23-Chem-A2 Unit Operations and Separation Processes · December 2015
Question 3 of 6: Time to Fill a Buffer Tank — Laminar vs. Turbulent Limits
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 A3: Time to Fill a Buffer Tank — Laminar vs. Turbulent Limits (25 marks)
Given. A fixed tank volume filled through a fixed-bore pipe; the "laminar/turbulent" wording constrains the pipe Reynolds number to each regime's boundary.
Quantity
Value
Pipe inside diameter $D$
3 cm = 0.03 m
Tank diameter
1.5 m
Tank height
3 m
Liquid density $\rho$
1040 kg/m³
Liquid viscosity $\mu$
$1.6\times10^{-3}$ Pa·s
Find. (a) the minimum fill time under laminar flow; (b) the maximum fill time under turbulent flow.
Figure A3 — Filling a fixed-volume buffer tank through a 3 cm line; the extreme times occur at the Reynolds-number boundaries of each named regime.
Approach. The fill time is $t=V_{\text{tank}}/(vA)$, monotone in the pipe velocity; since velocity is monotone in $\mathrm{Re}$, evaluate each regime at its boundary Reynolds number.
Fixed geometry. $$V_{\text{tank}}=\frac{\pi}{4}(1.5)^2(3)=5.301\ \text{m}^3,\qquad A=\frac{\pi}{4}(0.03)^2=7.069\times10^{-4}\ \text{m}^2.$$ Only throughput matters, so friction factor and pump head never enter.
Part (a): fastest laminar flow ($\mathrm{Re}=2100$). The minimum time comes at the top of the laminar range: $$v=\frac{\mathrm{Re}\,\mu}{\rho D}=\frac{2100(1.6\times10^{-3})}{1040(0.03)}=0.1077\ \text{m/s},\quad Q=vA=7.61\times10^{-5}\ \text{m}^3/\text{s}.$$ $$t_{\min}=\frac{V_{\text{tank}}}{Q}=\frac{5.301}{7.61\times10^{-5}}.$$ $t_{\min}=6.96\times10^{4}\ \text{s}\approx19.3\ \text{h}$
Part (b): slowest turbulent flow ($\mathrm{Re}=4000$). The maximum time while still turbulent comes at the onset of turbulence: $$v=\frac{4000(1.6\times10^{-3})}{1040(0.03)}=0.2051\ \text{m/s},\quad Q=vA=1.450\times10^{-4}\ \text{m}^3/\text{s}.$$ $$t_{\max}=\frac{5.301}{1.450\times10^{-4}}.$$ $t_{\max}=3.66\times10^{4}\ \text{s}\approx10.2\ \text{h}$
Consistency check. The transition band $2100<\mathrm{Re}<4000$ is excluded; the maximum turbulent time (10.2 h) is shorter than the minimum laminar time (19.3 h), as it must be, since turbulence only begins at the higher velocity.
