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Thread: Integration salt diffusion problem

  1. #1
    Jun 2009

    Exclamation Integration salt diffusion problem

    I worked through this an am not confident with my answer:

    An aquarium pool has volume 2x10^6 litres. The pool initially contains fresh water. At t=0 minuts, water containing 10 gms/litre of salt is poured into the pool at a rate of 60 litre/min. The salt water instantly mixes with the fresh water and excess mixture is drained out of the bottom of the pool at the same rate (60 litres/min). Let S(t) = mass of salt in pool at time t.

    a. solve for S(t)

    b. What happens to S(t) as t --> infinity?

    My answer;
    2x10^6 - 2x10^6 e^(-3t/100,000) = S(t)

    therefore as t approaches infinity, S(t) approaches 2x10^6, but I don't think this makes sense with the story problem...

    HELP please!!!
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  2. #2
    MHF Contributor Calculus26's Avatar
    Mar 2009
    first you would expect the final answer to be the volume times

    the concentration of salt being added

    2*10^6 * 10 = 2* 10^7gms

    The IVP is dS/dt = Salt in - Salt out

    dS/dt = 10gms/litre*60litres/min - S(t)gms/(2*10^6litres)*60(litres/min)

    dS/dt = 600 - 3*10^(-5)S to make things easier let a = 3*10^(-5)

    dS/dt = 600 - a S

    you should obtain S= 2*10^7 - 2*10^7 e^[-3*10^(-5)t]

    The limit is 2*10^7 as expected
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