Question 1

Purpose of a surface condenser in Rankine cycle plants: “A condenser maintains a pressure below atmospheric to extract maximum energy from steam and thereby improve efficiency.” State whether this statement is correct.

Accepted Answer

Introduction / Context: Condensing is a central feature of efficient Rankine cycle operation in power plants. By reducing the exhaust (back) pressure, the turbine or engine can expand steam through a higher pressure ratio, extracting more specific work from each kilogram of steam. Given Data / Assumptions: Surface condenser operated with cooling water, air extraction, and vacuum maintenance. Boiler pressure remains fixed by boiler/turbine settings; condenser affects the low-pressure end. Steady-state operation with adequate heat rejection capacity. Concept / Approach: Lowering back pressure increases the expansion ratio and the area under the expansion curve on p–v or T–s diagrams, which corresponds to higher work output per unit mass of steam. The condenser converts exhaust steam to condensate at low pressure and temperature, enabling feedwater recovery and improved cycle efficiency. Step-by-Step Solution: Identify objective: maximize specific work by extending expansion to low pressure.Condenser action: condense exhaust, remove latent heat, and create vacuum below atmospheric pressure.Effect on cycle: increased pressure ratio → higher work and thermal efficiency.Conclusion: the statement is correct. Verification / Alternative check: Comparing Rankine cycle efficiencies at p_cond = 0.1 bar vs 1.0 bar shows significant efficiency increase at lower condenser pressure. Why Other Options Are Wrong: “Disagree” and conditional variants contradict standard Rankine analysis; condenser action does not raise boiler pressure; it reduces exhaust pressure. Common Pitfalls: Attributing efficiency increase to boiler pressure alone; neglecting the role of air removal and cooling-water temperature in maintaining condenser vacuum. Final Answer: Agree

Question 2

Firebox design feature: “The crown (top) of certain furnace or firebox chambers is made hemispherical to better resist severe heat and pressure in the firebox region.” Evaluate this statement.

Accepted Answer

Introduction / Context: High-temperature, high-pressure furnace enclosures require shapes that minimize thermal stresses and resist pressure. Hemispherical and other curved surfaces are common in pressure-vessel design due to favorable stress distributions. Given Data / Assumptions: Hemispherical (or domed) crowns are used in some boiler furnaces (e.g., Scotch marine, certain shell boilers). Intense radiant heat exists in the firebox; pressure acts on the furnace surface from the water/steam side. Materials are adequately stayed or self-supporting depending on geometry. Concept / Approach: Curved shells like hemispheres distribute membrane stresses more uniformly than flat plates. Under pressure and severe thermal exposure, a hemispherical crown minimizes bending stress, reduces the need for extensive staying, and withstands cyclic thermal gradients more effectively than flat crowns. Thus, the statement that hemispherical crowns are used to resist intense service conditions is correct in principle for applicable boiler types. Step-by-Step Solution: Recognize pressure-vessel principle: spherical shapes offer highest strength for given thickness.Apply to firebox crown: curved crown reduces bending and local hot spots.Contrast with flat crown: requires dense stays (e.g., locomotive firebox) to handle pressure and distortion.Conclude: hemispherical crowns are chosen to resist heat and pressure more effectively. Verification / Alternative check: Design codes and historical practice show domed crowns in marine shell boilers where geometry allows, confirming the advantage. Why Other Options Are Wrong: Flat plates do not “always” perform better; they typically need extensive staying. Limiting to a specific boiler type or low pressure ignores the general structural advantage of curved shells. Common Pitfalls: Assuming all fireboxes are hemispherical (e.g., locomotive fireboxes are often flat but heavily stayed); conflating specific boiler types with the general design principle. Final Answer: True

Question 3

Thermal performance definitions: The indicated thermal efficiency of a steam engine is defined as the:

Accepted Answer

Introduction / Context: Multiple efficiencies are used when evaluating steam engines: indicated thermal, brake thermal, mechanical, boiler, and overall efficiencies. Clear definitions are essential for correct performance analysis and comparison across machines. Given Data / Assumptions: I.P. (indicated power) is power developed in the cylinder from the indicator diagram. Energy supplied in steam refers to the enthalpy flow available to the engine. Steady operation with measurable mass flow and state points. Concept / Approach: Indicated thermal efficiency η_it compares the useful power developed inside the cylinder (indicated power) to the rate of thermal energy supplied in the incoming steam. Mathematically, η_it = (heat equivalent of I.P.) / (steam energy rate). This differs from brake thermal efficiency (uses B.P.) and mechanical efficiency (B.P./I.P.). Step-by-Step Solution: Identify numerator: I.P. converted to kJ/s (1 kW = 1 kJ/s).Identify denominator: m_dot * (h_in − h_ref) for the engine boundary.Form the ratio: η_it = I.P. / steam heat rate.Select the option that states this definition explicitly. Verification / Alternative check: Cross-check with related definitions: η_mech = B.P./I.P.; η_bt = B.P./(steam heat rate); η_overall = B.P./(fuel heat rate). Why Other Options Are Wrong: Options (a) and (d) are not standard definitions; (b) is mechanical efficiency; (e) mixes unrelated plant-level metrics. Common Pitfalls: Using brake power instead of indicated power; mixing engine-boundary efficiency with plant-wide overall efficiency. Final Answer: ratio of heat equivalent to indicated power to the energy supplied in steam

Question 4

Boiler mounting function check: A fitting that empties the boiler when required and discharges mud, scale, or sediment from the lowest part of the shell is called the blow-off cock. Is this identification correct?

Accepted Answer

Introduction / Context: Boiler mountings are essential safety and control devices. The blow-off arrangement is particularly important for removing sludge and for draining the boiler during maintenance, washing, or inspection. Given Data / Assumptions: Mounting located at the lowest point of the boiler water space. Operated under proper procedures to avoid thermal shock and erosion. Applicable to both fire-tube and water-tube designs with appropriate fittings. Concept / Approach: The blow-off cock or blow-down valve provides a direct outlet from the mud drum or lowest shell point to discharge water and settled solids. Periodic short-duration blows maintain water quality and remove sludge. For full draining, the same connection is used under controlled conditions. This function is distinct from safety valves (overpressure protection) and fusible plugs (overheat protection). Step-by-Step Solution: Identify location: lowest point to capture sediments.Identify function: intermittent blow-down and complete draining.Separate from other mountings: not a level control, not an overpressure device.Conclude: the description correctly matches a blow-off cock. Verification / Alternative check: P&IDs show blow-down lines with valves and cooling/flash arrangements leading to blow-down tanks for safe disposal. Why Other Options Are Wrong: Fusible plug melts on overheat; safety valve relieves pressure; neither drains sludge. Limiting to water-tube boilers is incorrect; both main classes use blow-down devices. Common Pitfalls: Operating the blow-down valve for too long, risking level loss or tube exposure; blowing down at inappropriate pressures or without proper cooling arrangements. Final Answer: Yes

Question 5

Boiler classification by gas and water paths: In a fire-tube boiler, which flow arrangement correctly describes the relative paths of hot gases and water?

Accepted Answer

Introduction / Context: Boilers are broadly divided into fire-tube and water-tube types. The distinction rests on whether hot gases or water/steam occupy the tubes. This affects heat transfer area, response time, and allowable pressure. Given Data / Assumptions: Fire-tube: gas-in-tube, water-outside configuration. Water-tube: water-in-tube, gas-outside configuration. Typical fire-tube examples: Lancashire, Cornish, Locomotive, Scotch marine. Concept / Approach: In fire-tube boilers, combustion products pass through multiple small tubes immersed in the boiler shell water. Heat transfers from the hot gas inside the tubes to the surrounding water, generating steam. This configuration yields large water content and stable but slower response compared to water-tube designs. Step-by-Step Solution: Identify class: “fire-tube” means fire/gas in the tubes.Describe path: flames and hot gases inside tubes; water surrounds tubes in shell.Eliminate alternatives: water-in-tube describes water-tube boilers; forced circulation is not inherent to fire-tube design.Select the matching statement. Verification / Alternative check: Standard diagrams of locomotive and Scotch marine boilers show banks of gas-carrying tubes through a water-filled shell. Why Other Options Are Wrong: Option (b) is the water-tube arrangement; (c) confuses circulation methods; (d) and (e) do not match standard classifications. Common Pitfalls: Mixing up the two major classes or assuming forced circulation applies to fire-tube boilers. Final Answer: the flames and hot gases pass through the tubes which are surrounded by water

Question 6

Indicator diagram realism factor: The diagram factor (ratio of actual indicated work to ideal/rectangular work for the same peak pressure) depends primarily on which operating conditions?

Accepted Answer

Introduction / Context: The diagram factor accounts for departures of the real indicator diagram from an idealized rectangle of the same maximum pressure and stroke. It is used to correct theoretical power estimates to more realistic indicated power values. Given Data / Assumptions: Real steam expansion is not perfectly isentropic or rectangular. Admission, cut-off, release, and compression phases shape the diagram. Operating conditions such as initial pressure, quality, and back pressure influence the loop area. Concept / Approach: The diagram factor is sensitive to how early or late cut-off occurs, throttling at valves, wire-drawing, condensation and re-evaporation, and exhaust back pressure. Higher initial pressure/temperature, different steam quality, and condenser back pressure all alter the diagram shape and thus the factor. Step-by-Step Solution: Identify variables: initial steam conditions (pressure/temperature/quality) affect admission line and early expansion.Recognize back pressure: higher back pressure shrinks the expansion area, lowering the factor.Consider initial pressure specifically: sets the head of the diagram and influences valve losses.Therefore, the diagram factor depends on all listed items. Verification / Alternative check: Indicator cards under different loads show changing loop areas; computed factors vary accordingly. Why Other Options Are Wrong: Choosing only one variable ignores the multi-parameter nature of the indicator diagram. “None of these” is obviously false since the factor responds to operating conditions. Common Pitfalls: Using a fixed factor across widely different operating regimes; neglecting the effect of valve timing and wire-drawing. Final Answer: all of these

Question 7

Firing arrangement in water-tube boilers: Water-tube boilers (with water/steam inside tubes and hot gases outside) are generally classified as which type with respect to firing location?

Accepted Answer

Introduction / Context: Boiler classifications include not only the path of water and gas but also the location of the furnace relative to the pressure parts. This affects maintenance, access, and allowable firing rates. Given Data / Assumptions: Water-tube: water/steam inside tubes; hot gases flow outside over the tube banks. Furnace chamber is typically refractory-lined and separate from the pressure shell. Examples: Babcock & Wilcox, Stirling, modern utility boilers. Concept / Approach: Water-tube boilers place the furnace outside the main pressure shell, within a separate, water-cooled or refractory-lined chamber. This is termed externally fired. In contrast, many fire-tube boilers are internally fired because the furnace is inside the shell (e.g., Cornish, Lancashire, Scotch marine). Step-by-Step Solution: Identify furnace position: outside the pressure shell for water-tube boilers.Apply terminology: this is “externally fired.”Eliminate other options: internally fired describes many shell fire-tube units; “both” is not standard for classic water-tube designs.Choose the correct classification. Verification / Alternative check: Plant layouts show large membrane-wall furnaces external to drums/tubes pressure boundary, confirming external firing. Why Other Options Are Wrong: Internally fired applies to shell boilers; “both/none/top-fired only” do not reflect standard classification practice. Common Pitfalls: Confusing gas path classification (fire-tube vs water-tube) with firing location terminology. Final Answer: externally fired

Question 8

Steam engine compounding (Woolf type): mechanical arrangement In a Woolf-type compound engine used in steam practice, how are the high-pressure (HP) and low-pressure (LP) cylinders mechanically connected and arranged with respect to their pistons an…

Accepted Answer

Introduction / Context: The Woolf-type compound engine is a classic configuration in which steam expands in two stages: first in a high-pressure (HP) cylinder and then in a low-pressure (LP) cylinder. Understanding its distinctive mechanical linkage helps students differentiate it from cross-compound and tandem-compound layouts, which affects torque smoothness, footprint, and maintenance. Given Data / Assumptions: The question concerns the mechanical connection between HP and LP cylinders. Woolf compound engines are intended for two-stage expansion on a common axis. No numerical data is required; this is a conceptual design-identification item. Concept / Approach: Woolf compounding is essentially a tandem arrangement: the HP and LP pistons act in line on the same piston rod. Steam exhausts from the HP cylinder directly into the LP cylinder (sometimes via a minimal receiver), keeping the assembly compact and ensuring axial alignment of forces. Step-by-Step Solution: Identify Woolf type as tandem compounding.In a tandem engine, HP and LP cylinders are in line and act on a common rod and common crank.Therefore the pistons share the same piston rod; they are not mounted with separate cranks at 90° (that would be cross-compound).Conclude that the correct mechanical connection is a common piston rod. Verification / Alternative check: Historic diagrams of Woolf engines show the HP cylinder mounted ahead of the LP cylinder on the same axis, both attached to one rod and driving a single crosshead/crank system. This verifies the tandem, common-rod nature of the design. Why Other Options Are Wrong: Set at 90° with separate cranks describes cross-compounding, not Woolf. Separate piston rods with independent cranks again imply cross-compound or twin arrangements. V-configuration is a different geometric layout not typical of the Woolf tandem form. Common Pitfalls: Confusing “Woolf/tandem compound” with “cross compound.” Remember: tandem (Woolf) shares a rod; cross-compound uses separate cranks (often 90° apart) on the same shaft. Final Answer: have common piston rod

Question 9

Boiler efficiency: correct formula using steam and fuel data Given m = total mass of water evaporated into steam (kg/h), mf = mass of fuel fired (kg/h), C = calorific value of fuel (kJ/kg), h = total heat of the produced steam (kJ/kg), and hf1 = sen…

Accepted Answer

Introduction / Context: Boiler efficiency compares useful energy transferred to water/steam with the chemical energy supplied by the fuel. A precise formula is needed in boiler trials and energy audits to quantify performance and benchmark different units or fuels under comparable conditions. Given Data / Assumptions: m = mass of water evaporated into steam (kg/h). mf = mass of fuel consumed (kg/h). C = lower calorific value of the fuel (kJ/kg), unless otherwise noted. h = total specific enthalpy of steam at delivery state (kJ/kg). hf1 = sensible heat of feedwater at inlet conditions (kJ/kg). Concept / Approach: Useful heat added to water is (h − hf1) per kg of steam. Total useful heat per hour is m * (h − hf1). Fuel energy input per hour is mf * C. Efficiency is useful output divided by fuel input. This aligns with standard test codes and allows corrections for feedwater temperature and steam condition (saturated or superheated). Step-by-Step Solution: Useful heat rate Q_out = m * (h - hf1).Fuel heat rate Q_in = mf * C.η_boiler = Q_out / Q_in.Therefore η_boiler = (m * (h - hf1)) / (mf * C). Verification / Alternative check: Dimensional check: numerator and denominator are in kJ/h, yielding a dimensionless ratio between 0 and 1 (or 0–100%). This confirms plausibility. Why Other Options Are Wrong: Option (b) is the reciprocal; (c) ignores feedwater preheat; (d) adds hf1 instead of subtracting; (e) is dimensionally inconsistent for general cases. Common Pitfalls: Using gross calorific value without accounting for latent heat of condensation in flue gases; forgetting to correct h and hf1 to the same reference (kJ/kg). Final Answer: η_boiler = (m * (h - hf1)) / (mf * C)

Question 10

Definition check: efficiency of a steam boiler Which statement correctly defines boiler efficiency from an energy perspective?

Accepted Answer

Introduction / Context: Boiler efficiency is a fundamental metric for power and process plants. It quantifies how effectively the chemical energy in fuel is converted into thermal energy of steam delivered by the boiler. Given Data / Assumptions: Energy-in basis: fuel heat input as measured by calorific value and mass flow. Energy-out basis: enthalpy rise of water to steam at delivery conditions relative to feedwater. Steady-state operation during the measurement period. Concept / Approach: Boiler efficiency = useful energy to steam / fuel energy input. The useful energy is m * (h − hf1). The fuel energy input is mf * C. This definition reflects first-law efficiency and provides a basis for comparing different designs and fuels. Step-by-Step Solution: Identify useful heat: increase of water/steam enthalpy.Identify input: fuel chemical energy released in furnace.Form the ratio: η = useful heat / fuel heat.Select the statement that matches this definition: option (a). Verification / Alternative check: Heat balance sheets for boiler trials allocate losses (stack, moisture, radiation); efficiency equals 1 minus total fractional losses, consistent with option (a). Why Other Options Are Wrong: Option (b) compares masses, not energies. Option (c) is the inverse of the correct definition. Option (e) refers to a specific loss instead of overall efficiency; (d) is therefore incorrect. Common Pitfalls: Ignoring feedwater temperature in the useful heat term; comparing boilers without correcting for steam pressure and quality. Final Answer: ratio of heat actually used in producing steam to the heat liberated in the furnace

Question 11

Chimney draught: condition for maximum discharge For maximum discharge through a chimney, the height of the effective hot-gas column producing draught should be:

Accepted Answer

Introduction / Context: Natural draught in a chimney results from density differences between hot flue gases in the chimney and cooler ambient air outside. For a given chimney, there exists a condition that maximizes volumetric discharge by balancing buoyancy-induced pressure head with frictional and entry/exit losses. Given Data / Assumptions: Steady flow of flue gas under natural draught. Uniform gas properties approximated along the height for conceptual analysis. Losses due to friction and fittings scale with velocity head. Concept / Approach: The draught head is proportional to the height of the hot column and the density difference. The discharge increases with effective head but is reduced by friction as velocity rises. Maximum discharge occurs when the effective height of the hot gas column contributing to draught equals the actual chimney height, a textbook result obtained by differentiating the flow with respect to temperature drop/height distribution and setting the derivative to zero. Step-by-Step Solution: Draught head Δp ≈ g * H * (rho_out - rho_gas).Flow rate increases with Δp but experiences losses ∝ v^2 along H.Optimizing discharge leads to the condition: effective hot-column height ≈ chimney height.Therefore choose: equal to the height of the chimney. Verification / Alternative check: Many classic boiler engineering references state the condition for maximum discharge as effective hot-gas column height equal to chimney height, confirming option (b). Why Other Options Are Wrong: Fractions or multiples of chimney height do not satisfy the optimum derived from the balance of buoyancy and friction for the standard model; “independent” contradicts physics of draught formation. Common Pitfalls: Confusing draught head with static head alone; ignoring frictional losses that limit flow at higher velocities. Final Answer: equal to the height of the chimney

Question 12

Steam engine geometry: definition of clearance ratio Clearance ratio in a reciprocating steam engine is defined as the ratio of:

Accepted Answer

Introduction / Context: Clearance volume is the volume remaining in the cylinder when the piston is at the end of its stroke (dead center). The clearance ratio is a dimensionless parameter that strongly influences volumetric efficiency, compression work, and indicator diagram shape. Given Data / Assumptions: V_c = clearance volume. V_s = swept volume (displacement) between dead centers. Cut-off is a valve-timing event, not used in the definition. Concept / Approach: The standard definition is: Clearance ratio, C = V_c / V_s. This ratio is used in analytical expressions for compression and expansion parts of the cycle and in estimating the mass of residual (cushion) steam present at the beginning of the stroke. Step-by-Step Solution: Identify V_c as the volume at dead center when piston travel is zero.Identify V_s as the volume swept by the piston during one stroke.Form the ratio C = V_c / V_s.Therefore, option (a) precisely matches the accepted definition. Verification / Alternative check: Textbook formulae for volumetric efficiency include terms in C = V_c / V_s, confirming the definition used industry-wide. Why Other Options Are Wrong: Options (b) and (c) incorporate volume at cut-off, which is unrelated to the basic definition. Option (d) is the inverse of the standard ratio; (e) uses total cylinder volume, which is V_c + V_s, not part of the definition. Common Pitfalls: Mixing up clearance ratio with compression ratio; in steam engines they are related but distinct concepts. Final Answer: clearance volume to the swept volume

Question 13

Condenser performance term: selecting the correct definition The ratio defined as: (temperature rise of cooling water) / (vacuum-temperature corresponding to condenser pressure minus the inlet cooling-water temperature) is termed:

Accepted Answer

Introduction / Context: Surface condensers cool exhaust steam by transferring heat to circulating water. To evaluate performance, engineers use a non-dimensional efficiency comparing actual water temperature rise to the maximum theoretically attainable rise for the given terminal temperatures. Given Data / Assumptions: T_out and T_in are outlet and inlet temperatures of cooling water. T_sat is the saturation (vacuum) temperature corresponding to condenser pressure. Steady operation with negligible heat leaks except to cooling water. Concept / Approach: Condenser efficiency, η_cond, is commonly expressed as (T_out − T_in) / (T_sat − T_in). The denominator is the maximum possible rise if the water could approach the steam saturation temperature at the hot end. The actual rise divided by this maximum gives a convenient performance index between 0 and 1 (or 0–100%). Step-by-Step Solution: Identify numerator: temperature gain of water = T_out − T_in.Identify denominator: T_sat − T_in (vacuum temperature minus inlet water temperature).Form the ratio: η_cond = (T_out − T_in) / (T_sat − T_in).Match with the name: condenser efficiency. Verification / Alternative check: Used alongside “vacuum efficiency” and “air removal effectiveness,” this definition appears in standard thermal engineering texts and condenser test procedures. Why Other Options Are Wrong: Vacuum efficiency relates to how closely the measured vacuum approaches the ideal based on air leakage; nozzle and boiler efficiency are unrelated; “air cooler effectiveness” is not standard in this context. Common Pitfalls: Confusing T_sat with measured shell temperature; using gauge vacuum directly as temperature without conversion to saturation temperature. Final Answer: condenser efficiency

Question 14

Boiler mounting: purpose of a water level indicator State whether the statement is correct: A water level indicator is a device that shows the exact water level inside a boiler at any instant for safe operation and control.

Accepted Answer

Introduction / Context: Boiler mountings are essential devices necessary for safe and reliable operation. Among them, the water level indicator (gauge glass assembly) provides the operator with a direct visual indication of the liquid level to prevent overheating of heating surfaces and to maintain proper steam quality. Given Data / Assumptions: Direct-reading gauge glass or reflex/transparent type indicator. Boiler operating within design pressure and temperature. Proper isolation and blow-through cocks installed for maintenance and testing. Concept / Approach: Maintaining correct water level is critical. Too low a level risks exposing furnace crowns or tubes, leading to overheating and failure; too high a level risks water carryover and priming. The indicator provides continuous level information; multiple indicators may be used for redundancy and verification. Step-by-Step Solution: Steam space connects to the top cock; water space to the bottom cock of the gauge glass.The interface in the glass tracks the internal level due to hydrostatic communication.Operators routinely blow through the cocks to ensure the passages are clear and the reading is true.Thus, the device indicates the water level at any instant. Verification / Alternative check: Operating procedures require gauge-glass checks and compare with alarmed probes on modern units; both corroborate the function stated. Why Other Options Are Wrong: Pressure gauges measure pressure, not level; restrictions to “low pressure” or “twin glass only” are unnecessary, as level indicators are employed across pressures and in single or multiple arrangements. Common Pitfalls: Trusting a fouled or blocked gauge; not performing routine blow-through tests; misreading reflex versus transparent glass images. Final Answer: True

Question 15

Boiler heat losses: identifying principal components In a practical boiler, significant fractions of input heat are lost. Which of the following are recognized loss categories in a standard boiler heat-balance sheet?

Accepted Answer

Introduction / Context: Even a well-designed boiler cannot convert 100% of fuel energy into useful steam energy. Heat-balance sheets categorize losses to pinpoint improvements and compare performance across fuels and operating conditions. Given Data / Assumptions: Combustion products leave at a temperature above ambient, carrying sensible heat. Fuels may contain moisture, and hydrogen in fuel forms water vapor upon combustion. Additional losses (radiation, incomplete combustion) exist but are not listed in the options. Concept / Approach: Dry flue-gas loss is the sensible heat carried by nitrogen, carbon dioxide, and excess oxygen in the stack. Moisture-in-fuel loss accounts for heating and evaporating inherent water. Hydrogen-in-fuel loss includes the latent and sensible heat of water formed during combustion. Summing these with other minor losses yields 1 − η_boiler. Step-by-Step Solution: Identify each loss mechanism and its physical basis.Recognize that all listed categories appear in standard heat-balance tabulations.Therefore, the comprehensive answer is “all of the above.”This directly supports calculating efficiency via loss method: η = 1 − Σ(loss fractions). Verification / Alternative check: Boiler trial standards enumerate these losses explicitly, often the three largest categories besides radiation and unburnt losses, validating their inclusion. Why Other Options Are Wrong: Each single selection is incomplete; “none” contradicts established thermal accounting for boilers. Common Pitfalls: Ignoring latent heat in water-vapor losses; underestimating excess-air impact on dry flue-gas losses. Final Answer: all of the above

Question 16

Typical admission pressure to a steam engine chest In traditional reciprocating steam engines supplied with superheated steam, the steam from the boiler is commonly admitted to the steam chest at approximately what pressure (rounded to the nearest o…

Accepted Answer

Introduction / Context: Historic and instructional examples of simple or compound reciprocating steam engines typically cite moderate admission pressures compared with modern water-tube power boilers. Understanding the typical pressure band helps frame valve gear design, cylinder sizing, and lubrication needs. Given Data / Assumptions: Conventional steam engines for industrial or marine service (non-supercritical, non-ultra-high-pressure). Superheated steam improves dryness fraction and reduces cylinder condensation. Educational references often illustrate ranges near 7–14 bar (≈7–14 atmospheres). Concept / Approach: While higher pressures exist in specialized applications, many classical engines operated near 10 bar gauge, sometimes up to 12–16 bar. Selecting the closest discrete option reflects the prevalent instructional figure, recognizing that exact values depend on design and era. Step-by-Step Solution: Identify typical instructional range: roughly 7–14 atm.Compare with options: 10 atm aligns centrally with that range.Hence choose 10 atmospheres as the representative admission pressure.Note: actual installations might differ, but the question seeks a typical textbook value. Verification / Alternative check: Classic training manuals and solved examples often assume 10 atm admission for sizing exercises, corroborating the chosen option. Why Other Options Are Wrong: 20–40 atm are feasible but were less common for traditional reciprocating engines; 5 atm is rather low for efficient power output with practical engine sizes. Common Pitfalls: Assuming all boilers for engines operate at the much higher pressures used in large modern turbines; conflating absolute and gauge pressures when reading “atmospheres.” Final Answer: 10 atmospheres

Question 17

La-Mont boiler: typical working pressure range (forced-circulation water-tube) Select the usual working pressure range for a La-Mont boiler from the options below.

Accepted Answer

Introduction / Context: The La-Mont boiler is a high-pressure, forced-circulation, water-tube design. It circulates water through small-diameter tubes at high velocity using an external pump to prevent film boiling and ensure excellent heat transfer. Recognizing its pressure band distinguishes it from low-pressure shell boilers and from ultra-high-pressure once-through units. Given Data / Assumptions: Pressure units: 1 MN/m^2 = 1 MPa ≈ 10 bar. La-Mont design emphasizes high circulation ratios (typically >5) to avoid tube overheating. Industrial practice places La-Mont in a high but not supercritical pressure class. Concept / Approach: La-Mont boilers were developed to operate efficiently at high subcritical pressures. Typical ranges cited in teaching texts span roughly 2.5–15 MN/m^2 (≈25–150 bar), covering many high-pressure process and power duties without entering supercritical territory, which would be handled by Benson/once-through types. Step-by-Step Solution: Convert ranges to bar to build intuition: 2.5–15 MN/m^2 ≈ 25–150 bar.Compare with known La-Mont use: high-pressure but subcritical.Therefore, select 2.5 to 15 MN/m^2 from the choices.This fits the classic classification for La-Mont boilers. Verification / Alternative check: Design notes and exam references list La-Mont within this pressure span, emphasizing forced circulation to maintain stable boiling at elevated heat fluxes. Why Other Options Are Wrong: 0.5–10 MN/m^2 is too broad and includes very low pressures; 1–15 overlaps but includes many low-pressure cases atypical of La-Mont; 3.5–20 or “above 25” shift the range unnecessarily high, encroaching on regimes typical of other designs. Common Pitfalls: Confusing pressure capability with circulation method; assuming forced circulation implies supercritical operation—it does not. Final Answer: 2.5 to 15 MN/m^2

Question 18

Valve Gear Basics — Function of the Valve Rod In a reciprocating steam engine with a D-slide or piston valve, what is the primary mechanical function of the valve rod within the valve gear?

Accepted Answer

Introduction / Context: The valve gear of a steam engine times the admission and exhaust of steam. Core elements include the eccentric, eccentric rod, valve rod, and the valve itself (D-slide or piston valve). Understanding the distinct role of the valve rod helps learners correctly map each linkage to its kinematic function and avoid common mix-ups in mechanism questions. Given Data / Assumptions: A conventional slide-valve or piston-valve steam engine with crank, eccentric, eccentric rod, and valve rod. Continuous rotary motion at the crankshaft. Usual small clearances and rigid links assumed. Concept / Approach: The crankshaft carries an eccentric (or an equivalent cam/linkage) that generates an approximately simple-harmonic motion. This motion is transmitted by the eccentric rod to the valve rod. The valve rod’s job is straightforward: it directly drives the valve in a reciprocating manner relative to the valve face and ports. The conversion from rotary motion to reciprocating motion is accomplished by the eccentric and its rod; the valve rod then carries that reciprocation to the valve without changing motion type. Step-by-Step Solution: Identify the motion source: the crankshaft via an eccentric provides SHM-like displacement.Follow the transmission: eccentric → eccentric rod → valve rod.Determine the valve rod’s role: deliver reciprocating motion directly to the valve.Conclude: the valve rod primarily provides reciprocating motion to the slide/piston valve. Verification / Alternative check: Mechanism sketches show the valve rod collinear with the valve spindle, constrained to linear motion, confirming the function is transmission of reciprocation, not conversion. Why Other Options Are Wrong: Converting piston reciprocation to crank rotation is the connecting rod’s job, not the valve rod (option b).Converting crank rotation to reciprocation is done by the eccentric and eccentric rod (option c).Simple harmonic characteristics arise from eccentric geometry; the valve rod does not impose SHM by itself (option d). Common Pitfalls: Confusing the names “eccentric rod” and “valve rod”. Remember: eccentric rod converts and transmits; valve rod mainly transmits to the valve. Final Answer: to provide reciprocating motion to the slide valve

Question 19

Valve Gear Basics — Function of the Eccentric Rod In a typical slide-valve steam engine, what is the principal function of the eccentric rod within the valve gear mechanism?

Accepted Answer

Introduction / Context: The eccentric and its rod are the heart of classical valve gear. They take the crankshaft’s rotation and generate the alternating linear motion needed to actuate the D-slide or piston valve, controlling port opening and closing relative to piston position. Given Data / Assumptions: Crankshaft fitted with an eccentric (offset disk) at a prescribed angle. Eccentric rod connected between eccentric strap and valve rod. Negligible backlash and rigid connections for basic kinematic description. Concept / Approach: As the crankshaft rotates, the eccentric generates a sinusoidal displacement at its strap. The eccentric rod transmits this to the valve rod, converting rotary input into a nearly simple-harmonic, linear output. This reciprocation shifts the valve to alternately uncover and cover steam ports. Hence, the eccentric rod’s role is the conversion and transmission of motion from rotation to linear reciprocation for the valve train. Step-by-Step Solution: Start with rotation at the crankshaft.Eccentric creates offset circular motion → sinusoidal displacement at the strap.Eccentric rod conducts this motion to the valve rod linearly.Valve rod then directly drives the slide/piston valve to time admission and exhaust. Verification / Alternative check: Kinematic diagrams and indicator tests show admission timing changing with eccentric throw and angle, proving the eccentric rod’s conversion function. Why Other Options Are Wrong: Providing reciprocation to the valve is the immediate effect at the valve rod (option a is incomplete; the conversion is the eccentric rod’s key role).Piston-to-crank energy conversion is the connecting rod’s duty (option b).SHM arises from the eccentric geometry; the rod itself does not impose SHM beyond transmitting the motion (option d). Common Pitfalls: Mixing up connecting rod vs. eccentric rod functions; they serve different sub-systems (power vs. valve gear). Final Answer: to convert rotary motion of the crankshaft into to and fro motion of the valve rod

Question 20

Compound Steam Engines — Flywheel Size Requirement Among receiver, tandem, and Woolf compound steam engine arrangements, which generally requires the smallest flywheel for smooth running?

Accepted Answer

Introduction / Context: Flywheels compensate for cyclic torque fluctuation. Different compound engine layouts distribute expansion and phasing differently, altering the uniformity of torque and thus the necessary flywheel inertia for steady speed. Given Data / Assumptions: Receiver compound: HP and LP cylinders on separate cranks, typically about 90° apart, connected via a receiver. Tandem compound: HP and LP in line on a common rod, single crank. Woolf compound: cylinders share a common rod without an intermediate receiver. Concept / Approach: Torque ripple diminishes when power strokes are phase-shifted. The receiver type uses cranks set roughly at right angles; when one cylinder approaches dead center (low torque), the other is near mid-stroke (high torque). The receiver also buffers pressure pulsations between cylinders. This phasing improves turning-moment uniformity, allowing a smaller flywheel than single-crank tandem or Woolf types, where torque peaks coincide more and smoothing is poorer. Step-by-Step Solution: Compare crank phasing: receiver → two cranks ~90° apart; tandem/Woolf → single crank.Relate phasing to torque uniformity: orthogonal cranks even out net torque.Infer flywheel requirement: better uniformity → smaller required inertia. Verification / Alternative check: Indicator and turning-moment diagrams for receiver compounds show reduced amplitude of fluctuation compared to single-crank designs. Why Other Options Are Wrong: Tandem type: single crank → larger torque swings → larger flywheel.Woolf type: also single crank and no receiver → limited smoothing capacity.all of these: cannot all be smallest simultaneously; receiver type is best. Common Pitfalls: Confusing “more cylinders” with smoother torque irrespective of phasing; phasing is crucial. Final Answer: Receiver type

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