To find the volume that passes the filter in 1 h which is 3600 s, that is to find V for t = 3600. 3600 = 0.05( V/A ) 2 + 2.0( V/A ) and solving this quadratic equation, we find that V/A = 250 kg m -2 and so the slurry passing through 9.3 m 2 in 1 h would be: = 250 x 9.3 = 2325 kg . Filter-cake Compressibility With some filter cakes, the specific resistance varies with the pressure drop across it. This is because the cake becomes denser under the higher pressure and so provides fewer and smaller passages for flow. The effect is spoken of as the compressibility of the cake. Soft and flocculent materials provide highly compressible filter cakes, whereas hard granular materials, such as sugar and salt crystals, are little affected by pressure. To allow for cake compressibility the empirical relationship has been proposed: r = r' D P s where r is the specific resistance of the cake under pressure P , D P is the pressure drop across the filter, r ' is the specific resistance of the cake under a pressure drop of 1 atm and s is a constant for the material, called its compressibility. This expression for r can be inserted into the filtration equations, such as eqn. (10.14), and values for r ' and s can be determined by carrying out experimental runs under various pressures. Filtration Equipment The basic requirements for filtration equipment are: mechanical