2 days ago I came across patent CA2978935C, which seems to confirm something I have been thinking about for a couple of years – is it possible to extract more power from water using a siphon, than it is using a conventional penstock design? In a siphon, because of energy conservation laws, the velocity in the pipe is uniform throughout it’s length. Meaning the speed of water entering the pipe will be the same as that leaving it, so what if the generator is placed at, or near the input, as in the patent above. This still means, as long as the generator does not break the siphon, that there will still be a stream of water exiting the siphon, which could be used to generate, albeit reduced, power using say a pelton or similar turbine. For example, if we had a pipe of cross sectional area 1 m2, flow rate of 10 m/s (easy maths), using P = (mv2)/2 ( where m = density(in kg)*area*v) we get: at the input - (1000*1*10*10*10)/2 = 500,000 watts assuming the velocity has now been halved (from Betz’s law) we get: at the output – (1000*1*5*5*5)/2 = 62,500 watts Does this mean we have gained 62,500 watts, because using a penstock would only generate the same as the input? (I know these figures don't take into account any generator inefficiencies) If this is correct I think it may be possible to increase it further!