A swimming pool whose volume is 10 comma 000 gal contains water that is 0.03 % chlorine. Starting at tequals 0, city water containing 0.001 % chlorine is pumped into the pool at a rate of 6 gal/min. The pool water flows out at the same rate. What is the percentage of chlorine in the pool after 1 hour ? When will the pool water be 0.002 % chlorine?

C (60) = 2.7*10⁻⁴

t = 1870.72 s

Step-by-step explanation:

Let x (t) be the amount of chlorine in the pool at time t. Then the concentration of chlorine is

C (t) = 3*10⁻⁴*x (t).

The input rate is 6 * (0.001/100) = 6*10⁻⁵.

The output rate is 6*C (t) = 6 * (3*10⁻⁴*x (t)) = 18*10⁻⁴*x (t)

The initial condition is x (0) = C (0) * 10⁴/3 = (0.03/100) * 10⁴/3 = 1.

The problem is to find C (60) in percents and to find t such that 3*10⁻⁴*x (t) = 0.002/100.

Remember, 1 h = 60 minutes. The initial value problem is

dx/dt = 6*10⁻⁵ - 18*10⁻⁴x = - 6 * 10⁻⁴ * (3x - 10⁻¹) x (0) = 1.

The equation is separable. It can be rewritten as dx / (3x - 10⁻¹) = - 6*10⁻⁴dt.

The integration of both sides gives us

Ln |3x - 0.1| / 3 = - 6*10⁻⁴*t + C or |3x - 0.1| = e∧ (3C) * e∧ (-18*10⁻⁴t).

Therefore, 3x - 0.1 = C₁*e∧ (-18*10⁻⁴t).

Plug in the initial condition t = 0, x = 1 to obtain C₁ = 2.9.

Thus the solution to the IVP is

x (t) = (1/3) (2.9*e∧ (-18*10⁻⁴t) + 0.1)

then

C (t) = 3*10⁻⁴ * (1/3) (2.9*e∧ (-18*10⁻⁴t) + 0.1) = 10⁻⁴ * (2.9*e∧ (-18*10⁻⁴t) + 0.1)

If t = 60

We have

C (60) = 10⁻⁴ * (2.9*e∧ (-18*10⁻⁴*60) + 0.1) = 2.7*10⁻⁴

Now, we obtain t such that 3*10⁻⁴*x (t) = 2*10⁻⁵

3*10⁻⁴ * (1/3) (2.9*e∧ (-18*10⁻⁴t) + 0.1) = 2*10⁻⁵

t = 1870.72 s