Q: For a given required volumetric flow and fluid, how does increasing pipe diameter generally affect pumping (energy) cost in steady single-phase flow?

Introduction / Context: Pipeline design balances capital cost (larger diameter, more material) against operating cost (pumping energy). Understanding how frictional losses scale with diameter allows engineers to choose an economical size for a specified flow rate and fluid. Given Data / Assumptions: Single-phase, incompressible flow at steady state. Fixed volumetric (or mass) flow requirement and fluid properties. Fully developed flow; roughness and friction factor depend on Reynolds number as usual. Concept / Approach: Frictional pressure drop in a pipe of length L is ΔP = f * (L/D) * (ρ v^2 / 2). For fixed volumetric flow, velocity v scales as 1/D^2, so ΔP scales approximately as 1/D^5 in laminar and roughly between 1/D^4 and 1/D^5 in turbulent (since f varies mildly with Reynolds number). Hence, larger diameter dramatically reduces pressure drop and thus pump power P ≈ ΔP * Q / η. Step-by-Step Solution: Increase D → decrease v = Q / A → reduce velocity head ρ v^2 /2.ΔP ∝ (L/D) * v^2 → strong inverse dependence on D.Pump power P ∝ ΔP for constant Q → P decreases as D increases, holding all else constant. Verification / Alternative check: Numerical examples with common friction correlations (e.g., Blasius for smooth turbulent) confirm the steep drop in ΔP with increasing D at constant Q. Why Other Options Are Wrong: (b) and (c) contradict fluid mechanics; (d) Newtonian vs. non-Newtonian can change f(Re) but not the fundamental inverse trend; (e) is not representative for a fixed Q. Common Pitfalls: Ignoring capital cost increase with D; forgetting that very large D may lead to low velocities and solids deposition in slurries. Final Answer: Pumping cost decreases.

Q: Sieve tray hydraulics: typical perforation size—what is the usual hole diameter range used on sieve trays in distillation columns?

Introduction / Context: Sieve trays use perforations to distribute vapor through a liquid layer. Hole diameter affects pressure drop, weeping tendency, capacity, and fouling susceptibility. Selecting an appropriate range is a core part of tray specification and rating. Given Data / Assumptions: Conventional sieve trays for common refinery/chemical services. Moderate fouling tendencies; typical tray thickness and spacing. Perforation fraction and weir height within standard design bands. Concept / Approach: Holes that are too small increase pressure drop and are prone to plugging; holes that are too large encourage weeping at low vapor rates and can reduce contacting efficiency. A practical window widely used in industry is roughly 3–12.5 mm diameter, which balances hydraulic stability and fouling tolerance across a broad range of services. Step-by-Step Solution: Establish conventional practice: aim for mid-single-digit millimeter holes for general duty.Check extreme ranges: 12.5 mm risks weeping at turndown.Therefore, select the standard range 3–12.5 mm. Verification / Alternative check: Tray vendors and handbooks list similar ranges, with specific choices tuned to vapor load, fouling risk, and perforation percentage. Why Other Options Are Wrong: 1–3 mm is typically small for most hydrocarbon services; 4–8 mm is a subset, not the general range; 12.5–18.5 mm or 20–30 mm are large and unusual for sieve trays except special duties. Common Pitfalls: Ignoring turndown and fouling when picking hole size; overlooking interaction with tray thickness and opening fraction. Final Answer: 3 to 12.5 mm

Question 1

Pressure vessel heads and closures: identify the incorrect statement about flanged-and-dished, shallow-dished, flat heads, and stress considerations.

Accepted Answer

Introduction / Context: Pressure vessel ends (heads) must safely transmit internal pressure to the shell while controlling stress concentrations. Common head types include flat heads, flanged-and-dished, shallow-dished, hemispherical, and variants. Recognizing which statements reflect real mechanical behavior is vital for safe design and code compliance. Given Data / Assumptions: Thin to moderately thick shells designed under applicable codes (e.g., ASME) at typical process pressures. Standard geometries: flanged-and-dished (F&D), shallow-dished, flat heads. Comparisons are qualitative, not code formula derivations. Concept / Approach: Curved heads (dished/hemispherical) distribute membrane stresses efficiently and require less thickness than flat plates for the same diameter and pressure. Flat heads experience bending stresses and thus become impractically thick at large diameters or high pressures. Statements about crown radius relative to shell diameter reflect standard geometric definitions for dished heads. Step-by-Step Solution: Assess (a): For standard F&D heads, crown radius on the order of shell diameter is typical; statement is acceptable.Assess (b): Shallow dished heads indeed use larger crown radii than the shell diameter; acceptable.Assess (c): Flat heads being “most suitable” for large diameter at very high pressure is incorrect; they require extreme thickness and are rarely economical.Assess (d): Flanged-only head notionally reduces sharp discontinuity at the center; acceptable as qualitative stress reasoning.(e) is a true generalization: dished heads lower edge bending relative to flat plates. Verification / Alternative check: Code formulas show required thickness for flat heads grows rapidly with diameter and pressure, validating why curved heads are preferred for high pressure/large diameter applications. Why Other Options Are Wrong: (a), (b), (d), and (e) reflect accepted geometry/stress behavior; (c) contradicts fundamental bending vs. membrane stress efficiency. Common Pitfalls: Assuming flat covers are interchangeable with dished heads; overlooking nozzle-induced local stresses and the need for reinforcement regardless of head shape. Final Answer: Flat head covers are most suitable for larger vessels operating at very high pressure.

Question 2

Gas filtration through fabric (bag filter): how does clean-side pressure drop scale with gas properties in first-cut sizing?

Accepted Answer

Introduction / Context: Bag filters remove particulates by forcing gas through a porous fabric. The pressure drop (ΔP) across the medium is a key design parameter because it drives fan power and operating cost. Understanding which gas properties influence ΔP helps avoid incorrect scaling during preliminary selection. Given Data / Assumptions: Clean or lightly dust-loaded fabric (cake resistance negligible for the moment). Low to moderate face velocity; laminar flow through pores (Darcy regime). Ideal gas behavior; steady operation at a given temperature. Concept / Approach: In laminar flow through porous media, Darcy’s law gives ΔP = (μ * v * L) / k, where μ is dynamic viscosity, v is superficial velocity, L is thickness, and k is permeability. In this regime, ΔP is directly proportional to viscosity and velocity; gas density does not directly appear in the linear term (density mainly affects fan power through volumetric flow and can appear in non-Darcy/inertial terms at higher velocities). Hence, simple statements that assert a direct or inverse proportionality to density are not generally correct for the Darcy regime. Step-by-Step Solution: Start from Darcy’s law: ΔP ∝ μ * v for a given medium.Note that none of the options (a), (b), or (c) matches this clean dependence because each imposes a specific density proportionality.Therefore, the only accurate choice within the given list is “None of these.” Verification / Alternative check: At higher velocities, Forchheimer (inertial) effects add a term ∝ ρ * v^2. That would make ΔP partly depend on density, but not “directly only” as in options given; it becomes a mixed μ and ρ dependence, still inconsistent with (a)–(c). Why Other Options Are Wrong: (a) ignores viscosity dependence; (b) and (c) assign density dependencies that are not inherent in Darcy’s linear regime; (e) compressibility factor is not a primary parameter here. Common Pitfalls: Confusing fan power (which depends on ρ via volumetric flow) with filter ΔP scaling; neglecting cake growth, which later dominates ΔP and may change scaling. Final Answer: None of these.

Question 3

Tray internals: typical bubble-cap tray layout—what is the usual pitch (center-to-center spacing) as a multiple of the cap outside diameter?

Accepted Answer

Introduction / Context: Bubble-cap trays distribute vapor through risers and caps to contact the liquid on the tray deck. The geometric pitch of caps governs hydraulics, froth regime, tray pressure drop, and mass-transfer area per tray. Recognizing typical pitch ranges helps during preliminary layout and vendor discussions. Given Data / Assumptions: Conventional bubble-cap trays handling hydrocarbon or aqueous systems at moderate pressure. Standard cap sizes and risers; no special high-capacity designs. Uniform staggering pattern across the active area. Concept / Approach: Caps must be close enough to ensure uniform vapor distribution and adequate interfacial area, yet far enough apart to limit hydraulic interference and excessive pressure drop. Industry practice typically places caps at roughly 1.6–2.0 times the cap outside diameter (OD), balancing area with froth stability. Step-by-Step Solution: Identify the common heuristic: pitch ≈ (1.6–2.0) * OD.Compare options: the offered range in option (b) matches standard design practice.Reject ranges that are too tight (≈1.3 * OD) or too loose (>3 * OD), which risk maldistribution or excessive pressure drop. Verification / Alternative check: Vendor catalogs and tray rating tools provide similar pitch guidance, adjusted for cap size, vapour rates, and desired capacity/efficiency trade-offs. Why Other Options Are Wrong: (a) is generally tight and may increase interference; (c) “exactly 2.5” is too rigid and typically high; (d) is overly broad; (e) is excessive for most services. Common Pitfalls: Assuming one pitch works for all cap sizes; neglecting downcomer layout and active area fraction; ignoring pressure drop constraints when crowding caps. Final Answer: 1.6 to 2 times the cap OD

Question 4

Shell-and-tube bundle layout choice: compared with square pitch, what is a key hydraulic/thermal consequence of triangular tube pitch?

Accepted Answer

Introduction / Context: Tube layout impacts thermal performance, pressure drop, cleaning, and mechanical integrity. The two common patterns—square and triangular—offer different trade-offs. Understanding these helps during exchanger specification and when reconciling performance with fouling behavior. Given Data / Assumptions: Single-phase crossflow on shell side; comparable baffle arrangements. Same shell diameter and tube outside diameter across layouts. No exotic anti-fouling or low-fin tubes. Concept / Approach: Triangular pitch packs more tubes into a given shell diameter, increasing heat-transfer area per length and promoting mixing/turbulence due to narrower cross-flow lanes. The trade-off is higher shell-side pressure drop and more difficult mechanical rodding compared with square pitch. Square pitch is preferred for dirty shell-side service because straight cleaning lanes are available. Step-by-Step Solution: Compare layouts at fixed shell diameter: triangular places more tubes → tighter flow passages → higher velocity and turbulence.Higher turbulence typically raises shell-side h but also increases ΔP.Therefore, (a) is the best statement of consequence. Verification / Alternative check: Thermal rating software and TEMA guidance consistently show higher shell-side ΔP and h for triangular vs. square layouts, all else equal. Why Other Options Are Wrong: (b) is opposite of reality; (c) square—not triangular—eases cleaning; (d) turbulence is higher, not lower; (e) layout does not “eliminate” vibration concerns. Common Pitfalls: Optimizing only for area density without accounting for pump power; choosing triangular layout for very dirty shell-side services. Final Answer: Triangular pitch results in higher shell-side pressure drop (for the same shell diameter).

Question 5

Selecting tube material purely on thermal conductivity: which material offers the highest conductivity among the following common choices?

Accepted Answer

Introduction / Context: Thermal conductivity strongly influences the overall heat-transfer coefficient U when tube-side or wall resistance is significant. Although many other criteria matter (cost, corrosion, fouling, strength), this question isolates conductivity to identify the best heat conduction material among common choices. Given Data / Assumptions: Room-temperature, typical commercial alloys. No consideration of corrosion resistance, erosion, or mechanical strength. Comparison is qualitative but grounded in well-known conductivity rankings. Concept / Approach: Approximate conductivities (W/m·K): copper ≈ 380–400; aluminium ≈ 200–235; carbon steel ≈ 45–60; stainless steels ≈ 14–20; titanium ≈ 17–22. Thus, copper clearly leads, producing a smaller wall resistance R_wall = t / (k * A), which increases U for wall-limited cases. Step-by-Step Solution: List relative k values: Cu > Al > carbon steel > stainless ≈ titanium.Select the highest: copper.Acknowledge trade-offs: despite high k, copper may be unsuitable in corrosive chloride or ammonia environments; designers often pick Cu-Ni or titanium for seawater, stainless for corrosives, etc. Verification / Alternative check: Material property databases and vendor catalogs confirm this ranking over typical operating temperatures relevant to process heat exchangers. Why Other Options Are Wrong: Aluminium is good but below copper; steels and titanium have substantially lower k and are chosen for other attributes. Common Pitfalls: Optimizing solely for k without considering corrosion and mechanical design; ignoring fouling resistance which can dominate overall U. Final Answer: Copper

Question 6

Typical geometry ratios in shell-and-tube exchangers: which statement about tube length to shell diameter (L/Ds) is most appropriate for common services?

Accepted Answer

Introduction / Context: The length-to-diameter ratio (tube length L to shell diameter Ds) guides thermal and mechanical design, affecting support spacing, pressure drop, and fabrication practicality. Different services (liquid–liquid vs. gas–gas) typically employ different L/Ds ranges due to property contrasts and allowable pressure drops. Given Data / Assumptions: Conventional single-phase services. Standard TEMA construction without unusual constraints. Economically reasonable pressure drops. Concept / Approach: Liquid–liquid duties can tolerate moderate pressure drops and benefit from longer tubes, giving L/Ds of roughly 4:1 to 8:1. Gas–gas duties usually require larger flow areas and shorter lengths per pass to control ΔP, hence L/Ds often less than 4:1. Extremely long tubes (e.g., 12:1) are possible but can challenge fabrication, support, and vibration control; they are not “universal” across services. Step-by-Step Solution: Match typical practice: liquid–liquid → 4:1 to 8:1; gas–gas → <4:1.Eliminate the option claiming a single 8:1–12:1 band for all services.Therefore, the best comprehensive statement is “Both (b) and (c).” Verification / Alternative check: Thermal rating experience and TEMA guidelines support these indicative L/Ds values; final selection depends on allowable ΔP, shell passes, and layout constraints. Why Other Options Are Wrong: (a) is too broad and not representative; (e) is unnecessary since (b) and (c) are valid. Common Pitfalls: Forgetting that multi-pass arrangements alter effective length per pass; ignoring vibration and support span when pushing to very long tubes. Final Answer: Both (b) and (c).

Question 7

For a given required volumetric flow and fluid, how does increasing pipe diameter generally affect pumping (energy) cost in steady single-phase flow?

Accepted Answer

Introduction / Context: Pipeline design balances capital cost (larger diameter, more material) against operating cost (pumping energy). Understanding how frictional losses scale with diameter allows engineers to choose an economical size for a specified flow rate and fluid. Given Data / Assumptions: Single-phase, incompressible flow at steady state. Fixed volumetric (or mass) flow requirement and fluid properties. Fully developed flow; roughness and friction factor depend on Reynolds number as usual. Concept / Approach: Frictional pressure drop in a pipe of length L is ΔP = f * (L/D) * (ρ v^2 / 2). For fixed volumetric flow, velocity v scales as 1/D^2, so ΔP scales approximately as 1/D^5 in laminar and roughly between 1/D^4 and 1/D^5 in turbulent (since f varies mildly with Reynolds number). Hence, larger diameter dramatically reduces pressure drop and thus pump power P ≈ ΔP * Q / η. Step-by-Step Solution: Increase D → decrease v = Q / A → reduce velocity head ρ v^2 /2.ΔP ∝ (L/D) * v^2 → strong inverse dependence on D.Pump power P ∝ ΔP for constant Q → P decreases as D increases, holding all else constant. Verification / Alternative check: Numerical examples with common friction correlations (e.g., Blasius for smooth turbulent) confirm the steep drop in ΔP with increasing D at constant Q. Why Other Options Are Wrong: (b) and (c) contradict fluid mechanics; (d) Newtonian vs. non-Newtonian can change f(Re) but not the fundamental inverse trend; (e) is not representative for a fixed Q. Common Pitfalls: Ignoring capital cost increase with D; forgetting that very large D may lead to low velocities and solids deposition in slurries. Final Answer: Pumping cost decreases.

Question 8

Sieve tray hydraulics: typical perforation size—what is the usual hole diameter range used on sieve trays in distillation columns?

Accepted Answer

Introduction / Context: Sieve trays use perforations to distribute vapor through a liquid layer. Hole diameter affects pressure drop, weeping tendency, capacity, and fouling susceptibility. Selecting an appropriate range is a core part of tray specification and rating. Given Data / Assumptions: Conventional sieve trays for common refinery/chemical services. Moderate fouling tendencies; typical tray thickness and spacing. Perforation fraction and weir height within standard design bands. Concept / Approach: Holes that are too small increase pressure drop and are prone to plugging; holes that are too large encourage weeping at low vapor rates and can reduce contacting efficiency. A practical window widely used in industry is roughly 3–12.5 mm diameter, which balances hydraulic stability and fouling tolerance across a broad range of services. Step-by-Step Solution: Establish conventional practice: aim for mid-single-digit millimeter holes for general duty.Check extreme ranges: <3 mm often clogs; >12.5 mm risks weeping at turndown.Therefore, select the standard range 3–12.5 mm. Verification / Alternative check: Tray vendors and handbooks list similar ranges, with specific choices tuned to vapor load, fouling risk, and perforation percentage. Why Other Options Are Wrong: 1–3 mm is typically small for most hydrocarbon services; 4–8 mm is a subset, not the general range; 12.5–18.5 mm or 20–30 mm are large and unusual for sieve trays except special duties. Common Pitfalls: Ignoring turndown and fouling when picking hole size; overlooking interaction with tray thickness and opening fraction. Final Answer: 3 to 12.5 mm

Question 9

Axial dispersion model and reactor contacting: what flow idealization is represented by a Peclet number approaching infinity (Pe → ∞)?

Accepted Answer

Introduction / Context: The Peclet number (Pe) measures the ratio of convective transport to axial dispersion (mixing). It appears in the axial dispersion model for tubular reactors, packed beds, and mass transfer equipment. Recognizing the limiting behaviors helps link real devices to ideal reactors used in design calculations. Given Data / Assumptions: Pe = u * L / D_ax, where u is superficial velocity, L is characteristic length, and D_ax is axial dispersion coefficient. Steady operation; constant properties along the device. Comparisons with idealized flow models: Plug Flow Reactor (PFR) vs. Continuous Stirred Tank Reactor (CSTR). Concept / Approach: As Pe → ∞, axial dispersion D_ax → 0 relative to convection, meaning material moves axially with negligible back-mixing. This is the essence of plug flow: concentration and temperature vary only along the axis, not radially/axially due to mixing. Conversely, Pe → 0 represents dominant dispersion and approaches complete back-mixing (CSTR-like behavior). Step-by-Step Solution: Define Peclet number: Pe = convective transport / dispersive transport.Take limit Pe → ∞: dispersion vanishes → profiles are sharp → behavior tends to PFR.Map to reactor idealizations: PFR corresponds to zero axial mixing. Verification / Alternative check: Solutions of the axial dispersion model show that dimensionless exit age distributions narrow as Pe increases; in the limit, the distribution collapses to a delta function—characteristic of plug flow. Why Other Options Are Wrong: (a) and (b) imply strong back-mixing (low Pe). (d) mentions zero eddy diffusivity but equivocates; the ideal limit is indeed plug flow. (e) is unrelated to Peclet characterization. Common Pitfalls: Confusing laminar velocity profiles with axial mixing; laminar flow can still approach plug-flow behavior at very high Pe if dispersion is negligible compared with convection. Final Answer: Plug flow (negligible axial mixing)

Question 10

In a rotary dryer used for solids handling, what is the correct expression for the average solids retention time?

Use the following symbols:
Z = length of the dryer (m)
ρ = apparent solids density (kg/m^3)
A = dryer internal cross-sectional ar…

Accepted Answer

Introduction / Context: Rotary dryers are widely used in chemical and process industries to reduce moisture content of bulk solids. A key design and operating parameter is the average solids retention time, sometimes called residence time. Knowing how long solids remain in the dryer helps size the unit, set rotational speed, and balance moisture removal with throughput. This question asks for the correct formula based on standard mass and holdup concepts using commonly defined symbols. Given Data / Assumptions: Z is the dryer length in metres. ρ (rho) is the apparent bulk density of the solids in kg/m^3. A is the internal cross-sectional area available for solids in m^2. L is the mass flow rate of dry solids in kg/s. H is the volumetric solids hold-up fraction (dimensionless) representing the fraction of the internal volume that is occupied by solids on average. Concept / Approach: The average retention time equals the total mass of solids held inside the dryer divided by the mass flow rate of solids. Mass held inside equals bulk density times the actual solids volume inside the shell. That actual volume is the product of the geometric volume (A * Z) and the hold-up fraction H. Therefore, retention time follows directly from a mass balance on the control volume of the drum interior. Step-by-Step Solution: Total geometric volume, V_total = A * Z.Solids volume present, V_s = H * V_total = H * A * Z.Mass of solids held, M = ρ * V_s = ρ * H * A * Z.Average retention time, t = M / L = (ρ * A * Z * H) / L. Verification / Alternative check: Dimensional analysis: ρ (kg/m^3) * A (m^2) * Z (m) gives kg; multiplying by H (dimensionless) keeps kg. Dividing by L (kg/s) yields seconds, which confirms unit consistency. The relationship also aligns with general residence time formulas used in solids handling equipment (inventory divided by throughput). Why Other Options Are Wrong: (b) Places L in the numerator, which would imply residence time increases with throughput—physically incorrect. (c) Omits density ρ, giving wrong units and ignoring mass/volume conversion. (d) Puts area A in the denominator, reducing time with larger cross-section, contrary to the inventory principle. Common Pitfalls: Confusing H as a mass fraction instead of volumetric fraction, or forgetting to include density to convert volume to mass. Another frequent mistake is to overlook cross-section area, which directly scales inventory in long drums. Final Answer: t = (ρ * A * Z * H) / L

Question 11

Electrostatic precipitators (ESPs): which change increases the dust collection efficiency, all else equal?

Accepted Answer

Introduction / Context: Electrostatic precipitators remove fine particulate by charging particles and driving them to collecting plates. Performance depends on electric field strength, particle charge, migration velocity, gas residence time, and collecting area. Designers tune gas velocity and plate area to achieve a target overall efficiency for a given dust load. Given Data / Assumptions: Comparable dust properties and voltage conditions. Single-stage, dry ESP behavior. Focus on geometric and flow variables: gas rate and collecting area. Concept / Approach: A standard performance expression (e.g., Deutsch–Anderson model) relates efficiency to effective migration velocity (w), collecting area (A), and volumetric flow rate (Q): η = 1 − exp(−w * A / Q). For fixed w, increasing A raises wA/Q and thus increases efficiency, while increasing Q reduces residence time and lowers efficiency. Step-by-Step Solution: Start with η = 1 − exp(−w * A / Q).Holding w constant, ∂η/∂A > 0, so more area improves efficiency.Holding A constant, increasing Q makes exponent more negative (larger magnitude), hence efficiency decreases. Verification / Alternative check: Field experience confirms that adding fields/plates (area) boosts performance, while overloading an ESP with higher gas flow degrades capture due to reduced residence time and higher drift requirements. Why Other Options Are Wrong: (a) Increasing gas flow rate reduces efficiency for a given area and field. (c) Both changes together are not beneficial because (a) is detrimental. (d) Incorrect because area clearly matters through migration length and time. Common Pitfalls: Overlooking the impact of non-ideal factors like re-entrainment, rapper settings, and dust resistivity; while these matter in practice, the core trend with A and Q remains as above. Final Answer: Increasing the effective electrode (collecting) area

Question 12

Heat-exchanger tube layout terminology: the shortest clear distance between adjacent tubes is called the ________.

Accepted Answer

Introduction / Context: In shell-and-tube heat exchangers, tube layout (triangular, square, rotated square) is defined by tube pitch (center-to-center spacing) and by the minimum metal ligament between holes on the tubesheet. Correct naming avoids confusion during mechanical design and thermal rating. Given Data / Assumptions: Standard ASME/TEMA terminology. Comparison of geometric terms only (no thermal calculations involved). Concept / Approach: Tube pitch is the center-to-center spacing of neighboring tubes. The shortest clear distance between adjacent tube outer surfaces is commonly called tube clearance or ligament. For a given pitch, triangular layouts yield smaller minimum clearance than square layouts, improving compactness but making cleaning harder. Step-by-Step Solution: Identify the asked distance: the shortest clear metal gap between neighboring tubes.By definition, that is the tube clearance (not the pitch).Note on layouts: triangular pitch has less clearance than square at equal pitch. Verification / Alternative check: Look at tubesheet drilling patterns: for equal center spacing, the diagonal proximity in triangular arrangements reduces the smallest ligament compared with orthogonal square arrangements. Why Other Options Are Wrong: (a) Tube pitch is center-to-center, not the clear distance. (c) The statement is reversed; triangular layouts have smaller, not larger, minimum clearance for the same pitch. (d) Not applicable because a standard term exists. Common Pitfalls: Confusing “pitch” with “clearance,” and assuming triangular always increases all distances. In reality, triangular improves area density at the expense of cleaning access and ligament thickness. Final Answer: Tube clearance

Question 13

Bag filter (fabric filter) design: why is gas temperature a critical input for design and operation?

Accepted Answer

Introduction / Context: Fabric filters (baghouses) capture particulate by passing dirty gas through a porous medium. Gas temperature strongly influences both the fluid-dynamic behavior of the gas–dust mixture and the durability and suitability of filter media. Designers use temperature to set allowable operating windows and to prevent bag damage or condensation. Given Data / Assumptions: Normal dry filtration with pulse-jet or shaker cleaning. No chemical degradation beyond temperature-driven effects. Gas composition does not change temperature significantly across the filter. Concept / Approach: First, temperature affects gas density (ρ) and viscosity (μ), and thus Reynolds number and pressure drop through the filter cake and fabric. Second, filter fibers have maximum continuous service temperatures; exceeding these can shrink, embrittle, or melt media (e.g., polyester vs. PPS vs. PTFE). Finally, temperature impacts dew point margins; operating near or below acid dew points causes condensation and cake blinding or corrosion. Step-by-Step Solution: Relate temperature to ρ and μ → changes in face velocity, pressure drop, and cleaning frequency.Map temperature to fibre choices (e.g., polyester < ~150°C, aramids ~200–220°C, glass/PTFE higher).Ensure margin from dew points to avoid wetting and acid attack. Verification / Alternative check: Empirical correlations for filter resistance include μ and cake properties; vendor datasheets list maximum continuous temperatures for each fabric, confirming temperature’s central role. Why Other Options Are Wrong: (a), (b), (c) each capture only part of the influence; the complete and correct choice is that all apply. Common Pitfalls: Ignoring acid dew point and focusing only on media softening temperature. Both upper and lower temperature limits matter for reliable, low-maintenance operation. Final Answer: All of (a), (b) and (c)

Question 14

For turbulent liquid flow inside tubes, the tube-side heat-transfer coefficient varies approximately with mass velocity G as:

Accepted Answer

Introduction / Context: Correlations like Dittus–Boelter and Sieder–Tate are used to estimate convective heat-transfer coefficients for turbulent flow in tubes. These relate Nusselt number to Reynolds and Prandtl numbers, revealing how h changes with mass velocity G (or Reynolds number) at otherwise similar conditions. Given Data / Assumptions: Turbulent regime in smooth tubes (Re > ~10^4). Moderate Prandtl number liquids, heating or cooling without strong property variation. Comparable diameters and fluid properties for scaling arguments. Concept / Approach: Dittus–Boelter: Nu = 0.023 * Re^0.8 * Pr^n (n ≈ 0.3 for heating, ≈ 0.4 for cooling). Since h = Nu * k / D and Re ∝ G * D / μ, holding D and properties fixed implies h ∝ Re^0.8 ∝ G^0.8. Hence, the strongest dependence among common exponents listed is roughly the 0.8 power for turbulent single-phase flow in clean tubes. Step-by-Step Solution: Start with Nu ∝ Re^0.8 Pr^n.Hold Pr and geometry approximately constant → h ∝ Nu ∝ Re^0.8.Because Re ∝ G, conclude h ∝ G^0.8. Verification / Alternative check: Other correlations (e.g., Sieder–Tate) give similar exponents near 0.8 for Re dependence in smooth tubes, validating the trend. Why Other Options Are Wrong: 0.2 and 0.5 underpredict the dependence for turbulent regimes. 1.5 is unrealistically strong and inconsistent with established correlations. Common Pitfalls: Extending the 0.8 exponent to laminar or transitional regimes, or to fouled tubes where h is dominated by deposits and not strongly by G. Final Answer: h ∝ G^0.8

Question 15

Distillation hardware and practice: pick the correct statement about condensers, tray details, packing, and riser attachment.

Accepted Answer

Introduction / Context: Column internals—trays, downcomers, risers, and packings—must be designed for hydraulic performance and maintainability. Ancillary equipment like condensers also follows best practices depending on whether total or partial condensation and subcooling are required. This question asks for the correct statement among several commonly misunderstood points. Given Data / Assumptions: Conventional trayed distillation column. Standard random and structured packings. Normal operating ranges, not special cases. Concept / Approach: Weep holes are small openings that allow draining of residual liquid from trays during shutdown to prevent pooling and corrosion. Random packing generally has higher pressure drop than structured (regular/stacked) packing at similar capacity; structured packings excel in low pressure drop and high capacity factors. Preferences for condenser orientation depend on duty: horizontal condensers are often selected for condensing with significant non-condensables or where no subcooling is needed; vertical units are common too but not universally preferred in the stated way. Riser attachments are normally secure and positive; saying they are “generally not connected” is misleading. Step-by-Step Solution: Evaluate (b): weep holes indeed aid drainage; this is correct.Evaluate (c): structured packing usually yields lower pressure drop than random; so (c) is false.Evaluate (a): orientation “preference” is situational; the blanket statement is not reliable.Evaluate (d): risers must be mechanically attached; the statement is incorrect. Verification / Alternative check: Vendor manuals and TEMA/standards for columns corroborate weep-hole use and structured packing advantages in low ∆P service. Why Other Options Are Wrong: (a) Overgeneralizes condenser orientation without context. (c) Reverses the established pressure-drop comparison. (d) Misstates mechanical practice; secure attachment is standard. Common Pitfalls: Assuming one condenser orientation is always best; in reality, subcooling target, condensate drainage, and non-condensables govern selection. Final Answer: Weep holes on trays facilitate draining of residual liquid during column shutdown

Question 16

Among common dished heads used on pressure vessels, which head form is inherently the strongest for a given diameter and thickness?

Accepted Answer

Introduction / Context: Pressure vessel end closures include hemispherical, elliptical (2:1), and torispherical (flanged and dished) heads. Their strength and thickness requirements vary because of differences in stress distribution under internal pressure. Selecting the correct head type can significantly reduce required thickness and weight at a given design pressure. Given Data / Assumptions: Same nominal diameter and material. Heads designed for the same internal pressure using basic membrane stress principles. Concept / Approach: A sphere is the ideal geometry for equal biaxial membrane stresses, producing the lowest maximum stress for a given thickness and pressure. Hemispherical heads therefore require the least thickness compared with elliptical and torispherical heads of the same diameter. Elliptical heads are stronger than torispherical but weaker than hemispherical; torispherical heads are typically the most economical to fabricate but need higher thickness for the same pressure. Step-by-Step Solution: Compare membrane stress equations for spherical vs. knuckled heads.Recognize hemispherical head yields minimum stress for a given thickness.Conclude hemispherical is inherently strongest among the listed forms. Verification / Alternative check: ASME code formulas for required thickness show the smallest t for hemispherical heads at equal conditions, confirming superior strength. Why Other Options Are Wrong: Elliptical and torispherical have regions of higher stress (crown/knuckle) and need more thickness. “None of these” contradicts established design relationships. Common Pitfalls: Equating “strongest” with “cheapest.” Torispherical is often cheaper to fabricate despite needing more thickness. Final Answer: Hemispherical head

Question 17

Shell-and-tube heat exchanger detailing: which statements about corrosion allowance (CA) are appropriate?

Accepted Answer

Introduction / Context: Corrosion allowance (CA) is added metal thickness to compensate for expected corrosion during service. In shell-and-tube exchangers, CA is applied to pressure-retaining parts exposed to corrosive fluids, while non-pressure hardware typically does not receive CA. Proper allocation of CA helps maintain code compliance and service life without overdesigning every component. Given Data / Assumptions: Carbon steel or cast iron in general chemical service. Industry-typical CA values (order of millimetres) unless client specifies otherwise. Tube material and thickness often governed by separate rules and corrosion data. Concept / Approach: Pressure parts exposed to corrosion (shells, channels, heads, covers, tube sheets) require CA. Non-pressure parts like tie rods, spacers, and baffles are not pressure boundaries and normally do not receive CA. Typical baseline CA values around 1.5 mm are common for carbon steel in moderate service; severe conditions justify higher CA (~3 mm or as specified). Tube sheets and internal covers may see corrosive attack from both sides and therefore can need CA on both faces. Step-by-Step Solution: Identify pressure vs. non-pressure parts.Assign baseline CA values appropriate to materials and service severity.Apply CA to both sides of parts wetted on both faces (e.g., tube sheets). Verification / Alternative check: Project standards and TEMA practices generally align with these guidelines; client specifications may refine exact values. Why Other Options Are Wrong: Each of (a), (b), and (c) is individually correct; the best choice is the combined statement. Common Pitfalls: Applying CA uniformly to all components, which can add weight and cost without improving safety; or neglecting CA on the second face of a tube sheet that is exposed on both sides. Final Answer: All (a), (b) and (c)

Question 18

Pipework practice: flanges may be attached to pipes by which of the following methods?

Accepted Answer

Introduction / Context: Flanges provide detachable joints in piping systems. The method of attaching a flange to pipe depends on pressure–temperature rating, material, service, and installation constraints. Common attachment types include threaded, welded, and—where appropriate—brazed connections, though the latter is limited to certain alloys and lower-temperature services. Given Data / Assumptions: Industrial piping following typical codes and standards. Materials compatible with the chosen joining method. Service conditions within the allowable limits for each method. Concept / Approach: Threaded (screwed) flanges are used on small-bore, low-to-moderate pressure lines for ease of assembly. Welded flanges (weld-neck, slip-on, socket-weld) dominate in higher pressures and critical services thanks to structural integrity and leak tightness. Brazed flanges, while less common in heavy industry, are used in HVAC, refrigeration, and certain non-ferrous or stainless systems where brazing is standard practice. Step-by-Step Solution: List viable methods: threaded, welded, brazed.Confirm applicability of each to at least some use cases.Therefore, all listed methods can be valid depending on service. Verification / Alternative check: Piping catalogs and standards show threaded and welded flanges as mainstream; brazed connections appear in specialized systems and OEM equipment. Why Other Options Are Wrong: Selecting only one method ignores common practice variety; the comprehensive option is correct. Common Pitfalls: Assuming brazing is universally acceptable; it must meet code and service limitations. Critical and high-temperature services normally require welded flanges. Final Answer: All (a), (b) and (c)

Question 19

Cylindrical storage tanks for general service: which roof shape is most commonly used?

Accepted Answer

Introduction / Context: Fixed-roof cylindrical tanks store water, chemicals, and many non-volatile liquids. Roof geometry influences structural efficiency, drainage, and fabrication complexity. Among roof types, the conical (cone) roof is widely used for ambient-pressure tanks due to its simplicity and effectiveness. Given Data / Assumptions: Atmospheric or low-pressure fixed-roof service. Typical field-erected steel tanks per industry standards. Concept / Approach: Conical roofs are easy to fabricate from plate segments, shed rain and snow effectively, and can be self-supporting or supported by rafters and columns. Dome roofs (spherical segments) are common for low-pressure or vapor-tight service but are more complex. Flat roofs are rare for outdoor storage due to drainage issues. “Umbrella” is a descriptive variant, not the standard default choice across industries. Step-by-Step Solution: Compare typical applications of cone, dome, and flat roofs.Identify the most prevalent in general fixed-roof service → cone roof.Exclude specialty cases (floating roofs for volatile products) not asked here. Verification / Alternative check: Vendor literature and tank construction standards consistently present conical roofs as the default for many ambient-pressure tanks. Why Other Options Are Wrong: Dome: used, but not the most common across general services. Flat: drainage and ponding problems limit use. Umbrella: not a standard mainstream selection term. Common Pitfalls: Confusing floating roofs (for volatile hydrocarbons) with fixed roof types; floating roofs are a separate category. Final Answer: Cone (conical) roof

Question 20

In rotary dryer design and operation, what typical fraction of the drum volume is occupied by the solid hold-up (material actually resident inside the rotating shell during steady operation)?

Accepted Answer

Introduction / Context: Rotary dryers are widely used in chemical and process industries to reduce moisture in granular, crystalline, or lumpy solids. A key operating parameter is the solid hold-up, meaning the fraction of the dryer shell volume that is actually occupied by solids at any moment under steady operation. Correctly estimating hold-up helps size the dryer, set residence time, and predict gas–solid contacting efficiency. Given Data / Assumptions: Cocurrent or countercurrent direct-fired rotary dryer handling free-flowing solids. Flight (lifter) arrangement is typical for industrial units. Operation is steady with normal fill levels—not start-up or emergency conditions. Concept / Approach: Inside a rotary dryer, solids are lifted and showered through the hot gas stream. Only a relatively small fraction of the cylinder cross-section is filled because excessive fill would suppress cascading and curtain formation, hurt heat/mass transfer, and increase power draw. Industry practice and pilot data typically show hold-up values in the low percentage range of total shell volume. Step-by-Step Solution: Define solid hold-up H as volume of solids inside the shell divided by the internal shell volume.For flighted rotary dryers with proper speed and slope, H is intentionally kept low to sustain good showering and cascade.Empirical ranges from design practice indicate H ≈ 0.05–0.15 (i.e., 5–15% of shell volume) for free-flowing solids at typical throughputs. Verification / Alternative check: Mass balance on solids (throughput = hold-up / residence time) combined with measured residence times in test units yields hold-up values generally in this 0.05–0.15 range for most granular products. Higher values are exceptional and can indicate suboptimal lifter design or sluggish materials. Why Other Options Are Wrong: 0.25 to 0.50: Too high; would impair falling curtain formation and cause excessive power consumption. 0.50 to 0.80 and 0.80 to 0.90: Unrealistically high for rotary dryers; these fills are more like rotating kilns with bed flow, not efficient dryers. Common Pitfalls: Confusing mass hold-up with volumetric hold-up; ignoring the effect of flight design and rotational speed; assuming sticky or highly cohesive solids behave like free-flowing materials. Final Answer: 0.05 to 0.15

Process Equipment and Plant Design Questions