Bernouli rule and free energy

Hi
The sum energy of ideal liquid in pipe before crossecton of pipie is equal
sum energy of ideal liquid after this crossection of this pipe.
So we have
mgh + (mv^2)/2 =mgh/2 + (mV^2)/2
mgh/2 =(mV^2)/2 where v=0

V =sqre(gh) and this function is monotonously
increasing
WHEN WE REMOVE THIS LIQUID WITH VELOCITY EQUAL V we do the work W=mgh1,
and h1<<h
But the kinetical energy of this lifted liquid is equal
Ekin =(mV^2)/2 =m(gh)/2
Because h1<<h=H WE GET FREE ENERGY E
E = (Ekin - mgh1) =mg(h-h1)
Sincere E.Warenda


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