Question 1

Drying of slurries on a rotating drum: In a single-drum or twin-drum dryer, which phenomenon most commonly controls the overall drying rate for thin films?

Accepted Answer

Introduction / Context: Drum dryers spread a thin layer of slurry on a hot rotating cylinder. As the drum turns, heat conducts from the metal surface through the liquid film, causing surface evaporation and producing a dried sheet that is scraped off. Understanding the controlling resistance helps set steam pressure, drum speed, and film thickness. Given Data / Assumptions: Thin films with short residence times (seconds). Well-wetted metal surface, negligible internal diffusion limitations for many foods/chemicals. Evaporation limited by energy supplied through the drum wall. Concept / Approach: For thin films typical of drum drying, the rate is frequently governed by heat transfer (supply of latent heat) rather than slow diffusion through a thick solid. The overall resistance lies in the metal/condensate/liquid-film path, not deep mass diffusion like in thick cakes. Hence steam pressure, condensate removal, and film contact dominate performance. Step-by-Step Solution: Recognize that the wet film is thin and exposed to high temperature at the surface.Heat must be transmitted quickly from condensing steam → drum wall → liquid film.Drying rate aligns with available heat flux; thus heat transfer controls.Select the option highlighting heat transfer control. Verification / Alternative check: Plant tuning shows increases in steam pressure or improved condensate drainage raise throughput markedly, consistent with heat-transfer control; changing air-side conditions has smaller effects. Why Other Options Are Wrong: Internal diffusion (a) is minor for very thin films and short paths. Both/neither (c/d) are overly broad and do not reflect the dominant mechanism in typical drum dryers. Common Pitfalls: Allowing condensate pooling inside the drum, which reduces heat-transfer coefficient. Letting films get too thick, inadvertently introducing diffusion limits. Final Answer: Heat transfer to the wet film (evaporation controlled by heat supply)

Question 2

Cooling tower selection with favorable ambient air: When the ambient air temperature and humidity are both low, which type of cooling tower draft is typically preferred?

Accepted Answer

Introduction / Context: Cooling towers reject heat by evaporative cooling, drawing ambient air into contact with warm water. The air can be moved by natural buoyancy (natural draft) or by fans (forced or induced draft). Ambient conditions influence which configuration is most practical and efficient. Given Data / Assumptions: Ambient air is cool and dry, creating a strong driving force for evaporation. Large installations where a hyperbolic natural-draft tower is feasible. No special plume abatement or high turn-down requirements. Concept / Approach: Low temperature and low humidity increase the wet-bulb depression, enhancing buoyancy of the warm, moist exit air. Natural-draft towers capitalize on this density difference to move large air volumes without fans, reducing mechanical complexity and power consumption. Step-by-Step Solution: Identify favorable ambient conditions (cool, dry) → strong natural buoyancy.Natural draft uses a tall chimney/hyperbolic shell to sustain airflow.Therefore, when conditions are favorable, natural draft is typically preferred.Select 'Natural draft'. Verification / Alternative check: Power plants in temperate climates often employ natural-draft towers to avoid fan energy; under hot/humid conditions, mechanical draft is chosen to guarantee airflow. Why Other Options Are Wrong: Forced and induced draft rely on fans and are favored when ambient conditions are less favorable or when precise airflow control is needed; not necessary here. 'None' is inappropriate because a tower type is indeed applicable. Common Pitfalls: Assuming natural draft suits all climates; high humidity or high ambient temperatures weaken buoyancy. Ignoring site constraints that may necessitate mechanical draft despite favorable weather. Final Answer: Natural draft

Question 3

Interphase mass transfer fundamentals: The rate of mass transfer between two contacting fluid phases does not necessarily depend on which of the following characteristics?

Accepted Answer

Introduction / Context: Mass transfer across liquid–gas or liquid–liquid interfaces is central to absorption, stripping, extraction, and aeration. The overall rate depends on transport resistances and interfacial driving forces, but not all listed factors are essential in the absence of reaction. Given Data / Assumptions: No chemical reaction at the interface or in the phases (pure physical mass transfer). Binary or dilute multicomponent systems. Isothermal operation. Concept / Approach: For physical mass transfer, the key determinants are: (1) physical properties (diffusivity, viscosity, density) that set film coefficients; (2) hydrodynamics or turbulence that thin boundary layers; and (3) interfacial area, which multiplies the local flux. Chemical properties matter primarily when reaction modifies the effective driving force or introduces enhancement factors. Step-by-Step Solution: Write an overall rate form: N = K_overall * A * ΔC (or Δp).Recognize K_overall depends on physical properties and turbulence (film theory).A is the interfacial area created by dispersion/packing.Without reaction, specific chemical reactivity does not directly set the rate. Verification / Alternative check: In gas absorption of a sparingly soluble gas without reaction, altering viscosity, diffusivity, or agitation changes rates, while changing “chemical nature” without affecting solubility/physical parameters has little effect. Why Other Options Are Wrong: Physical properties determine film coefficients and solubility. Turbulence reduces film thickness and increases transfer coefficients. Interfacial area directly scales the total transfer rate. Common Pitfalls: Confusing chemical equilibrium (which sets driving force) with intrinsic chemical reactivity; here no reaction is considered. Neglecting that reactions can dramatically enhance transfer—an exception noted but not part of this basic case. Final Answer: Chemical properties of the phases

Question 4

Relative volatility versus pressure: For a typical binary mixture in distillation, how does the relative volatility α generally change with increasing system pressure?

Accepted Answer

Introduction / Context: Relative volatility α measures the ease of separating two components by distillation. Higher α indicates a larger volatility difference and easier separation. Pressure affects vapor–liquid equilibrium (VLE), which in turn influences α. Given Data / Assumptions: Binary, non-azeotropic mixture over practical pressure ranges. No strong associating or pressure-activated complex formation. Comparison at the same temperature approach (or along operating range). Concept / Approach: As pressure increases, vapor-phase non-ideality and saturation pressures shift such that K-values (y/x ratios) of components tend to converge. This convergence usually lowers relative volatility (α = K_A / K_B), making separation more difficult. Hence low-pressure operation often improves α and reduces stages. Step-by-Step Solution: Define α = (y_A/x_A) / (y_B/x_B) = K_A / K_B.Note that with increasing pressure, K-values often move closer to 1 and to each other.Therefore, α generally decreases as pressure rises.Select 'Decreases' as the most representative trend. Verification / Alternative check: VLE charts and process simulators commonly show increased column difficulty (more stages/higher reflux) at higher pressures for many hydrocarbon pairs, consistent with reduced α. Why Other Options Are Wrong: 'Increases' is uncommon except for special systems; 'remains the same' ignores VLE pressure sensitivity; 'either/or' is too vague for the general rule-of-thumb expected in fundamentals. Common Pitfalls: For azeotropic/associating systems, trends can deviate; the question seeks the general case. Confusing α changes due to pressure with temperature effects along the column. Final Answer: Decreases

Question 5

Batch distillation with constant reflux: How does the distillate (overhead) composition change with time during a batch run at constant reflux ratio?

Accepted Answer

Introduction / Context: In batch distillation, a still is charged once and operated while compositions evolve. Even with a fixed reflux ratio, the composition of the distillate stream changes over time as the charge in the pot becomes depleted in the more volatile component. Given Data / Assumptions: Constant reflux ratio (L/D fixed). Ideal operation without holdup anomalies. Binary or light-key focused multicomponent mixture. Concept / Approach: At the start, the batch contains a higher fraction of the light component, producing a richer top product. As distillation proceeds, the pot composition becomes heavier, reducing the equilibrium vapor richness and, consequently, the instantaneous distillate composition. Thus, x_D(t) monotonically declines. Step-by-Step Solution: Early time: pot rich in light key → high y at top → high distillate purity.As time passes: light key removed → pot x decreases.Vapor–liquid equilibrium ensures y at the top follows pot composition trend.Therefore, overhead composition decreases with time at constant reflux. Verification / Alternative check: Rayleigh equation for batch distillation predicts a changing pot composition with time; stagewise models show declining top composition for fixed reflux when the pot progressively heavies. Why Other Options Are Wrong: 'Increases' would require enrichment of the light key in the pot, which does not occur. 'Does not vary' is incompatible with changing pot composition. 'May increase or decrease' is not typical for the standard case; monotonic decrease is expected. Common Pitfalls: Confusing constant reflux with constant distillate composition; the latter requires composition control or variable reflux. Ignoring tray holdup or finite condenser dynamics—second-order effects that do not change the overall trend. Final Answer: Decreases

Question 6

Distillation design — Overall (column) efficiency: what is the correct definition? Choose the statement that correctly defines the overall efficiency of a distillation column in terms of plates (or trays).

Accepted Answer

Introduction: Overall (column) efficiency is a lumped measure that connects the number of ideal (theoretical) equilibrium stages predicted by equilibrium calculations to the number of real, physical trays (or height of packing) required in an actual distillation column. This question checks whether you know the correct definition that engineers use when converting theoretical designs into hardware. Given Data / Assumptions: A distillation column designed by McCabe–Thiele, Fenske–Underwood–Gilliland, or rigorous simulation yields an ideal plate count. Actual trays exhibit non-idealities (mixing, weeping, entrainment, finite mass transfer). We need a single factor to scale theoretical stages to actual trays. Concept / Approach: Overall efficiency, often denoted E_o, is defined as E_o = N_ideal / N_actual. Rearranged, the actual number of trays is N_actual = N_ideal / E_o. E_o consolidates all non-ideal effects into one multiplier for column-wide sizing. It is different from Murphree tray efficiency, which is defined per tray and can vary along the column, and different from point efficiency, which is local to the mass-transfer surface. Step-by-Step Solution: Identify the theoretical equilibrium stage count from design calculations (N_ideal).Recognize that real trays are less efficient → more trays are needed.Apply the definition E_o = N_ideal / N_actual.Conclude: overall efficiency equals the ratio of ideal plates to actual plates. Verification / Alternative check: O’Connell-type correlations provide empirical overall efficiencies based on system properties (e.g., viscosity), confirming the practical use of E_o = N_ideal / N_actual. Why Other Options Are Wrong: (b) Inverts the definition. (c) Murphree efficiency is per tray and not generally identical to E_o. (d) Point efficiency can exceed or be less than other efficiencies; no universal ordering applies. (e) Efficiency is affected by hydraulics and contacting, not just volatility. Common Pitfalls: Confusing overall efficiency with Murphree or point efficiencies; forgetting that efficiencies can vary between rectifying and stripping sections. Final Answer: The ratio of number of ideal (theoretical) plates to actual plates

Question 7

Applications of the Thiele–Geddes method — In which separation operation is the Thiele–Geddes equation specifically used for rigorous stage-to-stage calculations?

Accepted Answer

Introduction: Different mass-transfer operations use different mathematical frameworks. The Thiele–Geddes method is a classic approach used to perform tray-by-tray material balance and equilibrium calculations where multiple components distribute between vapor and liquid phases at each stage. This question asks you to identify the correct operation that relies on this method. Given Data / Assumptions: Multiple components present in both phases. Equilibrium relationships (K-values or activity–fugacity models) are known. Stagewise (equilibrium-stage) model is appropriate. Concept / Approach: The Thiele–Geddes method is a rigorous multicomponent distillation formulation. It sets up a system of difference equations that relate liquid and vapor compositions between adjacent trays subject to equilibrium, mass balance, and flow assumptions (e.g., constant molar overflow), enabling computation of composition profiles and tray requirements. While absorption and extraction also use stagewise models, their canonical design tools differ, and chemical reaction would require additional rate terms beyond the classical Thiele–Geddes formulation. Step-by-Step Solution: Recognize a stagewise vapor–liquid contacting operation with many components.Recall that Thiele–Geddes is developed for distillation column calculations.Select “Multicomponent distillation.” Verification / Alternative check: Textbook derivations of Thiele–Geddes appear in multicomponent distillation chapters, often paired with Underwood and Fenske methods for minimum trays and minimum reflux. Why Other Options Are Wrong: (a) and (b) use different equilibrium and mass-transfer frameworks. (c) Adds reaction coupling, not the original scope of Thiele–Geddes. (e) Crystallization uses supersaturation and population balances, not tray-by-tray VLE equations. Common Pitfalls: Assuming any stagewise operation uses Thiele–Geddes; it is specifically a multicomponent distillation tool. Final Answer: Multicomponent distillation

Question 8

Adiabatic humidification with liquid recirculation — Which equipment is frequently used to perform adiabatic humidification–cooling with a recirculating liquid stream?

Accepted Answer

Introduction: Adiabatic humidification–cooling is the process where a gas (often air) is contacted with a liquid (often water) so that the gas approaches an adiabatic saturation state while the liquid recirculates and warms slightly. The question asks which piece of equipment is frequently used for such operations with explicit liquid recirculation. Given Data / Assumptions: Gas–liquid contact with recirculating water. Negligible external heat addition (adiabatic). Objective: cool and humidify the gas while slightly heating the liquid. Concept / Approach: A spray chamber provides direct-contact mass and heat transfer by spraying liquid into a flowing gas stream; the sprayed liquid is collected at the bottom and recirculated to maintain adequate wetting and approach the adiabatic saturation temperature. Cooling towers are also adiabatic contactors, but the “textbook” example highlighting adiabatic humidification with controlled recirculation and adiabatic saturation charts is the spray chamber apparatus. Step-by-Step Solution: Identify adiabatic gas–liquid contacting with free-surface sprays.Recognize that recirculation is a standard feature of spray chambers for steady approach to saturation conditions.Select “Spray chamber.” Verification / Alternative check: Psychrometric analyses of spray chambers show gas outlet states approaching the adiabatic saturation line; recirculation maintains steady liquid conditions. Why Other Options Are Wrong: Cooling towers indeed perform similar duties but are generally categorized as tower contactors; the canonical adiabatic humidifier with controlled recirculation in design problems is the spray chamber. “None of these” and bubble columns do not match standard adiabatic humidifier descriptions for air–water service. Common Pitfalls: Equating all air–water contactors; while cooling towers are practical, the spray chamber is the classic teaching device for adiabatic humidification with recirculation. Final Answer: Spray chamber

Question 9

Effect of reflux ratio — As the reflux ratio in a distillation column is increased from its minimum value, which overall trend is correct?

Accepted Answer

Introduction: Reflux ratio is a key design and operating variable in distillation. Moving away from minimum reflux reduces the number of equilibrium stages, but also increases internal flows and energy use. Understanding these trade-offs is crucial for selecting an economical design point. Given Data / Assumptions: Constant product specifications. Feasible range: from minimum reflux (infinite stages) to very high reflux (near total reflux, minimal stages). Typical cost structure: capital linked to plate count/size; operating linked to reboiler/condenser duties. Concept / Approach: As reflux increases from the minimum, the number of stages decreases sharply at first and then with diminishing returns. Capital cost initially falls (fewer trays/shorter column), but energy and utility costs rise due to increased internal liquid and vapor rates. The combined total annual cost therefore shows a shallow minimum at an intermediate reflux ratio. Step-by-Step Solution: At minimum reflux: N → very large, capital high, energy low.Increase reflux slightly: N drops rapidly; capital decreases.Further increases: N decreases slowly; utilities rise significantly.Result: a cost optimum where total annual cost is minimized. Verification / Alternative check: Gilliland correlations (or modern simulators) show diminishing returns in stage reduction at high reflux, consistent with the cost-tradeoff curve. Why Other Options Are Wrong: (a) The stripping-line slope is not the primary monotonic trend with reflux; rectifying line slope increases. (b) Trend is opposite: decrease is rapid at first, then slower. (d) Both internal liquid and vapor generally increase with reflux for fixed separation. (e) Minimum plates occur at very high (approaching total) reflux, not at minimum reflux. Common Pitfalls: Assuming linear trade-offs; ignoring condenser and reboiler size impacts at high reflux. Final Answer: The total annual cost first decreases (as plates fall) and then increases (as utilities rise), giving an optimum

Question 10

Total reflux in distillation — What requirement reaches its minimum under total-reflux operation (no product withdrawal, maximum internal reflux)?

Accepted Answer

Introduction: Total reflux is a limiting operating condition where all condensed distillate is returned as reflux and no products are withdrawn. It is a valuable conceptual point for understanding column behavior and for establishing minimum stage count independent of feed and product rates. Given Data / Assumptions: No material is withdrawn as products; all condensed vapor returns as reflux. Internal liquid and vapor rates are at their maximum for a given column. Separation specified by compositions at the top and bottom. Concept / Approach: At total reflux, the operating lines in the McCabe–Thiele diagram coincide with the 45-degree diagonal, yielding the minimum theoretical number of stages needed for the specified separation. However, because internal flows are the largest, both reboiler and condenser duties are maximized, not minimized. Step-by-Step Solution: Set reflux to total (R → ∞), product rates → 0.Operating lines approach the diagonal → best separation per stage.Count stages: minimum theoretical count achieved.Recognize heat duties are high due to large internal circulation. Verification / Alternative check: Textbook derivations and simulations consistently show fewer stages are needed at higher reflux, with the minimum at total reflux. Energy balances show higher condenser and reboiler loads under total reflux since all vapor generated must be condensed and returned. Why Other Options Are Wrong: (a) and (c) Heat loads are not minimal; they are maximal. (d) Cannot be true given (a) and (c) are false. (e) Cooling-water use generally increases at total reflux due to higher condenser duty. Common Pitfalls: Confusing the “minimum stages” limit with “minimum energy” limit (which is attained at minimum reflux, not total reflux). Final Answer: Number of plates (theoretical stages) needed

Question 11

Psychrometrics — During sensible cooling of moist air (no condensation), which property(ies) remain constant at fixed total pressure?

Accepted Answer

Introduction: Sensible cooling means the temperature of moist air is reduced without changing its moisture content (no dehumidification). Recognizing which properties remain constant under this condition is fundamental to psychrometric chart work and HVAC calculations. Given Data / Assumptions: No condensation occurs (final dry-bulb temperature remains above the dew point). Total pressure remains constant (e.g., 1 atm). Air behaves as an ideal gas mixture; water vapor content is unchanged. Concept / Approach: The specific humidity (humidity ratio) w is proportional to the vapor partial pressure p_v via w = 0.622 p_v / (P − p_v). If w is constant and the total pressure P is constant, p_v must also remain constant. Meanwhile, relative humidity increases as temperature falls because the saturation vapor pressure decreases with temperature, narrowing the gap between p_v and p_sat(T). Step-by-Step Solution: Impose constraint: no phase change → water mass fraction constant → w constant.At fixed P, solve w relation → p_v remains constant.Conclude: both w and p_v are unchanged; only relative humidity changes. Verification / Alternative check: Plotting a sensible-cooling line on a psychrometric chart progresses horizontally to the left (constant humidity ratio), intersecting higher relative-humidity curves as temperature decreases. Why Other Options Are Wrong: (a) or (b) alone is incomplete; both remain constant. (d) contradicts the definition of sensible cooling. (e) Wet-bulb temperature also decreases during sensible cooling without moisture removal. Common Pitfalls: Confusing relative humidity (which rises) with specific humidity; assuming p_v changes because temperature changes, forgetting composition is unchanged. Final Answer: Both (a) and (b)

Question 12

Cooling at constant pressure — Moist air at 101.3 kPa, 40°C, and 50% relative humidity is cooled to 30°C with no condensation. The cooled air will have:

Accepted Answer

Introduction: Psychrometric changes under sensible cooling at constant pressure illustrate how relative humidity responds to temperature. This scenario begins at 40°C and 50% RH and cools to 30°C without reaching the dew point (so no moisture is removed). Given Data / Assumptions: Total pressure fixed at 101.3 kPa. No condensation occurs; moisture content is unchanged. Ideal-gas behavior for air–vapor mixture. Concept / Approach: Specific humidity w and vapor partial pressure p_v remain constant when no water is added or removed. As temperature decreases, the saturation vapor pressure p_sat(T) decreases. Relative humidity is RH = p_v / p_sat(T). Since p_v is constant while p_sat(T) decreases, RH must increase. Step-by-Step Solution: Note initial: T1 = 40°C, RH1 = 50% → dew point around the high 20s °C.Cool to T2 = 30°C > dew point → no condensation → w, p_v constant.Because p_sat(30°C) < p_sat(40°C), RH2 = p_v / p_sat(30°C) > RH1. Verification / Alternative check: On a psychrometric chart, a horizontal move left from 40°C to 30°C (constant humidity ratio) crosses to a higher RH curve, confirming the conclusion. Why Other Options Are Wrong: (a) Dew point is set by p_v; since p_v is unchanged, dew point is the same, not higher. (b) Absolute humidity is unchanged (no moisture removal or addition). (d) Wet-bulb typically falls with dry-bulb during sensible cooling absent evaporation. (e) RH does not remain the same; it increases. Common Pitfalls: Confusing dew point (depends on p_v) with dry-bulb temperature; assuming RH always decreases with cooling. Final Answer: A higher relative humidity

Question 13

Dew point behavior — For a fixed gas–vapor composition, how does the dew point temperature change with system pressure?

Accepted Answer

Introduction: Dew point is the temperature at which a gas–vapor mixture becomes saturated upon cooling at fixed total pressure and composition. Understanding how dew point varies with pressure is important for condensers, dryers, and compressed air systems. Given Data / Assumptions: Gas–vapor mixture with fixed vapor mole fraction (composition constant). Total pressure is varied. Equilibrium governed by vapor–liquid equilibrium and saturation properties. Concept / Approach: At dew point, the saturation pressure of the vapor at the dew-point temperature equals its partial pressure in the mixture: p_sat(T_dew) = y_v * P_total. Holding y_v constant, increasing P_total raises the vapor partial pressure at the dew point. Since p_sat(T) increases monotonically with T, a higher p_sat requirement implies a higher dew-point temperature. Conversely, lowering total pressure lowers dew point. Step-by-Step Solution: Write the dew-point condition: p_sat(T_dew) = y_v * P.Increase P → required p_sat increases.Because p_sat grows with T, T_dew must increase. Verification / Alternative check: Compressed air dryers experience condensation at warmer surfaces after compression because the dew point rises with pressure—operational evidence of this relationship. Why Other Options Are Wrong: (a) and (b) conflate dew point (a function of pressure and composition) with dry-bulb temperature changes. (c) True statement but opposite direction to (d); only one should be selected—(d) is the direct phrasing. (e) Dew point is not pressure independent. Common Pitfalls: Assuming dew point is a property of composition alone; ignoring the role of total pressure. Final Answer: It increases with an increase in total pressure

Question 14

Absorption factor control — To increase the absorption factor A (where A = S / (m * G), with S = solvent liquid rate, G = gas rate, and m = slope of the equilibrium line), what change is appropriate?

Accepted Answer

Introduction: The absorption factor A is a convenient nondimensional parameter used in gas absorption design to compare the operating line slope with the equilibrium line slope. Adjusting A can strongly influence stage requirements and approach to the loading point. Given Data / Assumptions: A = S / (m * G), where S is solvent molar (or mass) flow rate, G is gas molar (or mass) flow rate, and m is the slope of the equilibrium relation y* = m x (for dilute systems). Equilibrium slope m is taken as fixed by system chemistry and temperature. Objective: increase A. Concept / Approach: Because A is directly proportional to S and inversely proportional to G (for fixed m), the most direct way to increase A is to raise S and/or reduce G. Increasing A typically moves the operating line farther from equilibrium, improving driving force and reducing tray count, at the expense of larger solvent circulation and associated recovery costs. Step-by-Step Solution: Start from A = S / (m * G).Hold m constant (property of system and temperature).To increase A, increase S and/or decrease G → select “Increase S and decrease G”. Verification / Alternative check: McCabe–Thiele constructions for absorption show fewer stages and larger approach margins as the operating line is steepened by higher S/G ratios (equivalently, higher A). Why Other Options Are Wrong: (a) Increasing both S and G proportionally keeps A approximately constant. (b) Decreasing both reduces throughput without necessarily increasing A. (d) Opposite effect: lowers A. (e) Changing m by heating is system-specific and not a reliable lever; the prompt asks for certain increase. Common Pitfalls: Ignoring downstream solvent recovery/regeneration costs when increasing S; overlooking pressure drop and flooding limits when altering G. Final Answer: Increase S and decrease G

Question 15

Cooling tower terminology — The “cooling range” of a cooling tower refers to which temperature difference?

Accepted Answer

Introduction: Cooling tower performance is commonly described using two key temperature differences: the cooling range and the approach. Knowing the exact definitions helps in specifying and evaluating tower duty and effectiveness. Given Data / Assumptions: Hot water from process enters the tower; cooled water returns to the process. Ambient air provides evaporative cooling; its wet-bulb temperature sets a theoretical lower limit. Steady operation at a given load and ambient condition. Concept / Approach: The cooling range is defined as Range = T_hot,in − T_cold,out. It measures how much the process water is cooled across the tower. The approach is defined as Approach = T_cold,out − T_wb,air. Range reflects tower heat duty at given flow, while approach reflects proximity to the psychrometric limit and is a measure of tower size/efficiency. Step-by-Step Solution: Identify inlet hot-water temperature (to tower) and outlet cold-water temperature (from tower).Compute Range = T_hot,in − T_cold,out.Distinguish from Approach = T_cold,out − T_wb. Verification / Alternative check: Cooling tower specifications list both “range” and “approach”; example: range 10°C, approach 4°C at a given wet bulb—matching the definitions above. Why Other Options Are Wrong: (a) and (b) incorrectly combine tower water temperatures with ambient wet bulb to define range (these correspond to approach variants). (d) Incorrect since a standard definition exists. (e) Dry-bulb minus wet-bulb defines the psychrometric depression, not cooling range. Common Pitfalls: Confusing range with approach; assuming smaller approach can be achieved without increasing tower size, airflow, or water distribution quality. Final Answer: Hot water entering the tower minus cooled water leaving the tower

Question 16

Distillation design — In plate (tray) columns using bubble-cap trays, the number of caps installed on a tray is primarily determined by allowable phase velocities to ensure capacity without weeping/entrainment. Which parameter set governs this most …

Accepted Answer

Introduction: In tray distillation, bubble caps distribute vapor through the liquid on each tray to promote mass transfer. The number of caps per tray is a capacity and hydraulics decision: it must pass the required vapor flow without excessive pressure drop or entrainment and maintain adequate liquid holdup without weeping. This item tests understanding of which variables primarily control cap count and spacing. Given Data / Assumptions: Bubble-cap tray operating within acceptable pressure drop and efficiency ranges. Flooding, entrainment, and weeping constraints apply. Tray diameter, downcomer design, and weir height are designed consistently with the same duty. Concept / Approach: Vapor capacity is set by allowable gas velocity through active area and cap slots; liquid handling is set by allowable liquid velocity across the tray and downcomer. The number of caps divides vapor among outlets to keep slot velocities below entrainment limits and to provide uniform vapor distribution. Thus, both allowable gas and liquid velocities govern the final number and layout, not feed composition directly (composition affects rates indirectly via V and L but not the primary criterion). Step-by-Step Solution: Estimate design vapor rate V from column material balance and energy balance.Choose allowable gas velocity based on flooding correlations → compute required active area.Select cap size/slot area; determine number of caps so slot velocity stays below entrainment criterion.Check liquid handling: crossflow area, weir loading, downcomer backup → adjust cap count/spacing if liquid velocities are excessive. Verification / Alternative check: Compare calculated tray pressure drop and approach-to-flood for several cap counts; the acceptable design meets both gas and liquid constraints simultaneously, confirming the dependence on both velocities. Why Other Options Are Wrong: A or B alone: considering only one phase can cause either weeping (too few caps) or entrainment (too many/small slots). D: Feed composition affects V and L indirectly but is not the primary governing parameter. E: Downcomer residence time is a check, not the primary determinant of cap count. Common Pitfalls: Oversizing cap count to reduce gas velocity without re-checking liquid backup; ignoring tray active area loss due to downcomers; neglecting turndown requirements. Final Answer: allowable gas and liquid velocities

Question 17

Transport analogies — For which case are mass diffusivity, thermal diffusivity, and eddy momentum diffusivity effectively equal (i.e., NPr = NSc = ______ )?

Accepted Answer

Introduction: Transport property analogies (Reynolds, Chilton–Colburn) connect momentum, heat, and mass transfer. When dimensionless groups align, similar boundary-layer behavior allows property analogies to be applied cleanly. This question asks the special condition under which momentum, heat, and mass diffusivities are equal in a normalized sense. Given Data / Assumptions: Prandtl number NPr = ν/α (momentum diffusivity/thermal diffusivity). Schmidt number NSc = ν/D (momentum diffusivity/mass diffusivity). Fully turbulent regime with similar eddy diffusivities for momentum, heat, and mass. Concept / Approach: When NPr = 1 and NSc = 1, momentum, heat, and mass boundary layers have similar thicknesses because ν ≈ α ≈ D (or, in turbulent flow, their eddy counterparts are comparable). Under this condition, the Chilton–Colburn j-factor analogy reduces to direct equivalence, simplifying transfer coefficient estimation and permitting the use of a single Stanton number relation for all three modes. Step-by-Step Solution: Write definitions: NPr = ν/α; NSc = ν/D.Set NPr = NSc = 1 → ν = α = D (effective diffusivities).Conclude that the three transport processes behave analogously.Therefore fill the blank with 1. Verification / Alternative check: Textbook analogies yield j_H ≈ j_D when Pr = Sc, with best agreement near unity; many gases at moderate temperatures have Pr of order 1, supporting the approximation. Why Other Options Are Wrong: B, C, E: Values other than 1 imply unequal diffusivities and dissimilar boundary layers. D: NPr or NSc = 0 is non-physical for real fluids. Common Pitfalls: Assuming perfect equality is required; in practice, analogies work reasonably well for Pr and Sc not too far from unity. Final Answer: 1

Question 18

Psychrometry — For an unsaturated gas–vapor mixture, what is the value of (wet-bulb temperature − adiabatic saturation temperature)?

Accepted Answer

Introduction: Two cornerstone temperatures in psychrometry are the thermodynamic wet-bulb temperature and the adiabatic saturation temperature. They are often treated as interchangeable in engineering calculations, yet their definitions are different. This question probes your understanding of their relationship for unsaturated mixtures under ideal assumptions. Given Data / Assumptions: Ideal (equilibrium) contact between gas and liquid. Negligible heat capacity effects of wetted materials; negligible radiation. Lewis number close to 1 (typical for air–water), so heat and mass transfer analogies apply. Concept / Approach: The adiabatic saturation process brings an unsaturated gas into contact with liquid in an insulated saturator until the outlet gas is saturated and in equilibrium with the outlet liquid. The wet-bulb temperature is the steady temperature reached by a wetted thermometer bulb exposed to the same gas stream with sufficient airflow. Under idealized conditions and Le ≈ 1, energy and mass balances show these temperatures coincide, leading to an essentially zero difference for engineering use. Step-by-Step Solution: Write the adiabatic saturation enthalpy balance: h_g,in = h_g,out + (m_l,out − m_l,in) * h_fg,adjusted.Write the wet-bulb energy balance on the wetted bulb surface: convective heat loss equals evaporative latent gain.Under Le ≈ 1 and ideal contact, the two balances yield the same surface temperature.Therefore, Twb − Tas ≈ 0 for unsaturated mixtures in standard practice. Verification / Alternative check: Psychrometric charts for air–water list wet-bulb lines that are nearly coincident with adiabatic saturation lines, confirming the equivalence within typical accuracy requirements. Why Other Options Are Wrong: A/B imply a consistent positive or negative bias; in reality, the ideal result is essentially zero. D/E are incorrect because these temperatures are well defined and comparable. Common Pitfalls: Confusing thermometer (sling) wet-bulb readings with thermodynamic wet-bulb temperature when ventilation is insufficient; real instruments can deviate from the ideal. Final Answer: zero

Question 19

Gas absorption design — The absorption factor (A) is defined in terms of slopes as (S1 = slope of the operating line, S2 = slope of the equilibrium curve). Which expression is correct?

Accepted Answer

Introduction: Design of packed and plate absorbers relies on graphical methods (McCabe–Thiele) and dimensionless groups. The absorption factor A links operating conditions to equilibrium behavior and influences the number of transfer units or theoretical stages required for a given separation. Given Data / Assumptions: Operating line slope S1 = L/G for molar flows of lean solvent and gas (on appropriate basis). Equilibrium relation y* = m x with slope S2 = m for dilute systems. Linear equilibrium approximation in the region of interest. Concept / Approach: The absorption factor is defined as A = L/(m G), which compares solvent flow to the product of gas flow and equilibrium slope. In slope notation, A = S1/S2. Large A indicates a strong solvent or high solvent rate (steeper operating line relative to equilibrium), generally reducing the number of stages at the expense of higher solvent circulation. Step-by-Step Solution: Start from A = L/(m G).Identify S1 = L/G and S2 = m.Substitute: A = (L/G)/m = S1/S2.Thus the correct expression is S1/S2. Verification / Alternative check: Stage-to-height relationships and Kremser equations use A explicitly; comparing slopes on the McCabe–Thiele diagram matches the S1/S2 form. Why Other Options Are Wrong: A inverts the factor; C and D are not dimensionally meaningful in this context; E is unrelated to established definitions. Common Pitfalls: Confusing slope of equilibrium with its inverse; always check the diagram axes and definitions to avoid inversion mistakes. Final Answer: S1/S2

Question 20

Distillation — Fenske’s equation for minimum number of theoretical stages (at total reflux) is valid under which condition?

Accepted Answer

Introduction: Fenske’s equation provides the minimum number of equilibrium stages required for a specified key-component split at total reflux. It is a powerful shortcut for preliminary column design when its assumptions are met. Recognizing those assumptions ensures correct application to binary and pseudo-binary separations. Given Data / Assumptions: Total reflux (no product withdrawal, maximum internal reflux). Constant or average relative volatility between the light and heavy key components. Constant molar overflow not required at total reflux but equilibrium behavior must be consistent. Concept / Approach: Under total reflux, the composition profile is governed by equilibrium alone. If the relative volatility α is reasonably constant from tray to tray, the separation factor per stage is constant, allowing the logarithmic form of the Fenske equation to predict stage count directly. Significant variation in α due to non-ideality or large temperature/composition swings undermines the equation’s accuracy and calls for rigorous methods (e.g., stage-by-stage simulation). Step-by-Step Solution: Define keys: light key (LK), heavy key (HK) with overall relative volatility α_LK,HK.Use Fenske: N_min = log[(x_D,HK/x_B,HK) * (x_B,LK/x_D,LK)] / log(α).Assume α ≈ constant → equation holds.Conclude that constancy of α is the central condition. Verification / Alternative check: Process simulators show good agreement between Fenske and rigorous calculations for near-ideal binaries with mild α variation; discrepancies grow as α varies strongly. Why Other Options Are Wrong: B/C point to deviations that often make α composition-dependent; D: Fenske can be extended to multicomponent using effective keys but the core assumption remains constant α; E: Feed condition is irrelevant at total reflux. Common Pitfalls: Applying Fenske away from total reflux or to strongly non-ideal systems without checking α variability. Final Answer: relative volatility is reasonably constant

Mass Transfer Questions