Q: Colloidal mills (rotor–stator type) are primarily employed for which level of grinding/dispersing service in liquids?

Introduction / Context: Colloidal mills are high-shear rotor–stator devices used to reduce droplet/particle size in suspensions and emulsions. They complement homogenizers and high-shear mixers in the wet processing toolkit. Concept / Approach: The narrow gap and high tip speed create intense shear and turbulence that break droplets/agglomerates down to fine sizes (often a few microms). However, they are not designed to produce dry, sub-micron powders (that role belongs to jet mills or media mills). For liquid systems, “fine” is the correct level—smaller than intermediate dispersers, but not truly ultrafine dry comminution. Step-by-Step Solution: Identify mechanism: liquid-phase high shear.Relate achievable sizes: fine dispersions in the low-microm range.Select “Fine”. Why Other Options Are Wrong: Coarse/Intermediate: understate the shear capability.Ultrafine: refers to dry micronization; not the typical domain of colloid mills. Final Answer: Fine

Q: Screw conveyors: which statement about operating speed and material suitability reflects standard practice?

Introduction / Context: Screw conveyors are ubiquitous for short, controlled transfers of powders, granules, and pastes. Their enclosed design limits dusting and allows metering, heating/cooling, and mixing along the trough. Concept / Approach: Standard screws do not run at “very high” rpm; excessive speed causes undue wear and product degradation. With correct flight design, clearance, and low speed, screws can move sticky or pasty materials that would bridge on belts. Conversely, highly abrasive solids cause significant screw and trough wear; while abrasion-resistant liners and hardfacing help, screws are not the first choice for severe abrasion. Step-by-Step Solution: Assess speed claim: typical screws operate at modest rpm to prevent surging and wear → (a) is incorrect.Assess sticky handling: with proper design, screws are used for sticky materials → (b) is correct.Assess abrasion: screws suffer wear with highly abrasive solids → (c) is incorrect as stated. Why Other Options Are Wrong: (a) Overstates speed; (c) ignores wear realities; (d) aggregates incorrect claims. Final Answer: They are suitable for sticky or pasty materials (with proper design).

Q: Grinding in a ball mill: which combined statements about wet vs. dry grinding, capacity vs. fineness, and power/cost vs. fineness are correct?

Introduction / Context: Ball mill performance depends on mill charge, speed, classification, and whether the circuit is wet or dry. Fineness targets directly affect throughput and energy consumption. Concept / Approach: Wet grinding suppresses dust, improves classification, and reduces coating on media and liners, typically achieving finer sizes than dry grinding. As fineness increases (smaller P80), specific energy per tonne rises; at fixed power, throughput decreases. Consequently, operating cost and power per tonne also increase with tighter grinds. Step-by-Step Solution: Evaluate (a): Wet circuits commonly reach finer sizes due to better transport and classification.Evaluate (b): Higher fineness requires more energy/tonne → lower t/h at fixed power.Evaluate (c): More energy and media wear mean higher cost → true.Therefore, (d) all three are correct. Final Answer: All of the above (a), (b) and (c).

Q: For transporting hot, lumpy, and highly abrasive bulk materials, which conveyor type is generally preferred for durability and reliability?

Introduction / Context: Materials handling equipment must match the thermal, mechanical, and abrasive characteristics of the load. High temperatures, sharp lumps, and abrasion quickly damage flexible belts or screws. Concept / Approach: Apron conveyors use overlapping metal pans bolted to heavy chains. They tolerate heat, impact, and abrasion far better than fabric belts. Screws and drag conveyors can move abrasive solids but suffer significant wear; belts are generally avoided for very hot or sharp, abrasive lumps unless specially engineered. Step-by-Step Solution: Identify key stresses: heat + impact + abrasion.Match to rugged design: metal pan apron conveyor.Select apron as preferred solution. Final Answer: Apron (pan) conveyor

Q: Why are cartridge filters sometimes described as “edge” filters in clarification service?

Introduction / Context: Cartridge filters employ wound fiber or pleated media to provide a tortuous path with a large surface area. Their retention arises from both surface sieving and depth capture, often described using “edge” terminology. Concept / Approach: The geometry of pleats or wound layers presents many “edges” where flow converges and fine particles are trapped. The bulk of solids removal occurs along these peripheral paths rather than a single flat face, hence the term “edge” filter. The term does not imply razor-sharp metal edges; it describes the interception mechanism. Step-by-Step Solution: Recognize cartridge morphology: pleated/wound depth media.Understand capture: edges and depth passages collect solids throughout the media.Therefore the “edge” label refers to periphery-dominant capture. Final Answer: Because most solids are intercepted along the cartridge edges/periphery (depth and edge effect)

Q: Fick’s law in transport phenomena relates to which physical process in fluids or solids?

Introduction / Context: Fick’s law is a cornerstone of mass transfer. It states that diffusive flux is proportional to the negative gradient of concentration, analogous to Fourier’s law for heat conduction. Concept / Approach: Fick’s first law: J = −D * (dC/dx), where J is molar flux by diffusion, D is the diffusivity, and dC/dx is the concentration gradient. Fick’s second law accommodates transient diffusion by combining the first law with the continuity equation. These do not address particle size, feed size, or energy in crushing; they govern molecular transport in gases, liquids, and solids. Why Other Options Are Wrong: Energy/particle size items belong to comminution, not mass diffusion. Final Answer: None of these (it describes molecular diffusion)

Q: Constant-pressure cake filtration (incompressible cake): 5 m³ of filtrate are produced in 20 minutes. How much additional time is needed to produce the next 5 m³ under the same constant pressure? (Repaired options.)

Introduction / Context: In constant-pressure filtration with an incompressible cake and negligible medium resistance, filtrate volume V grows with time t according to t = K * V^2, where K bundles viscosity, specific cake resistance, solids concentration, pressure, and area. This quadratic relation implies progressively longer times for equal volume increments. Given Data / Assumptions: Constant pressure operation throughout. Incompressible cake; medium resistance neglected. First 5 m³ obtained in 20 minutes. Find time for V = 5 m³ → 10 m³ increment. Concept / Approach: With t = K * V^2, evaluate K from the first data point, then compute incremental time for the next equal volume. Step-by-Step Solution: Use t = K * V^2.At V = 5 m³, t = 20 min ⇒ K = 20 / 25 = 0.8 min·m^-6.Time to reach V = 10 m³: t_10 = K * 10^2 = 0.8 * 100 = 80 min.Additional time for next 5 m³: Δt = t_10 − t_5 = 80 − 20 = 60 min. Verification / Alternative check: For constant pressure, plotting t versus V^2 yields a straight line through the origin. Doubling volume quadruples time; thus, from 20 min at 5 m³, four-fold gives 80 min at 10 m³ → +60 min more. Why Other Options Are Wrong: 20/30/40 minutes underestimate because filtration rate falls as cake thickens; equal volumes take longer at later stages. Common Pitfalls: Using a linear t ∝ V model (constant rate) instead of quadratic t ∝ V^2 for constant-pressure cake filtration. Final Answer: 60 minutes

Question 1

For general-duty belt conveyors handling bulk solids without cleats or special surfaces, what is the typical maximum slope (inclination) that can be used reliably?

Accepted Answer

Introduction / Context: The allowable slope for a belt conveyor depends on friction between the belt and material, particle properties, and the presence of cleats/sidewalls. Exceeding the recommended angle risks rollback and spillage. Given Data / Assumptions: Smooth, standard belt without cleats or high-friction lagging. Typical bulk solids (e.g., ores, grains) in continuous service. Conventional industrial design practices. Concept / Approach: General practice limits smooth belts to about 12–15° for many materials. While special belts with ribs/sidewalls can climb much steeper (even >30°), the typical textbook maximum for standard belts is around 15°. This balances capacity, stability of the bed, and housekeeping considerations. Step-by-Step Solution: Identify belt type: smooth, general-duty.Check common design heuristics: 12–15°.Select the conservative standard maximum: 15°. Verification / Alternative check: Vendor handbooks list 12–18° ranges depending on material “angle of surcharge”; 15° is a widely taught value. Why Other Options Are Wrong: 30°/45°/60° require special cleats or corrugated sidewalls; not general-duty smooth belts. Common Pitfalls: Assuming angle of repose equals allowable conveyor slope; rolling behavior differs from static repose. Final Answer: 15°

Question 2

To remove a very small amount of tiny solid impurities from a liquid stream (polishing duty), which separation device is preferred?

Accepted Answer

Introduction / Context: When solids loading is very low but high clarity is required, a polishing filter is selected to capture fine particulates without excessive footprint or complexity. Given Data / Assumptions: “Very small amount” of suspended fines. Objective: produce a clarified liquid. Operating mode: practical industrial service. Concept / Approach: Pressure filters (e.g., cartridge, plate-and-frame operated at pressure, sparkler filters) provide high driving force and fine retention in compact housings, making them ideal for polishing duties. Vacuum filters are efficient for higher solids slurries and cake discharge, not trace removal. Centrifuges excel for immiscible liquid separation or moderate solids, but for tiny amounts of fine solids, filtration media achieve finer nominal ratings. Coagulants assist clarification but are not a separation device by themselves. Step-by-Step Solution: Match low-solids polishing → fine media under pressure.Select pressure filter as the most suitable unit operation. Verification / Alternative check: Water polishing, chemical/petrochemical side-streams, and beverages commonly use cartridge or precoat pressure filters for final clarity. Why Other Options Are Wrong: Vacuum filter: better for bulk cake formation, not tiny residuals.Centrifugal filter: less effective for the finest, trace solids in low-solids feeds.Coagulant: chemical aid, not a physical separator. Common Pitfalls: Confusing low-solids polishing with bulk dewatering applications. Final Answer: Pressure filter

Question 3

Considering energy per unit mass of material, which equipment typically requires the highest specific energy input to achieve its target product fineness?

Accepted Answer

Introduction / Context: Specific energy in comminution rises dramatically as target size moves into the micron and sub-micron regime. Equipment designed for ultrafine grinding naturally exhibits higher energy per mass than coarse crushers or conventional grinders. Concept / Approach: Jaw crushers perform coarse size reduction at relatively low specific energy because crack propagation requires less surface creation per unit mass. Rod and ball mills increase energy usage as they grind finer. Fluid energy (jet) mills use very high-velocity gas to induce interparticle collisions, achieving extremely fine sizes but at high energy cost per kilogram of product. Step-by-Step Solution: Compare duty: coarse (jaw) < intermediate (rod/ball) < ultrafine (jet).Recognize that ultrafine targets require much more new surface area → higher energy.Select fluid energy mill as highest specific energy device. Why Other Options Are Wrong: Jaw/Rod/Ball mills operate at coarser targets with lower specific energy than jet mills for the same mass. Final Answer: Fluid energy (jet) mill

Question 4

Colloidal mills (rotor–stator type) are primarily employed for which level of grinding/dispersing service in liquids?

Accepted Answer

Introduction / Context: Colloidal mills are high-shear rotor–stator devices used to reduce droplet/particle size in suspensions and emulsions. They complement homogenizers and high-shear mixers in the wet processing toolkit. Concept / Approach: The narrow gap and high tip speed create intense shear and turbulence that break droplets/agglomerates down to fine sizes (often a few microms). However, they are not designed to produce dry, sub-micron powders (that role belongs to jet mills or media mills). For liquid systems, “fine” is the correct level—smaller than intermediate dispersers, but not truly ultrafine dry comminution. Step-by-Step Solution: Identify mechanism: liquid-phase high shear.Relate achievable sizes: fine dispersions in the low-microm range.Select “Fine”. Why Other Options Are Wrong: Coarse/Intermediate: understate the shear capability.Ultrafine: refers to dry micronization; not the typical domain of colloid mills. Final Answer: Fine

Question 5

Screw conveyors: which statement about operating speed and material suitability reflects standard practice?

Accepted Answer

Introduction / Context: Screw conveyors are ubiquitous for short, controlled transfers of powders, granules, and pastes. Their enclosed design limits dusting and allows metering, heating/cooling, and mixing along the trough. Concept / Approach: Standard screws do not run at “very high” rpm; excessive speed causes undue wear and product degradation. With correct flight design, clearance, and low speed, screws can move sticky or pasty materials that would bridge on belts. Conversely, highly abrasive solids cause significant screw and trough wear; while abrasion-resistant liners and hardfacing help, screws are not the first choice for severe abrasion. Step-by-Step Solution: Assess speed claim: typical screws operate at modest rpm to prevent surging and wear → (a) is incorrect.Assess sticky handling: with proper design, screws are used for sticky materials → (b) is correct.Assess abrasion: screws suffer wear with highly abrasive solids → (c) is incorrect as stated. Why Other Options Are Wrong: (a) Overstates speed; (c) ignores wear realities; (d) aggregates incorrect claims. Final Answer: They are suitable for sticky or pasty materials (with proper design).

Question 6

Grinding in a ball mill: which combined statements about wet vs. dry grinding, capacity vs. fineness, and power/cost vs. fineness are correct?

Accepted Answer

Introduction / Context: Ball mill performance depends on mill charge, speed, classification, and whether the circuit is wet or dry. Fineness targets directly affect throughput and energy consumption. Concept / Approach: Wet grinding suppresses dust, improves classification, and reduces coating on media and liners, typically achieving finer sizes than dry grinding. As fineness increases (smaller P80), specific energy per tonne rises; at fixed power, throughput decreases. Consequently, operating cost and power per tonne also increase with tighter grinds. Step-by-Step Solution: Evaluate (a): Wet circuits commonly reach finer sizes due to better transport and classification.Evaluate (b): Higher fineness requires more energy/tonne → lower t/h at fixed power.Evaluate (c): More energy and media wear mean higher cost → true.Therefore, (d) all three are correct. Final Answer: All of the above (a), (b) and (c).

Question 7

For transporting hot, lumpy, and highly abrasive bulk materials, which conveyor type is generally preferred for durability and reliability?

Accepted Answer

Introduction / Context: Materials handling equipment must match the thermal, mechanical, and abrasive characteristics of the load. High temperatures, sharp lumps, and abrasion quickly damage flexible belts or screws. Concept / Approach: Apron conveyors use overlapping metal pans bolted to heavy chains. They tolerate heat, impact, and abrasion far better than fabric belts. Screws and drag conveyors can move abrasive solids but suffer significant wear; belts are generally avoided for very hot or sharp, abrasive lumps unless specially engineered. Step-by-Step Solution: Identify key stresses: heat + impact + abrasion.Match to rugged design: metal pan apron conveyor.Select apron as preferred solution. Final Answer: Apron (pan) conveyor

Question 8

Why are cartridge filters sometimes described as “edge” filters in clarification service?

Accepted Answer

Introduction / Context: Cartridge filters employ wound fiber or pleated media to provide a tortuous path with a large surface area. Their retention arises from both surface sieving and depth capture, often described using “edge” terminology. Concept / Approach: The geometry of pleats or wound layers presents many “edges” where flow converges and fine particles are trapped. The bulk of solids removal occurs along these peripheral paths rather than a single flat face, hence the term “edge” filter. The term does not imply razor-sharp metal edges; it describes the interception mechanism. Step-by-Step Solution: Recognize cartridge morphology: pleated/wound depth media.Understand capture: edges and depth passages collect solids throughout the media.Therefore the “edge” label refers to periphery-dominant capture. Final Answer: Because most solids are intercepted along the cartridge edges/periphery (depth and edge effect)

Question 9

Fick’s law in transport phenomena relates to which physical process in fluids or solids?

Accepted Answer

Introduction / Context: Fick’s law is a cornerstone of mass transfer. It states that diffusive flux is proportional to the negative gradient of concentration, analogous to Fourier’s law for heat conduction. Concept / Approach: Fick’s first law: J = −D * (dC/dx), where J is molar flux by diffusion, D is the diffusivity, and dC/dx is the concentration gradient. Fick’s second law accommodates transient diffusion by combining the first law with the continuity equation. These do not address particle size, feed size, or energy in crushing; they govern molecular transport in gases, liquids, and solids. Why Other Options Are Wrong: Energy/particle size items belong to comminution, not mass diffusion. Final Answer: None of these (it describes molecular diffusion)

Question 10

Constant-pressure cake filtration (incompressible cake): 5 m³ of filtrate are produced in 20 minutes. How much additional time is needed to produce the next 5 m³ under the same constant pressure? (Repaired options.)

Accepted Answer

Introduction / Context: In constant-pressure filtration with an incompressible cake and negligible medium resistance, filtrate volume V grows with time t according to t = K * V^2, where K bundles viscosity, specific cake resistance, solids concentration, pressure, and area. This quadratic relation implies progressively longer times for equal volume increments. Given Data / Assumptions: Constant pressure operation throughout. Incompressible cake; medium resistance neglected. First 5 m³ obtained in 20 minutes. Find time for V = 5 m³ → 10 m³ increment. Concept / Approach: With t = K * V^2, evaluate K from the first data point, then compute incremental time for the next equal volume. Step-by-Step Solution: Use t = K * V^2.At V = 5 m³, t = 20 min ⇒ K = 20 / 25 = 0.8 min·m^-6.Time to reach V = 10 m³: t_10 = K * 10^2 = 0.8 * 100 = 80 min.Additional time for next 5 m³: Δt = t_10 − t_5 = 80 − 20 = 60 min. Verification / Alternative check: For constant pressure, plotting t versus V^2 yields a straight line through the origin. Doubling volume quadruples time; thus, from 20 min at 5 m³, four-fold gives 80 min at 10 m³ → +60 min more. Why Other Options Are Wrong: 20/30/40 minutes underestimate because filtration rate falls as cake thickens; equal volumes take longer at later stages. Common Pitfalls: Using a linear t ∝ V model (constant rate) instead of quadratic t ∝ V^2 for constant-pressure cake filtration. Final Answer: 60 minutes

Question 11

In mineral classification during beneficiation, which fluid medium is commonly employed to carry and stratify particles?

Accepted Answer

Introduction / Context: Classifiers separate particles based on settling velocity in a fluid. Depending on the size range and the process, the fluidizing medium can be air or water. Concept / Approach: Hydraulic classifiers and hydrocyclones use water; they are ubiquitous for ores and sands. Air classifiers and cyclones use air for dry separations and are preferred when water is undesirable or when drying is costly. The choice depends on particle size, density, moisture sensitivity, and downstream processing. Final Answer: Either air or water (depending on device and size range)

Question 12

For clarifying lube oils and printing inks (very fine solids in viscous liquids), which high-g device is most effective?

Accepted Answer

Introduction / Context: Lube oils and printing inks often contain very fine particulates that are difficult to remove by filtration due to viscosity and particle deformability. High-g centrifuges exploit density differences to accelerate separation. Concept / Approach: Disc-stack centrifuges (often branded Sharples “supercentrifuges”) provide very high centrifugal fields and large clarifying area per footprint, enabling efficient removal of sub-microm to few-microm particles from viscous liquids. Sparkler and precoat filters work well for many duties but may clog rapidly on viscous streams and deformable fines. Standard disc-bowl centrifuges are effective, but supercentrifuges extend the g-range and throughput for challenging fluids like inks. Final Answer: Sharples (supercentrifuge, high-g disc-stack)

Question 13

The Weber number (We) is most relevant to which mixing/dispersion phenomenon in fluid systems?

Accepted Answer

Introduction / Context: The Weber number compares inertial forces to interfacial (surface tension) forces: We = ρ * u^2 * L / σ. It indicates the propensity for drop or bubble breakup in turbulent fields. Concept / Approach: When We exceeds a critical value, inertial stresses overcome surface tension, causing a dispersed liquid phase to break into smaller droplets. Thus We is central to predicting drop size in liquid–liquid dispersions and bubble breakup in gas–liquid systems. While it appears in broader mixing discussions, its signature application is liquid-in-liquid dispersion control. Final Answer: Dispersion of a liquid into another immiscible liquid (drop breakup)

Question 14

Additives that promote the flocculation (aggregation) of suspended particles are typically which of the following?

Accepted Answer

Introduction / Context: Flocculation combines fine particles into larger aggregates that settle or filter more readily. It is widely used in water treatment and mineral processing. Concept / Approach: Electrolytes compress the electrical double layer and reduce repulsive forces (DLVO theory), enabling van der Waals attraction and bridging by polymers. Certain surfactants modify surface energies and can induce hydrophobic interactions that promote aggregation. Commercial flocculants are often polyelectrolytes (anionic/cationic) that act via charge neutralization and polymer bridging. Final Answer: Both (a) and (b)

Question 15

Classifier capacity for solids throughput: which expression (in tons of solids per hour) is dimensionally and practically correct? (A = cross-sectional area, m^2; V = rising fluid velocity, m/s; S = volume fraction solids; ρ = solids density, kg/m^3…

Accepted Answer

Introduction / Context: Classifier capacity relates volumetric flux through cross-sectional area to solid content and density to yield mass rate. Converting SI to tons/hour introduces the factor 3.6 (since 1 kg/s = 3.6 t/h). Concept / Approach: Volumetric flow of mixture through area A at velocity V is Q = A * V (m^3/s). The solids volume flow is Q_s = S * Q = A * V * S (m^3/s). Multiplying by solids density ρ (kg/m^3) gives mass flow m_s = A * V * S * ρ (kg/s). Converting to tons/hour multiplies by 3.6, giving 3.6 * A * V * S * ρ (t/h). Step-by-Step Solution: Compute Q = A V.Compute solids volumetric flux: S * Q.Compute solids mass rate: multiply by ρ.Convert kg/s to t/h with factor 3.6. Why Other Options Are Wrong: Omitting S ignores the fraction of solids.Dividing by ρ or dropping V produces incorrect dimensions. Final Answer: 3.6 A V S ρ

Question 16

For a compressible filter cake, where along the cake thickness is porosity usually the minimum (i.e., most compacted)?

Accepted Answer

Introduction / Context: Compressible cakes compact under increasing effective stress. Since pressure drop accumulates through the cake, compaction varies with depth, affecting permeability and specific resistance. Concept / Approach: The local pressure is highest near the filter medium because the entire upstream pressure drop is transmitted through the underlying cake. Thus particles are packed tighter there, giving smaller voids (lower porosity) and higher local specific resistance. Upstream layers experience lower compressive stress and therefore exhibit higher porosity. Final Answer: At the filter medium (downstream face)

Question 17

Which conveyor type is generally NOT recommended for highly abrasive bulk materials due to rapid wear and maintenance concerns?

Accepted Answer

Introduction / Context: Abrasive materials (e.g., sharp ores, silica sand) erode contact surfaces. Conveyor selection must consider wear life and downtime. Concept / Approach: Fabric belts and even many rubber compounds wear quickly under abrasion, especially at loading points and along idlers. Apron (metal pan) and heavy chain/flight conveyors are far more wear-resistant and are typical in quarries and sinter plants. While belts can be engineered with abrasion-resistant covers, for very abrasive duties they are still less robust and often uneconomic versus aprons or drag chains. Final Answer: Belt conveyor

Question 18

In a gyratory crusher, what is the dominant mechanism by which rock is broken?

Accepted Answer

Introduction / Context: Gyratory crushers and jaw crushers are principal compression machines for primary crushing of hard rock. Concept / Approach: In a gyratory crusher, a gyrating mantle inside a concave bowl compresses rock in a narrowing annular space. The material is crushed by nipping and compressive squeezing as it moves downward under gravity. Why Other Options Are Wrong: Impact is secondary in gyratories; primary impactors are separate machines.Attrition and cutting do not describe the core action in this device. Final Answer: Compression

Question 19

In a cyclone separator used for removing dust from a gas stream, how does the gas enter and move within the vessel?

Accepted Answer

Introduction / Context: Cyclone separators use swirling gas flow to generate centrifugal forces that drive particles to the wall, where they fall into a dust hopper while cleaned gas exits via a vortex finder. Concept / Approach: The gas typically enters tangentially at high velocity, establishing a strong vortex. Inside, the flow follows a double helical path: a downward outer spiral near the wall and an inner upward spiral exiting through the central outlet. This swirl produces radial acceleration that separates particles by inertia. Final Answer: Both (b) and (c)

Question 20

Paint manufacturing practice: for blending a light paste (low-to-medium viscosity) in paint and coatings industries, which mixer is typically used on the production floor?

Accepted Answer

Introduction / Context: Paint and coatings plants handle a spectrum of rheologies, from thin resins to heavy pastes. Choosing the correct mixer depends on viscosity, solids content, shear requirement, and batch size. This question targets practical shop-floor knowledge: which machine is routinely used to blend light paste batches in pails or drums before let-down, shade adjustment, or packing. Given Data / Assumptions: "Light paste" implies low-to-medium viscosity, not stiff doughs. Operation is blending/homogenizing pigments and resin already predispersed. Typical packaging vessels (cans or small drums) are used at the end of production. Concept / Approach: Charge can mixers (also called can mixers or pail mixers) mount directly on cans or small drums and use a propeller or folding blade to homogenize low-to-medium viscosity paint pastes. They are ideal for tinting, shade correction, and final blending without transferring product to a separate tank. Masticators and kneaders are high-torque batch mixers for rubber or very heavy pastes (e.g., putties, mastics) and are overkill for "light paste." Step-by-Step Solution: Identify viscosity regime: "light paste" → pail/drum blending feasible.Match equipment to task: portable/over-the-vessel blender preferred.Charge can mixer is designed for this duty: choose it. Verification / Alternative check: Standard operating procedures in paint shops specify can-mounted mixers for tinting and finish blending because they minimize cleaning, product transfer losses, and air entrainment when used correctly with appropriate blade diameter and speed. Why Other Options Are Wrong: Masticator: used in rubber compounding and very high-viscosity mastics, not typical paint light pastes.Kneader: sigma/planetary kneaders are for highly viscous doughs, sealants, or heavy putties.None of these: incorrect because a standard, widely used tool exists. Common Pitfalls: Confusing dispersion (done on high-speed dissolvers or bead mills) with simple blending; here the pigment is already dispersed, so a gentle can mixer suffices. Final Answer: charge can mixer

Mechanical Operations Questions