Question 1

Evaporator hydraulics — Increasing the liquor level (higher submergence) in an evaporator primarily leads to which operational change?

Accepted Answer

Introduction:Evaporator performance depends on available temperature driving force and overall heat-transfer coefficient. The liquor level (submergence) affects hydrostatic head at the heating surface, which in turn alters the local boiling temperature and effective temperature difference. Understanding this relation helps optimize capacity and economy.Given Data / Assumptions:Single-effect or multiple-effect evaporator with submerged heating surface.Boiling under a given shell-side steam pressure.Physical properties change modestly with level.Concept / Approach:A higher liquor column increases hydrostatic head at the heating surface. The saturation temperature at depth rises with pressure head, so the liquid boils at a higher temperature there. This reduces the effective temperature difference between the condensing steam and the boiling liquor at the surface, lowering the driving force for heat transfer and hence the evaporation rate (capacity). Any modest increase in wetted area or film coefficient typically cannot offset the loss in driving force.Step-by-Step Solution:Increase level → higher static head at the tube wall.Higher local boiling temperature → smaller ΔT against steam.Smaller ΔT → lower heat flux q = U * A * ΔT.Therefore, overall capacity decreases.Verification / Alternative check:Design charts show capacity dropping with increased submergence for vertical-tube and rising-film units beyond an optimum wetting level.Why Other Options Are Wrong:B: Film coefficient may change but the dominant effect is reduced ΔT. C: Hydrostatic head increases, not decreases. D: True temperature drop decreases, not increases. E: No practical condition yields infinite U; fouling and resistances remain.Common Pitfalls:Confusing increased wetting with improved performance; driving force reduction is usually controlling.Final Answer:decreased capacity

Question 2

Film condensation on a vertical surface — The average heat-transfer coefficient for laminar film condensation varies with the condensate film temperature drop ΔT as:

Accepted Answer

Introduction:Nusselt’s analysis for laminar film condensation on a vertical plate provides a closed-form scaling for the average heat-transfer coefficient. It reveals how physical properties and the temperature driving force influence condensate film thickness and heat transfer.Given Data / Assumptions:Steady, laminar, gravity-driven condensate film.Negligible vapor shear, isothermal wall, constant properties.Temperature drop across the liquid film is ΔT = Tsat − Twall.Concept / Approach:The classical result gives h_avg ∝ [ (ρ_l (ρ_l − ρ_v) g k_l^3 h_fg) / (μ_l L ΔT) ]^(1/4). Because ΔT appears in the denominator inside the brackets, the overall dependence is h ∝ (ΔT)^(−1/4), i.e., inversely proportional to ΔT raised to the one-fourth power. Larger ΔT slightly thickens the film (via increased condensate flow), which reduces h modestly in this laminar regime.Step-by-Step Solution:Write the proportionality from Nusselt theory with ΔT in the denominator.Take the one-fourth power → h ∝ (ΔT)^(−1/4).Interpret physically: more condensation increases film thickness, reducing h.Select the option stating '(ΔT)^(1/4) (inversely proportional)'.Verification / Alternative check:Comparisons with experimental data validate the weak inverse quarter-power dependence over the laminar range; deviations arise with vapor shear or turbulence in the film.Why Other Options Are Wrong:A/B/D predict much stronger inverse dependences than theory supports; E is incorrect since ΔT does influence h via the film flow rate.Common Pitfalls:Confusing condensation with convection where h often grows with ΔT; here, film dynamics impose the inverse relationship.Final Answer:(ΔT)^(1/4) (inversely proportional)

Question 3

Evaporators — Liquid entrainment (carry-over of droplets) from an evaporator most commonly results from which condition?

Accepted Answer

Introduction:Entrainment refers to liquid droplets being carried over with the vapor leaving an evaporator. This is undesirable because it contaminates condensate, causes product losses, and can foul downstream equipment. Understanding root causes enables better mechanical separation and operational control.Given Data / Assumptions:Boiling pool or film with rising vapor flow.Some solutions (e.g., detergents, sugars, proteins) tend to foam.Mechanical separators may or may not be present.Concept / Approach:Foaming forms stable bubbles with liquid films that can burst and project droplets upward or be swept out intact. This dramatically increases entrainment compared to non-foaming liquors at the same boil-up rate. While high throughputs can aggravate entrainment, foaming is the primary, characteristic cause across many services.Step-by-Step Solution:Identify foaming tendency as a key property of the feed.Link foam stability to droplet carry-over and mist formation.Conclude that foaming is the principal mechanism behind entrainment.Select 'foaming of the solution'.Verification / Alternative check:Plant experience uses antifoam agents, proper disengagement height, and demisters specifically to combat entrainment caused by foaming feeds.Why Other Options Are Wrong:A: High vacuum alone does not cause entrainment; it may even reduce boil-up temperature. B and D: High rates can contribute but are not the primary, characteristic cause compared to foaming. E: Demisters reduce, not cause, entrainment.Common Pitfalls:Confusing vapor load effects with foam stability; both matter, but foaming is the hallmark driver.Final Answer:foaming of the solution

Question 4

Multiple-effect evaporators — Why is backward-feed arrangement preferred over forward feed for concentrating a cold feed?

Accepted Answer

Introduction:In multiple-effect evaporation, the arrangement of effects (forward, backward, mixed) influences temperature driving forces, pumping requirements, economy, and capacity. For cold feeds, backward feed often offers thermodynamic advantages despite added mechanical complexity.Given Data / Assumptions:Cold feed entering the last (lowest-temperature) effect in backward feed.Vapor from a higher-temperature effect heats the next lower-temperature effect.Comparable total heating surface and operating pressures for both schemes.Concept / Approach:Backward feed exposes the most concentrated, viscous liquor to the highest temperature difference (first effect), improving heat transfer under difficult conditions and enhancing steam economy for cold feeds. However, because pressure increases from later to earlier effects, liquor must be pumped from low to high pressure, increasing hydraulic work and potentially reducing throughput (capacity) for a given pump set compared with gravity-aided forward feed.Step-by-Step Solution:Cold feed enters the coldest effect → preheats progressively without external heaters.Concentrated liquor meets the hottest surface in the first effect → better heat transfer and economy.Pumping against pressure between effects is required → added head limits flow, lowering capacity.Therefore, both higher economy and lower capacity apply → choose 'Both (a) and (b).'Verification / Alternative check:Design guides list backward feed as favorable for cold feeds and viscous products; forward feed is simpler and often higher in capacity for hot feeds.Why Other Options Are Wrong:A alone ignores the capacity penalty; B alone ignores the economy benefit; D is opposite; E is false because pumps are typically needed in backward feed.Common Pitfalls:Assuming one arrangement dominates in all metrics; selection depends on feed temperature, viscosity profile, and utility costs.Final Answer:Both (a) and (b).

Question 5

Pick the wrong statement — Identify the incorrect assertion among the following heat-transfer/phase-change statements.

Accepted Answer

Introduction:Quick conceptual checks are common in exams to ensure familiarity with definitions and order-of-magnitude criteria in heat transfer and separation processes. Hydraulic (equivalent) diameter, convection criteria, and distinctions between unit operations are foundational ideas.Given Data / Assumptions:Hydraulic (equivalent) diameter Dh = 4A/P for internal flows.Forced vs. natural convection dominance gauged by Gr/Re^2.Evaporation vs. distillation distinction is understood.Concept / Approach:For a square duct with side x, area A = x^2 and wetted perimeter P = 4x, hence Dh = 4A/P = 4x^2/(4x) = x. Therefore, stating Dh = 4x is incorrect. The criterion Gr/Re^2 indicates whether buoyancy (natural convection) competes with or dominates forced convection; small values imply forced convection dominance. Distillation involves vapor–liquid equilibrium separation by relative volatility, distinct from simple evaporation (removing solvent).Step-by-Step Solution:Compute Dh for a square: Dh = 4A/P = x.Compare with statement (b): claims Dh = 4x → wrong.Confirm (a) and (e) reflect standard Gr/Re^2 guidance.Confirm (c) and (d) are accepted truths in unit operations and boiling.Verification / Alternative check:Textbook examples routinely derive Dh = x for a square duct; published correlations use Dh in Reynolds/Nusselt numbers accordingly.Why Other Options Are Wrong:Here 'wrong' applies only to (b). Options (a), (c), and (d) are correct statements; (e) is a reinforcing truth explaining (a).Common Pitfalls:Mixing up Dh with the actual diameter in circular pipes; for non-circular passages always use Dh = 4A/P.Final Answer:For a square duct of side x, the equivalent diameter for heat transfer equals 4x.

Question 6

Thermodynamics – Reversible devices operating between the same two reservoirs A reversible heat engine operates between a hot and a cold thermal reservoir and has a thermal efficiency of 0.40. If the same temperature limits are used to run a reversi…

Accepted Answer

Introduction / Context:This problem tests core relations among reversible heat engines, refrigerators, and heat pumps operating between the same two temperature reservoirs. For Carnot (reversible) devices, efficiency and coefficients of performance are dictated solely by the temperature ratio, so once one quantity is known, all others follow deterministically.Given Data / Assumptions:Reversible (Carnot) behavior between the same hot and cold reservoirs.Thermal efficiency of the heat engine, eta = 0.40.No irreversibilities or additional losses; all devices see identical Th and Tc.Concept / Approach:For a reversible heat engine: eta = 1 - Tc/Th. For a reversible refrigerator: COP_R = Tc/(Th - Tc). For a reversible heat pump: COP_HP = Th/(Th - Tc). These can be written in terms of the temperature ratio r = Th/Tc. Using eta, we compute r, then evaluate COP_R and COP_HP directly.Step-by-Step Solution:1) From eta = 1 - Tc/Th = 0.40 ⇒ Tc/Th = 0.60.2) Hence r = Th/Tc = 1 / 0.60 = 1.6667.3) COP_R = Tc/(Th - Tc) = 1/(r - 1) = 1/(1.6667 - 1) = 1/0.6667 ≈ 1.5.4) COP_HP = Th/(Th - Tc) = r/(r - 1) = 1.6667/0.6667 ≈ 2.5.5) Therefore, (COP)R = 1.5 and (COP)HP = 2.5.Verification / Alternative check:For any reversible cycle between fixed reservoirs, COP_HP = COP_R + 1. Since COP_R ≈ 1.5, COP_HP should be ≈ 2.5, confirming consistency.Why Other Options Are Wrong:(COP)R = (COP)HP = 0.6: Confuses efficiency with COP and violates COP_HP = COP_R + 1.(2.5; 1.5): Reverses refrigerator and heat pump values.(Equal 2.5): COPs cannot be equal; heat pump COP must exceed refrigerator COP by 1.Common Pitfalls:Mixing up efficiency (engine) with COP (refrigerator/heat pump), or forgetting the identity COP_HP = COP_R + 1 for the same temperature limits.Final Answer:(COP)R = 1.5; (COP)HP = 2.5

Question 7

Evaporation fundamentals – Relating capacity and economy to steam consumption In a single- or multiple-effect evaporator, the steam consumption in kg per hour is expressed in terms of the evaporator capacity C (kg of water evaporated per hour) and t…

Accepted Answer

Introduction / Context:Evaporator performance is often summarized by capacity and economy. Capacity tells how much water is evaporated per hour, whereas economy captures how effectively steam is used (how many kilograms of water removed per kilogram of steam supplied). Converting between these defines the steam consumption, critical for utility sizing and cost analysis.Given Data / Assumptions:C = capacity = kg water evaporated per hour.E = economy = kg water evaporated per kg steam.Steady operation; negligible heat leaks for the relation.Concept / Approach:By definition, economy E = (water evaporated) / (steam used). Rearranging yields steam used = (water evaporated) / E. In rate form, steam consumption (kg/h) = C / E. This relation is independent of the number of effects and directly ties utility usage to throughput and economy.Step-by-Step Solution:Define E = (kg water evaporated) / (kg steam used).Rearrange: kg steam used = (kg water evaporated) / E.Express as rates: Steam consumption (kg/h) = C / E.Check units: (kg/h)/(kg/kg) = kg/h, consistent.Select the option C/E.Verification / Alternative check:If E = 1 (single effect ideal), steam consumption equals capacity, as expected. If E increases, required steam per hour decreases proportionally.Why Other Options Are Wrong:E/C: Inverts the relationship and has incorrect units.C E: Multiplies rather than divides, grossly overpredicting steam usage.E - C: Nonsensical dimensionally.Common Pitfalls:Confusing economy with efficiency; economy is a ratio tied to mass balance, not a percentage.Final Answer:C/E

Question 8

Heat conduction in solids – Dependence of thermal conductivity For a conducting solid material (e.g., metals, ceramics, porous refractories), thermal conductivity may vary with which intrinsic or structural factors? Identify the best choice.

Accepted Answer

Introduction / Context:Thermal conductivity k is not necessarily constant. In real solids, k often depends on temperature (phonon and electron transport effects) and on microstructure, including porosity. Recognizing these dependencies is important for furnace linings, insulation, and high-temperature equipment design.Given Data / Assumptions:Homogeneous solid whose microstructure may include pores.Steady-state, one-dimensional conduction concepts apply.Temperature range may affect carriers (phonons/electrons).Concept / Approach:In metals, electron scattering increases with temperature, often reducing k at high T; in many ceramics, phonon scattering changes so k can decrease with T over practical ranges. Porosity introduces low-conductivity gas-filled regions and tortuous paths, decreasing effective k. Therefore, both temperature and porosity influence k in practice.Step-by-Step Solution:Assess temperature dependence of carriers (electrons/phonons) ⇒ k varies with T.Consider microstructure; pores interrupt heat paths ⇒ lower effective k.Conclude that both temperature and porosity matter.Choose the combined option (a) & (b).Apply this understanding in selecting materials for insulation vs heat sinks.Verification / Alternative check:Handbooks list k(T) curves; porous firebricks show order-of-magnitude lower k than dense ceramics at the same temperature.Why Other Options Are Wrong:Single-factor options ignore well-documented dependencies.“Neither” contradicts extensive experimental data.Common Pitfalls:Using constant-k assumptions outside narrow temperature ranges; for accuracy, incorporate k(T) and porosity corrections in design calculations.Final Answer:both (a) & (b)

Question 9

Multiple-effect evaporators – Factors affecting steam economy For a given number of effects, which listed factor does not significantly influence the steam economy (kg water evaporated per kg of steam) of a multiple-effect evaporator?

Accepted Answer

Introduction / Context:Steam economy, the hallmark performance metric for multiple-effect evaporators, measures how effectively each kilogram of steam drives evaporation across effects. Some parameters alter economy directly, while others mainly affect equipment size or area without much changing economy itself.Given Data / Assumptions:Number of effects fixed; forward/backward feed arrangement chosen.BPE (boiling point elevation) reduces available temperature driving force per effect.Feed condition (cold vs hot) changes how much sensible heat must be supplied.Concept / Approach:Steam economy primarily depends on the number of effects and total available temperature drop after accounting for BPE and feed heating requirements. The overall rate of heat transfer (through UA) influences required area and duty distribution, but with enough area the same economy can be reached; it does not fundamentally change the kilograms evaporated per kilogram of steam for a fixed configuration and steady operation.Step-by-Step Solution:Acknowledge BPE reduces effective ΔT across effects ⇒ lowers economy.Recognize colder feed consumes part of the first-effect duty ⇒ can reduce apparent economy.Note rate of heat transfer sets equipment size/time to reach duty, not economy at steady state.Conclude that “rate of heat transfer” has the least direct influence on economy.Select option accordingly.Verification / Alternative check:Design texts separate thermodynamic economy (governed by effects and ΔT availability) from economic sizing (UA). Increasing area raises throughput but not kg water per kg steam fundamentally.Why Other Options Are Wrong:BPE: Directly reduces ΔT per effect, lowering economy.Feed temperature: Cold feed increases steam use for sensible heating.Thin/thick liquor ratio (concentration factor): Influences sensible loads distribution and flashing, affecting economy.Common Pitfalls:Confusing capacity (throughput) with economy; adding area increases capacity but not the inherent economy for a set effect count and ΔT.Final Answer:rate of heat transfer

Question 10

Cost scaling in shell-and-tube exchangers – Purchased cost per unit area For a shell-and-tube heat exchanger, how does the purchased cost per unit heat-transfer area typically vary as the total area is increased (all else equal)?

Accepted Answer

Introduction / Context:Capital cost estimation often uses scaling laws. In heat exchanger design, understanding how the purchased cost per square meter changes with size influences selection between one large unit versus multiple smaller units and helps optimize total installed cost.Given Data / Assumptions:Similar design and materials as size scales up.No exotic constraints (e.g., pressure extremes) that would change scaling.Market conditions aside, typical vendor pricing behavior applies.Concept / Approach:Equipment exhibits economies of scale: total purchased cost increases sublinearly with area. Therefore, the purchased cost per unit area generally decreases as area increases. This reflects fixed costs (heads, channels) being spread over more tubes and larger shells, reducing cost per square meter of heat transfer surface.Step-by-Step Solution:Use typical cost law: Cost ∝ Area^n with n Compute unit cost: Cost/Area ∝ Area^(n-1), which decreases as Area grows.Interpretation: larger exchangers are more cost-efficient per m^2.Select “decreases.”Note that installed cost considerations (foundations, cranes) may modify total project cost.Verification / Alternative check:Vendor quotes and cost indices show diminishing /m^2 as exchanger surface area increases within a product family.Why Other Options Are Wrong:Increases/constant: Contradict typical economies of scale for standard designs.Passes through a maxima: Not characteristic of purchased cost per area; unusual.Common Pitfalls:Confusing purchased cost with installed or lifecycle costs; piping, plot space, and maintenance can alter the economic optimum.Final Answer:decreases

Question 11

Convective heat transfer inside tubes – Location of thermal resistance When a hot tube wall transfers heat to a fluid flowing inside, where is the bulk of the convective thermal resistance located?

Accepted Answer

Introduction / Context:Convective heat transfer involves fluid motion and thermal gradients. The velocity and temperature boundary layers near a wall strongly influence the overall resistance, which informs why surface enhancements (fins, roughness, inserts) and high turbulence intensities are used to improve heat transfer in exchangers and process tubing.Given Data / Assumptions:Single-phase flow inside a tube with a hot wall and cooler fluid.Fully developed behavior conceptually; no phase change.Newtonian fluid; standard Prandtl numbers.Concept / Approach:Thermal and hydrodynamic boundary layers form at the wall. Most of the temperature drop occurs across the near-wall fluid layer where conduction dominates due to suppressed turbulence and low relative velocity. Thus, the dominant thermal resistance resides in the thin film adjacent to the surface, not in the well-mixed core.Step-by-Step Solution:Recognize boundary layer formation: steep gradients near wall.Heat crosses the wall and then the near-wall fluid film primarily by conduction.Core region is relatively well mixed with smaller gradients.Therefore, most resistance is near the wall within a thin film.Select the corresponding option.Verification / Alternative check:Classic correlations for h depend on near-wall behavior (e.g., friction factor analogy) and surface roughness effects, underscoring the dominance of the wall-adjacent film.Why Other Options Are Wrong:Core/uniform distribution: Do not reflect the steep near-wall gradients observed experimentally and theoretically.None of these: Incorrect as the thin film description is standard.Common Pitfalls:Ignoring fouling layers; while separate from fluid film, deposits further increase effective near-wall resistance.Final Answer:mainly confined to a thin film of fluid near the surface.

Question 12

Shell-and-tube heat exchangers – Purpose of shell-side baffles What is the principal effect of installing baffles on the shell side of a shell-and-tube heat exchanger?

Accepted Answer

Introduction / Context:Baffles are common internals in shell-and-tube exchangers. Their geometry and spacing critically impact shell-side hydrodynamics, pressure drop, and heat transfer. Understanding their role guides trade-offs between performance and pumping cost.Given Data / Assumptions:Single-phase shell-side flow; no phase change considered.Reasonable baffle cut and spacing practices.Goal: enhance heat transfer without excessive vibration or bypassing.Concept / Approach:Baffles direct shell-side flow across the tube bundle, creating crossflow and local turbulence, interrupting boundary layers, and reducing bypass streams. This increases the shell-side heat transfer coefficient h_s. It does not increase cross-sectional flow area; rather, it constrains flow path and raises velocity, typically increasing pressure drop as a trade-off.Step-by-Step Solution:Add baffles ⇒ induce crossflow and turbulence.Crossflow interrupts thermal boundary layers ⇒ higher h_s.Parallel flow is minimized; leakage/bypass controlled with proper sealing strips.Therefore, the main effect is increased shell-side heat transfer coefficient.Select option (c).Verification / Alternative check:Bell–Delaware methods quantify h_s and pressure drop improvements due to baffles and associated correction factors for leakage and bypass streams.Why Other Options Are Wrong:Increase cross-section: Opposite; effective flow area is restricted.Force parallel flow: Incorrect; baffles create crossflow.Decrease h_s: Contradicts the purpose of baffles.Common Pitfalls:Over-tight spacing causing high pressure drop and tube vibration; optimal design balances heat transfer gains and mechanical limits.Final Answer:increase the shell side heat transfer co-efficient.

Question 13

Evaporators – Actual temperature drop across the heating surface The actual temperature drop across the heating surface of an evaporator is influenced by several practical factors. Which combination best represents the dependencies?

Accepted Answer

Introduction / Context:Design correlations often assume an available temperature difference based on steam and vapor pressures. In real evaporators, the effective temperature drop across the heating surface also depends on hydrodynamics, hydrostatic head, and operating pressures, which together alter boiling temperatures and driving forces.Given Data / Assumptions:Boiling occurs on the liquor side; condensing steam supplies heat.Feed condition can vary (cold, preheated, viscous).Liquid depth and pressure profiles are not negligible.Concept / Approach:The liquor boiling temperature rises with hydrostatic head (deeper liquid layers). Feed condition affects sensible heating requirement and local boiling point. The pressure difference between the steam chest and vapor space sets saturation temperatures on each side, thus directly changing ΔT available. Consequently, all listed factors contribute to the actual temperature drop across the heating surface.Step-by-Step Solution:Account for feed condition altering local temperature and viscosity.Recognize deeper liquid ⇒ higher static head ⇒ higher local boiling temperature.Relate steam chest/vapor space pressures to their saturation temperatures.Combine effects to determine actual ΔT across the surface.Therefore, all (a), (b), and (c) apply.Verification / Alternative check:Operating data show ΔT varies along the calandria; shallow pools and reduced pressure drop increase effective driving force.Why Other Options Are Wrong:Each single-factor choice omits other equally important influences, giving an incomplete picture of ΔT determination.Common Pitfalls:Using nominal steam–vapor temperature differences without correcting for BPE and hydrostatic head can overpredict capacity.Final Answer:all (a), (b) and (c).

Question 14

Selecting a heat exchanger to heat air using condensing steam – Best arrangement Air at low heat transfer coefficient must be heated by condensing steam. Two exchanger types are available with equal tube-side area: (i) shell-and-tube and (ii) finned…

Accepted Answer

Introduction / Context:When one fluid has a very low convective coefficient (air) and the other a very high coefficient (condensing steam), surface augmentation should be placed on the low-h side to balance overall resistance. Finned tubes are a standard solution for gas heating/cooling with condensing or high-h liquids on the opposite side.Given Data / Assumptions:Air has a low film coefficient; condensing steam has a very high film coefficient.Tube-side areas are equal between the two choices.Fins greatly increase external area and turbulence for the gas side.Concept / Approach:Overall resistance R_total ≈ R_air + R_wall + R_steam. Because R_steam is very small for condensation, the dominant resistance is on the air side. Therefore, fins must be on the air side. Practically, this means steam should be on the tube side (smooth side), while air flows across the finned outside surfaces, maximizing effective area and reducing the controlling resistance.Step-by-Step Solution:Identify controlling resistance: air side.Place fins on the air side to increase area and disturbance.Arrange steam inside tubes (no need for fins) and air outside over fins.Select the finned-tube exchanger with air outside and steam inside.Confirm tube-side areas are equal; finning compensates on shell/outside.Verification / Alternative check:Design handbooks recommend steam-in-tube and air-over-fins for air heaters, economizers, and HVAC coils due to the low air-side h.Why Other Options Are Wrong:Air inside, steam outside: Puts fins on the wrong side or underutilizes them.Shell-and-tube options: Without fins on the air side, overall U will be much lower for a given area.Common Pitfalls:Adding fins on the condensing side yields little benefit and complicates drainage; always augment the low-h side.Final Answer:finned tube heat exchanger with air outside and steam inside.

Question 15

Units and dimensions – Heat transfer coefficient in SI units Select the correct SI unit for the (overall) heat transfer coefficient h used in convection and overall heat transfer calculations.

Accepted Answer

Introduction / Context:The heat transfer coefficient h relates heat flux to a temperature difference at a surface. Using correct SI units avoids dimensional mistakes in design equations such as qʺ = h * (T_s - T_∞) and in overall heat transfer calculations for exchangers (Q = U * A * ΔT_lm).Given Data / Assumptions:Heat flux qʺ has units W/m^2.Temperature difference ΔT has units K.h is defined by qʺ = h * ΔT.Concept / Approach:From qʺ = h * ΔT, rearrange to h = qʺ / ΔT. Since qʺ is W/m^2 and ΔT is K, h must be W/m^2·K. This unit is also used for overall heat transfer coefficient U in exchanger analysis.Step-by-Step Solution:Start with qʺ = h (T_s - T_∞).Assign units: qʺ → W/m^2; ΔT → K.Compute h units: (W/m^2)/K = W/m^2·K.Confirm dimensional consistency in exchanger formulas: Q = U A ΔT_lm ⇒ U must be W/m^2·K.Select W/m^2·K.Verification / Alternative check:Textbooks and standards uniformly list h and U in W·m^-2·K^-1.Why Other Options Are Wrong:J/m^2·K or J/m·K: Energy units, not power; would imply time integration.W/m·K: Unit for thermal conductivity k, not h.Common Pitfalls:Confusing h with k; k appears in Fourier’s law (q = -k A dT/dx) with units W/m·K, whereas h multiplies a temperature difference per unit area.Final Answer:W/m^2·K

Question 16

In convective heat transfer across a composite resistance (e.g., a hot fluid–wall–cold fluid system), the “controlling” heat transfer film coefficient is associated with the layer that offers the greatest hindrance.\r Which film or side is referred …

Accepted Answer

Introduction:Whenever heat flows through several resistances in series (for example, a hot fluid film, a wall, and a cold fluid film), the overall rate is governed by the largest resistance in the path. In heat exchanger design this idea helps engineers focus on the “controlling” side for enhancement efforts such as adding fins, increasing velocity, or changing flow regimes.Given Data / Assumptions:Multiple resistances exist in series: hot-side film, wall conduction, and cold-side film.Film coefficients are inversely related to film resistances.We are identifying the controlling resistance and its qualitative magnitude.Concept / Approach:Overall heat transfer coefficient U for series resistances is determined by 1/U = R_total = R_hot film + R_wall + R_cold film. The largest of these dominates the total. A “controlling film coefficient” means the film with the lowest h (largest 1/h) and hence the greatest resistance, which throttles the heat flow.Step-by-Step Solution:Write total resistance: R_total = 1/h_hot + (δ/k) + 1/h_cold.Identify which term is largest; that term controls R_total and thus U.Translate to wording: the controlling film offers the largest resistance.Verification / Alternative check:Electrical analogy (series resistors) shows current is limited by the largest resistor; likewise, heat flow is limited by the largest thermal resistance.Why Other Options Are Wrong:No / negligible / lower / least: These imply small resistance, which would not control the overall process.Common Pitfalls:Confusing “largest coefficient” with “largest resistance.” Remember resistance is 1/h; a small h means a large resistance and hence controlling.Final Answer:the largest

Question 17

In pulp and paper manufacturing, the spent cooking liquor (“black liquor”) must be concentrated before recovery boiler firing.\r Which evaporator arrangement is conventionally used to concentrate black liquor efficiently?

Accepted Answer

Introduction:Black liquor from kraft pulping contains dissolved lignin, hemicelluloses, and inorganic chemicals. Before chemical recovery, it must be concentrated from low solids to much higher solids. Energy efficiency is crucial because evaporation is steam-intensive.Given Data / Assumptions:Feed is dilute black liquor requiring concentration.Target is an energy-efficient, industry-standard method.No crystallization step is routinely required for black liquor concentration.Concept / Approach:Multiple-effect evaporation reuses vapor from one effect as the heating medium for the next, dramatically cutting steam consumption relative to a single effect. Black liquor plants typically use several effects (often 5–7) in forward, backward, or mixed feed configurations, sometimes with falling film or forced-circulation effects to handle viscosity rise at high solids.Step-by-Step Solution:Identify energy goal: minimize steam per kg of evaporation.Compare options: single effect wastes latent heat; multiple effects cascade it.Select the standard: multiple effect evaporator trains are conventional.Verification / Alternative check:Mill process flow diagrams universally show multi-effect evaporator sets upstream of concentrators and recovery boilers.Why Other Options Are Wrong:Single effect (with or without crystallizer): Uneconomical and crystallization is not a normal step for black liquor.Crystalliser options: Black liquor solids are concentrated, not crystallized, prior to firing.Common Pitfalls:Ignoring viscosity and fouling; liquor properties dictate effect type and temperature profile but not the multi-effect principle.Final Answer:multiple effect evaporator.

Question 18

Leidenfrost phenomenon in boiling heat transfer: the “Leidenfrost point” refers to which process condition?

Accepted Answer

Introduction:The Leidenfrost point is a hallmark of boiling heat transfer behavior at very high surface temperatures. Above this temperature, liquid droplets levitate on a stable vapor layer, dramatically altering heat transfer mechanisms and visual appearance.Given Data / Assumptions:A hot solid surface is in contact with a cooler liquid.Surface temperature can exceed the liquid’s saturation temperature significantly.Boiling regimes transition from nucleate to transition and then to film boiling.Concept / Approach:When a surface is sufficiently hotter than the liquid, vigorous nucleate boiling gives way to unstable transition boiling and ultimately to film boiling. The Leidenfrost point marks the onset of a stable vapor film that insulates the liquid from the surface, leading to droplet “skating” and greatly reduced heat transfer coefficients compared with nucleate boiling.Step-by-Step Solution:Recognize the regime: very hot surface, droplet levitation observed.Identify the definition: the temperature at which a continuous vapor film forms beneath the liquid.Associate this strictly with boiling at a hot surface, not condensation or mixing.Verification / Alternative check:Classical boiling curves show a minimum in heat flux near the Leidenfrost region; beyond this point, heat transfer occurs through a vapor layer (film boiling).Why Other Options Are Wrong:Condensation or evaporation of corrosives: Not related specifically to Leidenfrost behavior.Viscous liquid heat transfer and natural convection: Different phenomena.Common Pitfalls:Assuming higher surface temperature always increases heat transfer; in film boiling, heat transfer can actually decrease due to insulation by vapor.Final Answer:boiling of a liquid on a hot surface.

Question 19

Reynolds analogy (simplest form) for turbulent forced convection over surfaces relates momentum and heat transfer.\r Which proportionality captures the essence of the analogy under its ideal assumptions?

Accepted Answer

Introduction:Reynolds analogy provides a bridge between momentum and heat transfer in turbulent boundary layers. Under simplifying assumptions (notably Prandtl number near unity and negligible pressure gradients), the analogy connects friction factor with Stanton number, enabling heat-transfer predictions from friction data.Given Data / Assumptions:High-Reynolds-number turbulent flow.Prandtl number approximately 1.No form drag and negligible pressure gradients.Concept / Approach:The simplest Reynolds analogy states: St = f/2, where St is the Stanton number (dimensionless heat transfer coefficient) and f is the Fanning friction factor. In proportional terms, St ∝ f, matching option (a). Extended analogies (Chilton–Colburn) generalize this to St * Pr^(2/3) = f/2 for a wider range of Pr.Step-by-Step Solution:Recall definition: St = h/(ρ * u * Cp).Apply the simple analogy: St = f/2 ⇒ St ∝ f.Map to offered statements: Nst α f aligns; other pairings are incorrect.Verification / Alternative check:Using friction-factor correlations, one can estimate heat transfer where direct measurement is hard, provided Pr ~ 1 and surface conditions are comparable.Why Other Options Are Wrong:Nst α NRe or NNu α f: These do not reflect the canonical simple-analogy relationship.NRe α f: Friction factor scales differently with Re; they are not simply proportional.Common Pitfalls:Applying the analogy far from its validity (very high or low Pr). Use Chilton–Colburn j-factors for broader applicability.Final Answer:Nst α f

Question 20

Heat exchanger analysis: identify the WRONG statement among the following commonly cited facts about LMTD, phase change, and controlling resistance during condensation heating of air.

Accepted Answer

Introduction:Thermal design relies on proper interpretation of temperature driving forces and resistances. Misunderstanding LMTD behavior or where the dominant resistance lies can lead to oversized or underperforming exchangers. This question probes common truths and one deliberate misconception.Given Data / Assumptions:Condensing steam heating air: very high steam-side h, low air-side h.LMTD comparison for counterflow vs. parallel flow with one fluid at constant temperature (phase change).LMTD correction factor F behavior in 1–2 shell-and-tube units when temperature cross occurs.Concept / Approach:(a) is correct: the air film usually controls because condensing steam has a very high film coefficient. (b) is correct: if one stream changes phase at constant temperature (e.g., saturated steam), the LMTD for parallel and counter arrangements can be identical for the same terminal temperatures. (d) is correct: pure-fluid phase change at fixed pressure occurs at the saturation temperature. The problematic statement is (c): when a temperature cross occurs, the correction factor F does not “increase sharply”; it typically drops, approaching zero as the temperature cross becomes severe, indicating an ineffective arrangement for the specified terminal temperatures.Step-by-Step Solution:Assess each statement with standard heat transfer principles.Identify (c) as contrary to the known trend of F decreasing with temperature cross.Choose (c) as the wrong statement.Verification / Alternative check:Design charts for F versus temperature ratios clearly show F decreasing when temperature cross is approached or exceeded.Why Other Options Are Wrong:(a), (b), and (d) are standard and consistent with textbooks.(e) “None” is invalid because (c) is indeed wrong.Common Pitfalls:Confusing large LMTD with high effectiveness; shell arrangements need both adequate F and feasible terminal temperatures.Final Answer:In case of a 1 - 2 shell and tube heat exchanger, the LMTD correction factor value increases sharply, when a temperature cross occurs.

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