Question

A saline solution with concentration 3 % flows at a constant rate into a tank with capacity 1500 gallons, initially full of a 7 % saline solution. The solutions remain well-mixed and the rate of flow of mixed solution out of the tank is twice the rate of flow into the tank.

What constant rate of flow into the tank will dilute the solution in the tank to 4 % in 16 hours?

How much solution will be in the tank at that time?

Answer 1

let VO= volumetric flow rate of tank

from mass balance ( assuming the density of solution through out the change in 1 g/cc)

flow rate of solution in= Flow rate of solution out+ rate of accumulation

let CA = Concentration at any time in % , Vo =Volumetric flow rate in Gallons/hr

Vo* 3/100 = 2VO*CA/100 + d/dt*(1500*CA/100)

Vo*3= 2Vo*CA+ 1500* dCA/dt

-1500* dCA/dt = Vo*(2CA-3)

when integrated

-1500*dCA/(2CA-3) = Vo*dt

dCA/(2CA-3)= -(VO/1500)*dt

when integrate ln (2CA-3)/2= -(VO/1500)*t+C, where C is integration constant

at t=0, CA=7

ln(2*7-3)/2= C, C=1.2

hence ln (2CA-3)/2= -(VO/1500)*t +1.2

at t=16 hr, CA= 4

hence ln(2*4-3)/2 = (VO/1500)*16 +1.2

(VO/1500)*16= 0.80

Vo= 0.8*1500/16 = 75 gallons/hr