Differential equation relating the height of the water to time.

Question

A toilet cistern was emptied and refilled and the following data obtained:
t = [ 0 2 3 4 5 8 14 23 28 34 41 53 74] time in seconds

h = [17.5 11 7 4 2 0.5 5 10 12 15 16 17 17.5]; height in centimetres 

You may assume the final height of the water when the valve closes after
the emptying and refilling cycle is 17.5 centimetres and that at t = 0 the
emptying sequence begins.

 

Consider the cistern emptying.
If it is assumed that the cistern has
a regular cross-section A cm2
, write down and solve the differential equation
relating the height of the water to time. Keep in mind that the rate of
change of volume is negative and proportional to the height of the water.

My equation A*dh/dt = -kh doesn't seem to be right

Answer 1

Basic equation:   A*dh/dt = Qin - Qout    where Q is volumetric flow rate.

 

Assumptions:  

1. Qin is constant from 0 to 74 seconds
2. At some point T1 between t = 5 and t = 14 seconds, h = 0
3. From 0 seconds to T1, Qout = k*h.  From T1 to 74 seconds, Qout = 0

 

Summary of solution procedure:

1. Write / Solve a differential equation to model the filling from T1 to 74 seconds.
2. Use the data pairs (t, h) after t = 14 seconds to find the parameters T1 and Qin in the solution.
3. Use these parameters in a differential equation to model the tank drainage from 0 to T1 seconds.
4. Solve this equation and find the parameter k.

 

As can be seen, I've still left plenty of work for any interested reader to complete for themselves.