A nitric acid solution flows at a constant rate of 5L/min into a large tank that initially held 200L of a 0.5% nitric acid solution. The solution inside the tank is kept well stirred and flows out of the tank at a rate of 10L/min. If the solution entering the tank is 10% nitric acid, determine the volume of nitric acid in the tank after t minutes.

x(t) = −39e

−0.03t + 40.

Explanation:

Let V (t) be the volume of solution (water and

nitric acid) measured in liters after t minutes. Let x(t) be the volume of nitric acid

measured in liters after t minutes, and let c(t) be the concentration (by volume) of

nitric acid in solution after t minutes.

The volume of solution V (t) doesn’t change over time since the inflow and outflow

of solution is equal. Thus V = 200 L. The concentration of nitric acid c(t) is

c(t) = x(t)

V (t)

=

x(t)

200

.

We model this problem as

dx

dt = I(t) − O(t),

where I(t) is the input rate of nitric acid and O(t) is the output rate of nitric acid,

both measured in liters of nitric acid per minute. The input rate is

I(t) = 6 Lsol.

1 min

·

20 Lnit.

100 Lsol.

=

120 Lnit.

100 min

= 1.2 Lnit./min.

The output rate is

O(t) = (6 Lsol./min)c(t) = 6 Lsol.

1 min

·

x(t) Lnit.

200 Lsol.

=

3x(t) Lnit.

100 min

= 0.03 x(t) Lnit./min.

The equation is then

dx

dt = 1.2 − 0.03x,

or

dx

dt + 0.03x = 1.2, (1)

which is a linear equation. The initial condition condition is found in the following

way:

c(0) = 0.5% = 5 Lnit.

1000 Lsol.

=

x(0) Lnit.

200 Lsol.

.

Thus x(0) = 1.

In Eq. (1) we let P(t) = 0.03 and Q(t) = 1.2. The integrating factor for Eq. (1) is

µ(t) = exp Z

P(t) dt

= exp

0.03 Z

dt

= e

0.03t

.

The solution is

x(t) = 1

µ(t)

Z

µ(t)Q(t) dt + C

= Ce−0.03t + 1.2e

−0.03t

Z

e

0.03t

dt

= Ce−0.03t +

1.2

0.03

e

−0.03t

e

0.03t

= Ce−0.03t +

1.2

0.03

= Ce−0.03t + 40.

The constant is found using x(t) = 1:

x(0) = Ce−0.03(0) + 40 = C + 40 = 1.

Thus C = −39, and the solution is

x(t) = −39e

−0.03t + 40.

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