Pomping
From: Eugeniusz (lala_at_magma-net.pl)
Date: 11/19/04
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Date: Fri, 19 Nov 2004 23:54:49 +0100
Hi
The tube L=1000m and crossection S = 1m^m ,is have put
perpendiculary to the deep water in big reservoir.
In order to remove the water from this tube ,we use compressed air.
When all tube will be empty , we disconnect the compressed air from
this tube
Then the tube is filled again
When the statik level of liquid in tube and in reservoir are balanced
Velocty V of water in this tube can be equal V = 100m/s .
because
W = Ekin
dW= F dL = rogLS dL M - massa of water in
tube
W = MgL/2 Ekin = M VV/2
From here
V = sqrt ( gL ) =~100m/s
There is problem; if water pump is used here ,can we ashive velocity V' =
sqrt ( 2 gL) ??
With kindly regards. Eugeniusz W.
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