. . Mixture problems Mixing problems occur quite frequently in chemical industry. Now we explain how to solve the basic model involving a single tank.
A substance S is allowed to flow into a certain mixture in a container at a constant rate, and the mixture is kept uniform by stirring. Further, in one such situation, this uniform mixture simultaneously flows out of the container at another rate. Now we seek to determine the quantity of the substance S present in the mixture at time t . Letting x to denote the amount of S present at time t and the derivative dx dt to denote the rate of change of x with respect to t .
If IN denotes the rate at which S enters the mixture and OUT denotes the rate at which it leaves, then we have the equation dx dt = IN OUT Example . A tank contains litres of water in which grams of salt is dissolved. Brine ( Brine is a high-concentration solution of salt (usually sodium chloride) in water ) runs in a rate of litres per minute, and each litre contains 5grams of dissolved salt. The mixture of the tank is kept uniform by stirring.
Brine runs out at litres per minute. Find the amount of salt at any time t . Let x t ( ) denote the amount of salt in the tank at time t . Its rate of change is dt = in flow rate out flow rate Now, grams times litres gives an inflow of grams of salt.
Also, the out flow of brine is litres per minute. This is / of the total brine content in the tank. Hence, the outflow of salt is . times x t ( ) , that is x t .
Thus the differential equation for the model is dx dt = − This can be written as dt = − ( . Integrating both sides, we obtain log = − input output Fig. . - - or x Ce or x Ce Initially, when t , so + C .Thus, C = − .
Hence, the amount of the salt in the tank at time t is x