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as shown. So the net force is

2
F  (p  p 2 )  z (7.20)
1
Since the element does not accelerate, this force must just balance

the viscous retarding force at the surface of this element. According
Newton's law of viscosity this force is equal
dv
F   S (7.21)
dz
. The area over which the viscous force acts is S  2 z  l  .Thus, the
viscous force is

dv
F   2  z  l  . (7.22)
dz
Equating this to the net force due to pressure on the ends and
rearranging, we find that
dv p ( 1  p 2 z )
  . (7.23)
dz l  2

This shows that the velocity changes more and. more rapidly as we
go from the center (z = 0) to the pipe wall (z = r). The negative sign
must be introduced because v decreases as z increases. Integrating, we

find
0 p  p r
 dv  1 2  zdz (7.24)

v l  2 z
and
p  p 2 2 2
1
v  (r  z ). (7.25)
4 l

The velocity decreases from a maximum value p( 1  p 2 r ) 2 l  4 /

at the center to zero at the wall. Thus, the maximum velocity is
proportional to the square of the pipe radius and is also proportional to

the pressure change per unit length p( 1  p 2 l / ) and is called the
pressure gradient. The curve in Fig. 7.9 is a graph of Eq. (7.25) with the
v-axis horizontal and the z-axis vertical. Equation (7.25) may be used to

find the total rate of flow of fluid through the pipe. The velocity at each
point is proportional to the pressure gradient , so the total flow rate must
also be proportional to this quantity. Let us consider the thin-walled

element in Fig. 7.9. The volume of fluid dV crossing the ends of this
element in a time dt is dV  v (  dt) dS , where v is the velocity at the

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