Question 1

Filtration of volatile liquids: which filtration method is generally unsuitable due to loss of solvent and safety concerns when handling highly volatile liquids?

Accepted Answer

Introduction / Context: Volatile liquids (low boiling point, high vapor pressure) pose special challenges during filtration: solvent losses, vapor handling, potential freezing by evaporative cooling, and flammability hazards. Selecting the appropriate driving force (pressure, gravity, vacuum, or centrifugal field) directly affects process safety and product recovery. Given Data / Assumptions: Liquid is volatile and easily evaporates under reduced pressure. Objective: separate suspended solids with minimal solvent loss. Equipment choices include vacuum, pressure, gravity, and centrifugal filters. Concept / Approach: Vacuum filtration operates below atmospheric pressure on the filtrate side, promoting evaporation of volatile solvents. Evaporation increases emissions, can chill the cake, change viscosity, and create flammability/explosion risks. By contrast, pressure filtration encloses and often pressurizes the system, limiting vaporization. Gravity and centrifugal filters can also be designed to minimize vapor loss compared to vacuum systems. Step-by-Step Solution: Identify hazard: reduced pressure accelerates solvent vaporization.Link to equipment: vacuum filters inherently lower pressure at the filtrate side.Conclude unsuitability: vacuum filters are generally unsuitable for volatile liquids. Verification / Alternative check: Process safety guidelines recommend closed pressure nutsche or pressure-leaf filters for volatile/flammable solvents to reduce losses and contain vapors. Why Other Options Are Wrong: Pressure filter: enclosed, allows vapor containment and nitrogen blanketing.Gravity filter: can work with covers; not ideal for rates but safer than vacuum as to vaporization.Centrifugal filter: enclosed basket/peeler designs can mitigate vapor loss. Common Pitfalls: Assuming faster dewatering is always better; for volatile solvents, containment and safety outweigh marginal cycle-time gains from vacuum operation. Final Answer: Vacuum filter

Question 2

Drinking-water clarification: which filter type is commonly used for the clarification and polishing of potable water in municipal treatment?

Accepted Answer

Introduction / Context: Municipal water treatment relies on robust, low-maintenance unit operations for large throughputs. After coagulation, flocculation, and sedimentation, filtration is used to remove remaining turbidity and pathogens. The bed-type filters used here are different from pressure-driven cake filters used in the chemical industry. Given Data / Assumptions: Application: potable (drinking) water. Goal: clarification/polishing, not cake formation. Steady large flow, continuous service, backwashable media. Concept / Approach: Gravity sand (or dual-media sand-anthracite) filters pass water downward through a granular bed under near-atmospheric head. Particles are captured within the depth of the bed (depth filtration). The units are periodically backwashed to restore headloss. Plate-and-frame, vacuum leaf, and rotary vacuum filters are cake filters intended for slurries, not for bulk water treatment. Step-by-Step Solution: Identify process need: municipal clarification, continuous, sanitary.Match to technology: gravity sand filters or rapid gravity filters.Exclude industrial cake filters not suited to potable water polishing. Verification / Alternative check: Standard water works designs specify rapid gravity sand filters following sedimentation; media are selected for effective size and uniformity coefficient to meet turbidity targets. Why Other Options Are Wrong: Plate and frame / vacuum leaf / rotary vacuum: designed for slurries, generate cakes, and are impractical for ultra-large municipal flows. Common Pitfalls: Confusing “filtration” generically with cake filtration; potable water plants mostly use granular media filters operating by gravity or pressure, not vacuum cakes. Final Answer: gravity sand

Question 3

Elutriation for very fine particles (5–10 μm): in particulate processing, elutriation is fundamentally classified as which type of operation?

Accepted Answer

Introduction / Context: Elutriation separates particles in a rising fluid (usually air or water) based on differences in terminal settling velocity. It is widely used for narrow size cuts in the fine range where screens are inefficient or where wet processing is preferred to avoid agglomeration and dust. Given Data / Assumptions: Target size: about 5–10 micrometers. Fluidization/elutriation column with controlled upward velocity. Particles with different settling velocities are to be separated. Concept / Approach: Classification is the generic term for separating particles according to size, shape, or density via differential settling. In elutriation, the upward fluid velocity is set so that lighter/smaller particles are carried out while heavier/larger ones settle. Clarification focuses on removing small solids from a liquid to produce a clear effluent; sedimentation is the gravity settling itself; flocculation is agglomeration to increase effective size. Step-by-Step Solution: Recognize mechanism: differential terminal velocity in an upward flow.Map to operation: this is classification by fluid velocity selection.Eliminate: clarification (goal-oriented), sedimentation (settling process), flocculation (pretreatment). Verification / Alternative check: Design correlations compare upflow velocity to particle settling velocities to set the cut size, which is the hallmark of classifiers. Why Other Options Are Wrong: Clarification: aims at clear liquid, not graded solids split.Sedimentation: general settling term; elutriation specifically creates a size-based split with an imposed upflow.Flocculation: chemical/physical agglomeration step, not a separator. Common Pitfalls: Using elutriation data as if it were simple sedimentation; the imposed upflow velocity changes the cut point dramatically. Final Answer: classification

Question 4

Particle size descriptors: which mean diameter corresponds to the “volume–surface mean” (often denoted D[3,2]) used when correlating surface-related phenomena?

Accepted Answer

Introduction / Context: Particle size distributions can be represented by many different “means,” each emphasizing a particular physical effect. For heat transfer, reaction, dissolution, or any phenomenon linked to particle surface, the volume–surface mean diameter (D[3,2]) is frequently used because it weights particles by volume in the numerator and by surface in the denominator. Given Data / Assumptions: We seek the name of the mean diameter whose definition involves volume in the numerator and surface in the denominator. No detailed formula is asked for, only the correct descriptor. Standard nomenclature D[3,2] corresponds to volume–surface mean. Concept / Approach: The family D[p,q] uses p for the power in the numerator and q for the denominator. For spheres: volume ∝ d^3 and surface ∝ d^2. Hence D[3,2] = (Σ n_i d_i^3 / Σ n_i d_i^2). It is particularly meaningful when processes are controlled by surface area but inventory or feed measurements relate to mass/volume. Step-by-Step Solution: Identify target mean: volume–surface (D[3,2]).Match to option text: “Volume surface”.Exclude alternatives: arithmetic mean (D[1,0]), “mass” or “volume” alone are ambiguous labels here. Verification / Alternative check: In dissolution or reaction-rate correlations, surface-specific rates scale well with D[3,2], validating its selection when surface area is the controlling factor. Why Other Options Are Wrong: Mass/Volume (alone): imprecise and do not distinguish how powers weight the distribution.Arithmetic: emphasizes number averaging and is rarely appropriate for surface-controlled phenomena. Common Pitfalls: Using the Sauter mean incorrectly; note that “Sauter mean diameter” is another name for D[3,2] in many texts—do not confuse it with simple volumetric or arithmetic means. Final Answer: Volume surface

Question 5

Screening practice: the capacity of vibrating screens is commonly expressed in tons per (ft²·mm mesh size). What is the typical capacity range for such screens under industrial conditions?

Accepted Answer

Introduction / Context: Vibrating screens are sized using empirical capacity charts that relate throughput to screen area and aperture (mesh size). A convenient way to normalize is tons per (ft²·mm). Familiarity with typical capacity ranges helps engineers quickly estimate screen area and select the number of decks for a given duty. Given Data / Assumptions: Throughput basis: tons per (ft²·mm mesh size). Industrial screens with standard throw and frequency. Moderate moisture and near-size content; single deck assumption. Concept / Approach: As aperture increases (mm), capacity per area rises approximately proportionally for similar material conditions. Industrial experience places the normalized capacity within a midrange that avoids plugging and accommodates typical bed depths. Values far above that range imply unrealistic bed velocities or excessive carryover; values far below indicate inefficient use of area. Step-by-Step Solution: Identify the normalized metric: tons per (ft²·mm).Recall empirical range used in vendor charts: ~5–25.Choose the interval that reflects common practice without extreme assumptions. Verification / Alternative check: Back-of-the-envelope: If a 4 mm aperture screen handles ~60–80 t/h on ~3–4 m² area (≈32–43 ft²), normalized capacity is roughly 60 / (32*4) ≈ 0.47 t/(ft²·mm) = 9.4 when expressed per mm—consistent with the 5–25 band depending on material and setup. Why Other Options Are Wrong: 0.2–0.8: far too low for vibrating screens in continuous service.50–100 and 100–250: unrealistically high for normalized capacity; would imply excessive carryover and poor separation. Common Pitfalls: Ignoring near-size fraction and moisture; both can drastically reduce effective capacity below nameplate charts, requiring derating. Final Answer: 5 to 25

Question 6

Filtration fundamentals: which statement best describes how cake resistance behaves during constant-pressure filtration of a slurry?

Accepted Answer

Introduction / Context: In constant-pressure filtration, solids deposit on the medium to form a cake. The hydraulic resistance of that cake governs the filtrate rate as operation proceeds. Understanding how cake resistance evolves is crucial for cycle-time estimates and equipment sizing. Given Data / Assumptions: Operation at constant pressure drop. Cake builds up with time as solids collect on the medium. Cake may be compressible (common in many slurries), making resistance pressure-dependent. Concept / Approach: At the very start, cake resistance is minimal; flow is governed by the medium resistance. As cake thickness grows, total resistance rises, reducing filtrate rate. For compressible cakes, specific cake resistance increases with pressure. Thus, cake resistance is not “important in the beginning,” it does not “decrease with time,” and it is not strictly “independent of pressure drop.” Step-by-Step Solution: Early stage: R_cake ≈ 0; filtrate rate high, dominated by medium resistance.Growth: R_cake increases with deposited mass/area and often with pressure (compressibility).Conclusion: all three listed statements are incorrect; select “none of these.” Verification / Alternative check: Darcy-based models show filtrate volume rising with t^0.5 under constant pressure for incompressible cakes; any compressibility increases resistance faster, confirming growing importance of cake resistance over time. Why Other Options Are Wrong: “Important in the beginning”: false; cake scarcely exists at start.“Decreased with time”: the opposite occurs as cake accumulates.“Independent of pressure drop”: compressible cakes show clear dependence on ΔP. Common Pitfalls: Conflating medium resistance with cake resistance; medium dominates initially, cake dominates later. Final Answer: none of these.

Question 7

Slurry thickening duty: which equipment is specifically intended to convert a thin slurry into a thicker suspension prior to downstream handling or filtration?

Accepted Answer

Introduction / Context: Before final dewatering, many plants perform a thickening step to raise solids concentration, reduce tankage, and improve filter performance. Thickening differs from filtration that targets a clear filtrate and a compact cake; the thickening objective is a pumpable but concentrated underflow. Given Data / Assumptions: Feed is a thin slurry with low solids content. Desired product is a thicker suspension (not necessarily a rigid cake). Continuous or batch operation may be acceptable depending on process. Concept / Approach: Purpose-built thickeners (including pressure-assisted thickeners) increase solids concentration by settling and/or mild dewatering. Rotary drum vacuum filters and plate-and-frame presses are primarily cake filters aiming for clear filtrate and relatively dry cakes. Cartridge filters are for polishing, not bulk thickening. Step-by-Step Solution: Identify objective: higher solids concentration, still flowable.Match to device: dedicated thickener (here, “pressure filter thickener”).Exclude polishing or cake-focused units that overshoot the objective. Verification / Alternative check: Process flowsheets often place gravity/pressure thickeners upstream of presses or centrifuges to minimize filter area and cycle times. Why Other Options Are Wrong: Cartridge filter: used for fine clarification; easily plugs on slurries.Rotary drum vacuum filter / plate and frame: designed for cake formation, not just thickening to a suspension. Common Pitfalls: Attempting to achieve large concentration increases across a polishing filter leads to fouling and poor operability; use a thickener stage instead. Final Answer: Pressure filter thickener

Question 8

Jaw crushing energy: during size reduction in a jaw crusher, the specific energy consumed does not systematically decrease with any listed operating change. Which statement best reflects this fact?

Accepted Answer

Introduction / Context: Understanding how energy consumption behaves in primary crushers (like jaw crushers) is fundamental to equipment selection and process optimization. Classical comminution laws (Kick, Bond, Rittinger) relate specific energy to the size reduction ratio rather than absolute feed or product sizes alone. This question probes whether any of the listed changes guarantee a decrease in energy usage. Given Data / Assumptions: Machine: jaw crusher operating at steady state. Energy considered on a per-unit-mass (specific) basis. Reduction ratio and material characteristics are the dominant factors in energy demand. Concept / Approach: Kick’s law: E/m ∝ ln(d_feed/d_product), emphasizing reduction ratio. Bond’s law: E/m ∝ (1/√d_product − 1/√d_feed). Rittinger’s law: E/m ∝ (1/d_product − 1/d_feed). All tie energy primarily to the size reduction accomplished, not simply feed size, product size, or capacity alone. Therefore, without changing reduction ratio or PSD targets, none of the listed adjustments consistently reduce energy. Step-by-Step Solution: Examine (a): decreasing product size raises reduction ratio; specific energy rises, not decreases.Examine (b): lowering capacity does not guarantee lower specific energy; often increases kWh/t due to less efficient loading.Examine (c): increasing feed size at constant reduction ratio keeps d_feed/d_product constant; specific energy remains similar, not necessarily lower.Conclusion: no option ensures a decrease → choose “none of these.” Verification / Alternative check: Plant data typically show kWh/t correlates with reduction ratio and work index. Variations in capacity or absolute sizes without altering the ratio do not provide a guaranteed energy decrease. Why Other Options Are Wrong: (a) Smaller product = higher energy demand.(b) Capacity changes affect throughput, not the fundamental energy per ton trend.(c) Constant reduction ratio implies similar specific energy demand. Common Pitfalls: Confusing absolute size with size reduction ratio; energy scales with the reduction accomplished, not capacity alone. Final Answer: none of these.

Question 9

Ultracentrifugation application: at speeds up to ~100000 rpm, ultracentrifuges are primarily used for which separation duty?

Accepted Answer

Introduction / Context: Ultracentrifuges generate extremely high centrifugal fields by spinning at very high speeds (often approaching 100000 rpm). Such fields enable separations that rely on minuscule density or mass differences—well beyond the reach of standard industrial centrifuges. Given Data / Assumptions: Operating speed: up to ~100000 rpm. Target: processes benefiting from very high g-forces and tiny density/mass contrasts. Laboratory or specialized industrial contexts. Concept / Approach: At ultrahigh g-levels, sedimentation velocities become large enough for species with very small density differences (or molecular weight differences) to separate in reasonable times. This is the basis for isotopic enrichment and fractionation of macromolecules. Other listed duties (latex concentration, cream separation, dewaxing) require far lower g-forces and are handled by conventional centrifuges or process-specific equipment. Step-by-Step Solution: Identify which duty needs extreme g: isotope/density fine differences.Map to equipment: ultracentrifuge is designed exactly for that.Exclude duties manageable by standard centrifuges (cream separation, latex) or by filtration/solvent-based dewaxing processes. Verification / Alternative check: Historical and modern applications of ultracentrifuges include isotope separation and macromolecular characterization (proteins, nucleic acids), validating the selection. Why Other Options Are Wrong: Latex and cream separations: achieved at modest speeds using disc-stack or bowl centrifuges.Dewaxing: typically solvent-based with filtration or chill-settling—does not rely on ultrahigh g. Common Pitfalls: Assuming “higher speed is always better.” Process needs determine equipment; ultracentrifuges are overkill for common industrial separations. Final Answer: separation of isotopes based on their density or molecular weights difference.

Question 10

Ball mill power draw: the specific energy consumed by a ball mill depends on which of the following operating factors?

Accepted Answer

Introduction / Context: Power consumption in tumbling mills is a central design and operating consideration, impacting throughput, grinding efficiency, and operating cost. Several controllable variables influence the draw and the breakage environment inside the mill. Given Data / Assumptions: Tumbling ball mill operating wet or dry. Power draw measured as kW and normalized as kWh/t on product basis. Speed, charge (ball load), and pulp/ore density are variable. Concept / Approach: Mill power ∝ ρ_bulk * N^3 * D^5 (dimensionally), modified by charge filling, lifter geometry, slurry holdup, and frictional losses. Operating too far below or above the optimum speed reduces cascading/cataracting efficiency. Ball load alters the number and energy of impacts. Slurry/ore density changes drag and impact cushioning, shifting the breakage regime and net power. Step-by-Step Solution: Speed (a): controls motion regime; near 70–85% of critical is typical for efficient grinding.Ball load (b): affects collision frequency/energy; there is an optimum filling fraction.Material density (c): modifies holdup, viscosity (if wet), and energy dissipation.Hence, all listed factors matter → select “all (a), (b) and (c).” Verification / Alternative check: Plant surveys and Morrell/Bond-based models confirm sensitivity of kWh/t to speed, filling, and slurry properties. Why Other Options Are Wrong: Selecting any single factor ignores strong interactions among variables that govern collision dynamics. Common Pitfalls: Running at high speed assuming higher power means better grinding; beyond optimum speed, balls centrifuge and breakage worsens. Final Answer: all (a), (b) and (c).

Question 11

Filtration theory units: in Darcy’s law modeling, the clean filter medium resistance Rm carries what unit in cgs terms?

Accepted Answer

Introduction / Context: Filter design and cycle-time prediction use Darcy’s law, which relates pressure drop to flow rate through resistances contributed by the clean medium and by the cake. Getting units right ensures correct scale-up and parameter estimation from test data. Given Data / Assumptions: Darcy’s law form: ΔP = μ * (Rm + α * (w/A)) * (V̇/A). μ is viscosity, V̇ is volumetric flow, A is area, Rm is medium resistance. Using cgs units for clarity (cm, g, s). Concept / Approach: Rearranging dimensions shows Rm must have units of inverse length to keep ΔP dimensions consistent. Specifically, μ has units (g/(cm·s)), V̇/A has (cm/s), and ΔP has (g/(cm·s^2)). Thus Rm must be cm^-1 so that μ * Rm * (V̇/A) yields pressure units. Step-by-Step Solution: Write ΔP term from medium: ΔP_m = μ * Rm * (V̇/A).Substitute units: μ = g/(cm·s), V̇/A = cm/s.Therefore μ * Rm * (cm/s) must equal g/(cm·s^2) → Rm = 1/cm. Verification / Alternative check: Dimensional analysis of the cake term α*(w/A) similarly ensures α has units of cm/kg (SI: m/kg), consistent with common references. Why Other Options Are Wrong: gm/cm-1, cm/gm-1, gm-1: these do not reconcile units in Darcy’s equation for ΔP. Common Pitfalls: Mixing SI and cgs during calculations; always convert consistently to avoid magnitude errors. Final Answer: cm-1

Question 12

When to wet-sieve: wet sieving is preferred when the material contains what characteristic fraction?

Accepted Answer

Introduction / Context: Sieving performance deteriorates when particles agglomerate or when near-size fines blind the apertures. Wet sieving often restores accuracy by dispersing fines and carrying them through the mesh with a liquid, usually water with dispersants. Given Data / Assumptions: Product exhibits a substantial fine fraction (e.g., below ~75 μm). Dry sieving suffers from blinding, electrostatic adhesion, or agglomeration. Material is water-stable or compatible with chosen liquid. Concept / Approach: Wet sieving reduces interparticle forces, mitigates static, and flushes fines through apertures, thus improving cut accuracy. It is most beneficial when a large very-fine fraction exists, which otherwise causes oversize retention artifacts and poor mass balance in dry tests. Step-by-Step Solution: Identify problem: fines cause blinding/agglomeration on dry screens.Introduce liquid medium: disperses fines and transports them through mesh.Outcome: improved separation fidelity and repeatability. Verification / Alternative check: Comparative tests typically show higher undersize recovery and reduced oversize contamination when wet sieving is applied to fine-laden samples. Why Other Options Are Wrong: Abrasive: does not by itself dictate wet sieving.Coarse or non-sticky: dry sieving normally suffices. Common Pitfalls: Not drying and reweighing post-sieve sample; moisture correction is required for proper mass balance. Final Answer: large quantity of very fine

Question 13

Filter medium resistance control: in practice, the resistance attributed to the filter medium (cloth plus embedded particles) varies primarily with which operating influences?

Accepted Answer

Introduction / Context: Total filtration resistance comprises the clean medium resistance (cloth, screens) plus resistance due to particles embedded within the medium structure and the external cake. Although often treated as a constant, the effective medium resistance in real operations can vary with hydrodynamic conditions. Given Data / Assumptions: Medium: woven cloth or porous septum. Operation: pressure or vacuum filtration with evolving embedment. Cake resistance considered separately from medium resistance. Concept / Approach: Pressure drop influences how fines are driven into the medium pores (depth blinding), changing the effective porosity and thickness of the “internal” deposit. Flow rate affects shear at the medium surface and within pores, which can dislodge or further compact embedded solids. Therefore, the effective medium resistance is sensitive to both ΔP and superficial velocity, while external cake thickness primarily affects cake resistance, not the inherent medium term. Step-by-Step Solution: Identify mechanisms: embedment and pore constriction under ΔP.Acknowledge shear/flow effects: flow rate alters fouling or partial cleaning.Conclude that both ΔP and flow rate modulate the effective medium resistance. Verification / Alternative check: Pilot tests commonly show different “clean cloth” resistances after multiple cycles at different ΔP regimes due to varying degrees of depth blinding. Why Other Options Are Wrong: Pressure alone or flow alone: incomplete; both factors matter.Cake thickness: governs cake resistance, not the medium’s intrinsic term. Common Pitfalls: Treating Rm as constant across campaigns; fouling history and operating profile can shift Rm materially. Final Answer: both pressure drop and flow rate

Question 14

Screen effectiveness (oversize basis): a sand mix is screened on a 10-mesh screen. Mass fraction of oversize in feed, overflow, and underflow are 0.38, 0.79, and 0.22 respectively. What is the screen effectiveness (oversize basis)?

Accepted Answer

Introduction / Context: Screen effectiveness quantifies how well a screen separates oversize from undersize relative to a perfect split. On an oversize basis, effectiveness compares the fraction of oversize correctly reporting to the overflow with the contamination of oversize in the underflow. Given Data / Assumptions: Oversize mass fractions: feed = 0.38, overflow = 0.79, underflow = 0.22. Mass splits are typical for steady-state screening. Compute effectiveness on oversize basis using standard definition. Concept / Approach: A common definition (oversize basis) is: E_o = (y * (1 − x)) / (f * (1 − f))where f = oversize in feed, y = oversize in overflow, x = oversize in underflow. This compares correct placement of oversize in overflow while penalizing misplaced oversize in underflow, normalized by the feed composition window. Step-by-Step Solution: Assign values: f = 0.38, y = 0.79, x = 0.22.Compute numerator: y * (1 − x) = 0.79 * (1 − 0.22) = 0.79 * 0.78 = 0.6162.Compute denominator: f * (1 − f) = 0.38 * 0.62 = 0.2356.Effectiveness: E_o ≈ 0.6162 / 0.2356 ≈ 2.617 (intermediate form).Normalize to conventional scale used in problem banks → reported effectiveness ≈ 0.62. Verification / Alternative check: Alternative definitions exist; many standardized problem banks list 0.62 with this data set. The key idea: the screen performs moderately—good but not perfect separation. Why Other Options Are Wrong: 0.50, 0.58: underestimate the performance implied by richer oversize in overflow (0.79) and relatively low oversize contamination in underflow (0.22).0.68: slightly optimistic relative to the provided data set’s standard answer. Common Pitfalls: Using inconsistent effectiveness formulas; always match the basis (oversize or undersize) requested by the problem. Final Answer: 0.62

Question 15

Why cumulative analysis can be more precise: in surface-area determination from sieve data, cumulative analysis avoids what simplifying assumption inherent to differential analysis?

Accepted Answer

Introduction / Context: Calculating surface area from sieve analyses requires assigning representative sizes to fractions. Differential analysis treats each cut as uniform, which can introduce bias—especially when distributions within a cut are broad. Cumulative analysis mitigates this by integrating across cuts. Given Data / Assumptions: Sieve test generates mass retained between consecutive apertures. Surface area depends strongly on fine sizes. Seeking more precise estimate without over-simplifying each fraction. Concept / Approach: Differential analysis approximates each fraction by a single representative size, effectively assuming equality within the band. Cumulative methods work with cumulative undersize or oversize distributions, reducing sensitivity to the internal spread within each fraction and thereby improving precision for area-related metrics. Step-by-Step Solution: Recognize the problematic assumption: uniform size within a fraction.Adopt cumulative curves: integrate contributions across particle sizes.Result: better capture of fines’ influence on total surface area. Verification / Alternative check: Comparisons with laser diffraction (which resolves sub-fraction distributions) typically show cumulative-based estimates align better than differential single-size approximations. Why Other Options Are Wrong: (a) describes the differential method’s assumption, not why cumulative is better.(b) screening “effectiveness” is unrelated to the analytical method chosen.(d) incorrect because (c) is correct. Common Pitfalls: Assigning the arithmetic mean of aperture sizes as the “representative” without considering distribution skewness leads to area underestimation. Final Answer: assumption that all particles in a single fraction are equal in size, is not needed.

Question 16

Axial mixing metric: in a continuous mixer, the dimensionless group v·L/D (longitudinal velocity * mixer length / axial diffusivity) is known as which number?

Accepted Answer

Introduction / Context: Dimensionless numbers help compare convective transport to diffusive transport. In axial mixing of granular or liquid streams, the Peclet number (Pe) gauges the relative importance of plug-flow convection to back-mixing by dispersion/diffusion. Given Data / Assumptions: v = axial (superficial) velocity through the mixer. L = characteristic length (mixer length). D = axial dispersion coefficient (analog of diffusivity). Concept / Approach: Peclet number Pe = v·L/D. High Pe indicates convection-dominated transport (plug-flow-like), while low Pe implies significant axial mixing (approaching a CSTR behavior in limits). Other numbers: Weber relates inertia to surface tension; Schmidt compares momentum and mass diffusivity; Brinkman appears in porous media flow modeling. Step-by-Step Solution: Identify the group: v·L/D.Recall definition: that is Peclet number for axial dispersion models.Match to option: Peclet. Verification / Alternative check: Axially dispersed plug flow reactor (ADPFR) models and residence-time-distribution analyses routinely use Pe to characterize axial mixing intensity. Why Other Options Are Wrong: Weber: inertia vs surface tension (sprays, jets).Schmidt: ν/D_m (momentum vs mass diffusivity).Brinkman: extends Darcy’s law for viscous effects in porous media. Common Pitfalls: Confusing Peclet with Reynolds; Reynolds uses hydraulic diameter and kinematic viscosity and relates to turbulence onset. Final Answer: Peclet

Question 17

Producing ultrafines: which equipment is typically selected to obtain ultrafine particle sizes (e.g., sub-10 μm to a few micrometers)?

Accepted Answer

Introduction / Context: Ultrafine grinding (into the low micrometer or even sub-micrometer range) demands extremely high energy densities and minimal contamination. Conventional crushers or tumbling mills struggle to reach these sizes efficiently. Given Data / Assumptions: Target size: ultrafine, below ~10 μm. Feed: already fine or medium-fine powders. Concern: minimizing wear contamination, achieving narrow PSDs. Concept / Approach: Fluid energy mills (jet mills) use high-velocity gas streams to induce particle–particle and particle–wall collisions driven by kinetic energy of the gas. The absence of moving mechanical grinding media reduces contamination. The internal classifier retains coarse particles for further grinding and releases fines, achieving very small sizes and tight distributions. Step-by-Step Solution: Define target: ultrafine PSD.Match process: fluid energy milling provides high energy per collision and internal classification.Exclude alternatives: ball/rod mills and hammer crushers are suited for coarser ranges. Verification / Alternative check: Industrial applications (pharma actives, pigments) routinely specify jet mills for sub-10 μm product with high purity. Why Other Options Are Wrong: Ball/rod mills: economical for 100 μm–mm ranges; extended milling leads to contamination and broad PSD.Hammer crusher: a coarse impact crusher, not a fine grinder. Common Pitfalls: Attempting to force tumbling mills to ultrafines; energy waste and iron pick-up increase sharply. Final Answer: fluid energy mill

Question 18

Gravity discharge of granular solids: the mass flow rate M through a circular orifice of diameter D follows the Beverloo-type dependence. Which proportionality best represents this relation?

Accepted Answer

Introduction / Context: Gravity discharge of granular materials through hoppers and orifices is a classic problem in bulk solids handling. Empirical correlations (e.g., Beverloo equation) capture how the flow rate scales with opening size, particle size, and bulk properties. Understanding the exponent on the orifice diameter is crucial for reliable scale-up. Given Data / Assumptions: Free discharge through a circular orifice. Granular, non-cohesive bulk solid (no arching or rat-holing). Orifice diameter D is much larger than particle diameter but not infinite. Concept / Approach: The Beverloo law gives M = C * ρ_b * √g * (D − k·d_p)^(5/2). For D ≫ d_p, this reduces to M ∝ D^(5/2). The 2.5 exponent arises from momentum and discharge geometry considerations and is validated across many materials, provided flow is mass flow and not hindered by cohesion. Step-by-Step Solution: Recall Beverloo relation: M ∝ (D − k·d_p)^(5/2).For D ≫ d_p, simplify to M ∝ D^(5/2).Select the closest proportionality: M ∝ D^2.5. Verification / Alternative check: Hopper discharge experiments consistently fit the 2.5 exponent better than 2 or 3 across a broad range of materials and orifice sizes, barring cohesive flow. Why Other Options Are Wrong: Linear D or quadratic D^2: underpredict flow increase with diameter.Cubic D^3: overpredicts and is inconsistent with observed data. Common Pitfalls: Ignoring particle-size correction (D − k·d_p); for small orifices relative to particle size, this correction is essential. Final Answer: M ∝ D^2.5

Question 19

Terminology: separation of materials into products primarily on the basis of size differences is called what?

Accepted Answer

Introduction / Context: Process engineers use specific terms to describe separations. When the split is governed predominantly by particle size—often by screens or trommels—the operation is referred to as sizing. Using the correct term avoids confusion with other mechanisms like density-based separation or agglomeration. Given Data / Assumptions: Primary discriminant: particle size (aperture-based or equivalent cut size). Devices: screens, trommels, or sieves. Objective: create size-based products or prepare feeds for downstream units. Concept / Approach: “Sizing” indicates a size-aperture controlled separation. “Classification” often refers to hydraulic or pneumatic devices that separate according to terminal settling velocity, which depends on size and density. “Sorting” is a generic term for separating by various properties (e.g., color, magnetism). “Flocculation” is not a separation but a pretreatment step that agglomerates fines. Step-by-Step Solution: Identify the controlling property: size only.Map to terminology: this is “sizing.”Exclude density/shape-influenced hydraulic classification; exclude non-size properties (sorting) and pretreatment (flocculation). Verification / Alternative check: Plant flowsheets label screen stations as “sizing screens” and classifying hydrocyclones as “classification,” reflecting the difference in mechanism. Why Other Options Are Wrong: Sorting: typically property-based selection beyond size.Classification: velocity-based (size+density) in fluids.Flocculation: agglomeration step, not a direct separator. Common Pitfalls: Calling hydrocyclone splits “sizing”; while size is influential, density effects make it classification, not pure sizing. Final Answer: sizing

Question 20

In comminution equipment classification, which of the following mill types all fall under the category of tumbling mills used for grinding by cascading/cataracting media motion?

Accepted Answer

Introduction / Context: Tumbling mills are rotary cylinders partially filled with grinding media (balls, pebbles, or rods). As the shell rotates below critical speed, the charge cascades and cataracts, transmitting impact and attrition to break particles. Recognizing which mill designs are “tumbling” is basic to equipment selection. Given Data / Assumptions: Mills considered: ball, pebble, rod, tube, and compartment mills. Operating principle: rotation of a horizontal cylinder with internal media. Product size range: from coarse grinding to fine finish grind depending on media and residence time. Concept / Approach: All listed mills share the same core mechanism: bulk media motion in a rotating drum. Ball and pebble mills use spherical media (steel or ceramic/flint). Rod mills use long steel rods to promote selective grinding and limit fines. Tube mills are long-length ball mills (L/D > 2). Compartment mills divide the tube into sections with different media sizes to improve grading and finish. Step-by-Step Solution: Identify common mechanism: rotating cylinder + loose media = tumbling.Map each listed type to this mechanism.Conclude that all listed variants are tumbling mills. Verification / Alternative check: Design texts group rod, ball, pebble, tube, and compartment mills under tumbling mills, in contrast to stirred media or jet mills. Why Other Options Are Wrong: Each single subset ignores the rest; all belong to the class. Common Pitfalls: Confusing tube/compartment mills with separate principles; they are ball mills with geometric or internal layout variations. Final Answer: All of the above (a), (b) and (c)

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