Question 1

Standard sieve terminology: a “200 mesh” screen corresponds to how many openings per linear inch in the U.S. standard sieve system?

Accepted Answer

Introduction: Understanding sieve nomenclature is essential for particle size analysis and specifying screen decks. “Mesh” indicates the number of openings per unit length, not per area, in the U.S. standard sieve series. Given Data / Assumptions: “200 mesh” is defined in the U.S. (Tyler/ASTM) practice. We are interested in the standard linear definition. Concept / Approach: Mesh number is the count of openings per linear inch measured from center-to-center of wires. While open area depends on wire diameter and spacing, the “mesh” designation itself is strictly linear. Therefore, “200 mesh” means 200 openings in one linear inch. Step-by-Step Solution: Identify definition: mesh = openings per linear inch.For 200 mesh, count is 200 per inch.Reject area-based interpretations (per inch² or per cm²). Verification / Alternative check: ASTM E11 and Tyler standards both define mesh as linear count; aperture size is then computed from mesh and wire diameter. Why Other Options Are Wrong: Per cm or per cm²: not the U.S. standard definition. Per inch²: would vastly overstate the count; mesh is not area-based. Per mm: not standard in U.S. mesh terminology. Common Pitfalls: Confusing mesh with aperture size; finer mesh does not uniquely determine opening size without knowing wire diameter. Final Answer: Per inch (linear)

Question 2

Pneumatic conveying design: the material throughput (capacity) of a pneumatic conveying system primarily depends on which of the following?

Accepted Answer

Introduction: Pneumatic conveying is widely used for powders and granules in chemical, food, and cement plants. Capacity prediction requires understanding how material properties, air conditions, and pipeline geometry interact to determine solids loading and pressure drop. Given Data / Assumptions: Steady dilute or dense phase conveying. Air as the conveying gas. Fixed blower/compressor performance curve and pipeline route. Concept / Approach: Throughput depends on both the driving force (air pressure/vacuum) and the resistances (friction in bends and straight runs) as well as the material’s conveyability (bulk density, particle size, cohesiveness). Larger diameters reduce gas velocity for a given volumetric flow, affecting saltation and pickup conditions. Bulk density alters solids loading ratio at the same volumetric feed. Step-by-Step Solution: Relate blower capacity and line resistance to available pressure drop.Note that diameter and routing (bends, vertical lifts) change frictional losses.Incorporate material factors (bulk density) into solids loading and pickup velocity.Therefore, capacity depends on all listed factors. Verification / Alternative check: Empirical design charts (e.g., Zenz–Othmer) and vendor software require inputs for material properties, air conditions, and pipeline geometry, confirming multi-factor dependence. Why Other Options Are Wrong: Any single-factor choice ignores the coupled nature of pressure drop and minimum conveying velocity. “None” contradicts well-established design practice. Common Pitfalls: Oversizing air velocity (causing attrition) or undersizing (leading to plugging); forgetting that elbows drastically increase pressure drop. Final Answer: All (a), (b) and (c)

Question 3

Agitation selection: which mixer type is generally most suitable for handling fibrous, dense slurries that require strong bulk circulation and solids suspension?

Accepted Answer

Introduction: Choosing an agitator involves matching impeller type to rheology and solids content. Fibrous, dense slurries need robust shear near the impeller and strong macro-circulation to prevent rope formation and dead zones. Turbine impellers are a common industrial choice for such duties. Given Data / Assumptions: Fibrous and dense slurries (higher apparent viscosity). Need for reliable suspension and dispersion. Conventional tank with baffles. Concept / Approach: Radial-flow turbines (e.g., Rushton) and pitched-blade turbines provide strong shear fields and top-to-bottom circulation, which help cut and disperse fibrous agglomerates while keeping solids suspended. Light-duty axial propellers are best for low-viscosity liquids; specialty mixers (helical ribbons, anchors) handle pastes but are not among the given options. Step-by-Step Solution: Match slurry characteristics to impeller duty: need shear and circulation.Propeller options are inadequate for dense/fibrous media.Turbine impellers deliver the required power per volume and mixing pattern.Hence, select turbine agitator. Verification / Alternative check: Handbooks recommend turbines for gas–liquid dispersion and solid suspension in moderate-to-high solids; field practice corroborates successful performance with fibrous slurries. Why Other Options Are Wrong: Propeller (axial): efficient for thin liquids, poor with fibers at high solids. Cone-type: not a standard high-shear impeller selection. Radial propeller: still light-duty relative to turbines. Common Pitfalls: Ignoring baffles (needed to prevent vortexing) and underestimating power number differences among impellers. Final Answer: Turbine agitator (radial or mixed-flow impeller)

Question 4

Use of filter aids (asbestos, kieselguhr/diatomaceous earth): how are they applied during cake filtration to raise porosity and reduce cake resistance?

Accepted Answer

Introduction: Filter aids are inert, highly porous powders that improve filtrate clarity and throughput by preventing fine solids from blinding the medium and by forming an open, permeable cake structure. Understanding application modes is central to troubleshooting slow filtrations. Given Data / Assumptions: Common aids: kieselguhr/diatomaceous earth, perlite, cellulose, asbestos (legacy). Goal: higher porosity, lower pressure drop across cake. Concept / Approach: There are two primary ways to use filter aids: as a precoat layer that protects the septum and as a body-feed mixed with the slurry so solids deposit interspersed with aid particles. After filtration, the aid is removed with the cake; depending on the product, it may be separated by dissolving product solids or thermally destroying the aid if compatible. Step-by-Step Solution: Precoat stage: form a uniform aid layer on the medium.Body feed: dose aid into slurry to keep pores open during run.Post-run: remove cake; choose separation method appropriate to chemistry (dissolution or burnout where safe). Verification / Alternative check: Industrial operating procedures (precoat + body feed) are standard for rotary vacuum filters and pressure leaf filters to delay terminal pressure rise and maintain flux. Why Other Options Are Wrong: Any single method alone is incomplete; all three statements are valid in practice. “None” contradicts routine filtration practice. Common Pitfalls: Overdosing aid (wasting capacity), failing to re-precoat after breakthrough, or choosing incompatible aids that contaminate the product. Final Answer: All ‘‘a’’, ‘‘b’’ & ‘‘c’’

Question 5

Diatomaceous earth in solid–liquid separation: which role best describes its common industrial use?

Accepted Answer

Introduction: Diatomaceous earth (DE), also called kieselguhr, consists of fossilized diatom skeletons rich in amorphous silica. Its micro-porous, lightweight structure makes it ideal in certain separation and purification operations. Given Data / Assumptions: Focus is on solid–liquid filtration applications. Product purity and clarity are important. Concept / Approach: DE is dosed as a filter aid (precoat and/or body feed) to create a highly permeable cake that captures fine particles while maintaining reasonable flow rates. It is not a structural filter medium by itself (a septum is still required), nor is it typically used as a catalytic active phase (though it can be a support in special cases). Step-by-Step Solution: Recognize DE’s properties: high porosity, low density, irregular microstructure.Map properties to function: forming a porous filtration layer.Conclude that its common role is as a filter aid, not as an explosive, structural medium, or catalyst. Verification / Alternative check: Brewing, edible oil, pharmaceuticals, and pool filtration widely use DE-based precoat/body-feed processes to achieve clarity with minimal blinding. Why Other Options Are Wrong: Explosive: DE is actually used as an absorbent to desensitize nitroglycerin (historically in dynamite), but by itself it is not the explosive. Permanent filter medium: DE requires a support such as a cloth or metal leaf. Catalyst: not active; at most a support. Abrasive only: while mildly abrasive, its core industrial role here is filtration aid. Common Pitfalls: Confusing DE’s role in dynamite as being the explosive; it merely acts as an absorbent/desensitizer. Final Answer: Filter aid to build a porous cake

Question 6

Mixing of highly viscous liquids and pastes in process vessels: which impeller choice from the list is generally preferred?

Accepted Answer

Introduction: Very viscous fluids and pastes require low-speed, high-torque impellers that can scrape, sweep, and move material without excessive shear heating or motor overload. Selecting the wrong impeller results in poor turnover and stagnant zones. Given Data / Assumptions: Target media: highly viscous liquids/pastes (high apparent viscosity, often non-Newtonian). Conventional baffled tanks may be used, but wall-sweeping is valuable. Concept / Approach: Multiple-blade paddles (and related anchor/helical ribbon mixers) provide bulk movement at low rpm and high torque, suitable for viscosities where propellers or turbines would simply churn locally without overall circulation. Turbines excel in moderate viscosities; propellers are for thin liquids with high flow numbers; gas-inducing designs address gas–liquid dispersion rather than paste handling. Step-by-Step Solution: Identify rheological demand: high torque, low speed.Map impeller types to viscosity windows.Choose multiple-blade paddles for pastes and very viscous fluids. Verification / Alternative check: Design charts (e.g., Metzner–Otto type correlations) show paddles/anchors/ribbons in the high-viscosity corner of selection maps, confirming suitability. Why Other Options Are Wrong: Propellers: inadequate pumping in high viscosity; poor turnover. Turbines: better than propellers, but still targeted to moderate viscosities. Gas-inducing impeller: used for gas–liquid mass transfer, not for pastes. Common Pitfalls: Running high rpm to compensate for poor circulation—this only increases power draw and shear without improving bulk mixing in viscous media. Final Answer: Multiple-blade paddles (slow-speed, high-torque)

Question 7

Size reduction using Bond’s law (energy estimate) Limestone grinding: The energy required to grind from a very large size to 100 μm is 12.7 kWh/ton. Using Bond’s law, estimate the energy needed to grind from a very large size down to 50 μm.

Accepted Answer

Introduction / Context: Bond’s law is widely used in mineral processing to estimate comminution energy. It relates specific energy consumption to the change in characteristic particle size and is especially useful for comparing energy needs at different target sizes for the same material and mill type. Given Data / Assumptions: Material: limestone (same ore, same work index). Initial size is effectively “very large,” so its contribution in the Bond expression is negligible. Energy to reach 100 μm from very large size: 12.7 kWh/ton. Target: estimate energy to reach 50 μm from very large size. Concept / Approach: Bond’s law (for very large feed) implies energy E is proportional to 1/√P80, where P80 is the product size. Therefore E2/E1 = (1/√P2)/(1/√P1) = √(P1/P2). If the product size is halved, energy increases by √2 because the inverse square-root term grows accordingly. Step-by-Step Solution: Let E1 correspond to P1 = 100 μm: E1 = 12.7 kWh/ton.For P2 = 50 μm, E2/E1 = √(100/50) = √2 ≈ 1.414.Compute E2: E2 = 12.7 * 1.414 ≈ 17.96 ≈ 18 kWh/ton. Verification / Alternative check: If we write E = k*(1/√P), then k = E1 * √P1 = 12.7 * 10 = 127. For P2 = 50 μm, 1/√50 ≈ 0.1414, so E2 ≈ 127 * 0.1414 ≈ 17.96 kWh/ton. Same result confirms the calculation. Why Other Options Are Wrong: 6.35 kWh/ton: would imply energy decreases when making a finer product, which contradicts Bond’s law. 9.0 kWh/ton: underestimates; still lower than the 100 μm case. 25.4 kWh/ton: corresponds to a doubling rather than √2 increase; too high for halving size. Common Pitfalls: Using a linear or inverse relationship with size instead of inverse square-root. Forgetting that “very large feed” makes the feed-size term negligible. Final Answer: 18 kWh/ton

Question 8

Mineral processing — principle of jigging Jigging (a stratification process in a pulsating fluid) primarily separates particles based on which property?

Accepted Answer

Introduction / Context: Jigging is a classic gravity concentration method used in coal and ore beneficiation. A pulsating fluid (usually water) stratifies a bed of particles so that those with different settling characteristics separate into layers for recovery. Given Data / Assumptions: Bed is pulsed vertically to promote stratification. Particles differ in density, size, and sometimes shape. Medium is water or air (less common). Concept / Approach: Although size can influence settling, jigging exploits differences in apparent specific gravity under pulsation. Denser particles settle more quickly and migrate downward, while lighter gangue rises. Size effects are mitigated by classification ahead of jigging and by the pulsation profile. Step-by-Step Solution: Recognize that jigging is a gravity separation method → density control.Denser valuable minerals settle to the bottom layers; lighter material forms upper layers.Thus, the primary separation basis is particle density. Verification / Alternative check: Industrial practice classifies feeds before jigging to narrow size range, confirming that density is the main discriminatory property. Why Other Options Are Wrong: Size and shape matter secondarily but are not the principal mechanism in well-operated jigs. “Mixed” is contrary to the goal of separation. Common Pitfalls: Confusing jigging with simple screening (size-based). Ignoring the need for proper pulsation and bed dilation for efficient density stratification. Final Answer: separated by particle density

Question 9

Centrifugation equipment selection Compared with disk-bowl centrifuges, tubular (tabular) bowl centrifuges typically:

Accepted Answer

Introduction / Context: For microbe and fine-solids separations, both disk-bowl and tubular-bowl centrifuges are used. Understanding their comparative features helps in choosing the right unit for bioprocessing and pharmaceutical clarification. Given Data / Assumptions: Tubular-bowl machines have narrow, long bowls with very high rotational speeds. Disk-bowl machines contain conical plates that create many parallel thin layers. Concept / Approach: Tubular-bowl centrifuges achieve extremely high g-forces thanks to higher rotational speed and small radius, making them suitable for ultrafine solid separation. Disk-bowl units are more versatile for continuous clarification but generally run at lower speeds than tubular bowls. Step-by-Step Solution: Compare speeds: tubular > disk-bowl → option (a) correct.Bowl diameter: tubular bowls are slender (smaller diameter), not larger → (b) false.Pressure capability: both designs have variants with sealing; blanket statements like (c) are not universally true → false.Fine solids: tubular bowls are actually excellent for very fine solids → (d) false. Verification / Alternative check: Manufacturer specs list tubular-bowl speeds often above 15,000–20,000 rpm, exceeding many disk-bowl speeds. Why Other Options Are Wrong: (b) Contradicts geometry of tubular bowls. (c) Overgeneralizes; sealing options exist. (d) Opposite of common application (cell debris, yeast, fine pigments). Common Pitfalls: Assuming larger diameter equals better g-force; actually g ∝ ω^2 r and tubular bowls compensate with very high ω. Final Answer: operate at higher speed

Question 10

Ball mill scaling — power requirement vs. mill diameter For a given ball load, the power required to drive a ball mill is proportional to which function of the mill diameter D?

Accepted Answer

Introduction / Context: Power draw correlations for tumbling mills are crucial for scale-up. A commonly used empirical relation shows a strong dependence of power on the mill diameter for mills with similar loading and speed fraction of critical. Given Data / Assumptions: Same ball load fraction, liner condition, and speed fraction. Geometric similarity is assumed. Concept / Approach: Empirical and semi-theoretical treatments (e.g., Bond/Rowland) indicate that mill power P scales approximately as D^2.5 * L, where L is length. Holding L proportional or fixed, the diameter exponent remains dominant, thus P ∝ D^2.5. Step-by-Step Solution: Recognize the standard scaling: P ∝ D^2.5 (for given load and operating conditions).Eliminate inverses 1/D and 1/D^2.5 as contrary to observed behavior.Linear D underpredicts power growth with size; D^2.5 matches practice. Verification / Alternative check: Industrial datasets corroborate the super-quadratic dependence of power on D for dynamically similar mills. Why Other Options Are Wrong: D or inverse powers do not reflect measured power draw trends in scale-up. Common Pitfalls: Using only cross-sectional area (∝ D^2) and ignoring charge lift and trajectory effects that steepen the exponent. Final Answer: D^2.5

Question 11

Ball mill operation: which statement is correct for selecting ball size and operating speed?

Accepted Answer

Introduction / Context: Ball mills reduce particle size through impact and attrition between tumbling balls and the charge. Matching media size and speed to feed size and desired product is central to efficient comminution. Given Data / Assumptions: Larger balls deliver higher impact energy suited to coarse particles. Critical speed defines the onset of centrifuging; operation is normally at a fraction of critical (e.g., 65–80%). Fine feeds benefit from smaller balls for increased surface contact and attrition. Concept / Approach: Impact energy scales with ball mass and velocity. Coarser particles need larger balls to fracture effectively. Operating above critical speed causes balls to stick to the shell (centrifuging), reducing grinding action; thus mills run below critical speed. For fine grinding, smaller media increases contact frequency and surface area. Step-by-Step Solution: Relate particle size to required impact: coarse → larger balls.Reject the claim that fine feed needs larger balls; inverse is true.Reject running above critical speed; typical operation is below. Verification / Alternative check: Design handbooks recommend graded charges with larger top sizes for coarser feeds and lower mill speeds as a fraction of critical. Why Other Options Are Wrong: (b) contradicts standard practice; (c) contradicts the definition of critical speed; (d) is invalid because (a) is correct. Common Pitfalls: Ignoring mixed-media charges; single-size charges are less efficient across wide feed distributions. Final Answer: Coarse feed requires a larger ball size.

Question 12

Froth flotation reagents: pine oil serves primarily as which type of reagent?

Accepted Answer

Introduction / Context: Froth flotation separates mineral phases by selectively attaching hydrophobic particles to air bubbles. Reagents include collectors (adsorb to make surfaces hydrophobic), frothers (stabilise froth and control bubble size), modifiers (pH regulators, depressants), and activators (enable collector adsorption). Given Data / Assumptions: Pine oil is a classic frother used historically in sulfide ore flotation. Its role is to produce a stable, fine-bubble froth. It does not function as a collector for specific mineral surfaces. Concept / Approach: Frothers reduce surface tension and prevent bubble coalescence, leading to persistent froth that carries hydrophobic mineral particles to the surface. Pine oil and modern alcohol/glycol frothers are selected based on froth stability and selectivity requirements. Step-by-Step Solution: Identify pine oil's function: stabilise froth and control bubble dispersion.Differentiate from collectors (e.g., xanthates), activators (e.g., copper sulfate), and modifiers (e.g., lime, NaCN).Choose 'Frother' accordingly. Verification / Alternative check: Flowsheets list pine oil under frothers in reagent addition schemes for rougher/cleaner circuits. Why Other Options Are Wrong: Collector/activator/modifier are distinct reagent classes with different mechanisms and chemistries. Common Pitfalls: Assuming all organic reagents are collectors; functionality depends on adsorption mechanism and hydrophile–lipophile balance. Final Answer: Frother

Question 13

Stirred tank power scaling at high Reynolds number: if the power number is constant, how does power input vary with impeller speed N?

Accepted Answer

Introduction / Context: In turbulent mixing (high Reynolds number), the power number Np becomes approximately constant for a given impeller–tank geometry. Power scaling with speed is fundamental for mixer design and scale-up. Given Data / Assumptions: Np = P / (rho * N^3 * D^5) is constant at high Re. Fluid density rho and impeller diameter D are fixed. We vary only the rotational speed N. Concept / Approach: Rearranging the definition of Np: P = Np * rho * N^3 * D^5. With Np, rho, D fixed, power P scales as N^3. This cubic dependence underpins motor sizing and energy-cost estimates during process optimization. Step-by-Step Solution: Start with Np = P / (rho * N^3 * D^5).Solve for P: P = Np * rho * D^5 * N^3.Hence, P ∝ N^3. Verification / Alternative check: Log–log plots of power versus speed show slope near 3 for fully turbulent regimes with constant Np. Why Other Options Are Wrong: N^0, N^1, N^2: underestimate the steep increase of power with speed in turbulent mixing. Common Pitfalls: Applying viscous regime scaling (P ∝ N^2 or N^1) outside of their valid Reynolds number ranges. Final Answer: N^3

Question 14

Size reduction of fibrous materials: which equipment is most suitable for breaking fibrous feed?

Accepted Answer

Introduction / Context: Material breakage behavior depends on structure. Fibrous materials (e.g., dehydrated plant matter, textiles) resist compressive fracture but can be effectively torn, sheared, and impacted using devices designed for such morphology. Given Data / Assumptions: Fibrous feed stock with high aspect ratio fibres or mats. Need for tearing/shearing rather than pure compressive crushing. Comparison among common comminution devices. Concept / Approach: Squirrel-cage disintegrators employ high-speed rotating beaters/cages that impart shear and impact, well-suited to ripping fibres apart. Roll crushers are best for friable, coarse size reduction by compression. Ball and tube mills grind brittle, granular solids by impact/attrition and are inefficient for fibrous tearing. Step-by-Step Solution: Match fibrous behavior to shearing/impact mechanisms.Select squirrel-cage disintegrator as the appropriate device.Eliminate compression-centric or attrition-centric mills for fibrous stock. Verification / Alternative check: Industrial practice uses hammer mills, cutter mills, and disintegrators for fibrous biomass; the squirrel-cage disintegrator is a classical example. Why Other Options Are Wrong: Roll crusher: compressive; Ball/Tube mill: suited to brittle granular feed, not fibrous tearing. Common Pitfalls: Assuming higher energy in ball mills guarantees efficiency regardless of material morphology; mechanism–material matching is crucial. Final Answer: Squirrel-cage disintegrator

Question 15

Grindability as a material property: which parameter listed does NOT affect a material’s inherent grindability?

Accepted Answer

Introduction / Context: Grindability describes how easily a material can be reduced in size under specified conditions. It depends on intrinsic mechanical properties, not on the external dimensions of the feed pieces (which affect equipment selection rather than inherent grindability). Given Data / Assumptions: Grindability relates to hardness, toughness, brittleness, and elasticity (affecting rebound). Feed 'size' is an external parameter that influences staging of crushers/mills but not the intrinsic property. Standard indices (e.g., Bond Work Index) are referenced to material behavior, not strictly to starting piece size. Concept / Approach: Intrinsic properties determine fracture initiation and propagation under impact, compression, and attrition. While large initial size may require pre-crushing, it does not change the underlying energy required per unit new surface area or the relative ease of grinding compared to another material. Step-by-Step Solution: Identify intrinsic versus extrinsic variables.Elasticity, hardness, toughness → intrinsic; size → extrinsic.Select 'Size (initial piece size)' as not affecting inherent grindability. Verification / Alternative check: Testing protocols normalise feed to set sizes to isolate material properties; the results are attributed to inherent grindability, not the starting dimension. Why Other Options Are Wrong: Elasticity affects rebound; hardness and toughness directly influence crack formation and energy consumption. Common Pitfalls: Confusing equipment capacity limits with material grindability; both matter but are distinct. Final Answer: Size (initial piece size)

Question 16

Air pollution control: which device is best suited to purify fine dust-laden air streams?

Accepted Answer

Introduction / Context: Industrial air streams often carry particulate matter. Device selection depends on particle size distribution, loading, gas temperature, and required outlet emissions. Fine dust removal requires high collection efficiency at small particle sizes. Given Data / Assumptions: Air stream contains fine dust (sub-micron to tens of microns). We seek high efficiency and low outlet particulate concentration. Comparing cyclone, bag filter, gravity settler, and tubular centrifuge. Concept / Approach: Bag filters remove fine particles by filtration through a fabric medium, achieving high efficiencies across a broad size range. Cyclones rely on inertial separation and are better for coarse particles. Gravity settlers work only for very coarse particles with long residence times. Tubular centrifuges are typically used for liquid–solid separation in small liquid volumes, not for large air flows. Step-by-Step Solution: Match device principle to particle size: fine dust → surface filtration.Select bag filter for high-efficiency fine particulate capture.Eliminate cyclone/settler for fine cut sizes and centrifuge for gaseous streams. Verification / Alternative check: Emission standards in many plants are met using baghouses or electrostatic precipitators (ESPs); baghouses are robust for fine dust. Why Other Options Are Wrong: Cyclone/settler are low-efficiency at fine sizes; tubular centrifuge is not typical for air service. Common Pitfalls: Assuming cyclones suffice for all dust problems; fine PM requires filtration or electrostatic capture. Final Answer: Bag filter (fabric filter)

Question 17

Definition of crushing efficiency: choose the correct ratio expressing crushing efficiency.

Accepted Answer

Introduction / Context: Efficiency metrics compare useful output to input energy. In comminution (crushing/grinding), only a tiny fraction of the input energy actually creates new surface area (useful work). Most is dissipated as heat, elastic/plastic deformation, and noise. Understanding this definition highlights why comminution is energy-intensive. Given Data / Assumptions: Useful work in crushing = surface energy associated with new surfaces created. Input energy = energy supplied to the machine (or absorbed by it). We ignore minor differences between motor input and energy delivered to breakage zone. Concept / Approach: Crushing efficiency η is defined as η = (Surface energy created) / (Energy supplied). Because the surface energy increment is small relative to total power draw, η is typically very low (well under 1%). This underscores the importance of choosing appropriate machines and operating conditions to minimise energy waste. Step-by-Step Solution: Identify the 'useful output': creation of new surface area.Identify the 'input': energy supplied to the machine.Form the ratio: η = Surface energy / Supplied energy. Verification / Alternative check: Classical texts contrast Rittinger's, Kick's, and Bond's laws for energy–size relationships; all imply low overall efficiencies in practical crushing operations. Why Other Options Are Wrong: They invert the ratio or substitute non-standard definitions that do not reflect the efficiency concept. Common Pitfalls: Confusing 'energy absorbed by the solid' with 'energy supplied to the machine'; losses occur between the two, but the standard efficiency reference uses supplied energy. Final Answer: Surface energy created by crushing / Energy supplied to the machine

Question 18

Jaw/gyratory crusher capacity correlation: capacity in tons per hour is approximately equal to which expression? (L = length of receiving opening, cm; S = greater width of discharge opening, cm)

Accepted Answer

Introduction / Context: Crusher sizing often uses empirical capacity correlations involving receiving and discharge opening dimensions. For jaw and gyratory crushers, a widely cited rule-of-thumb relates capacity to the product L * S multiplied by a constant. Given Data / Assumptions: L: length of receiving opening in cm. S: greater width of discharge opening in cm. Capacity in tons per hour is proportional to L * S with an empirical coefficient near 0.087. Concept / Approach: The correlation embodies average performance over typical operating ranges and bulk densities. It is not a replacement for detailed manufacturer curves, but it serves as a screening estimate in preliminary design and equipment comparison. Step-by-Step Solution: Write the empirical form: Capacity ≈ k * L * S.Use k ≈ 0.087 (ton·hr^−1 per cm^2) for jaw/gyratory crushers in standard units.Select '0.087 * L * S' as the correct expression. Verification / Alternative check: Comparisons with manufacturer specifications show reasonable agreement for mid-range settings; detailed sizing still requires feed characteristics and RPM. Why Other Options Are Wrong: 0.01 * L * S underestimates; L * S or L * S / 0.087 grossly overestimate capacity. Common Pitfalls: Treating the correlation as exact without considering material friability, moisture, and crusher speed, which all affect throughput. Final Answer: 0.087 * L * S

Question 19

Primary crushers: which equipment can accept feed sizes of at least 25 cm (≥ 250 mm)?

Accepted Answer

Introduction / Context: Primary crushing reduces run-of-mine rock to sizes suitable for secondary crushers or mills. Equipment choice depends on the maximum lump size the machine can accept and the required product size. Given Data / Assumptions: Feed size ≥ 25 cm (250 mm) implies large, coarse rocks or aggregate. We compare mills and crushers. Objective is initial size reduction, not fine grinding to powders. Concept / Approach: Jaw crushers, and similarly gyratory crushers, are designed to accept large lumps at the feed opening and reduce them by compressive force between a fixed and a moving jaw. Ball mills, rod mills, and fluid energy mills handle much smaller feed sizes; they are used after primary/secondary crushing. Step-by-Step Solution: Identify equipment meant for primary breakage: jaw crusher.Eliminate mills: they require pre-crushed feed well below 25 cm.Select 'Jaw crusher' as appropriate. Verification / Alternative check: Manufacturer data sheets list typical jaw crusher feed openings from a few hundred mm up to over a meter, confirming suitability for ≥ 25 cm. Why Other Options Are Wrong: Ball/rod mills accept much smaller feed and are used for grinding; fluid energy mills are for very fine powders and tiny feed sizes. Common Pitfalls: Confusing primary crushing with grinding; skipping the necessary staged reduction leads to poor efficiency and equipment damage. Final Answer: Jaw crusher

Question 20

Bulk solids handling — selecting a conveyor:
For transporting a pasty or sticky material over short-to-moderate distances in a plant, which mechanical conveyor is most suitable?

Accepted Answer

Introduction / Context: Pasty, sticky, or semi-solid materials pose challenges for bulk handling because they may adhere to belts, leak from buckets, or bridge in hoppers. Process engineers therefore choose conveying equipment that positively moves the material with an enclosed trough and rotating element that wipes and advances the mass. This question checks your ability to match material behavior with the most appropriate conveyor type used in chemical and process plants. Given Data / Assumptions: Material is pasty or sticky (non-free-flowing). Distance is plant-scale (not long overland), with moderate throughput. Cleanability and containment are desirable to avoid spills. Concept / Approach: Screw conveyors employ a helical flight on a rotating shaft inside a U-trough or tube. Rotation drags and pushes cohesive materials forward while the flights continuously scrape the trough surface, limiting build-up. Enclosure reduces spillage and odor, which is critical for sludges, filter cakes, and dough-like feeds. Belt conveyors need free-flowing solids to avoid carryback; buckets suit granular or lumpy solids, not pastes; apron conveyors target hot, coarse, and abrasive lumps rather than sticky pastes. Step-by-Step Solution: Identify material behavior: cohesive, sticks to surfaces, low flowability.Map to equipment that provides positive displacement-like movement and enclosure.Select the screw conveyor for reliable transport of pasty solids. Verification / Alternative check: Practical references list screw conveyors for sludge, wet filter cake, and dough-like feeds; belts and bucket elevators are reserved for granular, free-flowing materials. Why Other Options Are Wrong: Apron conveyor: suited to hot, coarse, abrasive lumps; inefficient for pastes. Belt conveyor: suffers from carryback and spillage with sticky pastes. Bucket elevator: meant for vertical lift of free-flowing particles, not cohesive pastes. Pneumatic pipeline: requires discrete particles, not cohesive pastes that can plug lines. Common Pitfalls: Assuming any enclosed conveyor works; without a rotating flight to shear and move the paste, build-up and blockages occur. Final Answer: Screw conveyor

Mechanical Operations Questions