On 12/04/2011 07:33, Home Guy wrote:
g wrote:


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.

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.
Power companies have means of adjusting line voltages 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?

On 4/12/2011 9:30 AM vaughn spake thus:

"David Nebenzahl" wrote in message
s.com...

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

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:

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wassup
nuttin
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On 4/12/2011 9:47 AM g spake thus:

On 11/04/2011 23:17, harry wrote:

On Apr 11, 8:01 pm, David
wrote:

You don't "push" electricity from your solar installation into the grid
by raising the voltage, as someone here postulated. It just don't work
that way.

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.

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
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On 4/12/2011 12:48 AM The Daring Dufas spake thus:

On 4/12/2011 1:47 AM, David Nebenzahl wrote:

On 4/11/2011 11:17 PM harry spake thus:

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.

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. ^_^

Read my posting a couple below this one.


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

yo
wassup
nuttin
wan2 hang
k
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"David Nebenzahl" wrote in message
s.com...
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/Brushle...electric_motor

Vaughn

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?

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.

On Apr 12, 7:50*pm, Home Guy wrote:
...
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.
...

Bad analogy. The 1V will be lost in the internal resistance of the
inverter connection, which is much higher than that of the grid. Think
of pouring water from a bucket into a lake. There's NO measurable rise
in the lake level.

jsw

Bruce Richmond wrote:

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?

On 12/04/2011 16:50, Home Guy wrote:
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.

The inverter must ensure that it transforms the DC from PV to the
frequency and voltage of the grid. To ensure flow of current into the
grid the voltage must be attempted to be raised. Because there are
losses between the inverter and the grid, the voltage will be higher
than 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.

Fairly good analogy, and due to internal resistance in the pipe then
that must be overcome by having a higher pressure. Don't forget that
somewhere someone else has to reduce the water flow into the pipe system
in order to avoid pressure buildup. Because the water in the pipe system
is used up as it is supplied, at the same rate.


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.

See the pipe analogy above, the power lines from the inverter has some
resistance, which results in a voltage drop. Therefore the voltage
measured at the inverter will be slightly higher than measured a
distance away.


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.

Why? take a hypothetical grid with 1 megawatt consumption. Generating
machinery produce that energy at a set voltage. Mr Homeowner connects to
the grid with a 10kW PV array. If no power utility adjustment took place
then the overall voltage of the grid will increase. OK for small
fluctuations, but if enough PV arrays came online, somewhere energy
production has to decrease or bad things will happen due to high grid
voltage.



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 cannot unless your local load is zero. You must subtract the local
load from the generated PV array power if the house load is lower. If
the house load is higher than the PV array output then you will use all
the PV array power with the difference supplied from the grid.


They're going to burn a little brighter -
Correct, due to a slightly raised voltage if there is a voltage drop
between the inverter and the grid. (There is some drop)

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.
Not possible, the current is controlled by the internal resistance in
the lamp. They will draw a current by the formula volt/resistance.
So when the PV array produces current, grid current is reduced.

The voltage increase you will see at the output of the inverter is very
small, but it does depend on the cables used.

An example: I have a 300 feet underground cable to the nearest utility
transformer and a 100A service panel.

If I max out the power, I will have a voltage drop over the cable of
about 6 Volts. Much higher than normal households.

When your PV array is producing full power, and your house load matches
that, then the voltage difference between the grid and inverter is zero.

at any other house load, current will flow in the power utility lines,
and the inverter voltage increase is a function of the loss in those
lines.

On Apr 4, 8:52*am, The Daring Dufas
wrote:
On 4/4/2011 4:35 AM, harry wrote:









On Apr 4, 3:17 am, *wrote:
* *We are talking about installations that can't

generate more than 10 kw - and more likely would only generate 5 or 6 kw
on a mid-summer day, with the bulk of that energy being consumed by the
home owner's own AC unit (I'm sure) with little or none to spare to be
injected back into the neighborhood grid.

Comments?

agreed, this is actually a bureaucratic *economics or business problem
and they are trying to call it a technical problem.

It isn't a technical problem.

Mark

The above comment is exactly right.
There is no technical problem. The PV array can be sized to overcome
any supply side issues. *What comes in can equally well go out.
But see diversity factor.
http://en.wikipedia.org/wiki/Diversity_factor
This has the biggest bearing on the matter.
Sounds tome that there are politcal/financial matters yet unresovled.

I am having a 4Kwp array fitted to my roof in two weeks time. *(UK)
It's just *a money thing. I shall have a 12% return on capital.

Keep an eye on your solar array my friend, those things like air
conditioners around here are being stolen at an increasing rate.
People install them at their remote cabins or camps only to return
to a powerless abode. Thieves will steel them while a home owner
is asleep at night!

http://www.usedsolarpowerpanels.com/...ng-solar-panel...

http://preview.tinyurl.com/3cc3rkl

TDD

A lot of them were sold for mountain vacation homes. Thieves steal
them often for the copper though there is not that much copper in
them.

On Tue, 12 Apr 2011 01:21:40 -0500, sno wrote:

On 4/11/2011 6:16 PM, zzzzzzzzzz wrote:
On Mon, 11 Apr 2011 17:54:33 -0400,
wrote:


wrote in message
...
On Apr 11, 1:32 pm, wrote:

Not he's got it right except for the fact that allr otating electric
lmachinary is AC.

Those two quite common examples seem to refute your
absolute determination that ALL rotating electric machinery
is operated with AC motors...

How so?


Look deeper in the motor. It's all AC on the inside. ;-)

I think you are pushing it....the brushes on a dc motor "guide" the dc
to different windings.....it is still dc...

No, it most definitely is *not*. Turn the motor to the next commutator step
and you'll see that the current reverses in the winding.

In an ac motor the windings are generally in parallel...all the ac is
applied at one time....

I think I got that right....is a long time since I covered motor
theory..grin...the ac is not "chopped up"....or guided anywhere....

That's why there are no brushes in AC motors? ;-)

You could say that all electric motors operate the same....as they all
depend on magnetism (all generally used motors...there are some operate
on static electricity, etc)
have fun...sno

....and you need a rotating magnetic field. That is, you need AC. ;-)

On Tue, 12 Apr 2011 15:24:52 -0400, "vaughn"
wrote:


"David Nebenzahl" wrote in message
rs.com...
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.

Correct. David should refrain from any EE lectures. He hasn't got it in him.

On Tue, 12 Apr 2011 18:50:22 -0700 (PDT), Bruce Richmond
wrote:

On Apr 11, 2:19*am, David Nebenzahl wrote:
On 4/10/2011 10:02 PM Bruce Richmond spake thus:





On Apr 10, 3:54 am, David Nebenzahl wrote:

That second statement is correct: you can't "push" electrons into
the grid. But it doesn't matter *how* your inverters are working;
it's a basic law of physics.

If you apply more volts to a line than what it is carrying what do you
think happens? *I run machines that use regenerative braking. *They
draw energy from the line to set things in motion. *To slow or stop
them the electric motor acts as a generator producing a higher voltage
than the grid, forcing power back into the grid. *An inverter can do
the same thing using solid state circuits. *The inverter in my Prius
takes DC current from the battery and converts it to whatever voltage
and frequency is needed at the time to run the variable frequency AC
motor. *When slowing down the motor becomes an AC generator and the
inverter converts the output to a DC voltage just a bit higher than
the battery, pumping charge back into it.

Sorry, I don't think you know what you're talking about.

If you do a little research I think you will change your mind.

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

At the bottom of that page you will find this link

http://www.solarpanelsplus.com/solar...lar-Panels.pdf

You seem to think that you can "force" or push "voltage" into a line, by
using a higher voltage than what's on the line.

More specificly I wrote, "forcing power back into the grid". Power is
watts or KW. That's volts times amps.

But watts is *not* volts times amps, in an AC circuit. There is a power
factor in there to worry about. In the capacitor example, watts dissipated is
zero (or close to it) but VA might be rather high.

The current will only flow if there is a difference in voltage.

Correct. Ohms Law.

In article ,
" wrote:

...and you need a rotating magnetic field. That is, you need AC. ;-)

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

On 4/12/2011 9:06 PM zzzzzzzzzz spake thus:

The current will only flow if there is a difference in voltage.

Correct. Ohms Law.

That is *not* Ohm's Law. Where do you get that? Sheesh--you're trying to
lecture *me* on this stuff???


--
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*?)

On Apr 12, 6:39*pm, g wrote:
On 12/04/2011 07:33, Home Guy wrote:





g wrote:

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.

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.
Power companies have means of adjusting line voltages 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?- Hide quoted text -

You don't. The supply company does that. The PV panel is insignifcant
compared with what they generate in any case.
If there got tobe huge ones or many small ones this would be more of a
problem.

On Apr 12, 7:44*pm, David Nebenzahl wrote:
On 4/12/2011 9:47 AM g spake thus:

On 11/04/2011 23:17, harry wrote:

On Apr 11, 8:01 pm, David
wrote:

You don't "push" electricity from your solar installation into the grid
by raising the voltage, as someone here postulated. It just don't work
that way.

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.

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*?)

The explanation for the above is that the battery too has internal
resistance. This why it's output is limited. The internal resistance
varies with the load and state of charge too.
"Potential" is an obsolete and confusing term, see EMF. (Electro-
Motive Force).

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

Emf is the driving force behind the whole system (measured in volts)
Voltage is the difference (or potential if you like) between any two
points in the system.

On Apr 13, 12:07*am, Home Guy wrote:
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?

Irrelevent.

On Apr 13, 1:05*am, Jim Wilkins wrote:
On Apr 12, 7:50*pm, Home Guy wrote:

...
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.
...

Bad analogy. The 1V will be lost in the internal resistance of the
inverter connection, which is much higher than that of the grid. Think
of pouring water from a bucket into a lake. There's NO measurable rise
in the lake level.

jsw

What happens to the "say" 1volt. It is only a local thing because
the utility drops it's output by 5Kw. Sure, there will be different
current flow around the system but nothing that can't be handled.

I don't know why you rabbit claiming it doesn't work when it clearly
does.

On Apr 13, 2:49*am, Home Guy wrote:
Jim Wilkins wrote:
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.

To continue that anology, don't forget someone somewhere takes a
bucket of water out of the lake.
Lots of buckets in. Lots of buckets out.

On Apr 13, 2:50*am, Bruce Richmond wrote:
On Apr 11, 2:19*am, David Nebenzahl wrote:





On 4/10/2011 10:02 PM Bruce Richmond spake thus:

On Apr 10, 3:54 am, David Nebenzahl wrote:

That second statement is correct: you can't "push" electrons into
the grid. But it doesn't matter *how* your inverters are working;
it's a basic law of physics.

If you apply more volts to a line than what it is carrying what do you
think happens? *I run machines that use regenerative braking. *They
draw energy from the line to set things in motion. *To slow or stop
them the electric motor acts as a generator producing a higher voltage
than the grid, forcing power back into the grid. *An inverter can do
the same thing using solid state circuits. *The inverter in my Prius
takes DC current from the battery and converts it to whatever voltage
and frequency is needed at the time to run the variable frequency AC
motor. *When slowing down the motor becomes an AC generator and the
inverter converts the output to a DC voltage just a bit higher than
the battery, pumping charge back into it.

Sorry, I don't think you know what you're talking about.

If you do a little research I think you will change your mind.

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

At the bottom of that page you will find this link

http://www.solarpanelsplus.com/solar...r-Inverters-Wo...

You seem to think that you can "force" or push "voltage" into a line, by
using a higher voltage than what's on the line.

More specificly I wrote, "forcing power back into the grid". *Power is
watts or KW. *That's volts times amps. *The current will only flow if
there is a difference in voltage.

That's not at all what's at work here when one has a photovoltaic system
and an intertie feeding power back into "the grid".

The intertie and the house's power connection are going to be at pretty
much exactly the same voltage. What happens is that the PV system is
connected *in parallel* with the grid; it's dumping more *current* into
the system, not more voltage.

It takes very little voltage difference to flow a lot of current when
the "load" is the grid.

You do understand the difference between current and voltage, don't you?

Yes, I do. *Now let's see if you can understand this.

Take two 12 volt car batteries with one discharged to 11 volts.
Connect them in parallel and check the voltage. *It will be some value
between what they measured seperately. *While they are connected like
this current is flowing from the charged battery to the discharged
battery. *Power is being forced into it raising its state of charge.
If the two batteries had been at exactly the same voltage there would
have been no current flow. *Now take 8 AAA batteries connected in
series to give 12 volts. *Disconnect the charged battery and connect
the AAAs to the 11 volt battery. *Measure the voltage. *It will be for
all practical purposes unchanged from 11 volts. *The AAA cells are
charging the bigger battery but they are so small compared to it that
they seem insignificant. *That is how your PV system looks to the
grid.

The voltage on the grid can vary by up to + or - 10%. *It is usually
kept withing + or - 5%. *Take a volt meter and check the voltage at
your wall outlet. *Check it several times during the day and you will
find that it varies. *Don't try to claim that it doesn't, check it and
you will find that it does. *As loads are put on the grid it drags the
voltage down. *The power company responds by generating more power to
bring the voltage back up. *As loads are taken off the voltage will
climb, and it is brought back down by producing less power.

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.- Hide quoted text -

- Show quoted text -

If you connect your two batteries in parallel, the 12v one will charge
the 11volt one up. The voltage at the terminals willbe the same but
current will be flowing.
In fact in this case, sufficient current would flow to irrepairably
damage the batteries.

On Apr 13, 2:50*am, Bruce Richmond wrote:
On Apr 11, 2:19*am, David Nebenzahl wrote:





On 4/10/2011 10:02 PM Bruce Richmond spake thus:

On Apr 10, 3:54 am, David Nebenzahl wrote:

That second statement is correct: you can't "push" electrons into
the grid. But it doesn't matter *how* your inverters are working;
it's a basic law of physics.

If you apply more volts to a line than what it is carrying what do you
think happens? *I run machines that use regenerative braking. *They
draw energy from the line to set things in motion. *To slow or stop
them the electric motor acts as a generator producing a higher voltage
than the grid, forcing power back into the grid. *An inverter can do
the same thing using solid state circuits. *The inverter in my Prius
takes DC current from the battery and converts it to whatever voltage
and frequency is needed at the time to run the variable frequency AC
motor. *When slowing down the motor becomes an AC generator and the
inverter converts the output to a DC voltage just a bit higher than
the battery, pumping charge back into it.

Sorry, I don't think you know what you're talking about.

If you do a little research I think you will change your mind.

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

At the bottom of that page you will find this link

http://www.solarpanelsplus.com/solar...r-Inverters-Wo...

You seem to think that you can "force" or push "voltage" into a line, by
using a higher voltage than what's on the line.

More specificly I wrote, "forcing power back into the grid". *Power is
watts or KW. *That's volts times amps. *The current will only flow if
there is a difference in voltage.

That's not at all what's at work here when one has a photovoltaic system
and an intertie feeding power back into "the grid".

The intertie and the house's power connection are going to be at pretty
much exactly the same voltage. What happens is that the PV system is
connected *in parallel* with the grid; it's dumping more *current* into
the system, not more voltage.

It takes very little voltage difference to flow a lot of current when
the "load" is the grid.

You do understand the difference between current and voltage, don't you?

Yes, I do. *Now let's see if you can understand this.

Take two 12 volt car batteries with one discharged to 11 volts.
Connect them in parallel and check the voltage. *It will be some value
between what they measured seperately. *While they are connected like
this current is flowing from the charged battery to the discharged
battery. *Power is being forced into it raising its state of charge.
If the two batteries had been at exactly the same voltage there would
have been no current flow. *Now take 8 AAA batteries connected in
series to give 12 volts. *Disconnect the charged battery and connect
the AAAs to the 11 volt battery. *Measure the voltage. *It will be for
all practical purposes unchanged from 11 volts. *The AAA cells are
charging the bigger battery but they are so small compared to it that
they seem insignificant. *That is how your PV system looks to the
grid.

The voltage on the grid can vary by up to + or - 10%. *It is usually
kept withing + or - 5%. *Take a volt meter and check the voltage at
your wall outlet. *Check it several times during the day and you will
find that it varies. *Don't try to claim that it doesn't, check it and
you will find that it does. *As loads are put on the grid it drags the
voltage down. *The power company responds by generating more power to
bring the voltage back up. *As loads are taken off the voltage will
climb, and it is brought back down by producing less power.

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.- Hide quoted text -

- Show quoted text -

Assuming someone is taking it out somewhere you can :-)

On Apr 13, 4:02*am, JIMMIE wrote:
On Apr 4, 8:52*am, The Daring Dufas
wrote:





On 4/4/2011 4:35 AM, harry wrote:

On Apr 4, 3:17 am, *wrote:
* *We are talking about installations that can't

generate more than 10 kw - and more likely would only generate 5 or 6 kw
on a mid-summer day, with the bulk of that energy being consumed by the
home owner's own AC unit (I'm sure) with little or none to spare to be
injected back into the neighborhood grid.

Comments?

agreed, this is actually a bureaucratic *economics or business problem
and they are trying to call it a technical problem.

It isn't a technical problem.

Mark

The above comment is exactly right.
There is no technical problem. The PV array can be sized to overcome
any supply side issues. *What comes in can equally well go out.
But see diversity factor.
http://en.wikipedia.org/wiki/Diversity_factor
This has the biggest bearing on the matter.
Sounds tome that there are politcal/financial matters yet unresovled.

I am having a 4Kwp array fitted to my roof in two weeks time. *(UK)
It's just *a money thing. I shall have a 12% return on capital.

Keep an eye on your solar array my friend, those things like air
conditioners around here are being stolen at an increasing rate.
People install them at their remote cabins or camps only to return
to a powerless abode. Thieves will steel them while a home owner
is asleep at night!

http://www.usedsolarpowerpanels.com/...ng-solar-panel...

http://preview.tinyurl.com/3cc3rkl

TDD

A lot of them were sold for mountain vacation homes. Thieves steal
them often for *the copper though there is not that much copper in
them.- Hide quoted text -

- Show quoted text -

They steal themto sell them.There is no copper in them apart from the
connecting wires.

On Apr 13, 5:23*am, David Nebenzahl wrote:
On 4/12/2011 9:06 PM spake thus:

The current will only flow if there is a difference in voltage.

Correct. *Ohms Law.

That is *not* Ohm's Law. Where do you get that? Sheesh--you're trying to
lecture *me* on this stuff???

--
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*?)

Yes it's Ohm's LAW. Law means no exceptions.Totally proven.

On 4/12/2011 10:58 AM, m II wrote:


"bud--" wrote in message ...

On 4/11/2011 4:31 PM, daestrom wrote:
On 4/10/2011 22:12 PM, m II wrote:


"daestrom" wrote in message ...

On 4/6/2011 19:31 PM, m II wrote:
The fault capacity of a household main breaker or fuses is not an
issue,
unless very old technology, like you.
One hundred feet of twisted triplex supply cable limits faults to well
within the fault tolerances.


Got some numbers/calculations to support that? Is that including the
next door neighbors with their PV installation?

daestrom

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

Sure! Basic Ohms lawa and a wire resistance table

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

A 200 ampere service running 240 Vac and only considering the straight
resistance of copper (many use AL outside conductors these days).
and considering the street transformer as an infinite current supply (0
Ohms impedance)

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.)


The chart shows we would use 2/0 copper (assuming solid copper, but it
won't be)

In a 100 feet of overhead run to a house, down the stack and through the
meter to the main panel, where the fuses or breakers are, not
considering the impedance of the overcurrent devices (that allegedly
cannot handle a fault this big) we come up a with a minimum copper
resistance of

200 feet (has to return) x 0.07793 x 10^-3 Ohms / foot (oh look ...your
old units too) = 0.015586 Ohms

Using 240 Vac as the fault supply (it won't be under a faulted
condition) the max fault current would be

240 Vac / 0.015586 Ohms = 15.4 kA.

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


Now we havent figured in any of the other impedances (very generous)
and any approved O/C device in a panel these days is rated at 100kA.

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.
http://static.schneider-electric.us/...ad-centers.pdf

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.

*--
bud--

-----------------
|Perhaps re-read ( or just read ) the last few posts. Your objection is
|mostly agreement with items already covered.

Perhaps you should take reading lessons. Maybe you and harry could get
group rates.

- You said "considering the street transformer as an infinite current
supply" which no one does.
- As a result your calculation is meaningless.
- You said Canadian house panels were protected with fuses. I disagree.
Perhaps a cite?
- You said "any approved O/C device in a panel these days is rated at
100kA". I asked for a cite - still missing.

- Daestrom said adding PV systems to residences could result in an
available fault current larger than the rating of existing service
panels. It is certainly an interesting point, but not likely for reasons
stated.

I did agree with daestrom that most US house panels are likely to have a
10kA IR.


|Can you cite the percent impedance of the transformers

5% impedance would be common

| or the code rules you discuss?

I didn't discuss code rules.


Your 'newsreader' is incompetent at treating sigs.

--
bud--

On 4/12/2011 11:59 AM, harry wrote:
On Apr 12, 4:44 pm, wrote:
On 4/11/2011 4:31 PM, daestrom wrote:





On 4/10/2011 22:12 PM, m II wrote:

"daestrom" wrote in ...

On 4/6/2011 19:31 PM, m II wrote:
The fault capacity of a household main breaker or fuses is not an issue,
unless very old technology, like you.
One hundred feet of twisted triplex supply cable limits faults to well
within the fault tolerances.

Got some numbers/calculations to support that? Is that including the
next door neighbors with their PV installation?

daestrom

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

Sure! Basic Ohms lawa and a wire resistance table

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

A 200 ampere service running 240 Vac and only considering the straight
resistance of copper (many use AL outside conductors these days).
and considering the street transformer as an infinite current supply (0
Ohms impedance)

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.)







The chart shows we would use 2/0 copper (assuming solid copper, but it
won't be)

In a 100 feet of overhead run to a house, down the stack and through the
meter to the main panel, where the fuses or breakers are, not
considering the impedance of the overcurrent devices (that allegedly
cannot handle a fault this big) we come up a with a minimum copper
resistance of

200 feet (has to return) x 0.07793 x 10^-3 Ohms / foot (oh look ...your
old units too) = 0.015586 Ohms

Using 240 Vac as the fault supply (it won't be under a faulted
condition) the max fault current would be

240 Vac / 0.015586 Ohms = 15.4 kA.

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



Now we haven’t figured in any of the other impedances (very generous)
and any approved O/C device in a panel these days is rated at 100kA.

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.
http://static.schneider-electric.us/...ad-centers.pdf

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.

--
bud--

In theory transformers can carry any load.

If you are talking about normal load ratings - what a useful revelation.
I am sure no one had any idea...

In practice there are losses due to resistance of the copper wire and
in the iron core that cause heating. How swiftly this heat can be
dissipated is one load carrying limtation. How much heat the
insulation can stand is another.

There is a limit on the normal current for a transformer?
I had no idea....

But heating is not a limit on fault current (which my post was almost
entirely about).


Fault currents are determined by assessing the "loop" resistance of a
worst case electrical fault to earth and also as a dead short. These
can be determined by calculation or by instruments.

Earth is not calculated because it is such a poor conductor. It may be
necessary in some of the screwier UK electrical systems with an RCD main.

So any switch or circuit breaker has two current ratings.

It's normal rating that it will carry continuously and interrupt many
thousands of times.

It's fault current rupturing capacity. Often several thousand amps. It
will only break this current a very limited number of times and carry
the current for milliseconds.

With minimal reading ability it is obvious that daestrom, mII (or
whoever) and I talked about the fault current ratings of circuit
breakers or fuses. Or did you think that houses have 10,000A services?


You really need to go for some instruction. These things can't be
worked out by lying on your bed and thinking about it.

I worked them out 40 years ago then worked with them the last 40 years.

You really should learn to read and think. Maybe when cows fly....

--
bud--

On Apr 12, 10:27*pm, Home Guy wrote:
Bruce Richmond wrote:
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.

Then you should look back and see that is precisely the issue being
discussed here. You wrote, "But still - you can't push more
electricity onto a network than the load is asking for (given that
your invertors are functioning correctly I guess)."

David wrote, "That second statement is correct: you can't "push"
electrons into the grid. But it doesn't matter *how* your inverters
are working; it's a basic law of physics."

I have explained in simple terms (without getting into power factors,
phase shifting, pulse width modulation, etc) the physics behind how
you can push your power onto the grid whether it is asking for it or
not.


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)

You do not raise it up as high as you can, you raise it just enough to
do the job. The ignition coil in your car raises the 12 volts of its
battery to many thousands of volts to force a spark across the gap of
its sparkplugs. You don't need thousands of volts to feed power into
the grid.

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).

Go back to the example I provided using two batteries. As I said,
when they are connected in parallel the 12 volt battery charges the 11
volt battery and the voltage across them will measure somewhere
between 11 and 12 volts. Connect a light to the batteries. You will
measure a slight drop in the voltage but it will still be over 11
volts. That means the 11 volt battery is still being charged and all
the power to light the light and charge the 11 volt battery is coming
from the 12 volt battery. Connect more lights (load) to the batteries
and you can drag the voltage down so that it is just over 11 volts.
So long as the voltage across the two batteries is higher than the
stand alone voltage of the 11 volt battery all the current going
through the lights will be coming from the 12 volt battery. And it
doesn't matter that the 12 volt battery has been dragged down to
within a small fraction of a volt over the 11 volt battery, the lights
see 11+ volts. Can you see now how the inverter can pump its power
into the system? By having its voltage just a bit higher than the
transformer, but well within the normal range of the line voltage, it
can take over feeding the local water heaters, cooking stoves, air
conditioners, lights, etc. No additional controlls are needed to
reduce the current coming from the transformer. The voltage
difference takes care of it.

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).

As you saw above the current flow through the transformer will be
reduced. The substation sees that as a reduced load and will behave
the same way it would any other time the load goes away. No
additional controls are needed.

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?

Yes, you have said enough to make it clear how little you know.

When an electric motor is running it produces a back EMF that counters
the flow of the current.

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

That AC compressor requires the same power it did before. Keeping it
simple that power is volts times amps equals watts. Divide the power
required by the higher volts now provided and you will find fewer amps
are required. The wasted heat is proportional to the square of the
current. So raising the voltage means there will be less waste heat
from the motor.

With resistance heaters the higher voltage flows more current, so you
get more heat, which is what you wanted anyway.

harry wrote:

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.

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

What happens to the "say" 1 volt. It is only a local thing because
the utility drops it's output by 5Kw.

I doubt that the regional sub-station is going to do that.

Sure, there will be different current flow around the system
but nothing that can't be handled.

I didn't say that it couldn't be handled.

I'm saying that a small-scale PV system is going to raise the local grid
voltage for the homes connected to the same step-down distribution
transformer. All the linear loads on the local grid will consume the
extra power (probably about 250 to 500 watts per home, including the
house with the PV system on the roof). The extra 250 to 500 watts will
be divided up between the various AC motors (AC and fridge compressors,
vent fans) and lights. They don't need the extra volt or two rise on
their power line supply - the motors won't turn any faster and the
lights will just convert those extra watts into heat more than light
output.

The home owner with the PV system will get paid 80 cents / kwh for the
40-odd amps he's pushing out into the grid, but that energy will be
wasted as it's converted disproportionately into heat - not useful work
- by the linear loads on the local grid.

I don't know why you rabbit claiming it doesn't work when it
clearly does.

I don't see a rabbit around here.

I'm not claiming that pushing current into the local grid by way of
raising the local grid voltage doesn't work.

I'm claiming that there won't be a corresponding voltage down-regulation
at the level of the neighborhood distribution transformer to make the
effort worth while for all stake holders.

On Apr 13, 12:06*am, "
wrote:
On Tue, 12 Apr 2011 18:50:22 -0700 (PDT), Bruce Richmond
wrote:





On Apr 11, 2:19 am, David Nebenzahl wrote:
On 4/10/2011 10:02 PM Bruce Richmond spake thus:

On Apr 10, 3:54 am, David Nebenzahl wrote:

That second statement is correct: you can't "push" electrons into
the grid. But it doesn't matter *how* your inverters are working;
it's a basic law of physics.

If you apply more volts to a line than what it is carrying what do you
think happens? I run machines that use regenerative braking. They
draw energy from the line to set things in motion. To slow or stop
them the electric motor acts as a generator producing a higher voltage
than the grid, forcing power back into the grid. An inverter can do
the same thing using solid state circuits. The inverter in my Prius
takes DC current from the battery and converts it to whatever voltage
and frequency is needed at the time to run the variable frequency AC
motor. When slowing down the motor becomes an AC generator and the
inverter converts the output to a DC voltage just a bit higher than
the battery, pumping charge back into it.

Sorry, I don't think you know what you're talking about.

If you do a little research I think you will change your mind.

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

At the bottom of that page you will find this link

http://www.solarpanelsplus.com/solar...r-Inverters-Wo...

You seem to think that you can "force" or push "voltage" into a line, by
using a higher voltage than what's on the line.

More specificly I wrote, "forcing power back into the grid". *Power is
watts or KW. *That's volts times amps. *

But watts is *not* volts times amps, in an AC circuit. *There is a power
factor in there to worry about. *In the capacitor example, watts dissipated is
zero (or close to it) but VA might be rather high.

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


The current will only flow if there is a difference in voltage.

Correct. *Ohms Law.- Hide quoted text -

- Show quoted text -

On Tue, 12 Apr 2011 21:23:03 -0700, David Nebenzahl
wrote:

On 4/12/2011 9:06 PM zzzzzzzzzz spake thus:

The current will only flow if there is a difference in voltage.

Correct. Ohms Law.

That is *not* Ohm's Law. Where do you get that? Sheesh--you're trying to
lecture *me* on this stuff???

E=IR, certainly *IS* Ohm's law. I and E are proportional. You can't increase
I without increasing E. Get it? I suppose not.

wrote in message
news

E=IR, certainly *IS* Ohm's law. I and E are proportional. You can't increase
I without increasing E.
Wrong. You CAN increase I without increasing E. You have 3 variables in that
formula, not just 2.


Get it? I suppose not.

Apparently not.

Vaughn

On 13/04/2011 17:59, Home Guy wrote:
harry wrote:


What happens to the "say" 1 volt. It is only a local thing because
the utility drops it's output by 5Kw.

I doubt that the regional sub-station is going to do that.

From Wikipedia:
"In an electric power distribution system, voltage regulators may be
installed at a substation or along distribution lines so that all
customers receive steady voltage independent of how much power is drawn
from the line."

Obviously when a local area is supplying power to the grid, power
generation elsewhere will be reduced. And any voltage changes that
results from that will be adjusted with line voltage regulators, if
necessary.




I'm saying that a small-scale PV system is going to raise the local grid
voltage for the homes connected to the same step-down distribution
transformer.All the linear loads on the local grid will consume the
extra power (probably about 250 to 500 watts per home, including the
house with the PV system on the roof). The extra 250 to 500 watts will
be divided up between the various AC motors (AC and fridge compressors,
vent fans) and lights. They don't need the extra volt or two rise on
their power line supply - the motors won't turn any faster and the
lights will just convert those extra watts into heat more than light
output.

How do you get the value 250-500W?

Motors will only increase their energy drain by raising the frequency,
Plus a small loss due to internal resistance in the windings.

As for a resistive load, increasing the voltage from 120 to 125 volt
will result in a power drain increase of about 8.5% or 8.5 W for a 100W
light bulb, assuming 120V is the nominal voltage.

Remember though that the voltage increase on the step-down side of the
transformer due to homeowners PV arrays will be less than 5 volt pretty
much guaranteed. Local codes state a maximum voltage drop (7V in BC)
over the lines to a house, at 80% load of service panel capacity.

Most households have a 200A service panel. A 10kW PV array is well below
the service panel capacity.

And you cannot just look at the PV array output. You must take into
account the local energy consumers as well. That will reduce the current
going into the grid, and thus the voltage increase.


The home owner with the PV system will get paid 80 cents / kwh for the
40-odd amps he's pushing out into the grid, but that energy will be
wasted as it's converted disproportionately into heat - not useful work
- by the linear loads on the local grid.

You claims are pretty vague, please explain what you mean by wasted.

By the way, there is some "waste" by just using the grid only as well.
Losses everywhere in the grid.


I'm claiming that there won't be a corresponding voltage down-regulation
at the level of the neighborhood distribution transformer to make the
effort worth while for all stake holders.

What is your definition of worth while? And what do you know about the
utility's voltage regulation policies?

The utilities _have_ to use voltage regulation due to demand changes.

"Home Guy" wrote in message ...

I'm claiming that there won't be a corresponding voltage down-regulation
at the level of the neighborhood distribution transformer to make the
effort worth while for all stake holders.

Oh but there is. Look around and you will see an occasional unit that looks
like a pole pig (transformer) with no low voltage wires attached, just the
high voltage ones. This is either a line reactor of some kind (inductor or
capacitor) or an on-load tap changer, which is an autotransformer that
automatically changes winding taps as needed to maintain the output voltage
within the 5 or 10% of spec.

If the load on it is say 50KVA and you switch on a GTI with a perfect 1.0
power factor and outputting 25KW the load on that transformer will drop to
25KVA, since all wiring has some resistance this will cause the voltage to
rise the same as if half of the load were switched off, the tap changer
responds after a delay period by changing taps which lowers the output
voltage back down.

On Wed, 13 Apr 2011 21:33:08 -0400, "vaughn"
wrote:


wrote in message
news

E=IR, certainly *IS* Ohm's law. I and E are proportional. You can't increase
I without increasing E.
Wrong. You CAN increase I without increasing E. You have 3 variables in that
formula, not just 2.

Dumbass, it's a fixed circuit.

Get it? I suppose not.

Apparently not.

You've only proved that you're just as stupid as David.

On Wed, 13 Apr 2011 18:06:45 -0700 (PDT), Bruce Richmond
wrote:

On Apr 13, 12:06*am, "
wrote:
On Tue, 12 Apr 2011 18:50:22 -0700 (PDT), Bruce Richmond
wrote:





On Apr 11, 2:19 am, David Nebenzahl wrote:
On 4/10/2011 10:02 PM Bruce Richmond spake thus:

On Apr 10, 3:54 am, David Nebenzahl wrote:

That second statement is correct: you can't "push" electrons into
the grid. But it doesn't matter *how* your inverters are working;
it's a basic law of physics.

If you apply more volts to a line than what it is carrying what do you
think happens? I run machines that use regenerative braking. They
draw energy from the line to set things in motion. To slow or stop
them the electric motor acts as a generator producing a higher voltage
than the grid, forcing power back into the grid. An inverter can do
the same thing using solid state circuits. The inverter in my Prius
takes DC current from the battery and converts it to whatever voltage
and frequency is needed at the time to run the variable frequency AC
motor. When slowing down the motor becomes an AC generator and the
inverter converts the output to a DC voltage just a bit higher than
the battery, pumping charge back into it.

Sorry, I don't think you know what you're talking about.

If you do a little research I think you will change your mind.

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

At the bottom of that page you will find this link

http://www.solarpanelsplus.com/solar...r-Inverters-Wo...

You seem to think that you can "force" or push "voltage" into a line, by
using a higher voltage than what's on the line.

More specificly I wrote, "forcing power back into the grid". *Power is
watts or KW. *That's volts times amps. *

But watts is *not* volts times amps, in an AC circuit. *There is a power
factor in there to worry about. *In the capacitor example, watts dissipated is
zero (or close to it) but VA might be rather high.

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

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

On Apr 13, 11:38*pm, "
wrote:
On Wed, 13 Apr 2011 21:33:08 -0400, "vaughn"
wrote:



wrote in message
news

E=IR, certainly *IS* Ohm's law. *I and E are proportional. *You can't increase
I without increasing E.
Wrong. *You CAN increase I without increasing E. *You have 3 variables in that
formula, not just 2.

Dumbass, it's a fixed circuit.

Get it? *I suppose not.

Apparently not.

You've only proved that you're just as stupid as David.

SO, NOW WE ALL SHOULD RESORT TO ANGRY NAMECALLING & FAGGOTY RESPONSES
HAHN???

GROW UP FOOL, OR DROP IT, YOUR STALE INSIGNIFICANT DULLARD POSTINGS
ARENT HELPING ANY.

PAT ECUM

On Wed, 13 Apr 2011 20:46:43 -0700 (PDT), The Ghost in The Machine
wrote:

On Apr 13, 11:38*pm, "
wrote:
On Wed, 13 Apr 2011 21:33:08 -0400, "vaughn"
wrote:



wrote in message
news

E=IR, certainly *IS* Ohm's law. *I and E are proportional. *You can't increase
I without increasing E.
Wrong. *You CAN increase I without increasing E. *You have 3 variables in that
formula, not just 2.

Dumbass, it's a fixed circuit.

Get it? *I suppose not.

Apparently not.

You've only proved that you're just as stupid as David.

SO, NOW WE ALL SHOULD RESORT TO ANGRY NAMECALLING & FAGGOTY RESPONSES
HAHN???

GROW UP FOOL, OR DROP IT, YOUR STALE INSIGNIFICANT DULLARD POSTINGS
ARENT HELPING ANY.

Speaking of dumbasses...

PAT ECUM

Here is Roy Queerjano, A.K.A. Pat E. Cum/

On Apr 13, 8:59*pm, Home Guy wrote:
...
The home owner with the PV system will get paid 80 cents / kwh for the
40-odd amps he's pushing out into the grid, but that energy will be
wasted as it's converted disproportionately into heat - not useful work
- by the linear loads on the local grid.

Plug a Kill-A-Watt (etc) voltmeter into an outlet and then switch on a
load, like a heater or iron. See what the voltage drop is for that
current. You can measure the drop at the breaker box by metering an
outlet on a different breaker on the same side of the line. This will
show you approximately how much your grid voltage changes with
current, from or to the grid.

jsw

wrote in message
...

E=IR, certainly *IS* Ohm's law. I and E are proportional. You can't
increase
I without increasing E.
Wrong. You CAN increase I without increasing E. You have 3 variables in that
formula, not just 2.

Dumbass

Names? Didn't your mother tell you how babyish that is?

, it's a fixed circuit.

You never defined the circuit, except perhaps in your own mind. I was
responding to your statement about E=IR.

Get it? I suppose not.

Apparently not.

You've only proved that you're just as stupid as David.

And you have proven yourself as a troll.

Bye
Vaughn