Question 1

Screening at fine sizes: as average particle size decreases during separation on a screen, what happens to the screening task?

Accepted Answer

Introduction / Context: Screens separate solids by size using apertures. At small particle sizes, several physical effects reduce efficiency and capacity. Recognising this trend helps engineers choose between fine screens and alternatives such as classification or filtration. Given Data / Assumptions: Dry or wet screening of granular solids. Particle sizes approach the aperture size or become very fine. No major changes in moisture other than typical fines behavior. Concept / Approach: As particles get smaller, they are more prone to agglomeration, electrostatic effects, and surface forces. Near-size particles bridge or peg the apertures. The probability of passage drops and blinding increases. Both capacity (throughput per area) and efficiency (sharpness of cut) tend to decrease. Step-by-Step Solution: Reduce size toward aperture → more near-size particles.Near-size particles lodge, agglomerate, or track apertures → blinding/pegging increases.Net effect → screening becomes progressively more difficult; capacity and efficiency fall. Verification / Alternative check: Plant experience shows that fine screening demands agitation aids (inclination, vibration intensity), spray water (wet screening), anti-blinding devices, or switching to classification when fines dominate. Why Other Options Are Wrong: “Easier” or “capacity/effectiveness increased” contradict the physical realities of fines handling.“None of these” is unnecessary because the correct trend is well established. Common Pitfalls: Ignoring moisture; even a little dampness can exacerbate fines agglomeration and blinding dramatically. Final Answer: screening becomes progressively more difficult.

Question 2

Rotary drum vacuum filter design: what percentage of the drum is typically submerged in slurry during operation?

Accepted Answer

Introduction / Context: Rotary drum vacuum filters are continuous filters where part of the drum is submerged to form cake and the remaining arc is used for washing, drying, and discharge. The submergence fraction balances cake build time against available drying arc. Given Data / Assumptions: Standard drum with sectoral compartments and vacuum boxes. Typical industrial slurries; no exotic extremely slow/fast caking systems. Concept / Approach: Common design practice uses approximately 25–40% submergence for many slurries to allow sufficient arcs for washing/drying and discharge. Too little submergence yields thin cakes; too much reduces drying time and can hinder discharge. Step-by-Step Solution: Identify typical range → about one-third of the circumference.Among the options, 30% best represents standard practice.Select 30% as the expected submergence. Verification / Alternative check: Operating manuals and vendor datasheets often show example layouts near 30–35% submergence for medium cake thicknesses and one wash zone. Why Other Options Are Wrong: 3% or 25%: too low for most duties (insufficient cake growth time).85%: unrealistically high; would leave minimal drying and discharge arc. Common Pitfalls: Assuming a fixed value for all slurries; in reality, optimal submergence shifts with filtrability, wash needs, and drum speed. Final Answer: 30

Question 3

Trommel critical speed (theoretical): express the critical rotational speed N (rps) in terms of g and r for a cylindrical trommel.

Accepted Answer

Introduction / Context: “Critical speed” for a rotating drum (trommel, mill) is the angular speed at which particles just cling to the wall and begin to centrifuge rather than tumble. This theoretical speed provides an upper bound for practical operation in screening or grinding equipment. Given Data / Assumptions: g = 9.81 m/s^2 (acceleration due to gravity). r = radius of the drum (metres) measured to the particle locus. Neglect lift and frictional effects; consider a single particle at the wall. Concept / Approach: At critical speed, centripetal acceleration equals gravitational acceleration at the top of the rotation: ω^2 r = g. Thus ω = (g/r)^0.5 (radians per second). Convert to revolutions per second using N = ω / (2π). Step-by-Step Solution: Set centripetal = gravitational: ω^2 * r = g.Solve for ω: ω = (g / r) ^ 0.5.Convert to rps: N = ω / (2π) = (1 / (2π)) * (g / r) ^ 0.5. Verification / Alternative check: Engineering handbooks also give the empirical rpm form N_rpm ≈ 42.3 / √D_m (for mills), which follows from the same derivation after unit conversions and diameter D = 2r substitution. Why Other Options Are Wrong: Dividing by 2r (option b) or changing the π factor (options c and d) stems from algebra/unit mistakes and does not match the physics balance. Common Pitfalls: Mixing diameter and radius, or rpm vs. rps; forgetting the 2π factor when converting angular velocity to revolutions per second. Final Answer: N = (1 / (2π)) * (g / r) ^ 0.5

Question 4

Filter types: which of the following is a continuous filter for large-scale operations?

Accepted Answer

Introduction / Context: Filtration equipment can be broadly classed as batch or continuous. Recognising which devices are inherently batch helps select the right hardware for steady, high-throughput duty. Given Data / Assumptions: Options limited to plate and frame, cartridge, and shell-and-leaf filters. “Continuous filter” refers to machines like rotary drum, rotary disc, or continuous vacuum belt filters operating without cycle interruptions. Concept / Approach: Plate-and-frame and cartridge filters are classical batch units: they must be stopped to discharge cake or replace elements. Leaf filters (shell-and-leaf) are also typically operated batchwise with cycles of filtration, cake washing, drying, and blowback discharge. Step-by-Step Solution: Identify each option as batch-operated.Conclude that none of the listed units is continuous in the strict sense.Therefore select “None of these.” Verification / Alternative check: Continuous filters prominently include rotary drum vacuum filters, rotary disc filters, and continuous belt filters—none appears among the options. Why Other Options Are Wrong: Plate-and-frame: batch by design.Cartridge: element replacement requires shutdown → batch.Shell-and-leaf: commonly batch cycles; not continuous discharge. Common Pitfalls: Assuming “leaf” implies continuous because of multiple leaves; despite large area, operation is cyclic. Final Answer: None of these

Question 5

Flow through filter cakes: for laminar filtrate flow through the deposited cake on a septum, which relation is applicable?

Accepted Answer

Introduction / Context: Modeling flow through porous media such as filter cakes requires empirical–theoretical correlations that link pressure drop, viscosity, velocity, and structural parameters (porosity, specific surface). Identifying the appropriate equation is foundational in filtration calculations. Given Data / Assumptions: Laminar flow regime within the cake pores. Porous cake characterized by porosity and specific surface. Incompressible or slowly compressible cake so that parameters are meaningful over small ranges. Concept / Approach: The Kozeny–Carman equation, derived from Darcy’s law with a capillary bundle model, relates pressure drop to superficial velocity and structural factors: ΔP/L = (180 * μ * (1 − ε)^2 / (ε^3 * d_p^2)) * v for beds of approximately spherical particles. The same form underpins the specific cake resistance used in filtration analysis. Step-by-Step Solution: Assume laminar flow through tortuous pores → Darcy-type behavior applies.Use Kozeny–Carman to connect ΔP, v, μ, ε, and particle size/specific surface.Conclude that Kozeny–Carman is the valid relation among the options. Verification / Alternative check: Other named equations are less standard for cake-flow laminar modeling; practical filtration textbooks reduce cake resistance to a Kozeny–Carman-based specific resistance α. Why Other Options Are Wrong: Leva’s and Blake–Plummer forms are not the canonical choice for laminar cake flow and are less widely used in filtration design. Common Pitfalls: Confusing bulk packed-bed pressure drop correlations with turbulent or transition-regime expressions; for cakes and low Reynolds numbers, the Kozeny–Carman framework is appropriate. Final Answer: Kozeny–Carman equation

Question 6

Trommel critical speed in rpm: give the formula in terms of trommel diameter D (ft).

Accepted Answer

Introduction / Context: Critical speed indicates the rpm at which particles would centrifuge against the shell rather than tumble. For screening action, operation is kept well below this limit. Engineers use a convenient rpm–diameter relation for quick estimates. Given Data / Assumptions: D = trommel diameter in feet. Standard gravity and unit conversions embedded in the constant. Ignoring liner thickness and particle lift. Concept / Approach: Starting from N (rps) = (1/(2π)) * √(g/r) with r = D/2, converting to rpm and feet leads to N_rpm ≈ 76.75 / √D. This is analogous to the mill critical-speed formula used in comminution texts. Step-by-Step Solution: Use r = D/2 and convert to rpm: N_rpm = 60/(2π) * √(g/r).Insert g and unit conversions (ft): yields ≈ 76.75 / √D.Select the option matching this relation. Verification / Alternative check: Handbooks quote 76.6–76.8 as the constant depending on rounding; any of these near 76.7 is acceptable in practice. Why Other Options Are Wrong: Multiplying by D or √D or dividing by D does not follow from the critical-speed derivation, which scales with 1/√D. Common Pitfalls: Using diameter in metres while keeping the imperial constant, or forgetting to subtract media diameter in grinding-mill cases. Final Answer: 76.75 / √D

Question 7

Screen performance issue: when is a screen said to be “blinded”?

Accepted Answer

Introduction / Context: Screen blinding is a common operational problem that lowers efficiency by blocking apertures. Recognising the correct definition helps troubleshoot capacity shortfalls and coarse carryover. Given Data / Assumptions: Sizing screen with fixed apertures. Mixture containing near-size particles and possibly sticky fines. Concept / Approach: “Blinding” refers specifically to clogging of apertures by particles or agglomerates. With holes blocked, true open area falls, the probability of passage drops, and both capacity and separation sharpness deteriorate. Misplacement of sizes between fractions is a symptom—not the definition—of blinding or other faults. Step-by-Step Solution: Define blinding: plugging/bridging of apertures.Understand consequence: reduced open area and efficiency.Select the option that states “apertures plugged with solids.” Verification / Alternative check: Anti-blinding devices (ball decks, sliders, ultrasonics) and wet screening mitigate true blinding by restoring open area. Why Other Options Are Wrong: Presence of wrong-size material in streams is an effect, not the definition.Increased capacity is the opposite of what blinding causes. Common Pitfalls: Confusing blinding with pegging (individual near-size wedges); both reduce open area, but blinding is more general pore blockage. Final Answer: the screen apertures are plugged with solid particles.

Question 8

Trommel operating speed: revolving screens (trommels) normally run in which rpm range for effective screening?

Accepted Answer

Introduction / Context: Trommels separate material by rotating a perforated cylinder. The operating speed must be high enough to tumble the material yet well below the critical speed to avoid centrifuging and loss of screening action. Given Data / Assumptions: Medium-size industrial trommels handling aggregates/ores. Target: typical rpm range rather than an exact design value. Concept / Approach: Typical practice places operating speed at a small fraction of critical (often 30–50% of N_crit depending on lifters and fill). For many diameters, this translates to the 15–20 rpm range, providing adequate cascading and retention without excessive carryover. Step-by-Step Solution: Consider typical diameters (1–3 m) and critical-speed relation → N_crit tens of rpm.Choose operating fraction (~0.3–0.5 of critical) → gives mid-teens to ~20 rpm.Select the 15–20 rpm range. Verification / Alternative check: Manufacturer curves and plant standards often list 12–22 rpm, depending on feed and screen design. Why Other Options Are Wrong: 1–2 rpm: too slow; poor cascading and throughput.40–50 or 60–75 rpm: near or above practical limits for most trommel sizes; risk of centrifuging. Common Pitfalls: Ignoring lifter bars or internal flights, which allow slightly lower speeds for the same performance. Final Answer: 15 – 20

Question 9

Fluid energy (jet) mill principle: which mechanism produces the greatest size reduction?

Accepted Answer

Introduction / Context: Fluid energy (jet) mills micronize powders by accelerating particles in high-velocity gas streams. Understanding the dominant breakage mechanism helps in predicting product fineness and choosing operating conditions. Given Data / Assumptions: Pulverization chamber with opposed or looped jets. Classifier outlet controlling top size. Hard, brittle feed suitable for micronization. Concept / Approach: In jet mills, particles are accelerated and repeatedly collide with each other at high relative velocities. The primary breakage is due to interparticle attrition and some impact during collisions; there are no mechanical grinding media, and wall impacts are secondary when properly tuned. Step-by-Step Solution: Note absence of grinding surfaces or media → compression/cutting are negligible.High-speed particle–particle collisions dominate → attrition/impact between particles.Among the options, “interparticle attrition” best captures the governing mechanism. Verification / Alternative check: Performance data correlate fineness with jet velocity and solids load, consistent with collision-driven breakage rather than compressive nipping. Why Other Options Are Wrong: Compression/cutting: typical in crushers or knife mills, not in jet mills.Impact (wall): present but not primary when the classifier is tuned; the dominant mechanism is particle–particle attrition. Common Pitfalls: Overfeeding to the point of excessive wall impact; inadequate classifier speed leading to coarse carryover. Final Answer: interparticle attrition

Question 10

Opposite of dispersion: in colloid and particle technology, which process is the reverse of dispersing particles?

Accepted Answer

Introduction / Context: Dispersion creates a stable distribution of fine particles within a fluid by overcoming attractive forces. The reverse idea is important for water treatment and solid–liquid separation, where aggregation improves settling and filtration. Given Data / Assumptions: Fine particles suspended in a liquid. Comparing unit operations and physicochemical processes. Concept / Approach: Flocculation is the process where destabilized particles aggregate into larger, loosely bound flocs under gentle mixing, often after coagulation (charge neutralization). These aggregates settle or filter more readily than the original dispersed particles. Sedimentation and filtration are separation steps that exploit the aggregates but are not the reverse of dispersion per se. Step-by-Step Solution: Define dispersion: creating/maintaining small, separate particles.Reverse action: aggregation → flocculation.Therefore, choose “flocculation.” Verification / Alternative check: Jar tests in water treatment demonstrate that after coagulant addition and flocculant polymer dosing, dispersed colloids aggregate and become separable by sedimentation/filtration. Why Other Options Are Wrong: Sedimentation/filtration: separation techniques, not the physicochemical reversal of dispersion.“None” is incorrect because flocculation is the established term. Common Pitfalls: Confusing coagulation (destabilization) with flocculation (growth by gentle mixing); both are typically used together. Final Answer: flocculation

Question 11

Magnetic separation applicability: which mineral among the following is generally not treated by magnetic separation?

Accepted Answer

Introduction / Context: Ore beneficiation relies on differences in physical properties. Magnetic separation exploits magnetic susceptibility contrasts. Knowing which minerals respond to magnetic fields is a frequent exam topic in mineral processing. Given Data / Assumptions: Low to high-intensity magnetic separators are available. Minerals listed: rutile (TiO2), galena (PbS), chromite (FeCr2O4), siderite (FeCO3). Concept / Approach: Galena (lead sulfide) is effectively non-magnetic; beneficiation typically uses froth flotation and gravity techniques. Chromite is strongly paramagnetic and separable in high-intensity magnetic fields. Siderite (iron carbonate) and some titanium minerals (rutile exhibits weak paramagnetism) can show measurable response in specialized magnetic separators, though rutile is more commonly separated electrostatically in practice. Step-by-Step Solution: Identify magnetic susceptibility: galena ≈ diamagnetic/negligible.Others exhibit paramagnetic behavior or contain iron enabling magnetic response.Select galena as not subjected to magnetic separation. Verification / Alternative check: Processing routes for lead ores emphasize flotation over magnetic methods, confirming galena’s unsuitability for magnetic separation. Why Other Options Are Wrong: Chromite: routinely beneficiated magnetically.Siderite: iron-bearing; magnetic upgrading possible after roasting or in high-intensity fields.Rutile: weakly paramagnetic; although often separated electrostatically, magnetic susceptibility exists, so “not subjected” fits galena best. Common Pitfalls: Equating common industrial route (electrostatic for rutile) with impossibility; the question asks which is not subjected at all—galena. Final Answer: Galena

Question 12

Milling of wheat: which equipment is typically used to grind wheat into flour in industrial practice?

Accepted Answer

Introduction / Context: Flour milling relies on staged size reduction and sifting to release endosperm and control particle size. The canonical machine is the roller mill, which uses pairs of counter-rotating rolls. In many question banks this is loosely referred to as a “roller crusher.” Given Data / Assumptions: Milling of tempered wheat kernels to produce flour and semolina. Terminology: options mention “roller crusher,” representing the roller-mill action. Concept / Approach: In roller milling, the first break rolls open the kernel; subsequent reduction rolls grind endosperm while plansifters classify the product. Hammer/impact mills produce too many fines and damage starch; jet mills are for ultra-fine chemical powders, not staple flour. Step-by-Step Solution: Match industry practice → multi-stage roller system.From the given options, “roller crusher” best corresponds to roller-mill technology.Select it as the correct choice. Verification / Alternative check: Modern flour mills universally use break and reduction rolls with controlled corrugations and speeds, confirming the selection. Why Other Options Are Wrong: Hammer/impact mills: excessive fines, heat, and starch damage for bread flour targets.Fluid energy mill: inappropriate energy-intensive micronization. Common Pitfalls: Confusing food-processing terminology with minerals terminology; in food, “roller mill” is standard. Final Answer: roller crusher

Question 13

Ultra-fine talc production: which mill is preferred to produce talcum powder of very fine particle size?

Accepted Answer

Introduction / Context: Talc is a very soft mineral but achieving cosmetic- or filler-grade fineness requires narrow top-size control well below 20–30 micrometres. Mill selection balances fineness, shape, and contamination risk. Given Data / Assumptions: Target: talcum powder with very fine size distribution. Purity and minimal contamination are important. Concept / Approach: Jet mills (fluid energy mills) generate ultra-fine particles through high-velocity particle–particle collisions with minimal metal contact. They deliver steep size cuts when combined with internal or external classifiers and avoid iron contamination that can be problematic in fillers and cosmetics. Step-by-Step Solution: Compare mills: ball/pin/hammer vs. jet.For sub-20 μm with tight top size and cleanliness → choose jet mill.Select “jet mill.” Verification / Alternative check: Industrial talc plants frequently use jet milling for highest grades; ball mills can serve coarser fillers but struggle to meet ultra-fine specs economically without advanced classification. Why Other Options Are Wrong: Ball/pin/hammer mills achieve coarser or broader distributions and have higher media/wear contamination risks. Common Pitfalls: Attempting extreme fineness in ball mills without adequate classification loops, leading to overgrinding energy penalties. Final Answer: jet mill

Question 14

Settling-based separations: what is the term for separating particles of various sizes, shapes, and densities by allowing them to settle in a fluid?

Accepted Answer

Introduction / Context: Solid–fluid separations employ different driving forces and mechanisms. When particles settle at different velocities in a fluid, engineers can exploit these differences to separate or grade materials. Given Data / Assumptions: Hindered or free settling occurs depending on solids concentration. Separation is based on differential settling velocities. Concept / Approach: Classification uses fluid flow and gravity to separate particles into size (and sometimes density) fractions. Hydraulic classifiers, cyclones, and elutriators are examples. Thickening and clarification focus on concentrating a slurry or producing clear overflow, not on sharp size grading; flotation relies on surface chemistry and air attachment, not settling alone. Step-by-Step Solution: Identify mechanism: settling-driven velocity differences.Name of process: classification.Select “classification.” Verification / Alternative check: Cut sizes in classifiers reflect balance between settling velocity and upward fluid velocity; process design targets desired size splits accordingly. Why Other Options Are Wrong: Froth flotation uses hydrophobicity, not settling.Thickening/clarification emphasize concentration/clarity rather than size separation into multiple fractions. Common Pitfalls: Assuming any gravity settler is a classifier; many settlers are designed for clarification rather than controlled cut points. Final Answer: classification

Question 15

Crushing stage identification: feed 100–300 mm reduced to product 10–50 mm corresponds to which stage of crushing?

Accepted Answer

Introduction / Context: Comminution circuits are divided into stages according to size ranges. Correctly labeling a reduction step helps in selecting appropriate crushers and screens and in estimating power and wear. Given Data / Assumptions: Feed size: 100–300 mm (typical product of primary crushers or run-of-mine with scalping). Product size: 10–50 mm. Concept / Approach: Primary crushers accept large lumps (0.5–1.5 m) and deliver products often in the 100–300 mm range. Secondary crushers then reduce this to 10–50 mm. “Fine” or “tertiary” crushing targets a few millimetres or less; “ultrafine” is in the micron range and belongs to grinding, not crushing. Step-by-Step Solution: Map size ranges to stages.100–300 mm → input compatible with secondary duty.10–50 mm → typical secondary product.Therefore, this step is secondary crushing. Verification / Alternative check: Flowsheets in aggregates and minerals show jaw (primary) → cone/roll (secondary) → screening to 10–40 mm products. Why Other Options Are Wrong: Primary: would start from much larger ROM sizes.Fine/ultrafine: refer to much smaller products, achieved by tertiary crushing or milling. Common Pitfalls: Using crusher names (e.g., “cone”) as proxies for stages; the same machine type can serve different stages at different sizes. Final Answer: secondary

Question 16

Filter press operation: when should filtration be stopped during a plate-and-frame (filter press) cycle?

Accepted Answer

Introduction / Context: Filter presses operate batchwise. Determining the proper end point of a cycle maximises throughput without wasting energy or risking cloth damage. Operators watch pressure, flow, and time trends. Given Data / Assumptions: Constant-pressure or ramped-pressure filtration in a plate-and-frame press. Observable variables: filtrate flow rate and pressure. Concept / Approach: As cake builds, resistance increases, flow declines, and pressure rises. The practical end point is when filtrate flow stops or becomes negligible and/or when pressure spikes, indicating the press is “choked” by the cake or cloth blinding. Stopping then prevents excessive mechanical stress with minimal gain in recovered liquor. Step-by-Step Solution: Monitor dV/dt → tends to zero near end of cycle.Observe ΔP → rises sharply as channels close.Stopping upon low/no flow and sudden overpressure is recommended. Verification / Alternative check: Standard operating procedures specify maximum time, minimum flow, or peak pressure criteria—any of which can trigger shutdown. Why Other Options Are Wrong: “Cake becomes very dense” is qualitative and not a reliable, instrumented trigger on its own. Common Pitfalls: Overextending cycles hoping for marginal extra filtrate; this increases downtime and risks cloth fouling. Final Answer: both (b) & (c).

Question 17

Gravity decantation: what phases can a gravity decanter separate effectively?

Accepted Answer

Introduction / Context: Decanters use gravity to separate liquid phases with a density difference. They are widely used after liquid–liquid extraction, in oil–water separation, and in biodiesel washing. Given Data / Assumptions: Two liquid phases are present. Phases are at least partially immiscible and have a density contrast. Concept / Approach: Separation occurs when immiscible phases form distinct layers under gravity. The denser phase settles; the lighter phase rises. If liquids are miscible, a single homogeneous phase forms and gravity cannot separate them. If immiscible but identical in density, stratification would not occur (a practical impossibility for distinct liquids). Step-by-Step Solution: Identify requirement: immiscibility + density difference.These conditions define decanter applicability.Select “immiscible liquids of different [density].” Verification / Alternative check: Chemical engineering textbooks model residence time using Stokes’ law for droplets, highlighting dependence on density difference and interfacial tension. Why Other Options Are Wrong: Miscible systems do not phase separate under gravity.“Immiscible of same density” would not stratify; separation would be impractical. Common Pitfalls: Neglecting emulsion formation; even with immiscibility and density contrast, surfactants can stabilize emulsions requiring coalescers. Final Answer: immiscible liquids of different

Question 18

Specific surface of spheres: how does specific surface vary with particle diameter Dp for spherical particles?

Accepted Answer

Introduction / Context: Specific surface (surface area per unit mass or per unit volume) is a key parameter in reaction rates, drying, and sintering. For spheres, a simple geometric relation links specific surface and particle diameter. Given Data / Assumptions: Spherical particles of diameter Dp. Specific surface per unit volume: S_v = surface area / volume. Concept / Approach: For a sphere, area A = πDp^2 and volume V = πDp^3/6. Thus S_v = A/V = 6/Dp. Per unit mass, S_m = S_v/ρ = 6/(ρ Dp). In both cases, specific surface is inversely proportional to Dp. Step-by-Step Solution: Write sphere geometry: A = πDp^2; V = πDp^3/6.Compute S_v = A/V = (πDp^2) / (πDp^3/6) = 6/Dp.Therefore, S ∝ 1/Dp. Verification / Alternative check: Dimensional analysis confirms inverse-length scaling; halving the diameter doubles specific surface. Why Other Options Are Wrong: Dp or Dp^2: imply larger particles have larger specific surface, contrary to geometry.1/Dp^2: corresponds to area per unit something else; not applicable for specific surface definitions. Common Pitfalls: Confusing specific surface per mass with per volume; both maintain 1/Dp dependence but differ by the density factor. Final Answer: 1/Dp

Question 19

Bulk solids handling facts: which statement below is correct regarding angle of repose/slide, storage vessels, and flotation reagents?

Accepted Answer

Introduction / Context: Plant layout and solids handling involve standard terminology (hopper, silo) and basic material properties (angles) as well as reagent roles in mineral processing. This question mixes these facts to test precise definitions. Given Data / Assumptions: Angle of repose vs. angle of slide. Definitions of hopper and silo in bulk storage. Role of pine oil in flotation circuits. Concept / Approach: A hopper is indeed a small bin or funnel-shaped vessel with a sloped/discharging bottom that promotes flow into feeders. Silos are typically tall, vertical storage vessels with relatively small footprints. In flotation, pine oil serves as a frother, not a modifying agent (modifiers include activators/depressants like copper sulfate or sodium cyanide). The angle of slide on a surface usually exceeds or equals the angle of repose of a free heap; it is not generally smaller. Step-by-Step Solution: Check (b): matches standard definition of a hopper → correct.Check (a): “repose > slide” is generally false; slide angle on a surface is typically greater or comparable.Check (c): silo described as “short and very wide” is incorrect; silos are commonly tall and slender.Check (d): pine oil is a frother, not a modifying agent → incorrect. Verification / Alternative check: Equipment glossaries and flotation reagent tables corroborate: hopper (sloping bottom), silo (tall), pine oil (frother). Why Other Options Are Wrong: (a) Misstates relationship of angles.(c) Mischaracterizes typical silo geometry and application.(d) Mislabels reagent function; modifiers are activators/depressants, not frothers. Common Pitfalls: Memorizing reagent lists without categories; conflating bins, hoppers, and silos based on shape rather than function. Final Answer: A hopper is a small bin with a sloping bottom.

Question 20

Unit operations check — pick the wrong statement:
Which of the following statements about separation and crushing/screening equipment is incorrect in standard chemical engineering practice?

Accepted Answer

Introduction / Context: This item tests basic vocabulary used in particle technology: magnetic separation modes, the definition of reduction ratio, the role of a gyratory crusher, and common screen motions. Getting these definitions correct is essential for flowsheet reading and quick equipment selection during preliminary design. Given Data / Assumptions: Conventional mineral/chemical processing definitions apply. Reduction ratio refers to a single crushing stage. “Gyratory crusher” implies the large primary machine used ahead of secondary/tertiary stages. Concept / Approach: Magnetic separators are indeed used in both dry and wet duty. Gyratory crushers are the classic primary coarse crushers. Screens can be stationary (grizzlies), moving/gyratory, or vibratory (linear/circular). The only problematic statement is the definition of reduction ratio: it is correctly defined as characteristic feed size divided by characteristic product size (e.g., F80/P80 or feed opening/product opening). The option given inverts and also misstates it (“minimum feed” over “maximum product”). Step-by-Step Solution: Recall correct definition: reduction ratio R = feed size / product size.Identify that the option states “minimum feed / maximum product,” which is incorrect.Confirm other options match standard practice. Verification / Alternative check: Handbooks define R using F80 and P80 (or nominal openings), never “minimum feed” over “maximum product,” which would contradict observed capacity trends. Why Other Options Are Wrong: (a) Correct: magnetic separation is used for both dry and wet ore processing. (c) Correct: gyratories are coarse/primary crushers. (d) Correct: all listed screen motions are widely used. Common Pitfalls: Confusing “openings” with statistical sizes (F80/P80); always state which sizes are used. Final Answer: Reduction ratio in crushing is the ratio of minimum feed size to maximum product size.

Mechanical Operations Questions