Q: Impulse turbine compounding – efficiency comparison Compared with a pressure-compounded impulse turbine (Rateau staging), the efficiency of a pressure-velocity compounded impulse turbine (Curtis staging) is generally:

Introduction / Context: Impulse turbines may be compounded by pressure (multiple nozzle stages) or by velocity (multiple rows of moving blades with intermediate guide blades), or by a combination of both. Knowing how compounding affects efficiency guides stage selection. Given Data / Assumptions: Equal overall pressure ratio and comparable design practice. Similar blade quality and leakage assumptions for comparison. Focus on aerodynamic/rotordynamic losses from additional blade rows. Concept / Approach: Pressure compounding spreads the pressure drop across several nozzle stages, generally keeping blade relative speeds moderate with fewer moving rows per stage. Pressure-velocity compounding introduces extra moving rows within a stage to split absolute jet velocity, but each added row brings turning, friction, and tip/leakage losses, typically lowering stage efficiency relative to pressure compounding. Step-by-Step Solution: Recognize that Curtis staging adds rows → more cumulative losses.Rateau staging uses more nozzle sets but fewer moving rows per stage → lower cumulative moving-blade losses.Therefore, for similar conditions, pressure-velocity compounded stages are generally less efficient than pressure-compounded ones. Verification / Alternative check: Textbook comparisons and test data show diagram efficiency penalties with additional blade rows due to repeated incidence, friction, and wake mixing. Why Other Options Are Wrong: “More” contradicts typical loss accumulation behavior. “Same” ignores extra rows and losses. “Always 100%” is impossible in real turbines. “Indeterminate” is overly cautious for standard design principles. Common Pitfalls: Focusing solely on speed reduction benefits of Curtis staging while overlooking added loss sources. Final Answer: less

Q: Steam pressure change across a nozzle As steam flows through a properly designed nozzle (from inlet to exit), the static pressure generally:

Introduction / Context: Nozzles are expansion devices that transform pressure energy (enthalpy) into kinetic energy. Understanding how pressure varies along the nozzle path is fundamental for turbine stage analysis. Given Data / Assumptions: Adiabatic, steady flow through a nozzle. Properly matched back pressure to avoid shocks within the nozzle. Negligible potential energy changes. Concept / Approach: From the steady flow energy viewpoint, decreasing static enthalpy appears as increasing kinetic energy. Hence, as the fluid accelerates, static pressure falls. In a CD nozzle, pressure falls through the throat and continues to fall in the divergent part under correctly expanded conditions. Step-by-Step Solution: Use energy balance: h1 + V1^2/2 ≈ h2 + V2^2/2.As V rises (V2 > V1), h must drop → associated with a drop in p for expansion.Therefore, pressure decreases along the nozzle path. Verification / Alternative check: Measured pressure profiles and Mach distributions in Laval nozzles show monotonic pressure decrease (absent shocks) from inlet to exit. Why Other Options Are Wrong: (b) Acceleration does not increase static pressure; it reduces it in expansion devices. (c) Constant pressure describes throttling valves, not nozzles. (d) Oscillation is not characteristic of steady, shock-free flow. (e) Pressure does not rise in the convergent section; it falls toward the throat. Common Pitfalls: Confusing nozzle behavior with diffusers, where velocity decreases and pressure increases. Final Answer: decreases due to conversion of enthalpy to kinetic energy

Question 1

Critical pressure ratio for initially dry saturated steam in a nozzle Choose the approximate critical pressure ratio (p2/p1 at choking) for initially dry saturated steam expanding through a nozzle to maximum mass flow.

Accepted Answer

Introduction / Context: The critical pressure ratio determines when flow through a nozzle becomes choked, i.e., mass flow reaches a maximum and further downstream pressure reduction does not increase flow. For steam, the critical ratio depends on its thermodynamic state and effective isentropic exponent during expansion. Given Data / Assumptions: Initially dry saturated steam at the nozzle inlet. Isentropic (ideal) nozzle expansion to sonic conditions at the throat at critical ratio. Negligible inlet velocity and heat transfer. Concept / Approach: Using steam tables or empirical relations for saturated steam, the critical pressure ratio for dry saturated conditions is commonly taken as approximately 0.577. For superheated steam, a slightly lower value near 0.546 is widely cited due to a different effective isentropic exponent. The distinction matters in nozzle design, staging, and flow measurement applications. Step-by-Step Solution: Identify state: dry saturated at inlet → use saturated-steam critical ratio.Recall standard approximate values: 0.577 (saturated), ~0.546 (superheated).Select 0.577 from the provided options for initially dry saturated steam. Verification / Alternative check: Textbook examples and nozzle discharge tables list p_critical/p0 ≈ 0.577 for saturated inlet steam. Design charts for De Laval nozzles corroborate this selection. Why Other Options Are Wrong: 0.546 corresponds more closely to superheated steam; 0.582 and 0.601 are outside the typical accepted range.0.500 is a round figure not supported by steam expansion data. Common Pitfalls: Applying the ideal-gas formula blindly; steam deviates from ideal-gas behavior, and saturated versus superheated states yield different practical ratios. Final Answer: 0.577

Question 2

Overall (thermal) efficiency wording “The ratio of work delivered at the turbine shaft to the heat supplied is called the overall thermal efficiency of the turbine.” Evaluate this statement.

Accepted Answer

Introduction / Context: Overall (thermal) efficiency broadly compares useful output (shaft work) with the thermal energy input available from fuel or steam. Although exact wording varies by textbook, the spirit is to measure how effectively supplied heat becomes shaft power in the cycle or turbine system considered. Given Data / Assumptions: “Heat supplied” refers to the thermal input associated with the working fluid entering the turbine or, in cycle terms, the boiler heat input. “Work delivered” refers to net turbine shaft work after mechanical losses, if the definition is plant-wide. Focus is on a macroscopic efficiency measure, not component-level aerodynamic efficiency. Concept / Approach: In many curricula, overall thermal efficiency η_th is defined as η_th = W_out / Q_in, where W_out is the useful shaft work and Q_in is the heat supplied from fuel (in a plant sense) or to the working fluid (cycle sense). The phrasing “of the turbine” is informal; rigorously, overall thermal efficiency usually applies to the cycle, but the ratio concept remains valid as a measure of heat-to-work conversion effectiveness. Hence the statement is accepted as true in the context of common exam terminology. Step-by-Step Solution: Define η_th = W_shaft / Q_supplied.Recognize that W_shaft is measured at the turbine output coupling.Use heat balances to determine Q_supplied from boiler/fuel input when evaluating a complete plant. Verification / Alternative check: Standard Rankine-cycle analyses compute η_th by dividing net cycle work by boiler heat input; component wording varies but the ratio mirrors the definition given. Why Other Options Are Wrong: Mechanical efficiency deals with internal power vs. shaft power, not heat-to-work conversion.Limiting to impulse turbines is unnecessary; definition is general. Common Pitfalls: Confusing internal (isentropic) efficiency with overall thermal efficiency; the former uses ideal enthalpy drops, the latter uses boiler heat input. Final Answer: True

Question 3

Speed control in steam turbines The engineering term for maintaining turbine speed essentially constant over a range of load conditions is what?

Accepted Answer

Introduction / Context: Power systems require prime movers to hold speed nearly constant despite load fluctuations. In steam turbines, specialized control systems modulate steam flow to maintain set speed, ensuring frequency stability for generators and mechanical drivetrains. Given Data / Assumptions: Turbine connected to a load (e.g., electrical generator) with variable demand. Control elements available: throttle valves, nozzle control valves, or governing valves. Feedback based on speed or frequency deviation. Concept / Approach: Governing is the collective term for speed-control methods in turbines. Common schemes include throttle governing (varying inlet pressure), nozzle control governing (varying the number of active nozzles), and bypass or combination systems. The governor senses speed error and adjusts steam admission to keep speed at the setpoint within a specified droop, maintaining stability and sharing load among parallel units. Step-by-Step Solution: Detect speed deviation from setpoint via a governor (mechanical, hydraulic, or electronic).Actuate valves to change steam flow: throttle, nozzle group control, or bypass.Restore speed near setpoint while accommodating new load. Verification / Alternative check: Frequency control in grids uses droop characteristics; turbine governing systems are tuned to share loads according to droop while keeping speed/frequency within standards. Why Other Options Are Wrong: Bleeding extracts steam for feed heating, not speed control.Reheating raises steam temperature between turbine sections, not a speed regulator.Condensing refers to the condenser process downstream, unrelated to speed regulation. Common Pitfalls: Assuming throttle governing is the only method; many large turbines use nozzle control or combination systems for better efficiency under part load. Final Answer: governing

Question 4

Steam nozzles – factors governing maximum (choked) discharge For a convergent–divergent steam nozzle operating under choked conditions, the maximum mass discharge rate depends primarily on which parameters?

Accepted Answer

Introduction / Context: In steam turbines and ejectors, convergent–divergent nozzles accelerate steam to high velocities. Once the flow is choked at the throat, understanding what controls the maximum discharge is vital for sizing and performance predictions. Given Data / Assumptions: Dry or superheated steam modeled with standard nozzle relations. Choked flow at the throat when the pressure ratio exceeds the critical value. Steady, one-dimensional flow with negligible inlet velocity. Concept / Approach: Under choked flow, the mass flow rate becomes independent of further reductions in back pressure. The maximum discharge is set by the sonic condition at the throat, which depends on the throat area and the inlet thermodynamic state (pressure and specific volume/temperature). Exit (back) pressure matters only until choking occurs. Step-by-Step Solution: Recognize choking: when p_exit ≤ p_critical, Mach number at throat = 1.Write mass flow per unit area in plain form: m_dot/A_throat = function(p0, v0, k) that is independent of p_exit once choked.Therefore, total m_dot ∝ A_throat and is set by inlet total state (p0, T0 or p0, v0).Hence, the governing parameters are throat area and initial pressure/volume, not the final (back) pressure after choking. Verification / Alternative check: Performance charts show a plateau of m_dot versus back pressure beyond the critical ratio; increasing A_throat or raising inlet pressure increases m_dot_max. Why Other Options Are Wrong: (c) Back pressure affects discharge only until choking; beyond, it does not raise m_dot. (e) Surface roughness influences losses and velocity slightly but does not set the maximum choked m_dot. Common Pitfalls: Assuming exit pressure always controls flow rate; once choked, it does not. Confusing throat and exit areas. Final Answer: both (a) and (b)

Question 5

Reaction turbines – where does the steam pressure drop occur? In a reaction steam turbine stage, the steam's static pressure is reduced in which blade rows?

Accepted Answer

Introduction / Context: Impulse and reaction turbines differ in how the pressure and velocity of steam change across stator and rotor passages. Correctly characterizing the distribution of pressure drop is fundamental to stage design and efficiency analysis. Given Data / Assumptions: Single reaction stage with fixed (stator) and moving (rotor) blades. Steady flow, negligible leakage for conceptual understanding. Nozzle action may occur in both blade rows for a reaction stage. Concept / Approach: In a reaction turbine, part of the enthalpy (pressure) drop occurs in the fixed blades and part in the moving blades. The moving blades act as expanding nozzles. In contrast, in a pure impulse stage the entire pressure drop occurs in the fixed nozzles, while the rotor ideally turns the jet without pressure change. Step-by-Step Solution: Define reaction: R = (enthalpy drop in rotor) / (total stage enthalpy drop).If R > 0, there is pressure drop in the moving blades.Therefore, in a reaction stage, both stator and rotor experience pressure reductions. Verification / Alternative check: Velocity triangles for reaction stages show acceleration in both blade rows, consistent with pressure decrease in each. Why Other Options Are Wrong: (a) Describes an impulse stage. (b) Not physically consistent for a standard stage design. (d) and (e) contradict expansion physics. Common Pitfalls: Equating all turbines with impulse behavior; many modern stages have finite reaction (e.g., 50% reaction in Parsons stages). Final Answer: Pressure is reduced in both fixed blades and moving blades

Question 6

Turbine stage efficiencies – definition check Is the following statement correct? “The ratio of energy supplied to the blades per kg of steam to the total energy supplied per stage per kg of steam is called mechanical efficiency.”

Accepted Answer

Introduction / Context: Turbine performance uses several efficiency definitions: nozzle efficiency, diagram (blading) efficiency, stage efficiency, and mechanical efficiency. Confusion among these terms leads to incorrect diagnostics. Given Data / Assumptions: Energy supplied to blades per kg = work transferred to the rotor (per kg steam). Total energy supplied per stage per kg = isentropic (or actual) stage heat drop basis, depending on convention. Mechanical efficiency accounts for rotor windage, bearing, and transmission losses. Concept / Approach: Mechanical efficiency is the ratio of brake (shaft) power to rotor power (power developed in the blades). The quoted ratio compares rotor work to stage energy input, which corresponds to diagram (blading) efficiency or stage efficiency, not mechanical efficiency. Step-by-Step Solution: Identify the numerator: energy to blades → rotor work.Identify the denominator: total stage energy supplied.This ratio reflects how effectively the stage converts available enthalpy drop into rotor work → diagram/stage efficiency.Mechanical efficiency = shaft (deliverable) power / rotor power, accounting for mechanical losses. Verification / Alternative check: Standard definitions list: η_mech = P_shaft / P_rotor; η_nozzle relates kinetic energy to isentropic drop; η_stage = W_rotor / Δh_stage, aligning with the statement's ratio. Why Other Options Are Wrong: (a) Mislabels the metric. (c) Stage type (impulse vs reaction) does not convert this into mechanical efficiency. (d) and (e) reference different efficiency types. Common Pitfalls: Using “mechanical efficiency” to describe fluid-dynamic conversion quality; that is a separate (diagram/stage) efficiency. Final Answer: No; that ratio defines blading/diagram or stage efficiency, not mechanical efficiency

Question 7

Critical (throat) pressure in steam nozzles – terminology The static pressure of steam at the throat of a steam nozzle (where the flow first reaches sonic velocity) is commonly termed the:

Accepted Answer

Introduction / Context: In nozzle flow, the term “critical” refers to the sonic condition at the minimum area (throat). Recognizing the associated pressure is essential for sizing and for understanding choking and discharge behavior. Given Data / Assumptions: Convergent–divergent steam nozzle. Isentropic idealization for terminology. Critical condition corresponds to Mach number = 1 at the throat. Concept / Approach: When the pressure ratio across the nozzle exceeds the critical value, the flow chokes at the throat and the static pressure there takes a specific value called the critical pressure. Downstream pressure reductions do not alter throat pressure under choked conditions. Step-by-Step Solution: Identify the throat as the minimum area where choking occurs.Associate the sonic condition with a unique pressure level for given inlet state.Name this pressure: critical pressure. Verification / Alternative check: T–s and p–v diagrams for nozzle flow show the state at the throat marked by the critical condition; empirical discharge curves confirm a plateau beyond the critical pressure ratio. Why Other Options Are Wrong: Stagnation pressure refers to the total pressure at zero velocity (inlet total state). Exit/back pressure is downstream of the nozzle. Condenser pressure pertains to the steam cycle sink, not the nozzle throat. Impulse pressure is not a standard term here. Common Pitfalls: Confusing stagnation (total) properties with static throat properties. Critical pressure is a static property at M = 1. Final Answer: critical pressure

Question 8

Effect of nozzle friction on jet speed How does internal friction (irreversibility) within a steam nozzle influence the exit velocity of the steam jet for a given inlet and back pressure?

Accepted Answer

Introduction / Context: Nozzle friction converts part of the available enthalpy drop into entropy rather than directed kinetic energy. Appreciating this effect is crucial for predicting jet speed and turbine work. Given Data / Assumptions: Same inlet state and back pressure for the comparison. Real nozzle with friction vs. ideal isentropic nozzle. Negligible heat transfer to surroundings. Concept / Approach: In an ideal nozzle, the isentropic enthalpy drop fully converts to kinetic energy: V^2/2 = Δh_isentropic. With friction, the actual kinetic energy equals η_nozzle * Δh_isentropic, where 0 < η_nozzle < 1. Thus, exit velocity is lower. Step-by-Step Solution: Write V_actual^2 / 2 = η_nozzle * Δh_isentropic.For fixed inlet and exit pressures, friction raises entropy and reduces effective Δh to kinetic energy.Therefore, V_actual < V_isentropic; exit velocity decreases. Verification / Alternative check: Measured nozzle coefficients (velocity coefficient < 1) quantify this reduction; test data show lower exit Mach numbers and speeds than ideal predictions. Why Other Options Are Wrong: (b) Choking does not eliminate friction losses; it limits mass flow but velocity still falls from the ideal. (c) Internal viscous heating increases entropy, not useful kinetic energy. (d) Mass flow changes are secondary and do not increase exit kinetic energy beyond ideal. (e) Friction affects magnitude strongly. Common Pitfalls: Assuming choked flow guarantees ideal velocity; choking fixes Mach at the throat, not exit losses. Final Answer: It decreases the exit velocity due to kinetic energy loss

Question 9

Impulse turbine compounding – efficiency comparison Compared with a pressure-compounded impulse turbine (Rateau staging), the efficiency of a pressure-velocity compounded impulse turbine (Curtis staging) is generally:

Accepted Answer

Introduction / Context: Impulse turbines may be compounded by pressure (multiple nozzle stages) or by velocity (multiple rows of moving blades with intermediate guide blades), or by a combination of both. Knowing how compounding affects efficiency guides stage selection. Given Data / Assumptions: Equal overall pressure ratio and comparable design practice. Similar blade quality and leakage assumptions for comparison. Focus on aerodynamic/rotordynamic losses from additional blade rows. Concept / Approach: Pressure compounding spreads the pressure drop across several nozzle stages, generally keeping blade relative speeds moderate with fewer moving rows per stage. Pressure-velocity compounding introduces extra moving rows within a stage to split absolute jet velocity, but each added row brings turning, friction, and tip/leakage losses, typically lowering stage efficiency relative to pressure compounding. Step-by-Step Solution: Recognize that Curtis staging adds rows → more cumulative losses.Rateau staging uses more nozzle sets but fewer moving rows per stage → lower cumulative moving-blade losses.Therefore, for similar conditions, pressure-velocity compounded stages are generally less efficient than pressure-compounded ones. Verification / Alternative check: Textbook comparisons and test data show diagram efficiency penalties with additional blade rows due to repeated incidence, friction, and wake mixing. Why Other Options Are Wrong: “More” contradicts typical loss accumulation behavior. “Same” ignores extra rows and losses. “Always 100%” is impossible in real turbines. “Indeterminate” is overly cautious for standard design principles. Common Pitfalls: Focusing solely on speed reduction benefits of Curtis staging while overlooking added loss sources. Final Answer: less

Question 10

Steam pressure change across a nozzle As steam flows through a properly designed nozzle (from inlet to exit), the static pressure generally:

Accepted Answer

Introduction / Context: Nozzles are expansion devices that transform pressure energy (enthalpy) into kinetic energy. Understanding how pressure varies along the nozzle path is fundamental for turbine stage analysis. Given Data / Assumptions: Adiabatic, steady flow through a nozzle. Properly matched back pressure to avoid shocks within the nozzle. Negligible potential energy changes. Concept / Approach: From the steady flow energy viewpoint, decreasing static enthalpy appears as increasing kinetic energy. Hence, as the fluid accelerates, static pressure falls. In a CD nozzle, pressure falls through the throat and continues to fall in the divergent part under correctly expanded conditions. Step-by-Step Solution: Use energy balance: h1 + V1^2/2 ≈ h2 + V2^2/2.As V rises (V2 > V1), h must drop → associated with a drop in p for expansion.Therefore, pressure decreases along the nozzle path. Verification / Alternative check: Measured pressure profiles and Mach distributions in Laval nozzles show monotonic pressure decrease (absent shocks) from inlet to exit. Why Other Options Are Wrong: (b) Acceleration does not increase static pressure; it reduces it in expansion devices. (c) Constant pressure describes throttling valves, not nozzles. (d) Oscillation is not characteristic of steady, shock-free flow. (e) Pressure does not rise in the convergent section; it falls toward the throat. Common Pitfalls: Confusing nozzle behavior with diffusers, where velocity decreases and pressure increases. Final Answer: decreases due to conversion of enthalpy to kinetic energy

Question 11

Impulse turbine inlet – steam velocity approaching the nozzle In a single-stage impulse turbine, the absolute velocity of steam immediately upstream of the nozzles (approaching the nozzles) is typically:

Accepted Answer

Introduction / Context: Impulse turbines rely on nozzles to convert pressure energy to high-speed jets that strike the rotor. The flow conditions just before the nozzles help set up the velocity triangles used for stage analysis. Given Data / Assumptions: Single-stage impulse arrangement with casing leading steam into stationary nozzles. Large pressure drop occurs in the nozzles, not in the inlet plenum. Negligible pre-nozzle acceleration. Concept / Approach: Because the pressure drop is arranged to occur across the nozzles, the steam in the casing upstream of the nozzles is nearly stagnant relative to the subsequent jet speed. Thus, inlet absolute velocity is taken as negligible when forming inlet velocity triangles. Step-by-Step Solution: Note design intent: maximize acceleration within the nozzle passages.Therefore, V_upstream ≈ 0 compared with V_exit from nozzle.Adopt this assumption for constructing velocity triangles used to compute blade work. Verification / Alternative check: Stage design literature treats inlet absolute velocity as small, ensuring that most kinetic energy appears at nozzle exit, where it is directed onto the rotor blades. Why Other Options Are Wrong: Equal to or greater than nozzle exit velocity contradicts the function of the nozzle. “Equal to blade speed” confuses gas velocity with rotor peripheral speed. Supersonic upstream of the nozzle is unrealistic and undesirable. Common Pitfalls: Mixing up “approaching the nozzles” with “approaching the moving blades.” The very high velocity appears at the nozzle exit, not before the nozzles. Final Answer: negligible compared with the nozzle exit velocity

Question 12

Identifying a convergent–divergent (Laval) nozzle by geometry A nozzle whose cross-section first decreases from its entrance to a minimum (throat) and then increases from the throat to the exit is:

Accepted Answer

Introduction / Context: Supersonic expansion of compressible fluids demands a geometry that accommodates the transition from subsonic acceleration to sonic conditions and then to supersonic speeds. The convergent–divergent (CD or Laval) nozzle provides exactly this shape. Given Data / Assumptions: Single-stream nozzle with smooth area variation. Choking occurs at the minimum area (throat). Back pressure is appropriately set to allow fully expanded or correctly expanded flow. Concept / Approach: In subsonic flow, converging passages accelerate the fluid; at the throat Mach = 1; beyond the throat, a diverging passage accelerates supersonic flow further. Thus, a geometry that first converges to a throat and then diverges is the textbook definition of a CD nozzle. Step-by-Step Solution: Identify geometric sequence: converge → throat → diverge.Map to compressible behavior: subsonic acceleration to sonic at throat, then supersonic acceleration downstream.Therefore, the device is a convergent–divergent (Laval) nozzle. Verification / Alternative check: Mach number distributions measured in Laval nozzles confirm the accelerate–choke–accelerate pattern consistent with the stated geometry. Why Other Options Are Wrong: (a) Directly contradicts the definition. (c) and (d) describe single-section devices lacking a throat followed by the opposite section. (e) A throttling valve is a constant-area orifice introducing an isenthalpic pressure drop, not a shaped nozzle. Common Pitfalls: Confusing diffusers (pressure-recovery devices) with nozzles (velocity-increase devices). Final Answer: a convergent–divergent nozzle (Laval nozzle)

Question 13

Defining a stage in a reaction turbine In the context of a reaction steam turbine, a “stage” is best represented by:

Accepted Answer

Introduction / Context: Stage counting differs in impulse and reaction turbines because of where the pressure drop occurs. Recognizing how a “stage” is represented aids communication between designers and operators and helps in performance bookkeeping. Given Data / Assumptions: In reaction turbines, both fixed and moving rows can contribute to pressure drop. Manufacturers and textbooks sometimes vary in counting conventions; the question uses the common pedagogical convention referenced in many exam syllabi. Focus is on the pressure-changing elements. Concept / Approach: Because a reaction turbine experiences pressure drop across both fixed and moving rows, many treatments describe each row as a “stage element,” and questions often phrase the representation as “each row of blades” for a stage in reaction type. This contrasts with impulse turbines, where one nozzle row plus one moving row typically forms a stage. Step-by-Step Solution: Note that pressure changes in both stator and rotor for reaction stages.Thus, each row functions as a pressure-changing component.Accordingly, stage representation is commonly taken as each row of blades in reaction turbines for counting purposes in such questions. Verification / Alternative check: Parsons-type 50% reaction stages split the enthalpy drop between stator and rotor; either row could be used as a stage count unit in some academic treatments. Why Other Options Are Wrong: (a) Casing count does not equal stage count; a casing can contain multiple rows. (b) and (c) refer to machine boundaries, not internal stage elements. (e) While many texts define a stage as a pair (fixed+moving), the question's standard answer emphasizes representation by each row for reaction stages per common exam convention. Common Pitfalls: Assuming impulse and reaction stage definitions are identical; terminology varies, but exam context often expects the stated interpretation. Final Answer: each row of blades (fixed or moving) considered as a pressure-changing element

Question 14

Steam Nozzles – Definition of a Divergent Nozzle In steam and gas dynamics, classify a nozzle as ‘‘divergent’’ when considering how its flow area varies along the flow direction from entrance, through any throat if present, to the exit. Choose the s…

Accepted Answer

Introduction / Context: A clear understanding of nozzle geometry is essential in steam turbines, gas turbines, ejectors, and rocket engines. The words divergent, convergent, and convergent–divergent (C–D) describe how the flow area changes along the nozzle centerline and directly control whether the flow accelerates subsonically, reaches sonic conditions, or expands to supersonic speeds. Given Data / Assumptions: Nozzle refers to a duct guiding flow with varying cross-sectional area. Steady, quasi-one-dimensional behavior is assumed for conceptual discussion. ‘‘Divergent’’ refers solely to geometry (area change), not to pressure ratio or Mach number outcome by itself. Concept / Approach: A divergent nozzle is one whose flow area increases monotonically from inlet to outlet. By contrast, a convergent nozzle has monotonically decreasing area. A convergent–divergent nozzle (de Laval) first converges to a minimum area (the throat) and then diverges. Whether a divergent section accelerates or decelerates the flow depends on the local Mach number: subsonic flow decelerates in a diverging passage, while supersonic flow accelerates in it. The geometric definition, however, remains independent of regime. Step-by-Step Solution: Identify purely geometric definitions: divergent = area increases along the flow path.Recognize other cases: convergent = area decreases; C–D = first decreases to throat, then increases.Match the given statements to these definitions.Select the statement that area ‘‘increases continuously’’ from entrance to exit as the correct description of a divergent nozzle. Verification / Alternative check: Standard nozzle sketches show a bell or flared shape for a divergent section. In C–D designs, the downstream portion is the divergent section following the throat; a purely divergent nozzle has no upstream convergent part by definition. Why Other Options Are Wrong: ‘‘Decreases continuously’’ defines a convergent nozzle. ‘‘First decreases then increases’’ defines a convergent–divergent nozzle. ‘‘Constant area’’ is a duct, not a nozzle, and ‘‘none of the above’’ is unnecessary given a correct option. Common Pitfalls: Equating ‘‘divergent’’ with ‘‘supersonic’’ unconditionally; the flow regime depends on the pressure ratio and presence of a throat, not solely on geometry. Final Answer: when the cross-section of the nozzle increases continuously from entrance to exit

Question 15

Steam Power Cycles – Name the Ratio: Isentropic Heat Drop Divided by Heat Supplied In an ideal steam power plant analysis, the ratio of the isentropic heat drop in the turbine to the total heat supplied in the boiler is referred to as which efficien…

Accepted Answer

Introduction / Context: Steam-plant performance is summarized by several efficiencies. It is important to distinguish component efficiencies (nozzle, blade, stage, internal/isentropic turbine) from cycle-level (thermal) efficiency. The question asks for the name of the ratio of the ideal (isentropic) turbine enthalpy drop to the total heat added in the boiler—an expression that mirrors the ideal Rankine thermal efficiency when pump work is small. Given Data / Assumptions: Ideal Rankine cycle with isentropic turbine and pump, negligible pump work. Heat supplied occurs primarily in the boiler from feedwater inlet to turbine inlet. Isentropic heat drop is h_in − h_out,s across the turbine. Concept / Approach: For the ideal Rankine cycle, thermal efficiency is eta_th ≈ (h1 − h2s)/(h1 − h_fw), i.e., the isentropic enthalpy drop across the turbine divided by the boiler heat input. This is what many textbooks call the Rankine efficiency for the idealized case. Component factors such as stage efficiency or internal efficiency compare actual to isentropic within the turbine only and do not involve boiler heat input in the denominator. Reheat factor is the ratio of the sum of isentropic drops in stages to the overall isentropic drop—again a different concept. Step-by-Step Solution: Write ideal turbine work per kg: w_t,ideal = h1 − h2s.Write heat supplied per kg: q_in = h1 − h_fw (neglecting pump work).Form the ratio: (h1 − h2s)/(h1 − h_fw).Recognize this as the ideal Rankine thermal efficiency and select ‘‘Rankine efficiency’’. Verification / Alternative check: Energy-balance diagrams for the ideal Rankine cycle show turbine work equaling the isentropic drop; with pump work very small, the ratio above matches eta_th used for cycle benchmarking. Why Other Options Are Wrong: Reheat factor relates cumulative stage drops, not boiler heat. Stage efficiency compares a single stage output to its isentropic drop. Internal efficiency compares the actual turbine work to the isentropic drop. Nozzle efficiency is local to nozzles only. Common Pitfalls: Confusing cycle efficiency with turbine internal efficiency; overlooking pump work (often negligible but not zero in precise calculations). Final Answer: Rankine efficiency

Question 16

Steam Turbines vs. Reciprocating Steam Engines – Comparative Advantages Which statements correctly compare steam turbines with reciprocating steam engines in terms of achievable speed, efficiency, and steam consumption?

Accepted Answer

Introduction / Context: Steam turbines displaced reciprocating engines in large-scale power generation because of superior performance characteristics. Understanding why helps in appreciating plant design choices and matching prime movers to applications ranging from power stations to marine propulsion. Given Data / Assumptions: Comparing mature designs operating near their practical scales. Steam quality and pressure ranges typical of utility plants. Focus on steady-state operation and overall plant metrics. Concept / Approach: Steam turbines are rotary, continuous-flow devices with minimal reciprocating masses. They readily achieve very high rotational speeds (thousands of rpm) with smooth torque. Their isentropic and internal efficiencies are generally higher than those of piston engines at comparable ratings due to reduced mechanical losses, continuous expansion, and better utilization of high inlet pressures/temperatures. Consequently, for a given shaft output, turbines typically require less steam flow (lower specific steam consumption), especially at utility scales with multi-stage expansions and reheats. Step-by-Step Solution: Assess speed: absence of reciprocating inertia enables high rpm in turbines.Assess efficiency: multi-stage expansion and optimized velocity triangles improve stage and overall efficiencies.Link to steam rate: lower losses and higher efficiency reduce kg of steam per kWh.Combine findings: all listed advantages apply to turbines over reciprocating engines. Verification / Alternative check: Historical data: utilities shifted from large piston engines to condensing steam turbines early in the 20th century because turbines offered better thermal efficiency, more compactness at high power, and smoother operation compatible with AC generators. Why Other Options Are Wrong: Choosing any single advantage ignores the others, while ‘‘none of these’’ contradicts well-established practice and performance records. Common Pitfalls: Comparing very small turbines with large slow-speed reciprocating engines; at tiny scales, economics and part-load behavior may differ, but the general statements remain true for power generation scales. Final Answer: all of these

Question 17

Impulse Turbine – Exit-Whirl Condition for Maximum Efficiency State whether, for maximum blade efficiency in a pure impulse turbine stage, the steam should leave the moving blades with a velocity component at right angles to blade motion (i.e., zero…

Accepted Answer

Introduction / Context: Blade or diagram efficiency in an impulse stage depends on how effectively the rotor converts jet kinetic energy to shaft work. The tangential (whirl) components of velocity at rotor inlet and exit determine the torque and thus the work per unit mass. The question tests the condition for maximizing this efficiency in a pure impulse stage. Given Data / Assumptions: Pure impulse stage: nearly all pressure drop in stator nozzles; negligible pressure change across rotor. Friction is present but small for conceptual analysis; optimal inlet angle and speed ratio are considered. Blade speed and nozzle angle are adjusted for best efficiency. Concept / Approach: Shaft work per unit mass in a turbine stage is proportional to the change in whirl velocity: w = U*(Vw1 − Vw2), where U is blade speed, Vw1 is inlet whirl, and Vw2 is exit whirl. For a given inlet condition and blade speed ratio, maximizing w (and diagram efficiency) requires minimizing the magnitude of exit whirl. The ideal target is Vw2 = 0, meaning the absolute exit velocity is orthogonal to the wheel motion (purely axial for an axial-flow stage). This condition eliminates kinetic energy associated with tangential swirl in the exhaust, thus minimizing losses. Step-by-Step Solution: Express stage work: w = U*(Vw1 − Vw2).For fixed U and Vw1, maximize w by minimizing Vw2.Set optimal Vw2 → 0 → exit absolute velocity normal to blade motion.Conclude: for maximum efficiency, steam leaves at right angles to blade motion (zero exit whirl). Verification / Alternative check: Classical velocity-triangle analysis and diagram-efficiency derivations show the maximum occurs near zero exit whirl when blade-friction losses are modest and incidence is optimized. Why Other Options Are Wrong: Reaction stages intentionally retain pressure drop in the rotor; the stated condition is specific to impulse optimization and is not restricted to partial admission or reheat considerations. Common Pitfalls: Confusing axial exit (Vw2 = 0) with zero absolute velocity; the magnitude of exit velocity is not zero—it is oriented axially to avoid tangential kinetic-energy carryover. Final Answer: Correct

Question 18

Nozzle with Friction – Primary Effects on Steam Flow and Quality What is the principal combined effect of internal friction on the flow of wet steam through a nozzle, considering mass flow and moisture content at the exit?

Accepted Answer

Introduction / Context: Real steam nozzles are not frictionless. Wall friction and boundary-layer growth introduce irreversibilities that alter discharge, velocity, pressure distribution, and condensation behavior. Recognizing the signs of these changes is important for diagnosing performance shortfalls on test stands and in service. Given Data / Assumptions: Flow of wet or potentially condensing steam through a converging or converging–diverging nozzle. One-dimensional, steady modeling with friction and associated entropy rise. Back pressure low enough to make compressibility effects relevant. Concept / Approach: Friction converts some of the available isentropic enthalpy drop into internal energy (heating) rather than kinetic energy. This reduces the achievable velocity at a given pressure ratio and effectively lowers the discharge for the same upstream stagnation conditions and back pressure. Moreover, in wet-steam conditions, irreversible mixing and wall heat transfer promote additional condensation, reducing dryness fraction and thus increasing wetness (mass fraction of liquid). The result is a lower mass flow than ideal and a wetter exhaust. Step-by-Step Solution: Account for entropy rise: friction → higher s than isentropic path.Relate to velocity: less of the enthalpy drop becomes kinetic energy → lower V and lower m_dot through a given throat/back pressure.Consider phase behavior: nonequilibrium effects and losses favor increased liquid content → higher wetness at exit.Conclude the combined effect: decreased mass flow and increased wetness. Verification / Alternative check: Empirical discharge coefficients for steam nozzles (Cd < 1) capture reduced mass flow relative to isentropic predictions. Moisture measurements downstream of high-load nozzles commonly show greater wetness than ideal models predict. Why Other Options Are Wrong: Options claiming increased mass flow contradict the universal trend of Cd < 1 with friction. Decreased wetness is unlikely in condensing flows with friction. A claim of ‘‘no change’’ ignores real, measurable losses. Common Pitfalls: Confusing static temperature trends (which can locally rise due to frictional dissipation) with overall discharge behavior; the key combined effect remains reduced mass flow rate and increased moisture. Final Answer: decrease the mass flow rate and to increase the wetness of steam

Question 19

Supersaturation (Supercooling) in Steam Nozzles – Net Effect on Discharge and Properties When condensation is delayed and the steam remains metastable (supersaturated) while expanding through a nozzle, what is the predominant effect?

Accepted Answer

Introduction / Context: During rapid nozzle expansions, steam can remain in a metastable, supersaturated state below the saturation temperature corresponding to local pressure. This delays droplet formation and shifts thermodynamic properties away from equilibrium values. Designers must understand whether this increases or decreases discharge and how it affects quality and velocity. Given Data / Assumptions: Fast expansion with delayed nucleation (Wilson line behavior). One-dimensional nozzle flow reaching or exceeding sonic conditions. Comparison with an equilibrium (fully condensed) expansion at the same upstream conditions. Concept / Approach: Supersaturation implies the steam remains drier than equilibrium at the same pressure—droplet formation is postponed. The metastable vapor has lower specific volume than the equilibrium wet mixture at that pressure, and the entropy rise due to phase change is delayed. Consequently, for the same pressure ratio, the nozzle can achieve a slightly higher mass flux and higher exit velocity than predicted by equilibrium wet-steam tables. This manifests as an apparent discharge coefficient greater than unity relative to equilibrium calculations. Step-by-Step Solution: Contrast equilibrium vs. metastable paths on an h–s plot.Note delayed condensation → reduced v compared to wet mixture at same p.Mass flux m_dot ∝ (p0 * A / sqrt(T0)) * function(gamma, M) increases when v decreases at the throat.Therefore, supersaturation increases discharged mass and tends to increase exit velocity. Verification / Alternative check: Classic Wilson line experiments show higher-than-equilibrium mass flows, requiring correction factors in nozzle testing. The metastable path features lower entropy than the equilibrium wet path at the same pressure, contradicting the notion that entropy increases in the supersaturated state. Why Other Options Are Wrong: (b) is incorrect because, at a given pressure, the supersaturated vapor typically exhibits lower entropy and smaller specific volume than the equilibrium wet mixture. (c) is wrong since exit velocity generally increases slightly. (d) cannot be true if (b) and (c) are false. Common Pitfalls: Assuming any condensation automatically reduces discharge; in metastable flow, the opposite trend is observed until nucleation occurs abruptly downstream. Final Answer: mass of the steam discharged increases

Question 20

Identify the De Laval Turbine – Classification by Principle Select the correct classification of the classic De Laval turbine based on where pressure drop occurs and how energy conversion is accomplished.

Accepted Answer

Introduction / Context: The De Laval turbine is one of the earliest practical steam turbines and underpins much of the elementary theory of impulse stages. Correctly identifying its type clarifies how the stage produces work and how velocity triangles are constructed. Given Data / Assumptions: Classic single-stage design with a convergent–divergent nozzle set and one row of moving blades. Large pressure drop through the nozzles; negligible pressure change across the rotor. High-velocity jet from the nozzles impinges on the moving blades. Concept / Approach: In an impulse turbine stage, nearly the entire pressure drop occurs in stationary nozzles, converting pressure energy into kinetic energy. The moving blades ideally only redirect the flow, reducing its absolute velocity and changing whirl to produce torque, with little static pressure change. This description matches the De Laval turbine. Reaction turbines, in contrast, split the pressure drop between stator and rotor, causing additional acceleration within the moving passages. Step-by-Step Solution: Identify nozzle function: convert enthalpy to jet kinetic energy.Identify rotor function: change jet direction and reduce absolute velocity to extract work.Check pressure distribution: large drop in stator, negligible in rotor → impulse.Conclude De Laval is a simple impulse turbine. Verification / Alternative check: Historical diagrams and laboratory test data for De Laval wheels show near-constant static pressure across blades with significant velocity changes, consistent with impulse behavior. Why Other Options Are Wrong: Reaction and impulse–reaction imply pressure drop in the rotor, which is not the De Laval case. ‘‘None of these’’ is invalid as a clear correct classification exists. Common Pitfalls: Confusing multi-row Curtis (velocity-compounded) stages with De Laval; Curtis involves multiple rotor rows but still follows impulse principles. Final Answer: simple impulse turbine

Steam Nozzles and Turbines Questions