Re: Bernouli rule and free energy


"Gieniu" <warendag@xxxxxxxxx> wrote in message
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"Gieniu" <warendag@xxxxxxxxx> wrote in message
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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. mgH +(mv^2)/2 =mgh/2 +
((m1+m)V^2)/2

2mgH +mv^ 2 =mgh +(m1 + m)V^2
V^2 =(2hSgH-hSgh)/(LS+hS)
V^2 =(2gH -gh)/(L/h +1)
V^2 =(2gH -gh )/(L/h+1)

V=0 if h=0
V =sqrt(gH) if h=H and L=0

Free energy exsist becouse the function
V=sqrt(2gH -gh)/(L/h+1) is function monotonuosly
increasing
Sincere E.W.













































































































































































































































































































































































































































































































































































































































































































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