The title of this topic might sound a bit uppity, but I was wondering how much electricity is lost when electricity is converted into a laser, and then into electricity again. I would also like to know the amount of resistance lost in a current channeled through a power line per distance and the energy loss per distance of a laser in a fiber-optic cable.

What I'm trying to figure out is if, although electricity-laser-electricity conversion is less efficient than normal power lines, if the energy loss of a fiber-optic cable per distance is less than that of a power line, fiber-optics might be a more efficient form of long distance energy transfer.

One way to visualize this badly described proposition would be to imagine a graph with distance on the x-axis and efficiency on the y axis. The slope of the electricity would be -1X + 50 and the slope of the laser would be 25. At a certain distance the lines would intersect and past that point the laser would be more efficient. (of course this is assuming that fiber-optics are more efficient than power-lines per distance, which is what i'm asking in this long-winded question.)

Thanks for taking the time to read this post and please don't flame me If you think I was trying to act smart or something, this post was only long because it was a badly written description of a fairly simple idea.

Yes, because major scientific breakthroughs and serious technological discussion will be discovered or occur in a forum where the mean age is 16.

Am I speaking to cleverbot?

man eric when you have kids I bet when they ask how something works you'll hand them schematics (I am making fun of you here because you are discouraging someone for at least making an attempt to learn about something while simultaneously being intensely passive aggressive with a superior tone to just about everyone on the forum)

Lasers are drastically inefficient (to generate) and sending one with actual power generation capacity through FiOS is nigh-impossible. Data is, relatively speaking, low intensity and light on attenuation, but power by nature requires more complex devices. You'd also need a reliable, safe, long-lasting means of re-converting the laser light into energy. All devices which can do so are also drastically inefficient. Oh, and you'd need signal boosters at some point along the line, which further reduces efficiency.

According to this, current US power transmission has losses of only 7.2%. Admittedly, that's millions of volts, but for any FiOS system to be worth both the extra cost and the reduced efficiency you'd need a massively increased range, which is not needed with current power generation technology. If someone built a massive fusion reactor in Kansas and was gonna ship power to China: ehhh, maybe it's useful.

Thanks! Although I'm disappointed, that was a very clear and useful reply.

EDIT: Just out of curiosity's sake, could you explain to me why elecritcity-laser-electricity conversion is so inefficient? Thanks.

FiOS is (c) Verizon Communications, Inc.

Also, Tucker he could stand to learn more if he actually took a course in electrical engineering. Energy loss and circuit resistance are pretty much ECE 1001 topics.

If my kid wants to know something, I will tell him to ask his mother, or consult wikipediaz. That is assuming I don't actually know the answer (which we know is of such unlikely probability to barely warrant contemplation).

Thinking about it, I don't see what would be so inefficient about the conversion I was talking about. After all, a laser is just concentrated light, so if you had a photocell on the receiving end you could absorb it and convert it to energy without losing very much. Alternatively, you could use the lasers to heat water for steam turbines if that would work better.

The real thing I wanted in this question, even though Grif's answers were good, were some numbers which I could plug into some equation to compare efficiency like:

1/ e (r^2) and L/l (r^2)

Where L is the percentage of energy maintained in a electricity-light-electricity conversion, in decimal form

e and l are ratios of energy lost per distance

and r is the distance between the power source and its receiver

If you had any info on those sorts of things, it would be really nice.

(also, in the first equation, there is a one instead of a variable because there is no conversion, therefore 100% efficiency

Also, Thor, I'm not trying to be the source of the next scientific breakthrough. I know being contrary and negative is super cool and all, but I actually want to have a serious discussion here.

Homework: Determine the total energy cost of generation of a carbon dioxide laser at 200kV. Be sure to include the energy to refine pure quantities of the gases involved. Next, factor in the practical efficiency of a thermophotovoltaic system designed to convert the infrared from your carbon dioxide laser into usable energy. Then add in the losses from voltage and amperage steps at the site of your generator and your receivers.

Yuck. Wish I had a half decent science teacher at my school (since its an arts focused school we have a college drop out biology teacher, a part-time teacher who forgets all of my calculus asignments, and a general math teacher who is also the dean of students) who could help me with this, since im writing an essay on alternative energy for the Telluride Association TASP program and I thought this idea might be a nice touch to throw in.

Thanks once again for the info Grif, studying right now.

One wording-related thing I ought to point out is that there is a difference between merely focusing light and actually lasing it into a coherent beam.

Second thing; every conversion you've mentioned has significant loss. Ever tried to charge a battery up with another battery of the same type? (Do it, if you graph the electrical and thermal energies you get from this over time you can have some fun).

Thirdly, you're missing the whole deal with light transmission in fibers. There's a limit you hit really fast when you start piling one or another form of energy into some transmissive material, at which you start to damage it. In the case of high-power laser transmission, that would very rapidly degrade, unless you were using laser-grade optic glass (far more expensive than the silica glass or plastics used currently), at which point you have defeated the point of this mental exercise.
Not to mention that you could not use the cheap and great plastic fibers with IR lasers.

Fun test, once you're done with all that assigned work; get a laser diode, a regulated 5v supply, and a photocell.
In a silvered vacuum tube, measure the power you get out at the photocell side, and the temperatures of the diode and cell.
It wil be significantly lower than the input, but the total energy (you'll need to do some conversion here) will obviously be almost exactly what you began with.

Short answer: solid-state laser tech needs a lot of work, and chemical lasers are not sustainable.


But hey, this all goes out the window when you can stretch a vacuum tube between the two points, or are transferring power between a satellite and say, a small ion-propelled spacecraft.

Edit:
Also, if you tack on some power regeneration, you can get useful power back from the heat you produce on both ends; for example, using a stirling cycle engine operating on the heat differential between the vacuum-insulated laser emitter and the cooling effects of decompressing gasses (or just rock). Generating steam is a pretty awesome way to turn heat into motion, but it's really inefficient at small scales.

I feel like I should be reporting this thread to the FBI or something.

(in Russian accent) NO report or I kill you!

I like that vacuum idea very much, but the rigidity of the vacuum pipes you are talking about would greatly decrease the application possibilities of the laser power. Perhaps you could make "vacuum cables" which were basically flexible vacuum pipes with reflective insides. I don't think you would have to worry about heat loss very much because of the speed of the laser. As I read in some equation on wikipedia I forget the name of, the loss in energy due to heat while transferring a force, like a laser or electrical current, is directly proportional to the amount of time it takes to transfer, or inversely proportional to its speed. And lasers are pretty fast.

Also, Grif, why do I need to make a gas laser and then use a thermovoltaic system to capture the heat? I might be mistaken, but it seems like these extra conversion steps would just decrease the efficiency of the laser.

Everything I just said might be very stupid and useless. If it is, remember that I'm just a high school student interested in something he doesnt know very much about.

you need to create your laser and then you need to reconvert it to electricity somehow

vacuum "wires" would suffer from the following-
incredibly complex to manufacture
difficult to maintain
subtext three, which is not as easily bulletable.

subtext three:
long distance power transmission in any form is going to require high amounts of energy. while lasers inside a fixed medium would suffer less attenuation and wouldn't need nearly as many interim steps (compared to raw wires), you'd still hit the problem of losses. unless you can cheaply manufacture nearly limitless quantities of something with a perfect mirror surface, you're going to be losing energy every single "bounce" of the laser. there's also the inevitable diffraction, tube or no tube.

I'm sorry I don't have anything to actually contribute to this, but thanks anyways for giving me something to sit here and read while I'm taking a break from MW2.
. . . It's so interesting to read about all this stuff, even though I have no knowledge or experience with any of it whatsoever.