Q: Rotary drum vacuum filter capacity: identify the primary factors governing filtrate throughput for a given installation and slurry.

Introduction / Context: Rotary drum vacuum filters are widely used for continuous solid–liquid separation in inorganic chemicals, fertilizers, and minerals. Filtration capacity (filtrate volume per time) depends on both operating setpoints and the nature of the slurry. Understanding the dominant levers helps troubleshoot low throughput and optimize cycle parameters (submergence, rotation speed, vacuum, wash). Given Data / Assumptions: Constant vacuum operation with periodic cake formation, dewatering, washing, and discharge. Slurry properties influence cake resistance and formation rate. Cake thickness directly changes resistance and drainage path length. Concept / Approach: Darcy's law for flow through porous media states that filtrate flux decreases with higher cake resistance and thickness. Slurry particle size distribution, compressibility, and solids loading govern cake structure and permeability. Thus both cake thickness and slurry characteristics are principal determinants of capacity, alongside secondary factors like cloth condition and vacuum level. Step-by-Step Solution: Relate filtrate flow to pressure drop and total resistance (medium + cake).Note that thicker cakes increase resistance linearly (for incompressible cakes) or even more (compressible cakes).Recognize slurry characteristics set the cake permeability/compressibility; both govern capacity. Verification / Alternative check: Startup optimization usually tunes drum speed (hence cake thickness) for a given slurry to maximize throughput while achieving target moisture—confirming the joint role of (a) and (b). Why Other Options Are Wrong: Cake thickness only or slurry only omits a coequal driver.“Neither”: contradicts fundamental filtration theory. Common Pitfalls: Running with excessively thick cakes to chase dryness, which collapses capacity; or ignoring slurry pretreatment (coagulation/conditioning) that can improve permeability dramatically. Final Answer: Both (a) & (b)

Q: Operating speed for efficient ball milling: relative to the critical speed, which regime is recommended for effective grinding in industrial ball mills?

Introduction / Context: Critical speed is the rotational speed at which centrifugal force holds the grinding media against the mill shell, preventing cataracting. Industrial mills are purposely run below this speed to maintain a cascading motion, which creates impact and abrasion necessary for size reduction. Given Data / Assumptions: Standard lifter-equipped tumbling ball mills. Objective: stable cataracting/cascading without centrifuging. Common practice targets 65–80% of critical speed. Concept / Approach: Below critical speed, media rise and then fall, generating impact on the toe of the charge. At or above critical speed, media cling to the shell, drastically reducing effective breakage. Therefore, efficient operation is “less than” critical speed, typically tuned with media size and mill filling to reach target grind. Step-by-Step Solution: Define critical speed condition (centrifuging at the wall).Relate grinding action to cascading trajectories below critical.Select the sub-critical regime as the efficient range. Verification / Alternative check: Empirical correlations for mill power draw and breakage rates show peak performance around 70–78% of critical, consistent with decades of operating experience. Why Other Options Are Wrong: Equal to or more than critical: causes centrifuging and poor breakage.“Minimum possible small balls”: ball size selection depends on feed top size; exclusively small balls may underperform on coarse feed. Common Pitfalls: Ignoring lifter design and charge filling; both change the effective trajectory and optimal percent of critical speed. Final Answer: At a speed less than the critical speed

Q: Agitation scale-up fundamentals: a vigorous mixing guideline is about 2–3 hp per gallon for very thin liquids. In dimensionless form, the impeller “Power number” (Po) is defined as which ratio of variables?

Introduction / Context: In mixing, the dimensionless Power number (Po) collapses impeller power draw across scales and fluids. Together with Reynolds number (Re = rho * N * D^2 / mu), it underpins geometric similarity scale-up. Knowing the correct Po definition ensures consistent design from lab to plant for blending, suspension, and gas–liquid contacting. Given Data / Assumptions: P is power (W), rho is fluid density (kg/m^3), mu is viscosity (Pa·s), N is rotational speed (s^-1), and D is impeller diameter (m). Thin (low-viscosity) liquids at high turbulence are assumed for the “2–3 hp per gallon” rule-of-thumb; however, Po is general. Concept / Approach: By dimensional analysis, P has units kg·m^2/s^3. The term rho * N^3 * D^5 also has units kg·m^2/s^3, so their ratio is dimensionless. Thus Po = P / (rho * N^3 * D^5). In turbulent regimes for a given impeller geometry, Po becomes nearly constant, enabling direct horsepower prediction at scale. Step-by-Step Solution: Write Po candidate forms and check dimensional consistency.Confirm rho * N^3 * D^5 has the same dimensions as P.Select the dimensionally correct and standard definition: Po = P / (rho * N^3 * D^5). Verification / Alternative check: Impeller datasheets list constant Po values (e.g., Rushton disk turbine Po ≈ 5–6 in turbulent flow), which are used with the above formula to size motors. Why Other Options Are Wrong: N^2 * D^3 or N * D^2 denominators do not render a dimensionless group equal to Po; those break dimensional consistency.Using viscosity (mu) in Po is incorrect; mu appears in Re, not Po. Common Pitfalls: Confusing Po with torque number (Np = T / (rho * N^2 * D^5)) or applying laminar correlations in turbulent regimes without checking Re. Final Answer: Po = P / (rho * N^3 * D^5)

Q: Work index consistency across circuits: if Wi is determined in closed-circuit grinding, what is the effective work index to use for comparable open-circuit grinding conditions?

Introduction / Context: Bond's Work Index (Wi) characterizes material grindability. While Wi is a material property, practical circuit performance differs between open and closed circuit operation due to classification efficiency and overgrinding. Empirical correction factors are therefore used to translate laboratory Wi to plant power estimates for specific flowsheets. Given Data / Assumptions: Wi was measured under closed-circuit conditions. We seek an equivalent for open-circuit grinding to the same product size range. Standard Bond practice recognizes open circuit typically requires more energy for the same size reduction target. Concept / Approach: Open-circuit grinding lacks the classification feedback loop, so fines remain in the mill longer, reducing breakage efficiency. A widely used translation is that open-circuit power demand is about 1.34 times that of closed circuit for the same P80, which maps to an “effective” Wi_open = 1.34 * Wi_closed for preliminary estimates. Step-by-Step Solution: Identify circuit type difference (open vs. closed).Recall empirical relationship: P_open ≈ 1.34 * P_closed.Translate to Wi scaling: Wi_open ≈ 1.34 * Wi_closed = 1.34 Wi. Verification / Alternative check: Plant benchmarking often shows substantial energy savings when converting open to closed circuits, which is consistent with the factor near 1.34 used in preliminary design and feasibility studies. Why Other Options Are Wrong: Wi or 0.5 Wi: underestimate open-circuit energy.3.34 Wi: overstates the difference far beyond typical observations. Common Pitfalls: Applying the factor blindly without considering classification efficiency, circulating load limits, or drastic changes in target fineness that can alter the effective factor. Final Answer: 1.34 Wi

Q: Screen analysis notation: interpret the size class denoted as “+5 mm / −10 mm”. Which statement correctly describes the particles in this fraction?

Introduction / Context: Screen analysis classifies particulate solids by their ability to pass or be retained on standard apertures. Understanding shorthand notations like “+a / −b” is essential when reading sieve analyses, designing screen decks, or comparing product specs. Given Data / Assumptions: “+a” means the particles are larger than a (retained on a sieve of aperture a). “−b” means the particles are smaller than b (pass through b). Here, a = 5 mm and b = 10 mm. Concept / Approach: Particles labeled “+5 mm / −10 mm” are those that pass a 10 mm sieve but are retained on a 5 mm sieve. This describes the size interval between 5 and 10 mm inclusive of screening conventions (with a practical tolerance due to near-mesh particles and shape effects). Step-by-Step Solution: Interpret “+5”: retained on 5 mm, so size ≥ 5 mm (by screening convention).Interpret “−10”: pass 10 mm, so size ≤ 10 mm.Combine: 5–10 mm class, i.e., pass 10 mm and retain on 5 mm. Verification / Alternative check: Check typical screen analysis tables: fractions are commonly labeled with an upper “minus” (passing) and lower “plus” (retained) bound to define the interval. Why Other Options Are Wrong: “Pass 5 mm and retain on 10 mm” is impossible because a particle cannot be larger than 10 mm and simultaneously pass it.“Pass both” or “neither” do not define a standard size interval. Common Pitfalls: Reversing the “+” and “−” meanings; always remember “+” ties to retention (coarser than) and “−” to passing (finer than) the stated size. Final Answer: Passing through 10 mm screen and retained on 5 mm screen

Q: Primary size reduction of hard rock: which equipment is typically selected for the preliminary breaking (coarse crushing) stage before secondary reduction?

Introduction / Context: Primary crushing handles run-of-mine hard rock, reducing it to manageable sizes for subsequent secondary/tertiary crushers or grinding mills. The machine must accept large lumps, deliver high throughput, and provide a controlled top size. Given Data / Assumptions: Very hard, abrasive rock in large lump sizes. Objective: coarse crushing to a product suitable for conveyors and secondary crushers. Continuous, high-capacity duty. Concept / Approach: Gyratory crushers provide continuous crushing action with high capacity and can accept very large feed sizes due to their open intake and robust design. They are the standard choice for primary crushing of hard rock in large mines and quarries, delivering a relatively uniform product for downstream processing. Step-by-Step Solution: Identify primary duty requirements: large feed, high throughput.Match equipment features: gyratory offers continuous compression and large gape.Select gyratory crusher over mills and specialized disintegrators intended for finer or softer materials. Verification / Alternative check: Flowsheets in hard-rock mining typically list “ROM → gyratory → conveyors → secondary cone/crusher,” confirming standard practice. Why Other Options Are Wrong: Ball/tube mills: grinding equipment for much finer sizes, not for ROM lumps.Squirrel-cage disintegrators: used for soft/brittle materials, not massive hard rock. Common Pitfalls: Attempting to feed very large boulders into jaw-only small primaries; gyratories are preferred for the largest capacities and feeds. Final Answer: Gyratory crusher

Q: Ultrafine grinders: identify the dominant size-reduction mechanism typically exploited to reach sub-100 micrometre and micron-scale product sizes.

Introduction / Context: Ultrafine grinding targets very small particle sizes with narrow distributions for pigments, pharmaceuticals, and specialty minerals. Different machines (jet mills, stirred media mills) employ different physics than coarse crushers. Recognizing the dominant mechanism aids in choosing the correct mill and media. Given Data / Assumptions: Target sizes: tens of micrometres down to a few micrometres or below. Equipment examples: fluid energy (jet) mills, attritors/stirred media mills, pin mills. Feed material is friable enough for fine comminution. Concept / Approach: At ultrafine scales, attrition (rubbing/shearing along surfaces) dominates because it can continuously erode particle asperities and break weak points, especially in high-shear zones between media or in opposed jets. Impact alone becomes less efficient for very small targets due to decreasing fracture probability at low mass and the need for high collision frequency. Step-by-Step Solution: Identify typical ultrafine equipment and their internal motion (e.g., stirred beads).Relate micro-scale breakage to shear/attrition prevalence.Select “Attrition” as the dominant mechanism for ultrafine grinders. Verification / Alternative check: Process descriptions for jet mills and stirred media mills emphasize particle–particle and particle–media shear interactions, with operating metrics (specific energy, tip speed) tuned for attrition-driven breakage. Why Other Options Are Wrong: Slow compression: characteristic of coarse crushers.Impact: important in some mills but not the primary driver for consistent sub-100 μm production across machines like attritors.Cutting action: relevant for fibrous materials, not typical for mineral ultrafines. Common Pitfalls: Expecting ball mills alone to achieve ultrafine sizes efficiently; specialized attrition-based mills are more suitable. Final Answer: Attrition

Q: Crusher identification: which machine is commonly described as a combination of a jaw crusher and a roller crusher due to its continuous compression and gyrating action?

Introduction / Context: Primary and secondary crushers use different mechanical principles. Jaw crushers apply intermittent compression with a moving plate; roll crushers apply continuous compression with two cylinders. Gyratory crushers blend features of both by using a gyrating cone within a stationary shell, creating continuous, high-capacity compression suitable for large lumps. Given Data / Assumptions: Objective: identify the crusher whose action resembles a combination of jaw and roll mechanisms. Feed: large, hard rock typical of primary stages. Requirement: continuous throughput and uniform product for downstream processing. Concept / Approach: A gyratory crusher uses a conical head that gyrates inside a concave bowl. Material is continuously nipped and compressed as it moves downward, akin to the continuous compression of rolls, while the nip and release across the crushing chamber share conceptual similarity with jaw crushers. Hence, it is often described as combining both jaw-like nip mechanics and roll-like continuous duty. Step-by-Step Solution: List mechanisms: jaw (intermittent compression), roll (continuous compression), gyratory (gyrating cone with continuous compression).Match to description: the “combination” characterization fits the gyratory crusher.Eliminate mills (rod/ball) and fluid energy mill—used for fine grinding, not primary compression crushing. Verification / Alternative check: Standard flowsheets in mine operations show gyratories as primary units where very high capacity and large feed openings are required—consistent with the combination description. Why Other Options Are Wrong: Rod mill, ball mill, fluid energy mill: these are grinding mills relying on media or jets; they do not combine jaw/roll crushing actions. Common Pitfalls: Associating “fluid energy” with high power and mistakenly assuming primary breakage capability; jet mills are for ultrafines, not coarse crushing. Final Answer: Gyratory crusher

Q: Crushing efficiency benchmark: the ratio of useful surface-creating work to total energy input in crushing equipment typically lies in which range?

Introduction / Context: Crushing and grinding systems are intrinsically energy intensive. “Efficiency” in comminution refers to the fraction of energy that actually creates new surface area or effectual breakage versus heat, noise, and nonproductive motion. Recognizing realistic efficiency bounds helps set expectations and identify improvement levers (pre-screening, choke feeding, proper closed-side setting, and crusher selection). Given Data / Assumptions: Conventional mechanical crushers (jaw, gyratory, cone, rolls). Efficiency definition: useful breakage work / electrical input power. Steady-state operation with standard feeds. Concept / Approach: Numerous plant audits and classical comminution studies show very small conversion of input energy into creating new surface. Crushers typically achieve efficiencies well below those of mills; even then, numbers remain in the low single digits, often as low as fractions of a percent for certain duty points. Step-by-Step Solution: Recall empirical ranges: crushing efficiencies are usually sub-2%.Compare to provided intervals; only 0.1–2% matches accepted benchmarks.Select this as the best representation for “typical” performance. Verification / Alternative check: Heat balances on crushers show substantial temperature rise and mechanical losses; moreover, attempts to raise efficiency rely on process design (e.g., pre-screening fines to reduce overgrinding) rather than expecting the machine alone to be efficient. Why Other Options Are Wrong: 5–10% or higher ranges contradict widely reported industrial data for crushing. Common Pitfalls: Confusing motor efficiency (often >90%) with comminution efficiency (often near 1%). They measure completely different phenomena. Final Answer: 0.1 to 2

Question 1

Ball mill energy efficiency: to the nearest order of magnitude, what percentage of input energy is typically converted into useful grinding work in an industrial ball mill?

Accepted Answer

Introduction / Context: Ball mills are ubiquitous in mineral processing and cement industries. Despite their versatility, they are energy-inefficient: only a small fraction of supplied power produces new surface area and specific breakage, while most becomes heat, liner/media friction losses, and nonproductive motion. Knowing the typical efficiency range guides energy audits and motivates alternatives (pre-crushing, classification, high-pressure grinding rolls). Given Data / Assumptions: Industrial-scale, wet or dry tumbling ball mills. Overall “grinding efficiency” refers to useful comminution work vs. total electrical input. Typical plant data and literature show very low conversion to new surface. Concept / Approach: Energy balance and comminution theory indicate that most input power dissipates as heat and vibration. Practical efficiency for size reduction in ball mills is commonly reported in the low single digits. Hence, “order of magnitude” best matches the 1–5% range rather than higher values that would contradict well-established plant experience. Step-by-Step Solution: Identify the definition of efficiency (useful breakage work / total input).Recall benchmark values from plant energy surveys: low single-digit percentages.Match to the closest option: 1–5%. Verification / Alternative check: Energy benchmarking studies of milling circuits consistently show that classification improvements and pre-crushing offer major savings precisely because ball mill inherent efficiency is low (near a few percent). Why Other Options Are Wrong: 40–50%, 75–80%, 90–95%: These would imply near-conservative systems, which is unrealistic for tumbling mills; such values are inconsistent with measured temperature rise and media losses. Common Pitfalls: Confusing motor efficiency (often >90%) with comminution efficiency (often <5%); these are very different metrics. Final Answer: 1–5

Question 2

Rotary drum vacuum filter capacity: identify the primary factors governing filtrate throughput for a given installation and slurry.

Accepted Answer

Introduction / Context: Rotary drum vacuum filters are widely used for continuous solid–liquid separation in inorganic chemicals, fertilizers, and minerals. Filtration capacity (filtrate volume per time) depends on both operating setpoints and the nature of the slurry. Understanding the dominant levers helps troubleshoot low throughput and optimize cycle parameters (submergence, rotation speed, vacuum, wash). Given Data / Assumptions: Constant vacuum operation with periodic cake formation, dewatering, washing, and discharge. Slurry properties influence cake resistance and formation rate. Cake thickness directly changes resistance and drainage path length. Concept / Approach: Darcy's law for flow through porous media states that filtrate flux decreases with higher cake resistance and thickness. Slurry particle size distribution, compressibility, and solids loading govern cake structure and permeability. Thus both cake thickness and slurry characteristics are principal determinants of capacity, alongside secondary factors like cloth condition and vacuum level. Step-by-Step Solution: Relate filtrate flow to pressure drop and total resistance (medium + cake).Note that thicker cakes increase resistance linearly (for incompressible cakes) or even more (compressible cakes).Recognize slurry characteristics set the cake permeability/compressibility; both govern capacity. Verification / Alternative check: Startup optimization usually tunes drum speed (hence cake thickness) for a given slurry to maximize throughput while achieving target moisture—confirming the joint role of (a) and (b). Why Other Options Are Wrong: Cake thickness only or slurry only omits a coequal driver.“Neither”: contradicts fundamental filtration theory. Common Pitfalls: Running with excessively thick cakes to chase dryness, which collapses capacity; or ignoring slurry pretreatment (coagulation/conditioning) that can improve permeability dramatically. Final Answer: Both (a) & (b)

Question 3

Operating speed for efficient ball milling: relative to the critical speed, which regime is recommended for effective grinding in industrial ball mills?

Accepted Answer

Introduction / Context: Critical speed is the rotational speed at which centrifugal force holds the grinding media against the mill shell, preventing cataracting. Industrial mills are purposely run below this speed to maintain a cascading motion, which creates impact and abrasion necessary for size reduction. Given Data / Assumptions: Standard lifter-equipped tumbling ball mills. Objective: stable cataracting/cascading without centrifuging. Common practice targets 65–80% of critical speed. Concept / Approach: Below critical speed, media rise and then fall, generating impact on the toe of the charge. At or above critical speed, media cling to the shell, drastically reducing effective breakage. Therefore, efficient operation is “less than” critical speed, typically tuned with media size and mill filling to reach target grind. Step-by-Step Solution: Define critical speed condition (centrifuging at the wall).Relate grinding action to cascading trajectories below critical.Select the sub-critical regime as the efficient range. Verification / Alternative check: Empirical correlations for mill power draw and breakage rates show peak performance around 70–78% of critical, consistent with decades of operating experience. Why Other Options Are Wrong: Equal to or more than critical: causes centrifuging and poor breakage.“Minimum possible small balls”: ball size selection depends on feed top size; exclusively small balls may underperform on coarse feed. Common Pitfalls: Ignoring lifter design and charge filling; both change the effective trajectory and optimal percent of critical speed. Final Answer: At a speed less than the critical speed

Question 4

Agitation scale-up fundamentals: a vigorous mixing guideline is about 2–3 hp per gallon for very thin liquids. In dimensionless form, the impeller “Power number” (Po) is defined as which ratio of variables?

Accepted Answer

Introduction / Context: In mixing, the dimensionless Power number (Po) collapses impeller power draw across scales and fluids. Together with Reynolds number (Re = rho * N * D^2 / mu), it underpins geometric similarity scale-up. Knowing the correct Po definition ensures consistent design from lab to plant for blending, suspension, and gas–liquid contacting. Given Data / Assumptions: P is power (W), rho is fluid density (kg/m^3), mu is viscosity (Pa·s), N is rotational speed (s^-1), and D is impeller diameter (m). Thin (low-viscosity) liquids at high turbulence are assumed for the “2–3 hp per gallon” rule-of-thumb; however, Po is general. Concept / Approach: By dimensional analysis, P has units kg·m^2/s^3. The term rho * N^3 * D^5 also has units kg·m^2/s^3, so their ratio is dimensionless. Thus Po = P / (rho * N^3 * D^5). In turbulent regimes for a given impeller geometry, Po becomes nearly constant, enabling direct horsepower prediction at scale. Step-by-Step Solution: Write Po candidate forms and check dimensional consistency.Confirm rho * N^3 * D^5 has the same dimensions as P.Select the dimensionally correct and standard definition: Po = P / (rho * N^3 * D^5). Verification / Alternative check: Impeller datasheets list constant Po values (e.g., Rushton disk turbine Po ≈ 5–6 in turbulent flow), which are used with the above formula to size motors. Why Other Options Are Wrong: N^2 * D^3 or N * D^2 denominators do not render a dimensionless group equal to Po; those break dimensional consistency.Using viscosity (mu) in Po is incorrect; mu appears in Re, not Po. Common Pitfalls: Confusing Po with torque number (Np = T / (rho * N^2 * D^5)) or applying laminar correlations in turbulent regimes without checking Re. Final Answer: Po = P / (rho * N^3 * D^5)

Question 5

Work index consistency across circuits: if Wi is determined in closed-circuit grinding, what is the effective work index to use for comparable open-circuit grinding conditions?

Accepted Answer

Introduction / Context: Bond's Work Index (Wi) characterizes material grindability. While Wi is a material property, practical circuit performance differs between open and closed circuit operation due to classification efficiency and overgrinding. Empirical correction factors are therefore used to translate laboratory Wi to plant power estimates for specific flowsheets. Given Data / Assumptions: Wi was measured under closed-circuit conditions. We seek an equivalent for open-circuit grinding to the same product size range. Standard Bond practice recognizes open circuit typically requires more energy for the same size reduction target. Concept / Approach: Open-circuit grinding lacks the classification feedback loop, so fines remain in the mill longer, reducing breakage efficiency. A widely used translation is that open-circuit power demand is about 1.34 times that of closed circuit for the same P80, which maps to an “effective” Wi_open = 1.34 * Wi_closed for preliminary estimates. Step-by-Step Solution: Identify circuit type difference (open vs. closed).Recall empirical relationship: P_open ≈ 1.34 * P_closed.Translate to Wi scaling: Wi_open ≈ 1.34 * Wi_closed = 1.34 Wi. Verification / Alternative check: Plant benchmarking often shows substantial energy savings when converting open to closed circuits, which is consistent with the factor near 1.34 used in preliminary design and feasibility studies. Why Other Options Are Wrong: Wi or 0.5 Wi: underestimate open-circuit energy.3.34 Wi: overstates the difference far beyond typical observations. Common Pitfalls: Applying the factor blindly without considering classification efficiency, circulating load limits, or drastic changes in target fineness that can alter the effective factor. Final Answer: 1.34 Wi

Question 6

Electrostatic precipitator (ESP) operating envelope: select the approximate maximum gas temperature and pressure for which dust-laden streams can be effectively cleaned at very high collection efficiencies (≈99.9%).

Accepted Answer

Introduction / Context: Electrostatic precipitators (ESPs) exploit electrostatic charging and drift to collect fine particulates from gas streams in power, cement, and metallurgical operations. Their usable temperature and pressure ranges are limited by materials, electrical breakdown characteristics, and gas properties that affect corona stability and resistivity. Given Data / Assumptions: Target collection efficiency on the order of 99.9% for fine dusts. We consider the upper practical bounds for industrial ESPs with appropriate materials. Gas composition and dust resistivity are within typical controllable ranges (conditioning possible). Concept / Approach: ESPs are routinely applied to hot flue gases; specialized designs can handle several hundred degrees Celsius with proper electrode and insulator materials. While most operate near atmospheric pressure, ESP principles are not inherently limited to low pressure; pressure vessels and process constraints may extend the pressure envelope. Among the options provided, the pair of about 500°C and tens of atmospheres reflects the most reasonable “maximum” within the set. Step-by-Step Solution: Eliminate unrealistic extremes: 1000°C severely challenges electrodes/insulators and corona.Recognize typical high-temperature ESP practice near a few hundred °C; 500°C is an upper bound in the given choices.Select the associated pressure bound (50 atm) as the best available match from the list. Verification / Alternative check: Vendor specifications often cap continuous operation substantially below 1000°C. Engineering references cite high-temperature ESPs up to several hundred °C; pressure capability depends on housing design rather than collection physics. Why Other Options Are Wrong: 1000°C options: impractical due to thermal and electrical constraints.200°C and 5 atm: feasible but not “maximum” envelope among the options. Common Pitfalls: Ignoring dust resistivity windows and gas conditioning (e.g., SO3 injection or humidification), which strongly affect capture independent of temperature/pressure limits. Final Answer: 500°C and 50 atm

Question 7

Screen analysis notation: interpret the size class denoted as “+5 mm / −10 mm”. Which statement correctly describes the particles in this fraction?

Accepted Answer

Introduction / Context: Screen analysis classifies particulate solids by their ability to pass or be retained on standard apertures. Understanding shorthand notations like “+a / −b” is essential when reading sieve analyses, designing screen decks, or comparing product specs. Given Data / Assumptions: “+a” means the particles are larger than a (retained on a sieve of aperture a). “−b” means the particles are smaller than b (pass through b). Here, a = 5 mm and b = 10 mm. Concept / Approach: Particles labeled “+5 mm / −10 mm” are those that pass a 10 mm sieve but are retained on a 5 mm sieve. This describes the size interval between 5 and 10 mm inclusive of screening conventions (with a practical tolerance due to near-mesh particles and shape effects). Step-by-Step Solution: Interpret “+5”: retained on 5 mm, so size ≥ 5 mm (by screening convention).Interpret “−10”: pass 10 mm, so size ≤ 10 mm.Combine: 5–10 mm class, i.e., pass 10 mm and retain on 5 mm. Verification / Alternative check: Check typical screen analysis tables: fractions are commonly labeled with an upper “minus” (passing) and lower “plus” (retained) bound to define the interval. Why Other Options Are Wrong: “Pass 5 mm and retain on 10 mm” is impossible because a particle cannot be larger than 10 mm and simultaneously pass it.“Pass both” or “neither” do not define a standard size interval. Common Pitfalls: Reversing the “+” and “−” meanings; always remember “+” ties to retention (coarser than) and “−” to passing (finer than) the stated size. Final Answer: Passing through 10 mm screen and retained on 5 mm screen

Question 8

Primary size reduction of hard rock: which equipment is typically selected for the preliminary breaking (coarse crushing) stage before secondary reduction?

Accepted Answer

Introduction / Context: Primary crushing handles run-of-mine hard rock, reducing it to manageable sizes for subsequent secondary/tertiary crushers or grinding mills. The machine must accept large lumps, deliver high throughput, and provide a controlled top size. Given Data / Assumptions: Very hard, abrasive rock in large lump sizes. Objective: coarse crushing to a product suitable for conveyors and secondary crushers. Continuous, high-capacity duty. Concept / Approach: Gyratory crushers provide continuous crushing action with high capacity and can accept very large feed sizes due to their open intake and robust design. They are the standard choice for primary crushing of hard rock in large mines and quarries, delivering a relatively uniform product for downstream processing. Step-by-Step Solution: Identify primary duty requirements: large feed, high throughput.Match equipment features: gyratory offers continuous compression and large gape.Select gyratory crusher over mills and specialized disintegrators intended for finer or softer materials. Verification / Alternative check: Flowsheets in hard-rock mining typically list “ROM → gyratory → conveyors → secondary cone/crusher,” confirming standard practice. Why Other Options Are Wrong: Ball/tube mills: grinding equipment for much finer sizes, not for ROM lumps.Squirrel-cage disintegrators: used for soft/brittle materials, not massive hard rock. Common Pitfalls: Attempting to feed very large boulders into jaw-only small primaries; gyratories are preferred for the largest capacities and feeds. Final Answer: Gyratory crusher

Question 9

Ultrafine grinders: identify the dominant size-reduction mechanism typically exploited to reach sub-100 micrometre and micron-scale product sizes.

Accepted Answer

Introduction / Context: Ultrafine grinding targets very small particle sizes with narrow distributions for pigments, pharmaceuticals, and specialty minerals. Different machines (jet mills, stirred media mills) employ different physics than coarse crushers. Recognizing the dominant mechanism aids in choosing the correct mill and media. Given Data / Assumptions: Target sizes: tens of micrometres down to a few micrometres or below. Equipment examples: fluid energy (jet) mills, attritors/stirred media mills, pin mills. Feed material is friable enough for fine comminution. Concept / Approach: At ultrafine scales, attrition (rubbing/shearing along surfaces) dominates because it can continuously erode particle asperities and break weak points, especially in high-shear zones between media or in opposed jets. Impact alone becomes less efficient for very small targets due to decreasing fracture probability at low mass and the need for high collision frequency. Step-by-Step Solution: Identify typical ultrafine equipment and their internal motion (e.g., stirred beads).Relate micro-scale breakage to shear/attrition prevalence.Select “Attrition” as the dominant mechanism for ultrafine grinders. Verification / Alternative check: Process descriptions for jet mills and stirred media mills emphasize particle–particle and particle–media shear interactions, with operating metrics (specific energy, tip speed) tuned for attrition-driven breakage. Why Other Options Are Wrong: Slow compression: characteristic of coarse crushers.Impact: important in some mills but not the primary driver for consistent sub-100 μm production across machines like attritors.Cutting action: relevant for fibrous materials, not typical for mineral ultrafines. Common Pitfalls: Expecting ball mills alone to achieve ultrafine sizes efficiently; specialized attrition-based mills are more suitable. Final Answer: Attrition

Question 10

Agitation in laminar regime: during liquid mixing, which variable does not make the power consumption proportional when flow is strictly laminar and impeller geometry is fixed?

Accepted Answer

Introduction / Context: Power draw in mechanically agitated vessels depends on flow regime. In the laminar regime (low Reynolds number), viscous forces dominate and the classical turbulent correlations (constant Power number) do not apply. Understanding which variables control power under laminar flow is essential for sizing lab mixers, dosing tanks, and high-viscosity blending operations without over- or underpowering the drive. Given Data / Assumptions: Flow is strictly laminar (Re << 10). Impeller geometry and tank dimensions are fixed (geometric similarity). Steady, single-phase liquid mixing; no gas dispersion or solids suspension. Concept / Approach: In laminar mixing, the Power number, Po = P / (rho * N^3 * D^5), varies inversely with Reynolds number, Re = rho * N * D^2 / mu. For many laminar impellers, Po = K / Re, where K is a geometry constant. Rearranging gives P = K * mu * N^2 * D^3. Thus, P is directly proportional to viscosity (mu), to the square of rotational speed (N^2), and to the cube of impeller diameter (D^3). Notably, density (rho) cancels from the laminar result; hence, power is not proportional to density in this regime. Step-by-Step Solution: Start with Po = P / (rho * N^3 * D^5).Use Po = K / Re and Re = rho * N * D^2 / mu.Therefore, P / (rho * N^3 * D^5) = K * mu / (rho * N * D^2).Solve for P: P = K * mu * N^2 * D^3 (no rho term remains). Verification / Alternative check: Practical scale-ups for viscous laminar systems hold N^2 * D^3 * mu constant to preserve shear fields and power density, further confirming the proportionalities above and the independence from density. Why Other Options Are Wrong: Viscosity: Appears linearly in P = K * mu * N^2 * D^3; power is proportional to mu.Impeller diameter cubed: Power scales with D^3 in laminar mixing.Rotational speed squared: Power scales with N^2 in laminar mixing. Common Pitfalls: Confusing turbulent and laminar correlations; using a constant Power number at low Re leads to major sizing errors. Also, overlooking temperature effects on viscosity can cause large power deviations in laminar service. Final Answer: density of the liquid

Question 11

Crusher identification: which machine is commonly described as a combination of a jaw crusher and a roller crusher due to its continuous compression and gyrating action?

Accepted Answer

Introduction / Context: Primary and secondary crushers use different mechanical principles. Jaw crushers apply intermittent compression with a moving plate; roll crushers apply continuous compression with two cylinders. Gyratory crushers blend features of both by using a gyrating cone within a stationary shell, creating continuous, high-capacity compression suitable for large lumps. Given Data / Assumptions: Objective: identify the crusher whose action resembles a combination of jaw and roll mechanisms. Feed: large, hard rock typical of primary stages. Requirement: continuous throughput and uniform product for downstream processing. Concept / Approach: A gyratory crusher uses a conical head that gyrates inside a concave bowl. Material is continuously nipped and compressed as it moves downward, akin to the continuous compression of rolls, while the nip and release across the crushing chamber share conceptual similarity with jaw crushers. Hence, it is often described as combining both jaw-like nip mechanics and roll-like continuous duty. Step-by-Step Solution: List mechanisms: jaw (intermittent compression), roll (continuous compression), gyratory (gyrating cone with continuous compression).Match to description: the “combination” characterization fits the gyratory crusher.Eliminate mills (rod/ball) and fluid energy mill—used for fine grinding, not primary compression crushing. Verification / Alternative check: Standard flowsheets in mine operations show gyratories as primary units where very high capacity and large feed openings are required—consistent with the combination description. Why Other Options Are Wrong: Rod mill, ball mill, fluid energy mill: these are grinding mills relying on media or jets; they do not combine jaw/roll crushing actions. Common Pitfalls: Associating “fluid energy” with high power and mistakenly assuming primary breakage capability; jet mills are for ultrafines, not coarse crushing. Final Answer: Gyratory crusher

Question 12

Liquid–solid separation terminology: the operation that splits a suspension into a clear supernatant liquid and a denser underflow slurry is called

Accepted Answer

Introduction / Context: Process engineers frequently handle suspensions that must be separated into a clear liquid and a concentrated slurry. Correct terminology avoids confusion when specifying thickeners, clarifiers, and conditioning steps. Two distinct ideas are particle aggregation (coagulation/flocculation) and separation by gravity (sedimentation/clarification). Given Data / Assumptions: Feed: a dilute suspension of solids in a liquid. Objective: obtain a clear supernatant and a concentrated underflow (slurry). No chemical addition implied unless stated. Concept / Approach: Sedimentation is the gravitational settling of particles, yielding a clear supernatant above and a denser slurry below. Clarification refers to the overall process of producing a clear liquid (often the supernatant product), sometimes used as a broader term. Coagulation and flocculation are pretreatments that alter particle interactions to enhance settling but are not the separation itself. Step-by-Step Solution: Identify the physical phenomenon: gravitational settling of suspended particles.Name the operation: sedimentation (in thickeners/clarifiers).Recognize that coagulation/flocculation are chemical conditioning steps, not the separation outcome. Verification / Alternative check: Thickener design equations (e.g., batch settling tests) use sedimentation velocities to size equipment for achieving the required split between supernatant and underflow—reinforcing that the separation is sedimentation. Why Other Options Are Wrong: Coagulation: neutralizes charges using coagulants; not the final separation step.Flocculation: builds larger flocs via polymers; a conditioning step.Clarification: often the name of the process goal (clear liquid), but the physical operation producing the split is sedimentation. Common Pitfalls: Using “clarification” and “sedimentation” interchangeably; while common in industry, exam questions typically reserve “sedimentation” for the gravity-driven split into supernatant and slurry. Final Answer: sedimentation

Question 13

Crushing efficiency benchmark: the ratio of useful surface-creating work to total energy input in crushing equipment typically lies in which range?

Accepted Answer

Introduction / Context: Crushing and grinding systems are intrinsically energy intensive. “Efficiency” in comminution refers to the fraction of energy that actually creates new surface area or effectual breakage versus heat, noise, and nonproductive motion. Recognizing realistic efficiency bounds helps set expectations and identify improvement levers (pre-screening, choke feeding, proper closed-side setting, and crusher selection). Given Data / Assumptions: Conventional mechanical crushers (jaw, gyratory, cone, rolls). Efficiency definition: useful breakage work / electrical input power. Steady-state operation with standard feeds. Concept / Approach: Numerous plant audits and classical comminution studies show very small conversion of input energy into creating new surface. Crushers typically achieve efficiencies well below those of mills; even then, numbers remain in the low single digits, often as low as fractions of a percent for certain duty points. Step-by-Step Solution: Recall empirical ranges: crushing efficiencies are usually sub-2%.Compare to provided intervals; only 0.1–2% matches accepted benchmarks.Select this as the best representation for “typical” performance. Verification / Alternative check: Heat balances on crushers show substantial temperature rise and mechanical losses; moreover, attempts to raise efficiency rely on process design (e.g., pre-screening fines to reduce overgrinding) rather than expecting the machine alone to be efficient. Why Other Options Are Wrong: 5–10% or higher ranges contradict widely reported industrial data for crushing. Common Pitfalls: Confusing motor efficiency (often >90%) with comminution efficiency (often near 1%). They measure completely different phenomena. Final Answer: 0.1 to 2

Question 14

Geometric identities for spheres: which expression corresponds to the surface-based shape factor for a sphere when using particle diameter D (with A = surface area, V = volume)?

Accepted Answer

Introduction / Context: Particle technology uses simple geometric relations for ideal shapes to define size descriptors, sphericity, and shape factors. For spheres, exact formulas exist linking surface area (A), volume (V), and diameter (D). Recognizing these identities is fundamental before extending to irregular particles and shape-corrected diameters (e.g., volume-surface mean diameter). Given Data / Assumptions: Spherical particle of diameter D. Surface area A and volume V denote geometric properties. We are asked specifically about a surface-based factor. Concept / Approach: For a sphere, A = π D^2 and V = (π/6) D^3. The commonly referenced “surface shape factor” in simple MCQs is A / D^2, which for a sphere equals π. The companion “volume shape factor” is V / D^3 = π/6. Both are exact identities that act as baselines for comparing non-spherical particles. Step-by-Step Solution: Recall area of sphere: A = 4 * π * r^2 = π * D^2.Compute A / D^2 = (π * D^2) / D^2 = π.Recognize that V = π/6 * D^3 is the volume identity, not the requested surface-based factor. Verification / Alternative check: Dimensional analysis confirms A / D^2 is dimensionless; for a sphere, this reduces exactly to the constant π, matching textbook geometry. Why Other Options Are Wrong: π/6 (= V / D^3) is the volume-based identity, not the surface shape factor asked for.“None of these” is incorrect because π (= A / D^2) is correct. Common Pitfalls: Mixing up surface and volume relations; ensure you identify whether the factor normalizes area or volume before selecting the expression. Final Answer: π (= A / D^2)

Question 15

Tube mill versus ball mill: select the correct combined statements about product fineness, aspect ratio, and ball size in tube mills.

Accepted Answer

Introduction / Context: Tube mills are a subtype of ball mills characterized by a much larger length-to-diameter ratio. They are widely used for finishing grinding (e.g., cement finish mills) where a finer product is required. The longer residence time and staged compartments allow the use of progressively smaller media to achieve the final fineness efficiently. Given Data / Assumptions: Comparing conventional ball mills (L/D ≈ 1) with tube mills (L/D > 2). Grinding duty emphasizes finer product sizes. Media grading in tube mills typically shifts to smaller balls downstream. Concept / Approach: Greater length promotes classification of media sizes by compartment, enabling coarse breakage up front and fine grinding at the back. Smaller balls provide higher surface area and more points of contact, aiding finish grinding. Hence, tube mills produce finer products, are longer relative to diameter, and use smaller balls compared with standard short ball mills. Step-by-Step Solution: Identify product target: finer finish cement or minerals.Relate L/D and compartment design to residence time and media grading.Conclude that all three statements (a), (b), and (c) are accurate. Verification / Alternative check: Classic cement circuits specify multi-compartment tube mills with media gradation from larger to smaller sizes, confirming the practice and outcomes listed. Why Other Options Are Wrong: Any single statement alone is incomplete; the combined statement best represents the standard differences. Common Pitfalls: Assuming one ball size fits all compartments; failing to grade media undermines the benefits of tube mill geometry. Final Answer: all (a), (b) & (c).

Question 16

End-of-cycle washing in constant-pressure filtration: when washing begins immediately after filtration at the same pressure, how does the initial washing rate compare to the filtration rate?

Accepted Answer

Introduction / Context: In constant-pressure filtration, cake forms and thickens with time, raising flow resistance. At the end of filtration, a washing step may begin at the same applied pressure to displace mother liquor from pores. Understanding the relation between the washing rate and the preceding filtrate rate helps estimate washing time and solvent usage. Given Data / Assumptions: Cake is already formed; filtration was at constant pressure. Washing starts immediately at the same pressure differential. Fluid properties during wash are similar, unless otherwise noted. Concept / Approach: At the instant washing begins, the hydraulic resistance of the cake and medium has not changed from the end of filtration. Therefore, with the same pressure drop, the volumetric flow rate initially equals the final filtration rate. As washing proceeds, composition and viscosity may change and concentration gradients disappear, often causing a gradual rate decrease as flow paths equilibrate. Step-by-Step Solution: Recall Darcy's law: Q = ΔP / (μ * R_total) * A.At switch-over, ΔP, μ (approximately), and R_total are initially the same as at end of filtration.Therefore, initial wash rate equals the rate at end of filtration; then it decreases as conditions evolve (e.g., cake consolidation, viscosity differences). Verification / Alternative check: Batch filtration wash curves commonly start at the end-of-filtration rate and then decline, matching industrial experience and standard design heuristics. Why Other Options Are Wrong: “At time zero” without context is ambiguous; the key is equality at the transition point.“At the end of filtration” alone omits the subsequent decrease.“Half the filtrate” has no basis in Darcy's law for this comparison. Common Pitfalls: Ignoring cake compressibility and fluid property changes during washing; both can accentuate the decline in rate compared with the end-of-filtration value. Final Answer: at the end of filtration, but decreases with time subsequently.

Question 17

Filter medium resistance: in practical filtration, what constitutes the resistance attributed to the “filter medium”?

Accepted Answer

Introduction / Context: In Darcy-type filtration models, total hydraulic resistance comprises the filter medium resistance and the cake resistance. Correctly attributing medium resistance is important for start-up behavior, cloth selection, and cleaning schedule since it affects the initial filtrate flux before appreciable cake has formed. Given Data / Assumptions: The medium is usually a fabric (cloth) or porous plate supporting the cake. Over time, particles embed within the pores of the cloth/septum. We consider steady, repeated cycles typical of industrial practice. Concept / Approach: The “filter medium resistance” is not just the virgin cloth resistance; it quickly includes the resistance of particles that lodge in the medium pores and remain after discharge. Thus, effective medium resistance equals the resistance of the cloth plus the embedded solids, forming a composite baseline independent of the removable cake layer. Step-by-Step Solution: Define R_total = R_medium + R_cake.Recognize R_medium = R_cloth + R_embedded (persistent fouling layer).Conclude that option describing both cloth and embedded particles is correct. Verification / Alternative check: Filter cycle data show that initial clean-bed flux declines over successive cycles unless cloth is backwashed or replaced, indicating accumulation of embedded resistance. Why Other Options Are Wrong: Cloth alone omits embedded fines, underestimating resistance.Embedded particles alone ignore the baseline cloth resistance.“None of these” contradicts standard filtration practice. Common Pitfalls: Assuming medium resistance is constant; in reality, it drifts upward with fouling and must be managed via cleaning protocols. Final Answer: filter cloth and the embedded particle collectively.

Question 18

Filtration mode identification: the operation in which inlet pressure is progressively increased to maintain throughput as cake builds up is termed

Accepted Answer

Introduction / Context: As a filter cake grows, its resistance rises, causing the filtrate rate to fall if driving pressure is held constant. Operators can choose different operating modes depending on product goals: maintain pressure (and accept declining rate) or maintain rate (and raise pressure). Correctly naming these modes is essential when specifying filter presses, rotary vacuum filters with boosters, or pressure nutsche systems. Given Data / Assumptions: Single-phase liquid through a compressible or incompressible cake. Operation seeks to keep filtrate flow approximately steady despite rising cake resistance. Pressure can be manipulated by the process (e.g., pump ramping). Concept / Approach: By Darcy's law, Q = ΔP / (μ * R_total) * A. If R_total increases with time due to cake buildup, keeping Q constant requires ΔP to increase correspondingly. This is called constant-rate filtration. Conversely, holding ΔP constant and allowing Q to decline is constant-pressure filtration. Step-by-Step Solution: Recognize the operational goal: maintain throughput as cake resistance rises.Infer the manipulated variable: increase inlet pressure over time.Name the mode: constant-rate filtration. Verification / Alternative check: Filter press cycles sometimes employ stepped pressure profiles to approximate constant rate early in the cycle, verifying the link between increasing pressure and rate stabilization. Why Other Options Are Wrong: “Varying pressure” is descriptive but not the standard mode name; the recognized term is “constant rate.”“Varying rate” and “constant pressure” describe the alternative mode where pressure is fixed and the rate declines. Common Pitfalls: Confusing “constant rate” with “constant velocity” of the cake front in dewatering beds; ensure you refer to filtrate volumetric flow rate. Final Answer: constant rate

Question 19

Grinder reduction ratio definition: for size reduction equipment handling fine products, the standard reduction ratio (based on characteristic diameters) is defined as

Accepted Answer

Introduction / Context: Reduction ratio is a concise descriptor of how much a grinder or crusher reduces a characteristic particle size from feed to product. It guides preliminary equipment selection and indicates whether single- or multi-stage reduction is needed to reach a target size distribution. Given Data / Assumptions: Df = characteristic (average) diameter of the feed. Dp = characteristic (average) diameter of the product. We adopt the conventional ratio definition used in comminution texts. Concept / Approach: The standard reduction ratio for grinders is defined as the feed size divided by the product size, i.e., Df / Dp. Values greater than 1 indicate size reduction; the larger the value, the greater the extent of reduction. This ratio can be defined on various characteristic sizes (e.g., average, median, or P80), as long as consistency is maintained across calculations. Step-by-Step Solution: Identify the conventional definition: Reduction ratio R = Df / Dp.Confirm that this gives R > 1 for actual size reductions.Select the option explicitly stating Df/Dp. Verification / Alternative check: Equipment datasheets and plant reports commonly state crusher and mill reduction ratios in this form, aligning with sizing methodologies and scale-up heuristics. Why Other Options Are Wrong: Dp/Df would give values less than 1 for size reduction and inverts the conventional definition.Df - Dp is dimensionally a length, not a dimensionless ratio, and is not standard.“None of these” is incorrect because Df/Dp is the accepted definition. Common Pitfalls: Mixing definitions when comparing vendors (e.g., some may quote R on P80 while others on median size). Always confirm which characteristic size is being used. Final Answer: Df/Dp

Question 20

In particle-size engineering, a fluid energy mill (also called a jet mill) belongs to which category of comminution equipment used for producing extremely fine powders?

Accepted Answer

Introduction / Context: A fluid energy mill, often called a jet mill or micronizer, accelerates particles in high-velocity gas streams where they collide and fracture. It is designed specifically for producing sub-50 microm and even sub-10 microm powders used in pharmaceuticals, pigments, ceramics, and specialty chemicals. Given Data / Assumptions: Equipment: fluid energy (jet) mill. Operating principle: gas jets, no mechanical media. Target: extremely fine product. Concept / Approach: Comminution devices can be grouped into crushers (coarse), grinders (intermediate), and ultrafine grinders. Jet mills create intense interparticle impact and attrition within a gas vortex. The absence of mechanical contact minimizes contamination and enables very fine, narrow cuts—hence the classification as an ultrafine grinder. Step-by-Step Solution: Identify function: produce very fine powders.Identify mechanism: high-velocity gas → interparticle collisions.Match category: ultrafine grinder. Verification / Alternative check: Commercial jet mills routinely achieve d50 values in the few microm range, far beyond typical ball or hammer mills. Why Other Options Are Wrong: Crusher/Cutter: suited for coarse or shape-cutting duties.General grinder: too broad; jet mills are specifically ultrafine. Common Pitfalls: Confusing “high energy” with mechanical media mills; jet mills rely on gas energy, not heavy grinding media. Final Answer: Ultrafine grinder

Mechanical Operations Questions