Electric Energy Economizer !!

[Keith R. Williams]

In article <REiRXTBjDm$$Ew$J@jmwa.demon.co.uk>,
jmw@jmwa.demon.contraspam.yuk says...

I read in sci.engr.electrical.compliance that Keith R. Williams
krw@attglobal.net> wrote (in <MPG.1a67d14da9b5b08498aa97@enews.newsguy.
com&gt about 'Electric Energy Economizer !!', on Thu, 8 Jan 2004:

Well, the idea is to show that significant energy is being
*wasted* and that it may cost some money to save money (and all
that other green stuff). The suggestions so far say that the
*power company* is at fault for generating this "wasting power".

I don't agree that they do say that, but in any case, what is your
suggestion for a new term?

I'm not sure I have a good "new" term. It's possible there is a
good term or another way of expressing the problem such that
politicians/bureaucrats can understand the issues.

AIUI, it's largely a transmission problem and not generation.
It's counter-intuitive for people to believe that "phantom
power"

That term is already used for something else.

Understand.

eats up "real" power in the transmission system.
Remember, we don't want to spend anything on transmission, since
it doesn't "do" anything. This problem is far beyond the 30sec
sound-bites the six-o-clock news gives it. It's this not
interesting to politicians. Perhaps if the *entire* system were
privatized (rather than the lip-service given to generation),
such that the transmission companies could *invest* in solutions.

In UK, we have complete privatization, but with a 'Regulator',
independent of government, who has power to prevent abuses, such as
unjustified price increases or failure to maintain a reliable supply.
The effect is to seriously restrict the type of investment that you
propose.

It's not all that much different here (perhaps even worse since
the generation, transmission, and distribution is often split).
The bureaucrats don't understand the technology for which they're
charged with regulating. The public understands even less.
"Privatization" and "deregulation" have become such bad words
that nothing is likely to get fixed.

--
Keith

 

[Keith R. Williams]

In article <a0TTLfBkLm$$Ew8a@jmwa.demon.co.uk>,
jmw@jmwa.demon.contraspam.yuk says...

I read in sci.engr.electrical.compliance that Keith R. Williams
krw@attglobal.net> wrote (in <MPG.1a67d2fd6df25e8198aa98@enews.newsguy.
com&gt about 'Electric Energy Economizer !!', on Thu, 8 Jan 2004:

I guess the only solution is to fight the NIMBYs,
force the generation into their back yard, and shrink the grids
to something more manageable.

Embedded generation is not all good news. It increases the fault level
(prospective fault current or MVA), which means the network may well
need to be reinforced (a thicker grade of damp string).

Sure, but a smaller network will have less inductance. When
talking with by brother he mentioned that short lines are
capacitive, quickly becoming inductive at some lenth (don't
remember numbers). This adds to the instability.

("thicker grade of damp string" - I like it! ;-)

DC interconnect would seem to help
here (transfer power, but not the uncontrolled phase problem).

Yes, but the up-front cost is not negligible.

Understand. Though the cost must have come down significantly
over the last 20 years or so when the DC interconnects were done
in the western US.

Sme of these were caused by sheer stupidity, as the '67(?) NE
blackout.

But I think they all began with small, and not unusual faults or
mistakes, that happened to be critical under the circumstances.

Smaller grids and more distributed/local generation should help
in mess prevention, no?

Sure, mistakes propagate fast in unstable systems, but
I believe the larger problem here was the lack of local
generation. Once the critical path blew, there was no chance to
anything other than disaster.

Unfortunately, the physical dimensions of the network are almost
irrelevant. In other words, a large network with 500 generators behaves
very similarly to a small one with 500 generators. There is less
transmission-line capacitance in the smaller network, but there may be
more capacitance elsewhere.

Again, I understood that short lines were slightly capacitive
(can be a good thing since large loads tend to be inductive) and
that long lines were highly inductive (bad). ...or did I miss
the boat here?

--
Keith

 

[Keith R. Williams]

In article <3ffff46c$1@news.rivernet.com.au>,
noemail@noemail.noemail.com says...

"Keith R. Williams" <krw@attglobal.net> wrote in message
news:MPG.1a67d2fd6df25e8198aa98@enews.newsguy.com...
In article <m5bMIdAHnb$$EwrC@jmwa.demon.co.uk>,
jmw@jmwa.demon.contraspam.yuk says...
I read in sci.engr.electrical.compliance that Keith R. Williams
krw@attglobal.net> wrote (in <MPG.1a669c8b86c0234c98aa8f@enews.newsguy.
com&gt about 'Electric Energy Economizer !!', on Wed, 7 Jan 2004:

Power systems a obviously above this microprocessor developer's
expertise, but why wouldn't point-source current control work?
I'm all ears here! ...or are you saying the problem isn't
manageable at all?

I certainly don't claim to be an expert, but power systems including
generators have complex dynamic characteristics. Oscillation with
various degrees of damping, both above and below critical, are certainly
possible. Generators can appear as inductors or capacitors, according to
the level of excitation, and can lose synch. Making a power-factor
correction at one point can precipitate a crisis elsewhere.

Yeah, I was thinking about the synchronous capacitors talked
about here a few months ago. They aren't free, but seem to be a
good idea to limit the "waste current". ;-) However, I can see
that there are some rather long time-constants with such
mechanical beasts (as with generators), so control of a large
system may be chaotic, thus not be possible under all
circumstances. I guess the only solution is to fight the NIMBYs,
force the generation into their back yard, and shrink the grids
to something more manageable. DC interconnect would seem to help
here (transfer power, but not the uncontrolled phase problem).


Quit blaming power factor for blackouts.

But there was significant useless current (switches are set for
line current) in the lines that faulted. I'm told the lines
heated enough to droop and then arced over to trees causing the
system to fail.

The blackouts are caused by overload , for which the cure is load shedding.

Sure. Those excessive AMPS didn't help the overload, now did
they?

The problem easily becomes unmanageable, as we saw last year, in USA,
Italy and London, England. These are not the only major outages in
living memory, NE USA and SE England have had major incidents in the
past, caused by relatively minor trigger events.

Sme of these were caused by sheer stupidity, as the '67(?) NE
blackout. Sure, mistakes propagate fast in unstable systems, but
I believe the larger problem here was the lack of local
generation. Once the critical path blew, there was no chance to
anything other than disaster.

The west coast of the USA had a similar blackout, same cause as the recent
NE USA blackout.

I'm not sure the cause was the same. According to my brother (he
was responsible for setting the switches on one edge of the
transmission lines that failed, saving his system and further
outages) it *was* VARS that caused the problem in the August Ohio
debacle. He wasn't there for the CA blackout.

--
Keith

 

[Keith R. Williams]

In article <3fffee5b@news.rivernet.com.au>,
noemail@noemail.noemail.com says...

"Keith R. Williams" <krw@attglobal.net> wrote in message
news:MPG.1a669adeab3c983498aa8d@enews.newsguy.com...
In article <L3NnrQAYJl+$EwGZ@jmwa.demon.co.uk>,
jmw@jmwa.demon.contraspam.yuk says...
I read in sci.engr.electrical.compliance that Keith R. Williams
krw@attglobal.net> wrote (in <MPG.1a63df869880ac1998aa6b@enews.newsguy.
com&gt about 'Electric Energy Economizer !!', on Mon, 5 Jan 2004:
In article <oj4v+tAOlR+$EwCo@jmwa.demon.co.uk>,
jmw@jmwa.demon.contraspam.yuk says...
I read in sci.engr.electrical.compliance that Ross Mac
this.is.a.mung@e
xample.invalid> wrote (in
yo0Kb.597415$0v4.23524771@bgtnsc04-news.ops.w
orldnet.att.net&gt about 'Electric Energy Economizer !!', on Sun, 4
Jan
2004:

Since power is measured by the meter, when the power factor is
corrected,
would there not be a savings of power ???

The ordinary rotating disc meters measure real power only. They do
not
measure 'reactive power', which is badly named anyway.

You prefer "imaginary" power? My brother, an engineer in the
power biz, *hates* that term (though it is real ;-). The media
(thus politicians) can't grasp the concept of the "imaginary", so
cannot spend money on the problem. According to him, the August
debacle wouldn't have happened if the power companies could
control the 'j'. ...it's not like "imaginary" power is free
either!

No, I can see the problems with 'imaginary', which is also a poor name
by itself. Both words, 'reactive' and 'power' are misleading to non-
technical people. I don't have a suggestion for a better name that I
feel is good enough to withstand the inevitable challenges.

Ok, but it's got to be called something! Perhaps we can call it
"terrorist power". The pols will do something then! ;-)

Seriously, perhaps there is a way of showing the wasted power in

Its NOT WASTED !

Sure it is. It's useless current in the transmission system that
causes heating. *That* is wasted energy.

Reactive power includes an imaginary component that is truely imaginary.

I know what reactive power is. I *are* an EE, though not a power
jock (thus asking questions here).

Maybe the better word is "imaginarily increased power". or "peak volts peak
amps power".. the power you get when you use peak voltage and peak amps to
calculate power.

Imaginary increased power? Doesn't that imply that you're
wasting something. But wait! You said there is nothing wasted.
Peak amps? Peak volts?

Eg The actual power is 10 kW, the 'reactive power' is 14 kW, then there is
4kW * time in second joules of imaginary power, and its perfectly imaginary
because there was no transfer of those joules from supply to consumer !

No, it was dissipated by the transmission system. You generated
amps (4kW worth) that the load didn't consume. That must be
dissipated somewhere, no? (wasted).

Large factories arent allowed to have bad PF because they would burn out
transformers and things "faster" (or "more often" , as the case may be) ,
and causes a problem with quality of supply to other consumers (and the
rest of the plant)

I thought you said there wasn't anything wasted. What is burning

out these "things"?

--
Keith

 

[John Woodgate]

I read in sci.engr.electrical.compliance that Keith R. Williams
<krw@attglobal.net> wrote (in <MPG.1a69fc3413cbc9d198aa9f@enews.newsguy.
com&gt about 'Electric Energy Economizer !!', on Sat, 10 Jan 2004:

Again, I understood that short lines were slightly capacitive
(can be a good thing since large loads tend to be inductive) and
that long lines were highly inductive (bad). ...or did I miss
the boat here?

No, you are right. But it's not a straightforward 'good/bad' thing. By
itself, the line inductance can't cause instability. It does cause a
voltage drop, and thus contributes to MVAr.
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!

 

[Keith R. Williams]

In article <SnQh$7BXKFAAFwqV@jmwa.demon.co.uk>,
jmw@jmwa.demon.contraspam.yuk says...

I read in sci.engr.electrical.compliance that Keith R. Williams
krw@attglobal.net> wrote (in <MPG.1a69fc3413cbc9d198aa9f@enews.newsguy.
com&gt about 'Electric Energy Economizer !!', on Sat, 10 Jan 2004:

Again, I understood that short lines were slightly capacitive
(can be a good thing since large loads tend to be inductive) and
that long lines were highly inductive (bad). ...or did I miss
the boat here?

No, you are right. But it's not a straightforward 'good/bad' thing. By
itself, the line inductance can't cause instability. It does cause a
voltage drop, and thus contributes to MVAr.

However those MVArs are wasteful of a critical resource
(transmission line capacity).

--
Keith

 

[Isaac]

On Sat, 10 Jan 2004 12:53:18 -0500, Keith R Williams <krw@attglobal.net> wrote:

In article <3fffee5b@news.rivernet.com.au>,
noemail@noemail.noemail.com says...

Sure it is. It's useless current in the transmission system that
causes heating. *That* is wasted energy.

Reactive power includes an imaginary component that is truely imaginary.

I know what reactive power is. I *are* an EE, though not a power
jock (thus asking questions here).

Maybe the better word is "imaginarily increased power". or "peak volts peak
amps power".. the power you get when you use peak voltage and peak amps to
calculate power.

Imaginary increased power? Doesn't that imply that you're
wasting something. But wait! You said there is nothing wasted.
Peak amps? Peak volts?

Eg The actual power is 10 kW, the 'reactive power' is 14 kW, then there is
4kW * time in second joules of imaginary power, and its perfectly imaginary
because there was no transfer of those joules from supply to consumer !

No, it was dissipated by the transmission system. You generated
amps (4kW worth) that the load didn't consume. That must be
dissipated somewhere, no? (wasted).

You are confusing some concepts here. If the transmission lines were
perfect zero ohm conductors, then the reactive power would not cause any
losses in the transmission lines. Yet the reactive power would still
exist.

The reactive power comes from energy that is alternately stored in
electrostatic and magnetic fields and then returned to the system when
those fields collapse. In an ideal system the energy stored and returned
each cycle adds up to zero and thus no net energy is required from the
source of supply.

The reactive power does increase the current transmitted over the system
and the increased current results in increased dissipation in the
transmission system. But that dissipated power is real not imaginary.
The extra power being dissipated is not numerically equal to the
reactive power (except by a sheer coincidence). Its size can be greater
or smaller than the size of the reactive power.

In short, you are confusing the system losses associated with reactive
power with the reactive power itself.

I'm sure you knew this stuff at one time, but apparently the knowledge
is fading from non use.

Isaac

 

[Peter Dettmann]

On Sat, 10 Jan 2004 23:21:34 +1100, "Leon."
<noemail@noemail.noemail.com> wrote:

not
measure 'reactive power', which is badly named anyway.

You prefer "imaginary" power? My brother, an engineer in the
power biz, *hates* that term (though it is real ;-). The media
(thus politicians) can't grasp the concept of the "imaginary", so
cannot spend money on the problem. According to him, the August
debacle wouldn't have happened if the power companies could
control the 'j'. ...it's not like "imaginary" power is free
either!

No, I can see the problems with 'imaginary', which is also a poor name
by itself. Both words, 'reactive' and 'power' are misleading to non-
technical people. I don't have a suggestion for a better name that I
feel is good enough to withstand the inevitable challenges.

Ok, but it's got to be called something! Perhaps we can call it
"terrorist power". The pols will do something then! ;-)

Seriously, perhaps there is a way of showing the wasted power in

Its NOT WASTED !

Reactive power includes an imaginary component that is truely imaginary.

Maybe the better word is "imaginarily increased power". or "peak volts peak
amps power".. the power you get when you use peak voltage and peak amps to
calculate power.


Eg The actual power is 10 kW, the 'reactive power' is 14 kW, then there is
4kW * time in second joules of imaginary power, and its perfectly imaginary
because there was no transfer of those joules from supply to consumer !

Yes you are half right, the reactive power is not wasted as such.
However you omitted to mention that there is an increase in losses due
to the higher line current which does cause loss of power which is
wasted.
But the worst aspect is that the generator has to have the additional
MVA capability to supply the load, and this is generally the most
important factor as with reactive load being supplied this can
seriously reduce the real power availabity from an alternator.

Peter Dettmann

 

[Repeating Rifle]

in article dpf10098qiikro71ksdb1m09g675fg5134@4ax.com, Peter Dettmann at
peter@aardvark.net.au wrote on 1/10/04 7:23 PM:

measure 'reactive power', which is badly named anyway.

This is not so badly named. It represents true power transfer, although is
not net power transfer. It is tranfered to the customer and back to the
source at double the line frequency.

Bill

 

[Charles Perry]

"Repeating Rifle" <SalmonEgg@sbcglobal.net> wrote in message
news:BC2608A6.555B%SalmonEgg@sbcglobal.net...

in article dpf10098qiikro71ksdb1m09g675fg5134@4ax.com, Peter Dettmann at
peter@aardvark.net.au wrote on 1/10/04 7:23 PM:

measure 'reactive power', which is badly named anyway.

This is not so badly named. It represents true power transfer, although is
not net power transfer. It is tranfered to the customer and back to the
source at double the line frequency.

Bill

Excuse me? "double the frequency"?

Charles Perry P.E.

 

[Keith R. Williams]

In article <slrnc00slc.213.isaac@latveria.castledoom.org>,
isaac@latveria.castledoom.org says...

On Sat, 10 Jan 2004 12:53:18 -0500, Keith R Williams <krw@attglobal.net> wrote:
In article <3fffee5b@news.rivernet.com.au>,
noemail@noemail.noemail.com says...

Sure it is. It's useless current in the transmission system that
causes heating. *That* is wasted energy.

Reactive power includes an imaginary component that is truely imaginary.

I know what reactive power is. I *are* an EE, though not a power
jock (thus asking questions here).

Maybe the better word is "imaginarily increased power". or "peak volts peak
amps power".. the power you get when you use peak voltage and peak amps to
calculate power.

Imaginary increased power? Doesn't that imply that you're
wasting something. But wait! You said there is nothing wasted.
Peak amps? Peak volts?

Eg The actual power is 10 kW, the 'reactive power' is 14 kW, then there is
4kW * time in second joules of imaginary power, and its perfectly imaginary
because there was no transfer of those joules from supply to consumer !

No, it was dissipated by the transmission system. You generated
amps (4kW worth) that the load didn't consume. That must be
dissipated somewhere, no? (wasted).

You are confusing some concepts here. If the transmission lines were
perfect zero ohm conductors, then the reactive power would not cause any
losses in the transmission lines. Yet the reactive power would still
exist.

No, the *current* has to be dissipated *somewhere*. The
generator is producing VA and the load is dissipating W. The
difference *must* be dissipated *somewhere*. ...and that is in
the transmission/distribution system.

The reactive power comes from energy that is alternately stored in
electrostatic and magnetic fields and then returned to the system when
those fields collapse. In an ideal system the energy stored and returned
each cycle adds up to zero and thus no net energy is required from the
source of supply.

Sure, I fully understand reactive power. However, the generator
is producing VA. The load is consuming W. If there is a
reactance inbetween, there is a loss in the wires caused by the
reactance.

The reactive power does increase the current transmitted over the system
and the increased current results in increased dissipation in the
transmission system. But that dissipated power is real not imaginary.

Good grief! That's my point! It may be "imaginary", but the
loss is *REAL*.

The extra power being dissipated is not numerically equal to the
reactive power (except by a sheer coincidence). Its size can be greater
or smaller than the size of the reactive power.

If the generator is producing 14kVA and the load consuming 10kW,
exactly where is the rest going?

In short, you are confusing the system losses associated with reactive
power with the reactive power itself.

Well, they're sorta related, no?

Ok, a generator *can* be used to assume a MVAr regulation role,
but it isn't putting out it's maximum power then either. Someone
else is getting a ride...

I'm sure you knew this stuff at one time, but apparently the knowledge
is fading from non use.

No, I'm not a power engineer (as I've said, I do high-end
microprocessor development) and I never did study power systems.
However, your arguments haven't convinced me yet. Please
continue to try though. ;-)

--
Keith

 

[Matthew Beasley]

"Repeating Rifle" <SalmonEgg@sbcglobal.net> wrote in message
news:BC2608A6.555B%SalmonEgg@sbcglobal.net...

in article dpf10098qiikro71ksdb1m09g675fg5134@4ax.com, Peter Dettmann
at
peter@aardvark.net.au wrote on 1/10/04 7:23 PM:

measure 'reactive power', which is badly named anyway.

This is not so badly named. It represents true power transfer,
although is
not net power transfer. It is tranfered to the customer and back to
the
source at double the line frequency.

Bill

For single phase it travels back and forth at twice the line frequency.
For BALANCED three phase, there is no back and forth... the
instantanious power flow is always to the load... YET there is still VAR
flow.

Matthew

 

[Matthew Beasley]

"Charles Perry" <pipesandtobacco@hotmail.com> wrote in message
news:btqhh0$a8nlm$1@ID-103962.news.uni-berlin.de...

"Repeating Rifle" <SalmonEgg@sbcglobal.net> wrote in message
news:BC2608A6.555B%SalmonEgg@sbcglobal.net...
in article dpf10098qiikro71ksdb1m09g675fg5134@4ax.com, Peter
Dettmann at
peter@aardvark.net.au wrote on 1/10/04 7:23 PM:

measure 'reactive power', which is badly named anyway.

This is not so badly named. It represents true power transfer,
although is
not net power transfer. It is tranfered to the customer and back to
the
source at double the line frequency.

Bill


Excuse me? "double the frequency"?

Charles Perry P.E.

For SINGLE phase, the power flow reverses at twice line frequency.

Assume a circuit of an AC generator connected to an ideal capacitor,
with the circuit in equilibrium.
Start at the positive going zero line crossing. From this point until
the line crest, the capacitor will charge. Energy flow :
Generator ----> Capacitor
From the crest until the negative going zero crossing the current
reverses, energy flow:
Capacitor ----> Generator
From the negative going zero crossing, the current will remain the same
but the sign of the voltage reversed. Energy flow:
Generator ----> Capacitor
From the negative crest until the positive going zero crossing, the
current will reverse. Energy flow:
Capacitor -----> Generator

For one cycle of the line the energy flow reverses twice, hence, twice
the frequency.

Matthew

 

[Matthew Beasley]

"Keith R. Williams" <krw@attglobal.net> wrote in message .
-snipped-

If the generator is producing 14kVA and the load consuming 10kW,
exactly where is the rest going?

I think you may be confusing that the generator's rating may be measured
in kVA, but that doesn't mean how much POWER it is producing - or draws.
If you measure the power input (mechanical) to the generator, it will
increase only a small amount when adding VAR's.
In your example if the generator is producing 10kW at unity power
factor, lets say for discussion sake it takes 10.3kW to spin.
Now lets add a 9.8kVAR load to the output to get to 14kVA. (BTW, since
real and reactive currents are in quadniture, the VA is the root mean
sum.) In this case, the input to the generator might raise to 10.4kW
input. The small increase is due to the extra losses from the
increasing current - but the input power doesn't increase to 14kW.
Carrying VARs only increases the fuel consumption that the plant owner
is paying for by a small amount. It's big impact is that often the
generator is only able to carry full load at .85 or .9 PF. Any lower
PF, and the output drops to keep from overheating the generator.

Matthew

 

[Repeating Rifle]

in article btqhh0$a8nlm$1@ID-103962.news.uni-berlin.de, Charles Perry at
pipesandtobacco@hotmail.com wrote on 1/10/04 7:56 PM:

Excuse me? "double the frequency"?

The direction of power transfer reverses evey half cycle.

Bill

 

[Repeating Rifle]

in article Qe6Mb.3528$1e.93@newsread2.news.pas.earthlink.net, Matthew
Beasley at beasleys@teleport.com wrote on 1/10/04 10:41 PM:

For single phase it travels back and forth at twice the line frequency.
For BALANCED three phase, there is no back and forth... the
instantanious power flow is always to the load... YET there is still VAR
flow.

What you say is true if you average over the phases. But for each phase
individually, there is a (reactive) power transfer at double frequency.

For real power, as for an incandescent lamp, power is also transferred at
double frequency.

 

[Matthew Beasley]

"Leon." <noemail@noemail.noemail.com> wrote in message
news:3ffff46c$1@news.rivernet.com.au...

Quit blaming power factor for blackouts.

The blackouts are caused by overload , for which the cure is load
shedding.

California had done a large amount of work at being able to shed load to
help in this regard. The 96 blackout occurred IN SPITE OF load
shedding.
If it was simply to much load, the frequency would have just dropped
until the system collapsed. It didn't.

The problem easily becomes unmanageable, as we saw last year, in
USA,
Italy and London, England. These are not the only major outages in
living memory, NE USA and SE England have had major incidents in
the
past, caused by relatively minor trigger events.

Sme of these were caused by sheer stupidity, as the '67(?) NE
blackout. Sure, mistakes propagate fast in unstable systems, but
I believe the larger problem here was the lack of local
generation. Once the critical path blew, there was no chance to
anything other than disaster.

The west coast of the USA had a similar blackout, same cause as the
recent
NE USA blackout.

Not quite the same, but many similarities.
Both involved lines sagging into trees, but that's where the main
similarity ends.
The official NE report isn't out. But it appears that leading up to the
NE blackout the system was already at the edge of stability. Due to
multiple reasons, the operators didn't know how much trouble they were
already in. Lines into trees pushed it over the edge. Lines opened
mainly due to out of step protection fairly quickly afterward.

In the '96 blackout, the initiating event was lines into trees. The
problem was exacerbated due to lack of VAR support in the northwest due
to some equipment failures and components out for maintenance. The
Oregon California tie lines were tripped on low voltage.. not from to
many watts, but to many VARs. The DC line was also dropped to half
capacity due to equipment failure resulting from the voltage
fluctuations. One the OCI was lost, the ties through Nevada and Utah
cascaded, much like the NE blackout.
There were problems associated with load shedding mentioned in the '96
report. They wouldn't have kept the system together, but they might
have kept more of the load in California on.

The 96 report is available at http://www.nerc.com/~filez/archives.html
look down to 11/96

 

[Keith R. Williams]

In article <GB6Mb.3635$1e.2645@newsread2.news.pas.earthlink.net>,
beasleys@teleport.com says...

"Keith R. Williams" <krw@attglobal.net> wrote in message .
-snipped-
If the generator is producing 14kVA and the load consuming 10kW,
exactly where is the rest going?


I think you may be confusing that the generator's rating may be measured
in kVA, but that doesn't mean how much POWER it is producing - or draws.

I don't think I'm confused at all (and I'm not talking about
"rating, per se". VARs cause wasted power in the distribution
system because , the *real* current that can't do "real" work.

If you measure the power input (mechanical) to the generator, it will
increase only a small amount when adding VAR's.

Try putting a transmission line inbetween the load and the
generator. You'll find that the reactive current is heating that
transmission line just as much as the real current.

In your example if the generator is producing 10kW at unity power
factor, lets say for discussion sake it takes 10.3kW to spin.
Now lets add a 9.8kVAR load to the output to get to 14kVA. (BTW, since
real and reactive currents are in quadniture, the VA is the root mean
sum.) In this case, the input to the generator might raise to 10.4kW
input. The small increase is due to the extra losses from the
increasing current - but the input power doesn't increase to 14kW.

The current being generated (squared) times the distribution (and
generator) resistance is the *real* power lost. If the phase
angle of the load is whacked, one must generate more current to
supply a constant power to the load. This increased current is
wasted in the distribution system. There is good reason why the
power company doesn't like crappy power factors.

Carrying VARs only increases the fuel consumption that the plant owner
is paying for by a small amount. It's big impact is that often the
generator is only able to carry full load at .85 or .9 PF. Any lower
PF, and the output drops to keep from overheating the generator.

....and just why is it that the generator can't carry the VARs?
Perhaps because it's *HEATING* (wasted power).

--
Keith

 

[daestrom]

"John Woodgate" <jmw@jmwa.demon.contraspam.yuk> wrote in message
news:SnQh$7BXKFAAFwqV@jmwa.demon.co.uk...

I read in sci.engr.electrical.compliance that Keith R. Williams
krw@attglobal.net> wrote (in <MPG.1a69fc3413cbc9d198aa9f@enews.newsguy.
com&gt about 'Electric Energy Economizer !!', on Sat, 10 Jan 2004:

Again, I understood that short lines were slightly capacitive
(can be a good thing since large loads tend to be inductive) and
that long lines were highly inductive (bad). ...or did I miss
the boat here?

No, you are right. But it's not a straightforward 'good/bad' thing. By
itself, the line inductance can't cause instability. It does cause a
voltage drop, and thus contributes to MVAr.

I don't completely agree with 'the line inductance cant cause instability'.

The line inductance is the major factor limiting the amount of *real* power
than can be transferred/carried through many lines. With the voltage
allowed at each end of a line determined by other considerations, the amount
of real power that a line carries is a function of voltage at each end,
phase difference and line impedance (classic four-terminal network problem).

If the line characteristic suddenly changes (such as a capacitor tripping
off or one of several parallel lines isolating), the max power that can be
transferred will drop sharply. If the power angle needed to carry the real
power exceeds 90 degrees, the system pulls out of sync and becomes very
unstable. And the line inductance is a major part of that equation.

Most people don't realize the line inductance for long lines is more of a
problem than the resistance. Lose the compensating capacitor banks, or the
voltage support equipment and the thing quickly becomes unstable.

daestrom

 

[Isaac]

On Sun, 11 Jan 2004 14:29:21 -0500, Keith R Williams <krw@attglobal.net> wrote:

Why don't you draw yourself an ac circuit with a resistor and an inductor in
it and calculate VA, watts and VAR. If you use ideal transmission
lines you will find no wasted power but plenty of imaginary power.

Isaac