Q: Reheat factor — definition check “The ratio of the cumulative heat drop (sum of effective enthalpy drops over all stages) to the overall isentropic heat drop is called the reheat factor.” Is this statement correct?

Introduction / Context: Because real stages have varying efficiencies and the specific volume increases as steam expands, the total of the per-stage heat drops may not exactly equal the single isentropic drop between inlet and exhaust. The reheat factor summarizes this difference and is used in detailed performance estimates. Given Data / Assumptions: Multi-stage turbine with known inlet and exit states. Per-stage enthalpy drops are measurable or estimable. Isentropic reference is taken between the same end states. Concept / Approach: Reheat factor RF = (cumulative heat drop over stages) / (overall isentropic heat drop). In many practical turbines RF is slightly greater than 1 because of the way velocity and volume change across stages and how losses distribute, though the exact value depends on inlet conditions, back pressure, and stage design. Step-by-Step Solution: Compute per-stage effective enthalpy drop from measurements/calculations.Sum these drops to obtain the cumulative heat drop.Determine the ideal isentropic drop from inlet to exit states.Form RF = cumulative drop / isentropic drop. Verification / Alternative check: Performance maps often report RF around 1.02–1.07; when multiplied by internal efficiency, it helps reconcile stage-wise sums with overall turbine output. Why Other Options Are Wrong: Choosing “No” would contradict standard steam-turbine terminology used in design texts and practice. Common Pitfalls: Confusing RF with “reheat” in the Rankine cycle hardware; RF does not require an actual reheater between turbine sections. Final Answer: Yes

Q: Choked flow in nozzles — terminology When a nozzle operates at the condition of maximum mass flow rate (for fixed upstream total state), the nozzle is said to be:

Introduction / Context: Designers of turbines, ejectors, and propulsion systems rely on the concept of choked flow. At a particular back pressure ratio, the sonic condition is reached at the throat and the mass flow rate stops increasing even if the downstream pressure is reduced further. Given Data / Assumptions: Steady, adiabatic nozzle without shaft work. Negligible potential-energy changes; kinetic energy rise dominates. Ideal gas/steam behavior for defining the critical pressure ratio. Concept / Approach: Choking occurs when Mach number at the minimum area section equals 1. At this point, the mass flux ṁ/A depends only on upstream stagnation temperature and pressure and on gas properties. Further reducing back pressure changes the downstream expansion pattern but does not increase ṁ for the given upstream state. Step-by-Step Solution: Decrease back pressure from near-stagnation; ṁ rises until the sonic condition is met.At the critical pressure ratio, throat reaches Mach 1 ⇒ ṁ is maximum.Beyond this point, ṁ remains fixed; only downstream shock/expansion structure is altered. Verification / Alternative check: For perfect gases, the critical pressure ratio p*/p0 is a function of γ. Measurements of ṁ vs. back pressure show a plateau beyond the critical point—classic evidence of choking. Why Other Options Are Wrong: “Underdamping/Overdamping” are vibration terms, unrelated to nozzle flow; “None of these” is incorrect because the proper term is “Choked.” Common Pitfalls: Assuming temperature must always drop in throttling; here we discuss nozzles (expansion with kinetic energy rise), not throttling valves. Final Answer: Choked

Q: Reheating in multi-stage steam turbines: What are the principal benefits of reheating the steam between turbine stages?

Introduction / Context: Large steam power plants commonly use reheat between high-pressure (HP) and intermediate/low-pressure (IP/LP) turbine sections. Reheating influences thermodynamic efficiency, turbine output, and blade durability by controlling moisture content in the last stages. Given Data / Assumptions: Expansion is split into at least two turbine sections. Steam is extracted from the HP exhaust, returned to the boiler reheater, and then admitted to the IP/LP section. Operating goal: limit wetness, increase average temperature of heat addition, and raise cycle efficiency. Concept / Approach: Reheating raises the temperature of steam before the second expansion. This increases the mean temperature at which heat is supplied to the Rankine cycle, improving thermal efficiency. By reheating, the quality (dryness fraction) at the LP exhaust increases, reducing liquid droplets that erode and corrode last-stage blades. Also, the second expansion begins at higher enthalpy, increasing the total turbine work output compared with the non-reheat case for similar boiler and condenser limits. Step-by-Step Solution: Insert a reheater between turbine stages to raise steam temperature at approximately constant pressure.Higher inlet temperature to the second stage → larger enthalpy drop overall → more turbine work.Higher average heat-addition temperature → improved cycle efficiency.Higher quality at exhaust → less moisture → reduced erosion and wear. Verification / Alternative check: T–s diagrams of reheat Rankine cycles clearly show increased area (work) and improved exhaust quality compared to simple Rankine cycles, validating all three benefits listed. Why Other Options Are Wrong: Selecting any single benefit understates reheat; in practice it yields all listed improvements. Common Pitfalls: Assuming reheat always increases condenser load; while mass flow may change, the dominant effect is improved efficiency and moisture control. Final Answer: all of these

Q: Curtis (velocity-compounded) impulse stage behavior: In a velocity-compounded impulse turbine, what happens to the steam velocity as it passes through the second row of moving blades?

Introduction / Context: Velocity compounding (Curtis staging) splits the large nozzle jet velocity across multiple moving rows with an intervening set of guide (fixed) blades. Understanding how velocity and pressure evolve through each row is key to analyzing impulse stages and estimating diagram efficiency. Given Data / Assumptions: Impulse principle: pressure drop occurs mainly in the nozzle and fixed guides; moving blades ideally see constant pressure. Two (or more) moving rows with a fixed guide (re-directing) row between them. Energy is extracted from the jet by turning it in the moving blades. Concept / Approach: In an impulse stage, moving blades convert part of the incoming kinetic energy into work. Thus, across each moving row, the absolute velocity magnitude decreases as energy is extracted, while the static pressure ideally remains approximately constant. The fixed guide between moving rows re-directs the flow (changing direction and sometimes redistributing velocity components) without adding net energy. Step-by-Step Solution: Across first moving row: velocity decreases due to work transfer to the rotor.Fixed guide re-directs the reduced velocity to the proper inlet angle for the second row.Across second moving row: velocity decreases further, again transferring energy to the rotor. Verification / Alternative check: Velocity triangles drawn for a Curtis stage show diminishing absolute outlet velocities after each moving row, while static pressure across moving rows is nearly constant, confirming the trend. Why Other Options Are Wrong: Velocity increases/constant: Opposite to energy extraction behavior. Pressure remains constant: True for moving rows, but the question asks about velocity change specifically. Common Pitfalls: Mixing the behavior of fixed and moving rows; assuming pressure drops in moving blades (that is characteristic of reaction stages, not pure impulse). Final Answer: velocity decreases

Q: Rateau staging identification: The Rateau turbine is best described as which type of impulse staging?

Introduction / Context: Impulse turbines can be compounded either by velocity (Curtis), by pressure (Rateau), or by combined pressure-velocity methods. Recognizing these distinctions is important for predicting stage count, efficiency, and blade loading in design and analysis. Given Data / Assumptions: Impulse staging: pressure drops primarily in fixed nozzles. Compounding approach defines how the total pressure drop is shared among stages. Rateau is a historical and widely taught scheme. Concept / Approach: In Rateau (pressure-compounded) turbines, the overall boiler-to-condenser pressure drop is divided across multiple nozzle–rotor pairs. Each stage has its own nozzle where a portion of the pressure drop occurs to produce a moderate velocity, followed by an impulse rotor that extracts work. This limits extreme velocities and losses while maintaining impulse characteristics in each stage. Step-by-Step Solution: Split the total pressure drop into several sequential stages.Each stage: nozzle pressure drop → rotor work at near-constant pressure.Outcome: lower velocities per stage compared to single-stage De Laval, reducing losses. Verification / Alternative check: Rateau’s scheme is contrasted with Curtis (velocity compounding) where a single nozzle gives a large velocity drop followed by multiple moving rows to share velocity changes within one stage. Why Other Options Are Wrong: Simple reaction turbine: Reaction involves pressure drop in moving blades, not impulse. Velocity compounded: That is Curtis staging, not Rateau. Pressure-velocity compounded: A hybrid method, not the classic Rateau approach. Common Pitfalls: Confusing the locus of pressure drop (nozzles vs blades) when distinguishing impulse from reaction forms. Final Answer: pressure compounded turbine

Q: Purpose of governing in steam turbines: What is the primary function of the governing mechanism in a steam turbine under varying load?

Introduction / Context: Steam turbines drive generators and mechanical loads that vary over time. A governing system adjusts steam flow to keep rotational speed near the set value, protecting equipment and maintaining grid frequency or process requirements. Given Data / Assumptions: Governor senses speed (or frequency) and actuates valves or nozzles. Load disturbances cause speed deviations without control. Primary goal is speed regulation, not thermodynamic improvement. Concept / Approach: When load increases, turbine speed tends to drop; the governor opens valves to admit more steam, raising torque and restoring speed. When load decreases, it throttles steam to prevent overspeed. While governing influences efficiency indirectly via throttling or nozzle control, its core purpose is speed control, especially critical for synchronous generators tied to a power grid. Step-by-Step Solution: Detect speed deviation via flyweights or electronic sensors.Modify steam admission (throttle or nozzle-control) to correct speed error.Stabilize around setpoint with proper droop or isochronous characteristics. Verification / Alternative check: Grid-connected units employ droop characteristics to share load among multiple turbines; process turbines use isochronous or cascaded controls to hold set speed within tight tolerance. Why Other Options Are Wrong: Reduce heat drop/reheat steam/balance end thrust: These are design or thermodynamic tasks unrelated to the primary control function of governing. Common Pitfalls: Confusing governing (control) with cycle enhancements like reheat/regeneration or with axial thrust balancing arrangements (dummy pistons, balance pipes). Final Answer: maintain the speed of the turbine

Q: Nozzle geometry terminology: If the flow area increases continuously from the inlet to the outlet, how is such a steam nozzle classified?

Introduction / Context: Nozzles are classified by how their cross-sectional area changes along the flow direction. The geometry dictates whether subsonic or supersonic acceleration occurs and where choking is possible. Correct naming avoids confusion in turbine and jet calculations. Given Data / Assumptions: Cross-sectional area increases monotonically from inlet to outlet. Working fluid is steam, but terminology is general. Flow regime depends on inlet state and pressure ratio. Concept / Approach: A nozzle with monotonically increasing area is a divergent nozzle. In compressible flow, a purely divergent duct decelerates subsonic flow but accelerates supersonic flow. Choking and acceleration from subsonic to supersonic require a convergent–divergent (C–D) shape (De Laval nozzle), not a purely divergent or convergent one. Step-by-Step Solution: Identify geometry: area increases → divergent.Recall subsonic vs supersonic response to divergence.Distinguish from C–D nozzle used for supersonic acceleration after choking at the throat. Verification / Alternative check: Classical area–Mach number relations show that dA > 0 accelerates M > 1 flow and decelerates M < 1 flow, consistent with the naming and applications. Why Other Options Are Wrong: Convergent: Area decreases. Convergent–divergent: Area first decreases to a throat, then increases. None of these: Incorrect because “divergent nozzle” precisely matches. Common Pitfalls: Assuming any divergent passage yields supersonic speeds; without a throat and sufficient pressure ratio, the flow may remain subsonic and decelerate. Final Answer: divergent nozzle

Q: Function of a steam nozzle in turbines and jets: What principal energy conversion takes place in a steam nozzle?

Introduction / Context: Steam nozzles are critical in turbines, ejectors, and other devices where a high-velocity jet is needed. They convert the thermodynamic availability in high-pressure, high-temperature steam into directed kinetic energy for doing mechanical work on turbine blades or entraining secondary flows. Given Data / Assumptions: Steam expands through a shaped passage (nozzle). Negligible shaft work; energy conversion is primarily flow work and enthalpy drop to kinetic energy rise. Adiabatic approximation is often used. Concept / Approach: Across a nozzle, the stagnation enthalpy decreases as static kinetic energy increases. The first law for steady flow reduces to h1 + V1^2/2 ≈ h2 + V2^2/2 (neglecting heat, shaft work, elevation). Thus, the enthalpy (thermal) content is converted to kinetic energy, producing a high-speed jet that can drive impulse stages or create suction in ejectors. Step-by-Step Solution: Begin with high-pressure, high-enthalpy steam at the inlet.Expand through the nozzle causing pressure and enthalpy to drop.Observe a large increase in V at the exit: thermal → kinetic conversion. Verification / Alternative check: Measured nozzle exit velocities correlate with enthalpy drops from steam tables, validating the energy conversion mechanism. Why Other Options Are Wrong: Kinetic → heat or potential → heat: Opposite of nozzle intent. Heat → potential energy: Potential energy changes are negligible compared to kinetic rise. Common Pitfalls: Ignoring losses; real nozzles have efficiency less than 100%, so actual velocity is less than the ideal value derived from the enthalpy drop. Final Answer: heat energy of steam into kinetic energy

Question 1

Nozzle efficiency — correct definition Select the correct definition of nozzle efficiency for a steam or gas nozzle operating between a given inlet state and a specified outlet pressure.

Accepted Answer

Introduction / Context: Nozzles convert enthalpy into directed kinetic energy. The nozzle efficiency indicates how effectively the available isentropic enthalpy drop is transformed into jet kinetic energy, a crucial metric for turbine stage matching and thrust devices. Given Data / Assumptions: Specified nozzle inlet state and outlet pressure. Adiabatic device with no shaft work. Negligible changes in potential energy. Concept / Approach: The ideal kinetic energy per kilogram obtainable from a given pressure drop equals the isentropic enthalpy drop h_in − h_out,s. The actual kinetic energy equals V_actual^2 / 2. Their ratio defines nozzle efficiency: η_nozzle = (V_actual^2 / 2) / (h_in − h_out,s). This measure excludes blade work and stage aggregation; those are covered by diagram and stage efficiencies. Step-by-Step Solution: Establish ideal reference using isentropic outlet: Δh_s = h_in − h_out,s.Measure actual outlet velocity V_actual; compute KE_actual = V_actual^2 / 2.Form η_nozzle = KE_actual / Δh_s.Interpret: η_nozzle < 1 due to friction, shock, and boundary-layer losses. Verification / Alternative check: The velocity coefficient φ_v = V_actual / V_ideal relates as η_nozzle = φ_v^2, providing an alternative way to determine nozzle efficiency from measured velocities. Why Other Options Are Wrong: Option A is the diagram (blading) efficiency. Option B is the stage efficiency. Option D is false because option C states the standard definition correctly. Common Pitfalls: Confusing nozzle efficiency with discharge coefficient (area flow losses) or with stage/overall efficiencies that include rotor effects. Final Answer: Nozzle efficiency = (actual kinetic energy at nozzle exit per kg) / (ideal isentropic heat drop across the nozzle per kg)

Question 2

Reheat factor — definition check “The ratio of the cumulative heat drop (sum of effective enthalpy drops over all stages) to the overall isentropic heat drop is called the reheat factor.” Is this statement correct?

Accepted Answer

Introduction / Context: Because real stages have varying efficiencies and the specific volume increases as steam expands, the total of the per-stage heat drops may not exactly equal the single isentropic drop between inlet and exhaust. The reheat factor summarizes this difference and is used in detailed performance estimates. Given Data / Assumptions: Multi-stage turbine with known inlet and exit states. Per-stage enthalpy drops are measurable or estimable. Isentropic reference is taken between the same end states. Concept / Approach: Reheat factor RF = (cumulative heat drop over stages) / (overall isentropic heat drop). In many practical turbines RF is slightly greater than 1 because of the way velocity and volume change across stages and how losses distribute, though the exact value depends on inlet conditions, back pressure, and stage design. Step-by-Step Solution: Compute per-stage effective enthalpy drop from measurements/calculations.Sum these drops to obtain the cumulative heat drop.Determine the ideal isentropic drop from inlet to exit states.Form RF = cumulative drop / isentropic drop. Verification / Alternative check: Performance maps often report RF around 1.02–1.07; when multiplied by internal efficiency, it helps reconcile stage-wise sums with overall turbine output. Why Other Options Are Wrong: Choosing “No” would contradict standard steam-turbine terminology used in design texts and practice. Common Pitfalls: Confusing RF with “reheat” in the Rankine cycle hardware; RF does not require an actual reheater between turbine sections. Final Answer: Yes

Question 3

Choked flow in nozzles — terminology When a nozzle operates at the condition of maximum mass flow rate (for fixed upstream total state), the nozzle is said to be:

Accepted Answer

Introduction / Context: Designers of turbines, ejectors, and propulsion systems rely on the concept of choked flow. At a particular back pressure ratio, the sonic condition is reached at the throat and the mass flow rate stops increasing even if the downstream pressure is reduced further. Given Data / Assumptions: Steady, adiabatic nozzle without shaft work. Negligible potential-energy changes; kinetic energy rise dominates. Ideal gas/steam behavior for defining the critical pressure ratio. Concept / Approach: Choking occurs when Mach number at the minimum area section equals 1. At this point, the mass flux ṁ/A depends only on upstream stagnation temperature and pressure and on gas properties. Further reducing back pressure changes the downstream expansion pattern but does not increase ṁ for the given upstream state. Step-by-Step Solution: Decrease back pressure from near-stagnation; ṁ rises until the sonic condition is met.At the critical pressure ratio, throat reaches Mach 1 ⇒ ṁ is maximum.Beyond this point, ṁ remains fixed; only downstream shock/expansion structure is altered. Verification / Alternative check: For perfect gases, the critical pressure ratio p*/p0 is a function of γ. Measurements of ṁ vs. back pressure show a plateau beyond the critical point—classic evidence of choking. Why Other Options Are Wrong: “Underdamping/Overdamping” are vibration terms, unrelated to nozzle flow; “None of these” is incorrect because the proper term is “Choked.” Common Pitfalls: Assuming temperature must always drop in throttling; here we discuss nozzles (expansion with kinetic energy rise), not throttling valves. Final Answer: Choked

Question 4

De Laval (single-stage impulse) steam turbine practice: In a De Laval impulse turbine, the nozzle that accelerates the steam jet is positioned how relative to the moving blades?

Accepted Answer

Introduction / Context: The De Laval turbine is a classic single-stage impulse turbine that uses a convergent–divergent nozzle to accelerate steam to very high velocity before it strikes the moving blades. Understanding why the nozzle is placed very close to the rotor blades helps minimize losses and improve stage efficiency in impulse machines. Given Data / Assumptions: Impulse principle: pressure drop occurs almost entirely in the nozzle; blades ideally see velocity change only. Nozzle produces a high-velocity jet at a designed exit angle α. Mixing with ambient and jet spreading increase with free-jet travel distance. Concept / Approach: In an impulse stage, the goal is to preserve as much nozzle exit kinetic energy as possible until it is transferred to the rotor. If the nozzle is far from the blades, the jet spreads, entrains surrounding steam, and slows down due to turbulence and viscous effects before reaching the blade inlet. Placing the nozzle very close to the blade inlet edge reduces these free-jet dissipation losses and maintains the desired incidence angle, improving diagram efficiency. Step-by-Step Solution: Recognize that pressure drop is concentrated in the nozzle; blades should only redirect velocity.Minimize jet travel in free space by mounting the nozzle adjacent to the rotor periphery.This preserves jet speed and direction at entry, improving energy transfer. Verification / Alternative check: Practical De Laval designs show nozzle rings closely hugging the blade periphery. Computational and experimental studies confirm reduced incidence losses and less jet dissipation when spacing is minimized. Why Other Options Are Wrong: “No”: Placing the nozzle far away allows jet decay, increased incidence losses, and lower efficiency. Common Pitfalls: Confusing impulse and reaction staging; in reaction stages, pressure drop also occurs across blades, but proximity of fixed and moving rows remains important to control losses. Final Answer: Yes

Question 5

Reheating in multi-stage steam turbines: What are the principal benefits of reheating the steam between turbine stages?

Accepted Answer

Introduction / Context: Large steam power plants commonly use reheat between high-pressure (HP) and intermediate/low-pressure (IP/LP) turbine sections. Reheating influences thermodynamic efficiency, turbine output, and blade durability by controlling moisture content in the last stages. Given Data / Assumptions: Expansion is split into at least two turbine sections. Steam is extracted from the HP exhaust, returned to the boiler reheater, and then admitted to the IP/LP section. Operating goal: limit wetness, increase average temperature of heat addition, and raise cycle efficiency. Concept / Approach: Reheating raises the temperature of steam before the second expansion. This increases the mean temperature at which heat is supplied to the Rankine cycle, improving thermal efficiency. By reheating, the quality (dryness fraction) at the LP exhaust increases, reducing liquid droplets that erode and corrode last-stage blades. Also, the second expansion begins at higher enthalpy, increasing the total turbine work output compared with the non-reheat case for similar boiler and condenser limits. Step-by-Step Solution: Insert a reheater between turbine stages to raise steam temperature at approximately constant pressure.Higher inlet temperature to the second stage → larger enthalpy drop overall → more turbine work.Higher average heat-addition temperature → improved cycle efficiency.Higher quality at exhaust → less moisture → reduced erosion and wear. Verification / Alternative check: T–s diagrams of reheat Rankine cycles clearly show increased area (work) and improved exhaust quality compared to simple Rankine cycles, validating all three benefits listed. Why Other Options Are Wrong: Selecting any single benefit understates reheat; in practice it yields all listed improvements. Common Pitfalls: Assuming reheat always increases condenser load; while mass flow may change, the dominant effect is improved efficiency and moisture control. Final Answer: all of these

Question 6

Curtis (velocity-compounded) impulse stage behavior: In a velocity-compounded impulse turbine, what happens to the steam velocity as it passes through the second row of moving blades?

Accepted Answer

Introduction / Context: Velocity compounding (Curtis staging) splits the large nozzle jet velocity across multiple moving rows with an intervening set of guide (fixed) blades. Understanding how velocity and pressure evolve through each row is key to analyzing impulse stages and estimating diagram efficiency. Given Data / Assumptions: Impulse principle: pressure drop occurs mainly in the nozzle and fixed guides; moving blades ideally see constant pressure. Two (or more) moving rows with a fixed guide (re-directing) row between them. Energy is extracted from the jet by turning it in the moving blades. Concept / Approach: In an impulse stage, moving blades convert part of the incoming kinetic energy into work. Thus, across each moving row, the absolute velocity magnitude decreases as energy is extracted, while the static pressure ideally remains approximately constant. The fixed guide between moving rows re-directs the flow (changing direction and sometimes redistributing velocity components) without adding net energy. Step-by-Step Solution: Across first moving row: velocity decreases due to work transfer to the rotor.Fixed guide re-directs the reduced velocity to the proper inlet angle for the second row.Across second moving row: velocity decreases further, again transferring energy to the rotor. Verification / Alternative check: Velocity triangles drawn for a Curtis stage show diminishing absolute outlet velocities after each moving row, while static pressure across moving rows is nearly constant, confirming the trend. Why Other Options Are Wrong: Velocity increases/constant: Opposite to energy extraction behavior. Pressure remains constant: True for moving rows, but the question asks about velocity change specifically. Common Pitfalls: Mixing the behavior of fixed and moving rows; assuming pressure drops in moving blades (that is characteristic of reaction stages, not pure impulse). Final Answer: velocity decreases

Question 7

Rateau staging identification: The Rateau turbine is best described as which type of impulse staging?

Accepted Answer

Introduction / Context: Impulse turbines can be compounded either by velocity (Curtis), by pressure (Rateau), or by combined pressure-velocity methods. Recognizing these distinctions is important for predicting stage count, efficiency, and blade loading in design and analysis. Given Data / Assumptions: Impulse staging: pressure drops primarily in fixed nozzles. Compounding approach defines how the total pressure drop is shared among stages. Rateau is a historical and widely taught scheme. Concept / Approach: In Rateau (pressure-compounded) turbines, the overall boiler-to-condenser pressure drop is divided across multiple nozzle–rotor pairs. Each stage has its own nozzle where a portion of the pressure drop occurs to produce a moderate velocity, followed by an impulse rotor that extracts work. This limits extreme velocities and losses while maintaining impulse characteristics in each stage. Step-by-Step Solution: Split the total pressure drop into several sequential stages.Each stage: nozzle pressure drop → rotor work at near-constant pressure.Outcome: lower velocities per stage compared to single-stage De Laval, reducing losses. Verification / Alternative check: Rateau’s scheme is contrasted with Curtis (velocity compounding) where a single nozzle gives a large velocity drop followed by multiple moving rows to share velocity changes within one stage. Why Other Options Are Wrong: Simple reaction turbine: Reaction involves pressure drop in moving blades, not impulse. Velocity compounded: That is Curtis staging, not Rateau. Pressure-velocity compounded: A hybrid method, not the classic Rateau approach. Common Pitfalls: Confusing the locus of pressure drop (nozzles vs blades) when distinguishing impulse from reaction forms. Final Answer: pressure compounded turbine

Question 8

Purpose of governing in steam turbines: What is the primary function of the governing mechanism in a steam turbine under varying load?

Accepted Answer

Introduction / Context: Steam turbines drive generators and mechanical loads that vary over time. A governing system adjusts steam flow to keep rotational speed near the set value, protecting equipment and maintaining grid frequency or process requirements. Given Data / Assumptions: Governor senses speed (or frequency) and actuates valves or nozzles. Load disturbances cause speed deviations without control. Primary goal is speed regulation, not thermodynamic improvement. Concept / Approach: When load increases, turbine speed tends to drop; the governor opens valves to admit more steam, raising torque and restoring speed. When load decreases, it throttles steam to prevent overspeed. While governing influences efficiency indirectly via throttling or nozzle control, its core purpose is speed control, especially critical for synchronous generators tied to a power grid. Step-by-Step Solution: Detect speed deviation via flyweights or electronic sensors.Modify steam admission (throttle or nozzle-control) to correct speed error.Stabilize around setpoint with proper droop or isochronous characteristics. Verification / Alternative check: Grid-connected units employ droop characteristics to share load among multiple turbines; process turbines use isochronous or cascaded controls to hold set speed within tight tolerance. Why Other Options Are Wrong: Reduce heat drop/reheat steam/balance end thrust: These are design or thermodynamic tasks unrelated to the primary control function of governing. Common Pitfalls: Confusing governing (control) with cycle enhancements like reheat/regeneration or with axial thrust balancing arrangements (dummy pistons, balance pipes). Final Answer: maintain the speed of the turbine

Question 9

Nozzle geometry terminology: If the flow area increases continuously from the inlet to the outlet, how is such a steam nozzle classified?

Accepted Answer

Introduction / Context: Nozzles are classified by how their cross-sectional area changes along the flow direction. The geometry dictates whether subsonic or supersonic acceleration occurs and where choking is possible. Correct naming avoids confusion in turbine and jet calculations. Given Data / Assumptions: Cross-sectional area increases monotonically from inlet to outlet. Working fluid is steam, but terminology is general. Flow regime depends on inlet state and pressure ratio. Concept / Approach: A nozzle with monotonically increasing area is a divergent nozzle. In compressible flow, a purely divergent duct decelerates subsonic flow but accelerates supersonic flow. Choking and acceleration from subsonic to supersonic require a convergent–divergent (C–D) shape (De Laval nozzle), not a purely divergent or convergent one. Step-by-Step Solution: Identify geometry: area increases → divergent.Recall subsonic vs supersonic response to divergence.Distinguish from C–D nozzle used for supersonic acceleration after choking at the throat. Verification / Alternative check: Classical area–Mach number relations show that dA > 0 accelerates M > 1 flow and decelerates M < 1 flow, consistent with the naming and applications. Why Other Options Are Wrong: Convergent: Area decreases. Convergent–divergent: Area first decreases to a throat, then increases. None of these: Incorrect because “divergent nozzle” precisely matches. Common Pitfalls: Assuming any divergent passage yields supersonic speeds; without a throat and sufficient pressure ratio, the flow may remain subsonic and decelerate. Final Answer: divergent nozzle

Question 10

Function of a steam nozzle in turbines and jets: What principal energy conversion takes place in a steam nozzle?

Accepted Answer

Introduction / Context: Steam nozzles are critical in turbines, ejectors, and other devices where a high-velocity jet is needed. They convert the thermodynamic availability in high-pressure, high-temperature steam into directed kinetic energy for doing mechanical work on turbine blades or entraining secondary flows. Given Data / Assumptions: Steam expands through a shaped passage (nozzle). Negligible shaft work; energy conversion is primarily flow work and enthalpy drop to kinetic energy rise. Adiabatic approximation is often used. Concept / Approach: Across a nozzle, the stagnation enthalpy decreases as static kinetic energy increases. The first law for steady flow reduces to h1 + V1^2/2 ≈ h2 + V2^2/2 (neglecting heat, shaft work, elevation). Thus, the enthalpy (thermal) content is converted to kinetic energy, producing a high-speed jet that can drive impulse stages or create suction in ejectors. Step-by-Step Solution: Begin with high-pressure, high-enthalpy steam at the inlet.Expand through the nozzle causing pressure and enthalpy to drop.Observe a large increase in V at the exit: thermal → kinetic conversion. Verification / Alternative check: Measured nozzle exit velocities correlate with enthalpy drops from steam tables, validating the energy conversion mechanism. Why Other Options Are Wrong: Kinetic → heat or potential → heat: Opposite of nozzle intent. Heat → potential energy: Potential energy changes are negligible compared to kinetic rise. Common Pitfalls: Ignoring losses; real nozzles have efficiency less than 100%, so actual velocity is less than the ideal value derived from the enthalpy drop. Final Answer: heat energy of steam into kinetic energy

Question 11

Maximum rotor (diagram) efficiency condition for an impulse stage: The impulse turbine rotor efficiency reaches a maximum value of 0.5 * cos^2(α), where α is the nozzle exit angle, under which blade condition?

Accepted Answer

Introduction / Context: Rotor (diagram) efficiency in impulse turbines depends on velocity triangles, blade speed ratio, and blade geometry. A key textbook result is the maximum efficiency achievable for given nozzle exit angle α when blades are designed to minimize losses and have symmetric (equiangular) inlet and outlet relative flow angles. Given Data / Assumptions: Impulse staging: pressure drop in nozzles, ideally constant pressure in moving blades. Nozzle exit absolute velocity makes an angle α to the wheel tangent. Blade friction neglected when velocity coefficient ≈ 1. Equiangular blades: inlet and outlet relative flow angles equal in magnitude. Concept / Approach: Analyzing the velocity triangles and optimizing with respect to blade speed ratio yields a closed-form expression for the maximum rotor (diagram) efficiency. For equiangular and frictionless blades, the optimum gives η_r,max = 0.5 * cos^2 α for the stated condition. This relation highlights the importance of keeping α modest (to increase cos α) and ensuring high-quality, low-loss blade passages to approach the theoretical limit. Step-by-Step Solution: Construct inlet and outlet velocity triangles with nozzle angle α and equiangular blades.Express work per unit mass in terms of whirl components and blade speed.Differentiate with respect to blade speed ratio and set derivative to zero to find optimum.Substitute optimum into efficiency expression to obtain η_r,max = 0.5 * cos^2 α. Verification / Alternative check: Standard turbine textbooks present this derivation; numerical checks with typical α (for example 20–30 degrees) give plausible maximum efficiencies in the 0.35–0.44 range, consistent with single-stage impulse expectations when only rotor efficiency is considered (overall stage efficiency will be lower after accounting for nozzle and mechanical losses). Why Other Options Are Wrong: Equiangular only: Without assuming frictionless flow (velocity coefficient unity), losses reduce the maximum below the ideal expression. Velocity coefficient unity only: Symmetry (equiangular) is also required to reach the stated form. Blade solidity 0.65: Solidity influences losses and flow turning but does not by itself produce the stated closed-form maximum. Common Pitfalls: Confusing rotor (diagram) efficiency with overall stage efficiency; mixing the result η_r,max = cos^2 α (a different assumption set) with the present 0.5 * cos^2 α expression. Final Answer: blades are equiangular and frictionless

Question 12

Supersaturated (meta-stable) steam property check: The density of supersaturated steam is approximately how large compared to the ordinary saturated vapour at the same pressure?

Accepted Answer

Introduction / Context: Supersaturated (meta-stable) flow can occur during rapid expansions where condensation lags thermodynamic equilibrium. Understanding how properties like density compare with saturated values is useful for nozzle calculations and interpreting departures from equilibrium charts. Given Data / Assumptions: Comparison at the same pressure. Supersaturation implies delayed condensation, not a vastly different equation of state at a given p. Small deviations from equilibrium in density are expected at identical pressure for steam near saturation. Concept / Approach: At a given pressure, the vapour density of a meta-stable (supersaturated) state is close to that of the equilibrium saturated vapour because the state point lies near the saturation dome but on a non-equilibrium path. The principal differences show up in temperature, wetness, and nucleation behavior rather than order-of-magnitude density changes. Hence, the density is approximately the same, not 2×, 4×, or 8× larger. Step-by-Step Solution: Fix the pressure p for both states.Recognize that supersaturation delays phase change but does not drastically alter vapour density at the same p.Conclude densities are approximately the same for engineering MCQ purposes. Verification / Alternative check: Experimental nozzle studies show slight departures in temperature and non-equilibrium wetness, but density changes at the same pressure are not multiples as large as 2× or more; they are comparatively small. Why Other Options Are Wrong: 2×, 4×, 8×: These would imply radical changes inconsistent with steam property behavior at fixed pressure. Common Pitfalls: Equating supersaturation with superheat; supersaturated (meta-stable) refers to delayed condensation below saturation temperature, not necessarily a large density jump. Final Answer: same as

Question 13

Location of supersonic flow in a steam nozzle: In a properly operating convergent–divergent (De Laval) nozzle, where is the steam flow supersonic?

Accepted Answer

Introduction / Context: Compressible-flow behavior in nozzles depends on area changes and pressure ratios. The De Laval nozzle accelerates subsonic flow to sonic at the throat and then to supersonic speeds in the diverging section when the back pressure is low enough to choke the flow. Given Data / Assumptions: Adequate pressure ratio to achieve choking at the throat. Adiabatic, frictionless approximation for basic explanation. Steam as the working fluid (concept applies to gases generally). Concept / Approach: Area–Mach number relations state that subsonic flows accelerate in converging passages; sonic Mach number M = 1 occurs at the minimum area (throat) when choked; supersonic flows accelerate in diverging passages. Thus, in a C–D nozzle with sufficient pressure ratio, the divergent section supports M > 1 flow. Step-by-Step Solution: Convergent section: subsonic acceleration toward throat.Throat: locally sonic (M = 1) when choked.Divergent section: supersonic acceleration (M > 1). Verification / Alternative check: Measured pressures and temperatures along instrumented nozzles match isentropic C–D predictions when properly expanded, confirming supersonic flow in the diverging section. Why Other Options Are Wrong: Entrance/convergent portion: Flow is typically subsonic there. Throat: At best sonic (M = 1), not supersonic. Common Pitfalls: Assuming supersonic flow can occur in a purely convergent nozzle; without a diverging section, maximum Mach is 1 at the exit when choked. Final Answer: in the divergent portion of the nozzle

Question 14

Curtis stage identification in steam turbines In steam turbine terminology, a Curtis turbine refers to which type of staging/compounding arrangement used to reduce rotor speed while extracting work efficiently?

Accepted Answer

Introduction / Context: The Curtis turbine (often called a Curtis stage) is a classic configuration in steam turbine design. It is used to handle a large available pressure drop by converting it into high jet velocity in one set of nozzles and then extracting work over two or more moving blade rows separated by one or more intermediate guide (fixed) rows. Recognizing what “Curtis” means helps students connect blade velocity triangles with practical compounding strategies for speed control. Given Data / Assumptions: A high-pressure steam supply expands through nozzles to produce high jet velocity. Multiple moving blade rows are arranged with intermediate fixed guide rows. Total pressure drop occurs mainly in the first nozzle set; subsequent rows primarily redirect and extract kinetic energy. Concept / Approach: In velocity compounding, the entire pressure drop is taken in one nozzle set, creating a very high jet velocity V. That kinetic energy is then extracted over multiple moving rows (e.g., two) so that each moving row operates at a lower relative velocity, keeping rotor speed within manageable limits. This is precisely the Curtis concept, distinguishing it from pressure compounding (Rateau) where pressure drop is split across several nozzle stages, and from reaction staging where pressure drops in both fixed and moving blades. Step-by-Step Solution: Identify that a Curtis stage uses one big pressure drop in nozzles to generate high-velocity jets.Recognize subsequent energy extraction over multiple moving rows with fixed guide rows in between, i.e., velocity compounding.Conclude that “Curtis turbine” denotes a velocity-compounded impulse arrangement. Verification / Alternative check: Blade-to-jet speed ratios and velocity triangles for the Curtis stage show reduction of jet velocity in steps across successive moving rows, consistent with velocity compounding principles. Why Other Options Are Wrong: Pressure compounded turbine: that is Rateau staging, not Curtis.Simple reaction turbine: involves pressure drop in moving blades, unlike Curtis.Pressure-velocity compounded: combines both methods; Curtis specifically refers to velocity compounding.Simple impulse without compounding: lacks multiple moving rows used in Curtis. Common Pitfalls: Confusing “Curtis” (velocity compounding) with “Rateau” (pressure compounding). Remember: Curtis = velocity-compounded impulse stage. Final Answer: velocity compounded turbine

Question 15

Definition check – diagram (blading) efficiency in steam turbines Which ratio correctly defines the diagram efficiency associated with turbine blades for a given stage?

Accepted Answer

Introduction / Context: Diagram efficiency (also called blading efficiency) quantifies how effectively the kinetic energy presented to a set of blades is converted into useful work on the rotor. It focuses on the velocity triangles at blade entry and exit, separating blade performance from other losses (nozzle, mechanical, and leakage losses). Given Data / Assumptions: Consider one turbine stage with known inlet and outlet velocity triangles. Energy supplied to the blades refers to the available jet kinetic energy at blade entry. Workdone on the blades is the specific work extracted from the jet by the rotor (per kg steam). Concept / Approach: Diagram efficiency η_d is defined as the ratio of specific work obtained from the blades to the specific kinetic energy supplied to the blades. In symbols: η_d = (work on blades per kg) / (energy supplied to blades per kg). This is distinct from stage efficiency (which uses isentropic heat drop for the whole stage) and from overall plant efficiency (shaft power over boiler heat input). Step-by-Step Solution: Compute energy supplied to blades: E_in = V1^2 / 2 (if V1 is the absolute velocity at blade inlet; generalized forms include whirl components).Compute rotor work per kg: W_blade = Vw1 * u − Vw2 * u (based on whirl components and blade speed u).Form the ratio η_d = W_blade / E_in. Verification / Alternative check: From velocity triangles, maximizing the exit whirl reduction (Vw2) for a given inlet whirl (Vw1) increases W_blade and thus η_d, consistent with the definition. Why Other Options Are Wrong: Option (b) mixes blade and stage quantities; stage energy includes nozzle and other drops.Option (c) inverts the ratio; η must be less than or equal to 1.Option (e) refers to overall thermal efficiency, not diagram efficiency. Common Pitfalls: Confusing diagram efficiency with stage or nozzle efficiencies; always check the numerator and denominator definitions. Final Answer: workdone on the blades to the energy supplied to the blades

Question 16

Reaction turbine behavior In a reaction steam turbine, does the steam velocity increase in both the fixed (stator) blades and the moving (rotor) blades due to pressure drop occurring across each?

Accepted Answer

Introduction / Context: Reaction turbines differ from pure impulse turbines in that pressure decreases across both stator and rotor passages. This design influences how absolute and relative velocities evolve through the stage and affects the distribution of work between fixed and moving rows. Given Data / Assumptions: Reaction turbine stage with finite degree of reaction (R between 0 and 1). Stator and rotor passages shaped as nozzles or diffusers to create pressure and velocity changes. Steady, quasi-one-dimensional flow assumptions for velocity-triangle analysis. Concept / Approach: In a reaction stage, a portion of the total enthalpy (pressure) drop occurs in the moving blades. The stator accelerates the steam (pressure drop → velocity rise), and the rotor also experiences a pressure drop causing an increase in relative velocity through the rotor passages, even though the absolute exit velocity may be moderated by blade speed and turning. Thus, considering the proper frame, steam is accelerated in both blade rows. Step-by-Step Solution: Define degree of reaction: R = (static enthalpy drop in rotor) / (total stage enthalpy drop).Because rotor has a static enthalpy drop, the relative velocity typically increases across the rotor (acceleration in blade frame).Stator also drops pressure, increasing absolute velocity into the rotor. Verification / Alternative check: Velocity triangles for Parsons-type (R ≈ 0.5) show acceleration in stator and rotor passages, consistent with positive reaction staging. Why Other Options Are Wrong: Statements restricting the effect to compounded or Curtis stages confuse impulse compounding with reaction behavior.Zero-degree nozzle angle is not a condition for reaction; geometry sets reaction via blade shapes and pressure distribution. Common Pitfalls: Equating “velocity increase” only with absolute velocity; in the rotor, it is the relative velocity that is key when a pressure drop occurs in moving passages. Final Answer: True

Question 17

Reducing rotor speed in impulse turbines Which compounding method(s) are employed to reduce the excessively high rotor speed that would result from a single-stage impulse expansion?

Accepted Answer

Introduction / Context: A single large pressure drop in impulse turbines creates very high jet velocities and would require unrealistically high rotor speeds for optimum energy extraction. Compounding methods split the work extraction to bring practical speeds while maintaining efficiency. Given Data / Assumptions: Impulse expansion of steam through nozzles produces high absolute velocities. Blade tip speed for best efficiency is a fraction of jet speed. Mechanical limits and generator synchronization restrict allowable rotor speeds. Concept / Approach: Velocity compounding (Curtis) converts the full pressure drop to a single high-velocity jet but extracts energy over multiple moving rows, lowering the required rotor speed. Pressure compounding (Rateau) divides the pressure drop across multiple nozzle-rotor stages, reducing jet velocity each time. Pressure-velocity compounding uses a hybrid approach, first splitting pressure, then using velocity compounding within each partial drop to further moderate speed. All three reduce rotor speed relative to a single-stage impulse arrangement. Step-by-Step Solution: Identify speed issue: u_opt ≈ fraction of V_jet → too high if V_jet is large.Apply velocity compounding: many moving rows per pressure drop → lower u.Apply pressure compounding: several stages reduce V_jet per stage.Combine (pressure-velocity) for additional flexibility. Verification / Alternative check: Comparing blade-speed ratios and power curves across compounded arrangements shows reduced optimal rotor speeds while maintaining acceptable efficiency bands. Why Other Options Are Wrong: Reaction staging is a different principle (pressure drop in moving blades) and is not required to solve an impulse rotor speed limitation. Common Pitfalls: Assuming one method is universally superior; selection depends on pressure ratio, power, and manufacturability. Final Answer: all of these

Question 18

Typical application of De Laval (single-stage impulse) turbines De Laval turbines are most commonly applied in which operating range, considering their single-stage impulse design and very high rotor speeds?

Accepted Answer

Introduction / Context: The De Laval turbine is a classic single-stage impulse turbine with a convergent-divergent nozzle that accelerates steam to very high velocities. Because only one rotor row extracts work, the optimum rotor speed is extremely high, making it suitable for specific applications rather than large power plants. Given Data / Assumptions: Single set of nozzles and a single moving blade row (no compounding). High jet velocity and corresponding high optimum blade speed. Mechanical constraints require gear reduction if driving moderate-speed machinery. Concept / Approach: Single-stage impulse machines extract only part of the kinetic energy efficiently unless the rotor spins very fast. Thus, De Laval turbines are best for small power duties where compactness and simplicity are desired and high-speed operation (often with reduction gears) is acceptable. For large power outputs, multi-stage (compounded) impulse or reaction turbines are preferred to achieve reasonable shaft speeds and higher overall efficiency. Step-by-Step Solution: Relate single-stage impulse to high jet velocity and high u for good efficiency.Recognize practicality: small power units can run very fast with gearing.Select the option describing small power and high speeds. Verification / Alternative check: Historical and modern micro-turbine applications of De Laval-type stages confirm their use as high-speed, small-output prime movers. Why Other Options Are Wrong: Large power service requires multi-staging to control speed and improve efficiency.“Low speed” options contradict the fundamental velocity-triangle requirements of single-stage impulse turbines. Common Pitfalls: Assuming all turbines are interchangeable across power ranges; staging dictates practical speed and scale. Final Answer: for small power purposes and high speeds

Question 19

Terminology match – blading efficiency is also called what? In steam turbine performance nomenclature, “blading efficiency” is commonly referred to as which of the following?

Accepted Answer

Introduction / Context: Different efficiency terms in turbines capture losses at different parts of the machine. Clear terminology helps you analyze where performance is gained or lost and avoid mixing blade aerodynamics with thermodynamic stage or cycle measures. Given Data / Assumptions: Focus on a blade row and the kinetic energy it receives. Blade-entry and blade-exit velocity triangles are known or measurable. Nozzle behavior and mechanical losses considered separately. Concept / Approach: Blading efficiency is the same concept as diagram efficiency: the ratio of work obtained from the blades to the energy supplied to them. Stage efficiency uses the isentropic heat drop for the entire stage (nozzle plus rotor) as the reference. Nozzle efficiency compares actual to isentropic velocity rise across the nozzle only. Therefore, the correct synonym for blading efficiency is diagram efficiency. Step-by-Step Solution: Write η_diagram = (work on blades per kg) / (kinetic energy supplied per kg).Contrast with η_stage = (actual stage work) / (isentropic stage heat drop).Identify that “blading efficiency” and “diagram efficiency” are interchangeable terms. Verification / Alternative check: Textbook definitions and solved velocity-triangle examples use η_d and “blading efficiency” synonymously. Why Other Options Are Wrong: Stage efficiency and nozzle efficiency reference different domains (stage-wide and nozzle-only).Mechanical efficiency covers shaft power versus internal power, not blade aerodynamics. Common Pitfalls: Using stage efficiency when the question demands a blades-only metric; always check what the denominator represents. Final Answer: diagram efficiency

Question 20

Rankine (turbine) efficiency dependence A statement about Rankine (turbine) efficiency says: it depends upon the total useful heat drop and the total isentropic heat drop across the turbine stage or machine. Is this statement correct?

Accepted Answer

Introduction / Context: Rankine efficiency (often called turbine efficiency or internal efficiency for the whole turbine or stage) compares the actual work output to the ideal (isentropic) work output between the same inlet and exit states. It provides a thermodynamic measure independent of mechanical losses. Given Data / Assumptions: Same inlet pressure/temperature and exit pressure for actual and isentropic reference processes. Negligible changes in kinetic and potential energy across the control volume for the basic definition. Moisture effects may exist but are folded into the enthalpy changes. Concept / Approach: The internal (Rankine) efficiency is defined as η_internal = (actual heat drop utilized) / (isentropic heat drop). Because actual turbine work equals the actual enthalpy drop (neglecting KE/PE), and the ideal work equals the isentropic enthalpy drop, the efficiency indeed depends on the ratio of these two “heat drops.” Hence the statement is correct and applies to both impulse and reaction turbines. Step-by-Step Solution: Write actual work per kg: W_actual ≈ h_in − h_out(actual).Write ideal work per kg: W_ideal ≈ h_in − h_out(isentropic).Form η_internal = W_actual / W_ideal = (useful heat drop) / (isentropic heat drop). Verification / Alternative check: Performance tests compute isentropic outlet enthalpy from measured exit pressure and assumed isentropic expansion; comparing with measured exit enthalpy yields turbine internal efficiency consistent with the definition. Why Other Options Are Wrong: Limiting the definition to impulse or reaction types is unnecessary; it is a general thermodynamic definition.Moisture content affects values but not the definition. Common Pitfalls: Mixing internal efficiency with mechanical efficiency (shaft vs. internal power) or with stage/diagram efficiency; each metric has different denominators. Final Answer: Correct

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