23-Chem-A2 Unit Operations and Separation Processes · December 2013
Question 3 of 6: Hydrometer Densities and Centrifuge Interface
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 A3: Hydrometer Densities and Centrifuge Interface (25 marks)
Given. Hydrometer rod: length $L=150$ mm, diameter $15$ mm, mass $20.800$ g; reads $0.8533$ (oil) and $0.8004$ (aqueous). Centrifuge overflow radii $r_1=10.160$ mm (light) and $r_2=10.414$ mm (heavy).
Quantity
Value
Rod cross-section $A=\tfrac{\pi}{4}(1.5\ \mathrm{cm})^2$
$1.7671\ \mathrm{cm^2}$
Rod mass $m$
$20.800$ g
Oil / aqueous reading
$0.8533$ / $0.8004$
Light / heavy overflow radius
$10.160$ / $10.414$ mm
Find. (a) the density of the oil and aqueous phases; (b) the radius of the liquid–liquid interface (neutral zone) in the bowl.
Figure A3 — A floating hydrometer sinks lower (larger submerged reading) in the lighter oil; the centrifuge interface (dashed red) sits at the neutral-zone radius $r_i$ where the centrifugal pressure columns from the two overflow weirs balance.
Approach. Get each density from a flotation (buoyancy) balance on the rod, then place the interface from the tubular-centrifuge neutral-zone balance.
Densities from flotation (part a). A floating rod displaces its own weight; the reading gives the submerged fraction of the 15-cm rod, so $\ell=\text{reading}\times15$ cm and $m=\rho A\ell$: $$\rho_{\text{oil}}=\frac{20.800}{1.7671(0.8533\times15)}=\boxed{0.920\ \mathrm{g/cm^3}=920\ \mathrm{kg/m^3}},$$ $$\rho_{\text{aq}}=\frac{20.800}{1.7671(0.8004\times15)}=\boxed{0.980\ \mathrm{g/cm^3}=980\ \mathrm{kg/m^3}}.$$ The lighter oil floats the rod lower, so the aqueous phase is the heavy liquid.
Interface (neutral-zone) radius (part b). In a tubular bowl the interface sits where the centrifugal pressures built from the two overflow radii are equal, $\rho_L(r_i^2-r_1^2)=\rho_H(r_i^2-r_2^2)$: $$r_i=\sqrt{\frac{\rho_H r_2^2-\rho_L r_1^2}{\rho_H-\rho_L}}=\sqrt{\frac{980(10.414)^2-920(10.160)^2}{980-920}}=\boxed{r_i\approx13.7\ \mathrm{mm}}.$$ The interface lies outboard of both weirs, as it must; the small $60\ \mathrm{kg/m^3}$ density difference makes $r_i$ very sensitive to the ring-dam settings.
