Is there some low frequency radiation or hysterisis loss or what?
Bret Cahill
Is there some low frequency radiation or hysterisis loss or what?
Bret Cahill
On a sunny day (Tue, 22 Apr 2008 12:13:06 -0700 (PDT)) it happened Bret Cahill wrote in :
It all depends, AC has inductive losses, and capacitive losses between lines. DC can be efficient at very high voltages, where the current is low, as losses in the wires are i^2.R, but requires big solid state AC/DC and DC/AC converters at each side.
Bret Cahill wrote: :10X More Efficient
Cite your sources for 920% efficiency.
Cite where you think he said that. It's not in this thread.
Where do you get your inital incorrect idea from ?
Graham
"Dope Bowey"
"JeffM"
** What is 10 times more efficient than 92 % ?? 920% ?The OP's question is an absurd troll.
Like you.
...... Phil
Are modern "scientists" really this dense? Ten times efficient isn't
920%! What you must have just graduated from a "modern" high school? Dig. If standard transmission is 92% efficient, then that means there is 8% of the energy lost. TWICE as efficient would only have 4% of the energy lost or would be 96%. The interested student can take it from there...Since 10X would be more than 99% efficient I am somewhat skeptical that even modern converters can produce so little loss.
Gee, thanks phil.
Going 10X further with the same 8% loss?
One tenth as lossy for the same distance?
And who mentioned 8%?
Bret Cahill
Europeans claim they can run a DC power line from solar thermal fields in the Sahara thousands of kms to N. Europe.
Bret Cahill
If you mean long-distance transmission lines, DC can be run at higher average voltages (less corona losses, relatively) and has no skin loss or inductive coupling to the world. I don't know about 10:1.
DC systems do need inverters and rectifiers on the ends, which have losses.
John
The losses in electrical transmission have increased from about 5% to 9.5% in recent years, so efficiency is 90.5% to 95%.
According to
I would assume that a tenfold increase in efficiency would be a tenfold reduction in losses, so the present 7.2% would be 0.72%, for 99.3% efficiency. That is very likely impossible outside of laboratory conditions, and would probably require impractical amounts of copper, or cryogenic means to achieve superconductivity, which would itself require power and reduce overall efficiency.
But in direct answer to your question, there is a substantial amount of radiated energy loss with AC that does not occur with DC. But open air DC transmission lines do have losses in the form of a flow of ions, and both AC and DC have losses due to corona.
Paul
No they don't.
Graham
"Benj"
** WHAT COMPLETE BOLLOCKS !!...... Phil
"Bret Cahill = TROLL "
Going 10X further with the same 8% loss?
One tenth as lossy for the same distance?
** No way either of those things is expressed the f****it way you came up with......... Phil
That's 8% INefficiency
Your 4% would be HALF as INefficient; not at all the same thing.
10 x 92% is still 920%.
Tom Davidson Richmond, VA
"Paul E. Schoen"
** Transmission * loss percentage * increases with the amount of load - plus the overall energy lost depends on load levels and time.The operating efficiency percentage of a transmission link cannot be known stated unless the load is too.
** Total BOLLOCKS !** Why ?????
It makes nonsense of the English language.
One can increase the efficiency percentage OR reduce the loss percentage by a number.
When the efficiency percentage is low, you use the former and when it is high, the latter.
..... Phil
Of course. There has been an increase in demand without a corresponding increase in the infrastructure, so the power grid is more heavily loaded. Being largely dependent on I^2R losses, a 50% increase in demand (current) results in a doubling of losses.
A tenfold increase means ten times better. 99% is twice as good as 98%, because it costs me half as much in losses. So it is a 100% increase in efficiency, or a 50% reduction of losses.
A similar situation exists with accuracy, but it is reversed in meaning. A
1% instrument is really 99% accurate, so an instrument that is twice as good is 0.5%. A 100% increase in accuracy, but the accuracy figure is really a statement of inaccuracy.So, when one tries to develop a more accurate instrument, the goal would more reasonably be to make it like 10% more accurate or 50% more accurate, which are easy to comprehend. To say one wanted to increase the accuracy of an instrument by 1% would be reasonable if one knew it to be 2%, but would be impossible if it were already 1% or better.
The same with efficiency, but it is better understood in terms of reducing losses.
But you are right that it does put strains on English language usage.
Paul
"Paul E. Schoen"
** Try learning to bloody READ - you imbecile !!!!The PERCENTAGE loss of the transmission line is a *varying quantity* with load.
The numbers you quoted are NOT measures of the energy efficiency of the transmission lines.
** Asinine, idiotic s**te.A " tenfold increase " can only mean a 10 times increase in a quantity.
LEARN to READ !!
** So must be stated as 50% reduction in losses.( snip more reams of mind numbing, f****it drivel )
** It is purest fuckwittery and a misuse of meaning....... Phil
It's not more efficient. That's why Edison's DC power idea failed and Westinghouse's AC prevailed.
-- Blattus Slafaly ? 3 :) 7/8
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.