23-Chem-A2 Unit Operations and Separation Processes · December 2015
Question 1 of 6: Voidage of a Segregated Two-Size Fluidized Bed
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 A1: Voidage of a Segregated Two-Size Fluidized Bed (25 marks)
Given. A liquid-fluidized column that has segregated into a fine-particle upper layer and a coarse-particle lower layer, both swept by one common superficial velocity.
Quantity
Value
Small-sphere diameter (upper) $d_s$
0.75 mm
Large-sphere diameter (lower) $d_l$
1.5 mm
Richardson–Zaki index $n$
4.6 (Stokes regime)
Expansion law
$u_c/u_0 = e^{\,n}$
Lower-bed voidage (part a) $e_L$
0.60
Find. (a) the upper-bed voidage $e_U$ when $e_L=0.6$; (b) the lower-bed voidage at which the 0.75 mm fines are just completely elutriated.
Figure A1 — Segregated liquid-fluidized bed: fine spheres float to the more open upper layer, coarse spheres settle to the lower layer; both layers see the one superficial velocity $u_c$.
Approach. Write the Richardson–Zaki law separately for each layer, exploit the fact that the superficial velocity is shared, and eliminate $u_c$ to leave a pure relation between the voidages.
Terminal-velocity ratio from Stokes' law. Stokes' law gives $u_0=\dfrac{d^2(\rho_s-\rho)g}{18\mu}\propto d^2$, so the two single-particle terminal velocities differ only through the diameter squared: $$\frac{u_{0,l}}{u_{0,s}}=\left(\frac{d_l}{d_s}\right)^2=\left(\frac{1.5}{0.75}\right)^2=4.$$
Couple the layers through the shared velocity. Both layers are swept by the same $u_c$, so applying $u_c=u_0\,e^{\,n}$ to each and dividing eliminates $u_c$: $$u_{0,l}\,e_L^{\,n}=u_c=u_{0,s}\,e_U^{\,n}\;\Longrightarrow\;\left(\frac{e_U}{e_L}\right)^{n}=\frac{u_{0,l}}{u_{0,s}}=4.$$
Part (a): upper-bed voidage. The voidage ratio is fixed by the size ratio alone: $$\frac{e_U}{e_L}=4^{1/4.6}=1.352\;\Longrightarrow\;e_U=1.352\times0.6.$$ $e_U = 0.811$ — the finer particles occupy the more open (higher-voidage) layer, as physically expected.
Part (b): elutriation threshold of the fines. The fines are just fully carried out when $u_c$ reaches their own terminal velocity, i.e. $e_U\to1$ and $u_c\to u_{0,s}$. Applying the law to the lower layer at that instant: $$u_{0,s}=u_{0,l}\,e_L^{\,n}\;\Longrightarrow\;e_L=\left(\frac{u_{0,s}}{u_{0,l}}\right)^{1/n}=\left(\tfrac14\right)^{1/4.6}=0.25^{\,0.2174}.$$ $e_{L,\min} = 0.740$ — once the lower bed expands to $e\approx0.74$ the flow elutriates every 0.75 mm sphere.
