Sciencemadness Discussion Board - global energy grid

global energy grid

xxxxx - 28-8-2007 at 10:15

for some types of electrical generation systems such as wind which is usually more powerful during the day and solar which is only useful during the day i was wondering if a global energy grid could be used so that a solar cell farm in abu dabi could power a vcr in quito. what i guess i am asking is how far it is economical to transport electricity and why.

Sauron - 28-8-2007 at 10:49

You have a brother in law in the high tension cable business?

Or do you plan to resuscitate Tesla's vision of wireless power transmission?

Last I looked there is an ocean between the UAE and Ecuador. Maybe therefore a bad choice of example.

12AX7 - 28-8-2007 at 12:11

Electricity goes about 500 miles economically. DC high voltage lines (equipped with converters at each end, feats of high voltage engineering that they are) are best for long distance, high capacity transmission -- especially between different grids, like 50 and 60Hz.

It's much more economical to store the energy locally, in batteries, water pumped up and down to and from an elevated holding pond, compressed air in a salt dome, etc.

Tim

Rosco Bodine - 28-8-2007 at 14:33

Superconductors could be used advantageously
for long distance power transmission ...but that technology is still in development .

DC long distance transmission would eliminate LC losses .

DC could be converted back to local
distribution AC by brush commutation inverters
or possibly by some industrial duty solid state inverters .

Blind Angel - 28-8-2007 at 18:42

Here most of the electricity is produced by Superdam which are about 900 miles up north, some farther, and we still are able to sell electricity to our friendly southern neighboors. I think that the high voltage line are around 735kV if I remember well, and it's still very economical I think. 500miles must be for lower production, so the point is to get the voltage the highest as possible for long distance travel. Using a principle like the internet (node and very big pipe in between), I don't think it would be impossible.

franklyn - 28-8-2007 at 19:39

@ xxxxx
While not global , north american electrical utilites already have their respective
power distribution infrastructure interconnected. The reason why there have
been multistate blackouts in the northeast when capacity cannot meet a surge
in demand causing a cascade of outages.

@ 12AX7
No no no , there is no DC transmission anymore anywhere. The legacy Edison
DC grid in New York city shut down forever in the early 1970's.
Europe uses 50 hz in the US its 60hz.

@ Rosco Bodine
LC also termed reactive power , is lossless , thats it's definition.
See => http://www.infoplease.com/ce6/sci/A0930671.html
http://en.wikipedia.org/wiki/AC_power
Much more complex explaination =>
http://www.eirgrid.com/EirGridPortal/uploads/Regulation%20an...
The only loss from transmission comes from heating the conductor a direct
result of its resistence value in ohms sometimes called DCR ( direct current
resistence ) expressed by Joule's law.
See => http://en.wikipedia.org/wiki/Joule%27s_laws
Given the relation of power consumed according to the equation Watts = I ² R
it makes sense to transmit at the highest attainable voltage in order to minimize
the current value ( I ). This is what is done. Superconduction requires nothing
different from what is in use now , except for the conductor cable.
See => http://dspace.lib.utexas.edu/bitstream/2152/235/1/hamwk032.p...

.

12AX7 - 28-8-2007 at 19:45

LOL

Rosco is stuck in the 19th century, figures. Motor-generator sets haven't been used for ages.

DC power transmission is relatively new, but it does exist, and moves substantial amounts of power. Shame on you for not consulting even Wikipedia first!

Superconductors for power transmission may never become economically viable. If a room-temperature superconductor can even be made, it will still require cooling, because the critical field at the critical temperature is nil. Current flow means magnetic field, so critical field strength is your maximum current density. Maybe in Alaska and other arctic countries, but not likely to ever work in any temperate region.

Tim

[Edited on 8-28-2007 by 12AX7]

not_important - 28-8-2007 at 19:46

The Pacific DC Intertie is roughly 1300 km long and has a capacity of 3 GW at 500 kV DC. The Inga-Shaba in central Africa is 1700 km long, again a 500 kV DC line.

The economic limit at that voltage may be 2 to 3 times as long.

Twospoons - 28-8-2007 at 19:52

@franklyn: Wrong, there is still DC transmission. DC is used crossing the Cook Strait in New Zealand, shipping power from the hydro stations in the South Island, to the bulk of consumers in the North Island. It gets converted back to AC for transmisson over land. I've seen the inverter station : those were the biggest thyristors I've ever seen!

Also, additional losses in AC distribution come from induced current in the surroundings : farmers have been forced to move fences for this reason (by the electricity companies - a smart farmer can used the induced current to power things like pumps

)

[Edited on 29-8-2007 by Twospoons]

[Edited on 29-8-2007 by Twospoons]

Rosco Bodine - 28-8-2007 at 20:55

Superconductors are already used in limited power applications .

AC on long transmission lines has in addition to the usual
IR losses , AC reactance losses due to the inductance and capacitance of the long transmission lines , as well as EM radiational loss . There is also ionization and leakage
through the air itself to the ground , which you can feel standing beneath the lines ....of course that will apply with DC as well .

And so far as a high current inverter being commutated mechanically or by solid state .....both work just fine , no matter the century . It's like they still make filament light bulbs for many uses ....even though some are not so bright .

[Edited on 28-8-2007 by Rosco Bodine]

franklyn - 28-8-2007 at 21:26

Very interesting I never new about this. http://en.wikipedia.org/wiki/Pacific_DC_Intertie

@ 12AX7
They're called rotary inverters not motor generators. The point is that it's
lower conversion efficiency precludes it's use. A switching converter will
acheive better efficiency. http://www.fact-sheets.com/science-nature/energy/inverters
Mindful that the rotary device is not also superconducting.
I don't know about " haven't been used for ages " see inverters here =>
http://hnsa.org/doc/pdf/aviationpower.pdf
Because of the pure sinewave output and compact size they remain the
method of choice for aircraft.

@ Rosco Bodine
I'm still wondering what you mean by reactance losses , inductance and
capacitance are lossless unless there exists an ohmic component. The
resistive loading from the compensating devices and switchgear are the
only actual sources of incidental losses. Transformers are only about 96%
efficient. As for 60 Hz radiation , a transmission line makes a poor antenna.
Inductive coupling with metal objects and enviornmental ground conduction
present more of a concern. A transmission line can be viewed as a low pass
filter having capacitive coupling to ground that is very small and only
significant given the size of the voltage and scale of the line some hundreds
of miles long

.

[Edited on 29-8-2007 by franklyn]

Rosco Bodine - 28-8-2007 at 21:42

Inductive Reactance (in ohms)
X sub L = 2 pi F L
where F is frequency in Hertz and L is inductance in henrys

Capacitive Reactance (in ohms)
X sub C = reciprocal of 2 pi F C
where F is frequency in Hertz and C is capacitance in farads

You would be surprised what the total losses actually are
on 60 Hz long transmission lines . Exactly why power plants are built near where they are needed when possible .

I never said motor - generator , I said brush commutation ...
a whole different thing where a very small DC motor having precision speed control simply turns a commutator for brushes changing the DC at thousands of Amps to square wave 60 Hz for the power transformers which largely smooth it to a sinewave output and step it down to distribution voltage levels at hundreds of thousands of Amperes .

[Edited on 29-8-2007 by Rosco Bodine]

Twospoons - 28-8-2007 at 21:53

Yes, but franklyn's point is that inductance and capacitance do not dissipate power, therefore contribute no loss. Don't confuse impedance with resistance. Both may be measured in ohms, but they are NOT the same.

Rosco Bodine - 28-8-2007 at 22:08

Don't dissipate power ?????

Try feeding power through an inductor which has a lot of reactance to the AC frequency being used and explain the heating .

If it walks like a duck and quacks like a duck .....

franklyn - 28-8-2007 at 22:27

Okay I think I understand now you refer to inductive eddy current
which is not coupled to an output. This is different duck though.
Mallard I think

Rosco Bodine - 28-8-2007 at 22:37

No ....Reactance X can be directly substituted in Ohms Law
calculations for the usual ohmic resistance R ,
and the effect on power dissipation holds as well ...
but instead of an IR loss it is an IX loss ?

Or is IX just a roman numeral ?

Don't stop now , I'm having great fun with this

tumadre - 29-8-2007 at 04:16

Stop!!!

An increase in IX causes an increase in IR heat loss, due to the power factor.

the reason DC is used is because at any given time the sum of the current in a 3 phase line is zero.

so use one line at ~110% the size of one of the 3 phase lines, use earth as a return, and now the copper/aluminum loss is one third.

Or use a 2 line system, and the transmission is 2/3 the loss of the 3-phase line

12AX7 - 29-8-2007 at 06:37

Rosco, you again demonstrate your ignorance of the subject. Would you PLEASE just step away from electronics and concentrate on the chemistry you ARE good at?

Reactance is completely and utterly lossless. The resistive component of your AC choke is what's dissipating power, and I will add, is dissipating a whole lot less power than the VAr's passing through the system. The same goes for any reasonable capacitor.

For instance, the capacitor bank on my induction heater hardly warms up (P < 10W, roughly) despite having around 100V at 280A = 28kVAr passing through it. I might like you to explain how I get V*A = 28k when this device is powered from an outlet delivering only 1kW, but then, you would be at a loss to do so without realizing that reactance does not in fact dissipate any power.

Power lines have more capacitance than inductance, with the result that reactors (i.e., inductors) are provided to terminate long runs, correcting both voltage (the off-tuned series resonance recovers apparent attenuation) and power factor (as the phase shift (between current and voltage) is returned to zero). Evidently these units are capable of producing a substantial discharge when attempting to break the circuit... These and other maladies are avoided with DC transmission, which is more expensive to install and hence is not used much.

Tim

unionised - 29-8-2007 at 10:31

Just for the record there has been a HV DC link between the UK and France for ages. According to this site it was set up in 1955
http://www.modernpowersystems.com/storyprint.asp?sc=2025560
So it's hardly that modern. The 2 countries use different frequencies so and AC link would have been impossible but for very high voltages DC has the advantage of needing less insulation for a given power (because there isn't a peak voltage to worrry about).

Oh, BTW, Rosco, they're right about the losses- in principle an inductor or capacitor doesn't dissipate any power, it stores it then gives it back in a confusing manner. The brush commutator would generate more radio interference etc than anyone would want, right up till the brushes burned out from the inductive arcing. It would be horribly inefficient to turn DC into AC that way- even a motor/generator rig might work better.

Rosco Bodine - 29-8-2007 at 11:59

Soooooo .... what do you suppose is the net effect of the
opposition to AC current flow resulting from the algebraic sum of reactance and resistance , along a simple straight
conductor , but a less efficiently operating conductor for
a given diameter as compared with DC current through the same conductor ? If you must raise the source voltage to push the same RMS value of AC current , because of impedance , which is derived from and greater than resistance and reactance alone ....then
you necessarily get greater IR losses as compared with DC even if the the reactances return their energy 100% on changing current with AC . So the net effect *is* a dissipation increase for having to overcome the reactance , even if the reactance itself conserves energy .
The purely resistive component is there all the time ,
but reactance combined with it , results in greater
dissipation by it at the same power transmission level
for AC as compared with DC . So strictly speaking , the effect of greater dissipation is there , caused by the
reactance however indirectly .

If I missed something there ...say where

Oh pleeease 12AX7 , don't drag out the phasors .

BTW , aren't the reactors on power grids actually used for transient surge protection ? ( Meaning they would be useful on a DC grid as well )

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

http://www.geocities.com/SiliconValley/2072/elecrri.htm

quote below is from

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

Quote:

In an alternating current transmission line, the inductance and capacitance of the line conductors can be significant. The currents that flow in these components of transmission line impedance constitute reactive power, which transmits no energy to the load. Reactive current flow causes extra losses in the transmission circuit. The ratio of real power (transmitted to the load) to apparent power is the power factor. As reactive current increases, the reactive power increases and the power factor decreases. For systems with low power factors, losses are higher than for systems with high power factors. Utilities add capacitor banks and other components throughout the system — such as phase-shifting transformers, static VAR compensators, and flexible AC transmission systems (FACTS) — to control reactive power flow for reduction of losses and stabilization of system voltage.

[Edited on 29-8-2007 by Rosco Bodine]

tumadre - 29-8-2007 at 12:54

The flame in this thread is bigger than any inductive kick I've seen.

Obviously the areal installation of HVAC works.

But add a foot of rubber, bury it and the capacitance skyrockets.

12AX7 - 29-8-2007 at 13:14

Quote:

Originally posted by Rosco Bodine
Soooooo .... what do you suppose is the net effect of the
opposition to AC current flow resulting from the algebraic sum of reactance and resistance ,

The sum is NOT algebraic, it is vectorial. For an inductor and resistor in series, Z = sqrt(R^2 + X_L^2).

Quote:

So the net effect *is* a dissipation increase for having to overcome the reactance , even if the reactance itself conserves energy .
The purely resistive component is there all the time ,
but reactance combined with it , results in greater
dissipation by it at the same power transmission level
for AC as compared with DC . So strictly speaking , the effect of greater dissipation is there , caused by the
reactance however indirectly .

If you in fact meant the currents drawn by reactive loading, then you should've said so in the first place!

Tim

Rosco Bodine - 29-8-2007 at 13:48

Vectorial indeed , it is the good old Pythagoraean theorem for the hypotenuse value . Pardon my algebra ..er I mean trig .

One thing I noticed is a good example of "reactive power"
as a loop current in a PSC motor , where I had more than one AC current meter monitoring the AC supply current and another meter in the capacitively coupled winding ,
as well as the run winding , all while varying the AC voltage and testing different capacitor values .....
There were some test setups where there was twice
as much reactive power circulating as a loop current ,
as there was AC current coming into the whole setup !

I tapped my meters and checked the range settings because it seemed so implausible at first glance , until I figured out this sort of "resonance" effect was good old reactive power ..which was more bad at this point than good ...and that circulating current definitely *exceeded*
the input current ...and it was making the motor warm ,
as even "reactive power" has to travel through the ordinary conductors which have resistance . So dissipation is "grandfathered in" by reactive power ....
simply because it is traveling through the same conductor , its phase shift notwithstanding ....it doesn't get a free pass through the conductors simply because
it is reactive power but pays its loss to IR just like any other current .

[Edited on 29-8-2007 by Rosco Bodine]

franklyn - 29-8-2007 at 20:30

Not being a power engineer the nuances of electric transmission are not immediately
self evident to me. It seems we're talking about different sides of the same coin.
You see a buffalo and I see an indian.
Reactance causes more current to be present than what actually corresponds to the
power delivered. It is superfluous and not consumed but is parasitized all the same by
line resistence thereby contributing to loss.
It's a subtle distinction but important in the economics of electric transmission.

From the text Basic Electricity by Van Valkenburgh , 1978 , OCLC: 14077988
Page 4-32
" In AC series circuits consisting of impedance and capacitance , with only negligible
resistence , the impedance is due only to inductive and capacitive reactance. Since
inductive and capacitive reactances act in opposite directions , the total effect of the
two is equal to the difference between them. / subtracting the smaller from the larger /
The circuit will then act as a capacitive or inductive reactance ( depending on which
is larger ) "
and on page 4-47
" inductance and capacitance are added to power transmission lines to correct the
power factor to unity to obtain the most efficient power transfer."

So then when XC = XL is made to happen , impedance becomes zero , only resistence
is present and the current is at the maximum possible value specified for that
transmission conductor.
Now one can see the rational of resorting to DC for long runs of transmission line , it also
eliminates the ancilliary equipment required for line conditioning ( impedance matching )
It's yet another consideration of the cost accounting when conceiving a project.

The following is excerpted from A Textbook of Electrical Technology : in S.I. system of units
by B L Theraja; A K Theraja , 21 edition 1988 , OCLC: 46340264
( An electrical engineering text of indian publication )

Capter /section 35-1 Transmission and distribution of DC power
" these days all production of power is as AC power and nearly all DC power is obtained
from large AC power systems by using converting machinery like synchronous or rotary
convertors , soild state convertors and motor generator sets etc."

Capter /section 36-1 General layout of the system
" Now a days generation and transmission is almost exclusively three phase."

" The transmission voltage is , to a very large extent , determined by economic
considerations. High voltage transmission / only / requires conductors of smaller cross
section which results in economy of copper or aluminum. But at the same time cost
of insulating the line and other expenses are increased. Hence , the economical voltage
of transmission is that for which the saving in copper or aluminum is not offset by the
increased cost of insulating the line , by the increased size of transmission line
structures and by the increased size of generating stations and substations. A rough
basis of determining the most economical transmission voltage is to use - 650 volt- per
kilometer of transmission line. For example , if transmission line is 200 kilometers , then
the most economical transmission voltge will be 200 X 650 or 132 kilovolts."

Capter /section 36-3 Systems of AC distribution
" As mentioned earlier , AC power transmission is always at high voltage and mostly
by 3 phase system. The use of single phase system is limited to single phase electric
railways. Single phase power transmission is used only for short distances and for
relatively low voltages. / / 3 phase power transmission requires less copper than either
single phase or 2 phase power transmission."

Capter /section 35-16 Comparison of 2 wire and 3wire Distribution System
" a 3 wire system / 3 phase / requires only 5/16 ( or 31.25 % ) as much copper as a
2 wire system. "

U P D A T E

Given that E = I Z , E being the station generated voltage which remains fixed , I is
the line current , Z is the total impedance of reactance including the line resistence.
When reactance ( Z ) increases what becomes of of the current ( I ) ? Being inversely
proportional to magnitude changes in Z , it will diminish of course. Increased reactance
then not only reduces the overall current , it also increasingly becomes a greater
portion of the already diminished current. What can be termed as the bandwidth of
a low resistence power line is charateristically narrow and consequently offers little
buffering to line fluctuations. This can be critical at times of peak demand , even if
adequate power generation exists it will be unavailable if the line is choked by
reactance reducing what can be provided.
DC transmitted current obviates this troubling possibility being devoid of reactance
it eliminates the cause of current fluctuation making it more reliable.

While I did say before there is no DC transmission ( I was uninformed about this )
there is no distribution of DC as such as this is converted to AC at the receiving
station. I am aware that end users such as aluminum smelters and electrolytic process
industries do require DC and these typically provide their own conversion infrastructure.
.

[Edited on 30-8-2007 by franklyn]

Twospoons - 29-8-2007 at 20:55

This is why factories invest in power factor correction - so that they present as close to a pure resistive load to the grid as possible. Electricity suppliers don't like supplying to reactive loads, and bill accordingly.

Rosco Bodine - 29-8-2007 at 22:08

It's not just a concern for power grids .....

Actually the same considerations are *very* important
in antenna circuits and their connecting cables being matched to the amplifier circuits , to minimize signal losses
receiving or transmitting .

And dittos for audio circuits . The quality of the audio
and the power transfer are better when there is an impedance matching front to back .

Same for data cables and phone lines .

It's all about having a phat pipe

12AX7 - 30-8-2007 at 08:46

Audio circuits are too short to give a damn about impedance. Telecom does.

Tim

Rosco Bodine - 30-8-2007 at 09:25

Electrovoice matching transformers must have been an anomaly huh ? Dittos for equal length stereo speaker wires , and "impedance spec" for the speaker coils
and crossover networks ....all that impedance stuff
just a total waste of time for audio . L-pads for balance control too I guess , just total nonsense .

12AX7 - 30-8-2007 at 10:46

Those are for maximizing power transfer, where either the power is low enough to matter (mics) or high enough that loss would be akward. I meant (forgetting to specify) in regards to transmission line theory, especially in line-level signals.

BTW, with modern amplifiers, does it really matter than a speaker is 8 ohms? Or 4, or 2, or 16, or 600 ohms? What would an actual impedance match be?

Tim

DerAlte - 30-8-2007 at 11:44

Matching load impedance to the conjugate of the load impedance gives you the maximum power transfer possible. In your audio amp cases this is not the case, such as your push-pull 6L6's. Here you are matching to the load-line impedance, which is determined by safe tube operating area. This is not a true impedance match.

The semiconductor amps nearly always use a emitter follow output and heavy NFB. The output impedance is near zero so you don't try to match that. Put an 8-ohm speaker on it and you get X watts, say. A 4-ohm will give you 2X. A 0.1 ohm speaker - weel, you'll probably blow the transistors.

The same reasoning applies to a power grid. Here you also have to take account of line losses. This design is a matter of engineering economics. Very complex.

In communication line systems, matching is not optional, it's essential to avoid echoes leading the pulse distortion or bad frequency respose. Here you have to match the source to the transmission medium impedance (possibly free space) at one end ( for maximum power transfer) and match the load to the medium ( for maximum power transfer and minimum reflection) The same priciple applies in optics to reducing reflection and increasing transmission by coating lenses and mirrors. Same mathematics, same theory as transmission lines. Learn the maths behind it and you'll be able to apply it generally...

Regards,

DerAlte

unionised - 30-8-2007 at 12:28

IIRC the original contention was "AC on long transmission lines has in addition to the usual
IR losses , AC reactance losses due to the inductance and capacitance of the long transmission lines , as well as EM radiational loss . There is also ionization and leakage"

The losses are not due to reactance. If the lines had zero resistance the losses would be near zero.
There are losses by inductive coupling, but these are small.
There are losses due to dissipation in the effective capacitance of the insulation, again these are small particularly where the insulation is mainly air.
The loss due to the fact that the power factor is less than 1 is easy to deal with; thats what PFC caps are for.
These equations
"
Inductive Reactance (in ohms)
X sub L = 2 pi F L
where F is frequency in Hertz and L is inductance in henrys

Capacitive Reactance (in ohms)
X sub C = reciprocal of 2 pi F C
where F is frequency in Hertz and C is capacitance in farads"
don't have any real bearing on these losses.

The output of modern amplifiers is (as has been said) near zero thanks to negative feedback. Connecting a matched load would cripple the amplifier. The "impedance" given for a loudspeaker is just a means to express the voltage required to drive the speaker to a given power. It is usually far from constant across the audio spectrum.

Twospoons - 30-8-2007 at 13:55

Matching an audio amp to loudspeaker impedance is rather non-sensical anyway. Loudspeaker rated impedance is only a nominal value - the actual impedance is complex (has real and imaginary components) and varies considerably across the audio band.