Question 1

Turbine agitators in mixing: identify the incorrect statement about speed, flow pattern, sizing, and placement in a stirred tank.

Accepted Answer

Introduction / Context: Turbine agitators (flat- and pitched-blade) are standard impellers for blending, gas dispersion, and heat/mass transfer in chemical and bioprocess tanks. Understanding their flow patterns and typical design rules improves scale-up and process reliability. Given Data / Assumptions: Pitched-blade turbines are compared to flat-blade (radial) turbines. Typical industrial design heuristics for diameter, placement, and speed apply. Concept / Approach: Flat-blade (Rushton) turbines generate highly radial flow. By contrast, pitched-blade turbines are intentionally angled (usually 30–45°) to produce a significant axial flow component for top-to-bottom circulation. Hence, any statement claiming “only radial” for a pitched blade is incorrect. Step-by-Step Solution: Check peripheral speed: 200–250 m/min is a common recommended range.Check sizing: D/T in the range 1/3 to 1/6 is typical; blade length heuristics are reasonable.Check placement: clearance ≥ impeller diameter and multiple impellers for deep liquids are standard practices.Therefore option (b) is wrong. Verification / Alternative check: Vendor literature shows pitched-blade turbines used when axial circulation is desired. Why Other Options Are Wrong: They reflect standard mixing design guidelines in common use. Common Pitfalls: Assuming all turbines are purely radial; blade pitch changes the flow to mixed or axial. Final Answer: A pitched-blade turbine produces only radial flow with no axial component at all.

Question 2

Rittinger number (new surface area created per unit mechanical energy absorbed during crushing) depends primarily on which material and loading characteristics?

Accepted Answer

Introduction / Context: Rittinger’s law correlates energy input with new surface area generated during comminution. The proportionality factor embodied in the Rittinger number reflects how readily a solid develops new fracture surfaces for a given mechanical energy. Given Data / Assumptions: Material properties (strength, elasticity) affect crack initiation/propagation. Loading mode (impact, compression, attrition) changes energy partitioning. Concept / Approach: Fracture mechanics indicates that both the energy to create new surfaces and the effectiveness of energy transfer depend on the material’s ultimate strength and elastic behavior as well as on how the force is applied. Hence, the Rittinger number is not a universal constant but varies with these factors. Step-by-Step Solution: Relate surface formation energy to fracture work.Acknowledge influence of strength and elasticity on crack growth.Recognize loading mode governs energy dissipation.Therefore, all listed factors matter. Verification / Alternative check: Empirical tests show different Rittinger constants for materials and for machines imposing different stress states. Why Other Options Are Wrong: Each single-factor choice omits other equally influential determinants. Common Pitfalls: Treating Rittinger’s constant as material-independent; in practice it varies widely. Final Answer: All of the above

Question 3

Roll crushers: give the correct geometric relationship for half the angle of nip α in terms of roll diameter d_r, feed particle diameter d_p, and roll gap d_f.

Accepted Answer

Introduction / Context: The angle of nip determines whether a particle can be nipped and drawn into the roll gap. It depends on roll geometry and the relative sizes of the particle and the set (gap). Correct geometry is essential for predicting operating windows and selecting roll diameters. Given Data / Assumptions: d_r: roll diameter; R = d_r/2. d_p: feed particle diameter. d_f: roll gap (distance between rolls). Rigid spheres and smooth rolls assumed for the geometric relation. Concept / Approach: From the triangle formed by roll center, particle center at contact, and the line of centers, the cosine of the half angle of nip equals the ratio of the distance from roll center to the gap midline over the distance from roll center to particle center at contact. This yields cos α = (R + d_f/2) / (R + d_p/2). Step-by-Step Solution: Set R = d_r/2.Distance to gap midline = R + d_f/2.Distance to particle center at contact = R + d_p/2.Therefore cos α = (R + d_f/2) / (R + d_p/2). Verification / Alternative check: For a very small gap (d_f → 0), α decreases; for a very small particle (d_p → 0), α → 0, consistent with geometry. Why Other Options Are Wrong: Using sin or tan here is inconsistent with the geometric construction.Expressions without the 1/2 factors do not reduce correctly in limiting cases. Common Pitfalls: Confusing the friction criterion tan α ≤ μ (separate condition) with the geometric relationship for α. Final Answer: cos α = (d_r/2 + d_f/2) / (d_r/2 + d_p/2)

Question 4

Identify the common filter aid used to build a porous precoat and prevent medium blinding during clarification.

Accepted Answer

Introduction / Context: Filter aids are inert, porous materials used either as a precoat or as a body feed to maintain permeability and avoid rapid pressure buildup in fine-solid filtrations. Given Data / Assumptions: Looking for the standard, widely used aid for precoat filtration. Applications include beverages, pharmaceuticals, and chemicals. Concept / Approach: Diatomaceous earth (kieselguhr) consists of fossilized diatom frustules with high porosity and complex microstructure, giving excellent permeability and fine particle capture. Canvas is a medium, not an aid; calcined lime is a reagent/filler and not a structural filter aid. Step-by-Step Solution: Match “filter aid” definition to inert porous solids.Select diatomaceous earth as the standard choice. Verification / Alternative check: Industry practice commonly references kieselguhr and perlite as primary aids; DE is the most common. Why Other Options Are Wrong: Canvas: medium support cloth, not an aid.Calcined lime: chemistry additive, not a precoat material. Common Pitfalls: Confusing the filter medium with filter aid. Final Answer: Diatomaceous earth

Question 5

Conveyors: the type commonly referred to as a “scraper” or “drag” conveyor is which of the following?

Accepted Answer

Introduction / Context: Different conveyors are selected based on the material handled and layout constraints. The term “scraper” refers to flights that drag material along a trough, distinct from belts, aprons, or screws. Given Data / Assumptions: We must identify which design is synonymous with “scraper”. Terminology: flight = scraper = drag conveyor. Concept / Approach: Flight (drag) conveyors have paddles or scrapers attached to a chain moving in a closed trough; these scrapers drag bulk solids along. Apron conveyors use overlapping pans; screw conveyors use a helical screw; belts use a flexible belt – none are termed “scrapers”. Step-by-Step Solution: Map the synonym: scraper = drag = flight conveyor.Choose the option explicitly naming flight (drag/scraper). Verification / Alternative check: Materials handling texts describe scraper conveyors as drag or flight conveyors. Why Other Options Are Wrong: Apron/Screw/Belt: different operating principles and standard names. Common Pitfalls: Assuming “flight” always means screw flight; here it means chain-mounted scrapers. Final Answer: Flight (drag/scraper) conveyor

Question 6

Screen capacity is commonly reported as a mass flux per screening area. Which unit correctly expresses screen capacity?

Accepted Answer

Introduction / Context: To compare screens of different sizes, capacity must be normalized by area. The industry standard metric is mass throughput per unit area of screening surface. Given Data / Assumptions: “Capacity” refers to the throughput rate normalized by area, not merely total feed. Concept / Approach: Total throughput (t/h) alone does not permit fair comparison across screens of different areas. The correct unit is t/h per ft^2 (or t/h per m^2 in SI), often called unit screen capacity. This helps in sizing and vendor comparison. Step-by-Step Solution: Define capacity basis: mass flow divided by screen area.Select the option that includes both time and area: t/h·ft^2. Verification / Alternative check: Design charts plot unit capacity versus aperture, slope, and material factors. Why Other Options Are Wrong: t/h or t/ft^2 alone omits the other needed dimension. Common Pitfalls: Using overall t/h for selection can oversize or undersize the screen when area changes. Final Answer: tons per hour per square foot (t/h·ft^2)

Question 7

A crushing duty with feed size 10–50 mm and product size 2–10 mm corresponds to which stage of comminution?

Accepted Answer

Introduction / Context: Comminution stages are broadly classified by size ranges: primary (very coarse), secondary (intermediate), and fine crushing (small, near mill feed). Given Data / Assumptions: Feed: 10–50 mm. Product: 2–10 mm. Concept / Approach: Primary crushing usually reduces run-of-mine sizes (0.5–1.5 m) to 50–150 mm. Secondary reduces roughly 50–150 mm to 10–20 mm. Fine crushing drives down to a few millimetres. The stated duty neatly fits the secondary regime. Step-by-Step Solution: Compare feed/product to typical stage ranges.10–50 mm → 2–10 mm aligns with secondary. Verification / Alternative check: Common secondary machines include cone crushers and roll crushers handling this band. Why Other Options Are Wrong: Primary: much coarser discharge.Fine/Ultrafine: target sizes smaller than stated. Common Pitfalls: Confusing fine crushing with grinding; grinding is a different unit operation. Final Answer: Secondary crushing

Question 8

How does screen effectiveness (separation efficiency) trend as screen capacity (throughput) is increased substantially?

Accepted Answer

Introduction / Context: Screening efficiency balances throughput against probability of particle presentation to the apertures. Pushing higher capacity typically compromises the probability that near-size particles find and pass through openings. Given Data / Assumptions: Same screen area and aperture. Higher feed rate increases bed depth and velocity. Concept / Approach: As loading rises, particle bed deepens and stratification worsens; near-size particles have fewer opportunities to contact openings, and blinding can increase. Thus, effectiveness (the sharpness/accuracy of separation) declines with increasing capacity at fixed area. Step-by-Step Solution: Increase feed → deeper bed, shorter residence.Lower probability of passage for near-size fractions.Efficiency decreases. Verification / Alternative check: Manufacturer capacity charts show trade-off between capacity and efficiency. Why Other Options Are Wrong: Unchanged/Increases: contradicts the fundamental capacity–efficiency trade-off.Exponential decrease: too specific; the trend is monotonic decrease, not necessarily exponential. Common Pitfalls: Confusing “capacity” with “area”; adding area can recover efficiency at higher total throughput. Final Answer: Decreases

Question 9

For removing dirt and debris from a flowing fluid stream, which device is used upstream to intercept large particles?

Accepted Answer

Introduction / Context: Process lines often need protection from debris to prevent fouling or damage to downstream equipment like pumps, valves, and exchangers. A strainer provides a simple mechanical barrier. Given Data / Assumptions: Dirt = relatively coarse solids carried in a liquid stream. Objective = quick removal with minimal complexity. Concept / Approach: Strainers are coarse filters (baskets, Y-strainers) that intercept larger particles. Coagulants aid in fine-particle agglomeration for sedimentation, not immediate debris interception. Gravity settlers and clarifiers are unit processes for bulk solids separation, not simple inline protection devices. Step-by-Step Solution: Match need (inline debris removal) with solution (strainer).Reject chemical/coarse clarification options. Verification / Alternative check: Plant P&IDs commonly place strainers upstream of pumps and control valves. Why Other Options Are Wrong: Coagulant: chemical aid, not a device.Settler/Clarifier: large equipment for bulk removal; not inline dirt interception. Common Pitfalls: Confusing strainers with fine filtration cartridges; strainers are coarse protection devices. Final Answer: Strainer (mechanical screen)

Question 10

Which of the following is a primary crusher commonly selected for hard and tough stones?

Accepted Answer

Introduction / Context: Primary crushers handle the largest feed sizes. For hard, tough rock, robust compressive crushers are required at the first stage. Given Data / Assumptions: Tough, hard stone feed. Primary stage selection. Concept / Approach: Jaw and gyratory crushers are the classic primary machines. Among the options provided, the jaw crusher is the most commonly specified for hard, tough stones in quarries due to simplicity, ruggedness, and ability to accept large feed with high reduction ratios. Cone crushers are used more often in secondary/tertiary stages. Step-by-Step Solution: Identify primary duty device: jaw/gyratory.From given choices, jaw is the standard selection. Verification / Alternative check: Quarry flowsheets typically start with a jaw (or gyratory) before cones. Why Other Options Are Wrong: Cone: usually secondary/tertiary.Gyratory could also be primary but is not the expected answer set here. Common Pitfalls: Assuming cone equals gyratory; while related, cones are typically downstream of the primary. Final Answer: Jaw crusher

Question 11

A fluid energy (jet) mill is primarily used for which level of grinding service?

Accepted Answer

Introduction / Context: Jet mills accelerate particles in gas streams to achieve very fine size reduction through interparticle collisions. They are chosen when contamination must be minimal and extremely fine sizes are needed. Given Data / Assumptions: No mechanical media or moving grinding parts. High gas velocities. Concept / Approach: The operating regime and achieved sizes place jet mills in the ultra grinding category. Typical outputs are in the few microm to sub-microm range with narrow size distributions, suitable for APIs, toner, and high-value pigments. Step-by-Step Solution: Relate mechanism (gas jets) to achievable fineness.Select “Ultra grinding”. Why Other Options Are Wrong: Cutting/Crushing: not applicable to jet mills.General grinding understates the capability. Final Answer: Ultra grinding (micron/submicron)

Question 12

On the Mohs hardness scale used in mineral processing, which of the following is the softest material?

Accepted Answer

Introduction / Context: The Mohs scale ranks minerals by scratch resistance from 1 (softest) to 10 (hardest). Knowledge of hardness guides equipment selection and wear considerations in comminution. Concept / Approach: On the Mohs scale: Talc = 1, Gypsum = 2, Calcite = 3, Feldspar ≈ 6, Corundum = 9. Therefore, talc is the softest listed mineral. Why Other Options Are Wrong: Calcite: hardness 3.Feldspar: hardness ~6.Corundum: hardness 9 (very hard). Final Answer: Talc

Question 13

For cake filtration, the instantaneous filtrate rate is directly proportional to which parameters and inversely proportional to which resistance/viscosity terms?

Accepted Answer

Introduction / Context: Filtration equations show how flow depends on driving force, medium/cake resistances, fluid viscosity, and area. Understanding proportionalities aids in debottlenecking and scale-up. Concept / Approach: For a given pressure drop, superficial flux is proportional to area and inversely proportional to total hydraulic resistance and viscosity. Hence, higher S increases throughput, while higher R or μ reduces it. Step-by-Step Solution: Identify variables: S, R, μ under constant ΔP.Rate ∝ S and ∝ 1/R and ∝ 1/μ.Therefore, all listed statements are correct. Final Answer: All of the above

Question 14

For a rotating trommel (cylindrical screen), the typical optimum operating speed as a fraction of critical speed is in which range?

Accepted Answer

Introduction / Context: Trommels are rotating screens where particle motion and residence time depend on rotational speed relative to the critical speed at which centrifuging would occur. Concept / Approach: To promote cascading and adequate presentation to apertures without centrifuging the bed, trommels are typically operated at about one-third to slightly less than one-half of critical speed. Why Other Options Are Wrong: Ratios above 1 indicate supercritical rotation causing centrifuging and no screening.Wider ranges include impractical speeds for effective screening. Final Answer: 0.33–0.45

Question 15

Separation of solid particles primarily by density differences (at similar sizes) is carried out by which device?

Accepted Answer

Introduction / Context: While many classifiers separate by size via settling velocity, some units exploit density differences to a greater extent, especially at overlapping sizes. Concept / Approach: Sorting or hydraulic classifiers stratify particles in a rising or falling fluid so that higher-density particles settle differently than lighter ones. Cyclones primarily classify by size (though density has an effect); filters and thickeners remove solids by phase separation or concentration, not by density discrimination at similar sizes. Final Answer: Sorting (hydraulic) classifier

Question 16

Centrifugal filtration without cake formation: for a liquid layer of thickness δ at radius R in a rotor spinning at angular speed ω, what is the initial pressure drop across the layer (liquid density ρ_L)?

Accepted Answer

Introduction / Context: In centrifugal filtration, the driving force is the radial pressure gradient generated by rotation. With no cake yet formed, the initial pressure drop arises from the centrifugal head across the liquid film. Concept / Approach: For a rotating fluid, pressure varies with radius as dP/dr = ρ_L * ω^2 * r. Integrating from r = R to r = R + δ gives ΔP = ∫RR+δ ρ_L * ω^2 * r dr = 0.5 * ρ_L * ω^2 * [(R + δ)^2 − R^2] = 0.5 * ρ_L * ω^2 * δ * (2R + δ). Step-by-Step Solution: Start from dP/dr relation.Integrate over the liquid thickness δ.Obtain ΔP expression shown above. Why Other Options Are Wrong: Forms using only R^2 or δ^2 ignore the cross term and do not result from the correct integral. Final Answer: 0.5 * ω^2 * δ * ρ_L * (2R + δ)

Question 17

For transporting dry, powdery solids horizontally over short distances, which conveyor type is most suitable?

Accepted Answer

Introduction / Context: Screw conveyors are widely used for powders, granules, and small particulates because they contain the material in a trough and meter it effectively. Concept / Approach: Enclosed screws minimize dusting, handle a range of bulk densities, and can incline modestly. Belts can spill fine powders and are less precise for short, enclosed transfers; bucket elevators are primarily vertical conveyors. Final Answer: Screw conveyor

Question 18

In a roll crusher, which statements about the two rolls are generally true?

Accepted Answer

Introduction / Context: Roll crushers consist of two cylindrical rolls set close together, drawing particles into the gap for compression and shearing. Concept / Approach: Standard practice uses equal-diameter rolls, rotating towards each other at the nip to draw in material. Speed can be equal or deliberately differential to increase shear and improve nip efficiency for certain materials. Final Answer: All of the above

Question 19

Critical rotation speed of a ball mill: calculate the critical speed (revolutions per second) for a mill of diameter 1.2 m charged with 70 mm balls.

Accepted Answer

Introduction / Context: The critical speed is the rotational speed at which the grinding media are centrifuged against the mill shell and cease to fall, eliminating grinding action. Mills are operated below this speed (typically 65–80% of critical). Given Data / Assumptions: Mill diameter, D = 1.2 m → radius R = 0.6 m. Ball diameter = 70 mm → ball radius r_b = 0.035 m. Acceleration due to gravity g ≈ 9.81 m/s^2. Concept / Approach: Critical speed (in rps) considering the ball radius is n_c = (1 / 2π) * sqrt( g / (R − r_b) ). Using the effective radius gives a more accurate estimate than using R alone. Step-by-Step Solution: Compute effective radius: R_eff = R − r_b = 0.6 − 0.035 = 0.565 m.Compute inside square root: g / R_eff = 9.81 / 0.565 ≈ 17.37.Take square root: sqrt(17.37) ≈ 4.168 s^-1.Divide by 2π: n_c ≈ 4.168 / 6.283 ≈ 0.664 rps. Verification / Alternative check: Converting to rpm, n_c ≈ 0.664 * 60 ≈ 39.8 rpm, a typical critical speed for this size of mill. Why Other Options Are Wrong: 0.50 or 1.00 rps: inconsistent with geometry and gravity.2.76 rps: far above critical for a 1.2 m mill. Common Pitfalls: Ignoring ball radius (using R only) slightly overestimates critical speed. Final Answer: 0.66 rps

Question 20

Plate-and-frame filtration concept: in constant-pressure operation, how does cake resistance change with filtration time?

Accepted Answer

Introduction / Context: This question checks understanding of how cake resistance evolves in a plate-and-frame filter and distinguishes between constant-pressure and constant-rate modes. Knowing which operating variable is held fixed is the key to predicting how the other variables and the cake resistance behave during a cycle. Given Data / Assumptions: Equipment: plate-and-frame filter (batch cake filtration). Operation mode: constant-pressure filtration (ΔP is held constant). Incompressible-to-moderately-compressible cake behavior typical of many slurries. Concept / Approach: Under constant-pressure filtration, Darcy’s law for laminar flow through a growing cake gives dV/dt = ΔP / (μ*(Rm + αCV/A)), where symbols have their usual meaning. As filtrate volume V increases, cake thickness grows and the term αCV/A increases, so the overall resistance rises steadily. Consequently, the filtrate rate falls with time even though the applied pressure is fixed. Step-by-Step Solution: Recognize mode: constant pressure → ΔP fixed by pump/vacuum.Cake builds with time → cake resistance Rc increases.Since ΔP is fixed, increasing Rc lowers the instantaneous filtration rate with time.Therefore, the statement “cake resistance increases steadily with time” applies to constant-pressure filtration, not constant-rate. Verification / Alternative check: In constant-rate operation, the rate is held fixed and the required pressure rises over time to overcome the growing cake resistance. That is the complementary scenario and contrasts with the present case. Why Other Options Are Wrong: Rate: incorrect—at constant-rate operation the pressure changes, not the held variable.Both / Neither: incorrect because the described behavior aligns specifically with constant-pressure filtration. Common Pitfalls: Confusing constant-pressure with constant-rate tests; forgetting that cake thickness and hence resistance increase as volume accumulates, regardless of mode. Final Answer: pressure

Mechanical Operations Questions