Differential Equations and Flow Rate

Date: 02/14/99 at 01:05:12
From: Stephen Johnson
Subject: Differential Equations
  
Each of two tanks contains 100 gal of pure water. A solution containing 
3 lb/gal of dye flows into Tank 1 at 5 gal/min. The well-stirred 
solution flows out of Tank 1 into Tank 2 at the same rate. Assuming the 
solution in Tank 2 is well-stirred and that this solution flows out of 
Tank 2 at 5 gal/min, determine the amount of dye in Tank 2 after 15 
minutes.

Date: 02/14/99 at 09:47:02
From: Doctor Anthony
Subject: Re: Differential Equations

Let  M1 = mass of dye in tank 1 at time t
     M2 = mass of dye in tank 2 at time t

The inflow to tank 1 is 3 * 5 lbs/min = 15 lbs/min into the 100 gallon 
tank. The outflow is 5 * M1/100 lbs/min = .05 M1 lbs/min from the 100 
gallon tank.                              

The differential equations are then:

   dM1/dt = 15 - 0.05M1  = .05(300-M1)

   dM1/(300-M1) = .05 dt    and integrating

   -ln(300-M1) = .05t + C

   ln(300-M1) = -.05t + C

   300-M1 = e^(-.05t+C)  = A e^(-.05t)

and so M1 = 300 - A e^(-.05t) when t = 0, M1 = 0. Thus A = 300 and

   M1 = 300[1-e^(-.05t)]

For tank 2 the differential equation is:

   dM2/dt = .05M1 - .05M2

   dM2/dt + .05M2 = .05M1 = 15[1-e^(-.05t)]

This is a linear equation and we multiply by the integrating factor
e^INT(.05dt)  = e^(.05t):

   e^(.05t) dM2/dt + .05e^(.05t) M2 = 15e^(.05t)[1-e^(-.05t)]

   d[e^(.05t) M2]/dt = 15[e^(.05t) - 1]    and integrating

   e^(.05t) M2 = 15[(1/.05)e^(.05t) - t] + C

   M2 = 15[(1/.05) - t e^(-.05t)] + C e^(-.05t)

at t = 0, M2 = 0, and so 0 = 15[1/.05] + C. So C = -300. Then

   M2 = 300 - 15te^(-.05t) - 300 e^(-.05t)

   M2 = 300[1-e^(-.05t)] - 15t. e^(-.05t)

When t -> infinity, M2 -> 300 which is correct.

When t = 15 this gives:

   M2 = 300[1-e^(-.75)] - 15 x 15 x e^(-.75)

      = 158.29 - 106.2825

      =  52.0075 lbs

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/