Question

Suppose water is leaking from a tank through a circular hole of area

A_h

at its bottom. When water leaks through a hole, friction and contraction of the stream near the hole reduce the volume of water leaving the tank per second to cA_h

2gh

, where

c (0 < c < 1)

is an empirical constant.

A tank in the form of a right-circular cone standing on end, vertex down, is leaking water through a circular hole in its bottom. (Assume the removed apex of the cone is of negligible height and volume.)

(a)

Suppose the tank is 20 feet high and has radius 8 feet and the circular hole has radius 2 inches. The differential equation governing the height h in feet of water leaking from a tank after t seconds is

dh

dt

= −

5

6h3/2

. In this model, friction and contraction of the water at the hole are taken into account with

c = 0.6,

and g is taken to be

32 ft/s².

See the figure below.A right-circular conical tank containing water is shown.

• The cone opens upward and the point of the cone at the bottom is labeled: circular hole.
• A dashed line extends horizontally from the circular hole.
• The surface of the water in the tank is labeled: A_w and a line segment from the dashed line to A_w is labeled: h.
• The radius at the top of the tank is labeled: 8 ft.
• The height of the tank is labeled: 20 ft.

If the tank is initially full, how long will it take the tank to empty? (Round your answer to two decimal places.)

14.31 minutes

(b)

Suppose the tank has a vertex angle of 60° and the circular hole has radius 3 inches. Determine the differential equation governing the height h of water. Use

c = 0.6

and

g = 32 ft/s².

dh

dt

=  

If the height of the water is initially 11 feet, how long will it take the tank to empty? (Round your answer to two decimal places.)

Answer 1

We have to find out the differential equation for t>0

The loss of water out at the bottom of the right-circular conical tank and the contraction rear the hole.

Assuming h is the height.

The velocity of the water which is leaving from tye tank (v)=

Area of the hole(Ah)

Volume=Ah x

Hence -(Ah x )

The negative sign indicates that volume is decreasing

Ignoring the friction of the hole. then the volume of water at time t is V(t)=Aw

X  

As an area of circular hole (Ah)= pi x r2

=

A.   as given in the question.

the tank is 20 feet high and has radius 8 feet and the circular hole has radius 2 inches.

Integrating both sides we get,

Applying initial condition h(0)=H

The tank is empty when the height is zero i.e h(t)=0

0=

putting H=20ft

=858.65 sec =14.31 Minutes.

B.As the vertex angle is 60 degree

h/d =1/2

As h=11 ft ,then d=11/2=5.5 ft

Radius(r) is 3 inches.

X  

The area of circular hole is(Ah)=

r=(3/12)=1/4 ft

Ah==

applying integration

Applying initial condition h(0)=11 ft

c=6.63

The tank is empty when the height is zero i.e h(t)=0

t=167.13 sec=2.79 minures