Cake filtration is widely used in industry to separate solid particles from suspension in liquid. It involves the build up of a bed or ‘cake’ of particles on a porous surface known as the filter medium. In cake filtration the pore size of the medium is less than the size of the particles to be filtered.

Filtration process is normally analysed in terms of flow of fluid through a packed bed of particles, the depth of which is increasing with time. In practice the voidage of the cake may also change with time (compressible cake). Incompressible cake is one, for which cake voidage is constant.

Ergun equation is given as: \[\frac{\Delta P}{\rho v^2}\frac{D_p}{L}\frac{\varepsilon^3}{1-\varepsilon} = \frac{150}{\text{Re}} + 1.75 \qquad \text{ (applicable for all the regions)}\] where \(\displaystyle \text{Re} = \frac{D_pv\rho}{\mu(1-\varepsilon)}\)

For filtration, pressure drop versus liquid flow relationship is described by the laminar part of Ergun equation as,

\[\frac{\Delta P}{L} = {\mathbf r} \mu v \tag*{(4)}\]

where \(\Delta P\) is the pressure drop across the cake; \(L\) is the thickness of cake; \(\mu\) is the viscosity of filtrate; \(v\) is the superficial velocity of filtrate; and, \(\mathbf r\) is the cake resistance. \[\mathbf{r} = \frac{150}{D_p^2}\frac{(1-\varepsilon)^2}{\varepsilon^3}\]

The Ergun equation uses the number 150 for laminar part. However, the most applicable equation for laminar part is given by Kozeny-Carman with 180 replacing 150. \[\frac{\Delta P}{L} = 180\frac{\mu v}{D_p^2} \frac{(1-\varepsilon)^2}{\varepsilon^3} \tag{Kozeny-Carman equation}\]

To avoid any further confusion, simply the term \(k\) will be used instead of 150 or 180. i.e., \[\mathbf{r} = \frac{k}{D_p^2}\frac{(1-\varepsilon)^2}{\varepsilon^3}\]

For flow through porous media, Darcy equation (similar to that of Kozeny-Carman or laminar part of Ergun equation) is often used. \[\Delta P = \frac{v \mu L}{\kappa} \tag{Darcy equation}\] where \(\kappa\) is permeability.

\(v\) is related to the volumetric flow rate of filtrate (\(dV/dt\)) as

\[v = \frac{1}{A}\frac{dV}{dt}\] where \(A\) is the filter area.

Using the above equation in Eqn.(4) we get \[\frac{\Delta P}{L} = {\mathbf r} \mu \frac{1}{A}\frac{dV}{dt} \tag*{(5)}\]

Each unit volume of filtrate is assumed to deposit a certain mass of particles, which form a certain volume of cake. This is expressed as \(\phi\), the volume of cake formed by the passage of unit volume of filtrate. \[\phi = \frac{\mbox{volume of cake}}{\mbox{volume of filtrate}} = \frac{LA}{V}\] i.e., \[L = \frac{\phi V}{A}\]

Using the above in Eqn.(5), we get

\[\frac{dV}{dt} = \frac{A^2 \Delta P}{{\mathbf r}\phi \mu V} \tag*{(6)}\]

If \(c\) is the mass of the particles deposited in the filter per unit volume of filtrate, then it is related to \(\phi\) as \[c = \phi \rho_p(1-\varepsilon) \tag*{(7)}\] where \(\rho_p\) is the density of particle.

Defining specific cake resistance (\(\alpha\)) as \[\alpha = \frac{{\mathbf r}}{\rho_p(1-\varepsilon)} = \frac{k(1-\varepsilon)}{\rho_pD_p^2\varepsilon^3} = \frac{k_1(1-\varepsilon)S_0^2}{\rho_p\varepsilon^3} \tag*{(8)}\] where

\(S_0\) = surface area per unit volume of particle
\(k_1\) = a dimensionless constant

From Eqns.(7) and (8), we get \[c\alpha = {\mathbf r} \phi\] Using this in Eqn.(6) we get \[\frac{dV}{dt} = \frac{A^2 \Delta P}{\mu \alpha c V} \tag*{(9)}\]

  • The pressure drop \(\Delta P\) may be constant or variable with time, depending on the characteristics of the pump used or on the driving force applied.