Re: Bernouli rule and free energy

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"Gieniu" <warendag@xxxxxxxxx> wrote in message
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"Gieniu" <warendag@xxxxxxxxx> wrote in message
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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 + (m+m1)(V^2)/2
mgH-mgh/2 =(m+m1)V^2)/2 where v=0
hS ro gH -hS ro gh/2 =(hS ro +LS ro)V^2)/2
hgH -hhg/2 =(h+L)(V^2)/2
V^2 =2gH-gh)/(L/h+1)
If h=0 then V=0
If h=H and L=0 then V=sqrt(gH)
The function V(h) is monotonously increasing.
The free energy is possimble
E.W.




















































































































































































































































































































































































































































































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