ON5MJ wrote:
Hi there,

A friend of mine asks what happens on high voltage distribution lines
about
SWR at the distribution frequency (50/60 Hz). I'm stuck.

I understand that HV generators/transformers behave like voltage
sources and
not power sources. This means that conjugate impedances don't apply
here but
what about the existence of SWR on those kind of lines and the
possible
consequences on very long lines.

Anyone has an idea ?

I understand that excessive SWRs on the long-haul transmission lines
have brought entire grids down.

73 de ON5MJ - Jacques.

w3rv

On Sat, 11 Dec 2004 20:54:07 -0600, "Nick Kennedy"
wrote:

Impedance matching a generator to the load would be a bad idea
....
Each station generates in phase with the voltage that happens to be present at its location.

Hi Nick,

Phase is necessarily found in Impedance. As you allow that multiple
generators share a line, they are also across the load as a load if
they do not present the right phase.

As for the "bad idea" of matching, you are appealing to Edison's old
and deliberate misreading of Thevenin. Matching does NOT require a
resistance, this is a mis-read of conjugate matching that follows the
fact (in antennas). It does not drive the need (in power delivery).
Power stations only need perform a Z Match, not a Conjugate Match.
Any form of X is sufficient to accomplish the task and they do it far
simpler through field excitation control.

Back when Edison was battling Westinghouse/Tesla in the DC vs. AC
distribution system wars; Edison tried to confuse the issue with his
munged up version of Thevenin's Theorem insisting that his competitors
would have to burn up half their power to deliver half their power.
He thus claimed his DC system to be more "efficient." New York
bankers didn't know the difference between Thevenin or Copernicus. In
fact, it was Thevenin's proof that crippled Edison in the marketplace.
The only way to cut losses was to lift potentials into the
stratosphere. Edison also had a campaign trying to prove AC was
lethal, but DC was survivable (largely true). But with a low loss
system running in the KV and no way to convert it to residential use,
the writing (about lethality) was on the wall. AC, on the other hand,
could deal with that easily.

Edison's business/technical logic would have to wait for nearly 100
years to be resurrected for the ENRON bubble to coincide with New York
banker IQ phasing.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
"---the distance was far enough to observe the effects of SWR - in
exactly the same manner we observe them at HF or VHF etc."

Yes, that was anticipated.

The wavelength is a little less than 186,000 miles per second divided by
60 cycles per second, or 3100 miles per cycle approximately. The
reduction in wavelength is due to the velocity factor of the
transmission line. Construction determines the velocity factor.

The phase delay ib transmission over the actual distance between Hoover
Dam near Las Vegas and Los Angeles is only a few degrees. For example, a
60 Hz transmission line slectrical length of 310 miles would be 1/10
wavelength or about 36 degrees. Surely noticible but not crippling.

Now, many high-voltage transmission lines are transporting d-c. The rule
of thumb is that you need a kilovolt per mile of trangmission line
length to get efficiency. So, hundreds of miles require hundreds of KV.
At these voltages, the difference between rms and peak voltage becomes
important. RMS = DC. Now, conveersion from a-c to d-c and back again is
fairly easy and efficient. So, we have HV, DC power transmission. Tesla
had the first laugh. Now, maybe Edison has the last laugh after a
hundred years of development.

Best regards, Richard Harrison, KB5WZI

On Sun, 12 Dec 2004 04:21:55 GMT, Richard Clark
wrote:

On Sat, 11 Dec 2004 20:54:07 -0600, "Nick Kennedy"
wrote:

Impedance matching a generator to the load would be a bad idea
...
Each station generates in phase with the voltage that happens to be present at its location.

Hi Nick,

Phase is necessarily found in Impedance. As you allow that multiple
generators share a line, they are also across the load as a load if
they do not present the right phase.

As for the "bad idea" of matching, you are appealing to Edison's old
and deliberate misreading of Thevenin. Matching does NOT require a
resistance, this is a mis-read of conjugate matching that follows the
fact (in antennas). It does not drive the need (in power delivery).
Power stations only need perform a Z Match, not a Conjugate Match.
Any form of X is sufficient to accomplish the task and they do it far
simpler through field excitation control.

Back when Edison was battling Westinghouse/Tesla in the DC vs. AC
distribution system wars; Edison tried to confuse the issue with his
munged up version of Thevenin's Theorem insisting that his competitors
would have to burn up half their power to deliver half their power.
He thus claimed his DC system to be more "efficient." New York
bankers didn't know the difference between Thevenin or Copernicus. In
fact, it was Thevenin's proof that crippled Edison in the marketplace.
The only way to cut losses was to lift potentials into the
stratosphere. Edison also had a campaign trying to prove AC was
lethal, but DC was survivable (largely true). But with a low loss
system running in the KV and no way to convert it to residential use,
the writing (about lethality) was on the wall. AC, on the other hand,
could deal with that easily.

Edison's business/technical logic would have to wait for nearly 100
years to be resurrected for the ENRON bubble to coincide with New York
banker IQ phasing.

73's
Richard Clark, KB7QHC

I had never thought of the power industry having "matching" problems
until I picked up a magazine ( the New Yorker of all things ) in a
doctor's office years ago and there was a story about fluctuating
SWR's etc and what the power company did to compensate. Unfortunately
I got called in to see the Doc before I could finish the article.
Thanks for sharing some light on the subject.

Gary K8IQ

The VERY first thing to remember about power transmission is that generators
must NOT be conjugate matched to loads.

We CANNOT have half the power dissipated in the generator!
===
Reg.

The following details will give you some idea of what you are waffling
about.

A 2-WIRE, WIDE-SPACED, POWER TRANSMISSION LINE.

At a frequency of 60 Hz

Length = 100 miles.
Wire diameter = 1 inch.
Wire spacing = 10 feet.

Nominal RF Zo = 650 ohms.
Actual Zo at 60 Hz = 650 ohms.
Angle of Zo = -2.6 degrees.

Velocity factor = 0.99
1/4-wavelength = 767 miles.
Resonant Q at 60 Hz = 11

Inductance = 3.54 milli-henrys per mile.
Capacitance = 0.00836 micro-farads per mile.

FOR A LOAD OF 500 OHMS -

Input impedance = 520 + j*51
Line loss = 0.1 dB.
Power Loss in line = An economical 2.3 percent.

Reflection coefficient = 0.133
Angle of reflection coefficient = 170 degrees.
VSWR = 10.5

Economics rules the roost at power frequencies.

The normal transmission voltage on such a line is measured in terms of
100,000 volts.

Note that, with a resonant Q of 11, should an open-circuit fault occur at a
distance of 1/4-wavelength the voltage at the fault can rise to a million
volts or more. Electrical power engineers have far worse problems than mere
radio engineers have on the popular 40m band. They too are concerned with
reflection coefficients and SWR. ;o)

But the technicalities were all exactly sorted out in the Victorian age by
the young, self-educated, recluse and hard-of-hearing genius like
eethoven - Oliver Heaviside who was derided by the old-wives and silly guru
university professors of his age.

The above technical details, and more, can be computed and studied, from
power frequencies up to UHF, by downloading, practical, easy-to-use programs
RJELINE2 or 3 from the following website. Download in a few seconds, not
zipped-up, and run immediately under common-or-garden DOS/Windows.
----
.................................................. ..........
Regards from Reg, G4FGQ
For Free Radio Design Software go to
http://www.btinternet.com/~g4fgq.regp
.................................................. ..........

Jaques, ON5MJ wrote:
"I am not aware of this but it makes sense that efficiency of AC/DC +
DC/AC conversion must be higher than the use of pure AC transmission."

Yes. It makes no sense to lose more in conversion than on the
transmission line.

The problem with extreme high voltage power transmission is insulator
flashover and corona. Alternating current and voltage have effective
values which are their peak values divided by the square root of two.

This means that peak volts times peak amps divided by two is the same
average power as rms volts times rms amps.

The same average power transmission requires peak values 1.414 times the
rms value, effective value, or d-c value.

In d-c tramnsmission, peak and effective values are the same 100% of the
tiime, so the required d-c voltage is only 0.707 times the a-c voltage
peak for the same power transmission.

Jaques also wrote:
"By definition loads vary all of the time but voltage must not vary
accordingly."

Use of d-c eliminates reactance as a cause of voltage variation. It also
eliminates "skin effect" as an impefance so that the entire
cross-section of the line is used.

The case for extremely high voltage d-c transmission is pretty good.

Best regards, Richard Harrison, KB5WZI

Note that, with a resonant Q of 11, should an open-circuit fault occur
at
a distance of 1/4-wavelength the voltage at the fault can rise to a
million
volts or more.

How do you compute this ?

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

It is computed in exactly the same way at 60 Hz as it is at 60 MHz.

A 1/4-wavelength line behaves as any other tuned circuit. The voltage at
the open end rises to Q times the voltage applied at the input end.

With a short-circuit line the current at the short circuit rises to Q times
the current at the input end.

The Q of a tuned circuit is the reactance of the inductance divided by the
wire resistance. Q = Omega*L / R.

The Q of a transmission line is the reactance of the line's overall
inductance divided by the line's overall resistance. And again, Q = Omega*L
/ R.

Since both L and R are directly proportional to line length, for a given
line, Q is a constant and is independent of length.

Omega = 2*Pi*Freq.

There are other more complcated ways of calculating the resonant rise in
open-circuit voltage from the line's transmission and propagation
properties. But they all give the same answer of course. Such calculations
provide a means of checking for program software bugs.

The hardest part of the exercise is calculating the inductance and
resistance from the line's physical dimensions and operating frequency.
Which are needed anyway to calculate all the many other output quantities
from the program. Q is just a spin-off.
----
Reg, G4FGQ

Richard Clark wrote:


Hi Nick,

Phase is necessarily found in Impedance. As you allow that multiple
generators share a line, they are also across the load as a load if
they do not present the right phase.

As for the "bad idea" of matching, you are appealing to Edison's old
and deliberate misreading of Thevenin. Matching does NOT require a
resistance, this is a mis-read of conjugate matching that follows the
fact (in antennas). It does not drive the need (in power delivery).
Power stations only need perform a Z Match, not a Conjugate Match.
Any form of X is sufficient to accomplish the task and they do it far
simpler through field excitation control.

Hello Richard,

Even with your redefined version of matching, individual generating
stations don't explicitly match to the load reactance or (certainly
not) load resistance. A little generation 101. (This goes a little
beyond the scope of discussion, but I throw it out because I think it's
interesting.):

Generators operate in two modes: Isochronous (single generator
supplying local bus) and parallel (multiple generators in parallel;
connected to a grid).
The operator has two main controls: Steam (or whatever) flow to the
turbine or other driver, and current to the field.

In the isochronous mode, varying steam flow causes the speed of the
generator to change. It also varies the amout of power delivered. In
the parallel mode, the generator can't measurably push the speed of the
grid, so increasing steam flow only increases the electrical power
output.

In the isochronous mode, varying field current changes the terminal
voltage of the generator. In parallel mode, varying field current
can't significantly change grid voltage. But it does change the
reactive power output (MVAR or kVAR) of the generator, as you said.

The normal mode of operation for a large generator is in parallel with
the grid, so the operator is using steam (diesel, water, hamsters,
etc.) to regulate real power output and field current to regulate
reactive power output.

Now some anecdotal stuff about how generators are operated. The system
dispatcher requests individual generators to adjust their power and
VARs to match load. This isn't impedance matching, it's simply
supplying the demand. In the case of VARs, the goal is both to supply
the demand and to equalize voltage across the system, not to cause any
kind of mathematical match between the generator's internal X and the
system's X. Oh yeah, I said earlier that individual generators don't
appreciably affect grid voltage. That's true, but locally they do have
an effect, like tent poles in a big canvas. So the local stations are
both supplying their share of the total reactive load and propping up
voltage in their area. (The operator increases VAR output by taking
his excitation switch to the "raise voltage" position.)

Anyway, I digressed from my anecdotal stuff. At my plant, the
generator puts out 1050 MW 24/7, but MVAR may vary between 0 (or
slightly negative) and 200 MVAR. So we're not matching to any specific
impedance, but supplying load and maintaining voltage.

A story transmission guys like to tell is how they may use open ended
transmission lines as a kind of capacitor bank. Say there's a line 100
miles long from my plant to somewhere that's not needed to carry load.
The system controller might connect it at my plant's end but leave the
breakers open at the far end. A line has both capacitive and inductive
reactance of course, but when unloaded, the capacitive dominates. So
the trick of the trade is to use it to supply reactive MVARs. The
point of the story in this context is that the controller isn't
concerned about SWR on this extremely mismatched line.

Another possibly relevant story. We connect our emergency diesel
generator to the grid for testing and load it to about 3000 kW and
typically from 0 to 100 kVAR. But to fully test the excitation system,
the kVAR is at some point raised to 1400. The point being that the
generator can be operated anywhere within its rating, with no need to
match to any mysterious impedances out there in the world. Makes sense
when you think about it. Who would want a generator that was
constrained to operate at some fixed ratio of real to reactive power?
73--Nick, WA5BDU


73's
Richard Clark, KB7QHC

Thanks very much for the interesting and informative tutorial from
someone in the industry. I have one question:

Nick wrote:
. . .
Another possibly relevant story. We connect our emergency diesel
generator to the grid for testing and load it to about 3000 kW and
typically from 0 to 100 kVAR. But to fully test the excitation system,
the kVAR is at some point raised to 1400. . .

If your customers' loads were, for the sake of argument, purely
resistive as seen at your power plant output, then the voltage and
current would be in phase at that point. But in order to make your
generator produce "reactive power", the voltage and current have to be
forced out of phase at the generator. How is this resolved? Is that
reactive power "delivered" to (actually swapped back and forth between)
other generators in the system -- that is, do the other generators in
the system shift their own phase angles so that the V and I can be at
some angle other than zero at your generator output (and, necessarily,
also at the outputs at other generators in the system) yet in phase at
your customers' loads? Or do you have some local bank of reactance that
you can switch in to feed the "reactive power" back and forth to when
you run this test?

Roy Lewallen, W7EL