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which at some initial time lies between the two cross sections S and S .
2
1
In a time interval t , the fluid that was initially atS moves to S , a
2
1
distance l  1 v 1 t  , where v is the speed at the section S of the
1
1
tube. In the same interval the fluid at section S passes a distance
2
 l 2 v 2 t  . Because of the continuity relation, the volume of fluid
 V passing any cross section in time t is

V  S  l  S 2 l  . (7.3)
1
1
2

We can compute the work done on this fluid during t . The force at
the first cross section is F  p 1 S , and at the second one
1 x
1
F 1  p 1 S 21 where p and p are the pressures of two section .
1
2
The net work done on the element during this displacement is therefore

A  F  l  F  l  p 1 S  l  p 2 S  l  ( p  p ) V (7.4)
1
2
2
2
1
2
2
1
1
1

The negative sign in the (7.3)arises because the force acts opposite in
direction to the displacement.
We now equate this work to the total change in energy, kinetic and
potential, of the element of mass m     V .
Thus the net change in kinetic energy is
 m v  2  m v  2 1
2
E  2  1     V (v  v 2 ) (7.5)
K 2 1
2 2 2
. The potential energy of the mass entering at section S in time t
1
is E P 1   m g  h  Similarly E P 2   m g  h  2
1
Therefore the change of potential energy is equal to
E  p    V g  (h  h 1 ) (7.6)
2

Combining (7.3),(7.5) and (7.5) we obtain

1 2 2
( p  p 2 ) V     V (v  v 1 )     V g  (h  h 1 ) (7.7)
2
2
1
2
or
1 2 2
( p  p 2 )    (v  v 1 )   g  (h  h 1 ) (7.8)
2
2
1
2

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