23-Chem-A2 Unit Operations and Separation Processes · December 2017
Question 5 of 6: Batch Drying Time of a Wet Slab
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 16-CHEM-A2 Unit Operations and Separation Processes, December 2017. 3 hours, open book (one text). Six problems — Part A Unit Operations (A1–A3) and Part B Separation Processes (B1–B3), each worth 25 marks; the rubric asks for at least two problems from each part (only the first two per part are marked). All six are worked below for completeness as a study resource.
Reference texts. McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Coulson & Richardson, Chemical Engineering Vol. 2 (particle technology, absorption, drying, crystallization); Treybal, Mass-Transfer Operations (3rd ed.); Perry's Chemical Engineers' Handbook (9th ed.).
Notes on chart / model reads. A1's minor-loss balance uses the Blasius smooth-pipe friction factor $f=0.079\,Re^{-0.25}$ (Fanning) — valid to $Re\approx10^5$, and all three test points fall in $3.8$–$8.9\times10^4$; the 1.5 m is taken as the head lost across the pipe-and-valve test section, because also charging an exit velocity head would make the measured open-valve flow physically impossible (see A1). B1(b) reads the Sherwood/Eckert flooding line supplied with the paper at abscissa $0.061$, giving flooding ordinate $\approx0.17$ (graphical, $0.16$–$0.18$), with the tabulated packing factor $F=160\ \text{ft}^{-1}$ converted to $525\ \text{m}^{-1}$ for the SI form of the ordinate. B3(a) sizes the vessel from its starting liquor volume, allowing for the water evaporated during the batch. Engineering choices are flagged in Check callouts.
Part A — Unit Operations
Question B2: Batch Drying Time of a Wet Slab (25 marks)
Figure B2 — Only the top face of area $A=0.35\ \text{m}^2$ loses moisture; the linear rate law makes this a pure falling-rate period.
Approach. The rate is linear in free moisture, so an unsteady moisture balance over the single face integrates to a log-mean of the driving force — convert the wet-basis moistures to dry basis and evaluate.
Convert moistures to dry basis. The rate law uses $X$ in kg water per kg dry solid; converting the wet-basis fractions via $X=w/(1-w)$: $$X_1=\frac{0.35}{1-0.35}=0.5385,\qquad X_2=\frac{0.05}{1-0.05}=0.0526.$$
Dry-solid mass and drying face. The bone-dry solid mass and (single) drying area are $$m_s=\rho V=1200(0.35\times0.007)=2.94\ \text{kg},\qquad A=0.35\ \text{m}^2.$$
Unsteady moisture balance. Equating moisture loss to the drying flux, $-m_s\dfrac{dX}{dt}=NA=kA(X-X^{*})$ with $k=0.95$, $X^{*}=0.01$; separating and integrating from $X_1$ to $X_2$: $$t=\frac{m_s}{kA}\ln\frac{X_1-X^{*}}{X_2-X^{*}}=\frac{2.94}{0.95(0.35)}\ln\frac{0.5385-0.01}{0.0526-0.01}=8.842\ln(12.40).$$
Evaluate. $$\boxed{t=8.842(2.517)=22.3\ \text{s}.}$$ The very short time follows directly from the large specified rate coefficient ($k=0.95$); the method — log-mean driving force in a linear falling-rate period — is what the question tests, and it is applied exactly as stated.
