Leaf Filters, Constant Pressure Filtration

Example 12.1 A leaf filter having 0.05 m² of filtering surface is operated under an absolute pressure of 30 kN/m².The volume of filtrate collected in the first 300 seconds is 250 cm³ and, after further 300 seconds, an additional 150 cm³ are collected. Assume the cake is incompressible. If the operation is run for additional 120 seconds, how much filtrate will be collected?

Solution:

Average filtration velocity can be calculated if resistance to filtration is known.Driving force is pressure differential. Filtration velocity, 1 over A dV over dt is equal to driving force (minus delta p) over resistance. Filtration resistance is offered by both cake (r mu l) and cloth (r μ L). V is volume of liquid flowing in time t, A is filtration area, l is thickness of filter cake, L is the thickness of filter cake offering same resistance as that of cloth, r is specific resistance of cake, μ is viscosity of fluid and minus delta p is total pressure drop.Filter cake thickness can is equal to V over A, where v is a ratio of volume of the deposited to the volume of filtrate. dV over dt is equal to A square time minus delta p over nu r μ time (V plus LA over nu).

Filtration is conducted in two stages. 1) Constant filtration rate, where pressure is gradually increased. 2) Constant pressure, where pressure is maintained till cake is completely built. In constant filtration step, dV over dt is equal to V over t. Pressure difference is built up gradually and during this period filtration is conducted at constant filtration rate for time t₁, then. B_1r V squared plus B_2rV is equal to t. For a period of constant pressure filtration, -delta P is constant, and B_1pV squared plus B_2pV is equal to t. B_1p is equal to r mu nu over 2 A square over delta p. B_2p is equal to r mu L over (A delta p).

For this operation, total filtrate collected in 300 seconds is 250 cc. Total filtrate collected in 600 seconds is 400 cc. Pressure diffential is 101.3 minus 30 or 71.3 kN per meter square. B_2pconstant can be calculated from two data points as V₁ V₂time (V₂ minus V₁) over (V₁t₂ minus V₂ t₁). Its value is 500 cm⁶ per sec. B_1p can be found as B₂t₁ minus V²₁ over V₁ or 350 cm³. For t₃ of 720 seconds, V₃ is 450 cm³.