Question 1

In industrial mixing, which application is the helical screw agitator best suited for? Select the scenario that matches its typical use.

Accepted Answer

Introduction / Context:Choosing the right impeller or agitator is critical in chemical, polymer, and food processing. The helical screw (also called helical ribbon) agitator is a specialty device for moving extremely viscous materials that conventional turbines cannot handle effectively.Given Data / Assumptions:Process fluid is highly viscous (pastes, gels, doughs, creams).Goal: bulk circulation with minimal dead zones in a tank.Helical screw/ribbon agitators sweep close to the tank wall.Concept / Approach:At high viscosity, inertial effects diminish and laminar flow dominates. Standard propellers lose efficiency because they rely on turbulence. Helical screw agitators create axial and radial flow in laminar regimes by physically conveying material along the vessel wall and back through the core, ensuring turnover without excessive shear heating.Step-by-Step Solution:Identify the rheology: pastes with very high viscosity require positive-displacement-like motion.Match impeller: a helical screw agitator provides wall-sweeping action, reducing stagnant zones.Rule out alternatives: turbines/propellers are suited to low-to-moderate viscosity and gas–liquid dispersion.Verification / Alternative check:Industry practice confirms ribbon/helical screw mixers in adhesive compounding, battery pastes, plastisols, sealants, and dough mixing, which are all high-viscosity duties.Why Other Options Are Wrong:Blending immiscible low-viscosity liquids usually uses Rushton turbines, pitched-blade turbines, or high-shear mixers, not helical screws.Very high temperature (> 250 °C) does not alone determine impeller type; material rheology is the key factor.High-turbulence gas–liquid dispersion needs radial turbines, not ribbon/screw agitators.Common Pitfalls:Assuming temperature dictates impeller choice; ignoring laminar vs turbulent regime; or selecting high-speed propellers for pastes (inefficient and heat-intensive).Final Answer:Mixing highly viscous pastes

Question 2

In smooth steel crushing rolls handling ordinary rock, what is the typical angle of nip (in degrees)? Choose the value most commonly cited for practical design.

Accepted Answer

Introduction / Context:The angle of nip in roll crushers is a foundational concept in comminution. It is the maximum angle between the two roll faces at which a particle can be nipped (gripped) for crushing without slipping. Properly estimating this angle ensures efficient crushing and prevents excessive recirculation.Given Data / Assumptions:Material: ordinary rock with typical frictional characteristics against steel.Rolls: smooth steel.We seek the commonly used design angle.Concept / Approach:The angle of nip depends on the coefficient of friction between particle and roll surface and the particle size relative to roll diameter. For smooth rolls and common rock, handbooks report a practical design range around 15°–20°, with 16° often quoted as the representative value for calculations.Step-by-Step Solution:Relate nip condition to friction: higher friction allows a larger nip angle.For smooth steel and ordinary rock, empirical data give a typical angle near 16°.Select the nearest standard value used in equipment sizing: 16°.Verification / Alternative check:Design examples for roll crushers frequently adopt 16° to ensure conservative sizing and reliable nipping for a wide range of feed particles.Why Other Options Are Wrong:32°, 40°, and 46° imply very high friction or special surfaces (e.g., corrugated), which are not applicable to smooth steel rolls for ordinary rock.22° is above the typical handbook value and risks slippage in smooth-roll practice.Common Pitfalls:Confusing corrugated/roughened rolls (higher nip angles) with smooth rolls; ignoring material friction; or assuming angle increases with feed size regardless of surface condition.Final Answer:16°

Question 3

In a jaw crusher under normal operation, which component experiences the maximum wear and tear over time? Select the part most frequently replaced during maintenance.

Accepted Answer

Introduction / Context:Jaw crushers are primary crushers that reduce large rocks by compressive action between a fixed and a moving jaw. Knowing which parts wear fastest helps with preventive maintenance planning, spares inventory, and lifecycle cost analysis.Given Data / Assumptions:Crusher type: single- or double-toggle jaw crusher.Material: typical mineral/rock feed with abrasiveness.Operating in standard duty.Concept / Approach:Wear correlates with rubbing, impact, and compression zones. The jaw plates (liners) form the crushing chamber where the feed is gripped and broken. They see the highest contact pressure, sliding abrasion, and impact—especially at the lower portion near the discharge where compressive work completes.Step-by-Step Solution:Identify zones of highest stress: the crushing faces on fixed and swing jaws.Relate to wear mechanisms: sliding abrasion and compressive micro-fracture concentrate on jaw plate surfaces.Conclude the most frequent replacement item: jaw plates.Verification / Alternative check:Maintenance manuals and OEM recommendations schedule regular inspections and replacement of jaw liners, often with options like manganese steel or composite profiles to extend life.Why Other Options Are Wrong:Check plates and toggles can wear or fail, but not as consistently as jaw plates.Drive shaft and flywheel are critical but generally experience fatigue rather than continuous abrasive wear.Common Pitfalls:Assuming toggles wear fastest because they are moving; overlooking that toggles are primarily load-transfer/relief components while jaw plates are the working wear surfaces.Final Answer:Jaw plates

Question 4

When mixing plastic solids (e.g., polymeric masses, dough-like materials) in processing, which operation most directly facilitates uniform mixing and dispersion of fillers or pigments?

Accepted Answer

Introduction / Context:Plastic and polymer processing frequently requires mixing solid or highly viscous masses with additives such as pigments, stabilisers, or fillers. The operation chosen must generate distributive and (when needed) mild dispersive mixing in a high-viscosity regime.Given Data / Assumptions:Material behaves as a plastic solid or very viscous paste.Goal: uniform incorporation of additives without degrading the base polymer.Equipment examples: sigma-blade mixers, internal (Banbury-type) mixers, twin-roller mills.Concept / Approach:Kneading applies strong shear and elongational deformation while continuously folding the mass. This promotes distributive mixing (spatial uniformity) and controlled dispersive mixing (breaking agglomerates). Mastication is more specific to raw rubber softening; high-speed propellers are ineffective in laminar, very viscous regimes.Step-by-Step Solution:Identify rheology: solid-like or doughy, implying laminar regime and poor flow with propellers.Match to unit operation: kneading devices exert strong shear at low speeds with intermeshing blades.Conclude: kneading is the correct mixing method for plastic solids.Verification / Alternative check:Industrial practice uses sigma or Banbury mixers for PVC plastisols, masterbatches, and thermoplastic compounds, confirming kneading as the standard choice.Why Other Options Are Wrong:Dispersion with propellers works for low-viscosity liquids, not for plastic solids.Mastication targets molecular breakdown/softening of natural rubber, not general polymer compounding.Spray atomization is unrelated to solid/paste mixing.Common Pitfalls:Confusing kneading with high-speed shear mixing; attempting propeller mixing in a paste, leading to overheating and poor uniformity.Final Answer:Kneading (using sigma or Banbury-type mixers)

Question 5

Considering shape factors in particle technology, what is the sphericity of Raschig rings (short hollow cylinders used as packings)? Choose the most appropriate qualitative value.

Accepted Answer

Introduction / Context:Sphericity is a dimensionless shape factor used to compare irregular particle shapes to a sphere. It affects surface area, drag, packing, and heat/mass transfer. Raschig rings are commonly used as tower packings and are distinctly non-spherical.Given Data / Assumptions:Sphericity, ψ, is defined as: surface area of a sphere with the same volume as the particle divided by the actual surface area of the particle.A perfect sphere has ψ = 1.Raschig rings are short hollow cylinders (non-spherical and open-ended).Concept / Approach:Because Raschig rings have more surface area for a given volume compared to a sphere, the ratio ψ is less than 1. This is consistent with other non-spherical packings such as saddles and rings where ψ typically falls below unity due to increased external and internal surface intricacies.Step-by-Step Solution:Recall ψ = A_sphere(V_particle)/A_particle.For ring-shaped bodies, A_particle is relatively large for the enclosed volume.Therefore ψ must be less than 1.Verification / Alternative check:Published data for industrial packings list ψ values below 1 for Raschig rings, Berl saddles, and Intalox types, confirming the qualitative choice.Why Other Options Are Wrong:1 corresponds to a perfect sphere, which a Raschig ring is not.Values like 3 or 0.5 are arbitrary here; only the inequality correctly captures the property broadly for this packing.Common Pitfalls:Assuming sphericity reflects “roundness” visually; or confusing ψ with porosity or packing factor.Final Answer:

Question 6

Among the following equipment, which one is not typically described as a “wet classifier” used for liquid–solid classification?

Accepted Answer

Introduction / Context:Liquid–solid classification separates particles based on settling behavior in fluids. Common wet classifiers include rake classifiers, spiral classifiers, hydrocyclones, and elutriators. Some high-speed centrifuges are primarily for phase separation rather than granular classification.Given Data / Assumptions:Devices listed include a high-speed centrifuge and hydro-mechanical classifiers.We distinguish devices meant for classification of particles in slurry from those mainly used for liquid–liquid separation or polishing.Concept / Approach:Hydrocyclones and rake classifiers are canonical wet classifiers. An elutriator uses upward flow to classify by size/settling rate. A Sharples (Sharples) supercentrifuge is primarily for liquid–liquid separation (e.g., cream separation) or very fine polishing, not the typical granular size classification duty performed by rake/hydrocyclone units.Step-by-Step Solution:Identify standard wet classifiers: hydrocyclone, rake classifier, elutriator.Differentiate centrifuges: Sharples supercentrifuges target phase separation at very high g, not routine granular classification.Select the outlier: Sharples supercentrifuge.Verification / Alternative check:Process flowsheets show hydrocyclones/rake classifiers between grinding stages, whereas supercentrifuges appear in dairy, biotech, or fine polishing contexts.Why Other Options Are Wrong:Hydrocyclones classify by centrifugal settling in slurry.Rake classifiers separate by hindered settling in a tank.Elutriators use counter-current flow for cut-size control.Common Pitfalls:Equating any centrifuge with a classifier; overlooking the typical duty of each device.Final Answer:Sharples (Sharples) supercentrifuge

Question 7

In a hammer mill, what is the dominant mechanism of size reduction? Consider the interactions among hammers, housing (liner), and the feed particles.

Accepted Answer

Introduction / Context:Hammer mills are widely used for coarse-to-intermediate grinding of minerals, biomass, and chemicals. Understanding the breakage mechanism helps optimize rotor speed, screen size, and liner design to balance capacity and product fineness.Given Data / Assumptions:Typical high-speed rotor with pivoted hammers.Perforated screens or grates control top size.Feed is friable to moderately hard.Concept / Approach:Hammer mills impart high-velocity impacts as feed contacts rapidly moving hammer tips and the breaker plates/liners. Secondary grinding occurs by attrition and inter-particle collisions as fragments circulate and decelerate against the housing and screen surface.Step-by-Step Solution:Primary breakage: impact by hammer tips against particles.Secondary breakage: attrition/scrubbing against liners and screens.Additional refinement: inter-particle impacts within the mill chamber.Verification / Alternative check:Design correlations for specific energy and tip speed (V_tip = π D N) assume impact dominance, while wear patterns on liners confirm attrition and rubbing contributions.Why Other Options Are Wrong:(a) alone ignores inter-particle interaction, which contributes significantly at higher loadings.(b) alone understates the primary role of hammer impact.Compression is not the main mechanism in a hammer mill.Common Pitfalls:Assuming only one mechanism; overlooking the role of screen clearance and recirculation in attrition.Final Answer:Both (a) and (b)

Question 8

Among the following bulk materials, which must be stored in silos (enclosed storage) and not in an open yard due to handling, loss, and safety considerations?

Accepted Answer

Introduction / Context:Storage strategy affects quality, loss, and safety of bulk solids. Fine fractions and self-heating materials require special handling to prevent wind loss, spontaneous combustion, or contamination. This question tests practical bulk storage knowledge.Given Data / Assumptions:Coke breeze: very fine coke fraction used in sintering or fuel blending.High volatile-matter (VM) bituminous coal: prone to oxidation; large stockpiles are common but need care.Sand: inert and non-reactive; open storage is typical.Concept / Approach:Fine, dusty materials such as coke breeze are highly susceptible to wind losses, environmental emissions, and non-uniform moisture pickup. Enclosed storage (silos/bins) minimizes loss and controls dust. While high VM coal requires fire-preventive practices, it is still commonly stored in engineered open yards with compaction, layering, and temperature monitoring.Step-by-Step Solution:Identify which material is finest and dustiest: coke breeze.Assess risk: wind loss, dust emission, contamination, and handling non-uniformity.Conclude enclosed storage (silos) is best for coke breeze.Verification / Alternative check:Iron and steel plants store coke breeze in bins/silos feeding sinter plants to ensure consistent dosing without environmental losses.Why Other Options Are Wrong:High VM coal can be stored in open yards with precautions (pile design, compaction, monitoring), though risks exist.Sand and clinker are commonly stored in open yards or sheds; dust and loss issues are manageable.Common Pitfalls:Assuming any oxidizable fuel cannot be stored outdoors; ignoring particle size and wind susceptibility as decisive factors for silo storage.Final Answer:Coke breeze

Question 9

For handling hot, abrasive “ash” and similar residues from boilers and furnaces, which conveyor type is generally the most suitable and robust choice?

Accepted Answer

Introduction / Context:Ash handling demands equipment that tolerates abrasion, elevated temperature, and sometimes wet/slurry conditions. The conveyor must provide positive displacement transport with minimal spillage and high wear resistance.Given Data / Assumptions:Material: hot, abrasive ash or clinker-like residues.Operating environment: harsh, often at boiler/furnace discharge.Goal: reliable, contained conveyance.Concept / Approach:Flight (scraper) conveyors drag material along the bottom of a trough using flights fixed to chains. This positive mechanical conveying is suited to abrasive, hot materials, allowing cooling sections or quench baths. Belt conveyors are vulnerable to abrasion and heat; ribbon conveyors suit mixing, not bulk transfer; dilute-phase pneumatic systems risk erosion and dust control issues.Step-by-Step Solution:Match material severity (hot, abrasive) to conveying principle: chain-and-flight design.Evaluate alternatives: belts wear rapidly; ribbons mix; pneumatic lines erode and dust handling is complex.Select the robust option: flight (scraper) conveyor.Verification / Alternative check:Power plants and foundries widely use submerged scraper conveyors or dry flight conveyors for bottom ash and clinkers, confirming the selection.Why Other Options Are Wrong:Drag/slat is related and can be used, but “flight (scraper)” is the canonical ash-handling choice in many designs.Belts lack heat and abrasion resistance for continuous ash duty.Ribbon conveyors are mixers, not bulk conveyors.Pneumatic dilute-phase risks severe pipeline erosion and dust control challenges.Common Pitfalls:Overlooking temperature limits of belts; underestimating abrasion on pneumatic elbows; confusing slat with scraper terminology without considering specific ash system design.Final Answer:Flight (scraper) conveyor

Question 10

In a rotating ball mill, are internal baffles typically installed to aid mixing and grinding, or is the mill operated without baffles?

Accepted Answer

Introduction / Context:Tumbling mills (ball, rod, pebble) rely on cascading and cataracting of the grinding media. The internal geometry—liners, lifters, and mill speed—governs the charge motion. Whether fixed baffles are used affects flow patterns.Given Data / Assumptions:Conventional cylindrical ball mill with lifter/liner profiles.Grinding media motion produced by rotation near a fraction of critical speed.We examine the need for “baffles.”Concept / Approach:Ball mills do not employ baffles the way agitated tanks do. Instead, lifter bars and liners create the desired media trajectory. Introducing baffles would obstruct rolling/cascading movement and reduce effective grinding action.Step-by-Step Solution:Recognize that media motion is governed by mill speed, fill level, and liner/lifter design.Understand that fixed baffles would interfere with free movement of balls and slurry/pulp.Conclude standard design: no baffles in ball mills.Verification / Alternative check:Equipment drawings show shell liners and lifters but no static baffles; process control relies on speed, load, and liner profile, not baffling.Why Other Options Are Wrong:“Horizontal baffles” and “only two baffles” are not standard features of ball mills.Baffles are neither universally required nor recommended in dry or wet grinding for tumbling mills.Common Pitfalls:Confusing mixing tanks (which often use baffles) with tumbling mills; assuming internal obstructions improve grinding.Final Answer:No baffles are provided

Question 11

Empirical sizing for ball milling relates the ball diameter (Db, metres) to the feed size (Df, metres) via a grindability constant K (≈ 0.9 to 1.4 for increasing hardness). Which relation is commonly used?

Accepted Answer

Introduction / Context:Selecting initial ball size is key for efficient breakage in a ball mill. A practical empirical guideline links ball diameter to the feed top size and grindability, ensuring sufficient impact energy without excessive wear.Given Data / Assumptions:Db = grinding media diameter (m), Df = characteristic feed size (m).K is a grindability constant (≈ 0.9–1.4, increasing with hardness).We seek a commonly cited proportionality.Concept / Approach:Energy transfer in impact breakage scales with ball mass and velocity; larger feed requires larger balls to generate required fracture energy. A frequently used rule is Db proportional to sqrt(Df), which can be written as Db^2 proportional to Df, with K capturing material effects.Step-by-Step Solution:Start from empirical observation: Db ∝ (Df)^0.5.Square both sides to express as Db^2 ∝ Df.Introduce K: Db^2 = K * Df.Verification / Alternative check:Industrial practice often begins with a top ball size derived from the square-root relation, refined by trials and Bond Work Index studies.Why Other Options Are Wrong:Db = K * Df suggests linear scaling, which overshoots at large feed sizes.Db^3 = K * Df or Db^2 = K * Df^2 do not reflect observed scaling in breakage tests.Common Pitfalls:Using oversized balls that increase wear and reduce fines production; ignoring material grindability (K).Final Answer:Db^2 = K * Df

Question 12

Which index most directly expresses the grinding characteristic (grindability) of solid fuels and minerals in routine mill selection and design?

Accepted Answer

Introduction / Context:Grinding indices quantify how easily a material can be ground. For coal and some minerals, a quick lab test provides a comparative measure that guides mill selection and expected power demand.Given Data / Assumptions:Coal and similar fuels are frequently specified by HGI.Other indices (angle of repose, abrasion, shatter) describe different properties.Concept / Approach:The Hardgrove Grindability Index (HGI) ranks materials on a relative scale—higher HGI indicates easier grinding. It is widely adopted in power and cement industries to compare coals and anticipate mill throughput and power needs.Step-by-Step Solution:Identify the property: grindability (ease of size reduction).Match to index: HGI directly reports grindability.Confirm industrial use: boiler coal specs include HGI targets.Verification / Alternative check:For mineral crushing/grinding circuits, Bond Work Index is also common; however, the question emphasizes routine characterisation where HGI is standard for coals.Why Other Options Are Wrong:Angle of repose measures flowability, not grindability.Shatter and abrasion indices address impact/abrasive wear, not ease of grinding.Common Pitfalls:Confusing HGI with Bond WI; using angle of repose to predict mill power.Final Answer:HGI (Hardgrove Grindability Index)

Question 13

A slurry contains 30% solids by volume (spherical sand) settling in a viscous oil. The measured hindered settling velocity is 4.44 μm/s. If the Richardson–Zaki index n = 4.5, what is the single-particle terminal velocity in the same fluid?

Accepted Answer

Introduction / Context:Hindered settling describes the reduced settling velocity of particles in concentrated suspensions. The Richardson–Zaki correlation relates hindered velocity to the single-particle terminal velocity and solids fraction. This is a staple calculation in sedimentation and classification design.Given Data / Assumptions:Solids volume fraction, φ = 0.30 (therefore fluid fraction = 1 − φ = 0.70).Hindered settling velocity, V_h = 4.44 μm/s.Richardson–Zaki index, n = 4.5 (appropriate to the regime and Re).Concept / Approach:The correlation is V_h = V_t * (1 − φ)^n, where V_t is the single-particle terminal velocity. Rearranging gives V_t = V_h / (1 − φ)^n. We substitute the given numbers and compute.Step-by-Step Solution:Write correlation: V_h = V_t * (1 − φ)^n.Insert data: 4.44 = V_t * (0.70)^4.5.Compute (0.70)^4.5 ≈ 0.2009; hence V_t ≈ 4.44 / 0.2009 ≈ 22.1 μm/s.Verification / Alternative check:The terminal velocity must exceed the hindered velocity because crowding impedes motion; 22.1 μm/s is indeed greater than 4.44 μm/s, consistent with physics.Why Other Options Are Wrong:0.90 and 1 μm/s are smaller than V_h, which is impossible for the same fluid/particle.0.02 μm/s is orders of magnitude too small.44.4 μm/s would imply (1 − φ)^n ≈ 0.10, inconsistent with φ = 0.30 and n = 4.5.Common Pitfalls:Using φ (solids) instead of (1 − φ); misapplying n for the wrong Reynolds regime; unit mistakes converting μm/s.Final Answer:22.1 μm/s

Question 14

In froth flotation, which reagent is added primarily to promote air-bubble attachment by rendering selected mineral surfaces hydrophobic?

Accepted Answer

Introduction / Context:Froth flotation separates valuable minerals from gangue using differences in surface hydrophobicity. Reagents play distinct roles: collectors make target mineral surfaces hydrophobic; frothers stabilize bubbles; modifiers adjust conditions; activators and depressants tune selectivity.Given Data / Assumptions:Goal: enhance air adhesion to mineral surfaces.Common collectors: xanthates, dithiophosphates, fatty acids (depending on mineral class).Concept / Approach:Collectors adsorb onto mineral surfaces, increasing contact angle so that bubbles attach and carry particles into the froth. Without adequate collector action, particles remain hydrophilic and do not report to the froth product.Step-by-Step Solution:Identify function: promote bubble attachment by increasing hydrophobicity.Match to reagent class: collectors perform this task directly.Cross-check roles of other reagents.Verification / Alternative check:Flowsheets and reagent schemes categorize xanthates/dithiophosphates as collectors; pine oil/MIBC as frothers; lime/soda ash as modifiers; CuSO4 as an activator for sphalerite.Why Other Options Are Wrong:Frothers mainly stabilize the froth rather than change surface hydrophobicity.Modifiers control pH/ionic strength; activators/depressants alter specific mineral responses but are not the primary hydrophobizing agents.Common Pitfalls:Confusing frother action with collecting; overdosing collectors leading to non-selective flotation.Final Answer:Collector

Question 15

In a cyclone separator, consider a small spherical particle of diameter Dp located at radius r from the rotation axis. For Stokes regime slip, what is the radial slip velocity of the particle (relative to gas) when the flow rotates at angular speed …

Accepted Answer

Introduction / Context:The motion of fine particles inside a cyclone involves a balance between outward centrifugal forcing and inward drag. In the laminar (Stokes) slip regime, an analytical expression relates the radial slip velocity to particle size, fluid properties, and rotational kinematics. This question tests recognition of that canonical relation.Given Data / Assumptions:Spherical particle of diameter Dp at radial location r.Angular speed of rotating flow field is ω (forced vortex idealisation locally).Low Reynolds number slip of the particle relative to gas; Stokes drag applies.Fluid viscosity μ; densities ρ_p and ρ_f.Concept / Approach:Radial slip arises from the net outward body force due to centrifugal acceleration acting on the excess mass of the particle over displaced fluid. For Stokes drag, drag force = 3 * π * μ * Dp * v_r. Effective outward body force per particle volume equals (ρ_p - ρ_f) * a_c, where a_c = ω^2 * r is centrifugal acceleration. Equating net body force to Stokes drag on a sphere yields the well known closed form for v_r.Step-by-Step Solution:Write force balance: Drag = Net body force.3 * π * μ * Dp * v_r = (π/6) * Dp^3 * (ρ_p - ρ_f) * ω^2 * r.Solve for v_r: v_r = ((ρ_p - ρ_f) * Dp^2 * ω^2 * r) / (18 * μ).Verification / Alternative check:Dimensional check: numerator has kg * m^-3 * m^2 * s^-2 * m = kg * m^0 * s^-2; denominator has Pa * s = kg * m^-1 * s^-1, and 18 is dimensionless. Result reduces to m * s^-1, consistent with velocity.Why Other Options Are Wrong:The form with ω (not ω^2) underestimates centrifugal forcing by an order in ω.The form linear in Dp ignores Stokes sphere scaling which is proportional to Dp^2.Omitting (ρ_p - ρ_f) neglects buoyancy correction.v_r = ω * r is the tangential gas speed in a forced vortex, not radial particle slip.Common Pitfalls:Confusing tangential gas velocity with radial slip; forgetting buoyancy correction; using turbulent drag instead of Stokes where Re_p is small.Final Answer:v_r = ((ρ_p - ρ_f) * Dp^2 * ω^2 * r) / (18 * μ)

Question 16

In comminution nomenclature, which mill type is commonly referred to as a “disintegrator” because it employs high-speed impact within a cage?

Accepted Answer

Introduction / Context:Grinding equipment is named based on the dominant breakage mechanism and internal geometry. Disintegrators are a class of high-speed impact mills in which feed is shattered by multiple impacts within a confined rotor and stator cage assembly.Given Data / Assumptions:We compare typical mills used in minerals and process industries.Focus on the term disintegrator used in many handbooks to denote cage mills.Concept / Approach:Cage mills consist of concentric cages rotating in opposite directions or a single cage against a stationary housing. Feed enters the center and is hurled outward, undergoing repeated high-energy impacts that rapidly disintegrate friable materials. This is distinct from tumbling mills where grinding media cascade under gravity.Step-by-Step Solution:Identify equipment where impact by bars within a cage is principal: cage mill.Exclude mills based on cascading media (pebble, compartment), which are tumbling types.Conclude the disintegrator corresponds to the cage mill.Verification / Alternative check:Manufacturers and classic references list cage mills under disintegrators for fertilizer, gypsum, and limestone size reduction at modest reduction ratios but high throughput.Why Other Options Are Wrong:Compartment and pebble mills are tumbling grinders using media; not disintegrators.“All tumbling mills” is overly broad and incorrect.Rod mill is a tumbling mill using rods and is not termed a disintegrator.Common Pitfalls:Confusing generic “impact mill” with all high-speed mills; overlooking the distinctive cage rotor geometry.Final Answer:Cage mill

Question 17

A Cottrell (electrostatic) precipitator removes particulate matter from gas streams primarily by using which phenomenon?

Accepted Answer

Introduction / Context:Electrostatic precipitators (often called Cottrell precipitators) are standard equipment for cleaning dusty flue gases in power, cement, and metallurgical industries. Understanding the underlying charging and collection mechanism is essential for design and troubleshooting.Given Data / Assumptions:Gas contains fine particulates requiring high collection efficiency.High-voltage discharge electrodes and grounded collection plates are present.Concept / Approach:Corona discharge is a non-thermal electrical discharge generated at sharp electrodes under high voltage. It creates a cloud of ions that charge particulates as they pass the electrostatic field. Charged particles then drift to oppositely charged or grounded plates under Coulombic forces and are removed by rapping.Step-by-Step Solution:Establish that particulate charging requires a stable ion source.Identify corona discharge as the steady ionisation mechanism used in ESPs.Recognize that particle migration happens due to electrostatic forces toward collection plates.Verification / Alternative check:Performance formulas include migration velocity proportional to electric field and particle charge acquired in the corona region, confirming the role of corona rather than spark breakdown.Why Other Options Are Wrong:Spark breakdown is disruptive and undesirable for continuous collection.Magnetic trapping applies only to ferromagnetic particles, not general dust.Inertial impaction and sonic agglomeration are different technologies with lower fine-particle efficiency compared to ESPs.Common Pitfalls:Confusing corona with arcing; assuming alternating current alone ensures collection without ion generation.Final Answer:Corona discharge with high-voltage electrodes

Question 18

For a solid particle that is a perfect cube, what is its sphericity ψ (defined as area of a sphere with equal volume divided by actual particle area)? Provide the canonical closed form.

Accepted Answer

Introduction / Context:Sphericity ψ is a fundamental shape factor in particle technology, influencing drag, heat and mass transfer, and packing. It is defined as the surface area of a sphere having the same volume as the particle divided by the actual surface area of the particle. Spheres have ψ = 1 and all other shapes have ψ Given Data / Assumptions:Particle shape: perfect cube with side length a.Definition: ψ = A_sphere(V_cube) / A_cube.Concept / Approach:Compute the area of the equivalent-volume sphere and divide by the area of the cube. For a cube, volume V = a^3 and surface area A_cube = 6 * a^2. The sphere with equal volume has radius r such that (4/3) * π * r^3 = a^3. Its surface area is A_sphere = 4 * π * r^2.Step-by-Step Solution:Equate volumes: (4/3) * π * r^3 = a^3 ⇒ r = a * (3/(4π))^(1/3).Compute A_sphere: 4 * π * r^2 = 4 * π * a^2 * (3/(4π))^(2/3).Compute ψ: ψ = A_sphere / A_cube = [4 * π * a^2 * (3/(4π))^(2/3)] / (6 * a^2) = (π/6)^(1/3).Verification / Alternative check:Numerically, (π/6)^(1/3) ≈ 0.806, which is less than 1 as expected for a cube and matches handbook tables.Why Other Options Are Wrong:π, π/3, and 1 are not dimensionless sphericities consistent with the definition for a cube.(π/6)^(1/2) corresponds to a different ratio and exceeds known cube sphericity values.Common Pitfalls:Using the wrong definition (e.g., volume shape factor) or forgetting to compute the equivalent-volume sphere correctly.Final Answer:(π/6)^(1/3)

Question 19

In filtration theory, what is the compressibility coefficient (s) for an absolutely compressible filter cake?

Accepted Answer

Introduction / Context:Filter cake compressibility is central to predicting how specific cake resistance changes with applied pressure in constant-pressure or constant-rate filtration. The compressibility coefficient s (sometimes n) characterises the sensitivity of cake resistance to pressure.Given Data / Assumptions:Definition: α = α_0 * (ΔP)^s, where α is specific cake resistance and s is the compressibility coefficient.Absolutely compressible cake implies maximum sensitivity to pressure.Absolutely incompressible cake implies no sensitivity to pressure.Concept / Approach:By convention, s = 0 corresponds to an incompressible cake (α independent of ΔP). As compressibility increases, s increases toward unity. An absolutely compressible cake exhibits a direct proportionality of resistance to pressure on a log-log basis with slope equal to 1.Step-by-Step Solution:Start from α = α_0 * (ΔP)^s.For absolutely compressible cake, resistance scales linearly with ΔP in log space, hence s = 1.For absolutely incompressible cake, s = 0 for comparison.Verification / Alternative check:Experimental plots of ln(α) versus ln(ΔP) give slope s. Highly compressible colloidal cakes show slopes close to 1, matching the limiting idealisation.Why Other Options Are Wrong:0 denotes incompressible cake.“Between 0 and 1” is true for partially compressible cakes but not for the absolute case.“∞” and “Depends only on pressure drop” are non-physical or vague in this context.Common Pitfalls:Confusing s with porosity exponent; misreading absolute compressibility as infinite resistance instead of maximum pressure sensitivity.Final Answer:1

Question 20

Roll crusher performance: which operating variables primarily affect both the specific power consumption and the production rate in a roll crusher?

Accepted Answer

Introduction / Context:Roll crushers are used for intermediate to fine crushing of ores, coal, and industrial minerals. Their economy depends largely on how much energy they draw per ton and how much material they can pass per hour. This question probes the two most influential variables engineers adjust during design and operation.Given Data / Assumptions:Two-roll crusher handling brittle, non-cohesive solids.Gap setting establishes target product top size.Speeds may be equal or intentionally differential to improve nip and pull-in.Concept / Approach:Specific power consumption and throughput are controlled by how much size reduction is imposed (reduction ratio) and by the material transport/biting mechanics at the rolls (influenced by roll speed and especially differential speed). Higher reduction ratios usually require more energy per ton and can lower capacity if the gap is tight. Differential roll speeds improve nip probability and shearing, often raising throughput but also impacting power draw.Step-by-Step Solution:Relate energy to size reduction: specific energy increases with increasing reduction ratio.Relate throughput to kinematics: differential speed enhances material draw-in, reducing slippage and increasing capacity.Conclude: both reduction ratio and differential roll speed affect power and rate.Verification / Alternative check:Pilot runs commonly show a capacity rise when differential speed is introduced at constant gap, with a corresponding change in kWh/t as the breakage mode shifts from pure compression toward combined compression-shear.Why Other Options Are Wrong:Reduction ratio alone (a): ignores strong influence of kinematics.Differential speed alone (b): ignores energy–reduction coupling.Neither (d): contradicted by operating data.Common Pitfalls:Assuming only the gap matters; in reality, speed ratio and surface condition also control performance.Final Answer:Both (a) and (b).

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