On Apr 13, 9:07*pm, "
wrote:
On Tue, 12 Apr 2011 21:23:03 -0700, 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???

I think he gets it directly from Ohms Law. V=IR.

Or, I = V/R

If V, the voltage is zero, then I, the current must be zero. Or, in
other words,
current will only flow if there is a difference in voltage.





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.

EFFECTS OF PHOTOVOLTAICS ON DISTRIBUTION SYSTEM VOLTAGE REGULATION

Peter McNute, Josh Hambrick, and Mike Keesee

National Renewable Energy Laboratory, Golden, Colorado
Sacramento Municipal Utility District, Sacramento, California

ABSTRACT

As grid-integrated photovoltaic (PV) systems become more prevalent,
utilities want to determine if, and at what point, PV systems might
begin to negatively impact the voltage regulation of their distribution
systems. This paper will briefly describe voltage regulation methods in
a utility distribution system. It will also take a preliminary look at
the distribution impacts of high penetrations of grid-integrated PV
systems being installed and operated in a Sacramento Municipal Utility
District community. In particular, the issue of excessive service
voltage and excessive substation voltage due to the reverse power flow
from exporting PV systems will be examined. We will also compare
measured data against modeled data.

Index Terms - Photovoltaic (PV), Distribution System, Voltage Regulation

INTRODUCTION

A three-year, joint project between the National Renewable Energy
Laboratory (NREL) and Sacramento Municipal Utility District (SMUD) began
in March 2008 to analyze the distribution impacts of high penetrations
of grid-integrated renewable energy systems, specifically PV-equipped
SolarSmartSM Homes found in the Anatolia III Residential Community
(hereafter referred to as Anatolia) in Rancho Cordova, California.
SolarSmart Homes combine high-efficiency features along with
rooftop-integrated 2.0-kWac PV systems with no energy storage. When
completely built out, Anatolia will have 795 homes, 600 of which will be
SolarSmart, eventually amounting to 1.2 MWac potential generation. (So
far, only 115 homes have been built for 238 kWac potential generation.)
In particular we are investigating if there will be excessive service
voltage or substation voltage due to reverse power flow from exporting
PV systems.

VOLTAGE REGULATION METHODS

The primary distribution voltage needs to remain within ANSI C84.1
limits: 114 V to 126 V or 11.4 kV to 12.6 kV, over the length of the
feeder. The service voltage is provided to the customer meter and
includes any voltage drop from the primary service through the
distribution transformer and service conductors. The electric supply
system is designed so that the voltage is within the 110 V to 126 V
range (Range A) most of the time and infrequently within the 106 V to
127 V range (Range B). The utilization voltage is the operating voltage
system and loads are designed to operate within. During heavy loading,
SMUD may adjust the substation voltage as high as 12.75 kV to account
for the voltage drop at the end of the feeder.

Load Tap Changer

A load tap changer (LTC) effectively varies the transformer turns ratio
to maintain the transformer secondary voltage at the substation as
primary-voltage changes occur due to changes in loading of the
transmission system, or as the load on the transformer itself varies.
The Anatolia substation voltage is automatically controlled by a LTC on
the secondary of the substation transformer, managed by a voltage
regulating relay. The band center is 123 V (12.3 kV lineto- line at the
substation). The bandwidth is +/- 1.5 V (or 3.0 V) total. The time delay
for adjustments is 60 seconds. The LTC compensates for added load by
increasing the band center linearly from 123 volts at no load to 126
volts at full load (about 20 MVA or 1,200 A).

Capacitors

Capacitors are reactive power sources that affect the voltage by
supplying leading reactive current that compensates for the lagging
reactive current of the load. The Anatolia substation has six
three-phase, 1,800kVAR capacitor banks (hereafter referred to as
capacitors). The capacitors are computer controlled using an algorithm
developed by SMUD called Capcon. During normal operations, one capacitor
is turned on each time the VAR flow exceeds 900 kVAR (from the
substation bank), and a capacitor is switched off each time the VAR flow
exceeds -1,200 kVAR. These set points are modified on days hotter than
100oF, but this setting can be manually overridden depending upon
weather forecasts. These settings automatically adjust for abnormally
high or low bulk system voltages.

SMUD ANATOLIA III OVERVIEW

Distribution System Description

Anatolia is a residential community served by individual single-phase
lateral circuits from a three-phase primary feeder which connects to a
20 MVA, 69 kV/12.47 kV delta-wye transformer at the Anatolia-Chrysanthy
Substation (hereafter referred to as Anatolia-Chrysanthy or substation).
The feeder has several points along the line that allow for switching.
The length of the feeder to the switching cubicle furthest from the
substation is about 18,895 feet and the furthest distribution
transformer, 5K7, is about 22,360 feet from the substation. In Anatolia,
there are a total of 85 singlephase pad-mounted distribution
transformers: 23 are 75 kVA and 62 are 50 kVA. Also connected to the
same substation feeder is a rendering plant fed by two 1,500kVA
transformers, consuming between 20 and 200 kW, a water storage plant,
and a residential community with an estimated 1,000 homes between the
substation and Anatolia. There are two additional feeders that also
connect to the 20-MVA transformer that have an additional estimated
1,000 customer loads each.

SolarSmart Homes Description

SolarSmart Homes are advertised as able to reduce annual residential
electric bills by 60%. They combine cost-effective, energy-efficient
features and a rooftop PV system. Typical SolarSmart Home features
include: radiant barriers to reflect summer heat that would otherwise
enter the attic and cause greater need for air conditioning;
90%-efficient furnaces; 14 SEER / 12 EER HVAC systems;
compact-fluorescent lighting; ENERGY STAR-qualified windows; and
independent third-party verification, required to confirm all
energy-efficiency measures are installed and operate correctly.

A 2.0-kWac PV system can generate a major portion of the electrical
energy consumed in the home. The majority of the Anatolia PV systems
comprise 36 SunPower 63-watt SunTile roof-integrated modules feeding
into a SunPower SPR-2800x positive-ground, grid-connected inverter. The
inverter can be remotely accessed by SunPower if the homeowner elects to
connect it via the internet. The orientation of the PV systems ranges
from southeast to southwest.

MONITORING

The distribution system is monitored at the substation feeder using a
DAQ Electronics ART-073-79-0 Supervisory Control and Data Acquisition
(SCADA). The SCADA is ANSI-class high-accuracy, better than 0.5%. The
sampling rate is every two seconds with an average recorded every five
minutes. Four distribution transformers are monitored using PMI Eagle
440 monitors having an accuracy better than 1.0%. The sampling rate is
256 samples per cycle with an average recorded every five minutes. Four
homes are monitored at the service panel using PMI iVS-2SX+ power
monitors having an accuracy better than 1.0%. The sampling rate is 128
samples per cycle with an average recorded every five minutes.

Solar and metrological data are also collected from a
solar/meteorological station installed at the substation.
One-minute-interval data from the solar/meteorological station are
collected automatically daily.

ANATOLIA RESULTS

SMUD wanted to investigate what effect reverse power flow from exporting
PV systems would have on service and substation voltage regulation.
Figure 1 shows the Home-to-Substation voltage difference (top) and solar
irradiance (bottom) on a clear, cool day, Saturday, March 7, 2009,
representative of a day with relatively low load and high local PV
penetration. Penetration is defined as the amount of PV output divided
by the load at a particular point in time.

At night the substation voltage ranged between 0.4 V to 0.7 V higher
than the home voltage. This is representative of a typical circuit with
voltage drops through the line and transformer impedances. During
daylight hours, this reversed and the home voltage rose as high as 0.7
V, or 0.6%, greater than the substation voltage. The voltage amplitude
was 124.5 V at its peak, so it remained well within ANSI C84.1 limits.
Due to heavy home loads in the morning, the PV system did not begin to
export to the distribution system until almost noon. In March 2009,
there was a 2.0-kWac PV system on the home, 30.1 kWac of PV on the
transformer (twelve SolarSmart Homes), and 238 kWac of PV on the
distribution system (115 SolarSmart Homes). Preliminary estimates of PV
penetration levels on the feeder were 11% to 13% under lightly-loaded
conditions (2.0 MW) and around 4.0% of the total substation transformer
load. Since the PV penetration levels are still relatively low, there
were no adverse effects on voltage regulation.

Modelling Anatolia

To measure the effects of PV on voltage regulation, a Distributed
Engineering Workstation (DEW) model was created that included the
distribution transformers, secondary, and service connections. Once
validated using measured field data, the model will be used to determine
acceptable levels of PV penetration.

The feeder contains significant loads which are not part of the system
under evaluation including a rendering plant, water storage facility and
approximately 1,000 residential customers. Additionally, the substation
transformer and LTC are shared by two unmonitored feeders.

For initial testing and model verification, the unknown residential
loads were represented as a lumped spot load. Eventually, the
residential loads will either be calculated from load research
statistics or distributed based on distribution transformer size. The
rendering plant load was represented as a spot load based on recorded
data.

The load on the other feeders was modeled using a spot load placed
immediately after the LTC. Since the additional feeders share the LTC,
the load on those feeders will affect the position of the tap as well as
the regulation set-point. While the load on the other feeders will not
greatly affect the voltage profile of the system under study, these
loads may affect the voltage regulation.

To verify the topology of the model and to ensure the switching devices
are represented in their correct states, the electrical distances from
the substation of the measurement points were compared against the
values estimated from a GIS map. Table 1 describes this comparison. All
modeled electrical distances are within 3.5% of the estimated distances
as determined from GIS drawings.

Next, the behavior of the secondary-side of the distribution transformer
was validated against real data. The voltage rise measured between the
secondary of the distribution transformers and the service entry to the
homes will be affected by the loads and generation of the other,
unmonitored homes that share the secondary connection.

Figure 2 shows the simulated and measured voltage rise from Home 3 to
the secondary of Transformer 3 (8K6). The data reflects roughly a month
of data where Home 3 is net exporting real power with an inductive load
between 0.18 kVAR and 0.22 kVAR. The secondary system was simulated
assuming uniform generation from all customers connected to the
secondary of the transformer. The generation from the homes was
increased while maintaining a constant inductive load of 0.2 kVAR. The
homes were modeled as constant current loads.

The variation in measured voltage rise data is largely due to
uncertainties with the other customers sharing the secondary connection
as well as the resolution of the voltage measurement device at the home.
Figure 2 indicates that the model reasonably reflects the behavior of
the actual system. With improved monitoring and meter accuracy, the
model could be better validated and, if necessary, adjusted to more
accurately reflect the system.

Currently, there are too many uncertainties on the circuit to perform
any meaningful whole system validation. Efforts are under way to
increase the monitoring on the system so that the overall model may be
better validated. This includes adding meters at critical points to
eliminate many of the unknown loads described above.

CONCLUSIONS

After one year of monitoring the Anatolia SolarSmart Homes Community,
there was no excessive service or substation voltage due to reverse
power flow from exporting PV systems. Preliminary estimates of PV
penetration levels on the feeder were 11% to 13% under lightly-loaded
conditions (2.0 MW) and around 4.0% of the total substation transformer
load. Since the PV penetration levels were relatively low, there were no
adverse effects on voltage regulation. It was possible to see the
effects of the PV systems on the voltage at the individual homes and the
distribution transformers. This slight voltage rise was approximately
0.6% on clear days in comparison to the normal drop of -0.6% without the
PV exporting. The DEW model that has been developed reasonably reflects
the behavior of the actual system. The model will be used to determine
acceptable levels of PV penetration.

================

Photovoltaic Specialists Conference (PVSC)
2009 34th IEEE
Issue Date: 7-12 June 2009
On page(s): 001914 - 001917
Date of Current Version: 17 February 2010

On Apr 14, 10:20*am, Home Guy wrote:
...
At night the substation voltage ranged between 0.4 V to 0.7 V higher
than the home voltage. This is representative of a typical circuit with
voltage drops through the line and transformer impedances. During
daylight hours, this reversed and the home voltage rose as high as 0.7
V, or 0.6%, greater than the substation voltage. ...

Currently, there are too many uncertainties on the circuit to perform
any meaningful whole system validation. Efforts are under way to
increase the monitoring on the system so that the overall model may be
better validated. This includes adding meters at critical points to
eliminate many of the unknown loads described above.

They have stated that the servo-loop stability of grid voltage
regulation -may- become questionable at over 20% uncontrolled solar
input.

http://eioc.pnnl.gov/research/gridstability.stm

jsw

On Apr 13, 4:15*pm, bud-- wrote:
On 4/12/2011 10:58 AM, m II wrote:





"bud--" wrote in ...

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--

-----------------
|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--- Hide quoted text -

- Show quoted text -

Impedance is the vector sum of resistance and reactance measured in
Ohms not a percentage.
Regulation of a transformer mau be measured in percentage terms.
What exactly are yo on about?

On Apr 14, 3:44*pm, Jim Wilkins wrote:
On Apr 14, 10:20*am, Home Guy wrote:

...
At night the substation voltage ranged between 0.4 V to 0.7 V higher
than the home voltage. This is representative of a typical circuit with
voltage drops through the line and transformer impedances. During
daylight hours, this reversed and the home voltage rose as high as 0.7
V, or 0.6%, greater than the substation voltage. ...
Currently, there are too many uncertainties on the circuit to perform
any meaningful whole system validation. Efforts are under way to
increase the monitoring on the system so that the overall model may be
better validated. This includes adding meters at critical points to
eliminate many of the unknown loads described above.

They have stated that the servo-loop stability of grid voltage
regulation -may- become questionable at over 20% uncontrolled solar
input.

http://eioc.pnnl.gov/research/gridstability.stm

jsw

The problem arises because the supply company is unable to make
voltage adjustments swiftly enough every time a cloud passes
overhead. The problem becomes more acute as more and more PV arrays
appear.
Wind turbines are even worse in this respect.

The most recent grid tie inverters switch off the array if over-
voltage or "islanding" is detected to help overcome this problem.
The display on them indicates what is going on.

On Apr 13, 3:24*am, harry wrote:
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.

For once I agree with Harry. We can forget about generators and
distributions systems. Just take two 12V batteries and connect
them in parallel to a 12ohm resistor. Under Homeguy's
theories, I don;t know what he thinks would happen. But
clearly he thinks if we put a second AC power source on
a distributions system, it has to be at a higher voltage to
"push" current out.

So, what happens with the two batteries? Under the laws
of physics the rest of us use the voltage would remain at
12 volts and BOTH batteries would be supplying part of
the 1 AMP flowing through the resistors.

On Apr 13, 8:59*pm, Home Guy wrote:
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.

You are claiming that any electricity produced by PV arrays that
goes onto the local grid just gets wasted because putting it on
the grid raises the voltage a tiny amount. I think that's what
he meant by saying "it doesn't work". That is you're saying
that PV arrays that have net current flowing into the grid
don't work, because the energy somehow just gets
dissapears.

There is SO much wrong in your analysis, that I don't know
where to begin. But here's a start. You claim that with
a slightly higher voltage, an AC motor in an HVAC
compressor won't turn any faster and hence the additional
power is wasted. What you've completely overlooked is
that power is P=VI, or power is voltage times current.
Give that motor an extra half a volt and I'll bet it's current
decreases by a corresponding amount.



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.

As Bud said a while back, you're new analysis must be devastating to
all the power companies in the world.

On Thu, 14 Apr 2011 08:26:22 -0400, "vaughn"
wrote:


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?

She taught me to tell the truth. I did.

, 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.

You're illiterate, then. Not surprising either.

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.

No, you insist on demonstrating just how stupid you really are, though.

On Thu, 14 Apr 2011 08:52:52 -0700 (PDT), "
wrote:

On Apr 13, 3:24*am, harry wrote:
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.

For once I agree with Harry. We can forget about generators and
distributions systems. Just take two 12V batteries and connect
them in parallel to a 12ohm resistor. Under Homeguy's
theories, I don;t know what he thinks would happen. But
clearly he thinks if we put a second AC power source on
a distributions system, it has to be at a higher voltage to
"push" current out.

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

So, what happens with the two batteries? Under the laws
of physics the rest of us use the voltage would remain at
12 volts and BOTH batteries would be supplying part of
the 1 AMP flowing through the resistors.

Not on Homeguy's and Vaughn's planet. One of the batteries will be charging
the other.

On 4/14/2011 5:48 AM spake thus:

On Apr 13, 9:07 pm, "
wrote:

On Tue, 12 Apr 2011 21:23:03 -0700, 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???

I think he gets it directly from Ohms Law. V=IR.

Or, I = V/R

If V, the voltage is zero, then I, the current must be zero. Or, in
other words, current will only flow if there is a difference in
voltage.

But that's not Ohm's Law (the statement "current will flow only if there
is a difference in voltage"). Actually, that is a *tautology* (look it
up). In other words, that's the very definition of current, which
requires a potential difference (voltage 0) to flow. Ohm's law didn't
establish that, because it was already established by the time he came
along.

You've correctly stated Ohm's Law, but that's not what it says. Strictly
speaking, what Ohm determined was that the current flowing in a circuit
is proportional to the voltage and inversely proportional to the
resistance--but only for certain resistors. Specifically, his carefully
calibrated metal resistances, at a certain temperature. So "Ohm's
law"--what he determined experimentally and published--is only this:

I = E / R

and that only at fixed temperature. Turns out "Ohm's law" does *not*
hold for a lot of things that look like resistances in the real world
(for example, any humble tungsten filament fails to observe it).

But that's going waaaaay deeper into it than we need to here ...


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

yo
wassup
nuttin
wan2 hang
k
where
here
k
l8tr
by

- from Usenet (what's *that*?)

" wrote:

We can forget about generators and distributions systems.
Just take two 12V batteries and connect them in parallel

There's your problem right there.

It's usually not a good idea to connect batteries together in parallel,
unless they are exactly of the same type, age, condition, etc. If you
get a weak cell in one of the batteries it will turn into a load.

And what happens when they are not exactly at the same voltage before
being connected together? How do you insure that you always get current
flowing out of both of them?

But clearly he thinks if we put a second AC power source on
a distributions system, it has to be at a higher voltage to
"push" current out.

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

So, what happens with the two batteries? Under the
laws of physics the rest of us use the voltage would
remain at 12 volts and BOTH batteries would be supplying
part of the 1 AMP flowing through the resistors.

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

Not on Homeguy's and Vaughn's planet. One of the batteries
will be charging the other.

If they are unequal in capacity, then yes that will eventually happen.

On Fri, 15 Apr 2011 00:27:36 -0400, Home Guy wrote:

" wrote:

We can forget about generators and distributions systems.
Just take two 12V batteries and connect them in parallel

There's your problem right there.

It's usually not a good idea to connect batteries together in parallel,
unless they are exactly of the same type, age, condition, etc. If you
get a weak cell in one of the batteries it will turn into a load.

And what happens when they are not exactly at the same voltage before
being connected together? How do you insure that you always get current
flowing out of both of them?

But clearly he thinks if we put a second AC power source on
a distributions system, it has to be at a higher voltage to
"push" current out.

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

What a dumbass!

So, what happens with the two batteries? Under the
laws of physics the rest of us use the voltage would
remain at 12 volts and BOTH batteries would be supplying
part of the 1 AMP flowing through the resistors.

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

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

Not on Homeguy's and Vaughn's planet. One of the batteries
will be charging the other.

If they are unequal in capacity, then yes that will eventually happen.

Wrong!

" unnecessarily full-quoted:

You are claiming that any electricity produced by PV arrays that
goes onto the local grid just gets wasted because putting it on
the grid raises the voltage a tiny amount. I think that's what
he meant by saying "it doesn't work". That is you're saying
that PV arrays that have net current flowing into the grid
don't work, because the energy somehow just gets dissapears.

I'm not saying that it dissapears.

I'm saying that if your local grid is sitting at 120V and your panels
come on and raise it to 121V, and if the utility company doesn't
down-regulate their side to bring the local grid back to 120V, then the
current that your panels are injecting is wasted. It's wasted because
all the linear loads on the grid that are designed for 120V will not
operate any better at 121 volts. Motors won't turn faster, lights won't
really burn brighter. They will just give off a little more heat thanks
to the extra current the panels are supplying to the grid.

But sure - electric heaters will get hotter. They're the only devices
on the grid that are intended to convert electrical energy into heat.

There is SO much wrong in your analysis, that I don't know
where to begin. But here's a start. You claim that with
a slightly higher voltage, an AC motor in an HVAC
compressor won't turn any faster and hence the additional
power is wasted. What you've completely overlooked is
that power is P=VI, or power is voltage times current.
Give that motor an extra half a volt and I'll bet it's
current decreases by a corresponding amount.

So why not run a 120V motor with 240 volts then?

AC Motors are not simple loads like a resistor, but they will still
"consume" power (V x I) as a function of their supply voltage.

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.

As Bud said a while back, you're new analysis must be
devastating to all the power companies in the world.

All the power companies in the world are in the business of generating
electricity in the thousands of volts and sending it out over
high-tension wires. That's what they'd rather do if they weren't being
hamstrung by crazy ideas and new rules / laws made by politicians about
small-scale co-generation.

Look at the microFIT program in Ontario. When the rules were changed to
allow local utilities to veto hookups based on "network capacity" or
"substation insufficiency", they were only too happy to start swinging
their veto left and right. They don't want to see this small-scale ****
coming on-line if they have a choice.

On Apr 15, 12:40*am, "
wrote:
On Thu, 14 Apr 2011 08:52:52 -0700 (PDT), "





wrote:
On Apr 13, 3:24*am, harry wrote:
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.

For once I agree with Harry. * We can forget about generators and
distributions systems. *Just take two 12V batteries and connect
them in parallel to a 12ohm resistor. *Under Homeguy's
theories, I don;t know what he thinks would happen. *But
clearly he thinks if we put a second AC power source on
a distributions system, it has to be at a higher voltage to
"push" current out.

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

So, what happens with the two batteries? * Under the laws
of physics the rest of us use the voltage would remain at
12 volts and BOTH batteries would be supplying part of
the 1 AMP flowing through the resistors.

Not on Homeguy's and Vaughn's planet. *One of the batteries will be charging
the other.- Hide quoted text -

- Show quoted text -

That is correct. One charges, the other discharges. Eventually
equilbrium will be achieved and the current will cease.
If there was a resistor in parallel with both,the higher voltage
battery would supply current to the other battery and the resistor
until the voltage fell to the lower voltage battery and than both
would supply current to the resistor.

On Apr 15, 2:57*am, David Nebenzahl wrote:
On 4/14/2011 5:48 AM spake thus:





On Apr 13, 9:07 pm, "
wrote:

On Tue, 12 Apr 2011 21:23:03 -0700, 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???

I think he gets it directly from Ohms Law. * V=IR.

Or, I = V/R

If V, the voltage is zero, then I, the current must be zero. *Or, in
other words, current will only flow if there is a difference in
voltage.

But that's not Ohm's Law (the statement "current will flow only if there
is a difference in voltage"). Actually, that is a *tautology* (look it
up). In other words, that's the very definition of current, which
requires a potential difference (voltage 0) to flow. Ohm's law didn't
establish that, because it was already established by the time he came
along.

You've correctly stated Ohm's Law, but that's not what it says. Strictly
speaking, what Ohm determined was that the current flowing in a circuit
is proportional to the voltage and inversely proportional to the
resistance--but only for certain resistors. Specifically, his carefully
calibrated metal resistances, at a certain temperature. So "Ohm's
law"--what he determined experimentally and published--is only this:

* * I = E / R

and that only at fixed temperature. Turns out "Ohm's law" does *not*
hold for a lot of things that look like resistances in the real world
(for example, any humble tungsten filament fails to observe it).

But that's going waaaaay deeper into it than we need to here ...

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

* *yo
* *wassup
* *nuttin
* *wan2 hang
* *k
* *where
* *here
* *k
* *l8tr
* *by

- from Usenet (what's *that*?)- Hide quoted text -

- Show quoted text -

Tautology? You are full of crap.
Resistance varies with temperature, cross sectional area length,
nature of material, even pressure in some cases.

Just stop wriggling. You are wrong. End of matter.

On Apr 15, 5:47*am, Home Guy wrote:
" unnecessarily full-quoted:

You are claiming that any electricity produced by PV arrays that
goes onto the local grid just gets wasted because putting it on
the grid raises the voltage a tiny amount. * I think that's what
he meant by saying "it doesn't work". * That is you're saying
that PV arrays that have net current flowing into the grid
don't work, because the energy somehow just gets dissapears.

I'm not saying that it dissapears.

I'm saying that if your local grid is sitting at 120V and your panels
come on and raise it to 121V, and if the utility company doesn't
down-regulate their side to bring the local grid back to 120V, then the
current that your panels are injecting is wasted. *It's wasted because
all the linear loads on the grid that are designed for 120V will not
operate any better at 121 volts. *Motors won't turn faster, lights won't
really burn brighter. *They will just give off a little more heat thanks
to the extra current the panels are supplying to the grid.

But sure - electric heaters will get hotter. *They're the only devices
on the grid that are intended to convert electrical energy into heat. *

There is SO much wrong in your analysis, that I don't know
where to begin. *But here's a start. *You claim that with
a slightly higher voltage, an AC motor in an HVAC
compressor won't turn any faster and hence the additional
power is wasted. * What you've completely overlooked is
that power is P=VI, or power is voltage times current.
Give that motor an extra half a volt and I'll bet it's
current decreases by a corresponding amount.

So why not run a 120V motor with 240 volts then?

AC Motors are not simple loads like a resistor, but they will still
"consume" power (V x I) as a function of their supply voltage. *

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.

As Bud said a while back, you're new analysis must be
devastating to all the power companies in the world.

All the power companies in the world are in the business of generating
electricity in the thousands of volts and sending it out over
high-tension wires. *That's what they'd rather do if they weren't being
hamstrung by crazy ideas and new rules / laws made by politicians about
small-scale co-generation.

Look at the microFIT program in Ontario. *When the rules were changed to
allow local utilities to veto hookups based on "network capacity" or
"substation insufficiency", they were only too happy to start swinging
their veto left and right. *They don't want to see this small-scale ****
coming on-line if they have a choice.

They don't want it because it costs them money. Nothing to do with the
technology.

On Apr 15, 5:46*am, "
wrote:
On Fri, 15 Apr 2011 00:27:36 -0400, Home Guy wrote:
" wrote:

We can forget about generators and distributions systems.
Just take two 12V batteries and connect them in parallel

There's your problem right there.

It's usually not a good idea to connect batteries together in parallel,
unless they are exactly of the same type, age, condition, etc. *If you
get a weak cell in one of the batteries it will turn into a load.

And what happens when they are not exactly at the same voltage before
being connected together? *How do you insure that you always get current
flowing out of both of them?

But clearly he thinks if we put a second AC power source on
a distributions system, it has to be at a higher voltage to
"push" current out.

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

What a dumbass!

So, what happens with the two batteries? * Under the
laws of physics the rest of us use the voltage would
remain at 12 volts and BOTH batteries would be supplying
part of the 1 AMP flowing through the resistors.

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

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

Not on Homeguy's and Vaughn's planet. *One of the batteries
will be charging the other.

If they are unequal in capacity, then yes that will eventually happen.

Wrong!- Hide quoted text -

- Show quoted text -

He said unequal voltage. In which case he's right.
Capacity is not voltage.

On Apr 15, 3:40*am, harry wrote:
On Apr 15, 5:46*am, "
...
He said unequal voltage. In which case he's right.
Capacity is not voltage.-


How to properly analyze the problem:
http://www.electronics-tutorials.ws/...its/dcp_4.html

Look at Example 1.

jsw

On Apr 14, 9:57*pm, David Nebenzahl wrote:
On 4/14/2011 5:48 AM spake thus:





On Apr 13, 9:07 pm, "
wrote:

On Tue, 12 Apr 2011 21:23:03 -0700, 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???

I think he gets it directly from Ohms Law. * V=IR.

Or, I = V/R

If V, the voltage is zero, then I, the current must be zero. *Or, in
other words, current will only flow if there is a difference in
voltage.

But that's not Ohm's Law (the statement "current will flow only if there
is a difference in voltage"). Actually, that is a *tautology* (look it
up). In other words, that's the very definition of current, which
requires a potential difference (voltage 0) to flow. Ohm's law didn't
establish that, because it was already established by the time he came
along.

I never said Ohm established it. Only that from Ohm's law for the
circuit
under discussion you can directly verify that with zero potential you
get
zero current.




You've correctly stated Ohm's Law, but that's not what it says. Strictly
speaking, what Ohm determined was that the current flowing in a circuit
is proportional to the voltage and inversely proportional to the
resistance--but only for certain resistors. Specifically, his carefully
calibrated metal resistances, at a certain temperature. So "Ohm's
law"--what he determined experimentally and published--is only this:

* * I = E / R


And if E is 0, what does this say I will be? Zero. Yes, Ohm isn;t
the
first guy to discover that current only flows from a potential
difference.
But his law clearly reflects it and shows it to be true.




and that only at fixed temperature. Turns out "Ohm's law" does *not*
hold for a lot of things that look like resistances in the real world
(for example, any humble tungsten filament fails to observe it).

Total nonsense. Just because a filament changes resistance with
temperature
does not mean Ohm's Law doesn't apply. Ohms law applies at every
discrete
temperature/resistance point the filament has. If we followed your
logic
almost nothing would behave according to Ohm's Law. Even the
simplest resistor changes resistance slightly as current flows
through it and it's temperature rises slightly. That means the
resistance
has changed, not that Ohm's Law no longer applies.




But that's going waaaaay deeper into it than we need to here ...

At least one step deeper than you should have gone.

On Apr 15, 3:40*am, harry wrote:
On Apr 15, 5:46*am, "





wrote:
On Fri, 15 Apr 2011 00:27:36 -0400, Home Guy wrote:
" wrote:

We can forget about generators and distributions systems.
Just take two 12V batteries and connect them in parallel

There's your problem right there.

It's usually not a good idea to connect batteries together in parallel,
unless they are exactly of the same type, age, condition, etc. *If you
get a weak cell in one of the batteries it will turn into a load.

And what happens when they are not exactly at the same voltage before
being connected together? *How do you insure that you always get current
flowing out of both of them?

But clearly he thinks if we put a second AC power source on
a distributions system, it has to be at a higher voltage to
"push" current out.

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

What a dumbass!

So, what happens with the two batteries? * Under the
laws of physics the rest of us use the voltage would
remain at 12 volts and BOTH batteries would be supplying
part of the 1 AMP flowing through the resistors.

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

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

Not on Homeguy's and Vaughn's planet. *One of the batteries
will be charging the other.

If they are unequal in capacity, then yes that will eventually happen.

Wrong!- Hide quoted text -

- Show quoted text -

He said unequal voltage. In which case he's right.
Capacity is not voltage.- Hide quoted text -

- Show quoted text -

This is what he said:

"If they are unequal in capacity, then yes that (one will be charging
the
other) will eventually happen. "

Which of course is total nonsense and I hope you agree.

On Apr 15, 7:54*am, Jim Wilkins wrote:
On Apr 15, 3:40*am, harry wrote:

On Apr 15, 5:46*am, "
...
He said unequal voltage. In which case he's right.
Capacity is not voltage.-

How to properly analyze the problem:http://www.electronics-tutorials.ws/...its/dcp_4.html

Look at Example 1.

jsw

Thank you Jim. I was sitting here wishing I could draw
a simple circuit into the newsgroup that models what we
are talking about and shows what really happens.
You've gone one better and found a perfect example
from an independent and credible source. That
circuit is EXACTLY the model for the dual battery
example I brought up. Each battery is modeled as
an ideal voltage source in series with a resistor. And
each supplies part of the current flowing through
the load. One never charges the other, nor does
one need to have a higher voltage to "push" current.

I suspect we'll be hearing soon from Homeguy
about how this isn't a valid way two batteries connected
in parallel to a load can be modeled.

" wrote:

But clearly he thinks if we put a second AC power source on
a distributions system, it has to be at a higher voltage to
"push" current out.

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

What a dumbass!

You're the dumbass!

Read the following:

=============
SMUD wanted to investigate what effect reverse power flow from exporting
PV systems would have on service and substation voltage regulation.
Figure 1 shows the Home-to-Substation voltage difference (top) and solar
irradiance (bottom) on a clear, cool day, Saturday, March 7, 2009,
representative of a day with relatively low load and high local PV
penetration. Penetration is defined as the amount of PV output divided
by the load at a particular point in time.

At night the substation voltage ranged between 0.4 V to 0.7 V higher
than the home voltage. This is representative of a typical circuit with
voltage drops through the line and transformer impedances. During
daylight hours, this reversed and the home voltage rose as high as 0.7
V, or 0.6%, greater than the substation voltage.
============

Did you read the last sentence dumbass?

The differential between substation and home voltage went from -.7 to
+.7 volts - a difference of almost 1.5 volts due to the effect of the
PV panels injecting current into the grid.

If, according to you, there was no such phenomena of an increase in
local grid voltage caused by PV panels, then there should be no basis
for the point of this IEEE research paper I posted yesterday.

But engineers know that there will be a voltage increase because of
these panels, and the excercise now is to figure out how much PV power
can come on-line before the substation becomes unable to properly
regulate it's output voltage levels.

==============
Since the PV penetration levels were relatively low, there were no
adverse effects on voltage regulation. It was possible to see the
effects of the PV systems on the voltage at the individual homes and the
distribution transformers.
==============

There will be more PV-equipped homes coming on-line in that project and
it's not yet known if in total their operation will cause poorly
controlled or unstable grid voltage.

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

They will share in proportion to their capacity.

Capacity is the "quantity of electrons" - which by itself tells you
nothing of the potential (voltage).

Electricity, water, nor **** flow uphill.

And height is exactly equivalent to voltage potential.

So PV panels can't push current into the grid unless the invertors raise
their voltage higher than the grid voltage. Just matching the grid
voltage gets you to the point where your current flow is ZERO. Every
millivolt you adjust your output voltage higher than the grid voltage
means some small increment in current outflow from your panels. Since
you can't store the energy coming from the panels via battery bank, then
it's in your best interest to always maximize the amount of current
you're injecting into the grid up to the full potential of the panel's
output capacity. That means raising the output voltage as high as you
need to so that every milliamp is squeezed out of them and onto the
grid.

" wrote:

How to properly analyze the problem:
http://www.electronics-tutorials.ws/...its/dcp_4.html

Look at Example 1.

Thank you Jim. I was sitting here wishing I could draw
a simple circuit into the newsgroup that models what we
are talking about and shows what really happens.
You've gone one better and found a perfect example
from an independent and credible source.

You dumbass.

Look more closely at the current flow in battery 1. It's NEGATIVE.

===================
The negative sign for I1 means that the direction of current flow
initially chosen was wrong, but never the less still valid. In fact, the
20v battery is charging the 10v battery.
==================

On Apr 15, 9:14*am, Home Guy wrote:
" wrote:
But clearly he thinks if we put a second AC power source on
a distributions system, it has to be at a higher voltage to
"push" current out.

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

What a dumbass!

You're the dumbass! *

Read the following:

=============
SMUD wanted to investigate what effect reverse power flow from exporting
PV systems would have on service and substation voltage regulation.
Figure 1 shows the Home-to-Substation voltage difference (top) and solar
irradiance (bottom) on a clear, cool day, Saturday, March 7, 2009,
representative of a day with relatively low load and high local PV
penetration. Penetration is defined as the amount of PV output divided
by the load at a particular point in time.

At night the substation voltage ranged between 0.4 V to 0.7 V higher
than the home voltage. This is representative of a typical circuit with
voltage drops through the line and transformer impedances. During
daylight hours, this reversed and the home voltage rose as high as 0.7
V, or 0.6%, greater than the substation voltage.
============

Did you read the last sentence dumbass?

The differential between substation and home voltage went from -.7 to
+.7 volts - a difference of almost 1.5 volts *due to the effect of the
PV panels injecting current into the grid.

If, according to you, there was no such phenomena of an increase in
local grid voltage caused by PV panels, then there should be no basis
for the point of this IEEE research paper I posted yesterday.

But engineers know that there will be a voltage increase because of
these panels, and the excercise now is to figure out how much PV power
can come on-line before the substation becomes unable to properly
regulate it's output voltage levels.

==============
Since the PV penetration levels were relatively low, there were no
adverse effects on voltage regulation. It was possible to see the
effects of the PV systems on the voltage at the individual homes and the
distribution transformers.
==============

There will be more PV-equipped homes coming on-line in that project and
it's not yet known if in total their operation will cause poorly
controlled or unstable grid voltage.

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

They will share in proportion to their capacity.

Capacity is the "quantity of electrons" - which by itself tells you
nothing of the potential (voltage).

Electricity, water, nor **** flow uphill.

And height is exactly equivalent to voltage potential.

So PV panels can't push current into the grid unless the invertors raise
their voltage higher than the grid voltage. *Just matching the grid
voltage gets you to the point where your current flow is ZERO.

Total nonsense and such a basic failure at elementary circuit basics
that it discredits just about anything else you have to say. The
voltage
of the array IS the voltage of the grid at the point it's connected.
How
can it be anything else, unless you want to include the resistance of
the wire used to make the connection, which is immaterial for the
discussion.



*Every
millivolt you adjust your output voltage higher than the grid voltage
means some small increment in current outflow from your panels. *Since
you can't store the energy coming from the panels via battery bank, then
it's in your best interest to always maximize the amount of current
you're injecting into the grid up to the full potential of the panel's
output capacity. *That means raising the output voltage as high as you
need to so that every milliamp is squeezed out of them and onto the
grid.

Take a look at the dual voltage source circuit diagram that Jim
Wilkins
supplied a couple posts back. It's example #1.

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

It's a simple diagram of two ideal voltage
sources with series resistors connected to a load. That serves as a
basic model for two batteries or two generators or two PV arrays, etc
connected to a load. The example gives the full equations for what
would be two batteries of differing voltages and internal resistance
connected in parallel to a load. Change the voltages so that they
are equal, make them 20V. Solve those equations and you'll find
that BOTH sources are supplying current to the load. The
voltage on the "grid", ie across the load resistor is just one value.
One source is not at a higher value to "push" current.

And I'll bet if you do the equations with the voltage sources at the
same value, you'll find that twice as much current flows from the
voltage source with the 10 ohm resistor as the one with the 20
ohm resistor.

On Apr 15, 3:30*am, harry wrote:
On Apr 15, 12:40*am, "





wrote:
On Thu, 14 Apr 2011 08:52:52 -0700 (PDT), "

wrote:
On Apr 13, 3:24*am, harry wrote:
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.

For once I agree with Harry. * We can forget about generators and
distributions systems. *Just take two 12V batteries and connect
them in parallel to a 12ohm resistor. *Under Homeguy's
theories, I don;t know what he thinks would happen. *But
clearly he thinks if we put a second AC power source on
a distributions system, it has to be at a higher voltage to
"push" current out.

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

So, what happens with the two batteries? * Under the laws
of physics the rest of us use the voltage would remain at
12 volts and BOTH batteries would be supplying part of
the 1 AMP flowing through the resistors.

Not on Homeguy's and Vaughn's planet. *One of the batteries will be charging
the other.- Hide quoted text -

- Show quoted text -

That is correct. One charges, the other discharges. Eventually
equilbrium will be achieved and the current will cease.
If there was a resistor in parallel with both,the higher voltage
battery would supply current to the other battery and the resistor
until the voltage fell to the lower voltage battery and than both
would supply current to the resistor.- Hide quoted text -

- Show quoted text -

Maybe it wasn't clear what I meant when I said:

"Just take two 12V batteries and connect
them in parallel to a 12ohm resistor. Under the laws
of physics the rest of us use the voltage would remain at
12 volts and BOTH batteries would be supplying part of
the 1 AMP flowing through the resistor."

I didn't mean two batteries labled nominally as 12V that are
actually at different voltages because one is fully charged, the
other only partially charged. We were trying to discuss a
simple a comparison as possible of using two power sources
on a circuit.

So, do you agree that under the condition of two identical
fully charged batteries at exactly 12V, connected in
parallel to a load, the current will flow from both
batteries through the load? I hope you do. As for Homeguy
he apparently believes one has to be at a higher voltage
to "push" current. I have yet to hear him explain how the
batteries then decide which one it will be and how they
will change their voltage to obtain the allegedly necessary
"push" to get the current flowing.

" wrote:

Take a look at the dual voltage source circuit diagram that Jim
Wilkins supplied a couple posts back. It's example #1.

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

I did, you dumbass #2.

Go and read my last post.

It's a simple diagram of two ideal voltage sources with
series resistors connected to a load.

And note what happens when the voltage sources have unequal voltages.

And note that we are not talking about batteries here in the case of a
municipal power grid and a PV system.

And I'll bet if you do the equations with the voltage sources
at the same value, you'll find that twice as much current
flows from the voltage source with the 10 ohm resistor as
the one with the 20 ohm resistor.

If that diagram shows reverse current flow because Battery 1 has a lower
voltage than Battery 2 (and current I1 is negative), then at what point
does current I1 become zero? What would the voltage of battery 1 have
to be for current I1 to be zero?

On 4/14/2011 10:33 AM, harry wrote:
On Apr 13, 4:15 pm, wrote:
On 4/12/2011 10:58 AM, m II wrote:





"bud--" wrote in ...

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--

-----------------
|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--- Hide quoted text -

- Show quoted text -

Impedance is the vector sum of resistance and reactance measured in
Ohms not a percentage.
Regulation of a transformer mau be measured in percentage terms.
What exactly are yo on about?

Use of "per-unit" values for voltage, current, impedance, ... is common
in the electric power field. It makes calculations easier, particularly
as the system gets more extensive. One of the "per-unit" values is "%
impedance".

A utility transformer is likely to be rated in "% impedance".

Daestrom has written about this. Looks like mII (or whoever) is familiar
with it. That leaves you. If you knew as much as you think you know you
would be familiar with % impedance.

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

--
bud--

On 4/14/2011 11:47 PM, Home Guy wrote:
" unnecessarily full-quoted:

There is SO much wrong in your analysis, that I don't know
where to begin.

(Which is why I didn't try.)

But here's a start. You claim that with
a slightly higher voltage, an AC motor in an HVAC
compressor won't turn any faster and hence the additional
power is wasted. What you've completely overlooked is
that power is P=VI, or power is voltage times current.
Give that motor an extra half a volt and I'll bet it's
current decreases by a corresponding amount.

So why not run a 120V motor with 240 volts then?

Duh...


AC Motors are not simple loads like a resistor, but they will still
"consume" power (V x I) as a function of their supply voltage.


A motor running at a constant RPM creates a fixed amount of mechanical
power for a given load. RPM of induction motors is not very sensitive to
voltage. The electrical power used is tied to the mechanical power
consumed. Raising the voltage a little lowers the current a little.

--
bud--

On Apr 15, 1:47*pm, "
wrote:
On Apr 15, 7:54*am, Jim Wilkins wrote:

On Apr 15, 3:40*am, harry wrote:

On Apr 15, 5:46*am, "
...
He said unequal voltage. In which case he's right.
Capacity is not voltage.-

How to properly analyze the problem:http://www.electronics-tutorials.ws/...its/dcp_4.html

Look at Example 1.

jsw

Thank you Jim. *I was sitting here wishing I could draw
a simple circuit into the newsgroup that models what we
*are talking about and shows what really happens.
You've gone one better and found a perfect example
from an independent and credible source. *That
circuit is EXACTLY the model for the dual battery
example I brought up. *Each battery is modeled as
an ideal voltage source in series with a resistor. *And
each supplies part of the current flowing through
the load. *One never charges the other, nor does
one need to have a higher voltage to "push" current.

I suspect we'll be hearing soon from Homeguy
about how this isn't a valid way two batteries connected
*in parallel to a load can be modeled.

I think I brought up Kirchoff way back in this post.
It's not possible to reason withe "self taught homeworkshop bungler".
They sit and think up erroneous conclusion sal lthe time.
They need to go to a place of education.

On Apr 15, 2:14*pm, Home Guy wrote:
" wrote:
But clearly he thinks if we put a second AC power source on
a distributions system, it has to be at a higher voltage to
"push" current out.

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

What a dumbass!

You're the dumbass! *

Read the following:

=============
SMUD wanted to investigate what effect reverse power flow from exporting
PV systems would have on service and substation voltage regulation.
Figure 1 shows the Home-to-Substation voltage difference (top) and solar
irradiance (bottom) on a clear, cool day, Saturday, March 7, 2009,
representative of a day with relatively low load and high local PV
penetration. Penetration is defined as the amount of PV output divided
by the load at a particular point in time.

At night the substation voltage ranged between 0.4 V to 0.7 V higher
than the home voltage. This is representative of a typical circuit with
voltage drops through the line and transformer impedances. During
daylight hours, this reversed and the home voltage rose as high as 0.7
V, or 0.6%, greater than the substation voltage.
============

Did you read the last sentence dumbass?

The differential between substation and home voltage went from -.7 to
+.7 volts - a difference of almost 1.5 volts *due to the effect of the
PV panels injecting current into the grid.

If, according to you, there was no such phenomena of an increase in
local grid voltage caused by PV panels, then there should be no basis
for the point of this IEEE research paper I posted yesterday.

But engineers know that there will be a voltage increase because of
these panels, and the excercise now is to figure out how much PV power
can come on-line before the substation becomes unable to properly
regulate it's output voltage levels.

==============
Since the PV penetration levels were relatively low, there were no
adverse effects on voltage regulation. It was possible to see the
effects of the PV systems on the voltage at the individual homes and the
distribution transformers.
==============

There will be more PV-equipped homes coming on-line in that project and
it's not yet known if in total their operation will cause poorly
controlled or unstable grid voltage.

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

They will share in proportion to their capacity.

Capacity is the "quantity of electrons" - which by itself tells you
nothing of the potential (voltage).

Electricity, water, nor **** flow uphill.

And height is exactly equivalent to voltage potential.

So PV panels can't push current into the grid unless the invertors raise
their voltage higher than the grid voltage. *Just matching the grid
voltage gets you to the point where your current flow is ZERO. *Every
millivolt you adjust your output voltage higher than the grid voltage
means some small increment in current outflow from your panels. *Since
you can't store the energy coming from the panels via battery bank, then
it's in your best interest to always maximize the amount of current
you're injecting into the grid up to the full potential of the panel's
output capacity. *That means raising the output voltage as high as you
need to so that every milliamp is squeezed out of them and onto the
grid.

I don't think there's any help for you. If you can't get your mind
round the topic by now, best give up. Colour in some pictures.

On Apr 15, 3:41*pm, "
wrote:
On Apr 15, 3:30*am, harry wrote:





On Apr 15, 12:40*am, "

wrote:
On Thu, 14 Apr 2011 08:52:52 -0700 (PDT), "

wrote:
On Apr 13, 3:24*am, harry wrote:
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.

For once I agree with Harry. * We can forget about generators and
distributions systems. *Just take two 12V batteries and connect
them in parallel to a 12ohm resistor. *Under Homeguy's
theories, I don;t know what he thinks would happen. *But
clearly he thinks if we put a second AC power source on
a distributions system, it has to be at a higher voltage to
"push" current out.

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

So, what happens with the two batteries? * Under the laws
of physics the rest of us use the voltage would remain at
12 volts and BOTH batteries would be supplying part of
the 1 AMP flowing through the resistors.

Not on Homeguy's and Vaughn's planet. *One of the batteries will be charging
the other.- Hide quoted text -

- Show quoted text -

That is correct. One charges, the other discharges. Eventually
equilbrium will be achieved and the current will cease.
If there was a resistor in parallel with both,the higher voltage
battery would supply current to the other battery and the resistor
until the voltage fell to the lower voltage battery and than both
would supply current to the resistor.- Hide quoted text -

- Show quoted text -

Maybe it wasn't clear what I meant when I said:

"Just take two 12V batteries and connect
them in parallel to a 12ohm resistor. *Under the laws
of physics the rest of us use the voltage would remain at
12 volts and BOTH batteries would be supplying part of
the 1 AMP flowing through the resistor."

I didn't mean two batteries labled nominally as 12V that are
actually at different voltages because one is fully charged, the
other only partially charged. *We were trying to discuss a
simple a comparison as possible of using two power sources
on a circuit.

So, do you agree that under the condition of two identical
fully charged batteries at exactly 12V, connected in
parallel to a load, the current will flow from both
batteries through the load? *I hope you do. *As for Homeguy
he apparently believes one has to be at a higher voltage
to "push" current. *I have yet to hear him explain how the
*batteries then decide which one it will be and how they
will change their voltage to obtain the allegedly necessary
"push" to get the current flowing.- Hide quoted text -

- Show quoted text -

What you say is correct. I was referring to some theory he was on
about with one battery @12V and one @11V

On Apr 15, 4:24*pm, bud-- wrote:
On 4/14/2011 10:33 AM, harry wrote:





On Apr 13, 4:15 pm, *wrote:
On 4/12/2011 10:58 AM, m II wrote:

"bud--" wrote in ...

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--

-----------------
|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--- Hide quoted text -

- Show quoted text -

Impedance is the vector sum of resistance and reactance measured in
Ohms not a percentage.
Regulation of a transformer mau be measured in percentage terms.
What exactly are yo on about?

Use of "per-unit" values for voltage, current, impedance, ... is common
in the electric power field. It makes calculations easier, particularly
as the system gets more extensive. One of the "per-unit" values is "%
impedance".

A utility transformer is likely to be rated in "% impedance".

Daestrom has written about this. Looks like mII (or whoever) is familiar
with it. That leaves you. If you knew as much as you think you know you
would be familiar with % impedance.

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

--
bud--- Hide quoted text -

- Show quoted text -

Link?
I used to work for a power company. Never once heard of Impedance
being expressed as a percentage.
Transformers are rated in Kva.
There may be a continuous KVA and a higher one for a specified period.
Regulation is sometimes expressed as a %. ie how the output voltage
varies between no load and full load conditions.
Here's my link:-
http://en.wikipedia.org/wiki/Impedance_(electrical)
Now, where's yours?

On Apr 15, 10:47*am, Home Guy wrote:
...
And note what happens when the voltage sources have unequal voltages.

And note that we are not talking about batteries here in the case of a
municipal power grid and a PV system.
...

Considering the level of this discussion when the solutions are simple
algebra, you aren't ready to handle the analytical geometry and
differential equations of AC circuit analysis.
http://www.allaboutcircuits.com/vol_2/index.html
http://www.circuit-magic.com/laws.htm

jsw

On 15/04/2011 07:41, wrote:


So, do you agree that under the condition of two identical
fully charged batteries at exactly 12V, connected in
parallel to a load, the current will flow from both
batteries through the load? I hope you do.

Your example, while it is correct, have little bearing on real life
problems. The grid can in no circumstances be looked at as identical to
a PV array voltage converted to AC voltage.

Two identical batteries have identical inner resistance, that is why
your example works.


As for Homeguy
he apparently believes one has to be at a higher voltage
to "push" current.

He is correct. Why? Because he wants to "push" current into the grid.
With your 2 batteries at the exact same voltage, how do you get current
flowing from the "home battery" to the "grid battery"?


I have yet to hear him explain how the
batteries then decide which one it will be and how they
will change their voltage to obtain the allegedly necessary
"push" to get the current flowing.

He does not have to explain that, it is self explanatory when one
understand that there are no perfect conductors with zero resistance in
a power distribution system.

To get power into the grid from a local generated power the voltage has
to be higher. HOW much higher depends on the impedance in the systems.

On Apr 15, 12:48*pm, Jim Wilkins wrote:
On Apr 15, 10:47*am, Home Guy wrote:

...
And note what happens when the voltage sources have unequal voltages.

And note that we are not talking about batteries here in the case of a
municipal power grid and a PV system.
...

Considering the level of this discussion when the solutions are simple
algebra, you aren't ready to handle the analytical geometry and
differential equations of AC circuit analysis.http://www.allaboutcircuits..com/vol...c.com/laws.htm

jsw

This is a good example:
http://www.allaboutcircuits.com/vol_2/chpt_5/4.html
"Now that we've seen how series and parallel AC circuit analysis is
not fundamentally different than DC circuit analysis,..."

Electrical engineers use j instead of i for the square root of -1
because i has long been the standard for Current (intensity). In this
instance the imaginary number is an excellent tool to analyze real-
world physical phenomena.
http://www.microwaves101.com/encyclo...Smithchart.cfm

jsw

On 15/04/2011 05:47, wrote:
On Apr 15, 7:54 am, Jim wrote:


How to properly analyze the problem:http://www.electronics-tutorials.ws/...its/dcp_4.html

Look at Example 1.

jsw



That
circuit is EXACTLY the model for the dual battery
example I brought up.

No it is not. It is an example with two different voltages.

It is a perfect example of how to supply power to the grid, resulting in
higher voltage at the "home-owners end"

hmmmmm. worked for a power company? Perhaps not an electrical "power "
company?
A weight lifter?
Meter Reader?

Sounds like a complete misnomer. Correctly that should be "energy" company.


All teasing aside, m II is quite correct in asking about percent impedance.
Have a look at any distribution or large power transformer and you will see
a rating called "%Imp."

I have seen pole transformers down to 1.2%Imp and up to 13.8%Imp.

This is where the electrical energy distribution companies get concerned
about losses. (see I even remembered my own comment. LOL)

Worked on a 2MVA Grounding Bank with a 89 %Imp. rating. (maybe it was 93???
Can't remember now)

As an aside... (utility guys will appreciate this one)
In the old days, before equipment was invented, we used to prove polarity of
CTs with a 6v lantern battery and an old Model T voltmeter coupled with a
variable resistor in series to avoid bashing the needle too hard. We would
generate positive and negative going pulse on the primary of the CT and see
the needle deflection on the secondary of the CT winding.

Trouble was this high impedance bitch would not pass any current from the 6V
lantern battery through the primary winding and through the CT primary
(bushing type). SO we get two 6V lantern batteries in series and try again
to no avail. hmmmmm...

Well we get talking about it and while still holding the leads on the
primary (14kV) bushings my partner begins to yell that the needle was now
deflecting upwards, so I pull it off!...Well! I drew about a 10" long arc
(yeah bare hands..no grid connection) and friggin' nearly fell off the
transformer, some 12-15' high, to the ground. WHEW!!!

Back we went to the 6V lantern battery, waited a little longer for this
impedance haystack to react and wore the 30kV gloves. Sitting on top was a
little more uncomfortable too...LOL



Here is your link with an oversimplified explanation.
http://www.transformerworld.co.uk/impedance.htm


---------------
"harry" wrote in message
...
Link?
I used to work for a power company. Never once heard of Impedance
being expressed as a percentage.
Transformers are rated in Kva.
There may be a continuous KVA and a higher one for a specified period.
Regulation is sometimes expressed as a %. ie how the output voltage
varies between no load and full load conditions.
Here's my link:-
http://en.wikipedia.org/wiki/Impedance_(electrical)
Now, where's yours?

On Apr 15, 1:03*pm, g wrote:
On 15/04/2011 07:41, wrote:



So, do you agree that under the condition of two identical
fully charged batteries at exactly 12V, connected in
parallel to a load, the current will flow from both
batteries through the load? *I hope you do.

Your example, while it is correct, have little bearing on real life
problems. The grid can in no circumstances be looked at as identical to
a PV array voltage converted to AC voltage.

Two identical batteries have identical inner resistance, that is why
your example works.

Wrong. Go back to the example in the link, as I suggested to
Homeguy. You have two ideal voltage sources with different voltages
connected through two resistors of different values, one 20 ohms,
the other 10 ohms. That is a basic model of a battery, where the
10 and 20 ohm resistors represent the different internal resistances
of the batteries. So they are batteries with identica voltage, but
different internal resistances.
Very basic stuff. The Kirchoff equations are right
there. All you have to do is change the voltages so that they
are equal.

Solve the equations for the case where the voltage sources
are at identical voltages and you will find that current flows from
BOTH sources. One is NOT at a higher potential than the
other, which is what Homeguy claims must exist for both
to provide current to the load.




As for Homeguy
he apparently believes one has to be at a higher voltage
to "push" current.

He is correct. Why? Because he wants to "push" current into the grid.
With your 2 batteries at the exact same voltage, how do you get current
flowing from the "home battery" to the "grid battery"?

I never said current flows from one battery to the other. It's
apparently
Homeguy and now you who are hung up on that for reasons unknown.
If the voltage sources are equal, no current flows between the two,
which
is the situation that is most desirable when powering
a load with two batteries in parallel. They just BOTH supply part of
the
current to the load resistor.


As I said, go back and solve the equations for the case where the
voltages are equal and you will find that they BOTH supply current
to the load. Because of the differing resistors which would represent
the internal resistance of the batteries, one supplies twice the
current of the other. Capiche?



I have yet to hear him explain how the
* batteries then decide which one it will be and how they
will change their voltage to obtain the allegedly necessary
"push" to get the current flowing.

He does not have to explain that, it is self explanatory when one
understand that there are no perfect conductors with zero resistance in
a power distribution system.

To get power into the grid from a local generated power the voltage has
to be higher. HOW much higher depends on the impedance in the systems.



Imperfections do not render a model useless or change the facts. If
you
want to model the transmission line, then insert some additional
resistors
after the 10 ohm and 20 ohm resistors to model the line. Make a
model
that includes capacitance and inductance too, and make it an AC
circuit . It changes
nothing with regard to the ridiculous requirement that in order to
supply
current, a source can't just be equal in voltageto another source
connected to the same load and that it has to be
higher. If you have a model that shows Homeguy's planet, we'd like
to see it. Until then, Jim's model is perfectly fine.

On Apr 15, 10:47*am, Home Guy wrote:
" wrote:
Take a look at the dual voltage source circuit diagram that Jim
Wilkins supplied a couple posts back. *It's example #1.

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

I did, you dumbass #2.

Go and read my last post.

I have read your posts and like most other people here
have concluded you are wrong, so I don't see why
you're calling ME the dumbass.



It's a simple diagram of two ideal voltage sources with
series resistors connected to a load.

And note what happens when the voltage sources have unequal voltages.

Uh, huh. I have noted that and if you have any questions about how
that works, I'll be happy to answer them. I'm still waiting for your
answer about the case with EQUAL voltages. That was the essence
of your argument, was it not? That a power source can't provide
power in parallel with another unless it raised the voltage? So,
leave everything else the same and just make the voltage sources
equal. Tell us what current flows through the load resistor and what
currents flow through each voltage source. This is EE course circuit
theory course 101, about the first week.

BTW, you have an engineering degree?



And note that we are not talking about batteries here in the case of a
municipal power grid and a PV system.

Yes and since you seem incapable of understanding a simple
circuit that represents 2 batteries connected in parallel to a
load, no need to add the additional complexities.



And I'll bet if you do the equations with the voltage sources
at the same value, you'll find that twice as much current
flows from the voltage source with the 10 ohm resistor as
the one with the 20 ohm resistor.

If that diagram shows reverse current flow because Battery 1 has a lower
voltage than Battery 2 (and current I1 is negative), then at what point
does current I1 become zero? *What would the voltage of battery 1 have
to be for current I1 to be zero?

I see you do have a question.
Simple. With no current flow through battery 1, then the circuit
is reduced to an ideal voltage source connected to two resistors
in series. One of these represents the internal resistance of
battery 2 and is 20 ohms. The other is the load resistor of 40
ohms. So, we have 20 volts across 60 ohms, giving a
current of .333Amps. That .333amps produces a voltage of
13.34 across the load resistor. (.333 A X 40ohms.) That means
voltage source V1 would have to be at the same potential,
13.34 volts and when it is, no current flows through what
represents battery 1.

I've answered your question, now answer mine:

What are currents I1, I2, I3 when the voltage sources V1 and V2
are both 20 volts.

On 15/04/2011 10:46, wrote:
On Apr 15, 1:03 pm, wrote:
On 15/04/2011 07:41, wrote:



So, do you agree that under the condition of two identical
fully charged batteries at exactly 12V, connected in
parallel to a load, the current will flow from both
batteries through the load? I hope you do.

Your example, while it is correct, have little bearing on real life
problems. The grid can in no circumstances be looked at as identical to
a PV array voltage converted to AC voltage.

Two identical batteries have identical inner resistance, that is why
your example works.

Wrong.

I am not sure what you think I am wrong about here. Could you please
indicate which of my above statements are wrong?

The topic is " Feeding solar power back into municipal grid". How are
the voltage levels when there is a flow into the grid from the PV array?

Maybe I missed a topic change somewhere in the previous posts, which is
easy to do since a few posters here does not bother to trim posts.


Go back to the example in the link, as I suggested to
Homeguy. You have two ideal voltage sources with different voltages
connected through two resistors of different values, one 20 ohms,
the other 10 ohms. That is a basic model of a battery, where the
10 and 20 ohm resistors represent the different internal resistances
of the batteries. So they are batteries with identica voltage, but
different internal resistances.

I did read the example, and yes, I understand Kirchoff's Law.

But you seem to be contradicting yourself, you state above the batteries
have different voltages, then you state they are batteries with
identical voltages. Maybe I am not understanding what you mean.


Solve the equations for the case where the voltage sources
are at identical voltages and you will find that current flows from
BOTH sources. One is NOT at a higher potential than the
other, which is what Homeguy claims must exist for both
to provide current to the load.

I have read all of Homeguy's messages, and I may have missed it, but
does he not claim the need to have a higher voltage to push current into
the GRID, not the LOAD.


I never said current flows from one battery to the other. It's
apparently
Homeguy and now you who are hung up on that for reasons unknown.

The reason of my "hangup" is the topic: Feeding solar power back into
municipal grid.

Your example of batteries with equal voltage might be interesting, but
irrelevant to the topic.