Question

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?

Answer 1

Answer:

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