Focused Microwaves Could Enable Wireless Power Transfer

esocid alerts us to news out of the University of Michigan, where physics researchers have found a way to focus microwaves to a point 20 times smaller than their wavelength using a new 'superlens'. Such resolution was thought to be impossible until recent years, and it could bring about the capability to transfer power wirelessly. "No matter how powerful a conventional lens, it cannot focus light down to more than about half its wavelength, the 'diffraction limit'. This limits the amount of data that can be stored on a CD, and the size of features on computer chips. The new lens is a 127-micrometer-thick plate of teflon and ceramic with a copper topping. 'The beauty of these is that they're planar,' Grbic says, 'they're easy to fabricate.' The lenses can be made through a single step of photolithography, the process used to etch computer chips."

Comments

Superlens = spillover = irradiation

[G4from128k]

What I remember from studying this technology 15 years ago was that it was possible to create a beam sharper than the diffraction limit, but the result was diffuse spill-over. That is, one could create an extremely sharp main lobe in the beam pattern, but one had to suffer higher side-lobes. That's OK for imaging and lithography applications -- the spill-over is diffuse enough not to cause too many problems. But for power applications it means both inefficiency (power lost to the side lobes) and irradiation for people who think they aren't in the beam.

You may have forgotten...

[MichaelSmith]

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Re:Nothing new here; still not a good idea

[khallow]

What I hear is that you can get 85% efficiency on a pass through Earth's atmosphere (between ground and orbit) (which is about equivalent to maybe 8-10 km of sea level atmosphere). That's pretty good and it improves as you increase in altitude. At 18,000 feet (or a bit over 5000 meters) the inefficiency is halved (to I suppose 92-93%). And I'm dubious about your claim that wire is more efficient. Sure running a microwave along the ground is crazy. But bouncing it off an orbital reflector is pretty efficient (or starting with a solar powered satellite in the first place).

Re:We tried that

[NeverVotedBush]

What I was referring to was current squared times resistance which equals power. The R was resistance and not radius. V = I * R, and W = I * V. Therefore, W = I * I * R.

Likewise, the IR drop is also just Ohm's law which equals voltage. The resistance will have some value per unit length and the longer the length, the more voltage drop.

The way to drop the current, so the I^2R (watts) losses can be reduced is to increase the voltage. But as you go to higher voltage, and higher altitude, where the air pressure starts getting low enough to support a plasma discharge, insulation starts getting important which just leads to more weight, etc.

Cutoff Point.

[pavon]

The energy in question is coming from the sun, and was going to enter the biosphere anyway. Some of it would have, but some of it would also be reflected. On average, the earth has an albedo (fraction of light reflected) of about 37%.

To a certain extent, the effect will be the exact opposite of what you are thinking, as the sunlight would have most assuredly heated the land, sea and air, but beamed down to the electrical grid, it will be stored in other forms In the long run it will all be converted to heat. Furthermore, there are very few uses of electricity that result in storage as potential energy of some form. Looking at California data, the Residential, Commercial, TCU and Streetlights will all be AC/lights/electronics which will be converted to heat immediately. The mining sector and industrial sectors will result in some potential (lifted mass, increased chemical potential of stable compounds, etc) But the machines they use to do this are not very efficient. Even if we are very generous and say that half of their energy is used for these purposes, and those machines are 50% efficient, that gives 5% of total energy use being converted to potential form.

So if the energy efficiency of the panel/beam is greater than about 100%-37%-5% = 58%, then this system will result in more heat than would normally occur from the sunlight.

Of course, even if it does significantly increase the amount of heat generated for the fraction of sunlight that it captures, that is still a tiny fraction of the sky that is covered, and the net result will be completely negligible compared to just about anything else.