Solve Diff. Eq. Problem on Symmetrical Tank with Aperture

[cefarix]

A symmetrical tank of height 15 feet, as shown in the figure on the next page, is leaking water through an aperture in its base. The aperture is a uniform circular hole with radius r_0 = 3 inches. The top and base radii of the tank are r_1 and r_2 respectively, where r_1 = r_2 = 2 feet. The curve passing through P1, P2 and P3 represents the arc of a parabola i.e. y = a + bx + cx^2 where a, b and c are suitable constants (to be found). The radius of the tank at P2 is r_3 and at P3 is r_4. It is also known that the vertical height of P2 from the top of the tank is h_4 such that h_4 = h_1/2.

a) Derive the height h of the water in the tank at any time t.

b) If the initial level of water in the tank is h_0, how long will it take to empty the whole tank?

c) How long will it take for half the water to leak out from the tank given the initial level h_0?

ASCII drawing of the tank:

************------------ <-- P1, h_1 = height from P1 to P3, rad.=r_1
************/********\
***********/**********\ This region is a parabolic curve outwards.
**********-************-
*********/**************\<-- P2, ht. h_4 = 0.5 * h_1, rad. = r_3
*******--****************--
******/********************\
***---**********************--- <-- P3, rad. = r_4, ht. from top h_1
***|*************************| ^
***|*************************| | <-- height is h_2
***|*************************| V
***\*************************/^
****\***********************/*| <-- height is h_3
*****\*********************/**| This region is a linear inwards.
******\*******************/***V
*******---------****---------- <-- bottom radius = r_2
**************^^^ radius of hole/aperture in bottom = r_0

The height values and radius values are:
h_0 = 10 feet
h_1 = 11 feet
h_2 = 2 feet
h_3 = 2 feet
r_3 = 3.5 feet
r_4 = 5 feet

Any help and guidance on this problem would be greatly appreciated.

 

[CarlB]

I'd write a computer program.

Carl

 

[quicknote]

a) To derive the height h of the water in the tank at any time t, we can use the conservation of mass principle. The rate of change of the water level in the tank is equal to the rate of water leaking out through the aperture. This can be represented by the following differential equation:

dV/dt = -A*dh/dt

Where V is the volume of water in the tank, A is the area of the aperture, and dh/dt is the rate of change of the water level. We can also express V in terms of h and the tank dimensions:

V = π/3 * (r_1^2 + r_2^2 + r_1*r_2) * h

Substituting this into the differential equation and solving for dh/dt, we get:

dh/dt = -π/3 * (r_1^2 + r_2^2 + r_1*r_2) * (1/A) * dh/dt

Where 1/A is the inverse of the area of the aperture. We can then solve this differential equation using separation of variables and find the height h as a function of time t.

b) To find the time it takes to empty the whole tank, we can set h equal to 0 and solve the differential equation for t. This will give us the time it takes for the water level to reach 0, which is when the tank is completely empty.

c) To find the time it takes for half the water to leak out from the tank given the initial level h_0, we can set h equal to h_0/2 and solve the differential equation for t. This will give us the time it takes for the water level to decrease to half its initial value.