23-Chem-A2 Unit Operations and Separation Processes · May 2014
Question 2 of 6: Turbine-Mixer Power and Scale-Up
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 2014 — open-book, 3 hours. Two sections: Section A (Mechanical Operations, A1–A3) and Section B (Thermal Operations, B1–B3); all problems are worth 25 marks. The rubric asks candidates to attempt two problems per section; all six are solved in full below as a study resource.
Reference texts: McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed., McGraw-Hill) — pipe-flow friction and loss coefficients, agitator power correlations (Fig. 9.13, Table 9.3), and filtration theory; de Nevers, Fluid Mechanics for Chemical Engineers (3rd ed.) and Brodkey & Hershey, Transport Phenomena — mechanical-energy balance, fitting equivalent lengths, and the appended Fanning chart; Coulson & Richardson, Chemical Engineering Vol. 2 — rotary-drum filtration; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (7th ed., Wiley) and Lienhard, A Heat Transfer Textbook — composite-wall resistance networks, the Colburn (Chilton–Colburn) analogy, and LMTD/ε–NTU cross-flow exchangers.
Property note. Water properties are as printed in each question; the appended Table A.2 loss coefficients ($k$) and Table A3 turbine constants ($K_T$) are used directly, and pipe-friction factors are taken from Colebrook (identical to the appended Fanning chart, Fig. A1) so the numbers are reproducible without reading a graph.
Question A2: Turbine-Mixer Power and Scale-Up (25 marks)
Find. (a) power per unit volume $P/V$; (b) motor power at 4.0 m tank for equal $P/V$; (c) the corresponding impeller speed.
Figure A2 — Baffled stirred tank with a six-blade disk (Rushton) turbine. Geometric similarity ($D_a/D_t$, $H/D_t$, $J/D_t$ all fixed) is preserved on scale-up.
Approach. Check the Reynolds number; in the fully-turbulent baffled regime the power number is the constant $K_T$, so $P=K_T\rho N^3 D_a^5$. For equal $P/V$ scale-up, the larger tank's power is set by its volume, and the speed follows from re-inverting the same power law.
Reynolds number — confirm turbulent. $$\mathrm{Re}=\frac{\rho N D_a^2}{\mu}=\frac{1000(1.417)(0.667)^2}{1.0\times10^{-3}}=6.3\times10^{5}\gg 10^4,$$ so flow is fully turbulent and the power number is constant at $K_T=5.75$ (Table A3, six-blade disk; Fig. A2 plateau).
(a) Power drawn. $$P=K_T\rho N^3 D_a^5=5.75(1000)(1.417)^3(0.667)^5=2.15\times10^{3}\ \mathrm{W}.$$ The liquid volume is $V=\tfrac{\pi}{4}D_t^2 H=\tfrac{\pi}{4}(2.0)^2(2.0)=6.28\ \mathrm{m^3}$, so $$\frac{P}{V}=\frac{2153}{6.28}=343\ \mathrm{W/m^3}.$$ $P/V\approx 343\ \mathrm{W/m^3}$ (2.15 kW in 6.28 m$^3$)
(b) Scale-up at equal $P/V$. Geometric similarity gives the new volume $$V_2=\frac{\pi}{4}D_{t2}^2 H_2=\frac{\pi}{4}(4.0)^2(4.0)=50.3\ \mathrm{m^3}.$$ Holding $P/V$ constant, $$P_2=\left(\frac{P}{V}\right)V_2=343(50.3)=1.72\times10^{4}\ \mathrm{W}\approx 17.2\ \mathrm{kW}.$$ Since the linear scale doubled, $V_2/V_1=2^3=8$, so $P_2=8P_1$ exactly. $P_2\approx 17.2\ \mathrm{kW}$
(c) Required impeller speed. Re-arranging the turbulent power law for the scaled impeller ($D_{a2}=4.0/3=1.333\ \mathrm{m}$): $$N_2=\left(\frac{P_2}{K_T\rho D_{a2}^5}\right)^{1/3}=\left(\frac{17{,}240}{5.75(1000)(1.333)^5}\right)^{1/3}=0.892\ \mathrm{rev/s}.$$ Equivalently $N_2=N_1(D_{a1}/D_{a2})^{2/3}=85(0.5)^{2/3}=53.5\ \mathrm{rpm}$ — equal $P/V$ turbulent scale-up requires $P/V\propto N^3 D_a^2$, so the bigger impeller turns slower. $N_2=0.892\ \mathrm{rev/s}=53.5\ \mathrm{rpm}$
