Question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 800 L of a dye solution with a concentration of 1 g/L. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 8 L/min, the well-stirred solution flowing out at the same rate. Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value. (Round your answer to one decimal place.)

Answer 1

t = 100 ln 100

Explanation:

D(t) : The amount of dye (in g) at time t (in min)

D(0) = 800 L * 1 g/L = 800 g

the change in D is:

`(dD(t))/(dt) =D_(in)- D_(out) \\D_(in): 0*8\ g/min \\D_(out): (D(t))/(800) *8\ g/min \\(dD(t))/(dt) = -(1)/(100)D(t)`

`(dD(t))/(D(t)) =-(1)/(100)dt \\\int\limits^(D(t))_(800) {(1)/(D(t)) } \, dD(t) =\int\limits^t_0 {t} \, dt \\ln((D(t))/(800))=-(1)/(100)t \\D(t) = 800e^{-(1)/(100)t} \\Solving\ D(t) = 0.01* D(0)=0.01*800 =8 \\8 = 800e^{-(1)/(100)t} \\ln ((1)/(100))=-(1)/(100)t \\100 ln 100 = t`