Compartmental Analysis

[starflakz]

A swimming pool whose volume is 10,000 gal contains water that is 0.01% chlorine. Starting at t=0 city water containing 0.001%chlorine is pumped into the pool at a rate of 5 gal/min. The pool water flows out at the same rate. What is the percentage of chlorine in the pool after 1h?

dx/dt=(5gal/min)(0.01%)-(5gal/min)(x/10000)
dx/dt+5x/10000=.005
calculated integrating factor e^(5t/10000)
e^(5t/10000)*x=int(.005e^(5t/10000))
e^(5t/10000)*x=10e^(5t/10000)+C
x(t)=10+Ce^(-5t/10000)
x(0)=0.01 --->C=-9.99
x(t)=10-9.99e^(-5t/10000)
x(60)=0.305
.305/10000 = .0000305% back of the book says is is 0.0097%

Where did I go wrong??

 

[Deleted member 4993]

A swimming pool whose volume is 10,000 gal contains water that is 0.01% chlorine. Starting at t=0 city water containing 0.001%chlorine is pumped into the pool at a rate of 5 gal/min. The pool water flows out at the same rate. What is the percentage of chlorine in the pool after 1h?

dx/dt=(5gal/min)(0.01%)-(5gal/min)(x/10000)
dx/dt+5x/10000=.005
calculated integrating factor e^(5t/10000)
e^(5t/10000)*x=int(.005e^(5t/10000))
e^(5t/10000)*x=10e^(5t/10000)+C
x(t)=10+Ce^(-5t/10000)
x(0)=0.01 --->C=-9.99
x(t)=10-9.99e^(-5t/10000)
x(60)=0.305
.305/10000 = .0000305% <<< You know it could not be less than 0.001%


back of the book says is is 0.0097%

Where did I go wrong??

 

[BigGlenntheHeavy]

$\displaystyle Dr. \ Hippo \ to \ the \ rescue \ with \ his \ amazing \ formula, \ to \ wit:$

$\displaystyle \frac{dy}{dt}+\frac{r_2y}{v_o+(r_1-r_2)t} \ = \ q_1r_1, \ note \ y \ = \ amt. \ of \ chlorine \ dissolved \ in \ pool.$

$\displaystyle Hence, \ we \ have \ \frac{dy}{dt}+\frac{5y}{10,000} \ = \ (.00001)(5)$

$\displaystyle \frac{dy}{dt}+\frac{y}{2,000} \ = \ .00005 \ \implies \ \frac{dy}{dt} \ = \ \frac{.1-y}{2,000}$

$\displaystyle Hence, \ \int\frac{dy}{.1-y} \ = \ \int\frac{dt}{2,000}, \ \implies \ ln|.1-y| \ = \ \frac{-t}{2,000}+C$

$\displaystyle |.1-y| \ = \ (e^{-t/2,000})(e^{C}), \ \implies \ .1-y \ = \ Ae^{-t/2,000}, \ A \ = \ \pm e^{C}$

$\displaystyle Ergo, \ y(t) \ = \ .1-Ae^{-t/2,000}, \ \implies \ y(0) \ = \ 1 \ = \ .1-A, \ \implies \ A \ = \ -.9$

$\displaystyle Therefore \ y(t) \ = \ .1+.9e^{-t/2,000}, \ \implies \ y(60) \ = \ .1+.9e^{-60/2000} \ \dot= \ .973 \ gal.$

$\displaystyle Ergo, \ \frac{.973}{10,000} \ = \ .0000973 \ = \ .00973\% \ of \ chlorine \ is \ in \ the \ pool \ after \ 1 \ hour \ (60 \ minutes).$