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23-Chem-A2 Unit Operations and Separation Processes · May 2013

Question 3 of 6: Packed-Bed Reactor — Porosity, Pellet Size, Area and Pressure Drop

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exam 04-Chem-A2 Mechanical and Thermal Operations, May 2013 — open-book, 3 hours. Two sections: Section A (Mechanical Operations, A1–A3) and Section B (Thermal Operations, B1–B3); all problems 25 marks. The rubric asks candidates to attempt two problems per section; all six are solved in full below.

Reference texts: McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed., McGraw-Hill) — pipe-flow friction, loss coefficients, sphericity and the Ergun equation (Tables 7.1, 5.1); de Nevers, Fluid Mechanics for Chemical Engineers (3rd ed.) and Brodkey & Hershey, Transport Phenomena — mechanical-energy balance and sudden expansion/contraction losses; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (7th ed., Wiley) — composite-wall resistance networks, LMTD/ε–NTU cross-flow exchangers and lumped radiative cooling; Lienhard, A Heat Transfer Textbook and Özişik, Radiative Transfer for the appended correction-factor and emissivity charts.

Compressible-flow note. Water properties at 180 °F are taken as $\rho=60.55\ \mathrm{lb/ft^3}=970\ \mathrm{kg/m^3}$ and $\mu=2.32\times10^{-4}\ \mathrm{lb/(ft\cdot s)}=3.45\times10^{-4}\ \mathrm{Pa\cdot s}$; commercial-steel roughness $\varepsilon=0.0457$ mm (Table A2).

Section A — Mechanical Operations

Question A3: Packed-Bed Reactor — Porosity, Pellet Size, Area and Pressure Drop (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $\rho_p=1800$, $\rho_b=1170\ \mathrm{kg/m^3}$; $h=d=4$ mm cylinders; bed $A_{\text{bed}}=0.15\ \mathrm{m^2}$, $L=2.5$ m; superficial $u=1.5$ m/s; vapour $\rho=0.62\ \mathrm{kg/m^3}$, $\mu=1.75\times10^{-5}\ \mathrm{kg/(m\,s)}$.

Find. (a) porosity $\varepsilon$; (b) effective (Ergun) diameter $D_p$; (c) specific surface $a$ (m$^{-1}$); (d) pressure drop $\Delta P$.

u = 1.5 m/s (superficial) L = 2.5 m, A = 0.15 m² cylinder h = d = 4 mm φₛ = 0.87
Figure A3 — Bed of equal-height cylindrical pellets ($h=d$). The bulk-to-pellet density ratio gives the void fraction; the surface-to-volume mean diameter feeds the Ergun pressure-drop correlation.

Approach. Get porosity from the density ratio; the effective diameter from the surface-to-volume mean of an $h=d$ cylinder (equivalently $\varphi_s\times$ the equal-volume sphere); the specific surface from $(1-\varepsilon)\,6/D_p$; and the pressure drop from the Ergun equation.

  1. (a) Porosity from densities. The bed is pellet plus void; solids fill $\rho_b/\rho_p$ of the volume: $$\varepsilon=1-\frac{\rho_b}{\rho_p}=1-\frac{1170}{1800}=\boxed{0.35}.$$
  2. (b) Effective diameter. For the cylinder ($h=d$), volume $V_p=\tfrac{\pi}{4}d^2\!\cdot\! d$ and surface $A_p=\pi d^2+\tfrac{\pi}{2}d^2=\tfrac32\pi d^2$. The surface-to-volume (Ergun) diameter is $$D_p=\frac{6V_p}{A_p}=d=4.0\ \mathrm{mm}.$$ Equivalently, the equal-volume sphere has $d_v=(6V_p/\pi)^{1/3}=4.58$ mm and sphericity $\varphi_s=\pi d_v^2/A_p=0.874$ (Table A4 lists 0.87), giving $D_p=\varphi_s d_v=\boxed{3.98\ \mathrm{mm}\approx4.0\ \mathrm{mm}}$.
  3. (c) Surface area per unit bed volume. Each unit of particle volume carries $A_p/V_p=6/D_p$ of surface, and particles occupy $(1-\varepsilon)$ of the bed: $$a=(1-\varepsilon)\frac{6}{D_p}=0.65\times\frac{6}{0.004}=\boxed{975\ \mathrm{m^{-1}}}.$$
  4. (d) Pressure drop — Ergun equation. $$\frac{\Delta P}{L}=\underbrace{\frac{150\,\mu u(1-\varepsilon)^2}{\varepsilon^3 D_p^2}}_{2425\ \mathrm{Pa/m}}+\underbrace{\frac{1.75\,\rho u^2(1-\varepsilon)}{\varepsilon^3 D_p}}_{9252\ \mathrm{Pa/m}}=11\,677\ \mathrm{Pa/m}.$$ Over $L=2.5$ m, $$\Delta P=11\,677\times2.5=\boxed{29.2\ \mathrm{kPa}}.$$ The inertial (Burke–Plummer) term dominates, consistent with the particle Reynolds number $\mathrm{Re}_p=D_p u\rho/[\mu(1-\varepsilon)]\approx3.3\times10^{2}$.
QuantityResult
(a) Porosity $\varepsilon$0.35
(b) Effective diameter $D_p$3.98 mm ($\approx$ 4.0 mm)
(c) Specific surface $a$975 m$^{-1}$
(d) Pressure drop $\Delta P$29.2 kPa