23-Chem-A2 Unit Operations and Separation Processes · December 2015
Question 2 of 6: Optimum Cycle Time of a Constant-Rate / Constant-Pressure Filter
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 04-CHEM-A2 Mechanical and Thermal Operations, December 2015. 3 hours, open book (one text). Six problems (Section A Mechanical Operations: A1–A3; Section B Thermal Operations: B1–B3), each 25 marks; candidates attempt at least two from each section. All six are worked below for completeness.
Reference texts. Coulson & Richardson, Chemical Engineering Vol. 1 (fluid flow, heat transfer) and Vol. 2 (particle technology, filtration, evaporation); McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.); Perry's Chemical Engineers' Handbook (9th ed.).
Note on the figureThe only figure in the paper is the LMTD correction-factor chart on page 5 (used in B3). It is read at the plotted parameters $P=0.25$, $R=2.0$; the value obtained ($F\approx0.94$) is confirmed analytically from the closed-form 1-outer-pass / 2-tube-pass expression.
Question A2: Optimum Cycle Time of a Constant-Rate / Constant-Pressure Filter (25 marks)
Given. Two-stage batch filtration — a pump-limited constant-rate start-up followed by a constant-pressure period — with the cloth resistance neglected.
Quantity
Value
Terminal pressure difference $\Delta P$
400 kN/m²
Constant-rate duration $t_1$
900 s
Filtrate in constant-rate stage $V_1$
$\tfrac13 V_f$
Cake-removal / re-dressing time $t_d$
1200 s
Medium resistance
neglected
Find. (a) the total cycle time for the run as described; (b) the cycle time when the constant-pressure period is chosen to maximize daily throughput.
Figure A2 — Pressure history: a linear rise during the pump-limited constant-rate stage (to 400 kN/m² at $t_1$), then a constant-pressure plateau for the rest of the filtration.
Approach. Use the incompressible-cake filtration equation with medium resistance dropped; fix the lumped constant from the constant-rate stage, integrate through the constant-pressure stage for (a), then maximize filtrate-per-cycle for (b).
Filtration equation (medium neglected). With cloth resistance zero, $$\frac{dV}{dt}=\frac{A^2\,\Delta P}{r\mu v\,V}\;\Longleftrightarrow\;V\,dV=\frac{\Delta P}{B'}\,dt,\qquad B'\equiv\frac{r\mu v}{A^2},$$ where $V$ is cumulative filtrate and $B'$ collects the cake/slurry constants.
Fix $B'V_1^2$ from the constant-rate stage. During constant rate $dV/dt=V_1/t_1$; at the end $V=V_1,\ \Delta P=400$, so $$\Delta P=B'V_1\frac{V_1}{t_1}\;\Longrightarrow\;B'V_1^2=\Delta P\,t_1=400\times900=3.6\times10^{5}\ \tfrac{\text{kN}\cdot\text{s}}{\text{m}^2}.$$ Since $V_1=\tfrac13V_f$, the full run collects $V_f=3V_1$.
Part (a): constant-pressure time and cycle. Holding $\Delta P=400$, integrate from $V_1$ to $3V_1$: $$t'=\frac{B'}{2\,\Delta P}\big[(3V_1)^2-V_1^2\big]=\frac{8\,B'V_1^2}{2(400)}=\frac{8(3.6\times10^{5})}{800}=3600\ \text{s}.$$ Total filtration $t_f=900+3600=4500$ s; adding the 1200 s down-time, $$t_{\text{cycle}}=4500+1200.$$ $t_{\text{cycle}}=5700\ \text{s}\approx1.58\ \text{h}$
Part (b): run length for maximum throughput. The start-up is fixed, but the batch size $V$ (hence the constant-pressure time) is free. With $c\equiv B'/(2\Delta P)$ and $t_{\text{cycle}}(V)=t_1+t_d+c(V^2-V_1^2)$, setting $\dfrac{d}{dV}\!\big(V/t_{\text{cycle}}\big)=0$ gives $t_{\text{cycle}}=2cV^2$, i.e. $cV_{\text{opt}}^2=t_1+t_d-cV_1^2$. Using $cV_1^2=B'V_1^2/800=450$ s, $$t'_{\text{opt}}=(t_1+t_d)-2cV_1^2=2100-900=1200\ \text{s}=t_d.$$ The optimum filtration time equals the down-time, so $t_f=900+1200=2100$ s and $$t_{\text{cycle,opt}}=2100+1200.$$ $t_{\text{cycle,opt}}=3300\ \text{s}\approx0.92\ \text{h}$
