Mixing Solutions: Problems & Solutions Explained

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SUMMARY

The discussion focuses on solving mixing solution problems, specifically using a swimming pool example to illustrate the concept. A pool with a volume of 10,000 gallons contains water with 0.01% chlorine, while city water with 0.001% chlorine is pumped in at a rate of 5 gallons per minute. The differential equation dy/dt = (5 gal/min * 0.001%) - (y(t) / 10000 * 5 gal/min) is established to model the change in chlorine concentration over time. The goal is to determine the chlorine percentage after one hour and when it reaches 0.002%.

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TheLaughingMan
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Can anyone help me out with mixing solution problems, I know its not that hard I just am having trouble understanding what to do. Sometimes its X sometimes its X/volume of the tank. Can anyone explain it to me in a simple way?
 
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If you give us a sample problem maybe we could help break it donw for you?
 
john_in_pdx said:
If you give us a sample problem maybe we could help break it donw for you?
Sample:
A swimming pool whose volume is 10,000 gal contains water that is .01% chlorine. Starting at t=0, city water containing .001% Cl is pumped in at 5 gal/min. The pool water flows out at the same rate. What is the % of chlorine in the pool after 1 hr? When will the pool be .002% Cl ?
 
Can you post an attempt?, it'll figure out what you are not understanding.
 
it seems you don't know how to set up the prob. you need to find y(60).
to do this you let y(t)=the amount of cl in the pool at time t. you need to find an expression for dy/dt.

dy/dt = rate in - rate out

rate in =5 gal / min * .001%

rate out = y(t) / 10000 * 5 gal / min

so dy /dt = (5 gal/min * .001%) - (y(t) / 10000 * 5 gal/ min)

you then find the I.F. and intergrate
 

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