1 Mixing Problem

Background

Mixing problems occur quite frequently in chemical industry. We explain here how to solve the basic model involving a single tank (see the figure on the right). The tank contains 1000 gal of water in which initially 100 lb of salt is dissolved. Brine – i.e. salt water – runs in at a rate of 10 gal/min , and each gallon contains 5 lb of dissolved salt. The mixture in the tank is kept uniform by stirring. Brine runs out at 10 gal/min .

Problem

Find the amount of salt in the tank at any time t

Solution Step 1: Setting up a model

Let y ( t ) denote the amount of salt in the tank at time t . Its time rate of change is

y = salt inflow rate - salt outflow rate (1)

5 lb times 10 gal gives an inflow of 50 lb of salt. Now, the outflow is 10 gal of brine. This is 10 / 1000 = 0.01 (=1%) of the total brine content in the tank, hence 0.01 of the salt content y ( t ) , that is, 0.01 y ( t ) . Thus, from (1) we obtain the following ODE as a model:

y = 50 - 0.01 y = - 0.01 ( y - 5000 ) . (2)

Solution Step 2: Solution of the Model

The ODE (2) is separable. Separation, integration, and taking exponents on both sides gives

d y y - 5000 = - 0.01 d t , ln | y - 5000 | = - 0.01 t + c , y - 5000 = c e - 0.01 t .

Initially, the tank contains 100 lb of salt. Hence y ( 0 ) = 100 is the initial condition that will give the unique solution. Substituting y = 100 and t = 0 in the last equation gives 100 - 5000 = c e 0 = c . Hence c = 4900 . Hence the amount of salt in the tank at time t is

y ( t ) = 5000 - 4900 e - 0.01 t . (3)

This function (see the graph on the right) shows an exponential approach to the limit 5000 lb . Can you explain physically that y ( t ) should increase with time? That its limit is 5000 lb ? Can you see the limit directly from the ODE?