Question

A large tank, Volume = V (m³), is initially filled with V m³ of dark blue colored solution of Cu₂SO₄ of concentration C_O (Kg/m³). At what we will call time t=0 (s) very dilute light blue solution of solute Cu₂SO₄ at concentration C_F is fed into the bottom of the tank at volume flow rate of Q_in (m³/s). The incoming aqueous solution is completely mixed with the solution in the tank thus diluting the concentration of Cu₂SO₄ in the tank from C_O to C_I, the instantaneous concentration of the solution in the tank. Diluted Cu₂SO₄ solution with the instantaneous concentration of C_I flows out of the top of the tank at flow rate Q_out. Note: Q_in=Q_out.

a. Determine and plot (C_I-C_F)/(C_O-C_F)=f(t).

Answer 1

The concentration of of CuSO₄ at any time t can be given as,

Rate of change of concentration in the given Volume = Rate of in flow of CuSO₄ - Rate of out flow of CuSO₄

VdCI/dt = Qin*CF - Qout*CI

Qin = Qout;

VdCI/dt = Qin( CF- CI)

dCI/(CF- CI) = Qin/V * dt

On integrating,

ln(1/(CF - CI) = Qin/V * t + A

A = integration constant.

at t=0 ; CI = CO

ln(1/(CF - CO)) = A

ln(1/(CF - CI) = Qin/V * t + ln(1/(CF - CO))

ln(1/(CF - CI) - (1/(CF - CO) = Qin/V* t

ln[(CF-CO)/CF-CI)] = Qin/v * t

Rearranging and bringing CI term to the numerator,

ln[(CI-CF)/(CO-CF)] = -Qin/V * t

let Qin/V = k , whose units will be time^-1

(CI-CF)/(CO-CF) = e^-kt

Following is a plot of (CI-CF)/(CO-CF) Vs Time ,t

for k = 0.1 min^-1 ; CO = 5 kg/m³ ; CF = 0.1.