A swimming pool whose volume is 10 comma 000 gal contains water that is 0.03% chlorine. Starting at tequals0, 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?
Answers
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
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