Environmental Engineering Reference
In-Depth Information
body exerts on the fluid and, according to Newton's Third Law, the fluid must
exert an equal and opposite force upon the body. In this way the body experiences
a resistive “viscous” force. The details depend upon the nature of the fluid and on
the efficiency with which the body interacts with it. The latter is in turn dependent
on the geometry of the body and on the nature of the surfaces that drag the fluid, as
well as the velocity of the object relative to the fluid. Empirical evidence supports
the following expression for the viscous force:
Dv
2
.
=−
−
F
v
Cv
(2.32)
≈−
For motion in liquids
C
Cv
, at least for low speeds. For motion
in air the quadratic term dominates and
F
v
≈−
D
and
F
v
Dv
2
.
Example 2.3.8
A ball bearing is released at rest in a tall cylinder of glycerol and
falls under the influence of gravity. Show that the speed tends to a limiting value.
Solution 2.3.8
The forces acting on the ball bearing are gravity and the viscous
force. The Second Law gives
m
d
v
ma
=
d
t
=
mg
−
Cv.
This is a first order differential equation in v. Rearranging gives
d
v
=
g
d
t
C
mg
v
−
1
C
which can be integrated (substitute for
1
−
mg
v) to yield
ln
1
mg
v
C
C
m
t
−
=−
+
B,
where B is the constant of integration. Rearrangement, and fixing the constant of
integration by the requirement that v
=
0
at t
=
0
, gives
1
m
t
.
mg
C
C
e
−
v
=
−
The above equation contains the essential features of an object falling in a fluid.
There is a limiting (or terminal) velocity v
t
=
mg
C
that the speed tends towards for
m
large t . The time taken to reach a speed of (e
C
, which is shorter the
more viscous the medium. Note that objects of different mass but the same value of C
have different terminal speeds. This is in contrast to the behaviour of falling objects
in vacuum, where the acceleration g is constant and independent of the mass. For
a medium of vanishing small viscosity, or for very small t , a Taylor expansion of
the exponential leads to
−
1
)/e of v
t
is
1
...
mg
C
C
m
t
v
≈
−
1
+
+
≈
gt,
which is as expected.
