Special Topic 2 of the Registered Chemical Engineer Professional Examination: Separation of Heterogeneous Systems, Solid Fluidization

(4) Perform constant-pressure filtration of a certain suspension using a plate and frame filter. The filtration time was 20 minutes, yielding 20 m3 of filtrate; the filter cake was not washed. The assembly and disassembly time was 15 minutes. The filter cake was incompressible, and the medium resistance could be ignored. Try to determine the production capacity of this machine ； If the filtration pressure increases by 20% while the assembly and disassembly time remains unchanged, what will be the production capacity? 【Analysis】 tw=0, t=1/3 h, td=1/4 h; Q=V/tw+t+td=20/(1/3+1/4)=34.3 m3/h. Medium resistance is neglected, so V^2=KA^2*T, and KA^2=400/(1/3)=1200 m6/h. Since K=2k*delta P^(1-S) with S=0, K is proportional to the pressure difference; thus K’=1.2K. V^2=K’A^2*T’, and T’=0.2778 h. Therefore, Q’=20/(0.2778+0.25)=37.89 m3/h. (5) A filtrate containing 3% solid phase by volume is filtered under constant pressure using a small plate-and-frame filter. The dimensions of each filter frame are 200 mm*200 mm*20 mm, with a total of 10 frames. It takes 2 hours for the filter residue to fill the frames. The solid fraction of the filter cake is 60%, the filter cake is incompressible, and medium resistance can be ignored. Assuming that the viscosity of the washing water, its gauge pressure, as well as the viscosity of the filtrate and the filtration pressure difference are all the same, the volume of washing water is 10% of the volume of the filtrate. The time required for each cycle of residue removal and cleaning is 0.5 hours. Determine the filtration constant and the production capacity of the filter. 【Analysis】The volume per unit filtrate per unit filter cake, r = LA/V = Q/(1 – S – Q) = 0.03/0.6 – 0.03 = 0.053. The filtration area A = 0.2 * 0.2 * 10 * 2 = 0.8 m³. The volume of the filter residue V = V_residue / r = 0.2 * 0.2 * 0.02 * 10 = 0.008; thus V/ r = 0.008/0.053 = 0.15 m³. The resistance due to the medium can be ignored. Using the formula V2 = KA2T, we get K = V2/A2T = 0.15²/0.8²/2 = 0.01758 m²/h. The rate of change of volume with respect to time is given by dV/dT = KA2/2(V + Ve) = KA2/2V = 0.01758 * 0.8²/2/0.15 = 0.0375 m³/h. The value of (dV/dT)w is 1/4(dV/dT)e = 0.0375/4 = 9.38*10^-3 m³/h. Finally, tw = Vw/(dV/dT)w = 0.1 * 0.15/9.38*10^-3 = 1.6 hours. Therefore, the production capacity Q = V/(t + tw + td) = 0.15/(2 + 1.6 + 0.5) = 0.0366 m³/h

This post was last edited by zhanghp30 on 2020-1-16 at 10:37 (6). A certain suspension contains 20% (by mass) solids and 80% (by mass) water; after filtration at 4°C, the filter cake consists of 50% (by mass) solids and the remainder being water. The water removed (the filtrate) contains no solids; all of it remains in the filter cake. When the volume of the filtrate is 1 m3, the mass of the solids is ( ) A 333.3 kg B 500 kg C 200 kg D 666.6 kg [Solution] Let the mass of solids in the filter cake be M when 1 m3 of filtrate is obtained each time. Then, M/0.2 = 1 * density_of_water + M/0.5 = 1000 + 2M = 5M, so M = 333.3 kg. (7) In a fluidized bed with an inner diameter of 1.5 m, coal dust obtained through dry distillation is cooled directly using gas. Given a semi-coke processing rate of 4200 kg/h, a density of 1120 kg/m3, a porosity of 0.6 for the bed layer, and a residence time of 10 minutes for the semi-coke particles in the bed layer, what is the pressure drop across the bed layer? 【Analysis】 Let the volume of the bed layer be V and its diameter be D; then Lmf = V / (0.785 × D²). Since V = 4000 × 10 / 60 / 1120 × (1 – 0.6) = 1.49 m³, it follows that Lmf = 1.49 / (0.785 × 1.5²) = 0.844 m. The pressure drop DetaP is given by DetaP = Lmf × (1 – s) × (ps – p) × g = 0.844 × 0.4 × 1120 × 9.81 = 3709 Pa. [Summary】 The difficulty level of the questions in this section is moderate; the proportion of topics covered in the exam questions isn’t large. However, there are also difficult questions. Topics such as pressure drop in fixed-bed reactors, solid particles, plate-and-frame filters, and constant-pressure filtration are frequently tested. To tackle complex problems, it is necessary to master the most basic conceptual formulas as well as the problem-solving approaches and methods.

This chapter always confuses the formulas! Are there any good methods?

The main focus is on Stokes’ formula, the constant-pressure filtration formula, and the filter cake washing formula; remember these most basic ones! For the others, just know where to look!