Question

A swimming pool whose volume is 10,000 gal contains water that is 0.01% chlorine. Starting at t = 0, city water containing0.001% chlorine is pumped into the pool at a rate of 5 gal/min. The pool water flows out at the same rate. What is thepercentage of chlorine in the pool after 1 h? When will the pool water be 0.002% chlorine?

Answer 1

Answer:

a) The percentage of chlorine after 1 hour is 0.00973%.

b) The pool water will habe a concentration of 0.002% chlorine at 4394 minutes (or 73.24 hours).

Step-by-step explanation:

We can define as X(t) the amount of chlorine that is in the pool at time t.

Then, the rate of change of X can be written as

$\frac{dX}{dt}=(rate\,in)-(rate\,out)\\ \\\frac{dX}{dt}=C_i*f_{in}-\frac{X(t)}{10,000}*f_{out}$

being Ci the concentration of chlorine of the inflow (0.001%), f_in the inflow (5 gal/min) and f_out the outflow (the same as the inflow, 5 gal/min).

$\frac{dX}{dt}=\frac{0.001}{100}*5-\frac{X}{10000}*5=\frac{1}{20000}-\frac{X}{2000}=-\frac{X-0.1}{2000}   \\\\\frac{dX}{X-0.1} =\frac{-dt}{2000}\\\\ln|X-0.1|=\frac{-t}{2000} +C\\\\X-0.1=C*e^{-t/2000}\\\\X=0.1+C*e^{-t/2000}$

When t=0, the concentration is 0.01%, so the amount of chlorine X is

$X=\frac{0.01}{100}*10000=1\,gal$

Replacing in the equation, we have

$X=0.1+C*e^{-t/2000}\\\\1=0.1+C*e^{-0/2000}=0.1+C*1\\\\C=0.9$

The amount of chlorine for any time t is then

$X(t)=0.1+0.9*e^{-t/2000}$

a) At one hour (t=60 min), the amount of chlorine is

$X(60)=0.1+0.9*e^{-60/2000}=0.1+0.9*0.9704=0.973$

This amount means a concentration of

$C=A/10000=0.973/10000=0.00973 \%$

b) A concentration of 0.002% of chlorine means an amount of chlorine of

$X=(0.002/100)*10000=0.2\,gal$

Then we can calculate

$X(t)=0.2=0.1+0.9*e^{-t/2000}\\\\e^{-t/2000}=(0.2-0.1)/0.9=0.1111\\\\-t/2000=ln(0.111)\\\\t=-2000*ln(0.1111)=4394\, min$

Answer 2

Answer:

The answer above is correct, it proves that the chlorine would take an estimate time.

Step-by-step explanation: