Filtration - Constant Pressure & Constant Rate Operation

A filtration is carried out for 10 min at a constant rate in a leaf filter and thereafter it is continued at constant pressure. This pressure is that attained at the end of the constant rate period. If one quarter of the total volume of the filtrate is collected during the constant rate period, what is the total filtration time? Assume that the cake is incompressible and the filter medium resistance is negligible.

Calculations:

For filtration operation,

${\displaystyle \frac{��}{��}=\frac{2�}{�}+\frac{2{�}_{�}}{�}(1)}$

where C is a function of ΔP and mean specific cake resistance (α).

i.e., C = C₁ΔP/α   → 2

C₁ is a constant.

Specific cake resistance α is related to the compressibility coefficient (s) by the relation

α = α_o(ΔP)^s

where α_o is a constant.

Equn.2 can be written as C = C₂ΔP where C₂ is a constant

For an incompressible cake, s = 0; in other words, α is a constant.

If filter medium resistance is negligible, then V_f = 0.

Now the Equn.1 can be written in the form,

${\displaystyle \begin{array}{rl}\frac{��}{��}& =\frac{2�}{�}\\ \frac{��}{��}& =\frac{2�}{{�}_2\mathrm{\Delta }�}(3)\end{array}}$

For constant rate filtration, dq/dV = Constant = θ/V

Given: θ = 10 min for V = 0.25 V_T

Where V_T = total volume of filtrate (filtrate collected in constant rate operation and constant pressure operation)

From Equn.3

2V/(C₂ΔP) = 10/(0.25V_T) = 40/V_T

i.e., ΔP = 2V/(40C₂/V_T) = V V_T/(20C₂)

DP at the end of constant rate operation = 0.25 V_T² / (20 C₂) = 0.0125 V_T²/C₂

For the given problem conditions C₂ = Constant.,

and ΔP for the constant pressure operation is that obtained at the end of the constant rate operation.

From Equn.3,

dθ/dV = 2V/(C₂ΔP) = 2V/(0.0125 V_T²) → 4

Integrating Equn.4 between the limits V = 0.25V_T to V_T and θ = 10 min to θ_T

∫ dθ = (2/0.0125V_T²) ∫ VdV

θ_T - 10 = (80/V_T²) [V_T² - (0.25V_T)²]

θ_T = 10 + (80/V_T²) x 0.9375 x V_T2

θ_T = 10 + 75 = 85 minutes

Total filtration time = 85 minutes