Instrumental technique

A problem for which the one-compartment system provides a useful representation is the mix-ing of fluids in a tank. Let represent the amount of a substance in a tank (compartment) at time t.To use the compartmental analysis model, we must be able to determine the rates at

xAtB

which this substance enters and leaves the tank. In mixing problems one is often given the rate at which a fluid containing the substance flows into the tank, along with the concentration of the substance in that fluid. Hence, multiplying the flow rate (volume/time) by the concentration (amount/volume) yields the input rate (amount/time).

The output rate of the substance is usually more difficult to determine. If we are given the exit rate of the mixture of fluids in the tank, then how do we determine the concentration of the substance in the mixture? One simplifying assumption that we might make is that the concen-tration is kept uniform in the mixture. Then we can compute the concenconcen-tration of the substance in the mixture by dividing the amount by the volume of the mixture in the tank at time t.

Multiplying this concentration by the exit rate of the mixture then gives the desired output rate of the substance. This model is used in Examples 1 and 2.

Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to flow at a constant rate of 6 L/min. The solution inside the tank is kept well stirred and is flowing out of the tank at a rate of 6 L/min. If the concentration of salt in the brine entering the tank is 0.1 kg/L, determine when the concentration of salt in the tank will reach 0.05 kg/L (see Figure 3.2).

We can view the tank as a compartment containing salt. If we let denote the mass of salt in the tank at time t, we can determine the concentration of salt in the tank by dividing by the volume of fluid in the tank at time t.We use the mathematical model described by equation (1) to solve for .

First we must determine the rate at which salt enters the tank. We are given that brine flows into the tank at a rate of 6 L/min. Since the concentration is 0.1 kg/ L, we conclude that the input rate of salt into the tank is

(2)

We must now determine the output rate of salt from the tank. The brine solution in the tank is kept well stirred, so let’s assume that the concentration of salt in the tank is uniform.

That is, the concentration of salt in any part of the tank at time tis just divided by the vol-ume of fluid in the tank. Because the tank initially contains 1000 L and the rate of flow into the tank is the same as the rate of flow out, the volume is a constant 1000 L. Hence, the output rate of salt is

(3)

The tank initially contained pure water, so we set . Substituting the rates in (2) and (3) into equation (1) then gives the initial value problem

(4)

as a mathematical model for the mixing problem.

dx

dt 0.6 3x

500 , xA0B0 ,

xA0B0 A6 L/minBc₁₀₀₀^{xAtB kg/L}d 3xAtB

500 kg/min .

xAtB A6 L/minB A0.1 kg/ LB0.6 kg/min .

xAtB

xAtB xAtB

xAtB

92 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

x(t) 1000 L x (0) = 0 kg 6 L/min

0.1 kg/L

6 L/min

Figure 3.2 Mixing problem with equal flow rates

Example 1

Solution

The equation in (4) is separable (and linear) and easy to solve. Using the initial condition to evaluate the arbitrary constant, we obtain

(5)

Thus, the concentration of salt in the tank at time tis

To determine when the concentration of salt is 0.05 kg/ L, we set the right-hand side equal to 0.05 and solve for t.This gives

and hence

min .

Consequently the concentration of salt in the tank will be 0.05 kg/ L after 115.52 min. _◆ From equation (5), we observe that the mass of salt in the tank steadily increases and has the limiting value

kg .

Thus, the limiting concentration of salt in the tank is 0.1 kg/ L, which is the same as the con-centration of salt in the brine flowing into the tank. This certainly agrees with our expectations!

It might be interesting to see what would happen to the concentration if the flow rate into the tank is greater than the flow rate out.

For the mixing problem described in Example 1, assume now that the brine leaves the tank at a rate of 5 L/min instead of 6 L/min, with all else being the same (see Figure 3.3). Determine the concentration of salt in the tank as a function of time.

The difference between the rate of flow into the tank and the rate of flow out is

1 L/min, so the volume of fluid in the tank after tminutes is L. Hence, the rate at which salt leaves the tank is

A5 L/minBc₁₀₀₀^xAtB_t ^{kg/ L}d 5xAtB

1000t kg/min .

A1000tB 65

tSqlim xAtB lim

tSq 100A1e^3t

/

t 500 ln 2

3 115.52

0.1

A

/

B

/

xAtB

10000.1

A

/

B

xAtB100A1e^3t

/

xA0B0

Section 3.2 Compartmental Analysis 93

x(t)

? L x (0) = 0 kg 6 L/min

0.1 kg/L

5 L/min

Figure 3.3 Mixing problem with unequal flow rates

Example 2

Solution

Using this in place of (3) for the output rate gives the initial value problem (6)

as a mathematical model for the mixing problem.

The differential equation in (6) is linear, so we can use the procedure outlined on page 48 to solve for . The integrating factor is Thus,

Using the initial condition , we find and thus the solution to (6) is

Hence, the concentration of salt in the tank at time tis

(7) kg/ L . ◆

As in Example 1, the concentration given by (7) approaches 0.1 kg/ L as . How-ever, in Example 2 the volume of fluid in the tank becomes unbounded, and when the tank begins to overflow, the model in (6) is no longer appropriate.