Download Compartmental Analysis

3
Applications of 1st-Order Equations
3.2 – Compartmental Analysis
For a first example we shall analyze a one-compartment system, and then later consider a
two-compartment system.
Example 3.1. A brine solution (a mixture of salt and water) flows at a constant rate of 4
L/min into a tank that initially contains 100 L of pure water. The solution inside the tank is
kept well-stirred and flows out of the tank at a rate of 3 L/min. If the concentration of salt
in the brine entering the tank is 0.2 kg/L, determine the mass of the salt in the tank after t
minutes. Find when will the concentration of salt in the tank be 0.1 kg/L.
Solution. Let x(t) be the mass of salt, in kilograms, that is in the tank at time t. Since the
tank is initially filled with fresh water we know that x(0) = 0. See Figure 1.
In order to determine x(t) we will use what we know about the rate at which x(t) changes
over time, which is x0 (t). An expression for x0 (t) will be given as the rate salt enters the tank
minus the rate salt leaves the tank (in kilograms per minute). To do this we first need to know
the volume of the solution in the tank at time t. The volume is given to be 100 L initially, and
since 4 L of liquid enters the tank and 3 L leaves with each minute, we can see that at time t
the volume of solution in the tank must be 100 + t.
Now, the rate salt enters the tank is easily reckoned: 4 liters of brine is entering per minute,
there’s 0.2 kg of salt per liter, and so a total of 0.8 kg of salt is entering per minute. As for the
rate salt leaves, at time t there’s x(t) kg of salt in the tank, and we assume that it is uniformly
dissolved throughout the 100 + t liters of solution to give a concentration of x(t)/(100 + t) kg of
salt per liter. Since 3 liters of solution is leaving per minute, we conclude that 3x(t)/(100 + t)
4 L/min
0.2 kg/L
x(t) kg salt
x(0) = 0
Figure 1
3 L/min
2
kg of salt is leaving per minute. The derivation of x0 (t) is as follows:
x0 (t) = (rate salt enters Tank 1) − (rate salt leaves Tank 1)
0.2 kg
4L
x(t) kg
3L
=
−
1L
1 min
100 + t L
1 min
3x(t)
= 0.8 −
.
100 + t
Thus we have a linear first-order ODE:
3
x = 0.8.
x0 +
t + 100
To solve this equation, we multiply by the integrating factor
Z
3
dt = e3 ln |t+100|+c = (t + 100)3
µ(t) = exp
t + 100
to obtain
(t + 100)3 x0 + 3(t + 100)2 x = 0.8(t + 100)3 ,
which becomes
0
(t + 100)3 x = 0.8(t + 100)3
and thus
Z
0.8
(t + 100)4 + c.
4
From this we get a general explicit solution to the ODE,
1
c
x(t) = (t + 100) +
.
5
(t + 100)3
3
(t + 100) x =
0.8(t + 100)3 dt =
To determine c we use the initial condition x(0) = 0:
c
c
1
⇒
= −20 ⇒ c = −2 × 107 .
0 = (0 + 100) +
3
3
5
(0 + 100)
100
Thus, the mass of salt in the tank at time t is given by
1
2 × 107
x(t) = (t + 100) −
.
5
(t + 100)3
The concentration, C(t), of salt at time t is given by the mass x(t) at time t divided by the
volume t + 100 at time t. That is, C(t) = x(t)/(t + 100), so that
C(t) = 0.2 −
2 × 107
.
(t + 100)4
Now, to find when the concentration of salt is 0.1 kg/L we solve C(t) = 0.1, giving the equation
0.1 = 0.2 −
2 × 107
.
(t + 100)4
√
From this we find that t = 100 4 2 − 100 ≈ 18.9 minutes.
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The following example illustrates a physical system consisting of two compartments. The
general approach is to analyze the first compartment, then use the information garnered to
analyze the second compartment.
Example 3.2. Beginning at time t = 0, fresh water is pumped at a rate of 3 L/min into a
well-stirred tank that is initially filled with 60 L of brine. The increasingly less concentrated
salt solution flows at a rate of 3 L/min out a drain that feeds into a second tank initially filled
with 60 L of pure water. The resultant mixture of water and salt in the second tank, which
is also well-stirred, is pumped into the ocean at a rate of 3 L/min. Find the time when the
concentration of salt in the second tank is greatest, and compare the maximal concentration in
the second tank to the initial concentration in the first tank.
Solution. It is not actually necessary to know how much salt is initially in Tank 1 to determine
when the concentration of salt in Tank 2 is greatest. Let x(t) be the number of kilograms of
salt in Tank 1 at time t, and let y(t) be the number of kilograms of salt in Tank 2 at time t.
We have x(0) = x0 for some constant x0 , and also y(0) = 0 since the water in Tank 2 is initially
pure. See Figure 2. Now, noting that the volume of solution in Tank 1 is a constant 60 L, we
have
x0 (t) = (rate salt enters Tank 1) − (rate salt leaves Tank 1)
x(t) kg
3L
3x(t)
=0−
,
=−
60 L
1 min
60
1
which yields the equation x0 = − 20
x, also written as dx/dt = −x/20. This equation is separable,
giving
Z
Z
20
dx = − dt,
x
and hence
ln x20 = −t + c0
for arbitrary constant c0 . Exponentiating both sides and letting c1 = ec0 be an arbitrary positive
constant, we obtain
x20 = e−t+c0 = c1 e−t ,
and then x(t) = c1 e−t/20 . Using the initial condition x(0) = x0 , we substitute t = 0 and x = x0
into the equation to get x0 = c1 e0 = c1 , and thus
x(t) = x0 e−t/20 .
(1)
Now we turn our attention to Tank 2. Since the volume of solution in Tank 2 is always
60 L, we have
y 0 (t) = (rate salt enters Tank 2) − (rate salt leaves Tank 2)
x(t) kg
3L
y(t) kg
3L
−
=
60 L
1 min
60 L
1 min
=
x(t) y(t)
x0 e−t/20 − y(t)
−
=
20
20
20
4
3 L/min
0 kg/L
3 L/min
1
x(t) kg/L
60
60 L solution
x(t) kg salt
x(0) = x0
60 L solution
y(t) kg salt
y(0) = 0
3 L/min
1
y(t) kg/L
60
Figure 2
where the last equality follows from (1). Hence we have the equation
x0
1
(2)
y 0 + y = e−t/20 ,
20
20
which is a first-order linear ODE and so can be solved by finding an appropriate integrating
factor µ(t). We have
Z
1
µ(t) = exp
dt = et/20
20
t/20
and so, multiplying (2) by e , we obtain
1
x0
y 0 et/20 + yet/20 = ,
20
20
which can be written
0 x0
yet/20 = ,
20
and therefore
Z
x0
x0
yet/20 =
dt = t + c.
20
20
Using the initial condition y(0) = 0, we substitute t = 0 and y = 0 into this equation to find
that c = 0, and at last we have an expression for y(t):
x0
y(t) = te−t/20 .
20
The concentration of salt in Tank 2 at time t, C(t), is given by C(t) = y(t)/60; that is,
x0 −t/20
C(t) =
te
.
1200
To determine when the concentration is greatest, we must find t > 0 for which C(t) attains a
global maximum value on (0, ∞).1 This entails finding t for which C 0 (t) = 0; that is, we must
solve
x0
x0 −t/20
e
−
te−t/20 = 0.
1200
24, 000
But this equation immediately implies that 20 − t = 0, and hence t = 20 minutes.
1Note
that, since the volume of solution in Tank 2 is a constant 60 L, we could just as well determine when
y(t) attains a maximum.
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Finally, at time t = 20 minutes we find that the mass of salt in Tank 2 is
x0
1
y(20) = (20)e−20/20 = x0 .
20
e
That is, Tank 2 is at most 1/e times as salty as Tank 1 was initially.
Observe that x(t) → 0 and y(t) → 0 as t → ∞, as is to be expected since fresh water is
ultimately displacing brine in the system.
Problems concerning the decay of one or more radioactive isotopes also lend themselves to
compartmental analysis. In the simplest case there is a single radioactive substance that decays
into some other element, so that a “compartment” is a sample of the substance. No quantity
of the substance is entering the compartment, but gradually over time atoms of the substance
are leaving (i.e. fissioning into other material). Empirical data make one thing clear: the rate
of decay of a radioactive substance is directly proportional to the amount of the substance that
is present. Thus, if x(t) is the amount of the substance (in grams, say), then x0 (t) is the rate
of change of the amount of the substance, with
x0 (t) = kx(t)
for some constant of proportionality k < 0 that depends on the isotope under consideration.
The half-life of a radioactive isotope, which is the time it takes for half of it to decay, can vary
from nanoseconds to gigayears.
Example 3.3. Cobra Commander has 260 grams of kaboomium-320 (chemical symbol 320 Ka)
in the basement of his secret hideout. Upon returning from a carefree five-hour drive with
Destro in the countryside in his spiffy new Nissan Cube, he finds that 192 grams remain. After
how many hours will only 10 grams remain?
Solution. Let x(t) be the amount of 320 Ka in grams at time t in hours, where x(0) = 260 and
x(5) = 192. The rate at which 320 Ka decays is proportional to the amount present, which is to
say that x0 (t) = kx(t) for some constant k. This is a separable ODE which becomes
Z
Z
1
dx = k dt,
x
and thus ln |x| = kt + c. Of course, x(t) is never negative, so we may write ln(x) = kt + c,
whence we obtain x(t) = ekt+c = x0 ekt (letting x0 = ec ). From the initial condition x(0) = 260
comes 260 = x0 e0 , so that x0 = 260 and hence
x(t) = 260ekt .
(3)
It remains to find k. Fortunately we have another bit of information available: x(5) = 192.
Putting this into (3) gives 192 = 260e5k , whence
k = 0.2 ln(192/260) ≈ −0.0606
and we fully determine x(t) to be
x(t) = 260e−0.0606t .
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With this model in hand we can ascertain how many hours it will be before only 10 g of
remains for Cobra Commander to play with: set x(t) = 10 to get
320
Ka
260e−0.0606t = 10,
and thus
1
1
t=
ln
≈ 53.8.
−0.0606
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That is, after about 53.8 hours only 10 g of kaboomium-320 will remain. We assume that Cobra
Commander knows his business. And knowing is half the battle!