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.