & build -
as if those panels didn't exist.
.. summer air conditioner peaks would be aces !
Air quality - big plus !
.. but the distribution system would still need to be built to
supply the peak at 5:45 pm December 20 -
.. solar output = zero. grid demand = near max.
John T.

H

#### Home Guy

Jan 1, 1970
0
bud-- full-quoted:
A devastating analysis.

I am sure when the utilities read it they will stop paralleling
generators,

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?

H

#### Home Guy

Jan 1, 1970
0
g said:
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.
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
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
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.
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

H

#### Home Guy

Jan 1, 1970
0
Jim said:
Bad analogy. The 1V will be lost in the internal resistance of the
inverter connection, which is much higher than that of the grid.

If that were the case, then your 42 amps would be converted into a
tremendous amount of heat as it burns up that internal resistance, and
there would be no measurable current for your revenue meter to measure.
Think of pouring water from a bucket into a lake.

For me to pour water into a lake, I have to raise it higher than the
lake level.

Think of height as eqivalent to voltage potential.
There's NO measurable rise in the lake level.

Unless water is compressible, there has to be a change in lake level.
The fact that I may not have a meter sensitive enough to measure it
doesn't mean there's no change in the level.

H

#### Home Guy

Jan 1, 1970
0
Bruce said:
The point here is that there is no such thing as the grid not being
able to accept the power you have produced. As long as you are
connected you can always force your KW in.

I don't think the issue is whether or not you can force current into the
grid via your 120/208 VAC service connection.

The question is:

a) does your power source need to overcome the instantaneous line
voltage in order to achieve a flow of current (answer: yes, and to the
extent that your power source has the capacity to do so, you raise the
output voltage as high as you can, because if you don't - then you have
excess capacity that is not going to make it out to the grid and hence
you won't gain revenue for the entire potential of your generating
system)

b) by raising the voltage on your local 120/208 grid, can your local
stepdown transformer adjust it's own operation by sensing that higher
voltage and reduce it's own output voltage in an attempt to regulate the
system back down to the desired setpoint? (answer: I don't know -
probably not. The neighborhood stepdown transformers probably weren't
designed to compete with sources of current being connected to their
distribution outputs).

c) So if the voltage on your local 120/208 grid is being raised slightly
because of your PV system and it's desire to push as much current back
into the grid as it can generate, then will this actually reduce the
amount of current that the regional sub-station is sending to your local
step-down transformer? (answer: the substation probably doesn't have a
direct line to your local stepdown transformer, and any alterations it
can make to it's output voltage is probably seen by many step-down
transformers including yours that are all wired to the same circuit. So
in reality it's doubtful that the regional substation would even sense
that your PV system has raised the local grid voltage).

d) So your PV system is raising the local grid voltage, and you're
probably pushing out 40 amps at 120 VAC or 20 amps at 240 VAC on a sunny
summer day. So what is that extra juice doing? Well, it's flowing
through the compressor motors of 10 to 20 of your neighbor's AC units -
whether they need it or not. Because you've raised the local grid
voltage slightly, that translates into a few extra watts (maybe 250
watts for each house that's fed from the same stepdown transformer). So
all the fridge compressors and AC compressor motors, lights - all linear
loads are going to blow away that extra line voltage as heat - instead
of useful work.

Nuf said?

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