4 ms·
40 nL of desalted water per minute, for a minute, using 3.0V voltage (from article) and 0.7 A current (standard AA, "store-bought battery" from article), yields
by solox3 13y ago
40 nL of desalted water per minute, for a minute, using 3.0V voltage (from article) and 0.7 A current (standard AA, "store-bought battery" from article), yields 3.2 GJ/L.
Even if we give it the benefit of the doubt and say we're overestimating the current by a factor of 100, 32 MJ/L still a ridiculous amount of energy, and ~10x more than just distilling water through evaporation, which is 2.23 MJ/L.
- sbierwagen 13y agoThe actual paper ( http://libgen.org/scimag5/10.1002/anie.201302577.pdf http://libgen.org/scimag5/10.1002/anie.201302577.pdf ) says they manage an efficiency of 25 mWhL^-1 (90 joules per litre) Where did you get that 0.7A number?
- solox3 13y ago0.7A is the current of the "store bought" (AA) battery, suggested by the article. To make this technology energetically favourable to boiling, it must not exceed ~700μA in current, and it has to be far less than that to be worth its complexity.
- sbierwagen 13y agoUh? No battery[1] delivers a constant amount of current: http://en.wikipedia.org/wiki/Ohm%27s_law http://en.wikipedia.org/wiki/Ohm%27s_law Solving for current, Ohm's law is I = V/R. You only deliver .7 amps of current into a 4.285 ohm load. If you connect a 3 volt battery to a 1,000 ohm resistor, then only 0.003 amps of current will flow, which is 9 milliwatts. Connect a 3 volt battery to a .01 ohm load (a dead short, almost) and 300 amps of current will flow, (900 watts!) very briefly, until the battery voltage sags under load. 1: Or any voltage source in general, barring some trickery that you can't do with a simple electrochemical battery.
- HCIdivision17 13y agoI had an electrical engineering old timer explain the misconception as the Current Push Theory. (His industrial war stories were pretty good.)
- jonmrodriguez 13y agoGoogling for "Current Push Theory" only gets 5 results, none of them relevant: https://www.google.com/search?num=100&q=%22Current+Push+Theory%22 https://www.google.com/search?num=100&q=%22Current+Push+Theo... Can you please define the Current Push Theory?
- sbierwagen 13y agoHe might be referring to the "hydraulic analogy" of electricity: A wire is a water pipe, voltage is water pressure, amperage is total water flow. A battery is a pump pushing water through the pipes, voltage is how hard the pump pushes, and amperage is limited by how wide the pipes are, and thus how much resistance they impose on the flow of water. http://en.wikipedia.org/wiki/Hydraulic_analogy http://en.wikipedia.org/wiki/Hydraulic_analogy http://hyperphysics.phy-astr.gsu.edu/hbase/electric/watcir.html http://hyperphysics.phy-astr.gsu.edu/hbase/electric/watcir.h...
- HCIdivision17 13y agoThe basic idea was that you can't put a 5kW drive on a 2.5kW motor: it would shove too much power into the motor and break it. This is not quite at all how those systems work.
- jonmrodriguez 13y agoDoesn't it depend on the situation? E.g., if you have a resistive load and an inductive power source, the power source will keep upping its voltage until the load accepts the current that's being shoved down its throat.
- ScottBurson 13y ago> Connect a 3 volt battery to a .01 ohm load (a dead short, almost) and 300 amps of current will flow Except it won't, of course, because all real batteries have internal resistance. (How much depends on the battery chemistry and size.) Probably the 700mA rating for the AA battery is into a dead short.
- cmsmith 13y agoThis seems somewhat hard to believe. If you were actually pumping that much energy into the chip, wouldn't you end up boiling the water anyways?