Question 1

Equivalent Evaporation — “From and At 100°C” Definition The term “equivalent evaporation from and at 100°C” is defined as the amount of water evaporated from feedwater at __________ into dry saturated steam at 100°C at normal atmospheric pressure (1…

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Introduction / Context: Boiler performance is compared using a common reference called “equivalent evaporation from and at 100°C.” This normalizes different steam conditions and feedwater temperatures to a standard basis so that efficiencies and steaming capacities can be compared meaningfully across plants and fuels. Given Data / Assumptions: Reference pressure is atmospheric (approximately 1.013 bar). Reference steam state is dry saturated steam at 100°C. Reference feedwater temperature is also at the saturation temperature of 100°C. Concept / Approach: “From and at 100°C” literally means the water begins at 100°C and is evaporated to dry saturated steam at 100°C. On this basis, the latent heat of vaporization used for comparison is the standard value at 100°C. Real boilers seldom operate exactly at these conditions; the definition provides a calculation method to convert the actual heat absorbed by feedwater into an equivalent mass that would have been evaporated under the standard reference condition. Step-by-Step Solution: Define the reference latent heat: h_fg at 100°C.Compute actual heat added to water/steam in the boiler per kilogram under operating conditions.Divide actual heat added by h_fg at 100°C to obtain equivalent kilograms evaporated “from and at” 100°C.Hence, the feedwater reference temperature in the definition is 100°C. Verification / Alternative check: Textbook boiler trials convert the measured enthalpy rise of feedwater to this standard using the factor of evaporation, which is based on the same 100°C reference. Why Other Options Are Wrong: 0°C, 40°C, 60°C: These are common feedwater temperatures in practice, but they are not the reference for the standardized “from and at 100°C” definition. Common Pitfalls: Confusing “equivalent evaporation” with actual mass flow; it is a normalized reference for comparison, not necessarily the real evaporated mass at operating conditions. Final Answer: 100°C

Question 2

Boiler Draught — Identify the Incorrect Statement Which one of the following statements about boiler draught systems is WRONG?

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Introduction / Context: Draught is the slight pressure difference that draws combustion air into the furnace and expels flue gases. It may be produced naturally by a chimney (buoyancy) or mechanically by fans (forced, induced, or balanced draught). Understanding what draught does—and does not do—helps in choosing efficient and safe systems. Given Data / Assumptions: Comparing natural and mechanical draught arrangements for conventional boilers. Considering typical impacts on fuel use, stack height, and system design. Concept / Approach: Natural draught depends on temperature difference and stack height, and is relatively weak and difficult to control; it often requires higher excess air and results in higher stack temperatures, which can increase fuel consumption. Mechanical draught uses fans to control and increase draught, allowing better air–fuel control, lower excess air, and reduced stack temperature, thereby reducing specific fuel consumption at a given output. Step-by-Step Solution: Evaluate (a): Mechanical draught reduces reliance on tall chimneys because fans provide the needed pressure head. This is correct.Evaluate (b): Claiming natural draught reduces fuel consumption is generally incorrect. Natural draught frequently leads to less efficient operation than well-designed mechanical systems.Evaluate (c): Balanced draught combines forced-draught (FD) and induced-draught (ID) fans to maintain near-atmospheric furnace pressure while ensuring adequate flow; also correct.Therefore, the wrong statement is (b). Verification / Alternative check: Plant performance data consistently show improved efficiency with controlled mechanical draught due to reduced excess air and improved heat recovery (economiser/air-preheater effectiveness). Why Other Options Are Wrong: (a) and (c) are standard practices; “all of the above” cannot be correct since at least two are true. Common Pitfalls: Equating “free” natural draught with lower fuel use; in reality, poor control and higher stack losses can negate any perceived benefits. Final Answer: The natural draught reduces the fuel consumption.

Question 3

Boiler Mountings — Typical Number of Water Level Indicators (Gauges) In standard practice for boilers, how many independent water level indicators (gauge glasses) are generally provided for safety and monitoring?

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Introduction / Context: Maintaining correct water level is critical to boiler safety. Too low a level risks overheating of heating surfaces; too high a level risks water carryover to steam lines and turbines. Therefore, codes and good engineering practice require redundancy in water level indication. Given Data / Assumptions: Conventional shell or water-tube boilers in industrial service. At least one direct-reading gauge glass plus redundancy. Auxiliary devices like remote level transmitters may also be installed but do not replace direct indicators in many jurisdictions. Concept / Approach: Two independent water level indicators are generally provided so that one can corroborate the other. This redundancy helps detect instrument failure, blockage, or misreading, and supports safer operation and compliance with regulatory inspections. Step-by-Step Solution: Identify the safety requirement: clear, reliable indication of drum or shell level.Apply redundancy principle: two gauge glasses are customary and often mandated.Therefore, the typical number is two. Verification / Alternative check: Many boiler standards and inspection authorities specify two direct level indicators on high-pressure boilers, with additional remote indicators or alarms as supplements. Why Other Options Are Wrong: One: lacks redundancy; not preferred for safety. Three or four: unusual and unnecessary in most designs, though multiple indicators may be used in special cases. Common Pitfalls: Relying solely on remote transmitters; direct visual indication is critical during commissioning and abnormal events. Final Answer: two

Question 4

La-Mont Boiler — Circulation Method In a La-Mont water-tube boiler, the circulation of water and steam through the evaporator tubes is primarily:

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Introduction / Context: The La-Mont boiler is a notable high-pressure water-tube unit designed to prevent film boiling and overheating of evaporator tubes through vigorous circulation. Understanding its circulation principle explains its suitability for higher heat fluxes and rapid load changes compared with natural circulation designs. Given Data / Assumptions: Water-tube geometry with a steam–water separator or drum. External circulation pump recirculates water through the evaporator tubes. Operation at pressures where natural buoyancy may be insufficient for required mass flux. Concept / Approach: La-Mont boilers employ a high-capacity recirculation pump to force water through the evaporator circuit at rates several times the steam generation rate. This stabilizes tube wall temperature, avoids local dry-out, and improves heat transfer. The forced flow also enables uniform distribution among parallel tube circuits. Step-by-Step Solution: Identify hardware: circulation pump, downcomers, risers, separator.Trace flow: pump drives water through evaporator tubes; mixture returns to separator; steam is separated while water recycles.Conclude: circulation is forced, not natural. Verification / Alternative check: La-Mont is categorically listed among forced-circulation boilers alongside Benson (once-through) and Velox types in standard boiler classifications. Why Other Options Are Wrong: Natural circulation: would rely on buoyancy only; La-Mont explicitly uses a pump. Common Pitfalls: Confusing La-Mont with drum-type natural circulation boilers because both may include a separator drum; the critical distinction is the circulation pump. Final Answer: forced

Question 5

Thermodynamic Basis — What Fundamental Law Do Steam Engines Rely On? All steam engines (prime movers converting heat to work) fundamentally operate in accordance with which law of thermodynamics?

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Introduction / Context: Steam engines convert thermal energy of steam into mechanical work using a thermodynamic cycle. While all physical processes must obey energy conservation (first law), the directionality and limits of heat-to-work conversion are governed by the second law. Recognizing this distinction is essential for understanding why no engine can be 100% efficient. Given Data / Assumptions: Heat engine operating between a hot reservoir (boiler) and a cold reservoir (condenser/ambient). Cycle includes expansion doing work and heat rejection. Concept / Approach: The first law states energy conservation: Q_in − Q_out = W_net. However, it does not restrict the fraction of heat that can be converted to work. The second law introduces entropy and the necessity of heat rejection; it establishes that complete conversion of heat into work in a cyclic process is impossible and sets the theoretical efficiency ceiling via the Carnot limit. Therefore, the very possibility and limitation of a heat engine depend on the second law. Step-by-Step Solution: Identify engine as a heat engine: requires a temperature difference.Apply first law: balances energy but allows, in principle, any efficiency if taken alone.Apply second law: dictates irreversibility and that some heat must be rejected; efficiency is fundamentally limited.Hence, steam engines “work on” (are constrained by) the second law. Verification / Alternative check: Ideal Rankine/Carnot analyses show that as sink temperature approaches source temperature, efficiency tends to zero, a direct consequence of the second law. Why Other Options Are Wrong: Zeroth law: defines temperature and thermal equilibrium, not engine operation. First law: necessary but not sufficient to explain direction and limits of conversion. None: incorrect because thermodynamic laws govern all heat engines. Common Pitfalls: Attributing engine feasibility solely to energy conservation; without the second law, perpetual heat-to-work conversion would not be prohibited. Final Answer: Second law of thermodynamics

Question 6

Boiler Trials — Factor of Evaporation (FoE) For practical boilers operating at pressures above atmospheric and with feedwater below saturation temperature, the factor of evaporation (FoE) is generally:

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Introduction / Context: The factor of evaporation (FoE) is used to convert actual steam generation under given conditions to the standardized “equivalent evaporation from and at 100°C.” It allows fair comparison of different boilers and operating states by normalizing to a common reference enthalpy rise. Given Data / Assumptions: Operating pressure above atmosphere (so saturation temperature > 100°C). Feedwater temperature typically below 100°C. FoE = (actual enthalpy rise of feedwater to delivered steam) / (latent heat at 100°C). Concept / Approach: Because the actual boiler must raise feedwater from below 100°C to saturation at a higher temperature and perhaps superheat it, the enthalpy rise per kilogram is usually greater than the latent heat at 100°C. Therefore, when we divide this larger enthalpy rise by the 100°C latent heat, the resulting ratio (FoE) exceeds 1. Only in the special case where feedwater enters at 100°C and steam is produced as dry saturated at 100°C would FoE be exactly 1. Step-by-Step Solution: Define FoE = (h_steam_actual − h_feed_actual) / h_fg_100.For typical plants: h_steam_actual includes sensible heating to saturation at elevated pressure plus latent heat and perhaps superheat.Since h_feed_actual is below 100°C and h_steam_actual is at conditions hotter than 100°C, numerator > h_fg_100.Therefore, FoE > 1 in normal practice. Verification / Alternative check: Worked examples routinely yield FoE values like 1.05 to 1.3 or higher depending on pressure and feedwater heating. Only laboratory reference conditions make FoE approach unity. Why Other Options Are Wrong: Equal to unity: true only for the special “from and at 100°C” case. Less than unity: would imply doing less enthalpy rise than the reference, contrary to most real operating states. Common Pitfalls: Confusing FoE with boiler efficiency; FoE is a normalization factor, not an efficiency measure by itself. Final Answer: greater than unity

Question 7

Boiler Heat Recovery — Name the Device Identify the device used to heat the feedwater by utilizing sensible heat in the exhaust flue gases before they leave through the chimney.

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Introduction / Context: Enhancing boiler efficiency requires recovering as much heat as practical from the flue gases. One key accessory in the gas path is designed specifically to preheat feedwater, reducing the fuel needed to raise it to saturation. Correctly identifying and applying this device is central to energy-saving retrofits and modern boiler designs. Given Data / Assumptions: Hot flue gases exit the boiler at elevated temperature. Feedwater is cooler than saturation temperature. A heat exchanger is placed in the gas path before the stack. Concept / Approach: An economiser is a feedwater heater that transfers sensible heat from exhaust gases to incoming feedwater. By raising feedwater temperature, it reduces the enthalpy rise required in the evaporator and superheater, thereby decreasing fuel consumption. This is distinct from a superheater, which raises steam temperature above saturation, and from safety or control fittings like fusible plugs and stop valves. Step-by-Step Solution: Locate the device in the flue-gas train: typically downstream of the evaporator/superheater and upstream of the air preheater or stack.Understand duty: increase feedwater temperature without changing its phase.Name the device: economiser. Verification / Alternative check: Boiler heat-balance calculations show a direct reduction in stack losses and fuel use when an economiser is installed and properly controlled. Why Other Options Are Wrong: Fusible plug: a safety device to prevent crown-sheet overheating. Superheater: heats steam, not feedwater. Stop valve: isolates steam flow; no heat recovery function. Common Pitfalls: Confusing economisers with air preheaters; the latter heat combustion air, not feedwater. Both may be used together for maximum efficiency. Final Answer: economiser

Question 8

Steam engine fundamentals: In a classic reciprocating steam engine, fuel is burnt in a separate furnace/boiler and steam is then admitted into the cylinder for expansion. Combustion of fuel takes place __________ the engine cylinder of a steam engin…

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Introduction / Context: A steam engine is a prime example of an external combustion system. Understanding where combustion occurs helps distinguish steam engines from internal combustion engines such as diesel and petrol engines, and is foundational for boiler–engine system design and safety. Given Data / Assumptions: The working fluid in the engine cylinder is steam generated in a separate boiler. Fuel (coal, oil, gas, biomass) is burned in a furnace that is part of the boiler, not the cylinder. The engine cylinder only receives steam for doing mechanical work on the piston. Concept / Approach: External combustion means the chemical energy release from fuel happens outside the power cylinder. Heat from combustion converts water to steam in the boiler. The steam then expands in the engine cylinder, creating pressure forces on the piston without any flame or combustion taking place inside the cylinder itself. Step-by-Step Solution: Identify the energy conversion locations: boiler (chemical to thermal) and engine (thermal to mechanical).Note the cylinder environment: contains steam only, plus exhaust to condenser or atmosphere.Recognize safety and design implications: no spark, injector, or mixing of air–fuel inside the cylinder.Therefore, combustion occurs outside the cylinder, in the boiler furnace. Verification / Alternative check: Compare with diesel/petrol engines where fuel–air burns inside the cylinder; the steam engine lacks an internal ignition event. Why Other Options Are Wrong: Inside/partly inside: contradicts external combustion definition. Inside only during starting/high pressure: there is never in-cylinder combustion in a true steam engine. Common Pitfalls: Confusing steam engines with internal combustion engines due to similar piston–crank mechanisms. Final Answer: outside

Question 9

Boiler feed system: Before feedwater enters the boiler, its pressure must be raised to exceed boiler pressure for safe delivery. Name the device primarily used to raise the feedwater pressure.

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Introduction / Context: Every steam generator requires a mechanism to deliver water into the boiler against the internal steam pressure. Selecting and understanding the correct device ensures reliable operation and prevents starvation or water hammer. Given Data / Assumptions: Boiler operates at positive pressure relative to atmosphere. Feedwater source is at low pressure (tank or condenser hotwell). Continuous, controllable delivery into the boiler is required. Concept / Approach: A feed pump (reciprocating, rotary, or centrifugal multistage) raises the pressure of feedwater to a value higher than boiler pressure so the feed check valve opens and water enters the drum or once-through circuit. An injector can also lift and force feedwater using a steam jet, but the generic and primary device specified by practice and codes is the feed pump. The feed check valve merely prevents backflow; it does not raise pressure. A pressure gauge only indicates pressure; it adds no energy to the water. Step-by-Step Solution: Identify need: p_feed > p_boiler at entry.Select device that adds hydraulic head to water: feed pump.Confirm ancillary roles: feed check valve prevents backflow; injector is an alternative jet device, not the baseline answer.Hence, the correct choice is feed pump (feedpump). Verification / Alternative check: Plant P&IDs show feed pumps upstream of the feed check valve; injectors are used on some locomotives and small boilers but are still categorized as feed-pumping devices driven by steam jets. Why Other Options Are Wrong: Injector: can serve the same purpose in specific setups but the generic term requested is feed pump. Feed check valve: non-return isolation, no pressure rise. Pressure gauge/air preheater: measurement and air-heating devices, not pumping equipment. Common Pitfalls: Assuming the check valve raises pressure or that the injector is the universal solution across all plants. Final Answer: feedpump

Question 10

Draught production in boilers and furnaces: Which of the following methods can be used to produce draught (the driving pressure difference) in a combustion system?

Accepted Answer

Introduction / Context: Draught (draft) is the small pressure difference that draws air through the grate and carries flue gases through the passes to the stack. Knowing how draught can be created is a key aspect of boiler operation and design. Given Data / Assumptions: Combustion needs steady airflow through fuel bed or burners. Pressure differences are small (commonly a few mm to a few tens of mm of water column). Multiple technologies exist to generate this pressure difference. Concept / Approach: Natural draught is created by a chimney due to density differences between hot gases and ambient air. Mechanical draught is provided by fans (forced-draught, induced-draught, or balanced-draught arrangements). Ejector (steam-jet) draught uses a high-velocity steam jet to entrain flue gases and create suction, historically important in locomotives and small boilers. Step-by-Step Solution: List available methods: chimney, fans, steam jet.Recognize all three produce a pressure difference across the furnace/grate.Select the combined option that includes all methods.Therefore, answer: all of these. Verification / Alternative check: Practical plants may use tall chimneys or FD/ID fans; steam jet ejectors are integral to locomotive boiler exhaust systems. Why Other Options Are Wrong: Each single-method option is incomplete; air preheater only recovers heat and does not generate draught. Common Pitfalls: Confusing air heating with pressure generation; ignoring ejector-based draught in historical designs. Final Answer: all of these

Question 11

Clearance volume effect in steam engines: Evaluate the statement: “The clearance in the engine cylinder has no effect on steam consumption.”

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Introduction / Context: Clearance volume is the space remaining when the piston is at top dead centre. In reciprocating steam engines, this volume significantly influences indicated work and steam consumption because a portion of the cylinder volume is filled with residual steam and fresh steam each cycle. Given Data / Assumptions: Cylinder has finite clearance volume V_c. Steam admission is cut off before end of stroke to allow expansion. Exhaust and compression/rew-expansion processes occur near dead centres. Concept / Approach: Clearance steam remains in the cylinder at the end of exhaust and expands during the compression and early expansion phases. This trapped steam reduces net fresh steam admitted for a given cut-off but also reduces effective expansion ratio. The presence of clearance alters the mean effective pressure and, for a given load and speed, affects the mass of steam required per cycle, hence the specific steam consumption. Step-by-Step Solution: Recognize that a non-zero V_c changes indicator card shape (compression and re-expansion loops).For the same nominal cut-off, effective admission of fresh steam changes due to residual steam occupying V_c.Mean effective pressure and indicated work depend on clearance; to meet the same brake load, steam rate adjusts.Therefore, clearance does affect steam consumption; the given statement is false. Verification / Alternative check: Indicator diagrams with and without clearance show different areas; textbooks provide correction factors for clearance effects. Why Other Options Are Wrong: “True” variants ignore compression/rew-expansion phenomena that are directly tied to clearance volume. Common Pitfalls: Assuming clearance only affects volumetric efficiency in I.C. engines; in steam engines it also alters expansion and m.e.p. Final Answer: False

Question 12

Locomotive boiler construction detail: Typical small-diameter fire tubes used in a locomotive boiler have approximately what outside diameter?

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Introduction / Context: Locomotive boilers are multi-tubular fire-tube boilers employing many small tubes to obtain large heating surface and rapid steam-raising capability. Knowing the typical order of tube diameters helps differentiate between fire-tube and water-tube proportions. Given Data / Assumptions: Fire tubes carry hot gases through the boiler shell water space. Practical locomotive practice uses tube diameters of a few centimetres, not a few millimetres. Multiple parallel tubes are fitted to maximize heat transfer area. Concept / Approach: Fire-tube diameters are selected to balance gas-side friction losses, heat transfer, and soot deposition. Values on the order of tens of millimetres are common. Options of 4.75 mm, 5.47 mm, and 7.45 mm are unrealistically small for locomotive boilers; approximately 47.5 mm aligns with textbook and historic practice ranges (about 38–51 mm). Step-by-Step Solution: Reject sub-centimetre options as impractical for gas flow in locomotives.Compare with known range: around 40–50 mm typical.Select nearest listed value: 47.5 mm.Conclude that 47.5 mm is the correct scale for locomotive fire tubes. Verification / Alternative check: Historical locomotive boiler drawings and standard problems cite tube diameters roughly 40–50 mm, with large counts to achieve the required heating surface. Why Other Options Are Wrong: Very small diameters would create excessive gas-side pressure drop and maintenance issues; 75.0 mm is larger than typical locomotive fire-tube practice. Common Pitfalls: Confusing fire-tube diameters with superheater elements or economiser tubes; mixing outside and inside diameters. Final Answer: 47.5 mm

Question 13

Compound engine configuration (Woolf type): In a Woolf type compound steam engine, what is the usual angular relationship between the high-pressure and low-pressure cranks?

Accepted Answer

Introduction / Context: Compounded steam engines expand steam in stages. The Woolf type is a classic arrangement historically noted for its direct HP-to-LP connection. Recognizing crank phasing clarifies torque characteristics and valve timing. Given Data / Assumptions: Woolf type indicates direct coupling of HP exhaust to LP admission without a large external receiver. Often constructed in tandem with a common piston rod and a single crank. Goal: smooth expansion with minimal receiver volume. Concept / Approach: In the Woolf type, the HP and LP pistons are mechanically aligned such that both move together; effectively the cranks are coincident. This corresponds to 0° phase difference. The arrangement simplifies flow between cylinders and reduces inter-stage storage volume, affecting pressure diagrams and avoiding pulsations associated with widely phased cranks. Step-by-Step Solution: Identify Woolf type → typically tandem construction.Common piston rod and single crank → same angular position for both cylinders.Therefore, crank angle difference = 0°.Select the option “0° to each other”. Verification / Alternative check: Engineering texts contrast Woolf (0°) with cross-compound (90°) for smoother torque but with a receiver. Why Other Options Are Wrong: 180° and 90° commonly describe multi-cylinder phasing for balance, not the Woolf tandem arrangement. “None of these” is unnecessary because 0° is explicitly listed. Common Pitfalls: Confusing “Woolf” with “receiver type” where a receiver allows other crank phasings (e.g., 90°). Final Answer: 0° to each other

Question 14

Steam power plant efficiencies: The ratio of the heat equivalent of brake power (useful output at the shaft) to the energy supplied in the steam to the engine is called the __________.

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Introduction / Context: Performance metrics for steam engines and turbines include several efficiencies that compare inputs and outputs at different boundaries: thermal, mechanical, and overall. Selecting the correct definition ensures meaningful analysis and benchmarking. Given Data / Assumptions: Brake power B.P. is measured at the shaft output. Energy supplied in steam is based on mass flow and steam enthalpy rise relative to feedwater. Steady-state operation. Concept / Approach: Brake thermal efficiency η_bt is defined as: η_bt = (heat equivalent of brake power) / (rate of heat supplied in steam). Indicated thermal efficiency uses indicated power in the numerator, while mechanical efficiency is B.P. / I.P. Overall efficiency often includes the boiler, i.e., useful power divided by fuel energy rate. Therefore, when comparing B.P. directly to steam energy supplied to the engine, we refer specifically to brake thermal efficiency. Step-by-Step Solution: Identify numerator: B.P. converted to heat equivalent (B.P. in kW corresponds to kJ/s).Identify denominator: steam energy input rate = m_dot * (h_in − h_ref).Form ratio: η_bt = (B.P. in kW) / (steam heat rate in kW).Select the term matching this ratio: brake thermal efficiency. Verification / Alternative check: Cross-check with definitions: η_it uses I.P.; η_mech = B.P./I.P.; η_overall = B.P./(fuel heat rate). Why Other Options Are Wrong: Mechanical efficiency does not reference steam energy; it compares B.P. to I.P. Indicated thermal efficiency uses I.P. as output. Overall efficiency uses fuel energy as input, not steam energy. Nozzle efficiency is for nozzles, not whole engines. Common Pitfalls: Mixing up “thermal” qualifiers at indicated vs brake levels; forgetting boundary definitions. Final Answer: brake thermal efficiency

Question 15

Role of a condenser in a steam power plant: Which of the following effects are achieved by employing a surface condenser in the Rankine cycle?

Accepted Answer

Introduction / Context: Condensers are integral to modern steam power plants. By condensing exhaust steam, they influence thermodynamic performance and facilitate feedwater recovery. Understanding their multiple simultaneous effects is crucial for Rankine cycle analysis. Given Data / Assumptions: Surface condenser operating under vacuum. Exhaust steam from turbine or engine enters the condenser. Cooling water removes latent heat, converting steam to water. Concept / Approach: Condensation reduces the exhaust pressure (back pressure) below atmospheric, thereby increasing the pressure ratio across the turbine/engine and the overall expansion ratio. At the same time, it lowers the temperature of the exhaust steam to near the saturation temperature at condenser pressure and returns condensate for reuse, improving cycle efficiency and reducing makeup water needs. Step-by-Step Solution: Condenser action: condense exhaust → create vacuum → lower p_back.Lower p_back increases expansion from boiler pressure to condenser pressure → higher specific work output.Heat rejection to cooling water reduces exhaust steam temperature during phase change.Collectively, these are described by the option “all of these”. Verification / Alternative check: Rankine T–s and p–v diagrams show increased area (work) when condenser pressure drops; condenser outlet condensate temperature matches saturation at vacuum pressure. Why Other Options Are Wrong: Any single effect description is incomplete; the condenser simultaneously reduces back pressure, increases expansion ratio, and reduces exhaust temperature. It does not directly raise boiler pressure (that is a function of boiler and control systems). Common Pitfalls: Attributing increases in boiler pressure to condenser performance; overlooking cooling water constraints and air removal for maintaining vacuum. Final Answer: all of these

Question 16

Cylinder sizing in compound engines: In a compound steam engine the first stage of expansion occurs in the high-pressure (H.P.) cylinder and the final expansion in the low-pressure (L.P.) cylinder. The diameter of the H.P. cylinder is __________ the…

Accepted Answer

Introduction / Context: Compound engines divide expansion between cylinders operating at different pressures. The cylinder sizes are chosen so that each stage contributes appropriately to the total work with reasonable steam velocities and mechanical stresses. Given Data / Assumptions: H.P. cylinder handles high inlet pressure and small specific volume. L.P. cylinder handles lower pressure and larger specific volume after partial expansion. Similar piston speeds are desirable across stages. Concept / Approach: Because steam specific volume increases as pressure falls during expansion, the L.P. cylinder must have a larger volume (and hence larger diameter for similar stroke) to pass the increased volumetric flow. Therefore, the H.P. cylinder diameter is less than that of the L.P. cylinder. This sizing keeps velocities and pressure drops within acceptable limits and balances indicated power between cylinders. Step-by-Step Solution: Relate specific volumes: v_LP > v_HP for the same mass flow.To accommodate higher volumetric flow at low pressure, cylinder area must be larger.For equal strokes, larger area implies larger diameter for L.P. cylinder.Hence, D_HP < D_LP → “less than”. Verification / Alternative check: Design examples and textbooks show L.P. cylinder diameter commonly 1.5–2 times the H.P. cylinder diameter depending on pressure ratios and receiver effects. Why Other Options Are Wrong: Equal/more than/double: contradict basic volumetric reasoning unless strokes differ greatly; typical practice uses larger L.P. diameter. Indeterminate: although exact ratio needs data, the directional relation is well established. Common Pitfalls: Assuming equal diameters or ignoring volumetric flow increase during expansion. Final Answer: less than

Question 17

Steam generation rate comparison: For a given size, the rate of flow of steam (steam-raising capacity per unit time) is generally __________ in fire-tube boilers compared with water-tube boilers.

Accepted Answer

Introduction / Context: Boilers are broadly classified as fire-tube (hot gases in tubes, water outside) and water-tube (water/steam in tubes, hot gases outside). Their steam-raising rates differ due to heat-transfer area, water content, and permissible furnace rates. Given Data / Assumptions: Like-for-like overall size comparison. Similar fuel and firing conditions. Standard design practices for safe operation. Concept / Approach: Water-tube boilers have many small-diameter water-filled tubes exposed to hot gases, yielding high heat-transfer area, rapid heat absorption, and lower water inventory. They can safely operate at higher pressures and respond quickly to load changes, giving higher steam-generation rates for a given size. Fire-tube boilers, with larger water content and fewer, larger gas tubes, raise steam more slowly and are suited to moderate rates. Step-by-Step Solution: Compare heat-transfer surfaces per unit volume: water-tube > fire-tube.Consider water inventory and response time: water-tube is quicker and supports higher steaming rates.Therefore, for the same size, steam flow rate of fire-tube boilers is less.Select “less” as the correct descriptor. Verification / Alternative check: Industry practice uses water-tube units for high-capacity, high-pressure power generation; fire-tube units serve smaller plants with lower steaming rates. Why Other Options Are Wrong: “More/the same/independent/unpredictable” ignore the fundamental geometry and heat-transfer differences that favor water-tube designs for higher rates. Common Pitfalls: Equating pressure capability with steaming rate without considering surface area-to-volume ratios and circulation characteristics. Final Answer: less

Question 18

Condenser classification in steam plants Identify the condenser type in which the circulating water flows inside a bundle of tubes while steam surrounds the outside of those tubes and condenses on their outer surfaces.

Accepted Answer

Introduction / Context: Condensers are used in steam power plants to convert exhaust steam back to water (condensate), improving cycle efficiency and enabling feedwater recovery. Two broad types are non-contact (surface) and contact (jet, barometric). Knowing which fluids mix—or do not mix—helps identify the condenser from its flow arrangement. Given Data / Assumptions: Steam is on one side; cooling water is on the other, possibly separated by tube walls. Flow description: water inside tubes; steam outside around the tube bundle. Goal: condense steam without contaminating the condensate. Concept / Approach: A surface condenser is a non-mixing heat exchanger. Cooling water flows through tubes; exhaust steam condenses on the outer tube surfaces. Because the two streams are separated by metal walls, the condensate remains essentially pure and can be pumped back as feedwater. Jet or barometric condensers intentionally mix steam and water; evaporative condensers reject heat by partial evaporation to ambient air, not by a closed water circuit. Step-by-Step Solution: Identify whether fluids mix: here, they do not—this implies a surface (non-contact) type.Check flow path: water inside tubes, steam outside around the tube bank.Match the description with textbook configurations: shell-and-tube surface condenser.Conclude that the correct name is a surface condenser. Verification / Alternative check: Plant drawings show exhaust steam entering the condenser shell, contacting tube exteriors; circulating water is pumped through the tubes to a cooling tower or river, confirming non-contact operation. Why Other Options Are Wrong: Jet and barometric condensers mix cooling water with steam, contaminating condensate. An evaporative condenser rejects heat to air via evaporation and is typical in refrigeration, not steam-cycle exhaust service. A “contact-mixing condenser” is just a generic term for jet types. Common Pitfalls: Confusing “barometric” (a jet condenser with a high tailpipe) with “surface.” Assuming any tube bundle automatically means evaporative type. Final Answer: surface condenser

Question 19

Boiler mountings for safe and proper operation Which of the following are standard boiler mountings fitted to ensure correct indication, pressure control, and safety?

Accepted Answer

Introduction / Context: Boiler mountings are the essential fittings provided on a boiler to ensure safe operation, measurement, and control. Examinations often test whether learners can distinguish “mountings” (mandatory safety/operational devices) from “accessories” (performance enhancers such as superheaters or economizers). Given Data / Assumptions: A working boiler producing saturated or superheated steam. Need for level monitoring, pressure indication, and overpressure protection. Standard codes require particular devices to be present and functional. Concept / Approach: Typical mountings include: water level indicators (to show safe level and avoid overheating), pressure gauges (to display operating pressure), safety valves (to relieve excess pressure), steam stop valve, feed check valve, blow-off cock, and fusible plug (on certain boilers). The items listed—water level indicator, pressure gauge, safety valve—are all canonical mountings. Step-by-Step Solution: Identify category: are these devices mountings or accessories?Water level indicator: mandatory for safe level control.Pressure gauge: mandatory to monitor internal pressure.Safety valve: mandatory to prevent overpressure.Therefore, selecting “all of these” is correct. Verification / Alternative check: Regulatory boiler codes and standard textbooks list these under “mountings,” separate from accessories like economizers and superheaters. Why Other Options Are Wrong: Choosing any single device omits other equally mandatory fittings; “none” is incorrect because these are universally required. Common Pitfalls: Confusing mountings with accessories; overlooking that multiple indicators/valves are often required in parallel for redundancy. Final Answer: all of these

Question 20

Engine performance terminology The ratio of brake power (measured at the shaft) to indicated power (calculated from the indicator diagram) is called:

Accepted Answer

Introduction / Context: Performance of reciprocating engines is expressed using several efficiencies. Distinguishing between indicated and brake quantities is fundamental: indicated power is the power developed in the cylinder, while brake power is the usable output at the crankshaft after mechanical losses. Given Data / Assumptions: Indicated power comes from pressure–volume (indicator) diagrams. Brake power is obtained from dynamometer measurements. Mechanical losses include friction, pumping losses, and auxiliaries. Concept / Approach: Mechanical efficiency quantifies how effectively the engine converts the gross cylinder power into net shaft power: η_mech = BP / IP. Overall or brake thermal efficiency compares brake power with fuel energy input; indicated thermal efficiency compares indicated power with fuel energy input. Volumetric efficiency concerns air-charge filling, not power conversion. Step-by-Step Solution: Define IP from area of indicator card and engine geometry.Measure BP using a dynamometer: BP = 2 * π * N * T.Compute η_mech = BP / IP.Recognize that η_mech < 1 due to friction and accessory loads. Verification / Alternative check: Test results typically show η_mech between 0.75 and 0.9 for healthy engines, aligning BP lower than IP. Why Other Options Are Wrong: Overall and brake thermal efficiencies relate to fuel energy; indicated thermal uses indicated power; volumetric efficiency is unrelated to power transmission losses. Common Pitfalls: Confusing brake and indicated values or mixing thermal and mechanical efficiency definitions. Final Answer: mechanical efficiency

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