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