Example 12.2: A slurry is filtered in a plate and frame press, containing 12 frames, each having a square cross section with 0.3 m in length and width, and 25 mm thick. During the first 200 seconds, the filtration pressure is slowly raised to a final value of 500 kilo Newton per m², and during this period the rate of filtration is kept constant. After the initial period, filtration is carried out at constant pressure, and the cake is completely formed in a further 900 seconds. The cakes are then washed at 375 kilo Newton per m² for 600 seconds using thorough washing. What is the volume of the filtrate collected per cycle and how much wash water is used? Given L over v as 0.35 cm, and r μ v as 7.13 time10^-4 kilo Newton second m² per cm⁶.

Solution:

Cross-sectional area of bed, A, is equal to 2 n L_F W_F or 2.16 m². And total pressure drop is 500 minus 101.3 or 398.7 kilo Newton per m². The constant B_1r for constant rate operation is A L over nu or 7.56 time 10³ cm³.

The constants B_2r is equal to minus delta p² over r mu nu. Its value is found to be 2.609 time 10⁶ cm⁶ per second. And volume of filtrate collected at the end of constant filtration rate period is 1.937 time 10⁴ cm³.

In the second step, filtration is conducted at constant pressure. Properties of cake and cloth and filtrate do not change.The values of B_1p for constant pressure operation is 2 B_1r and is equal to 1.512 time 10⁴ cm³. The values of B_2p for constant pressure operation is 2 B_2r and is equal to 5.218 time 10⁶ cm⁶ per second. This volume contains filtrate from both kinds of operations. Volume and t relationship written for two time t₁ and t can be used to write equation such as (V₂ minus V₁²) plus B₁ time (V minus V₁) equals B₂ time (t minus t₁).The above equation is quadratic in V and can be rewritten as V₂ plus B₁V minus C is equal to zero. The constant C is equal to V₁² plus B₁V plus B₂ time (t minus t₁). In this equation t₁ represents the time constant filtration rate operation during which volume of filtrate cllected is V₁. After a total time, t₂, of 1100 s, V₂ can be calculated as 6.607 time 10⁴ cm³.

Final rate of filtration is B_2p over (2V₂ plus B_1p). Its value is found to be 35.43 cm³ per sec. Cake washing is done at a pressure less than the filtration pressure. Cake is washed at a rate that is proportional to the final filtration rate. If washing pressure is P₃, then washing rate can be calculated as (375 minus 101.03) over (500 minus 101.03) time 35.43 or 24.32 cm³ per sec. If cake is washed for a period of t₃, water requirement can be estimated as 24.32 time 600 or 1.459 time 10⁴ cm³.