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Sahara Solar To Power Half the World By 2050
eldavojohn writes "A Japanese/Algerian effort called The Sahara Solar Breeder Project employs a simple concept revolving around the pure silica in the sand of the Sahara Desert. The silica can be used to build vast solar arrays which will then provide the power and means to build more solar arrays in a classic breeder model. They would then use DC powerlines utilizing high temperature superconductors. The lead of the project points out that silica is the second most abundant resource in the Earth's crust. The project's lofty goals to harness the Sahara's energy has a few requirements — including 100 million yen annually — but also the worldwide cooperation of many nations and the training of the scientists and engineers to create and man these desert plants. The once deadly wasteland of the Sahara now looks like a land rich in an important resource: sunlight."
Now all we have to do is build a massive worldwide network of new transmission lines, stabilize the governments of Africa, and get every country in the world to agree on how the power is to be shared.
SJW: Someone who has run out of real oppression, and has to fake it.
100million yen is 1 million dollars...That really isn't much money.
That said, the project is incredibly unrealistic, or at least the stated goal is.
AC is more stable over distance because DC has to compete against natural differences in ground voltage, but DC is better for really long distances as it is theoretically nearly lossless while AC loses proportional to the length of the cable.
Couldn't we do this in the US deserts?
The Mojave, the Sonoran?
I can imagine the Native Americans feeling a sense of deja vu. There's gold in those Indian lands.....
Distributing energy from the Sahara to all the world will meet some resistance.
Sandstorm + Solar Array = ???
AC is actually NOT better for long (and I mean LONG) distances. Short to medium runs (dozens of miles) it's not too bad and the ability to efficiently change voltages with AC using transformers means you can keep current down and wire sizes small.
AC won out in the beginning because there was no cost effective means to alter DC voltage between efficient transmission voltage and safe/practical usage voltage.
However, wires have capacitance. Overcoming that capacitance requires energy, which is an inefficiency. When your cable goes from dozens of miles to hundreds of miles these losses become significant. DC doesn't have to deal with the capacitance issue, so it is actually more efficient here. Modern solid state power electronics also make changing DC voltages efficient and practical enough to use HVDC across long distances and Medium-Low Voltage AC for local distribution.
Add superconductors to the mix and the advantage of DC increases substantially.
Lastly, transmitting in DC solves problems with synchronizing and matching AC frequencies where otherwise independent grids interconnect. Each end of the DC link doesn't "see" or care about the frequency/timing of the other end.
=Smidge=
AC and DC power lines both loose energy to resistance. AC power looses energy in another fashion due to capacitance and inductance called reactive power. By using superconductors (0 ohm resistance) for the power lines, you eliminate all losses for DC, most losses for AC, and introduce new losses for the cooling equipment. Of course, with superconductors the formula isn't as simple as V=IR because then you could get infinite current. (V/0 = I) With superconductors, there is a maximum current density (Amps per m^2 as the area of the cross section of the wire) before the wire starts to produce resistance.
"For every complex problem, there is a solution that is simple, neat, and wrong." - H.L. Mencken
Long distance power transmission is almost always DC for a number of reasons The first of these is are in DC there are no induction losses. The alternating current along transmission lines will inductively couple to each other and provide a loss in a similar manner to the crosstalk you get in everything from digital circuits to audio cables. DC just doesn't lose power this way.
AC is great for easy step up and step down in voltage, but it has a number of problems. In a transmission system you have two main limits, the maximum voltage you can use (limited by insulators used) and the I^2R losses in the cabling. Let's first assume no resistive losses or at least that you're not limited by heat loss: For a given cable and insulators you can therefore either run e.g. 1000V DC or 1000V AC Remember though that the AC is 1000V peak so the actual RMS voltage is effectively 1/sqrt(2) so 707V. Therefore for a given cable and insulator pair AC can carry less power.
The only way to reduce the I^2R losses is to run at higher voltages where currents required are less so DC will always be superior here provided your inverter technology is sufficiently efficient. Which for lengths of more than about 20km starts to happen.
What was true for the electrical systems of 20 years ago never mind back in the days of Edison is no longer the case, the AC vs DC situation is not as simple as it used to be.
"The weirdest thing about a mind, is that every answer that you find, is the basis of a brand new cliche" -
That would be pretty awesome. Instead of building casinos, get them building power plants. New, better-quality jobs for them, and clean electricity for everyone. It's a win-win situation.
Disclaimer: I'm not a power engineer but I am an electrical engineer, so while the principles I state are probably correct there is some guesswork as I apply it to power transmission.
However, wires have capacitance.
Yes, all wires have a certain capacitance and inductance per length. Given the very wide separation between the power line and ground the capacitance per length should be very small, since it is inversely proportional per distance. Given the wide area the current encloses, the inductance will conversely be large. Therefore I would guess that the inductance of the transmission line is more important than its capacitance, and that it can be modeled primarily as a resistance in series with an inductance.
Overcoming that capacitance requires energy, which is an inefficiency.
Correct, though I'd like to add to that. An ideal capacitor is a lossless device (if you bring up the "two capacitor problem" note that by definition the capacitors or the wires connecting them cannot be lossless or it would never reach steady state). However in any AC transmission there will be conductor loss due to the resistance of the wire, possibly increased by skin effect, and dielectric loss due to the changing polarization of polar molecules in the dielectric surrounding the wire. Air is a virtually lossless dielectric. The wire's insulation is not going to be lossless, but since dielectric loss is usually proportional to frequency and 60Hz is very low frequency, and the insulator is small, I would guess that conductor loss dominates. And since frequency is very low skin effect may be negligible and we can just use the DC resistance.
One more note is that most loads are inductive (ballasts for flourescent lights, motors for air conditioners, motors for industrial equipment, transformers, etc) and this is probably going to dominate the power factor of power transmission much more than the reactance of the power transmission lines themselves. That's why most load compensation is in the form of added shunt capacitance.
Of course there are still many advantages of DC transmission, but for power lines on poles I wouldn't be convinced of frequency-dependent loss playing a large role unless I saw a full analysis.