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Question

A rectangular vessel is divided into two equal compartments by a vertical porous membrame. Liquid in one compartment. intially at a depth of 20cm, passes into the other compartment, intially empty, at a rate proportional to the difference in levels.

If the level in the second compartment rises 2cm in the first 5 minutes, after what time with the difference in levels be 2cm.

Answer 1

let y = height of liquid in first compartment

(20 - y) = ht of liquid in 2nd

difference = 2y - 20

thus

dy/dt = k(2y - 20) = 2k(y - 10) , y ≥ 10

dy/(y - 10) = 2k dt

ln(y - 10) = 2kt + C

t = 5, y = 18

thus

ln(8) = 10k + C

t = 0, y = 20

ln(10) = C

ln(y - 10) = (ln8 - ln10)t/5 + ln10

if the difference in levels is 2cm , y = 11

solve for t .. . .