Here in the US midwest we've been having a drought lately, and it got me thinking of ways that we might be able to generate water more efficiently. Right now we're pumping an enormous amount of water out of the Ogallala Aquifer, beyond the replacement rate even in a rainy year I think. My understanding (correct me if I'm mistaken) of current desalinization plants is that they use a molecular sieve or reverse osmosis to separate seawater from fresh water, or a partial vacuum to flash boil the water. I don't know how much energy that sort of thing requires, but I know it is undoubtedlly more efficient than electrolyzing the water and combusting the H2 and O2 gases and condensing the water vapor immediately thereafter to obtain fresh water (otherwise they'd be doing that instead). But what if the energy from that combustion was used to aid the electrolysis? Obviously, you couldn't use it to power the entirety of the electrolysis, but it would definitely reduce the input energy needed while operating continuously. You could use the heat to power a small turbine and generate supplemental energy for the electrolysis. I'm a mechanical engineer by training, and so my chemistry is a little (okay, a lot) rusty, and I've forgotten how the math on that works. Here's my attempt so far: From Wikipedia, assume 50% efficient energy to H2 gas conversion (chemistry folks: is this a good assumption?): 1 m3 of H2 gas generation requires 7 kWh of energy Combustion of those gases would release 286 kJ/mol of H2. Assume 30% of that energy can be channeled back to electrolysis (rosy assumption): (1 m3 H2 @ STP)/(0.0224 m3/mol @ STP) = 44.643 mol H2 Energy obtained from combusting above quantity of H2 gas: (286 kJ/mol)(44.643 mol) = 3.55 kWh (3.55 kWh)(30%) = 1.06 kWh The total energy required to obtain 44.643 mol H2O (Hydrogen gas and water are in equal mole amounts on both sides of the chemical equation for combustion of hydrogen and oxygen gas): 7 kWh - 1.06 kWh = 6.94 kWh 44.643 mol H2O in liters: (18.0153 g/mol H2O)(44.643 mol H2O)(1 L/kg) = 0.804 L Crap. I don't like where this is going: (0.804 L)/(6.94 kWh) = 0.1159 L/kWh Well dang. Does all that math look right? My thinking was that the vast majority of the energy spent during electrolysis could be made up by burning the resulting gases and recapturing that energy. So I guess a better question would be how accurate these efficiency assumptions are. If my math above is solid (correct me if it's not), efficiency losses would have to be pretty close to zero for this to even maybe work. Any thoughts? I guess I answered my own question. Any creative fresh water generation ideas?