"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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| Photo by Math Principles in Everyday Life |
Since the given word problem involves the mixture of non-reactive fluids, the working equation will be as follows
where
r1 = volumetric flow rate at the entrance
c1 = concentration of substance at the entrance
r2 = volumetric flow rate at the exit
c2 = concentration of substance at the exit
Since c2 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
V0 is the initial volume of solution at t = 0
Therefore, the final working equation will be
In the given word problem, we know that
r1 = 3 gals/min
c1 = 2 lbs/gal
r2 = 2 gals/min
V0 = 100 gals
x0 = 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
