Thread: Integration salt diffusion problem

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

Question;
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!!!

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