Question 1

Boiler Classification — Benson Boiler Pressure Level Benson boiler is a __________ water-tube steam boiler that employs forced circulation (once-through) without a steam drum. Choose the correct pressure range classification.

Accepted Answer

Introduction:The Benson boiler is a once-through, forced-circulation boiler widely cited for supercritical and high-pressure operation. Understanding its classification clarifies why it omits a steam drum and how it achieves high efficiency with sliding pressure control.Given Data / Assumptions:Water-tube, once-through design; feedwater flows through economiser, evaporator, and superheater in series.Forced circulation by pumps; no steam drum in supercritical versions.Designed for high and supercritical pressures.Concept / Approach:“High pressure” boilers operate far above simple shell-boiler pressures. The Benson concept is synonymous with high-pressure and, in many installations, supercritical operation where phase change occurs without a distinct boiling interface. Hence, it is correctly classified as a high-pressure water-tube boiler using forced circulation.Step-by-Step Solution:Identify design: once-through path with pumps ⇒ forced circulation.Recognize pressure regime: designed for very high to supercritical pressures.Therefore, select “high pressure”.Verification / Alternative check:Plant engineering references list Benson as a canonical high-pressure, once-through boiler, contrasting with drum-type subcritical units.Why Other Options Are Wrong:Low pressure or vacuum: contrary to intended service.“Subcritical only”: many Benson units are explicitly supercritical.Common Pitfalls:Confusing forced circulation (Benson) with natural circulation (e.g., Babcock & Wilcox) and assuming a steam drum is universal.Final Answer:high pressure

Question 2

Steam Plant Basics — Definition of a Steam Boiler A closed vessel constructed of steel and used for the generation of steam by applying heat to water is known as a:

Accepted Answer

Introduction:Clear terminology prevents design and operation errors in steam plants. The component that converts the chemical energy of fuel into thermal energy of steam by heating water in a pressurized, steel enclosure is the boiler.Given Data / Assumptions:Enclosed, pressure-retaining steel vessel.Receives water and produces steam.External heat source via combustion or waste heat.Concept / Approach:A steam boiler is distinct from the turbine (which converts steam energy to mechanical work), the condenser (which condenses exhaust steam back to water), and injectors (which use steam jet to feed water). The definition focuses on the pressure vessel function and steam generation process.Step-by-Step Solution:Identify function: generate steam within a closed, pressurized vessel.Map to plant components: boiler matches this; turbine, condenser, injector have different roles.Therefore, choose “steam boiler”.Verification / Alternative check:Standards and codes (e.g., boiler regulations) define boilers as pressure vessels for steam generation, confirming the terminology.Why Other Options Are Wrong:Steam turbine: energy conversion from steam to shaft power, not generation.Condenser: recovers condensate by heat rejection.Injector: feeds water using a steam jet, not a vessel for steam production.Common Pitfalls:Using “boiler” generically for the entire steam plant rather than the specific pressure vessel and heating surfaces.Final Answer:steam boiler

Question 3

Compound Engines — Effective Crank Relationship in Tandem Type In a tandem-type compound steam engine, the high-pressure (HP) and low-pressure (LP) cylinders are aligned on the same piston rod and act on the same crank. The effective crank positions…

Accepted Answer

Introduction:Compound engines can be arranged as cross-compound (separate cranks) or tandem (same crank). The geometric relation between crank positions determines torque phasing and influences vibration and flywheel sizing. This item asks about the special case of tandem compounding.Given Data / Assumptions:Tandem compound configuration with a common piston rod.Both cylinders drive the same crank pin.Standard double-acting steam cycle.Concept / Approach:If both cylinders act on the same crank, their effective crank angles are coincident. Hence the phase difference is 0°. This contrasts with cross-compound engines where the cranks are often set 90° apart to even out turning moment.Step-by-Step Solution:Note mechanical link: one crank pin shared.Deduce geometry: both piston motions reference the same angular position.Therefore, effective crank positions are 0° apart.Verification / Alternative check:Mechanism schematics for tandem compounds show the HP cylinder bolted to the LP cylinder in line, sharing rod and crosshead, which necessarily implies the same crank phasing.Why Other Options Are Wrong:90° or 180°: applicable to multi-crank layouts, not tandem.None: unnecessary since the correct relation is definitive.Common Pitfalls:Confusing tandem compounding with cross-compounding where phase angles are chosen to smooth torque.Final Answer:0° to each other

Question 4

Defining a “Good” Steam Boiler — Steam Output for Given Fuel State whether the following statement is accurate: “A good steam boiler is one which produces the maximum quantity of steam for a given amount of fuel.”

Accepted Answer

Introduction:At its core, boiler performance is about converting fuel energy into steam energy efficiently and reliably. A concise way to express this is to maximize steam output (or heat to steam) per unit of fuel, which correlates with high thermal efficiency and effective heat recovery.Given Data / Assumptions:Comparable steam conditions (pressure/temperature) when comparing outputs.Boiler operates safely within design limits.Quality of steam (dryness/superheat) is maintained.Concept / Approach:For a fixed fuel input, higher useful heat absorbed by water/steam means higher efficiency. Practical designs achieve this via adequate heating surface, good circulation, low excess air, economisers/air preheaters, and minimized stack and radiation losses. Thus, “maximum steam for given fuel” sensibly captures efficiency as a key attribute of a “good” boiler.Step-by-Step Solution:Tie statement to efficiency: η_boiler ≈ useful heat to steam / fuel heat input.For constant firing rate, larger useful heat implies more steam generation.Therefore, the statement aligns with maximizing η and is acceptable.Verification / Alternative check:Heat balance diagrams show that reducing stack temperature and optimizing excess air directly increase steam generation per unit of fuel, supporting the definition.Why Other Options Are Wrong:Disagree: would imply efficiency is not a primary measure, which contradicts standard practice.Load- or fuel-specific qualifiers are unnecessary; the principle applies broadly.Common Pitfalls:Ignoring steam condition differences; equal kilograms of low-pressure saturated steam are not energetically identical to high-pressure superheated steam—comparisons must be like-for-like.Final Answer:Agree

Question 5

Locomotive Boiler — Correct Classification by Construction and Service Identify the correct description of a locomotive boiler from the options below.

Accepted Answer

Introduction:Locomotive boilers powered rail transport for decades and remain classic examples in boiler textbooks. Knowing their construction features helps distinguish them from stationary plant units and clarifies why they were effective in mobile service.Given Data / Assumptions:Fire-tube design with numerous small-diameter tubes (multitubular).Horizontal orientation to fit locomotive frames.Internal firebox (internally fired) within the boiler shell.Mobile application mounted on locomotive chassis.Concept / Approach:A locomotive boiler is a horizontal, multitubular, internally fired, mobile fire-tube boiler. The many tubes provide large heating surface per unit length, while the internal firebox allows compact integration with the engine and draft control via blastpipe and chimney.Step-by-Step Solution:Check tube arrangement: many small tubes ⇒ multitubular.Check firing: firebox inside the shell ⇒ internally fired.Check service: locomotive ⇒ mobile.Therefore, option C matches all attributes.Verification / Alternative check:Historical locomotive boiler diagrams confirm the internal firebox, tube nest, and compact horizontal layout optimized for mobile steam generation.Why Other Options Are Wrong:Single-tube or vertical: incorrect for locomotive boilers.Externally fired/stationary: do not describe locomotives.Common Pitfalls:Confusing marine scotch boilers (also fire-tube) with locomotive types; the latter are specifically mobile with distinct draughting arrangements.Final Answer:multitubular, horizontal, internally fired and mobile boiler

Question 6

Chimney Draught — Exit Velocity Under a Given Static Draught Head For a chimney providing a static draught head of H' metres (effective head), the theoretical velocity V of flue gases at the chimney exit is given by which expression?

Accepted Answer

Introduction:Natural draught chimneys create a small pressure difference that accelerates flue gases upward. When this draught is expressed as an effective static head H' (metres of gas), we can estimate the ideal exit velocity using energy principles. This relationship aids preliminary sizing and performance checks.Given Data / Assumptions:H' denotes effective static draught head expressed in metres of the flue gas.Inviscid, one-dimensional flow; friction and thermal mixing neglected for the ideal estimate.g is the gravitational acceleration.Concept / Approach:The pressure energy per unit weight corresponding to head H' is converted to kinetic energy at the exit. Using Bernoulli-style reasoning for the ideal case, the velocity head V^2/(2 * g) equals H'. Therefore, V = sqrt(2 * g * H'). If H' is instead given indirectly from density difference and chimney height, then H' = H * (rho_a - rho_g) / rho_g and the velocity expression can be written in extended form. Here, H' is already the effective head, so the compact square-root relation applies.Step-by-Step Solution:Start with energy balance: velocity head at exit = available head.Set V^2 / (2 * g) = H'.Solve for V: V = sqrt(2 * g * H').Verification / Alternative check:If H' is derived from chimney height H via H' = H * (rho_a - rho_g) / rho_g, substitution yields V = sqrt(2 * g * H * (rho_a - rho_g) / rho_g), consistent with the extended formula.Why Other Options Are Wrong:sqrt(2 * g * H' * (rho_a - rho_g) / rho_g): double-counts density ratio when H' already represents effective head.2 * g * H' and H'/(2g): dimensionally incorrect for velocity.Common Pitfalls:Mixing “height of chimney” H with “effective head” H'; always check which head the problem states.Final Answer:V = sqrt(2 * g * H')

Question 7

Boiler Accessories — What Does an Economiser Do? Assess the statement: “An economiser is used to increase the temperature of saturated steam without raising its pressure.”

Accepted Answer

Introduction:Boiler accessories improve plant efficiency by recovering heat from flue gases. Two common devices are the economiser and the superheater. Confusing their roles leads to incorrect system design and poor performance. The statement given confuses these functions.Given Data / Assumptions:Conventional subcritical drum boiler context.Economiser placed in the flue gas path upstream of the air preheater and downstream of the evaporator/superheater bank order may vary by design.Saturated steam refers to steam at the saturation temperature for its pressure.Concept / Approach:An economiser heats feedwater before it enters the boiler drum/evaporator, raising feedwater temperature and reducing fuel required. A superheater, not an economiser, raises the temperature of steam above saturation at essentially the same pressure. Therefore, the statement about an economiser increasing saturated steam temperature is incorrect.Step-by-Step Solution:Define economiser duty: raise feedwater temperature using waste heat from flue gases.Define superheater duty: raise steam temperature above saturation at near-constant pressure (superheat).Compare to statement: it describes superheater function, not economiser ⇒ statement is false.Verification / Alternative check:P&I diagrams show economiser in the feedwater line; thermal calculations show reduced fuel to reach boiling when feedwater is preheated.Why Other Options Are Wrong:“Correct” or qualifiers involving reheat/subcritical do not change the fundamental device function.Common Pitfalls:Using the term “economiser” generically for any heat recovery; always distinguish feedwater heating from steam superheating or reheating.Final Answer:Incorrect

Question 8

Speed Regulation — Primary Function of a Governor Evaluate the statement: “The function of a governor is to keep the engine speed uniform at all load conditions.” Choose the most appropriate assessment.

Accepted Answer

Introduction:Governors are automatic speed-control devices used on engines and turbines. Their role is to modulate the energy input when load changes so that speed remains essentially constant, protecting generators, pumps, and driven equipment from harmful deviations.Given Data / Assumptions:Engine or prime mover with variable load.Governor senses speed (or frequency) and adjusts fuel/steam/flow accordingly.Steady-state control objective: maintain near-constant speed within allowable band.Concept / Approach:In control terms, the governor closes the loop between measured speed and actuator position. When load increases, speed tends to fall; the governor increases input to restore speed. When load decreases, the governor reduces input. Idealized wording says “keep speed uniform,” which practically means maintaining speed within tight limits across load changes.Step-by-Step Solution:Define objective: regulate mean speed against disturbances (load steps).Mechanism: centrifugal/speed-sensing element produces an error signal; linkage or servos adjust input valve or rack.Outcome: speed remains nearly constant, meeting the intent of “uniform.”Verification / Alternative check:Speed–load curves with governing show small droop or isochronous behavior depending on design; both aim at near-uniform speed despite load variations.Why Other Options Are Wrong:“Incorrect”: dismisses the accepted purpose of governors.“Only for isochronous”: non-isochronous governors also keep speed effectively uniform with permissible droop.“Only during starting”: incorrect; governing is continuous during operation.Common Pitfalls:Equating “uniform” with mathematically constant. Real governors permit small speed droop or deadband; this does not negate their function.Final Answer:Correct

Question 9

In steam generators, the Benson (once-through) boiler operates at supercritical pressure and does not require steam–water separation within a drum. Identify how many drums are required in a Benson boiler construction.

Accepted Answer

Introduction:The Benson boiler is a once-through, supercritical-pressure boiler widely discussed in power plant engineering. Unlike subcritical drum boilers, the Benson design avoids steam–water separation hardware. This question tests your understanding of why the Benson boiler requires no steam drum and how that affects operation and control.Given Data / Assumptions:Benson boiler operates near or above the critical pressure of water.Once-through flow path from feedwater inlet to superheated steam outlet.No phase-separation hardware is needed along the flow path.Modern controls manage sliding pressure and variable load.Concept / Approach:At or above the critical pressure, the distinction between liquid water and steam disappears; there is no two-phase mixture with a free surface. Consequently, there is no need for a steam drum to separate steam from water. Even when operating slightly below critical pressure (sliding pressure at lower loads), a Benson-type once-through arrangement still does not rely on a fixed drum for separation; moisture control is achieved via throttling, separators, or bypass arrangements in ancillary systems, not a conventional steam drum within the boiler circuit.Step-by-Step Solution:1) Recognize that drum boilers use a steam drum to separate steam from water under subcritical conditions.2) In a Benson boiler, water passes through economizer → evaporator zone → superheater in a single pass.3) Operating near/above critical pressure removes the need for gravity separation; therefore no drum is required.4) Final answer: the Benson boiler construction requires no drum.Verification / Alternative check:Control philosophy and thermal-hydraulic analyses for once-through units consistently show absence of a steam drum; transient separators, if any, are outside the boiler pressure part classification.Why Other Options Are Wrong:One/two/three drums: these are characteristic of subcritical water-tube designs (e.g., Babcock–Wilcox, Lancashire shell type uses two large flue tubes but drum context differs).Drum with external separator: not integral to the once-through principle and not required.Common Pitfalls:Confusing Benson with subcritical drum boilers; assuming every boiler must have a steam drum for level control.Final Answer:no drum

Question 10

Classification of prime movers: A steam engine burns fuel in a separate boiler and admits steam to the cylinder for expansion. State whether the following is correct: “Steam engines are external combustion engines.”

Accepted Answer

Introduction:Prime movers are broadly divided into internal combustion (combustion inside the working cylinder) and external combustion (combustion outside the working cylinder). Understanding this distinction is foundational to thermodynamics and power plant engineering.Given Data / Assumptions:Steam engine uses steam produced in a boiler.Fuel is burnt in the boiler furnace, not inside the engine cylinder.Steam expands in the engine cylinder doing work on the piston.Concept / Approach:In external combustion engines, the working fluid (steam) is heated by combustion occurring in a separate chamber. The engine cylinder never experiences combustion; it only receives high-energy steam as the working medium. This is in contrast to internal combustion engines (e.g., Diesel, Otto) where the fuel–air mixture ignites within the cylinder itself.Step-by-Step Solution:1) Identify the location of combustion: in a boiler furnace, external to the engine.2) Recognize the working fluid path: boiler → steam line → engine cylinder.3) Note that energy conversion in the cylinder is purely due to steam expansion, not combustion.4) Conclude that a steam engine is an external combustion engine.Verification / Alternative check:Historical and modern references consistently classify steam engines and steam turbines as external combustion due to separate combustion chambers (boilers).Why Other Options Are Wrong:“No” and conditional variants contradict the clear definition; fuel type, compounding, or pressure level do not change the combustion location.Common Pitfalls:Confusing steam engines with internal combustion reciprocating engines due to similar piston–crank mechanisms.Final Answer:Yes

Question 11

Engine testing: To compute indicated power (I.P.) from mean effective pressure (m.e.p.), what additional engine data are required? Select the most appropriate set for a single-cylinder calculation.

Accepted Answer

Introduction:Indicated power (I.P.) measures the power developed inside the engine cylinder based on pressure–volume work. It is central to performance analysis and efficiency calculations, distinct from brake power and friction power.Given Data / Assumptions:Mean effective pressure p_m is known from an indicator diagram or pressure transducer.Single-cylinder engine with known geometry.Speed N in revolutions per minute (or power strokes per second) is available.Concept / Approach:The standard relation is I.P. = p_m * L * A * n, where L is stroke length, A is piston area, and n is the number of power strokes per second. For a four-stroke single-cylinder engine, n = N / 2 (in rev/s); for a two-stroke, n = N. Thus, in practical terms, you need bore (or piston diameter, to get A), stroke length L, and rotational speed N to convert p_m to power.Step-by-Step Solution:1) Compute piston area A = (π/4) * D^2 using piston diameter D.2) Use given stroke length L.3) Determine power strokes per second from speed N and cycle type.4) Evaluate I.P.: I.P. = p_m * L * A * n (units consistent; divide by 60 if N is in rpm).Verification / Alternative check:Cross-check with brake power and mechanical efficiency: η_mech = B.P. / I.P.Why Other Options Are Wrong:Options with calorific value and specific fuel consumption relate to fuel and brake power/thermal efficiency, not directly to indicated power from m.e.p.Torque appears in brake power, not indicated power determination from p_m.Common Pitfalls:Mismatching units (Pa vs bar), forgetting cycle factor (2-stroke vs 4-stroke), or using rpm directly without converting to strokes per second.Final Answer:piston diameter, length of stroke and speed of rotation

Question 12

Boiler accessories: What is the primary purpose of installing an economiser in a steam generator? Choose the option that best reflects its main function in thermal performance.

Accepted Answer

Introduction:An economiser is a heat-recovery device located in the boiler flue gas path. By preheating boiler feedwater using exhaust gas sensible heat, it boosts overall efficiency and reduces the specific fuel consumption of the boiler plant.Given Data / Assumptions:Feedwater enters economiser before the boiler drum/evaporator.Flue gases upstream of the air preheater still have recoverable sensible heat.No phase change occurs in the economiser (ideally subcooled liquid at outlet).Concept / Approach:Thermal efficiency of a boiler increases when stack losses are reduced. Preheating feedwater decreases the enthalpy rise required in the evaporator and superheater, thereby reducing fuel input for the same steam generation. Although economisers influence steam conditions indirectly, they do not superheat steam (that is the superheater’s job) nor do they directly raise boiler pressure (that depends on firing rate, controls, and turbine demand).Step-by-Step Solution:1) Identify heat source: hot flue gases leaving superheater.2) Transfer heat to cooler feedwater: increase feedwater temperature.3) Reduced heat duty in evaporator: lower fuel rate for same steam flow.4) Net effect: reduced fuel consumption and improved efficiency.Verification / Alternative check:Energy balance on boiler island shows stack temperature reduction and lower fuel input at constant steam production when an economiser is added.Why Other Options Are Wrong:Superheat the steam: handled by superheater, not economiser.Increase steam pressure: pressure is a control variable, not a direct economiser outcome.All of these: incorrect because only fuel reduction is the primary direct effect.Common Pitfalls:Allowing economiser outlet to reach saturation causing steaming; misplacing it relative to air preheater and SCR units.Final Answer:reduce fuel consumption

Question 13

Natural draught design: For maximum discharge through a chimney, should the flue gas temperature in the chimney be only slightly greater than atmospheric temperature? Judge the correctness of this statement.

Accepted Answer

Introduction:Chimneys create natural draught due to density differences between hot flue gases and cooler ambient air. The achievable mass flow (discharge) depends on both the buoyancy-induced pressure difference and the gas density. Understanding the trade-off between temperature (density) and pressure head is key to optimizing discharge.Given Data / Assumptions:Chimney of height H with uniform cross-section.Ambient air at temperature T_a; flue gas inside at average temperature T_g.Ideal gas behavior and steady flow.Concept / Approach:The pressure head (draught) is proportional to (ρ_a − ρ_g) * g * H. Increasing T_g reduces ρ_g, increasing the pressure difference. However, the mass flow rate is ρ_g * velocity * area; extremely high T_g lowers ρ_g and hence mass flow even if draught rises. The optimum for maximum discharge occurs at a temperature significantly higher than ambient, not just “slightly more”. Therefore, the statement claiming “slightly more than atmospheric” for maximum discharge is incorrect.Step-by-Step Solution:1) Draught ∝ (ρ_a − ρ_g) * H: higher T_g (lower ρ_g) improves pressure difference.2) Mass flow m_dot = ρ_g * A * √(2 * Δp / ρ_g) = A * √(2 * Δp * ρ_g).3) Substitute Δp ∝ (ρ_a − ρ_g) * H to see m_dot depends on both ρ_g and (ρ_a − ρ_g).4) The optimal T_g is well above T_a; hence “slightly more” is not the maximizing condition.Verification / Alternative check:Differentiating the idealized expression for m_dot with respect to ρ_g shows an internal extremum at a ρ_g value below ρ_a, not near-equal densities (i.e., not “slightly above” atmospheric temperature).Why Other Options Are Wrong:Conditional variants (very tall chimneys, air preheaters, low barometric pressure) do not change the fundamental physics; maximum discharge still requires a sizable temperature difference.Common Pitfalls:Assuming larger draught always means larger mass flow; ignoring the counteracting effect of reduced gas density.Final Answer:Incorrect

Question 14

Boiler performance terminology: The quantity of water evaporated (or steam generated) in kilograms per kilogram of fuel fired is called the evaporation ratio. State whether the statement “This is known as the power of a boiler” is correct.

Accepted Answer

Introduction:Boiler performance uses several distinct terms: evaporation ratio (kg steam per kg fuel), equivalent evaporation (from and at 100°C), boiler efficiency, and boiler power (or capacity). Confusing these definitions can lead to serious errors in design and reporting.Given Data / Assumptions:Evaporation ratio ER = kg of steam produced per kg of fuel fired.Boiler power/capacity normally stated as mass flow of steam (e.g., kg/s, t/h) or heat release rate.Standard conditions may define “equivalent evaporation” but not “power”.Concept / Approach:The statement equates evaporation ratio with “power of a boiler,” which is incorrect. Evaporation ratio is a specific output per unit fuel input, essentially a measure akin to efficiency influence. Boiler power refers to the absolute output rate (steam generation rate or thermal duty), not normalized by fuel mass.Step-by-Step Solution:1) Define ER = (mass of steam produced) / (mass of fuel fired).2) Define boiler power/capacity = mass flow rate of steam (e.g., t/h) or equivalent thermal duty.3) Note that ER is dimensionless; power has dimensions (kg/s or kW).4) Hence, calling ER the “power of a boiler” is false.Verification / Alternative check:Test sheets list ER separately from capacity; improving efficiency raises ER but does not change the name of the metric to “power”.Why Other Options Are Wrong:Conditional “True” statements confuse ER with equivalent evaporation or specific conditions; none make ER equal to “power”.Common Pitfalls:Using ER to compare boilers of different sizes; ER does not reveal absolute production capability.Final Answer:False

Question 15

Air-standard Diesel cycle parameter: The cut-off ratio r_c is defined using cylinder volumes at the start and end of the constant-pressure heat addition. Select the correct definition below.

Accepted Answer

Introduction:In the Diesel cycle, heat addition occurs at approximately constant pressure from the end of compression (top dead centre) to the cut-off point during the expansion. The ratio of volumes at these two instants is called the cut-off ratio, a key parameter that governs efficiency and work output.Given Data / Assumptions:Ideal air-standard Diesel cycle.State 2: end of compression at top dead centre (volume V_2).State 3: end of heat addition at cut-off (volume V_3).Concept / Approach:By definition, r_c = V_3 / V_2. At the end of compression, the piston is at TDC and the cylinder volume equals the clearance volume V_c. Therefore, V_2 = V_c. Hence, r_c = V_3 / V_c. Recognizing this identity resolves many confusions between swept volume, clearance volume, and instantaneous volumes during the combustion process in Diesel engines.Step-by-Step Solution:1) Identify the two states spanning constant-pressure heat addition: start at V_2, end at V_3.2) At TDC, V_2 equals the clearance volume V_c.3) Write r_c = V_3 / V_2 = V_3 / V_c.4) Match with the option that states “volume at cut-off to the clearance volume”.Verification / Alternative check:Other cycle relations: compression ratio r = V_1 / V_2; expansion ratio relates to V_4 / V_3; these are distinct from r_c.Why Other Options Are Wrong:Using swept volume is incorrect because swept volume does not equal V_2 or V_3 at the instants of interest.Clearance to cut-off (inverse) is not the standard definition.Start-of-compression volume is V_1, not V_2; r_c uses V_2 at TDC.Common Pitfalls:Mixing up compression ratio r with cut-off ratio r_c; forgetting that V_2 equals the clearance volume at TDC.Final Answer:volume at cut-off to the clearance volume

Question 16

Boiler construction: The Lancashire boiler (a horizontal, internally fired, fire-tube boiler) is characterized by multiple large internal flue tubes running through the shell. How many internal flue tubes does a Lancashire boiler have?

Accepted Answer

Introduction:The Lancashire boiler is a classic shell boiler design used historically in small power and process plants. Recognizing its geometry is fundamental for distinguishing it from Cornish and other fire-tube boilers.Given Data / Assumptions:Horizontal shell boiler with internal furnaces.Fire and flue gases pass through large-diameter tubes within a common water shell.Natural circulation around the tubes.Concept / Approach:The Cornish boiler has one large internal flue tube; the Lancashire boiler evolved the concept to two parallel furnace flues to increase heating surface and firing capacity while keeping manageable furnace dimensions. Both flues run from front to back and often split into smaller passes (Galloway tubes may be fitted as cross water-tubes for enhanced heat transfer).Step-by-Step Solution:1) Identify boiler type by furnace count.2) Cornish → one internal flue; Lancashire → two internal flues.3) Conclude: Lancashire boiler has two internal flue tubes.4) Extra features (Galloway tubes) do not change the count of main flues.Verification / Alternative check:Historical drawings and textbooks consistently depict two large parallel furnace flues in Lancashire boilers.Why Other Options Are Wrong:One/three/four: do not match the canonical Lancashire layout.Two with a central superheater flue: superheaters are not part of the original configuration and do not constitute a third main furnace flue.Common Pitfalls:Confusing with Cornish boiler or miscounting cross water tubes (Galloway) as main flues.Final Answer:two

Question 17

Compound steam engines: When the high-pressure (HP) and low-pressure (LP) cylinders are arranged in line and operate with a common piston rod (one rod connecting both pistons), what is this configuration called?

Accepted Answer

Introduction:Compounding expands steam in stages across multiple cylinders to improve efficiency and reduce thermal stresses. Different mechanical arrangements exist depending on how the cylinders are aligned and interconnected.Given Data / Assumptions:Two cylinders: HP and LP.Common piston rod indicates in-line mechanical connection.Shared crank or colinear arrangement depending on design.Concept / Approach:In the tandem compound engine, HP and LP cylinders are on the same axis and their pistons are mounted on a single, continuous piston rod. Steam exhausts from HP directly into the LP cylinder without an intermediate receiver volume in some variants. This is distinct from cross-compound (separate cylinders on different cranks), receiver type (explicit intermediate receiver between cylinders), and Woolf type (a particular form of compound without a receiver, but the defining feature asked here is the common piston rod aligned tandemly).Step-by-Step Solution:1) Identify key mechanical feature: common piston rod for HP and LP.2) Map to configuration names: “tandem” implies in line, same rod.3) Therefore, the correct term is Tandem type compound engine.4) Other names correspond to different physical arrangements.Verification / Alternative check:Classic steam engine texts categorize tandem (same rod), cross-compound (separate rods/cranks), receiver (with receiver), and Woolf (no receiver) types distinctly.Why Other Options Are Wrong:Receiver type: emphasizes an intermediate volume, not a common rod.Woolf type: lacks a receiver but does not necessarily imply a common piston rod configuration as the defining trait.Cross-compound: cylinders drive different cranks; not a common rod.Common Pitfalls:Mixing functional steam path terms (receiver/Woolf) with mechanical alignment terms (tandem/cross-compound).Final Answer:Tandem type compound engine

Question 18

Flywheel sizing: Compared with a simple (single-expansion) steam engine of similar rating, does a compound steam engine require a heavier flywheel? Select the correct statement.

Accepted Answer

Introduction:The purpose of a flywheel is to smooth cyclic fluctuations in torque and speed caused by the intermittent nature of piston work. The degree of fluctuation depends on how uniform the turning moment diagram is over a cycle.Given Data / Assumptions:Simple steam engine: single expansion with one cylinder (or identical rating).Compound engine: multi-stage expansion (HP and LP) sharing load across phases.Comparable mean torque and operating speed.Concept / Approach:Compounding divides the expansion across cylinders timed to produce more uniform torque. The HP and LP cylinders contribute work at different crank positions, filling in low-torque portions of the cycle. Therefore, the turning moment is smoother, the coefficient of fluctuation of speed is reduced, and the energy that must be stored and released by the flywheel each cycle is smaller. Hence, for comparable conditions, a compound engine typically requires a lighter flywheel, not a heavier one.Step-by-Step Solution:1) Consider turning moment diagrams for single vs compound expansion.2) Observe increased uniformity in compound due to phase-shifted cylinder contributions.3) Lower fluctuation of energy → lower flywheel moment of inertia required.4) Conclude the statement “requires a heavier flywheel” is false.Verification / Alternative check:Empirical practice and design handbooks recommend smaller flywheel sizes for multicylinder and compounded engines versus single-cylinder counterparts at similar ratings.Why Other Options Are Wrong:“True” and conditional forms misinterpret the effect of compounding; unbalance affects vibration, not primarily energy fluctuation addressed by the flywheel.Common Pitfalls:Conflating counterweighting for reciprocating mass balance with flywheel energy storage requirements.Final Answer:False

Question 19

Boiler basics: identify the configuration Statement: In a fire-tube (smoke-tube) boiler, the water is contained inside the tubes while the flames and hot gases flow outside the tubes.

Accepted Answer

Introduction / Context:The classification of boilers into fire-tube (smoke-tube) and water-tube types is fundamental in thermal engineering. Many design, safety, and performance features hinge on which medium—hot gas or water/steam—occupies the tubes. This question tests your ability to recall the correct configuration and its implications for pressure levels, response, and maintenance.Given Data / Assumptions:Conventional stationary boilers used for steam generation in power and process applications.Comparison between fire-tube and water-tube constructs only.No superheater, reheater, or economizer complications are required to answer.Concept / Approach:In a fire-tube boiler, hot combustion gases pass through tubes that run inside a water-filled shell. Heat transfers through the tube walls to the surrounding water, generating steam in the shell. In a water-tube boiler, the arrangement is reversed: water flows inside the tubes, and hot gases wash over the outside of the tubes. This inversion drives major differences in allowable pressure and response time.Step-by-Step Solution:Identify the claim: “water is contained inside the tubes” for a fire-tube unit.Recall: fire-tube → hot gases in tubes; water/steam around tubes in the shell.Therefore, the statement reverses the true arrangement.Conclusion: the statement is false.Verification / Alternative check:Classic examples include the Cochran and Lancashire boilers (fire-tube), where multiple smoke tubes carry flue gases through a water-filled shell. Water-tube examples (Babcock & Wilcox) clearly show water/steam inside tubes with external hot gas flow.Why Other Options Are Wrong:“True only for water-tube boilers” mislabels the class; water inside tubes defines water-tube, not fire-tube. Superheaters and once-through boilers are different subsystems or modern designs and do not make the original statement correct for a fire-tube boiler.Common Pitfalls:Memorizing names without associating them to geometry; assuming “tube” always contains water; conflating shell-and-tube heat exchanger intuition with boiler construction.Final Answer:False

Question 20

Cochran (vertical multi-tubular) boiler: typical count of fire tubes Select the commonly cited number of small-diameter fire tubes used in the standard textbook Cochran boiler description.

Accepted Answer

Introduction / Context:The Cochran boiler is a classic vertical, multi-tubular, fire-tube design widely discussed in boiler courses. Remembering its typical dimensions and tube counts is helpful for quick identification and for estimating heating surface area and gas-side pressure drop in conceptual problems.Given Data / Assumptions:Standard textbook representation of a Cochran boiler (hemispherical crown, multi-tubular fire-tube bank).Small-diameter smoke tubes carrying hot gases through a water-filled shell.We refer to the typical “exam schematic” configuration rather than proprietary variants.Concept / Approach:While practical designs may vary with manufacturer and rating, widely used teaching diagrams list a fire-tube bank numbering around the mid-hundreds. The commonly quoted figure in many exam problems and solved examples is approximately 165 tubes for a representative size, balancing heat transfer area and draft resistance.Step-by-Step Solution:Identify that this is a recall item from standard boiler descriptions.Recognize Cochran = vertical shell + multiple small smoke tubes.Recall the typical tube count cited: about 165 tubes.Select 165 as the most standard textbook value among the options provided.Verification / Alternative check:Many classroom notes and question banks align around 165 for the common schematic. Real units can have different counts depending on duty and shell size, but exam convention typically expects 165 here.Why Other Options Are Wrong:Values 75, 95, and 115 are below typical counts for the standard teaching example; 225 is on the higher side of certain larger variants and is not the commonly cited number for the canonical diagram.Common Pitfalls:Assuming a single “true” number across all Cochran boilers; neglecting that exam problems often use a representative standard count for consistency.Final Answer:165

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