__Category__: Chemical Engineering Math, Differential Equations"Published in Newark, California, USA"

Consider a tank that initially contains 100 gallons of a solution in which 50 pounds of salt are dissolved. Suppose that 3 gallons of brine, each gallon containing 2 pounds of salt, run into the tank each minute, and that the mixture, kept uniform by stirring, runs out at the rate of 2 gallons per minute. Find the amount of salt in the tank at time t.

__Solution__:

The first thing that we have to do is to analyze and illustrate the given word problem as follows

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Since the given word problem involves the mixture of non-reactive fluids, the working equation will be as follows

where

r

_{1}= volumetric flow rate at the entrance

c

_{1}= concentration of substance at the entrance

r

_{2}= volumetric flow rate at the exit

c

_{2}= concentration of substance at the exit

Since c

_{2}is usually not given in the problem, we can rewrite the above equation as follows

where

x = the amount of salt at time t

V = final volume of a solution at time t

but

where

V

_{0}is the initial volume of solution at t = 0

Therefore, the final working equation will be

In the given word problem, we know that

r

_{1}= 3 gals/min

c

_{1}= 2 lbs/gal

r

_{2}= 2 gals/min

V

_{0}= 100 gals

x

_{0}= 50 lbs

then the above equation becomes

Since the above equation is a first order, first degree linear equation, the integrating factor will be equal to

The general solution of the above equation is

If x = 50 lbs of salt at t = 0, then the value of C is

Therefore, the particular solution of the above equation or the amount of salt in the tank at time t is