Filtration - Constant Pressure & Constant Rate Operation

Home -> ChE Learning Resources -> Solved Problems -> Mechanical Operations->

For filtration operation,

\(\displaystyle \frac{d\theta}{dV} = \frac{2V}{C} + \frac{2V_f}{C} \qquad (1) \)

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,

\(\displaystyle \begin{align*}\frac{d\theta}{dV} &= \frac{2V}{C} \\ \frac{d\theta}{dV} &= \frac{2V}{C_2\Delta P} \qquad (3) \end{align*} \)

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

Last Modified on: 04-Feb-2022

Chemical Engineering Learning Resources - msubbu
e-mail: learn[AT]msubbu.academy
www.msubbu.in