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Question: How do you determine what percentage of chlorine is in the pool after \(1\) hour if a pool whose vol...

How do you determine what percentage of chlorine is in the pool after 11 hour if a pool whose volume is 1000010000 gallons contains water the is 0.01%0.01\% chlorine and starting at t=0t = 0 city water containing 0.001%0.001\% chlorine is pumped into the pool at a rate of 5gal  min15\,gal\;{\min ^{ - 1}} while the pool water flows out at the same rate?

Explanation

Solution

To solve this question we have to assume the amount of chlorine in the pool at the time tt be xx. By using the rate of change of chlorine in the pool, we will find the amount of the chlorine. After that we find the required answer using it.

Complete answer:
Let us take xx the amount of chlorine that is in the pool at time tt.
Then the rate of change of xx is:
dxdt=(rate  in)(rate  out)\dfrac{{dx}}{{dt}} = \left( {rate\;in} \right) - \left( {rate\;out} \right)
dxdt=ci×finx(t)10000×fout\dfrac{{dx}}{{dt}} = {c_i} \times {f_{in}} - \dfrac{{x\left( t \right)}}{{10000}} \times {f_{out}}
ci{c_i} is the concentration of chlorine of the inflow, 0.001%0.001\% . fin{f_{in}} is the inflow, 5  galmin15\;gal\,{\min ^{ - 1}}and fout{f_{out}} is the outflow, 5  gal  min15\;gal\;{\min ^{ - 1}}.
dxdt=0.001100×5x10000×5\dfrac{{dx}}{{dt}} = \dfrac{{0.001}}{{100}} \times 5 - \dfrac{x}{{10000}} \times 5
=120000x2000= \dfrac{1}{{20000}} - \dfrac{x}{{2000}}
x0.12000\Rightarrow - \dfrac{{x - 0.1}}{{2000}}
dxx0.1=dt2000\Rightarrow \dfrac{{dx}}{{x - 0.1}} = \dfrac{{ - dt}}{{2000}}
lnx0.1=t2000+c\Rightarrow \ln \left| {x - 0.1} \right| = \dfrac{{ - t}}{{2000}} + c
x0.1=c×et2000\Rightarrow x - 0.1 = c \times {e^{ - \dfrac{t}{{2000}}}}
n t=0t = 0, the concentration is 0.01%0.01\% , so amount of chlorine xx is
0.01100×10000=1  gal\dfrac{{0.01}}{{100}} \times 10000 = 1\;gal
Accordingly in the equation, we have
x=0.1+C×et2000x = 0.1 + C \times {e^{\dfrac{{ - t}}{{2000}}}}
1=0.1+c×e020001 = 0.1 + c \times {e^{ - \dfrac{0}{{2000}}}}
1=0.1+c×1\Rightarrow 1 = 0.1 + c \times 1
c=0.9\Rightarrow c = 0.9
The amount of chlorine for any time tt is then
x(t)=0.1+0.9×et2000x\left( t \right) = 0.1 + 0.9 \times {e^{ - \dfrac{t}{{2000}}}}
At one hour (t=60  min)\left( {t = 60\;\min } \right), the amount of chlorine is,
x(60)=0.1+0.9×e  602000x\left( {60} \right) = 0.1 + 0.9 \times {e^{ - \;\dfrac{{60}}{{2000}}}}
=0.1+0.9×0.9704= 0.1 + 0.9 \times 0.9704
=0.973= 0.973
This amount means a concentration of
c=A10000=0.97310000c = \dfrac{A}{{10000}} = \dfrac{{0.973}}{{10000}}
=0.00973%= 0.00973\%
This is the required answer.

Note:
Remember to check if there is no error in the calculation. Also look at the value of each and every term whether it is correct or wrong. It is also important to check whether the formula we are using is correct or wrong. After checking each of the above-mentioned problems we can get the right answer.