Question

A flow rate of Va (in) = 300 [mL / min] of water is mixed in a static mixer with an aqueous solution containing an active ingredient B, at the concentration Cb (in) = 0.2 [mM]. Water density = 1003 [kg / m ^ 3], density of the aqueous solution = 1010 [kg / m ^ 3]

Calculate the volumetric flow rate, V (in) in [mL / min] of the aqueous solution containing the active ingredient B entering the static mixer, to obtain, upon exiting the mixer, an aqueous solution with a final concentration of active principle equal to C (out) = 9.5 [microM]

**i assume density mixed solution =1010 kg/m^3**

Answer 1

let the molecular weight of B is M_B.

mass = no of moles x molecular weight.

molarity M = no of moles / Litre of solution.

given volumetric flowrate of water = 300 mL / min

mass flowrate of water entering = 300 x 10^-6 x 1003 = 0.3009 kg/min.

by mass balance :

mass entering the mixer = mass leaving the mixer.

mass entering the mixer = water stream + ingredient stream. = 0.3009 + V_in x 1010

mass flowrate = volumetric flowrate x density

mass leaving the mixer = V (out) x 1010

0.3009 + V_in x 1010 = V (out) x 1010 --------------eq-1

since no reaction is happening,

ingredient B is also conserved,

mass of B entering = mass of B leaving

C_b x V(in) x M_B x 1010 = C(out) x V(out) x  M_B x 1010

0.2 mM is equal to 0.2 mmols/liter of solution = 0.2 x 10^-3/10^-3 m³ = 0.2 mol/m³.

9.5 micro M is equal to 9.5 x 10^-6/10^-3 m³ = 9.5 x 10^-3 mol/m³.

C_b x V(in) = C(out) x V(out)

0.2 x V(in) = 0.0095 x V(out)

V(out) = 21.053 V(in) ----------------eq-2

from eq-1 and eq-2, we get

0.3009 + V_in x 1010 = V (out) x 1010

0.3009 = 20.053 V(in) x 1010

we get volumetric flowrate V(in) = 1.486 x 10^-5 m³/min = 0.0148 L/min = 14.86 mL/min.