23-Chem-A2 Unit Operations and Separation Processes · December 2018
Question 5 of 6: Batch Drying Time (Constant + Falling Rate)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-CHEM-A2, Unit Operations and Separation Processes — December 2018, 3-hour open-book exam. Six 25-point problems: Part A (Unit Operations) — A1 continuous-thickener sizing, A2 series-pipe friction pressure drop, A3 packed/fluidized-bed pressure drop & power; Part B (Separation Processes) — B1 flash vs. differential distillation, B2 batch-drying time, B3 adsorption isotherm fitting & surface area. The rubric asks for two problems per part; all six are worked in full below.
Reference texts (subject). McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Coulson & Richardson, Chemical Engineering Vol. 2 (5th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Treybal, Mass-Transfer Operations (3rd ed.); Perry's Chemical Engineers' Handbook (§6 fluid & particle mechanics, §12 drying, §16 adsorption, §18 solid–liquid separation).
Find. Total drying time from $X_1=0.20$ to $X_2=0.02$, splitting into the constant-rate and falling-rate periods.
Figure B2 — Rate is constant (1.71 kg/hr·m²) down to the critical moisture $X_c=0.14$, then falls. Drying time = analytic constant-rate term + graphical (trapezoidal) integral of $dX/R$ over the falling-rate period.
Approach. Time $=\dfrac{m_s}{A}\displaystyle\int_{X_2}^{X_1}\dfrac{dX}{R}$; the dry-solid loading per area is $m_s/A=1/(\text{effective area})$. Above $X_c=0.14$ the rate is constant (closed-form term); below it, integrate $dX/R$ by the trapezoidal rule on the tabulated points.
Dry-solid loading per unit area. Taking the effective area as per kg of dry solid (the same basis as the moisture contents), the batch mass cancels: $\dfrac{m_s}{A}=\dfrac{1}{6.15\times10^{-2}}=16.26\ \text{kg dry/m}^2$ (equivalently $1400\ \text{kg}\times6.15\times10^{-2}=86.1\ \text{m}^2$ of drying area).
Constant-rate period ($X_1=0.20\to X_c=0.14$). Rate held at $R_c=1.71$: $$t_c=\frac{m_s}{A}\cdot\frac{X_1-X_c}{R_c}=16.26\cdot\frac{0.20-0.14}{1.71}=\boxed{0.571\ \text{h}}.$$
Falling-rate period ($X_c=0.14\to X_2=0.02$). Interpolate the rate at $X_2=0.02$ between (0.026, 0.54) and (0.016, 0.376): $R(0.02)=0.376+\tfrac{0.02-0.016}{0.026-0.016}(0.54-0.376)=0.442$. Trapezoidal integration of $1/R$ over the falling points $\{0.14,0.096,0.056,0.042,0.026,0.02\}$ gives $\int dX/R=0.1068\ \text{m}^2\text{hr/kg}$, so $$t_f=\frac{m_s}{A}\int\frac{dX}{R}=16.26(0.1068)=\boxed{1.74\ \text{h}}.$$
Total drying time. $$t=t_c+t_f=0.571+1.74=\boxed{2.31\ \text{h}}.$$ The falling-rate period dominates (≈75% of the time) even though it spans a smaller moisture change, because the rate collapses toward the end.
