After you burn the entire 0.035lb of HHO, you only get 0.035lb of liquid water. You've applied enough power to lift 0.035lb up 144ft, and at the same time the water wheel generates as much power as you can get from dropping 0.035lb down 144ft. It's the same!!
You'd need a 100% efficient water wheel just to provide enough power to pump that gas up pipe B2. So there's nothing gained at all.
yes the system is sealed between the point where the HHO is generated and the point where the HHO is used as a fuel.
I forgot to include this in the image but the larger pipe
(B2) has air inside it , thats why its larger.
the HHO rises through the air due to its density.
air cannot flow through the pipe (B1) because of the
pressure inside (A1).
I did include the calculations to determine the amount of force required to just lift the HHO up the 144 ft distance.but I noted that this was if you needed to use it to determine
the amounts of force
to do so , not that it should be done.
since the HHO would simply float up the 144 ft distance
and you have agreed that the water falling would supply the energy that would be used up if it were lifted up 144 ft , the only cost in HHO production would be the cost at (A1).
this cost is recovered as the fuel is burned at the top.
and since HHO floats on air due only to its density as you
have admitted to in the previous thread , there is no reason that the system could not be higher than 144 ft.
and the higher it is the more energy you can produce as the water falls.
even if the water falls slowly , the higher it is the more torque the weight of the water will present to a shaft.
and high torque low rpm can be converted to low torque high rpm.
why would I go to all the trouble to just have a incompetent engineer work all the efficientcies out of the system , no I wouldnt hire any engineers.