Q: Steam distribution design: what is the typical velocity range for steam in plant piping under normal practice?

Introduction / Context: Steam line sizing balances pressure drop, noise/erosion risk, water-hammer potential, and installed cost. Plant standards specify target velocities for mains and branches to maintain dryness and controllability. Given Data / Assumptions: Saturated or slightly superheated steam in carbon-steel piping. General utility distribution, not high-velocity injection steam services. Goal: “normal” practice values. Concept / Approach: Common guidance recommends about 25–35 m/s for long mains and 15–25 m/s for branches; conservative designs may choose lower ends to limit noise and entrainment. Among the options, 10–20 m/s best matches mainstream practice as a safe, typical range used in many plants. Step-by-Step Solution: Compare listed ranges with recommended practice.Eliminate very low values (0.1–5 m/s) which inflate pipe sizes and cost unnecessarily.Eliminate very high values (80–100 m/s) which risk erosion, pressure drop, and noise.Select 10–20 m/s as a realistic standard range for many steam systems. Verification / Alternative check: Vendor charts and steam handbooks present similar ranges, with adjustments based on pressure level and allowable pressure drop per 100 m. Why Other Options Are Wrong: 0.1–0.5 or 1–5 m/s: uneconomical pipe diameters and poor responsiveness.80–100 m/s: excessive drop and noise; not typical for distribution mains. Common Pitfalls: Ignoring condensate carryover at high velocities; neglecting slope and drip legs that interact with velocity choice. Final Answer: 10–20 m/s

Q: Thin-walled cylindrical vessel under internal pressure: the circumferential (hoop) stress compared to the longitudinal stress is

Introduction / Context: For thin cylindrical shells under internal pressure, two primary membrane stresses arise: hoop (circumferential) and longitudinal (axial). Understanding their relationship is essential for safe pressure vessel design and sizing of thickness. Given Data / Assumptions: Thin-walled assumption: wall thickness t << radius r. Uniform internal pressure p; end caps integral. Elastic, isotropic material; negligible external loads. Concept / Approach: The classic thin-shell formulas are: hoop stress sigma_h = p * r / t, longitudinal stress sigma_l = p * r / (2 * t). These follow from free-body equilibrium of a longitudinal cut (for hoop) and a transverse cut (for longitudinal). Step-by-Step Solution: Write hoop stress: sigma_h = p * r / t.Write longitudinal stress: sigma_l = p * r / (2 * t).Take ratio: sigma_h / sigma_l = (p r / t) / (p r / (2 t)) = 2.Hence, hoop stress is twice the longitudinal stress. Verification / Alternative check: Textbook derivations and code background equations (e.g., for cylindrical shells) reproduce this 2:1 relationship under the thin-wall assumption. Why Other Options Are Wrong: Half or equal: contradicts the derived ratio.Eight times: far exceeds the thin-wall relationship and would imply grossly different mechanics. Common Pitfalls: Applying thin-wall formulas to thick shells (t not small vs r), or forgetting that external attachments can alter local stresses. Final Answer: Twice

Q: In mixing, how does impeller diameter affect power draw? (Assume fixed speed, turbulent regime, and constant fluid properties.)

Introduction / Context: Agitator power scaling is fundamental in mixer selection and scale-up. For geometrically similar systems under turbulent flow at fixed rotational speed, the dependence of power on impeller diameter is strong and guides both energy estimates and motor sizing. Given Data / Assumptions: Newtonian liquid, fully turbulent regime. Geometric similarity (constant impeller type, tank/impeller ratios). Fixed rotational speed N and constant density rho. Concept / Approach: The power number correlation gives P = Np * rho * N^3 * D^5 for turbulent mixing (where Np ≈ constant for a given impeller in fully turbulent flow). With speed fixed, P scales as D^5, showing why seemingly small diameter increases drastically raise power requirements. Step-by-Step Solution: Start from P ∝ rho * N^3 * D^5 (Np constant).Hold rho and N constant → P ∝ D^5.Compare with options: D^5 is the correct dependence. Verification / Alternative check: Pilot data commonly collapse onto P ∝ N^3 D^5 in turbulent regions; deviations occur in laminar/transition regimes where P ∝ mu * N^2 * D^3. Why Other Options Are Wrong: D, D^2: underpredict power; applicable to no standard turbulent scaling.D^9: vastly overpredicts; not supported by dimensional analysis. Common Pitfalls: Scaling diameter without adjusting speed or checking regime; assuming Np is constant outside fully turbulent conditions. Final Answer: Proportional to D^5

Q: In a shell-and-tube exchanger, as the overall temperature drop between hot and cold streams increases (other conditions similar), how does the LMTD correction factor FT typically change?

Introduction / Context: The log-mean temperature difference (LMTD) correction factor FT accounts for deviation from true countercurrent flow in multi-pass shell-and-tube exchangers. Its magnitude determines the effective temperature driving force used for sizing. Given Data / Assumptions: Comparison at fixed flow arrangement (e.g., 1–2 exchanger) and similar flow rates. We vary the overall temperature drop (i.e., less “pinch” at one end). Interest is the qualitative trend in FT. Concept / Approach: FT approaches 1.0 for large terminal temperature differences and moves significantly below 1.0 as the exchanger approaches temperature cross (small approach temperatures). Therefore, increasing the overall temperature drop (making approaches less tight) reduces the severity of the departure from true countercurrent behavior, and FT increases toward unity—often rapidly as the approach “opens up.” Step-by-Step Solution: With small temperature differences at one end (near pinch), FT is low.As the available ΔT increases, the temperature profiles separate further.This reduces multipass penalties → FT rises, trending quickly toward 1.0. Verification / Alternative check: Charts of FT vs. the temperature ratio parameters (R and S) show FT climbing steeply away from the cross region. Why Other Options Are Wrong: Decrease rapidly: opposite of observed behavior away from pinch.Remain constant or linear increase: FT response is non-linear and strongly dependent on R and S. Common Pitfalls: Confusing FT with LMTD itself; overlooking that FT is a correction for configuration, not for area or fouling. Final Answer: Increases very rapidly

Q: On a distillation plate, which change most directly increases liquid-side resistance to mass transfer and thus lowers plate efficiency?

Introduction / Context: Tray efficiency depends on vapor-liquid interfacial area and the resistances to mass transfer on the gas and liquid sides. Understanding which variables worsen liquid-side resistance helps diagnose low efficiency and guides solvent and operating condition choices. Given Data / Assumptions: Conventional tray column (distillation as the main context). Focus is the effect on liquid-film resistance. No flooding, weeping, or foaming issues assumed. Concept / Approach: Liquid-film mass-transfer coefficients generally decrease as viscosity rises, because higher viscosity dampens turbulence and reduces diffusivity. Hence, higher viscosity increases the liquid-side resistance, lowering stage efficiency. By contrast, higher relative volatility improves separation driving force in distillation, not resistance. Gas solubility pertains more to absorption/stripping gas-side phenomena than to liquid-side resistance on distillation trays. Step-by-Step Solution: Identify the direct influencer of liquid film properties: viscosity.Recognize that increased viscosity → lower liquid-side mass-transfer coefficient.Infer lower Murphree/overall efficiency for the same contacting time and area. Verification / Alternative check: Correlations for kL often show inverse dependence on viscosity (e.g., via dimensionless groups like Sc and Re), confirming the trend. Why Other Options Are Wrong: Higher relative volatility increases thermodynamic driving force; it does not increase resistance.Lower gas solubility affects gas-side transfer in absorbers, not directly the liquid-film resistance on trays in distillation. Common Pitfalls: Blaming tray design alone when liquid properties (viscosity, foaming) limit transfer; ignoring temperature changes that reduce viscosity and improve efficiency. Final Answer: An increase in liquid viscosity

Q: McCabe–Thiele analysis: what is the typical slope of the operating line in the stripping section (assume constant molal overflow)?

Introduction / Context: In binary distillation with constant molal overflow, the rectifying and stripping sections have straight operating lines on the y–x diagram. Their slopes depend on internal flow ratios and inform the number of stages and feasibility. Given Data / Assumptions: Constant molal overflow holds (enthalpy effects modest). Definitions: Rectifying slope = L/V above the feed; Stripping slope = L̄/V̄ below the feed. Typical column no side draws; total condenser and partial reboiler. Concept / Approach: In the stripping section, the internal vapor flow V̄ is commonly larger than the liquid flow L̄ (because of vapor generation in the reboiler), so L̄/V̄ is less than 1, giving an operating line with slope under unity. Only in unusual configurations (heavy reflux of bottoms, side draws) could the slope exceed 1. Step-by-Step Solution: Write stripping operating line: y = (L̄/V̄) x − (B/V̄) xB (form depends on definitions).Recognize slope = L̄/V̄.For typical designs, L̄ 1: contradicts the usual L̄/V̄ relationship in the stripping section. Common Pitfalls: Mixing rectifying and stripping definitions; forgetting side draw effects on internal flows. Final Answer: < 1

Q: Calendria (calandria) type evaporator design: the typical ratio of down-take area to total tube cross-sectional area is

Introduction / Context: In calandria (calendar/calandria) type evaporators, the circulation relies on vapor-liquid lift in the tubes and a central or peripheral down-take for the return of the heavier, cooler liquid. Proper sizing of the down-take area relative to the tubes’ total flow area is essential for stable natural circulation. Given Data / Assumptions: Natural-circulation calandria with central down-take. Single-effect operation at moderate pressure. Target is a typical design range for area ratio. Concept / Approach: If the down-take is too small, circulation chokes and the hydrostatic head cannot drive adequate return flow, risking boiling instability and tube dry-out. If it is too large, tube bundle area is sacrificed and velocities fall. Empirical design ranges often place down-take area roughly one-half to equal the total cross-sectional area of the tubes to balance head losses and maintain velocity. Step-by-Step Solution: Define Atube_total = sum of all tube flow areas.Choose Adowntake such that 0.5 * Atube_total ≤ Adowntake ≤ 1.0 * Atube_total.This yields sufficient return leg capacity without starving vapor-lift circulation. Verification / Alternative check: Many evaporator handbooks present similar ratios and provide head-loss calculations (friction + acceleration) to fine-tune the final design. Why Other Options Are Wrong: 1–1.5 or higher: typically excessive down-take area, penalizing bundle area and velocities.Ratios above 1.5–2.5: uncommon for natural-circulation calandrias. Common Pitfalls: Ignoring two-phase pressure drops, neglecting flashing in the disengagement space, or using the same ratio for very viscous liquors where larger areas may be needed. Final Answer: 0.5 to 1

Q: Distillation tray hydraulics: when the liquid gradient (height difference across a tray) becomes excessive, which malfunctions are most likely to occur?

Introduction / Context: Successful tray operation requires a controlled interaction between rising vapor and flowing liquid. Excess liquid gradient—large height differences from inlet to outlet—creates uneven hydraulic conditions that compromise vapor distribution and contacting, damaging capacity and efficiency. Given Data / Assumptions: Conventional sieve or cap trays with standard weirs and downcomers. Vapor and liquid loads within normal design ranges but with an excessive hydraulic gradient across the tray. No unusual internals beyond standard calming zones. Concept / Approach: A large gradient raises local liquid head near the outlet and lowers it near the inlet, skewing vapor-liquid contact. The inlet may weep while the outlet area froths excessively. Vapor seeks the path of least resistance, causing maldistribution and potential gas bypass beneath caps or through perforations in undesired zones, and liquid can back-trap in eddies near downcomers. Step-by-Step Solution: Recognize that gradient = higher static head at the outlet than the inlet.Higher outlet head increases local pressure drop demand on vapor, diverting vapor to lower-head regions → maldistribution.Unequal heads also promote recirculation pockets and back-trapping of liquid.Local vapor jets can blow beneath skirts/perforations where seal conditions degrade, creating erratic froth and entrainment. Verification / Alternative check: Tray rating tools and field troubleshooting guides correlate high gradients with patterns of weeping near inlets, jet flooding near outlets, and erratic pressure-drop profiles—consistent with all effects listed. Why Other Options Are Wrong: Each of (a), (b), and (c) can occur together; choosing any subset understates the risk. Therefore (d) is most accurate. Common Pitfalls: Ignoring inlet calming sections; over-tight weir/flow path causing excessive path length; neglecting tray leveling which can mimic high gradient effects. Final Answer: All (a), (b) and (c)

Q: Thin-walled cylindrical vessel under internal pressure: what is the relationship between circumferential (hoop) stress and longitudinal stress?

Introduction / Context: For thin-walled cylinders under internal pressure, two principal membrane stresses arise: the hoop (circumferential) stress around the circumference and the longitudinal stress along the axis. Understanding their ratio is essential for safe shell design, nozzle reinforcement, and head selection. Given Data / Assumptions: Thin-wall assumption: wall thickness t ≪ diameter d. Uniform internal pressure p; negligible external loads. Linear elastic, isotropic material; small deformation. Concept / Approach: Balance forces on free-body sections. For hoop stress, cut the cylinder axially: the pressure acting on the projected area balances the tensile forces in the two walls. For longitudinal stress, cut a transverse section: pressure on the head balances the tensile ring in the shell. These classical force balances yield closed-form expressions. Step-by-Step Solution: Hoop stress: σh = p d / (2 t).Longitudinal stress: σL = p d / (4 t).Therefore, σh = 2 σL.This ratio guides which failure mode is critical; typically hoop governs thickness sizing for long cylinders. Verification / Alternative check: Dimensional checks and energy methods (e.g., strain energy minimization) confirm the same results under thin-shell theory. Codes and handbooks list these identities as baseline shell formulas. Why Other Options Are Wrong: (b) and (c) contradict the derived ratio; (d) is false because a fixed ratio exists for thin shells; (e) has no theoretical basis in this context. Common Pitfalls: Applying thin-wall equations to thick shells; omitting external loads or bending from supports; misusing diameter (inside vs. outside); ignoring joint efficiency in code calculations. Final Answer: σh = 2 σL

Question 1

Packed towers: liquid distribution difficulties are accentuated when the ratio (tower diameter / packing size) is especially small. Which range indicates greatest difficulty?

Accepted Answer

Introduction / Context: Uniform liquid distribution at the top of a packed tower is crucial to achieve design mass-transfer performance. A key geometric parameter is the ratio of tower diameter (Dt) to packing characteristic size (dp). When Dt/dp is too small, edge effects and insufficient drip points create severe maldistribution. Given Data / Assumptions: Random or structured packings used in countercurrent gas–liquid contact. Conventional top liquid distributor. We assess qualitative difficulty as Dt/dp decreases. Concept / Approach: Rules of thumb suggest that to minimize wall effects and ensure multiple liquid drip points across the cross-section, Dt should be many times larger than packing size. If Dt/dp falls below roughly 8–10 for random packings, uniform distribution becomes difficult; fewer drip points and proximity to the wall lead to channeling and reduced efficiency. Step-by-Step Solution: Relate difficulty to Dt/dp: smaller ratios mean fewer packing elements across the diameter.Identify the “worst” among choices: values below ~7 are most prone to maldistribution.Select the range indicating greatest difficulty: < 7. Verification / Alternative check: Design guides specify minimum tower diameters for a given packing to keep Dt/dp above a threshold, and recommend special distributors or re-distributors if the ratio is small. Why Other Options Are Wrong: < 30 or ranges > 7: these provide many packing elements across the diameter, easing distribution compared with < 7. Common Pitfalls: Using standard trough distributors in small columns; not accounting for wall flow; ignoring need for re-distribution on tall beds. Final Answer: < 7

Question 2

Steam distribution design: what is the typical velocity range for steam in plant piping under normal practice?

Accepted Answer

Introduction / Context: Steam line sizing balances pressure drop, noise/erosion risk, water-hammer potential, and installed cost. Plant standards specify target velocities for mains and branches to maintain dryness and controllability. Given Data / Assumptions: Saturated or slightly superheated steam in carbon-steel piping. General utility distribution, not high-velocity injection steam services. Goal: “normal” practice values. Concept / Approach: Common guidance recommends about 25–35 m/s for long mains and 15–25 m/s for branches; conservative designs may choose lower ends to limit noise and entrainment. Among the options, 10–20 m/s best matches mainstream practice as a safe, typical range used in many plants. Step-by-Step Solution: Compare listed ranges with recommended practice.Eliminate very low values (0.1–5 m/s) which inflate pipe sizes and cost unnecessarily.Eliminate very high values (80–100 m/s) which risk erosion, pressure drop, and noise.Select 10–20 m/s as a realistic standard range for many steam systems. Verification / Alternative check: Vendor charts and steam handbooks present similar ranges, with adjustments based on pressure level and allowable pressure drop per 100 m. Why Other Options Are Wrong: 0.1–0.5 or 1–5 m/s: uneconomical pipe diameters and poor responsiveness.80–100 m/s: excessive drop and noise; not typical for distribution mains. Common Pitfalls: Ignoring condensate carryover at high velocities; neglecting slope and drip legs that interact with velocity choice. Final Answer: 10–20 m/s

Question 3

Thin-walled cylindrical vessel under internal pressure: the circumferential (hoop) stress compared to the longitudinal stress is

Accepted Answer

Introduction / Context: For thin cylindrical shells under internal pressure, two primary membrane stresses arise: hoop (circumferential) and longitudinal (axial). Understanding their relationship is essential for safe pressure vessel design and sizing of thickness. Given Data / Assumptions: Thin-walled assumption: wall thickness t << radius r. Uniform internal pressure p; end caps integral. Elastic, isotropic material; negligible external loads. Concept / Approach: The classic thin-shell formulas are: hoop stress sigma_h = p * r / t, longitudinal stress sigma_l = p * r / (2 * t). These follow from free-body equilibrium of a longitudinal cut (for hoop) and a transverse cut (for longitudinal). Step-by-Step Solution: Write hoop stress: sigma_h = p * r / t.Write longitudinal stress: sigma_l = p * r / (2 * t).Take ratio: sigma_h / sigma_l = (p r / t) / (p r / (2 t)) = 2.Hence, hoop stress is twice the longitudinal stress. Verification / Alternative check: Textbook derivations and code background equations (e.g., for cylindrical shells) reproduce this 2:1 relationship under the thin-wall assumption. Why Other Options Are Wrong: Half or equal: contradicts the derived ratio.Eight times: far exceeds the thin-wall relationship and would imply grossly different mechanics. Common Pitfalls: Applying thin-wall formulas to thick shells (t not small vs r), or forgetting that external attachments can alter local stresses. Final Answer: Twice

Question 4

In mixing, how does impeller diameter affect power draw? (Assume fixed speed, turbulent regime, and constant fluid properties.)

Accepted Answer

Introduction / Context: Agitator power scaling is fundamental in mixer selection and scale-up. For geometrically similar systems under turbulent flow at fixed rotational speed, the dependence of power on impeller diameter is strong and guides both energy estimates and motor sizing. Given Data / Assumptions: Newtonian liquid, fully turbulent regime. Geometric similarity (constant impeller type, tank/impeller ratios). Fixed rotational speed N and constant density rho. Concept / Approach: The power number correlation gives P = Np * rho * N^3 * D^5 for turbulent mixing (where Np ≈ constant for a given impeller in fully turbulent flow). With speed fixed, P scales as D^5, showing why seemingly small diameter increases drastically raise power requirements. Step-by-Step Solution: Start from P ∝ rho * N^3 * D^5 (Np constant).Hold rho and N constant → P ∝ D^5.Compare with options: D^5 is the correct dependence. Verification / Alternative check: Pilot data commonly collapse onto P ∝ N^3 D^5 in turbulent regions; deviations occur in laminar/transition regimes where P ∝ mu * N^2 * D^3. Why Other Options Are Wrong: D, D^2: underpredict power; applicable to no standard turbulent scaling.D^9: vastly overpredicts; not supported by dimensional analysis. Common Pitfalls: Scaling diameter without adjusting speed or checking regime; assuming Np is constant outside fully turbulent conditions. Final Answer: Proportional to D^5

Question 5

In a shell-and-tube exchanger, as the overall temperature drop between hot and cold streams increases (other conditions similar), how does the LMTD correction factor FT typically change?

Accepted Answer

Introduction / Context: The log-mean temperature difference (LMTD) correction factor FT accounts for deviation from true countercurrent flow in multi-pass shell-and-tube exchangers. Its magnitude determines the effective temperature driving force used for sizing. Given Data / Assumptions: Comparison at fixed flow arrangement (e.g., 1–2 exchanger) and similar flow rates. We vary the overall temperature drop (i.e., less “pinch” at one end). Interest is the qualitative trend in FT. Concept / Approach: FT approaches 1.0 for large terminal temperature differences and moves significantly below 1.0 as the exchanger approaches temperature cross (small approach temperatures). Therefore, increasing the overall temperature drop (making approaches less tight) reduces the severity of the departure from true countercurrent behavior, and FT increases toward unity—often rapidly as the approach “opens up.” Step-by-Step Solution: With small temperature differences at one end (near pinch), FT is low.As the available ΔT increases, the temperature profiles separate further.This reduces multipass penalties → FT rises, trending quickly toward 1.0. Verification / Alternative check: Charts of FT vs. the temperature ratio parameters (R and S) show FT climbing steeply away from the cross region. Why Other Options Are Wrong: Decrease rapidly: opposite of observed behavior away from pinch.Remain constant or linear increase: FT response is non-linear and strongly dependent on R and S. Common Pitfalls: Confusing FT with LMTD itself; overlooking that FT is a correction for configuration, not for area or fouling. Final Answer: Increases very rapidly

Question 6

On a distillation plate, which change most directly increases liquid-side resistance to mass transfer and thus lowers plate efficiency?

Accepted Answer

Introduction / Context: Tray efficiency depends on vapor-liquid interfacial area and the resistances to mass transfer on the gas and liquid sides. Understanding which variables worsen liquid-side resistance helps diagnose low efficiency and guides solvent and operating condition choices. Given Data / Assumptions: Conventional tray column (distillation as the main context). Focus is the effect on liquid-film resistance. No flooding, weeping, or foaming issues assumed. Concept / Approach: Liquid-film mass-transfer coefficients generally decrease as viscosity rises, because higher viscosity dampens turbulence and reduces diffusivity. Hence, higher viscosity increases the liquid-side resistance, lowering stage efficiency. By contrast, higher relative volatility improves separation driving force in distillation, not resistance. Gas solubility pertains more to absorption/stripping gas-side phenomena than to liquid-side resistance on distillation trays. Step-by-Step Solution: Identify the direct influencer of liquid film properties: viscosity.Recognize that increased viscosity → lower liquid-side mass-transfer coefficient.Infer lower Murphree/overall efficiency for the same contacting time and area. Verification / Alternative check: Correlations for kL often show inverse dependence on viscosity (e.g., via dimensionless groups like Sc and Re), confirming the trend. Why Other Options Are Wrong: Higher relative volatility increases thermodynamic driving force; it does not increase resistance.Lower gas solubility affects gas-side transfer in absorbers, not directly the liquid-film resistance on trays in distillation. Common Pitfalls: Blaming tray design alone when liquid properties (viscosity, foaming) limit transfer; ignoring temperature changes that reduce viscosity and improve efficiency. Final Answer: An increase in liquid viscosity

Question 7

McCabe–Thiele analysis: what is the typical slope of the operating line in the stripping section (assume constant molal overflow)?

Accepted Answer

Introduction / Context: In binary distillation with constant molal overflow, the rectifying and stripping sections have straight operating lines on the y–x diagram. Their slopes depend on internal flow ratios and inform the number of stages and feasibility. Given Data / Assumptions: Constant molal overflow holds (enthalpy effects modest). Definitions: Rectifying slope = L/V above the feed; Stripping slope = L̄/V̄ below the feed. Typical column no side draws; total condenser and partial reboiler. Concept / Approach: In the stripping section, the internal vapor flow V̄ is commonly larger than the liquid flow L̄ (because of vapor generation in the reboiler), so L̄/V̄ is less than 1, giving an operating line with slope under unity. Only in unusual configurations (heavy reflux of bottoms, side draws) could the slope exceed 1. Step-by-Step Solution: Write stripping operating line: y = (L̄/V̄) x − (B/V̄) xB (form depends on definitions).Recognize slope = L̄/V̄.For typical designs, L̄ < V̄ → slope < 1. Verification / Alternative check: Worked examples commonly show rectifying slope near R/(R+1) (below 1) and stripping slope also below 1 under standard conditions. Why Other Options Are Wrong: 0 or ∞: would imply no mass exchange or a vertical line; not representative of operating sections.> 1: contradicts the usual L̄/V̄ relationship in the stripping section. Common Pitfalls: Mixing rectifying and stripping definitions; forgetting side draw effects on internal flows. Final Answer: < 1

Question 8

Calendria (calandria) type evaporator design: the typical ratio of down-take area to total tube cross-sectional area is

Accepted Answer

Introduction / Context: In calandria (calendar/calandria) type evaporators, the circulation relies on vapor-liquid lift in the tubes and a central or peripheral down-take for the return of the heavier, cooler liquid. Proper sizing of the down-take area relative to the tubes’ total flow area is essential for stable natural circulation. Given Data / Assumptions: Natural-circulation calandria with central down-take. Single-effect operation at moderate pressure. Target is a typical design range for area ratio. Concept / Approach: If the down-take is too small, circulation chokes and the hydrostatic head cannot drive adequate return flow, risking boiling instability and tube dry-out. If it is too large, tube bundle area is sacrificed and velocities fall. Empirical design ranges often place down-take area roughly one-half to equal the total cross-sectional area of the tubes to balance head losses and maintain velocity. Step-by-Step Solution: Define Atube_total = sum of all tube flow areas.Choose Adowntake such that 0.5 * Atube_total ≤ Adowntake ≤ 1.0 * Atube_total.This yields sufficient return leg capacity without starving vapor-lift circulation. Verification / Alternative check: Many evaporator handbooks present similar ratios and provide head-loss calculations (friction + acceleration) to fine-tune the final design. Why Other Options Are Wrong: 1–1.5 or higher: typically excessive down-take area, penalizing bundle area and velocities.Ratios above 1.5–2.5: uncommon for natural-circulation calandrias. Common Pitfalls: Ignoring two-phase pressure drops, neglecting flashing in the disengagement space, or using the same ratio for very viscous liquors where larger areas may be needed. Final Answer: 0.5 to 1

Question 9

Forced-circulation long-tube vertical (LTV) evaporators vs. natural circulation: which statements are correct?

Accepted Answer

Introduction / Context: Evaporators remove solvent (often water) by boiling. Two common types are natural-circulation long-tube vertical (LTV) and forced-circulation LTV. The choice impacts heat-transfer coefficients, fouling behavior, boiling control, and installed/operating costs. Given Data / Assumptions: Comparison focuses on hydraulic/thermal features, not specific duty economics. Forced-circulation units have an external pump loop. Vapor disengagement occurs in a vapor drum or separator. Concept / Approach: Forced-circulation evaporators drive liquid through tubes at high velocity using a circulation pump. This suppresses nucleate boiling inside the tubes (often boiling is allowed mainly in the separator), increases heat-transfer coefficients, and reduces fouling. As a result, for the same duty, less heating surface may be required than in natural circulation. The hydraulic loop is typically arranged with a pump drawing from the vapor drum down-take and discharging to the tube-bundle inlet. Step-by-Step Solution: Evaluate (b): High velocity → higher h and often less area → correct.Evaluate (c): Loop arrangement with centrifugal pump between the down-take and bundle inlet → correct.Evaluate (a): “More economical in operation” is not universally true; pumps add power and maintenance, though benefits may offset in fouling/viscous services. Hence (a) is not a blanket statement.Therefore, select “Both (b) & (c).” Verification / Alternative check: Design texts and vendor datasheets consistently depict the external recirculation pump loop and report higher film coefficients with forced circulation. Why Other Options Are Wrong: (a) Over-generalizes economics; depends on service.Single selections (b) or (c) omit the other true statement; (d) captures both correct points. Common Pitfalls: Allowing boiling inside tubes in forced-circulation units with fouling fluids; undersizing the pump causing insufficient velocity and poor performance. Final Answer: Both (b) & (c)

Question 10

In a double-pipe concentric heat exchanger, the annulus carries one of the fluids between the outer pipe I.D. (D2) and the inner pipe O.D. (assumed approximately equal to its I.D. for thin walls). For heat-transfer calculations, what is the appropri…

Accepted Answer

Introduction / Context: The equivalent (hydraulic) diameter is used to convert non-circular flow passages into an “equivalent” circular size so that standard convection correlations can be applied. For the annulus in a double-pipe heat exchanger, knowing this diameter directly impacts Reynolds number, Nusselt number, and the predicted heat-transfer coefficient, all of which are central to exchanger sizing and rating. Given Data / Assumptions: The passage of interest is an annulus formed by two concentric pipes. D1 is the inner pipe inside diameter (taken as the relevant inside boundary of the annulus); D2 is the outer pipe inside diameter. Walls are thin enough that O.D. and I.D. distinction on the inner pipe does not change the result materially. Concept / Approach: The hydraulic diameter is defined as Dh = 4 * (flow area) / (wetted perimeter). For an annulus, area = (π/4) * (D2^2 - D1^2), and wetted perimeter = π * (D2 + D1). Substituting gives Dh = (D2^2 - D1^2) / (D2 + D1) = D2 - D1. This compact result is widely used in heat-transfer and pressure-drop calculations for annular flow. Step-by-Step Solution: Write Dh = 4A/P.Compute A_annulus = (π/4) * (D2^2 - D1^2).Compute P_wetted = π * (D2 + D1).Form Dh = [4 * (π/4) * (D2^2 - D1^2)] / [π * (D2 + D1)] = (D2^2 - D1^2)/(D2 + D1) = D2 - D1. Verification / Alternative check: Dimensional consistency confirms the result has units of length. Practical calculations using Dh = D2 - D1 reproduce standard handbooks and software outputs for annular flow correlations. Why Other Options Are Wrong: (a) is the definition, not the closed-form result; (c) simplifies to D2 + D1, which is incorrect for annuli; (e) doubles the correct value; (d) is unnecessary since a correct closed form exists. Common Pitfalls: Confusing (D2 - D1) with (D2 + D1); using the wrong boundary (e.g., outer O.D. rather than I.D.); forgetting that Dh is for convection/pressure-drop correlations, not geometric diameter for area calculations. Final Answer: D2 - D1

Question 11

Distillation tray hydraulics: when the liquid gradient (height difference across a tray) becomes excessive, which malfunctions are most likely to occur?

Accepted Answer

Introduction / Context: Successful tray operation requires a controlled interaction between rising vapor and flowing liquid. Excess liquid gradient—large height differences from inlet to outlet—creates uneven hydraulic conditions that compromise vapor distribution and contacting, damaging capacity and efficiency. Given Data / Assumptions: Conventional sieve or cap trays with standard weirs and downcomers. Vapor and liquid loads within normal design ranges but with an excessive hydraulic gradient across the tray. No unusual internals beyond standard calming zones. Concept / Approach: A large gradient raises local liquid head near the outlet and lowers it near the inlet, skewing vapor-liquid contact. The inlet may weep while the outlet area froths excessively. Vapor seeks the path of least resistance, causing maldistribution and potential gas bypass beneath caps or through perforations in undesired zones, and liquid can back-trap in eddies near downcomers. Step-by-Step Solution: Recognize that gradient = higher static head at the outlet than the inlet.Higher outlet head increases local pressure drop demand on vapor, diverting vapor to lower-head regions → maldistribution.Unequal heads also promote recirculation pockets and back-trapping of liquid.Local vapor jets can blow beneath skirts/perforations where seal conditions degrade, creating erratic froth and entrainment. Verification / Alternative check: Tray rating tools and field troubleshooting guides correlate high gradients with patterns of weeping near inlets, jet flooding near outlets, and erratic pressure-drop profiles—consistent with all effects listed. Why Other Options Are Wrong: Each of (a), (b), and (c) can occur together; choosing any subset understates the risk. Therefore (d) is most accurate. Common Pitfalls: Ignoring inlet calming sections; over-tight weir/flow path causing excessive path length; neglecting tray leveling which can mimic high gradient effects. Final Answer: All (a), (b) and (c)

Question 12

Thin-walled cylindrical vessel under internal pressure: what is the relationship between circumferential (hoop) stress and longitudinal stress?

Accepted Answer

Introduction / Context: For thin-walled cylinders under internal pressure, two principal membrane stresses arise: the hoop (circumferential) stress around the circumference and the longitudinal stress along the axis. Understanding their ratio is essential for safe shell design, nozzle reinforcement, and head selection. Given Data / Assumptions: Thin-wall assumption: wall thickness t ≪ diameter d. Uniform internal pressure p; negligible external loads. Linear elastic, isotropic material; small deformation. Concept / Approach: Balance forces on free-body sections. For hoop stress, cut the cylinder axially: the pressure acting on the projected area balances the tensile forces in the two walls. For longitudinal stress, cut a transverse section: pressure on the head balances the tensile ring in the shell. These classical force balances yield closed-form expressions. Step-by-Step Solution: Hoop stress: σh = p d / (2 t).Longitudinal stress: σL = p d / (4 t).Therefore, σh = 2 σL.This ratio guides which failure mode is critical; typically hoop governs thickness sizing for long cylinders. Verification / Alternative check: Dimensional checks and energy methods (e.g., strain energy minimization) confirm the same results under thin-shell theory. Codes and handbooks list these identities as baseline shell formulas. Why Other Options Are Wrong: (b) and (c) contradict the derived ratio; (d) is false because a fixed ratio exists for thin shells; (e) has no theoretical basis in this context. Common Pitfalls: Applying thin-wall equations to thick shells; omitting external loads or bending from supports; misusing diameter (inside vs. outside); ignoring joint efficiency in code calculations. Final Answer: σh = 2 σL

Question 13

Rotary dryers: within what range of number of heat transfer units (HTU or NTU) are these dryers most economically designed for typical services?

Accepted Answer

Introduction / Context: Rotary dryers are widely used for bulk solids. Economic design weighs capital cost (length, diameter, internals) against operating cost (fuel, air handling) and performance (moisture removal). The number of transfer units (NTU/HTU) is a convenient performance metric expressing how effectively the driving force for heat/mass transfer is utilized. Given Data / Assumptions: Direct-heated rotary drum with typical flights. Steady operation with reasonable inlet/outlet moisture targets. Ambient or preheated air; conventional burner duty. Concept / Approach: NTU relates the integrated driving force to the actual temperature/moisture approach accomplished. Very low NTU implies short residence and poor approach to equilibrium, inflating fuel use; very high NTU requires long drums, high capital, and diminishing returns. Industry practice and design heuristics place economic NTU for rotary dryers roughly around 1.5–2.5 for many non-critical solids. Step-by-Step Solution: Define NTU from heat/mass transfer balances for the drum.Evaluate cost vs. benefit: as NTU increases, outlet moisture decreases but length and pressure drop rise.Find the “knee” in the curve—commonly 1.5–2.5 NTU—for economic sizing.Confirm with vendor charts or past plant data for the specific solid/air conditions. Verification / Alternative check: Equipment vendor guidelines and empirical correlations align with this range for many granular products; special materials (heat-sensitive, highly cohesive) may require adjustments. Why Other Options Are Wrong: (a) under-sizes and wastes energy; (c)–(d) increase length and cost with limited added benefit; very large NTU values (e) are rarely economical. Common Pitfalls: Ignoring particle entrainment limits; neglecting dew point control; applying the same NTU to very different solids without pilot tests. Final Answer: 1.5 to 2.5

Question 14

Drying heat-transfer coefficient scaling with air mass velocity: for constant-rate drying over a surface, hG = 0.0176 * G^0.8 for parallel flow. What correlation applies when air impinges perpendicular to the surface?

Accepted Answer

Introduction / Context: During the constant-rate period of drying, convective heat transfer from air to a wet surface dominates. Empirical correlations relate the heat-transfer coefficient per unit mass flux (hG) to air mass velocity G. Flow orientation matters: parallel flow over a plane differs from perpendicular impingement jets hitting the surface. Given Data / Assumptions: Constant-rate drying region; surface is saturated with liquid water or solvent. Air properties are evaluated at film conditions; negligible radiation. Correlations are empirical and dimensionally consistent for hG in Kcal/hr·m^2·°C and G in kg/hr·m^2 (units as stated in the stem). Concept / Approach: Parallel flow correlations often show a stronger dependence on G (exponent near 0.8), while impingement jets provide high local turbulence but the scaling exponent tends to be lower, reflecting different boundary-layer development and stagnation-point heat transfer behavior for the metric hG defined here. A widely cited switch is from 0.0176 * G^0.8 (parallel) to 1.004 * G^0.37 (perpendicular). Step-by-Step Solution: Identify the given base: hG_parallel = 0.0176 * G^0.8.Apply orientation change to perpendicular impingement: hG_perp = 1.004 * G^0.37.Confirm units and exponents align with empirical data sets for impinging flow conditions. Verification / Alternative check: Drying literature and equipment vendor data show reduced exponent values for impingement configurations when hG is expressed per unit G in the given units, matching the presented correlation. Why Other Options Are Wrong: (b) keeps the 0.8 exponent, which suits parallel flow; (c) uses the small coefficient with the impingement exponent; (d) is unrealistic; (e) contradicts extensive empirical evidence. Common Pitfalls: Mixing SI and English units; applying correlations outside their validity ranges (very low or high G); ignoring humidity effects on properties. Final Answer: hG = 1.004 * G^0.37

Question 15

Evaporator selection: which type of liquor is best handled in a long-tube vertical (rising-film) evaporator from an operability standpoint?

Accepted Answer

Introduction / Context: Long-tube vertical (LTV) rising-film evaporators rely on vapor generation within vertical tubes to accelerate liquid upward, creating thin films and high heat-transfer coefficients. Equipment choice depends on fouling, foaming, viscosity, and crystallization tendencies of the liquor. Given Data / Assumptions: LTV (rising-film) operation: nucleate boiling at tube bottoms, rapid film acceleration, short residence times. Normal tube lengths and diameters; reasonable heating steam pressures. Steady production without deliberate crystallization inside tubes. Concept / Approach: Foaming liquids actually benefit from the shear and rapid vapor disengagement in rising-film tubes; the motion suppresses persistent foam accumulation compared with quiescent pools. Conversely, scaling, salting, or highly viscous liquors hamper LTV: scale blocks tubes, salts crystallize in films, and high viscosity damps acceleration and lowers heat transfer. Such services are better suited to forced-circulation or falling-film designs with better control over velocities and shear profiles. Step-by-Step Solution: List behaviors: foaming vs. scaling vs. salting vs. high viscosity.Match to LTV strengths: strong vapor-driven film motion, short holdup, good heat-transfer coefficients → favorable for foamy, non-fouling feeds.Eliminate problematic feeds: heavy scaling/salting or very viscous streams that need forced circulation or scraped surfaces. Verification / Alternative check: Industry practice (e.g., paper mill black liquors) and standard design texts list foaming liquids among the common LTV applications (with antifoam control as needed). Why Other Options Are Wrong: (b) and (d) invite tube plugging; (c) lowers film Reynolds number excessively; (e) risks deposition and erosion; none exploit LTV advantages like foamy, relatively clean liquids. Common Pitfalls: Underestimating foaming tendencies at higher solids; overlooking need for proper vapor-head disengagement; ignoring recirculation alternatives for foulers. Final Answer: Foamy liquor

Question 16

Pressure testing (hydrotest) of storage tanks and pressure vessels as per Indian practice: the recommended test pressure is approximately what multiple of design pressure?

Accepted Answer

Introduction / Context: Hydrostatic testing verifies the structural integrity and leak tightness of pressure containment equipment before commissioning. Standards specify test pressures above design to create a safety margin and reveal fabrication defects without overstressing the material. Given Data / Assumptions: Conventional atmospheric storage tanks and pressure vessels in process plants. Indian codes/practice; typical materials (carbon steel, low alloy). Elastic behavior during test; temperature near ambient. Concept / Approach: Design pressure reflects maximum allowable operating pressure with code-mandated safety factors. Hydrotest uses liquid (usually water) to avoid stored elastic energy hazards of gas testing. A multiplier of roughly 1.5 times design pressure is commonly adopted for pressure vessels; some storage tank cases may approach up to 2 times depending on code and service, hence a practical band of 1.5–2 is recognized. Step-by-Step Solution: Identify typical vessel code guidance: test pressure ≥ 1.5 * design pressure (adjusted for temperature & allowable stress at test conditions).Acknowledge tank specifics: geometry and shell courses can shift exact multipliers; upper bound near 2 can appear in certain cases.Select the range that best captures Indian practice: 1.5 to 2. Verification / Alternative check: Project specifications and vendor data sheets across refineries and chemical plants in India align with 1.5–2× ranges, subject to code clauses and allowable stress ratios at test vs. design temperatures. Why Other Options Are Wrong: Higher multipliers (3–5 or more) risk plastic deformation; testing at exactly design pressure offers insufficient margin. Common Pitfalls: Ignoring allowable stress temperature corrections; not venting air pockets; testing with cold water on warm vessels causing thermal shock. Final Answer: 1.5 to 2

Question 17

Shell-and-tube heat exchangers: what is the recommended minimum baffle spacing relative to shell inside diameter (Ds)?

Accepted Answer

Introduction / Context: Baffles in shell-and-tube exchangers control crossflow and support tubes. Baffle spacing strongly affects shell-side pressure drop, heat-transfer coefficient, vibration risk, and mechanical support. A practical lower bound prevents excessive pressure drop and flow maldistribution. Given Data / Assumptions: Single-phase shell-side service with conventional segmental baffles. Standard TEMA construction and allowable pressure-drop guidelines. Goal: avoid vibration and keep reasonable pressure drop while maintaining heat transfer. Concept / Approach: Too small a baffle spacing increases the number of crossflow windows, driving up pressure drop and potential erosion while offering diminishing returns on heat transfer. Too large a spacing lowers heat transfer and can permit flow-induced vibration. A widely used rule of thumb sets minimum baffle spacing at about 0.4 * Ds (often bounded also by an absolute minimum such as 50 mm). Step-by-Step Solution: Normalize spacing to shell ID: S_min ≈ 0.4 * Ds.Check pressure-drop and vibration constraints; increase spacing if pressure drop is too high, decrease if heat transfer is insufficient within mechanical limits.Confirm against project specs or TEMA-derived practices. Verification / Alternative check: Thermal rating tools show steep pressure-drop penalties as spacing drops below ~0.4 Ds, with modest gains in h, supporting the rule. Why Other Options Are Wrong: (a) and (c) are fixed values and not scalable; (d) is excessively large and defeats baffling purpose; (e) would cause extreme pressure drops and risk tube damage. Common Pitfalls: Ignoring two-phase shell-side behavior; misapplying the rule to helical or no-tubes-in-window designs; neglecting tube support criteria in vibration-sensitive services. Final Answer: Approximately 40% of the shell inside diameter

Question 18

Mechanical properties: the units of Young’s modulus of elasticity are identical to the units of which other quantity?

Accepted Answer

Introduction / Context: Young’s modulus E relates stress to strain in linear elastic materials. Recognizing its dimensional form helps interpret test data, validate formulas, and ensure unit consistency in calculations involving deflection, vibration, and pressure‐boundary design. Given Data / Assumptions: Linear elastic behavior: σ = E * ε. Stress σ carries units of force per area; strain ε is dimensionless. Standard SI or engineering unit systems are acceptable. Concept / Approach: From σ = E * ε, and ε being dimensionless, E must have the same units as σ. Thus, E is expressed in Pascals (N/m^2), MPa, GPa, or in psi/ksi in US customary units. This identity ensures dimensional consistency in beam bending, pressure vessel theory, and finite-element formulations. Step-by-Step Solution: Write the linear Hooke’s law: σ = E * ε.Since ε is unitless, E must have units identical to σ.Therefore, E shares the same units as stress: force per unit area. Verification / Alternative check: Check common values: steel E ≈ 200 GPa, which is indeed a pressure unit; typical working stresses are also stated in MPa—consistent units. Why Other Options Are Wrong: Strain and Poisson’s ratio are dimensionless; modular ratio is a ratio of moduli and dimensionless; therefore only stress shares the units with E. Common Pitfalls: Confusing engineering stress with true stress; mixing units (psi vs. MPa) without proper conversion; misreading GPa as a force or energy unit rather than pressure. Final Answer: Stress

Question 19

Liquid–liquid shell-and-tube exchangers: what is the typical range of the ratio of tube length to shell diameter (L/Ds) used for design?

Accepted Answer

Introduction / Context: The length-to-diameter ratio (tube length L versus shell inside diameter Ds) influences heat-transfer area packing, pressure drop, vibration control, and fabrication practicality in shell-and-tube exchangers. Liquid–liquid services typically allow moderate pressure drops and benefit from longer tubes for compact designs. Given Data / Assumptions: Conventional TEMA construction with segmental baffles. Single-phase on both sides; non-fouling to moderately fouling liquids. Economically acceptable pressure-drop constraints. Concept / Approach: Short tubes increase shell diameter or number of shells for a given area, raising capital cost. Very long tubes raise vibration concerns and fabrication complexity. A balanced design for liquid–liquid duties often uses L/Ds ≈ 4–8, enabling practical layouts, good area density, and manageable pressure drops. Step-by-Step Solution: Consider trade-offs: longer tubes → fewer shells and heads but higher risk of vibration if unsupported; shorter tubes → larger shell diameter for same area.Industry heuristics place liquid–liquid duties in the mid-range of length ratios.Select 4 to 8 as the representative range. Verification / Alternative check: Design guides and vendor ratings frequently show liquid–liquid exchangers in L/Ds ranges of 4–8, with gas–gas duties often below 4 to limit pressure drop. Why Other Options Are Wrong: (a)–(b) are on the short side, leading to bulky shells; (d)–(e) are longer than commonly economical and can complicate supports and vibration control. Common Pitfalls: Ignoring allowable nozzle loads and support spans; overlooking cleaning access for longer tubes; not checking acoustic/vortex-induced vibration criteria. Final Answer: 4 to 8

Question 20

In a washing-type plate-and-frame filter used for solid–liquid separation, what is the typical ratio of the washing rate to the final (end-of-cycle) filtration rate under comparable operating conditions?

Accepted Answer

Introduction / Context: In chemical engineering practice, plate-and-frame filters are frequently operated in a sequence that includes initial filtration, washing of the filter cake to remove mother liquor, and then air-drying or blowdown. A common rule-of-thumb concerns the recommended washing rate relative to the final filtration rate near the end of the cycle. This ratio helps ensure that fresh wash liquid effectively displaces interstitial liquor within the porous cake without excessive channeling or breakthrough. Given Data / Assumptions: Washing is performed after cake formation using the same hydraulic path. “Final filtration rate” refers to the filtrate rate towards the end of cake build-up (highest resistance period). Comparable viscosity and temperature of the filtrate and wash liquid. Concept / Approach: Washing effectiveness depends on achieving sufficient superficial velocity through the cake to minimize axial dispersion and bypassing. Empirical design practice typically recommends a modestly higher washing rate than the terminal filtration rate so the wash front advances uniformly, improves displacement, and reduces residual solute in the cake pores. Step-by-Step Solution: Identify target: ratio of wash rate to final filtration rate.Industry guidance recommends a slightly higher wash superficial velocity than during the final filtration phase.A commonly used value is about 1.25 times the final filtration rate. Verification / Alternative check: Operating manuals and classical filtration heuristics indicate 1.2–1.3 times as a practical range that balances improved displacement with pump energy and mechanical constraints of the filter pack. Why Other Options Are Wrong: 0.25 or 0.50: Too low; insufficient velocity causes poor displacement and channeling. 0.75: Still below final rate; rarely adequate for effective displacement washing. Common Pitfalls: Assuming much higher rates always improve washing; excessive flow can fluidize or crack the cake, lowering washing efficiency. Temperature and viscosity matching between filtrate and wash also matter. Final Answer: 1.25

Process Equipment and Plant Design Questions