Q: Quantifying blade friction effect in impulse turbines By approximately what percentage does blade friction reduce the steam velocity as it passes over the moving blades of an impulse turbine?

Introduction / Context: In impulse turbine rotors, ideally there is no pressure drop; the rotor simply turns the jet and extracts work. In practice, surface roughness and viscous effects cause friction, reducing the relative velocity across the blade passage and thus decreasing the work recovered and the diagram efficiency. Given Data / Assumptions: Impulse-type moving blades with negligible pressure drop across the rotor in the ideal model. Friction present due to surface finish, incidence, and secondary flows. Typical well-designed stage, not severely eroded or fouled. Concept / Approach: Engineering practice often assumes a relative-velocity reduction factor across the rotor (sometimes denoted by k or a blade friction coefficient). For clean, well-designed blades, the relative velocity at exit is commonly taken as about 0.85 to 0.90 of the inlet relative velocity, implying a reduction on the order of 10 to 15 percent. This rule-of-thumb is used in preliminary design and performance calculations. Step-by-Step Solution: Let w1 be inlet relative velocity and w2 be exit relative velocity.Assume w2 ≈ 0.85 to 0.90 * w1 for typical blades.Percentage reduction ≈ 10% to 15%. Verification / Alternative check: Velocity-triangle based design examples and textbook problems frequently adopt 10–15% reduction, matching test data for smooth, modern impulse blades at design incidence. Why Other Options Are Wrong: Higher ranges (20–40%) imply unusually high losses inconsistent with good design and would severely depress diagram efficiency.Below 5% is overly optimistic and rarely achieved in practice. Common Pitfalls: Confusing relative velocity reduction (rotor passage) with absolute velocity changes across the stage; the former is the correct metric for blade friction in impulse rotors. Final Answer: 10 to 15%

Q: Axial discharge condition For a turbine designed for axial discharge at the stage exit, what is the value of the whirl (tangential) component of absolute velocity at outlet?

Introduction / Context: Velocity triangles in turbines decompose absolute velocity into axial (flow), whirl (tangential), and sometimes radial components. Power transfer depends on the change in whirl component across the rotor. Some stages are designed for axial discharge to simplify downstream flow and reduce exit swirl losses. Given Data / Assumptions: Axial discharge means the absolute exit velocity is aligned with the turbine axis. Steady flow and well-designed outlet guide geometry. No significant swirl-producing leakage or secondary flows at design point. Concept / Approach: If the absolute exit velocity vector is purely axial, its tangential (whirl) component is zero. This choice reduces kinetic energy loss in the exhaust (swirl loss) and simplifies matching to the next stage or diffuser. In the Euler turbine equation, the work per unit mass is proportional to the difference in whirl components (Vw1 − Vw2); setting Vw2 = 0 makes the work equal to Vw1 * u, a common design target for impulse stages. Step-by-Step Solution: Define axial discharge: exit flow direction parallel to axis.Therefore, tangential component at outlet Vw2 = 0.This minimizes exit swirl losses and eases stage matching. Verification / Alternative check: Measured exit swirl angles near zero in well-tuned stages confirm negligible whirl; residual small swirl may exist off-design but the design intent is Vw2 ≈ 0. Why Other Options Are Wrong: Maximum/minimum (non-zero) contradict the definition of axial discharge.Equal to blade speed is unrelated; blade speed is a rotor property, not a flow component at exit. Common Pitfalls: Confusing relative velocity (in the blade frame) with absolute exit velocity; axial discharge refers to the absolute velocity direction. Final Answer: zero

Q: Relative velocity in ideal impulse rotors In an impulse turbine neglecting friction, the relative velocity of steam at the exit tip of the moving blade is ________ the relative velocity at the inlet tip.

Introduction / Context: Impulse rotors ideally experience no pressure change across the moving blades; they simply turn the jet and extract work via change in whirl component. Understanding how relative velocity behaves in this ideal case is critical for constructing correct velocity triangles and estimating diagram efficiency limits. Given Data / Assumptions: Impulse rotor: negligible pressure drop across moving blades. No friction (idealization), so no viscous dissipation in the rotor passage. Steady flow and correct incidence at blade entry. Concept / Approach: With no pressure change and no friction, the magnitude of the relative velocity vector should remain constant through the rotor passage; only its direction changes due to blade curvature. Hence w1 (at inlet) equals w2 (at exit) in magnitude. Real blades show a reduction (e.g., 10–15%) due to friction, but the ideal model preserves the magnitude. Step-by-Step Solution: Adopt impulse rotor assumptions: Δp_rotor ≈ 0, friction ≈ 0.Without friction, there is no loss of mechanical energy in the relative frame.Therefore, |w2| = |w1|; direction changes, magnitude stays the same. Verification / Alternative check: Euler's turbine equation shows rotor work depends on whirl change in absolute velocity, not on a change in relative speed magnitude for the ideal impulse rotor. Why Other Options Are Wrong: Less/greater than: would imply frictional loss or energy input, contradicting the ideal assumption.Half or twice: arbitrary and inconsistent with energy conservation in the relative frame. Common Pitfalls: Mixing real and ideal behavior; in practice, friction lowers w2, but the question explicitly neglects friction. Final Answer: equal to

Q: Effect of nozzle friction on heat drop Due to friction between steam and nozzle surfaces, the useful (available) heat drop across an expanding nozzle is typically reduced by about how much?

Introduction / Context: Nozzle efficiency is a key parameter in turbine performance, translating enthalpy drop into kinetic energy. Surface friction and boundary-layer growth reduce the effective heat drop that converts to jet velocity, lowering the discharge speed and stage power. Given Data / Assumptions: Smooth, well-designed steam nozzles operating near design conditions. Losses dominated by wall friction and minor shock/expansion mismatches. Comparison with ideal isentropic expansion between the same inlet and outlet pressures. Concept / Approach: Nozzle (velocity) efficiency is typically around 0.85–0.90 for good designs. That means the actual kinetic energy at exit is 85–90% of the ideal value, implying an effective reduction in the usable heat drop of roughly 10–15%. Designers account for this when computing throat area and exit velocity from the isentropic enthalpy drop, then applying efficiency to obtain the real exit speed. Step-by-Step Solution: Let Δh_is be isentropic heat drop and V_is the corresponding ideal exit speed.With nozzle efficiency η_n ≈ 0.85–0.90, actual V ≈ √(η_n) * V_is.Effective heat drop utilized ≈ η_n * Δh_is → reduction ≈ 10–15%. Verification / Alternative check: Experimental nozzle tests commonly report efficiencies in the above range unless there is erosion, fouling, or severe off-design operation; these conditions could worsen losses but are not typical for design values. Why Other Options Are Wrong: Higher reductions (25–60%) indicate very poor nozzles or off-design shocks; not representative of standard practice.Less than 5% is overly optimistic and rarely sustained across operating ranges. Common Pitfalls: Confusing reduction in heat drop with reduction in velocity directly; because kinetic energy scales with velocity squared, be careful when converting between efficiency definitions and percentage speed shortfall. Final Answer: 10 to 15 percent

Question 1

Stage count comparison: For the same overall pressure ratio and output, a pressure-compounded impulse turbine requires ________ number of stages compared with a velocity-compounded (Curtis) turbine.

Accepted Answer

Introduction / Context: Impulse turbines can be compounded by dividing the total expansion by pressure (Rateau compounding) or by velocity (Curtis compounding). The compounding method influences stage count, blade loading, and efficiency. Knowing which arrangement needs more stages helps in conceptual design and cost estimation. Given Data / Assumptions: Comparable total pressure drop and desired output. Acceptable blade tip speeds and nozzle angles. Classical idealizations for compounding. Concept / Approach: In pressure compounding, the overall pressure drop is split into multiple sequential nozzle–rotor stages, each with modest velocity. This typically requires several stages to share the expansion. In velocity compounding, a single large nozzle drop produces a high jet speed, and multiple moving rows within the same stage (with intervening fixed guides) distribute the velocity change. Consequently, for the same duty, pressure-compounded arrangements generally need more separate stages than velocity-compounded ones. Step-by-Step Solution: Identify compounding method: pressure-compounded → many nozzle–rotor pairs.Velocity-compounded → one large nozzle drop with 2–3 moving rows in one stage.Conclude pressure compounding requires more distinct stages. Verification / Alternative check: Historical turbine designs (Rateau vs Curtis) exhibit exactly this difference in stage allocation and blade speed constraints. Why Other Options Are Wrong: Same/Less: Contradicts standard compounding practice and design trade-offs. Indeterminate: While details vary, the general rule remains robust for comparable duties. Common Pitfalls: Confusing “stage” with “row”: velocity compounding places multiple moving rows in a single stage. Final Answer: more

Question 2

Necessary condition for flow through a nozzle: If the inlet steam pressure equals the exit pressure, what happens to pressure drop and flow in the nozzle?

Accepted Answer

Introduction / Context: Nozzles convert pressure energy into kinetic energy. A fundamental requirement for any flow through a nozzle is a driving potential—typically a pressure difference. Without a pressure differential, the nozzle cannot accelerate the fluid and sustained flow will not occur (ignoring transient effects). Given Data / Assumptions: Steady operation, no external body forces causing flow. Inlet static pressure equals exit static pressure. No pump or external device imparting momentum directly within the nozzle. Concept / Approach: From basic fluid mechanics, steady flow requires a gradient in total pressure (or head) to overcome friction and to produce acceleration. If inlet and outlet pressures are equal, there is no net driving force; in the ideal, frictionless limit, there is no acceleration; with real friction, any attempt at flow would immediately dissipate energy and eliminate motion, settling at zero flow. Therefore, when p_in = p_out, there is no pressure drop and no flow. Step-by-Step Solution: Check driving force: Δp = p_in − p_out = 0.With Δp = 0, Bernoulli predicts no increase in kinetic head absent other inputs.Thus, no sustained flow occurs through the nozzle. Verification / Alternative check: Energy equation for steady flow shows V_out^2/2 − V_in^2/2 ≈ (p_in − p_out)/ρ minus losses; with Δp = 0 and nonzero losses, only the trivial solution V_in ≈ V_out ≈ 0 remains steady. Why Other Options Are Wrong: Suggesting pressure drop or flow exists contradicts the premise of equal inlet and outlet pressures under steady, passive conditions. Common Pitfalls: Confusing instantaneous transients (e.g., opening a valve briefly) with steady-state; sustained flow requires a pressure difference. Final Answer: there is no pressure drop and fluid does not flow through the nozzle

Question 3

Reaction staging concept in steam turbines In an impulse–reaction (reaction-type) steam turbine stage, the static pressure of the steam decreases gradually and continuously over which blade rows?

Accepted Answer

Introduction / Context: Steam turbines are broadly categorized as impulse or reaction depending on where the pressure (enthalpy) drop occurs. Identifying whether the pressure falls in fixed blades, moving blades, or both is fundamental for drawing velocity triangles, estimating stage work, and understanding efficiency and blade loading. Given Data / Assumptions: The stage in question is a reaction-type (impulse–reaction) turbine stage. Flow is steady and the stage operates close to its design point. Heat transfer across the short stage is negligible; changes are mainly due to flow acceleration and expansion. Concept / Approach: In a pure impulse stage, almost all the pressure drop occurs in the fixed nozzles; moving blades ideally see no pressure change and only turn the jet (with losses due to friction). In a reaction stage, the total pressure drop is shared between stator (fixed) and rotor (moving) passages. The moving blades are shaped as expanding passages (nozzle-like), so static pressure decreases and relative velocity increases within the rotor, while the stator also accelerates the flow with its own pressure drop. Step-by-Step Solution: Recognize the definition of reaction: part of the enthalpy drop occurs in the rotor.Therefore, static pressure must decrease in the moving blades, not just in the fixed blades.Because the stator also expands the steam, pressure decreases in the fixed blades as well.Hence, pressure falls gradually and continuously across both fixed and moving blade rows. Verification / Alternative check: Parsons reaction stages (degree of reaction near 0.5) show matched velocity triangles with acceleration through both stator and rotor. Measured static pressure taps confirm a distributed pressure drop across both rows in such stages. Why Other Options Are Wrong: Fixed blades only: describes impulse behavior, not reaction.Moving blades only: ignores stator expansion, which is incorrect.None of these / nozzle throat only: contradicts reaction-stage fundamentals. Common Pitfalls: Confusing absolute velocity (stator frame) with relative velocity (rotor frame). In a reaction rotor, relative velocity tends to increase due to internal pressure drop even if absolute exit velocity is moderated by blade speed and turning. Final Answer: both fixed and moving blades

Question 4

Nature of expansion in reaction turbines As steam flows over the blades of a reaction turbine stage, which ideal thermodynamic process best represents the expansion within the blade passages?

Accepted Answer

Introduction / Context: When analyzing steam turbine stages, we often idealize the flow through fixed and moving blades as adiabatic and reversible to establish limits and to construct velocity triangles. This makes it crucial to know which thermodynamic process serves as the ideal reference for the expansion through blades. Given Data / Assumptions: Reaction turbine stage with pressure drop across both stator and rotor. Idealized analysis (no heat transfer, no friction), i.e., adiabatic and reversible flow. Steam properties referenced on an absolute temperature scale. Concept / Approach: The ideal flow through a nozzle or an expanding blade passage is taken as isentropic. Isentropic means adiabatic and reversible, resulting in constant entropy while static enthalpy drops and kinetic energy rises. Real blades incur losses (non-isentropic), but the isentropic model provides the benchmark for efficiency calculations (nozzle efficiency, stage efficiency, and turbine internal efficiency). Step-by-Step Solution: Model the blade passage as an adiabatic control volume.Assume negligible heat transfer and negligible external work other than flow work.For reversible flow, the entropy remains constant: the ideal expansion is isentropic. Verification / Alternative check: Nozzle and blade efficiency definitions compare actual enthalpy drop with the isentropic enthalpy drop between the same inlet and outlet pressures. This confirms the isentropic process is the intended idealization. Why Other Options Are Wrong: Isothermal: would require heat transfer to maintain constant temperature, not typical in fast turbine passages.Throttling: involves pressure drop at essentially constant enthalpy (not expansion work producing kinetic energy).Free expansion: irreversible, with no useful work and undefined path for turbine modeling. Common Pitfalls: Assuming “adiabatic” automatically means “isentropic.” Adiabatic plus reversible is required; in practice, friction makes real expansion non-isentropic, but the isentropic model remains the standard reference. Final Answer: isentropic process

Question 5

Pressure compounding (Rateau staging) in impulse turbines In pressure compounding of an impulse turbine, is the total pressure drop divided among multiple nozzle rings rather than occurring entirely in the first nozzle ring?

Accepted Answer

Introduction / Context: Impulse turbines can be compounded to manage speed and efficiency. Understanding the difference between velocity compounding (Curtis) and pressure compounding (Rateau) helps in matching turbine design to pressure ratio and power requirements. Given Data / Assumptions: Impulse turbine architecture (no pressure drop ideally in moving blades). Multiple stages of stationary nozzles and rotors. Goal: reduce jet velocity per stage and achieve practical rotor speed. Concept / Approach: In pressure compounding (Rateau), the total pressure drop from boiler to exhaust is split across several nozzle–rotor stages. Each nozzle ring takes only a portion of the pressure drop, creating a moderate jet velocity for its rotor, thereby reducing the required blade speed. This contrasts with Curtis (velocity) compounding, where most pressure drop occurs in a single nozzle set and multiple moving rows extract kinetic energy. Step-by-Step Solution: Define pressure compounding: divide total pressure drop among several nozzle rings.Each stage: modest nozzle pressure drop → manageable jet velocity.Rotor extracts work from that jet; repeat across stages. Verification / Alternative check: Stage-by-stage enthalpy–pressure diagrams and measured velocities show reduced jet speeds compared to a single-stage impulse design, validating the principle. Why Other Options Are Wrong: Curtis stages are velocity-compounded, not pressure-compounded.Impulse rotors ideally have negligible pressure drop; saying pressure drops only in moving blades is incorrect. Common Pitfalls: Confusing “compounding” types. Remember: Rateau = pressure compounding (many nozzle rings); Curtis = velocity compounding (many moving rows after one big nozzle drop). Final Answer: Agree

Question 6

Name of the process using bled steam to heat feedwater What is the process called when steam is extracted from intermediate turbine points during expansion, used to heat feedwater in feedwater heaters, and then the warmed feedwater is returned to th…

Accepted Answer

Introduction / Context: Large steam power plants improve cycle efficiency by raising the average temperature at which heat is added in the boiler. A standard method is to preheat the feedwater using steam extracted (bled) from the turbine. Understanding the terminology distinguishes component actions from the overall thermodynamic strategy. Given Data / Assumptions: Steam is extracted from turbine stages at controlled pressures. Feedwater heaters (open or closed) transfer heat from bled steam to feedwater. Heated feedwater returns to the boiler, reducing required boiler duty for a given cycle output. Concept / Approach: The overall process is called regenerative heating (or regeneration). The act of withdrawing steam is commonly termed bleeding. Reheating is different: it reheats partially expanded steam between turbine sections to reduce moisture and increase work, not to warm feedwater. Intercooling and desuperheating are also distinct processes used in other contexts (gas turbines, conditioning steam, etc.). Step-by-Step Solution: Identify purpose: raise feedwater temperature before boiler entry.Method: extract steam at intermediate pressures and transfer heat in feedwater heaters.Terminology: the cycle improvement is regenerative heating; steam withdrawal is bleeding. Verification / Alternative check: Rankine cycle T–s diagrams with multiple feedwater heaters show increased mean temperature of heat addition and improved thermal efficiency due to regeneration. Why Other Options Are Wrong: Reheating of steam: reheats main flow between turbine stages, not feedwater.Intercooling: used in multi-stage compressors, not in steam Rankine feedwater preheating.Desuperheating: reduces steam superheat, not the same as feedwater preheating. Common Pitfalls: Using “bleeding” as the name of the thermodynamic improvement. Bleeding is the action; the process goal and cycle strategy is regeneration (regenerative heating). Final Answer: regenerative heating

Question 7

Stage count comparison For the same required power output and inlet/exhaust conditions, an impulse turbine generally requires how many rows/stages of blades compared with a reaction turbine?

Accepted Answer

Introduction / Context: The number of stages in a steam turbine affects shaft speed, efficiency, cost, and length. Impulse and reaction staging distribute the same overall enthalpy drop differently between fixed and moving rows, which influences how many rows are required for a given duty. Given Data / Assumptions: Comparable inlet pressure/temperature and exhaust pressure for both turbine types. Similar design limits on blade speed and stage loading. Target: same power output. Concept / Approach: Impulse stages take nearly all pressure drop in fixed nozzles and mostly turn jets in the rotor; each stage extracts a limited work per unit pressure ratio without excessive exit losses. Reaction stages share pressure drop between stator and rotor, typically achieving higher stage loading for similar flow conditions. Hence, to absorb the same total enthalpy drop, an impulse machine usually needs more stages (more blade rows) than a reaction machine. Step-by-Step Solution: Set total enthalpy drop requirement from inlet to exhaust.Recognize impulse stages have smaller optimal work per stage.Conclude more impulse stages are needed than reaction stages for the same duty. Verification / Alternative check: Practical turbine lineups show many impulse stages for high pressure ratios, while reaction machines can reach similar duties with fewer rows at comparable speeds, confirming the trend. Why Other Options Are Wrong: Less/equal: contradicts typical stage-loading capabilities.“Cannot be compared”: broad trends are well established in design texts.“Zero”: impulse turbines still require stages to extract work. Common Pitfalls: Ignoring mechanical limits on blade speed and length; single-stage impulse devices (De Laval) run at very high speeds and cannot absorb large drops alone in practical large-power applications. Final Answer: more

Question 8

Critical pressure ratio comparison For flow of initially superheated steam through a nozzle, the critical pressure ratio (p2/p1 at choking) is how compared with the value for initially dry saturated steam?

Accepted Answer

Introduction / Context: The critical pressure ratio indicates the pressure ratio at which mass flow through a nozzle becomes choked. For steam, this ratio depends on the inlet state. Designers must know how it changes between saturated and superheated inlet conditions to size throats correctly and predict mass flow. Given Data / Assumptions: Isentropic (ideal) expansion to sonic velocity at the throat. Initially superheated versus initially dry saturated inlet states. Steam's effective isentropic exponent varies with state. Concept / Approach: Empirical/thermodynamic data show that the critical pressure ratio for initially dry saturated steam is about 0.577, whereas for initially superheated steam it is lower, commonly near 0.546 under typical conditions. Hence, for superheated inlet, choking occurs at a smaller downstream-to-upstream pressure ratio; equivalently, a larger pressure drop is needed to reach sonic conditions compared with the saturated case. Step-by-Step Solution: Recall benchmark values: ~0.577 (saturated), ~0.546 (superheated).Compare: 0.546 < 0.577.Conclude the critical ratio is less for superheated steam. Verification / Alternative check: Design charts and nozzle discharge tables corroborate the lower critical ratio for superheated conditions, reflecting different thermodynamic response (effective k) during expansion. Why Other Options Are Wrong: More/approximately equal: conflict with standard reference values.Twice or undefined: physically and dimensionally incorrect. Common Pitfalls: Applying the ideal-gas formula without adjusting for real-steam behavior and state dependence, which leads to inaccurate nozzle sizing. Final Answer: less

Question 9

Quantifying blade friction effect in impulse turbines By approximately what percentage does blade friction reduce the steam velocity as it passes over the moving blades of an impulse turbine?

Accepted Answer

Introduction / Context: In impulse turbine rotors, ideally there is no pressure drop; the rotor simply turns the jet and extracts work. In practice, surface roughness and viscous effects cause friction, reducing the relative velocity across the blade passage and thus decreasing the work recovered and the diagram efficiency. Given Data / Assumptions: Impulse-type moving blades with negligible pressure drop across the rotor in the ideal model. Friction present due to surface finish, incidence, and secondary flows. Typical well-designed stage, not severely eroded or fouled. Concept / Approach: Engineering practice often assumes a relative-velocity reduction factor across the rotor (sometimes denoted by k or a blade friction coefficient). For clean, well-designed blades, the relative velocity at exit is commonly taken as about 0.85 to 0.90 of the inlet relative velocity, implying a reduction on the order of 10 to 15 percent. This rule-of-thumb is used in preliminary design and performance calculations. Step-by-Step Solution: Let w1 be inlet relative velocity and w2 be exit relative velocity.Assume w2 ≈ 0.85 to 0.90 * w1 for typical blades.Percentage reduction ≈ 10% to 15%. Verification / Alternative check: Velocity-triangle based design examples and textbook problems frequently adopt 10–15% reduction, matching test data for smooth, modern impulse blades at design incidence. Why Other Options Are Wrong: Higher ranges (20–40%) imply unusually high losses inconsistent with good design and would severely depress diagram efficiency.Below 5% is overly optimistic and rarely achieved in practice. Common Pitfalls: Confusing relative velocity reduction (rotor passage) with absolute velocity changes across the stage; the former is the correct metric for blade friction in impulse rotors. Final Answer: 10 to 15%

Question 10

Axial discharge condition For a turbine designed for axial discharge at the stage exit, what is the value of the whirl (tangential) component of absolute velocity at outlet?

Accepted Answer

Introduction / Context: Velocity triangles in turbines decompose absolute velocity into axial (flow), whirl (tangential), and sometimes radial components. Power transfer depends on the change in whirl component across the rotor. Some stages are designed for axial discharge to simplify downstream flow and reduce exit swirl losses. Given Data / Assumptions: Axial discharge means the absolute exit velocity is aligned with the turbine axis. Steady flow and well-designed outlet guide geometry. No significant swirl-producing leakage or secondary flows at design point. Concept / Approach: If the absolute exit velocity vector is purely axial, its tangential (whirl) component is zero. This choice reduces kinetic energy loss in the exhaust (swirl loss) and simplifies matching to the next stage or diffuser. In the Euler turbine equation, the work per unit mass is proportional to the difference in whirl components (Vw1 − Vw2); setting Vw2 = 0 makes the work equal to Vw1 * u, a common design target for impulse stages. Step-by-Step Solution: Define axial discharge: exit flow direction parallel to axis.Therefore, tangential component at outlet Vw2 = 0.This minimizes exit swirl losses and eases stage matching. Verification / Alternative check: Measured exit swirl angles near zero in well-tuned stages confirm negligible whirl; residual small swirl may exist off-design but the design intent is Vw2 ≈ 0. Why Other Options Are Wrong: Maximum/minimum (non-zero) contradict the definition of axial discharge.Equal to blade speed is unrelated; blade speed is a rotor property, not a flow component at exit. Common Pitfalls: Confusing relative velocity (in the blade frame) with absolute exit velocity; axial discharge refers to the absolute velocity direction. Final Answer: zero

Question 11

Relative velocity in ideal impulse rotors In an impulse turbine neglecting friction, the relative velocity of steam at the exit tip of the moving blade is ________ the relative velocity at the inlet tip.

Accepted Answer

Introduction / Context: Impulse rotors ideally experience no pressure change across the moving blades; they simply turn the jet and extract work via change in whirl component. Understanding how relative velocity behaves in this ideal case is critical for constructing correct velocity triangles and estimating diagram efficiency limits. Given Data / Assumptions: Impulse rotor: negligible pressure drop across moving blades. No friction (idealization), so no viscous dissipation in the rotor passage. Steady flow and correct incidence at blade entry. Concept / Approach: With no pressure change and no friction, the magnitude of the relative velocity vector should remain constant through the rotor passage; only its direction changes due to blade curvature. Hence w1 (at inlet) equals w2 (at exit) in magnitude. Real blades show a reduction (e.g., 10–15%) due to friction, but the ideal model preserves the magnitude. Step-by-Step Solution: Adopt impulse rotor assumptions: Δp_rotor ≈ 0, friction ≈ 0.Without friction, there is no loss of mechanical energy in the relative frame.Therefore, |w2| = |w1|; direction changes, magnitude stays the same. Verification / Alternative check: Euler's turbine equation shows rotor work depends on whirl change in absolute velocity, not on a change in relative speed magnitude for the ideal impulse rotor. Why Other Options Are Wrong: Less/greater than: would imply frictional loss or energy input, contradicting the ideal assumption.Half or twice: arbitrary and inconsistent with energy conservation in the relative frame. Common Pitfalls: Mixing real and ideal behavior; in practice, friction lowers w2, but the question explicitly neglects friction. Final Answer: equal to

Question 12

Effect of nozzle friction on heat drop Due to friction between steam and nozzle surfaces, the useful (available) heat drop across an expanding nozzle is typically reduced by about how much?

Accepted Answer

Introduction / Context: Nozzle efficiency is a key parameter in turbine performance, translating enthalpy drop into kinetic energy. Surface friction and boundary-layer growth reduce the effective heat drop that converts to jet velocity, lowering the discharge speed and stage power. Given Data / Assumptions: Smooth, well-designed steam nozzles operating near design conditions. Losses dominated by wall friction and minor shock/expansion mismatches. Comparison with ideal isentropic expansion between the same inlet and outlet pressures. Concept / Approach: Nozzle (velocity) efficiency is typically around 0.85–0.90 for good designs. That means the actual kinetic energy at exit is 85–90% of the ideal value, implying an effective reduction in the usable heat drop of roughly 10–15%. Designers account for this when computing throat area and exit velocity from the isentropic enthalpy drop, then applying efficiency to obtain the real exit speed. Step-by-Step Solution: Let Δh_is be isentropic heat drop and V_is the corresponding ideal exit speed.With nozzle efficiency η_n ≈ 0.85–0.90, actual V ≈ √(η_n) * V_is.Effective heat drop utilized ≈ η_n * Δh_is → reduction ≈ 10–15%. Verification / Alternative check: Experimental nozzle tests commonly report efficiencies in the above range unless there is erosion, fouling, or severe off-design operation; these conditions could worsen losses but are not typical for design values. Why Other Options Are Wrong: Higher reductions (25–60%) indicate very poor nozzles or off-design shocks; not representative of standard practice.Less than 5% is overly optimistic and rarely sustained across operating ranges. Common Pitfalls: Confusing reduction in heat drop with reduction in velocity directly; because kinetic energy scales with velocity squared, be careful when converting between efficiency definitions and percentage speed shortfall. Final Answer: 10 to 15 percent

Question 13

Reaction turbines – interpreting degree of reaction When the degree of reaction (R) in a reaction steam turbine stage is zero, what does this imply about where the enthalpy (heat) drop occurs?

Accepted Answer

Introduction / Context: The degree of reaction is a key metric that distinguishes impulse from reaction behavior in turbine stages. It partitions the total stage enthalpy drop between fixed (stator) and moving (rotor) blades. Understanding extreme values of this ratio helps identify limiting cases such as pure impulse staging. Given Data / Assumptions: Degree of reaction R is defined on a single stage. Total stage heat (enthalpy) drop = drop in fixed blades + drop in moving blades. Steady, adiabatic flow is assumed for the ideal discussion. Concept / Approach: The degree of reaction is R = (enthalpy drop in moving blades) / (total enthalpy drop in the stage). When R = 0, the numerator is zero, meaning the rotor contributes no enthalpy drop. All the pressure/enthalpy drop then occurs in the fixed nozzles; the moving blades merely deflect the jet without further expansion. This is the textbook definition of a pure impulse stage (De Laval type). Step-by-Step Solution: Write R = Δh_rotor / (Δh_stator + Δh_rotor).Set R = 0 → Δh_rotor = 0.Therefore, the entire stage enthalpy drop occurs in the fixed (stator) blades.Conclusion: there is no heat (enthalpy) drop in the moving blades. Verification / Alternative check: Velocity triangles for a pure impulse stage show large nozzle exit velocity and predominantly constant static pressure through the rotor. Any kinetic-to-work conversion happens via momentum change, not further pressure drop in the moving row. Why Other Options Are Wrong: No heat drop in fixed blades: opposite of impulse behavior. Maximum heat drop in moving blades: corresponds to high reaction, not zero. Maximum heat drop in fixed blades: while true qualitatively, the precise implication of R = 0 is “no drop in moving blades,” which is the most accurate statement among the options. Common Pitfalls: Confusing momentum exchange (which still produces rotor work) with enthalpy drop. Zero reaction does not mean zero power; it means zero rotor enthalpy drop. Final Answer: no heat drop in moving blades

Question 14

Degree of reaction – precise definition for a turbine stage Select the correct expression for the degree of reaction (R) of a turbine stage in terms of enthalpy drops.

Accepted Answer

Introduction / Context: The degree of reaction quantifies how a stage divides its total enthalpy drop between rotor and stator. It informs blade design (e.g., Parsons 50% reaction) and affects efficiency, loading, and velocity triangles. Given Data / Assumptions: Total stage enthalpy drop Δh_stage = Δh_stator + Δh_rotor. All drops are defined per unit mass of working fluid. Steady, adiabatic flow assumed for ideal definition. Concept / Approach: By definition, the degree of reaction R is the fraction of the stage enthalpy drop that occurs in the rotor (moving blades): R = Δh_rotor / Δh_stage. With Δh_stage = Δh_stator + Δh_rotor, the denominator is the sum of drops across both blade rows. Step-by-Step Solution: State formula: R = Δh_rotor / (Δh_stator + Δh_rotor).Match to options: “heat drop in moving blades to the heat drop in fixed blades plus heat drop in moving blades.”Therefore, option (c) is the correct definition. Verification / Alternative check: Special cases: R = 0 → pure impulse; R = 0.5 → Parsons reaction stage; 0 < R < 1 → mixed impulse-reaction behavior with pressure drop in both rows. Why Other Options Are Wrong: (a) and (b) are ratios between blade-row drops, not normalized by the stage total. (d) is the reciprocal of the desired form. (e) omits the moving-blade drop and is not the standard definition. Common Pitfalls: Using energy (power) ratios instead of enthalpy-drop ratios; always use per-kg enthalpy drops for R. Final Answer: heat drop in moving blades to the heat drop in fixed blades plus heat drop in moving blades

Question 15

De Laval (impulse) turbine – maximum diagram efficiency For a single-stage De Laval impulse turbine with nozzle angle α (at rotor inlet), the maximum possible diagram (blading) efficiency is:

Accepted Answer

Introduction / Context: In a simple impulse stage (De Laval turbine), the fixed nozzles convert all pressure drop into a high-velocity jet. The rotor extracts work by turning this jet. For given inlet flow angle α (from the tangent), there exists an optimum blade-speed ratio that maximizes the diagram (blading) efficiency. Given Data / Assumptions: Impulse stage (no pressure drop in rotor). Symmetrical or suitably chosen blade angles; negligible losses for the ideal result. Velocity coefficient taken as unity for the theoretical maximum. Concept / Approach: The diagram efficiency η_d is the ratio of work obtained on the blades to the kinetic energy supplied at rotor inlet. Analysis of velocity triangles yields an expression for η_d as a function of the blade-speed ratio and α. Optimizing with respect to blade speed gives the well-known ideal limit η_d,max = cos^2 α. Step-by-Step Solution: Start from inlet velocity triangle: absolute jet makes angle α with wheel tangent.Express work per kg as product of blade speed and change in whirl component of velocity.Form η_d = (blade work per kg) / (V_inlet^2 / 2).Differentiate with respect to blade-speed ratio and set derivative to zero → optimum condition leads to η_d,max = cos^2 α. Verification / Alternative check: For α = 20°, cos^2 α ≈ 0.883; for α = 30°, cos^2 α = 0.75, matching classic charts for impulse stage efficiency ceilings. Why Other Options Are Wrong: sin^2 α, tan^2 α, cot^2 α, sec^2 α do not arise from the optimization; they misrepresent the trigonometric dependence. Common Pitfalls: Confusing α (nozzle angle) with blade inlet/exit angles; the maximum efficiency result is sensitive to which angle is used. Final Answer: cos^2 α

Question 16

Steam turbine types – power and process heat together Identify the turbine in which part of the steam, after partial expansion, is extracted for process heating while the remaining steam expands further for power generation.

Accepted Answer

Introduction / Context: Combined heat and power (CHP) applications use turbine configurations that supply both shaft power and process steam. Understanding the naming and function of these turbines is essential in plant design and energy audits. Given Data / Assumptions: A controlled extraction (pass-out) point exists at an intermediate pressure. Remaining steam continues through later stages for additional power. Process requires steady steam flow for heating or industrial use. Concept / Approach: A pass-out (also called extraction-condensing) turbine deliberately bleeds steam at a set intermediate pressure for process use. The rest of the flow is expanded further to the condenser pressure to generate power efficiently. This differs from a back-pressure turbine where all exhaust steam leaves at process pressure and no condenser is used. Step-by-Step Solution: Recognize the phrase “part used for process heating, remainder for power” → controlled extraction.Map terminology: pass-out/extraction turbine provides this function.Therefore, select “pass-out (extraction) turbine.” Verification / Alternative check: Plant heat-balance diagrams for CHP units show extraction lines to process headers and continued expansion lines to the condenser, characteristic of extraction turbines. Why Other Options Are Wrong: Back-pressure turbine exhausts all steam for process at high back pressure; it does not continue expansion to a condenser. Low-pressure and impulse are generic descriptors without the extraction feature. Reheat-only refers to reheating between turbine sections, not extraction for process heat. Common Pitfalls: Confusing “bleed” with uncontrolled leakage; pass-out is a controlled, valved extraction at a defined pressure. Final Answer: pass-out (extraction) turbine

Question 17

Critical pressure ratio for nozzles – proper expression Choose the correct expression for the critical pressure ratio in a steam nozzle, given inlet pressure p1 and throat (critical) pressure p2.

Accepted Answer

Introduction / Context: The critical pressure ratio characterizes the onset of choking in compressible nozzle flow. It compares the throat static pressure at sonic conditions to the inlet (stagnation/total or chamber) pressure and determines whether further reduction in back pressure will increase mass flow. Given Data / Assumptions: p1 is the inlet (upstream) total pressure used for the ratio. p2 is the static pressure at the throat when Mach = 1. Ideal isentropic assumption for the critical condition. Concept / Approach: By convention, the critical pressure ratio is defined as the ratio of throat pressure to inlet pressure, i.e., p2/p1. For a given working fluid, this ratio has a specific value that triggers choked flow. For perfect gases it depends on the specific heat ratio; for steam, steam-table relations or empirical coefficients are used. Step-by-Step Solution: Identify the two pressures: p1 (inlet), p2 (throat at sonic conditions).Form the ratio as throat over inlet → p2/p1.This ratio is less than 1 and marks the limit beyond which mass flow becomes insensitive to back pressure. Verification / Alternative check: Performance curves display a plateau in mass flow once p_back/p1 falls below p2/p1. For air with γ = 1.4, the ideal ratio is around 0.528; steam values differ but the definition remains p2/p1. Why Other Options Are Wrong: p1/p2 is the reciprocal of the standard definition. p1 p2 and p1 + p2 have incorrect dimensions or meaning. p1 − p2 indicates a pressure drop, not a ratio. Common Pitfalls: Mixing up back pressure with critical throat pressure; the critical ratio specifically uses throat quantities at Mach 1. Final Answer: p2/p1

Question 18

Effect of nozzle friction on steam quality (dryness fraction) How does internal friction within a steam nozzle influence the dryness fraction (quality) of wet steam at the exit?

Accepted Answer

Introduction / Context: When wet steam expands through a nozzle, the thermodynamic path may lead to further condensation or partial re-evaporation depending on losses and non-equilibrium effects. Friction alters the available enthalpy drop and the entropy rise, which impacts the exit quality. Given Data / Assumptions: Wet steam enters; exit is within the wet region for the comparison. Adiabatic nozzle with internal irreversibility (friction). Same inlet state and back pressure for the scenarios compared. Concept / Approach: In an ideal (isentropic) nozzle, the enthalpy drop converts to kinetic energy, with a certain quality at exit. Friction increases entropy, reducing the portion of enthalpy drop available to accelerate the steam. The actual exit state shifts to a higher entropy, lower quality (more moisture) condition than the isentropic case, so the dryness fraction decreases. Step-by-Step Solution: Reference isentropic exit: s2s = s1; determine x2s from tables (conceptually).Include friction: s2 > s1; at similar pressure, higher s in the wet region implies lower x (more moisture).Therefore, nozzle friction decreases the exit dryness fraction. Verification / Alternative check: Nozzle efficiency η_v = (actual KE gain) / (isentropic KE gain) < 1. Mollier (h–s) diagrams show the actual path deviating to the right (higher entropy), intersecting lower x lines at the same exit pressure. Why Other Options Are Wrong: No effect: contradicts entropy increase due to friction. Increases: would require entropy reduction or heat input, not present here. Non-monotonic change is not characteristic for steady adiabatic nozzle friction. Common Pitfalls: Confusing quality trends in reheating or superheating devices with nozzles; here, friction raises moisture content. Final Answer: decreases

Question 19

Reaction turbine rotor behavior – pressure and velocity trends As steam flows through the moving (rotor) blades of a reaction turbine stage, how do pressure and velocity change?

Accepted Answer

Introduction / Context: Unlike a pure impulse stage, a reaction stage experiences expansion (pressure drop) in both stator and rotor passages. The rotor thus acts as a nozzle, accelerating the fluid while extracting work through momentum change. Given Data / Assumptions: Finite degree of reaction R > 0. Well-designed blade passages with smooth area variation. Steady, adiabatic flow for conceptual analysis. Concept / Approach: In the moving blades of a reaction stage, part of the enthalpy drop occurs, resulting in a fall in static pressure and a corresponding increase in relative (and typically absolute) velocity. This converts pressure energy into kinetic energy within the rotor, contributing to torque via change in whirl velocity. Step-by-Step Solution: Define reaction: portion of total stage enthalpy drop in rotor.Pressure drop in rotor → fluid accelerates through blade passages.Hence, pressure decreases while velocity increases through the moving blades. Verification / Alternative check: Velocity triangles for 50% reaction (Parsons) show acceleration in both stator and rotor; measured static taps confirm pressure reduction across both rows. Why Other Options Are Wrong: (a) Reverse of nozzle-like behavior in rotor. (c) and (d) contradict energy conversion trends. (e) Constant pressure would imply impulse behavior, not reaction. Common Pitfalls: Assuming the rotor only turns the jet; that is valid for ideal impulse stages, not for reaction stages. Final Answer: pressure decreases while velocity increases

Question 20

Flow characteristics in turbines – typical speed levels In steam or gas turbines operating under design conditions, the working fluid undergoes continuous steady flow with a speed of flow that is generally:

Accepted Answer

Introduction / Context: Turbines are energy-conversion devices that accelerate fluids through nozzles and blade passages to extract mechanical work. Recognizing the typical magnitude of flow speeds helps set expectations for Reynolds number, Mach number, and loss mechanisms. Given Data / Assumptions: Well-designed axial or radial turbines with correctly matched pressures. Continuous steady flow, not intermittent like in reciprocating machines. Compressible effects relevant in many stages (especially gas turbines and high-pressure steam). Concept / Approach: Turbines convert pressure/enthalpy drop into kinetic energy and then into rotor work. Consequently, blade-relative and absolute velocities are generally high (tens to hundreds of metres per second in steam turbines; even higher in gas turbines). “Very high” might suggest extreme supersonic conditions everywhere, which is not generally true across entire stages, while “low” or “very low” contradicts the acceleration purpose of nozzling. Step-by-Step Solution: Recognize nozzle role → accelerates fluid to large velocities.Rotor extracts work via change in whirl component → still requires high-through-flow speeds.Therefore, typical turbine flow speeds are high. Verification / Alternative check: Design data show nozzle exit velocities often 150–400 m/s in steam stages and 300–600 m/s (or higher locally) in gas turbines, confirming “high.” Why Other Options Are Wrong: “Very low/low” mischaracterize turbine operation. “Very high” suggests an overgeneralization; not all passages are supersonic or at extreme speeds. “Approximately static” is incorrect for flowing turbomachinery. Common Pitfalls: Equating “high” with “always supersonic”; many efficient stages remain subsonic or transonic outside special nozzles. Final Answer: high

Steam Nozzles and Turbines Questions