Suppose we have an inverted conical tank with height H and radius R
Suppose fluid is flowing through a hole in the bottom with cross sectional area a
with velocity given by V(t) = k [2 g h(t)]1/2
where h(t) is the height of fluid in the tank.
Find the time required to empty the tank
Since V( t)
k 2g h ( t)
the flux through the circle of crossection a is
Recall the flux is just the volume of fluid per unit time flowing
k 2g h a
through the cross-section at the bottom
Let v be the volume then:
dv
k 2g h a
dt
(1)
(It is important to note for later that k < 0 as the velocity must be downward).
On the other hand :
v
1
3
2
r h
Using similar triangles
We obtain
Where
dv
dt
1
v
3
1
3
(H h )
r
R
2
R
H
2
(H h ) h
H
R
( H h ) h
2
2
H
dh
( H 3 h ) ( H h )
(2)
dt
Set (1) = (2) We obtain the IVP
( H 3 h ) ( H h )
dh
dt
k 2g h a
with h(0) = H
Separating
( H 3 h ) ( H h )
h
Where
k 2g a
dh
k 2g a
dt
dt
5
2 H2
h
2
2
h
5
3 h
6
5
t C
5
2 H2
h
2
8 H
2
h
8 H
3 h
6
5
t
8 H
2
15
Now to get the time required to empty the tank set h = 0 and remember k and therefore
α are negative
5
t
8 H
5
2
15
2
4 2 H
15 a g k
4 2 H R
2
45 a g k
For example
H
t
5
5.64
a
1
R
2
k
0.2
0
