Question 7 of 7: Batch Ultrafiltration — Membrane Area
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Format: open-book, three-hour paper; seven questions in three parts (Part A Q1–2, Part B Q3–4, Part C Q5–7). A candidate answers one from A, one from B and two from C (four questions, equal value). All seven are solved here as a complete study resource.
Reference texts: R.E. Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — film theory, wetted-wall and flat-plate convective mass transfer, packed-tower absorption, distillation; Coulson & Richardson, Chemical Engineering Vol. 1 (6th ed.) and Vol. 2 (5th ed., Richardson, Harker & Backhurst) — diffusion, drying/evaporation, absorption with reaction, adsorption, membrane separations; C.J. Geankoplis, Transport Processes and Separation Process Principles (4th ed.) — Sherwood-number correlations, McCabe–Thiele with side streams; property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 7: Batch Ultrafiltration — Membrane Area (Part C — equal value)
Given. $V_0=10$ m³, $C_0=20$ kg/m³, concentrate to $C_f=200$ kg/m³ (enzyme fully retained). Flux $J=0.04\ln(250/C_f)$ m/hr. Time $t=4$ h.
Find. (a) the batch membrane area; (b) whether the linear average-flux approximation is adequate.
Figure 7 — Permeate flux collapses as the retentate concentrates from 20 to 200 kg/m³ (it would reach zero at the gel point $C_f=250$).
Approach. A retained-solute mass balance links volume to concentration; integrating the permeate rate over the concentration range gives the area, which is then compared with the linear-flux estimate.
End-point volumes. Enzyme mass $=V_0C_0=200$ kg is conserved, so $V_f=200/200=1$ m³ and the permeate collected is $V_p=V_0-V_f=9$ m³.
Area from the exact integral. With $V=200/C_f$ (so $dV=-200/C_f^{2}\,dC_f$) and $-dV/dt=JA$, $$A=\frac{1}{t}\int_{V_f}^{V_0}\frac{dV}{J}=\frac{1}{t}\!\int_{20}^{200}\!\frac{200/C_f^{2}}{0.04\ln(250/C_f)}dC_f=\frac{157.7}{4}=\boxed{39.4\ \text{m}^2}.$$
Initial and final fluxes. $J_i=0.04\ln(250/20)=0.101$ m/hr and $J_f=0.04\ln(250/200)=0.0089$ m/hr.
Linear approximation (part b). $J_{av}=J_f+0.27(J_i-J_f)=0.0089+0.27(0.0921)=0.0338$ m/hr, giving $A_{approx}=\dfrac{V_p}{J_{av}t}=\dfrac{9}{0.0338(4)}=66.6$ m².
Compare with the true average. The exact area corresponds to a true average flux $J_{av}^{\ast}=\dfrac{V_p}{A\,t}=\dfrac{9}{39.4(4)}=0.057$ m/hr. The 0.27-weighting predicts only 0.034 m/hr — it oversizes the membrane by ~69%, so the approximation is not suitable for design.
Check (approximation form): the approximation is taken as $J_{av}=J_f+0.27(J_i-J_f)$. Any 0.27-weighting (either direction) misses the true average (0.057 m/hr) by 30–40% because the flux–concentration curve is strongly non-linear.
