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Author Topic: Can you use a optical fiber for electricity?  (Read 18993 times)

Offline Vern

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Can you use a optical fiber for electricity?
« Reply #25 on: 08/03/2009 00:22:35 »
From what I can glean out of the article MPT would be a trade off. It would cost much more per KWH but would not require the infrastructure connection. So you save on initial cost of the infrastructure, then pay for it forever :)
 

Offline wolfekeeper

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Can you use a optical fiber for electricity?
« Reply #26 on: 08/03/2009 02:53:45 »
It's been proposed to use across the Straits of Gibralta for example- laying cables is very expensive. Beyond a certain point it's cheaper to just waste some of the power than try to deploy a very expensive system due to ROI issues- if it takes longer than 5 years to break even, you're usually not going to, ever, due to compound interest; and then a slightly less efficient system like MPT can win.
 

Offline erickejah

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Can you use a optical fiber for electricity?
« Reply #27 on: 08/03/2009 22:01:56 »
 

Offline Vern

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Can you use a optical fiber for electricity?
« Reply #28 on: 08/03/2009 22:29:55 »
Return On Investment
 

Offline erickejah

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Can you use a optical fiber for electricity?
« Reply #29 on: 08/03/2009 22:41:27 »
:) tx
 

lyner

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Can you use a optical fiber for electricity?
« Reply #30 on: 08/03/2009 22:43:14 »
It's been proposed to use across the Straits of Gibralta for example- laying cables is very expensive. Beyond a certain point it's cheaper to just waste some of the power than try to deploy a very expensive system due to ROI issues- if it takes longer than 5 years to break even, you're usually not going to, ever, due to compound interest; and then a slightly less efficient system like MPT can win.


How about the ships passing through the straits? Isn't there some serious hazard involved?
 

Offline wolfekeeper

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Can you use a optical fiber for electricity?
« Reply #31 on: 08/03/2009 23:22:58 »
How about the ships passing through the straits? Isn't there some serious hazard involved?
No, not at all. If the beam is quite diffuse (if you have relatively big collectors at each end) it's no more dangerous than the radio waves emitted by your cell phone.

In fact, you mustn't make it too strong, otherwise you would fry all the birds that fly past- normal radio transmitters have the same problem if you think about it.
 

lyner

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Can you use a optical fiber for electricity?
« Reply #32 on: 09/03/2009 13:45:26 »
How big would you need these collectors (and transmitting array ) to be - bearing in mind the diffraction limit etc.? If cost is a factor then you couldn't be talking in terms of hundreds of metres of aperture. A cable would be cheaper than that!
Actually, 'normal' transmitters don't have the same problem at all. The beam from most transmitting antennae is very wide because the power budget of a comms or broadcast system doesn't require spreading losses of (as you imply) a very few dB.
Broadcast transmitters may use many kW of power but the actual power density is very low.
The only really large reflectors are used for radiotelescopes.
Basically, you would have to produce more calculated figures before you could come up with a convincing argument about possibilityof a competetive wireless sytem which provides tens of kW rather than mW of received power.
Transmitting amplifier efficiency, even, is unlikely to be much better than 60% and I have no idea what form of 'detector' circuit could produce useful output power.
 

Offline wolfekeeper

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Can you use a optical fiber for electricity?
« Reply #33 on: 12/03/2009 04:02:06 »
The demonstrated power with this kind of system was a few tens of kilowatts IRC.

It's similar tech to radio telescopes except you don't even need to make it steerable. The transmitter dish can be made out of little more than chicken wire, and doesn't have to be a single dish or anything (although they need to be contiguous due to the 'thinned array curse'). It probably needs to be quite tall (30 meters ish), but otherwise there's nothing particularly notable.

The receiver can be made of something called a 'rectenna'; it's pretty much just wires strung less than 1 wavelength apart with capacitors and diodes on it that rectify it and output a DC signal when you shine microwaves on it.

It's all really simple stuff. The overall efficiency can fairly easily be over 50%, even at extreme ranges, which is low by powerline standards, but then again you don't need to worry about people cutting the cables by mistake.

The sizes as a function of distance are diffraction limited and fairly straightforward to calculate.
« Last Edit: 12/03/2009 04:04:46 by wolfekeeper »
 

lyner

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« Reply #34 on: 12/03/2009 11:06:17 »
I can't believe that such a system could be as good as 50% efficient (and that, I'm sure, can't include the efficiency of the transmitter).
What frequency  is used? 'Chicken wire' implies that the wavelength would need to be in the region of 50cm (if you are not to experience some loss in the reflector.
An aperture of 30m would only represent 60 wavelengths - this has a diffraction limit of about 1 degree. The pattern of the reflector couldn't just be sin(x)/x - it would need to be tailored to suppress sidelobes - just like a good radiotelescope - the lost power problem is directly equivalent to the noise performance for radio astronomy. The actual aperture would need to be considerably oversize, to allow for tailoring of the dish illumination. Remember that the vast majority of the power has to fall within the area of the receiving array.
The straits of Gib are 14km wide - that is a required beamwidth of 0.1 degrees ('Most of' the  power not 3dB points) for a 30m receiving array.

The receiving array can't be treated as an independent set of elements - they will interact and, unless you get things right, they will fail to extract all the power from the incident beam - a dipole doesn't take all the power of a passing wave.
The figures really don't seem to add up. It's, in fact, far from "simple stuff". I imagine the project needs financial support from someone - don't let it be you!
 

Offline Vern

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Can you use a optical fiber for electricity?
« Reply #35 on: 12/03/2009 11:34:41 »
A few minutes with Google turned up a few articles claiming pretty high power transmission efficiencies. Much higher than I would have imagined. I think though that they are not considering all the losses; they are only measuring the efficiency at the antenna.

This link is to one of the papers

Abstract:
Quote from: the Abstract:
A new concept for solid state wireless microwave power transmission is presented. A
2.45 GHz rectenna element that was designed for over 85%0 RF to dc power conversion
efficiency has been used to oscillate at 3.3 GIIz with an approximate 10/0 dc to RF
conversion efficiency. The RF radiation was obtained from the same circuit by supplying
the dc output with reverse polarity dc power.


The bolding of the text was my edit.
Quote from: the link
                                    I. Introduction
The rectenna is a rectifying antenna operating in a receiving mode for reception of microwave power and subsequent conversion to dc by a diode rectifier. However if an IMPATT diode is used to rectify microwave power, the same diode can also operate in the avalanche region to generate and radiate RF power in the same circuit. Thus, the circuit can convert RF to dc and vice versa but not concurrently. The polarity of the dc voltage determines the operating mode. The dc current flow is in the same direction for
either mode of operation.

Based upon the conjecture of one of the authors ( Dickinson) [4], the concept is realized using a rectenna element obtained from the JPL Goldstone microwave power transmission experiment in 1975 [5]. The Goldstone rectenna array consisted of 4,590 elements that delivered up to 34 kW of output dc power from a2.388 Ghz microwave beam. This rectenna array demonstrated an average 82.5% collection and conversion efficiency whereas selected rectenna elements were tested at a 87/0 conversion efficiency
level [6].
« Last Edit: 12/03/2009 14:54:15 by Vern »
 

lyner

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Can you use a optical fiber for electricity?
« Reply #36 on: 12/03/2009 14:16:02 »
Receiving:
The rectenna looks quite useful as a receiver. However, in the paper, their idea of using the same device to transmit seems dodgy. You would need the oscillators all to be in phase lock if a proper beam is to be produced from a n array of 4000+  coherent elements.
The idea of using klystrons and magnetrons as RF to DC convertors is also novel but I think this would be more of an 'in principle' idea rather than an 'in practice' one. God knows what electron optics would be needed to cause a beam of 'bunched' electrons  to sort themselves out and to arrive at an electrode (collector) in such a way as to get a DC  potential which is elevated above the original bias voltage.

The rectenna elements each seem to handle at least 10W and the array  deals with 34kW.  The prototype array would, presumably, have been something under 7m square so a 30m array could handle about 600kW. That sounds like quite an expensive setup for such a small power output.  And it's far from being a few bits of wire and some diodes!

Transmitting:
I don't know the maximum achievable output power for current 3GHz devices but I should think that the power available at the transmitting end would be limited to less than that unless  they used the output from several devices.

Wikkers seems to reckon that the distance of a link is limited to about a km - the antenna design starts to be a bit more  reasonable then.

As the literature suggests, this sort of system is worth considering if you are talking about a fairly remote site with low power needs, with no alternative but there are ports in Morocco which  could supply the oil for that sort of power (they already do, of course) and a 1MW generating set is a tiddler. Load balancing between countries would require a link capacity of hundreds of MW to be effective.

I did see a suggested figure of up to £13M for an undersea cable to carry power from a wind farm over distances from 10 to 30km distances.  That doesn't sound a ridiculous cost for a 2GW(!!) power link.

RF compatibility would still be an issue in a busy shipping area - even though the power density may not be lethal.
 

Offline wolfekeeper

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Can you use a optical fiber for electricity?
« Reply #37 on: 12/03/2009 14:41:11 »
I can't believe that such a system could be as good as 50% efficient (and that, I'm sure, can't include the efficiency of the transmitter).
Proof by incredulity huh?

Nope, everything (DC-DC power transmission). Actually it's often over 65%, and nearly all of the losses are typically in the electronics at each end.

Quote
What frequency  is used? 'Chicken wire' implies that the wavelength would need to be in the region of 50cm (if you are not to experience some loss in the reflector.
Whatever you want, but the higher the better. 2.54 Ghz microwaves are 12cm and aren't used for other things.

Quote
An aperture of 30m would only represent 60 wavelengths - this has a diffraction limit of about 1 degree.
I haven't bothered to check it, but clearly you're out by a factor of 0.5/0.12, so the angle would be 1.2/5 * 1/360 * 2 * PI radians = 0.0042 radians. At 14000 that's 58m, so my guess of 30m was about right.

Quote
The pattern of the reflector couldn't just be sin(x)/x - it would need to be tailored to suppress sidelobes - just like a good radiotelescope - the lost power problem is directly equivalent to the noise performance for radio astronomy. The actual aperture would need to be considerably oversize, to allow for tailoring of the dish illumination.
You just build the receiver to catch as much as you can be bothered to really.
Quote
Remember that the vast majority of the power has to fall within the area of the receiving array.
The straits of Gib are 14km wide - that is a required beamwidth of 0.1 degrees ('Most of' the  power not 3dB points) for a 30m receiving array.
It's not that big; and increasing the frequency to say, 20 Ghz helps even more.
Quote
The receiving array can't be treated as an independent set of elements - they will interact and, unless you get things right, they will fail to extract all the power from the incident beam - a dipole doesn't take all the power of a passing wave.
The figures really don't seem to add up. It's, in fact, far from "simple stuff". I imagine the project needs financial support from someone - don't let it be you!
It's not so very complicated, and you seem to have plugged in an unrealistically low frequency of about 0.5 Ghz or something.
 

lyner

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Can you use a optical fiber for electricity?
« Reply #38 on: 12/03/2009 15:21:50 »
My assumption about frequency was based on the "chicken wire" thing.
Sorry about the "proof by incredulity" but the beamwidth needed for a 14km link is what I said - that's just 52X30/14000 degrees, if you want to get all your energy into the receive array. As I said, the 3dB width won't do or you're down to 50% efficiency for a start.

 Can you be sure of higher efficiency then 75% for a transmitter? The RF/DC conversion efficiency was claimed to be 85% (but how was that defined? - did it include the spillage from the transmitting horn or was it calculated on the basis of flux?). That's about 65% between the two.
Those figures (which are generous) were the cause of my initial incredulity. Then there is the pointing problem.

Are you really really serious when you say that you can rely on the transmitting array putting 76% of its input RF power onto an array 14km(+) away?

When you talk of increasing your operating frequency to get more directivity you have to change the devices at each end. Can you get 20GHz devices with the same efficiency as 2.5GHz devices? Also, the chicken wire idea goes straight out of the window. Your reflector will need to have 1/10 the error in its construction (+/-a quarter wavelength) and the receive array will need nearly a hundred times as many receptors.

Just one other point - did you consider atmospheric effects? Refraction is a significant issue in UHF TV reception at times; you might need to steer the beam. What do you do when it's raining 0.2dB/km loss at 20GHz for light rain.
 

Offline wolfekeeper

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« Reply #39 on: 12/03/2009 15:34:07 »
There's a worked example here done by the US DOE, over 38000 km range (!)

http://www.permanent.com/p-sps-tc.htm

You're bang on about the rain effects though, that is one issue particularly on with 14km sideways possibly through thick rain; but you can partly avoid or reduce it by choosing the frequency you use, some are better than others; there's quite a lot about the basic technology and issues and solutions on that site, and the wikipedia has some as well.
 

lyner

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« Reply #40 on: 12/03/2009 17:23:46 »
That link is very interesting and it deals with my (purely practical) objections.
In space, the size of an array is as big as you want; I had in mind that a huge balloon, silvered on one side (internally) could be inflated to act as the reflector.
Interestingly, they talk in terms of offshore receivers and cables to bring the power to land. I am not sure that a large array would be rugged enough for seaborne operation - we are seriously not that embarrassed for space that we couldn't find the odd few sq km somewhere in the UK.
I don't know about using flywheels up there for energy storage. Any 'massive' energy storage system would surely be better done on Earth. Batteries would maintain essential circuits during an eclipse, surely. You could even beam some power up to the satellite from Earth to keep it going. With an array as big as that, it would be easy to get power up to it when necessary using a similar system.
 

Offline yor_on

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Can you use a optical fiber for electricity?
« Reply #41 on: 13/03/2009 20:41:34 »
As we are talking about electricity in general and storage of it.
How about this? Would that be possible??
http://www.physorg.com/news156011642.html

------------
this was done with zinc-blende-structured MnAs quantum nanomagnets.
(Manganese Arsenide?)

Here is a PDF about making and characterizing MnAs nanostructures:
http://www.nnin.org/doc/NNINreu06Toyli.pdf
« Last Edit: 14/03/2009 12:51:27 by yor_on »
 

Offline wolfekeeper

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Can you use a optical fiber for electricity?
« Reply #42 on: 13/03/2009 21:18:12 »
Yeah, looks interesting.

FWIW there aren't any really stupidly significant eclipses in GEO though; the plane of the equator is angled significantly to the orbital plane, and this means that GEO sats only see eclipses for an hour or two during each of the equinoxes. There's not really any need for power storage, you just kick in your backup power supplies at those times (and you always need at least one backup anyway).
 

lyner

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Can you use a optical fiber for electricity?
« Reply #43 on: 13/03/2009 22:47:07 »
But you don't need continuity of service, like a TV satellite. Storage of energy is best done on Earth for the brief times of eclipses. Merely keeping the basic circuits running up there is trivial.
 

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