MODELING CONCENTRATION OF A SOLUTION (# 2.3.4)
A GUIDE BY CHRISTIAN WOODS
The Problem
A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt
gal
in solution. Water containing 1lb of salt per gal is entering at a rate of 3 min and the
gal
mixture is allowed to flow out at 2 min . Find the amount of salt in the tank at any time
prior to the instant when the solution begins to overflow.
1. Creating the Differential Equation Model
Since the question asks only about the tank before it overflows, we will develop a model
that assumes there is always space in the tank.
First, let’s figure out what quantities are in play in this system, and what units they
use:
lb C : concentration of salt water solution in the tank
gal
t : time (min).
Now we want to figure out how the concentration of the solution is changing over time.
We use our go-to strategy:
dC
= rate in − rate out.
dt
Let’s think about the “rate in.” In this scenario, rate in refers to the positive change in
concentration. That is, what factors are causing the concentration of the salt water to
increase, and how can we quantify them? You’ll notice that the only thing adding salt to
the system is the solution being pumped into the tank. Since the solution being pumped
gal
in has concentration 1 lb and it is being pumped in at a rate of 3 min , this tells us that
gal
the rate of the salt entering the tank is
lb
gal
lb
×3
=3
.
1
gal
min
min
Date: July 8, 2012.
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A GUIDE BY CHRISTIAN WOODS
But this describes the amount of salt entering the system. The “rate in” that we need
should describe the concentration. To get this, we need to divide the amount of salt entering
the tank by the volume of water already in the tank. So what is the volume of water in
the tank at any given time t?
To figure this out, we need to compare how much fluid is entering the tank and how much
is leaving the tank. In the course of a minute, we are told that 3 gal of solution enters
the tank and 2 gal of solution is expelled. This means that for each minute that passes,
the volume increases by 1 gal. So, if we wait t minutes, there should be t more gallons of
solution in the tank than there originally was. In other words, at time t there is 200 + t
gal of solution in the tank. Therefore, our differential equation becomes
3
dC
=
− rate out.
dt
200 + t
Now let’s find the “rate out.” How is the concentration of solution in the tank decreasing?
The most obvious contributing factor is the solution being pumped out of the tank. This
gal
happens at a rate of 2 min . To figure out how this changes the solution in the tank, we
need to find out how much salt is leaving and then convert this into a concentration. But
the amount of salt leaving the tank depends on the concentration of the solution in the
tank, the quantity we defined as C. So the rate of the amount of salt leaving the tank
is
gal
lb
lb
×2
= 2C
.
C
gal
min
min
We are faced with the same problem as above: we want to know how the concentration is
changing. Thus we divide this quantity by the volume of the solution in the tank, which
is again 200 + t gal.
However, the concentration is decreasing in another way. Besides the fact that salt is
leaving the tank, extra water is entering the tank (one gallon extra per minute, to be
exact). This will dilute the solution. Each gallon of the solution in the tank will have
to “share” its current concentration with a new gallon each minute, so the rate out also
comprises
lb
lb
C
gal
C
gal
gal
×1
=
200 + t gal
min
200 + t min
Plugging this information in, we complete our differential equation model:
dC
3
2C
C
3 − 3C
=
−
−
=
.
dt
200 + t 200 + t 200 + t
200 + t
All that is left to do is find an initial condition. Since we are told that the tank starts out
with 200 gal of solution, containing 100 lb of salt, the initial concentration is
100lb
1 lb
=
.
200gal
2 gal
MODELING CONCENTRATION OF A SOLUTION (# 2.3.4)
3
So C(0) = 12 .
2. Solving the Differential Equation
We are faced with the initial value problem
dC
3 − 3C
1
=
, C(0) = .
dt
200 + t
2
Notice that this differential equation is separable. Applying the solution technique, we
get
3 − 3C
dC
=
dt
200 + t
dC
−3(C − 1)
=
dt
200 + t
dC
−3
=
dt
C −1
200 + t
ln |C − 1| = −3 ln |200 + t| + k
−3
ln |C − 1| = ln |200 + t|
=⇒
=⇒
=⇒
=⇒
+ k,
where k is some arbitrary constant of integration. Now we can exponentiate both sides,
and we get the following general solution:
C(t) = 1 + K(200 + t)−3 ,
where K is also an arbitrary constant. To find the value of K that applies to our system,
3
2
we use the initial condition C(0) = 12 . This gives us K = − 200
2 = −100(200 ).
So the concentration of the solution in the tank at any time t (before the tank overflows)
is
C(t) = 1 − 100(2002 )(200 + t)−3 .
Since the amount of salt in lbs (call this value A(t)) is just the concentration of the solution
times its volume, we get
A(t) = (200 + t)C(t)
= (200 + t) 1 − 100(2002 )(200 + t)−3
= 200 + t − 100(2002 )(200 + t)−2 .
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A GUIDE BY CHRISTIAN WOODS
3. Behavior of the System
Equipped with formulas for C(t) and A(t), we can answer some questions about the system.
What is the concentration of the solution in the tank just before it overflows?
Since the tank can hold 500 gallons, it will begin to overflow when the volume is exactly
500 gal. We noticed before that the volume of the solution at time t is 200 + t. Solving
the equation
200 + t = 500
tells us that the tank will begin to overflow at 300 minutes. Thus the concentration at that
time is
C(300) = 1 − 100(2002 )(200 + 300)−3
= 1 − 100(2002 )(500)−3
121 lb
.
=
125 gal
What is the theoretical limiting concentration if the tank has infinite capacity?
If we assume that the tank can hold an infinite volume of solution, then we can use the
formula for C(t) for any nonnegative value of t, no matter how large. Therefore the
“limiting concentration” of the solution would be
i
h
lim C(t) = lim 1 − 100(2002 )(200 + t)−3
t→∞
t→∞
100(2002 )
t→∞ (200 + t)3
=1−0
lb
=1 .
gal
= 1 − lim
Notice that this is exactly the concentration of the solution being pumped into the tank.
Also notice that the concentration of the solution in the tank after 300 minutes is pretty
close to this theoretical limiting value.