23-Chem-A2 Unit Operations and Separation Processes · May 2018
Question 1 of 6: Washing of a Filter Cake — Porosity, Density and Solute Loading
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-CHEM-A2, Unit Operations and Separation Processes — May 2018, 3-hour open-book exam. Six 25-point problems: Part A (Unit Operations) — A1 filter-cake washing, A2 pipe flow & pump power, A3 free-settling of a particle; Part B (Separation Processes) — B1 packed-tower flooding/diameter, B2 adsorption isotherms, B3 multistage liquid–liquid extraction. The rubric asks for two problems per part; all six are worked in full below.
Reference texts (subject). McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Treybal, Mass-Transfer Operations (3rd ed.); Perry's Chemical Engineers' Handbook (§6 fluid & particle mechanics, §14 packed towers, §15 extraction, §16 adsorption).
Question A1: Washing of a Filter Cake — Porosity, Density and Solute Loading (25 points)
Find. (a) porosity $\varepsilon$; (b) final salt volume fraction on a dry basis; (c) cake density; (d) inert loading per unit area; (e) initial & final salt per unit area; (f) void volume per unit area.
Figure A1 — The voids of the inert skeleton hold the salt solution; porosity is the fraction of cake volume occupied by that solution (water + dissolved salt).
Approach. Work per 1 kg of as-formed cake: each component's volume is $m_i/\rho_i$; porosity is the solution (water + salt) volume fraction, density is 1 kg over total volume, and all per-area quantities scale with the cake thickness, with the invariant inert loading anchoring the solute bookkeeping.
Component volumes per kg of cake. $V_i=m_i/\rho_i$: $$V_\text{inert}=\tfrac{0.4789}{1420}=3.373\times10^{-4},\quad V_\text{water}=\tfrac{0.4641}{1000}=4.641\times10^{-4},\quad V_\text{salt}=\tfrac{0.0570}{1363}=4.182\times10^{-5}\ \text{m}^3,$$ summing to $V_\text{total}=8.432\times10^{-4}\ \text{m}^3/\text{kg}$.
(a) Average porosity. The solid skeleton is the inert phase; the voids hold the salt solution, so the salt volume counts as void: $$\varepsilon=\frac{V_\text{water}+V_\text{salt}}{V_\text{total}}=\frac{4.641\times10^{-4}+4.182\times10^{-5}}{8.432\times10^{-4}}=\boxed{0.600}.$$
(c) Cake density, then (d) inert per area. One kilogram occupies $8.432\times10^{-4}\ \text{m}^3$, so $\rho_\text{cake}=1/8.432\times10^{-4}=\boxed{1186\ \text{kg/m}^3}$. Over unit area the cake mass is $\rho_\text{cake}L=1186(0.051)=60.49\ \text{kg/m}^2$, of which the inert fraction is $$m''_\text{inert}=0.4789(60.49)=\boxed{28.97\ \text{kg/m}^2}.$$
(b) Salt volume fraction, dry basis (end of washing). After washing the dry solids are 0.24% salt / 99.76% inert by mass; per 1 kg dry solids, $$\phi_\text{salt}=\frac{0.0024/1363}{0.0024/1363+0.9976/1420}=\frac{1.761\times10^{-6}}{7.043\times10^{-4}}=\boxed{0.0025\ (0.25\%)}.$$
(e) Initial and final salt per unit area. Washing removes salt but not inert, so the salt loading is (salt/inert mass ratio)$\times m''_\text{inert}$; the ratio is $0.0570/0.4789$ initially and $0.0024/0.9976$ finally: $$m''_\text{salt,0}=\tfrac{0.0570}{0.4789}(28.97)=\boxed{3.45\ \text{kg/m}^2},\qquad m''_\text{salt,f}=\tfrac{0.0024}{0.9976}(28.97)=\boxed{0.070\ \text{kg/m}^2}.$$ About $3.38\ \text{kg/m}^2$ (98%) of the salt is washed out.
(f) Void volume per unit area. Porosity times thickness: $$V''_\text{void}=\varepsilon L=0.600(0.051)=\boxed{0.0306\ \text{m}^3/\text{m}^2}.$$ This is the theoretical wash-liquor volume for one displacement.
