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