Page 32 - Advanced engineering mathematics

12 CHAPTER 1 First-Order Differential Equations

We will derive a differential equation to model this problem. We need two principles from
physics. The first is that the rate of discharge of a fluid flowing through an opening at the bottom
of a container is given by
dV
=−kAv(t),
dt
in which V (t) is the volume of fluid remaining in the container at time t; v(t) is the velocity of
the discharge of fluid through the opening; A is the constant cross sectional area of the opening;
and k is a constant determined by the viscosity of the fluid, the shape of the opening, and the
fact that the cross-sectional area of fluid pouring out of the opening is in reality slightly less
than the area of the opening itself. Molasses will flow at a different rate than gasoline, and
the shape of the opening will obviously play some role in how the fluid empties through this
opening.
The second principle we need is Torricelli’s law, which states that v(t) is equal to the
velocity of a free-falling body released from a height equal to the depth of the fluid at time t.
(Free-falling means influenced by gravity only.) In practice, k must be determined for the
particular fluid, container, and opening and is a number between 0 and 1.
The work done by gravity in moving a body downward a distance h(t) from its initial
2
position is mgh(t), and this must equal the change in the kinetic energy, which is m(v(t) )/2.
Therefore,

v(t) = 2gh(t).
Put the last two equations together to obtain
dV
=−kA 2gh(t). (1.2)
dt
To illustrate these ideas, consider the problem of draining a hemispherical tank of radius 18
feet that is full of water and has a circular drain hole of radius 3 inches at the bottom. How long
will it take for the tank to empty?
Equation (1.2) contains two unknown functions, so we must eliminate one. To do this, let
r(t) be the radius of the surface of the fluid at time t, and consider an interval of time from t 0
to t 0 + t.The volume V of water draining from the tank in this time equals the volume of a
disk of thickness h (the change in depth) and radius r(t ) for some t between t 0 and t 0 + t.
∗
∗
Therefore,
2
∗
V = π(r(t )) h.
Then
V h
∗
= π(r(t )) 2 .
t t
In the limit as t → 0, we obtain
dV 2 dh
= πr .
dt dt
Substitute this into equation (1.2) to obtain
dh
2
πr =−kA 2gh.
dt
Now V (t) has been eliminated, but at the cost of introducing r(t). However, from Figure 1.4,
2
2
2
2
r = 18 − (18 − h) = 36h − h .
Then
dh
2
π(36h − h ) =−kA 2gh.
dt

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October 14, 2010 14:9 THM/NEIL Page-12 27410_01_ch01_p01-42