23-Chem-A2 Unit Operations and Separation Processes · December 2013
Question 2 of 6: Packed-Bed Reynolds Number
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, December 2013 — open-book, 3 hours, any non-communicating calculator. Two sections: Section A (Mechanical Operations, A1–A3) and Section B (Thermal Operations, B1–B3); every problem is 25 marks. The rubric asks candidates to attempt two problems per section, but all six are solved in full below.
Reference texts: McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed., McGraw-Hill) — pipe friction, loss coefficients, packed beds and centrifugal separation (Ch. 5–7); de Nevers, Fluid Mechanics for Chemical Engineers and Brodkey & Hershey, Transport Phenomena: A Unified Approach — the mechanical-energy balance and the appended friction-factor chart and fitting table; Geankoplis, Transport Processes and Separation Process Principles — tubular-centrifuge neutral-zone analysis; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (7th ed., Wiley) and Çengel, Heat and Mass Transfer — conduction with generation, annular-fin efficiency (Fig. B1) and LMTD/correction-factor exchanger design (Fig. B2). Loss-coefficient, friction-factor, fin-efficiency and correction-factor data are read from the appended Table A1, Fig. A1, Table B1 and Figs. B1–B2.
Section A — Mechanical Operations
Question A2: Packed-Bed Reynolds Number (25 marks)
Given. Air at $T=27\ \mathrm{^\circ C}=300.15$ K, $P=107.5$ kPa; spherical particles $d_p=5$ mm; bed height $5.0$ m, voidage $\varepsilon=0.33$; $\mu=1.5\times10^{-5}\ \mathrm{Pa\cdot s}$; $M_{\text{air}}=0.029\ \mathrm{kg/mol}$.
Find. the packed-bed Reynolds number $\mathrm{Re}_p=\rho u_s d_p/\mu$ (and the Ergun/bed form $\mathrm{Re}_p/(1-\varepsilon)$).
Check — missing datum (state-your-assumptions rubric) The bed Reynolds number needs a gas flow rate (superficial velocity $u_s$ or mass velocity $G=\rho u_s$), but none is printed on the 2013-December paper — a genuine exam defect. The complete method is given below, expressed per unit superficial velocity and evaluated at a representative $u_s=1.0\ \mathrm{m/s}$; the result scales linearly with whatever velocity the grader supplies.
Figure A2 — Fixed bed of 5-mm spheres. The Reynolds number uses the superficial velocity (volumetric flow ÷ empty-tower area) and the particle diameter as length scale; the bed height enters only when the problem is extended to an Ergun pressure drop.
Approach. Get the gas density from the ideal-gas law, then form the particle Reynolds number per unit superficial velocity and evaluate at the assumed $u_s$.
Gas density from the ideal-gas law. At $T=300.15$ K, $P=107\,500$ Pa: $$\rho_{\text{air}}=\frac{PM}{RT}=\frac{107\,500(0.029)}{8.314(300.15)}=\boxed{1.248\ \mathrm{kg/m^3}}.$$
Particle Reynolds number. Based on the superficial velocity, $$\mathrm{Re}_p=\frac{\rho\,u_s d_p}{\mu}=\frac{1.248(0.005)}{1.5\times10^{-5}}\,u_s=416\,u_s,\qquad \mathrm{Re}_{\text{bed}}=\frac{\mathrm{Re}_p}{1-\varepsilon}=\frac{416}{0.67}\,u_s=621\,u_s.$$ Evaluating at the representative $u_s=1.0\ \mathrm{m/s}$: $$\boxed{\mathrm{Re}_p\approx416\quad(\mathrm{Re}_{\text{bed}}\approx621)}.$$ A value of a few hundred places the flow in the transitional/inertial regime of the Ergun equation, where both the viscous ($150$) and kinetic ($1.75$) terms matter.
