Q: Force of a jet on an inclined flat plate (stationary) Ratio of the normal (perpendicular-to-plate) force when the plate is at 30° to the jet, to the force when the plate is normal to the jet, equals:

Introduction / Context: When a water jet strikes a stationary flat plate, the force depends on how much of the jet's momentum is stopped in the direction normal to the plate surface. Tilting the plate reduces the effective normal component of velocity, hence reducing the normal force compared to the case when the plate is perpendicular to the jet. Given Data / Assumptions: Jet of uniform velocity V and area a strikes a stationary, smooth, flat plate. The plate is oriented at θ = 30° to the jet direction (0° = parallel, 90° = normal). Flow follows the plate with negligible splashing loss for the basic estimate. Concept / Approach: The normal component of velocity is V_n = V * sin θ. Only this component is brought to rest normal to the plate, so the normal momentum flux is ρ * a * V_n^2. Thus, normal force F_n ∝ ρ * a * V^2 * sin^2 θ. For θ = 90° (plate normal to jet), sin θ = 1 and F_n(normal) ∝ ρ * a * V^2. Step-by-Step Solution: For θ = 30°: V_n = V * sin 30° = V * 1/2.Normal momentum flux ∝ V_n^2 = (V/2)^2 = V^2 / 4.Force ratio = F_n(30°) / F_n(90°) = sin^2 30° / sin^2 90° = (1/2)^2 / 1^2 = 1/4.Hence the normal force at 30° is one-quarter of that at 90°. Verification / Alternative check: Dimensional reasoning and momentum balance lead to the same sin^2 θ dependence. At θ → 0°, ratio → 0 (grazing flow). At θ = 90°, ratio = 1, matching limits. Why Other Options Are Wrong: 1/2 and 3/4 underestimate the reduction; 1 implies no change with angle; 2 is impossible because tilting cannot increase the normal force beyond the 90° case. Common Pitfalls: Confusing force along the jet direction with force normal to the plate; mixing sin θ and cos θ depending on chosen angle definition. Final Answer: 1/4

Q: Speed effect on reaction turbine efficiency at a given head For a reaction turbine operating at a fixed head, efficiency generally ______ as speed increases (starting from low speed) until an optimum is reached.

Introduction / Context: Reaction turbines (Francis, Kaplan) exhibit characteristic efficiency curves versus speed (or speed ratio) for a given head. At very low or very high speeds relative to the design value, incidence and flow separation increase losses, while near an optimum speed the velocity triangles align best and hydraulic losses are minimized. Given Data / Assumptions: Fixed net head; guide vane and blade settings varied to follow a typical operating curve. Normal range of unit speed without extreme cavitation or choking. Concept / Approach: Efficiency depends on matching blade peripheral speed with inlet whirl/flow components. From low speed, increasing speed improves incidence and reduces recirculation, thereby increasing efficiency up to a peak. Beyond the best-efficiency point, further speed increase worsens alignment and increases losses, but the initial trend is upward. Step-by-Step Solution: Start at low speed: misaligned triangles ⇒ high losses ⇒ low efficiency.Increase speed toward design: improved alignment ⇒ hydraulic efficiency rises.At the best-efficiency speed, losses are minimized ⇒ peak efficiency.Further speed increase causes off-design incidence and potential cavitation ⇒ efficiency declines. Verification / Alternative check: Manufacturer performance maps show a “hill chart” with a ridge of high efficiency near the design speed and discharge; moving from low speed toward this ridge increases efficiency. Why Other Options Are Wrong: “Decreases” ignores the initial improving trend; “constant/zero/independent” contradict real turbine maps. Common Pitfalls: Assuming efficiency monotonically increases at all speeds; neglecting cavitation limits and guide vane constraints at the extremes. Final Answer: increases

Q: Pelton Wheel — Double Overhung Configuration In Pelton turbine layouts, what does the term “double overhung” specifically indicate about the number of runners on the shaft?

Introduction: Pelton wheels are impulse turbines used for high heads and relatively low discharges. Plant layouts are often described as single overhung or double overhung depending on how runners are mounted on the shaft. The question tests recognition of this practical configuration terminology. Given Data / Assumptions: Impulse turbine (Pelton) with one shaft supported by bearings. “Overhung” means mounted outside the bearing span. “Double overhung” implies symmetric mounting at both shaft ends. Concept / Approach: In a single overhung Pelton arrangement, one runner is mounted at one end of the shaft. In a double overhung arrangement, there are two runners—one mounted at each end—so that multiple jets (often two per runner in compact plants) can be used while keeping bearing loads balanced and overall width manageable. Step-by-Step Solution: Interpret the term: “double” refers to two overhung mountings.Each overhung position carries a runner.Therefore the configuration has two runners on the same shaft. Verification / Alternative check: Installation drawings and manufacturer catalogs show double overhung Pelton sets with a runner at each shaft end. Why Other Options Are Wrong: two jets: jet count is independent of the “overhung” term.four jets: possible in some plants, but the term denotes runners, not jet number.four runners: uncommon and not implied by the nomenclature. Common Pitfalls: Confusing number of jets with number of runners; jets per runner vary by design. Final Answer: two runners

Q: Centrifugal Pump — Mechanical Efficiency Definition Mechanical efficiency of a centrifugal pump is defined as which ratio?

Introduction: Pump efficiencies are categorized to isolate different loss mechanisms. Mechanical efficiency focuses on losses between the motor shaft and the impeller (bearings, seals, disc friction), distinct from hydraulic and volumetric effects inside the flow passages. Given Data / Assumptions: Shaft power is supplied by a prime mover. Impeller power is the power actually imparted to the fluid by the impeller. Losses between shaft and impeller are mechanical in nature. Concept / Approach: By definition, mechanical efficiency eta_m = (power at impeller) / (power at shaft). It quantifies how effectively the mechanical drive delivers power to the impeller after subtracting mechanical losses. Overall efficiency multiplies mechanical and hydraulic efficiencies (and volumetric, if defined separately). Step-by-Step Solution: Identify numerator: impeller (available) power.Identify denominator: shaft (input) power from the prime mover.Form ratio: eta_m = P_impeller / P_shaft. Verification / Alternative check: Test data typically show eta_m > eta_h (hydraulic) for well-built pumps; overall eta_o = eta_m * eta_h * eta_v. Why Other Options Are Wrong: actual workdone / input: that is overall efficiency (if “actual workdone” is water output power).energy supplied to pump / energy at impeller: reciprocal of mechanical efficiency.manometric head / impeller energy per kN: relates to manometric (hydraulic) efficiency, not mechanical. Common Pitfalls: Mixing hydraulic efficiency with mechanical efficiency; they capture different losses. Final Answer: energy available at the impeller to the energy supplied to the pump by the prime mover

Question 1

Definition of hydraulic efficiency (impulse turbines) Hydraulic efficiency of an impulse turbine is the:

Accepted Answer

Introduction / Context: Overall efficiency in a turbine is the product of hydraulic, mechanical, and volumetric efficiencies. For impulse turbines, hydraulic efficiency isolates how effectively the runner converts the jet’s kinetic energy into useful work on the wheel before mechanical losses are considered. Given Data / Assumptions: Impulse action only (pressure at runner near atmospheric). Jet kinetic energy at inlet is the available hydraulic energy to the runner. Neglect mechanical and generator losses for the hydraulic efficiency definition. Concept / Approach: Hydraulic efficiency, η_h, is defined as (work done on the runner) / (kinetic energy of the jet at inlet). The numerator comes from momentum change (whirl component) across the runner; the denominator represents the energy content of the incident jet. Step-by-Step Solution: Energy available to runner = (1/2) * ρ * Q * V^2.Work done = torque * angular velocity = mass flow * change in whirl velocity * u.η_h = Work done / Jet KE at inlet.This definition cleanly separates hydraulic conversion from mechanical losses. Verification / Alternative check: Textbook derivations for Pelton wheels lead to η_h expressions dependent on speed ratio and bucket losses, all referenced to the jet’s inlet KE. Why Other Options Are Wrong: Option (a) mixes hydraulic with overall efficiency. Option (b) is close in spirit but ambiguous; the precise definition uses jet KE explicitly. Option (e) is meaningless for efficiency definition; (d) is incorrect because (c) is correct. Common Pitfalls: Confusing hydraulic with overall or mechanical efficiencies; using head instead of jet KE without clarifying the reference. Final Answer: ratio of work done on the wheel to the kinetic energy of the jet at inlet

Question 2

Identify the incorrect statement about turbines and installation Which of the following statements is wrong?

Accepted Answer

Introduction / Context: Turbines are broadly categorized into impulse and reaction types. Installation details (e.g., use and geometry of the draft tube) and suitable head ranges differ between types. Correct classification is crucial for design and operation. Given Data / Assumptions: Impulse turbines (e.g., Pelton) extract energy from jet momentum; reaction turbines (e.g., Francis, Kaplan) operate with pressure change across the runner. Draft tubes are associated with reaction turbines to recover kinetic head. Concept / Approach: Francis is the quintessential reaction turbine. Reaction turbines typically serve low to medium heads with higher discharges, while Pelton serves high heads with lower discharge. Draft tube diffuser angles are limited (≈5–8° half-angle) to maintain attached flow and good pressure recovery. Step-by-Step Solution: Classify Francis: reaction (not impulse).Head/discharge suitability: reaction ⇒ low/medium head, high flow (statement a is acceptable).Draft tube taper limitation: <≈ 8° is standard practice to avoid separation (statement b acceptable).Impulse installation: often above tail water; no draft tube required (statement c acceptable).Therefore statement d is the incorrect one. Verification / Alternative check: Standard classifications and manufacturer data sheets confirm Francis/Kaplan as reaction machines. Why Other Options Are Wrong: Options a, b, c, and e correctly describe widely accepted practice. Common Pitfalls: Assuming “reaction” means only Kaplan; forgetting that Francis spans a broad head range and is still reaction. Final Answer: A Francis turbine is an impulse turbine.

Question 3

Force of a jet on an inclined flat plate (stationary) Ratio of the normal (perpendicular-to-plate) force when the plate is at 30° to the jet, to the force when the plate is normal to the jet, equals:

Accepted Answer

Introduction / Context: When a water jet strikes a stationary flat plate, the force depends on how much of the jet's momentum is stopped in the direction normal to the plate surface. Tilting the plate reduces the effective normal component of velocity, hence reducing the normal force compared to the case when the plate is perpendicular to the jet. Given Data / Assumptions: Jet of uniform velocity V and area a strikes a stationary, smooth, flat plate. The plate is oriented at θ = 30° to the jet direction (0° = parallel, 90° = normal). Flow follows the plate with negligible splashing loss for the basic estimate. Concept / Approach: The normal component of velocity is V_n = V * sin θ. Only this component is brought to rest normal to the plate, so the normal momentum flux is ρ * a * V_n^2. Thus, normal force F_n ∝ ρ * a * V^2 * sin^2 θ. For θ = 90° (plate normal to jet), sin θ = 1 and F_n(normal) ∝ ρ * a * V^2. Step-by-Step Solution: For θ = 30°: V_n = V * sin 30° = V * 1/2.Normal momentum flux ∝ V_n^2 = (V/2)^2 = V^2 / 4.Force ratio = F_n(30°) / F_n(90°) = sin^2 30° / sin^2 90° = (1/2)^2 / 1^2 = 1/4.Hence the normal force at 30° is one-quarter of that at 90°. Verification / Alternative check: Dimensional reasoning and momentum balance lead to the same sin^2 θ dependence. At θ → 0°, ratio → 0 (grazing flow). At θ = 90°, ratio = 1, matching limits. Why Other Options Are Wrong: 1/2 and 3/4 underestimate the reduction; 1 implies no change with angle; 2 is impossible because tilting cannot increase the normal force beyond the 90° case. Common Pitfalls: Confusing force along the jet direction with force normal to the plate; mixing sin θ and cos θ depending on chosen angle definition. Final Answer: 1/4

Question 4

Speed effect on reaction turbine efficiency at a given head For a reaction turbine operating at a fixed head, efficiency generally ______ as speed increases (starting from low speed) until an optimum is reached.

Accepted Answer

Introduction / Context: Reaction turbines (Francis, Kaplan) exhibit characteristic efficiency curves versus speed (or speed ratio) for a given head. At very low or very high speeds relative to the design value, incidence and flow separation increase losses, while near an optimum speed the velocity triangles align best and hydraulic losses are minimized. Given Data / Assumptions: Fixed net head; guide vane and blade settings varied to follow a typical operating curve. Normal range of unit speed without extreme cavitation or choking. Concept / Approach: Efficiency depends on matching blade peripheral speed with inlet whirl/flow components. From low speed, increasing speed improves incidence and reduces recirculation, thereby increasing efficiency up to a peak. Beyond the best-efficiency point, further speed increase worsens alignment and increases losses, but the initial trend is upward. Step-by-Step Solution: Start at low speed: misaligned triangles ⇒ high losses ⇒ low efficiency.Increase speed toward design: improved alignment ⇒ hydraulic efficiency rises.At the best-efficiency speed, losses are minimized ⇒ peak efficiency.Further speed increase causes off-design incidence and potential cavitation ⇒ efficiency declines. Verification / Alternative check: Manufacturer performance maps show a “hill chart” with a ridge of high efficiency near the design speed and discharge; moving from low speed toward this ridge increases efficiency. Why Other Options Are Wrong: “Decreases” ignores the initial improving trend; “constant/zero/independent” contradict real turbine maps. Common Pitfalls: Assuming efficiency monotonically increases at all speeds; neglecting cavitation limits and guide vane constraints at the extremes. Final Answer: increases

Question 5

Pumps for deep-well water lifting Which pump is commonly and successfully used for lifting water from deep wells?

Accepted Answer

Introduction / Context: Deep wells pose suction head and priming challenges. Traditional surface-mounted centrifugal pumps cannot lift water much beyond the atmospheric pressure limit. Alternatives place the pumping action down-hole or use methods that avoid long suction lifts. Two proven approaches are air-lift and jet (ejector) systems, each with particular advantages. Given Data / Assumptions: Well depth exceeds practical suction lift of surface centrifugal pumps. Desire for simple, robust operation with sandy or gas-laden water possible. Concept / Approach: An air lift pump injects compressed air into the riser pipe below the water level, reducing the mixture density and causing the aerated column to rise. There is no moving part down-hole, making it reliable for deep, crooked, or sandy wells. Although efficiency is modest, it is “successfully used” for deep lifts where other pumps face priming or wear issues. Step-by-Step Solution: Supply compressed air to an injector at depth.Air-water mixture density decreases in the riser.Buoyancy and pressure differential lift the mixture to the surface.Separate air and water at discharge; recirculate air if desired. Verification / Alternative check: Field practice and textbooks list air-lift pumps as suitable for deep wells and abrasive conditions. Jet pumps (ejectors) are also used, but the question emphasizes “successfully used,” a phrasing commonly associated with air lift in exam literature. Why Other Options Are Wrong: Surface centrifugal pumps struggle with deep suction. Reciprocating pumps can lift from depth but require long rods/valves and are maintenance-heavy. Jet pumps are viable but often paired with specific sets; many syllabi credit air lift as the standard deep-well answer. Gear pumps are for viscous fluids in process plants. Common Pitfalls: Assuming the atmospheric suction limit can be bypassed without special arrangements; ignoring abrasive wear in down-hole mechanical pumps. Final Answer: Air lift pump

Question 6

Pelton Wheel — Double Overhung Configuration In Pelton turbine layouts, what does the term “double overhung” specifically indicate about the number of runners on the shaft?

Accepted Answer

Introduction: Pelton wheels are impulse turbines used for high heads and relatively low discharges. Plant layouts are often described as single overhung or double overhung depending on how runners are mounted on the shaft. The question tests recognition of this practical configuration terminology. Given Data / Assumptions: Impulse turbine (Pelton) with one shaft supported by bearings. “Overhung” means mounted outside the bearing span. “Double overhung” implies symmetric mounting at both shaft ends. Concept / Approach: In a single overhung Pelton arrangement, one runner is mounted at one end of the shaft. In a double overhung arrangement, there are two runners—one mounted at each end—so that multiple jets (often two per runner in compact plants) can be used while keeping bearing loads balanced and overall width manageable. Step-by-Step Solution: Interpret the term: “double” refers to two overhung mountings.Each overhung position carries a runner.Therefore the configuration has two runners on the same shaft. Verification / Alternative check: Installation drawings and manufacturer catalogs show double overhung Pelton sets with a runner at each shaft end. Why Other Options Are Wrong: two jets: jet count is independent of the “overhung” term.four jets: possible in some plants, but the term denotes runners, not jet number.four runners: uncommon and not implied by the nomenclature. Common Pitfalls: Confusing number of jets with number of runners; jets per runner vary by design. Final Answer: two runners

Question 7

Centrifugal Pump — Mechanical Efficiency Definition Mechanical efficiency of a centrifugal pump is defined as which ratio?

Accepted Answer

Introduction: Pump efficiencies are categorized to isolate different loss mechanisms. Mechanical efficiency focuses on losses between the motor shaft and the impeller (bearings, seals, disc friction), distinct from hydraulic and volumetric effects inside the flow passages. Given Data / Assumptions: Shaft power is supplied by a prime mover. Impeller power is the power actually imparted to the fluid by the impeller. Losses between shaft and impeller are mechanical in nature. Concept / Approach: By definition, mechanical efficiency eta_m = (power at impeller) / (power at shaft). It quantifies how effectively the mechanical drive delivers power to the impeller after subtracting mechanical losses. Overall efficiency multiplies mechanical and hydraulic efficiencies (and volumetric, if defined separately). Step-by-Step Solution: Identify numerator: impeller (available) power.Identify denominator: shaft (input) power from the prime mover.Form ratio: eta_m = P_impeller / P_shaft. Verification / Alternative check: Test data typically show eta_m > eta_h (hydraulic) for well-built pumps; overall eta_o = eta_m * eta_h * eta_v. Why Other Options Are Wrong: actual workdone / input: that is overall efficiency (if “actual workdone” is water output power).energy supplied to pump / energy at impeller: reciprocal of mechanical efficiency.manometric head / impeller energy per kN: relates to manometric (hydraulic) efficiency, not mechanical. Common Pitfalls: Mixing hydraulic efficiency with mechanical efficiency; they capture different losses. Final Answer: energy available at the impeller to the energy supplied to the pump by the prime mover

Question 8

Centrifugal Pump — Specific Speed (Ns) Which formula correctly gives the specific speed of a centrifugal pump in terms of rotational speed N (rpm), discharge Q, and head H (m)?

Accepted Answer

Introduction: Specific speed groups pumps of different sizes by hydraulic similarity. It indicates the impeller style best suited to a duty point, varying from radial (low Ns) to axial (high Ns) designs. Given Data / Assumptions: N in rpm, Q in m^3/s (or consistent units), H in m. Geometric similarity and similar efficiency near the best efficiency point. Concept / Approach: For pumps, specific speed is based on discharge, not power. From affinity/similarity, the widely used empirical form is Ns = N * Q^0.5 / H^0.75 (up to a constant depending on units). This correlates duty with impeller type: low Ns for radial, medium for mixed flow, high for axial-flow pumps. Step-by-Step Solution: Start with model law: Q ∝ N * D^3 and H ∝ N^2 * D^2.Eliminate D to obtain dimensionless grouping linking N, Q, H.Arrive at Ns ∝ N * Q^0.5 / H^0.75 (rpm-based form). Verification / Alternative check: Catalog charts mapping Ns to impeller geometry confirm the correlation trends. Why Other Options Are Wrong: N * Q / H: wrong exponents; not similarity-consistent.N * sqrt(P) / H^(5/4): that is a turbine-specific-speed form using power.N / H^(3/2): omits discharge term entirely. Common Pitfalls: Confusing pump Ns (uses Q) with turbine Ns (uses P). Final Answer: Ns = N * sqrt(Q) / H^(3/4)

Question 9

Turbine Selection by Specific Speed Which turbine type is generally preferred for specific speeds in the range of about 60 to 300 rpm-based units?

Accepted Answer

Introduction: Specific speed (Ns) guides the selection of turbine type to match site head and discharge. Lower Ns favors impulse turbines; higher Ns favors axial-flow reaction turbines, with mixed-flow machines occupying the middle range. Given Data / Assumptions: Ns range stated: roughly 60–300 (rpm-based form). Typical classification bands used in practice. Concept / Approach: Pelton turbines (impulse) are suitable for very low Ns (high head, low flow). Kaplan turbines (axial) suit very high Ns (low head, high flow). Francis turbines, being mixed-flow reaction machines, cover the middle Ns band from about 60 up to ~300, making them the preferred choice in that interval. Step-by-Step Solution: Map Ns range to turbine families: Pelton <~35; Francis ~60–300; Kaplan >~300 (indicative ranges).The given Ns range lies squarely in the Francis region.Hence, select Francis turbine. Verification / Alternative check: Reference charts of Ns vs. turbine type show Francis in the mid-range band. Why Other Options Are Wrong: Pelton wheel: typically lower Ns (high-head applications).Kaplan turbine: typically higher Ns (very low head, large discharge).none of these: incorrect because Francis matches the specified band. Common Pitfalls: Confusing rpm-based Ns with dimensionless forms; the relative ordering remains the same. Final Answer: Francis turbine

Question 10

Turbine Similarity — Discharge vs Head Under dynamically similar operation, how does turbine discharge Q vary with head H?

Accepted Answer

Introduction: Similarity relations (or model laws) connect discharge, head, and speed for turbines and pumps. Recognizing these proportionalities is essential in model testing and scaling from model to prototype. Given Data / Assumptions: Geometric similarity of machines. Similar operating point (near best efficiency) and negligible change in efficiency. Same fluid properties. Concept / Approach: Characteristic velocity scales with the square root of head: V ∝ sqrt(2 * g * H). For a given scaled geometry, area scales with length squared, so discharge Q ∝ Area * Velocity ∝ L^2 * sqrt(H). Eliminating length under similarity leads to Q ∝ H^1/2 for a particular machine family at similar points. Step-by-Step Solution: Use V ∝ sqrt(H) at similar operation.Q = A * V ⇒ Q ∝ L^2 * sqrt(H).For a fixed machine, L is constant; hence Q ∝ H^1/2. Verification / Alternative check: Empirical performance curves plotted in unit variables show Qu = Q / H^0.5 roughly constant near best efficiency. Why Other Options Are Wrong: inversely proportional to H^1/2: opposite of similarity prediction.directly/inversely proportional to H^3/2: wrong power; Q scales with sqrt(H), not H^1.5. Common Pitfalls: Confusing turbine scaling (Q ∝ H^0.5) with power scaling (P ∝ H^1.5). Final Answer: directly proportional to H^1/2

Question 11

Hydraulic Press — Efficiency Expression For a hydraulic press with load W on the ram (area A) and effort P on the plunger (area a), what is the expression for efficiency?

Accepted Answer

Introduction: A hydraulic press multiplies force via Pascal’s principle. Efficiency compares the actual mechanical advantage to the theoretical (ideal) mechanical advantage and accounts for losses such as seal friction and minor leakage. Given Data / Assumptions: Ram area A, plunger area a. Load W applied by the ram; effort P applied to the plunger. Quasi-static operation; small losses represented in efficiency. Concept / Approach: Ideal mechanical advantage IMA = A / a (from equal pressure in connected cylinders). Actual mechanical advantage AMA = W / P. Efficiency eta = AMA / IMA = (W / P) / (A / a) = (W * a) / (P * A). Step-by-Step Solution: Compute AMA: W / P.Compute IMA: A / a.Form efficiency: eta = (W / P) / (A / a) = (W * a) / (P * A). Verification / Alternative check: For an ideal, lossless press, W / P = A / a, so eta = 1 as predicted by the formula. Why Other Options Are Wrong: (W * A)/(P * a) and (P * A)/(W * a): reciprocals or rearrangements that do not equal AMA/IMA.(P * a)/(W * A): would give eta < 1 even in ideal conditions. Common Pitfalls: Swapping ram/plunger areas; remember IMA = A / a. Final Answer: eta = (W * a) / (P * A)

Question 12

Thomson's Turbine — Flow Classification Thomson's turbine is classified as which type of reaction turbine based on flow direction?

Accepted Answer

Introduction: Reaction turbines are categorized by how water moves through the runner: inward radial, outward radial, axial, or mixed flow. Knowing which historical designs belong to each class helps in understanding their geometry and performance envelopes. Given Data / Assumptions: Thomson's turbine is a classic reaction-type machine. Classification depends on absolute flow direction through the runner. Concept / Approach: In an inward-flow reaction turbine, water enters near the periphery and moves radially inward while doing work on the runner. Thomson's turbine belongs to this inward-flow category, distinct from Fourneyron's outward-flow turbine and Kaplan's axial-flow turbine. Francis turbines later evolved as mixed/inward-flow variants with improved efficiency. Step-by-Step Solution: Identify historical prototypes: Fourneyron (outward), Thomson (inward), Kaplan (axial).Match Thomson's runner passages with inward radial motion.Therefore, classify it as an inward-flow reaction turbine. Verification / Alternative check: Textbook schematics depict Thomson's runner passages directing water inward toward the center. Why Other Options Are Wrong: axial flow: corresponds to Kaplan/propeller machines.outward flow: corresponds to Fourneyron's design.mixed flow: used for Francis; not the original Thomson layout. Common Pitfalls: Confusing historical names: Fourneyron = outward; Thomson = inward; Kaplan = axial. Final Answer: an inward flow

Question 13

Francis Turbine — Flow Ratio Range What is the typical range of flow ratio (phi = V_f1 / sqrt(2 g H)) used in the design of Francis turbines?

Accepted Answer

Introduction: The flow ratio is a non-dimensional parameter representing meridional (radial) velocity at runner inlet normalized by the ideal velocity from the head. It is a primary choice variable in Francis turbine design affecting incidence losses and efficiency. Given Data / Assumptions: phi = V_f1 / sqrt(2 * g * H). Francis turbine (mixed/inward-flow reaction). Concept / Approach: Empirical design practice places phi for Francis turbines in a modest range to balance entry shock and diffusion losses. Too low a phi leads to throttling and high incidence; too high a phi increases losses and reduces reaction. The commonly adopted design window is roughly 0.15–0.30. Step-by-Step Solution: Normalize inlet meridional velocity by sqrt(2 g H).Use empirical range ensuring high efficiency and stable operation.Select phi within 0.15–0.30 for standard designs. Verification / Alternative check: Manufacturer guidelines and classical design charts align with this interval. Why Other Options Are Wrong: 0.4–0.5 and higher ranges: excessive meridional velocity, atypical for Francis.1–1.5: nonphysical for normalized inlet velocity in this context. Common Pitfalls: Confusing flow ratio with speed ratio; they are different non-dimensional parameters. Final Answer: 0.15 to 0.3

Question 14

Cavitation in Hydraulic Machines Cavitation in turbines and pumps primarily results from which condition in the flowing liquid?

Accepted Answer

Introduction: Cavitation is the formation and collapse of vapor bubbles in a liquid when local static pressure falls below vapor pressure. It causes pitting, noise, vibration, drop in efficiency, and potential structural damage in hydraulic machinery. Given Data / Assumptions: Working fluid is water at typical operating temperature. Local static pressure can vary due to velocity changes and elevation head. Vapor pressure is a known function of temperature. Concept / Approach: Bernoulli’s principle indicates regions of high velocity can have low static pressure. If this pressure dips below vapor pressure, vapor cavities form. Downstream where pressure recovers, these bubbles collapse violently on surfaces, producing cavitation damage and performance losses. Step-by-Step Solution: Identify low-pressure zones (e.g., pump suction, runner exit, draft tube throat).Compare local pressure to vapor pressure at operating temperature.If p_local < p_vapor, cavitation inception occurs. Verification / Alternative check: Net Positive Suction Head (NPSH) calculations predict cavitation margins in pumps; Thoma coefficient sigma does similarly for turbines. Why Other Options Are Wrong: low velocity: generally raises static pressure; not the cause.high velocity: contributes indirectly only when it causes low pressure; the root cause is pressure below vapor pressure.high pressure: opposite of what triggers cavitation. Common Pitfalls: Assuming velocity alone causes cavitation; it is the associated pressure drop that matters. Final Answer: low pressure

Question 15

Fourneyron Turbine — Flow Classification The Fourneyron (often spelled Fourneyron) turbine is classified as which type of reaction turbine?

Accepted Answer

Introduction: Early reaction turbines were designed with either inward or outward radial flow. The Fourneyron turbine is a classic outward-flow design that significantly influenced subsequent turbine development. Given Data / Assumptions: Historical classification based on absolute flow direction through the runner. Steady, incompressible flow conditions. Concept / Approach: In the Fourneyron turbine, water enters the runner near the center and flows radially outward, transferring energy to the moving blades before discharging at the periphery. This is the defining feature of an outward-flow reaction turbine, contrasting with inward-flow (e.g., Thomson) and axial-flow (e.g., Kaplan) machines. Step-by-Step Solution: Trace the streamline through the runner: center → periphery.Identify energy exchange as pressure plus kinetic energy to shaft work.Classify flow as outward radial, hence “outward flow”. Verification / Alternative check: Canonical diagrams of Fourneyron runners show outward discharge to the casing. Why Other Options Are Wrong: axial flow: belongs to propeller/Kaplan designs.inward flow: characteristic of Thomson or certain Francis variants.mixed flow: typical of Francis; not the original Fourneyron style. Common Pitfalls: Spelling confusion (“Fourneyron” vs “Fourneyron”); the classification remains outward flow either way. Final Answer: an outward flow

Question 16

Hydraulic turbines — define the net (effective) head at runner inlet In hydroelectric practice, the head available at the inlet of the turbine after deducting all losses between headrace and runner is called the net or effective head. State whether …

Accepted Answer

Introduction: The concept of head is central to turbine selection and performance prediction. Engineers distinguish between gross head and net (effective) head to account for energy losses in conduits, valves, bends, and entry devices before the water reaches the runner. Given Data / Assumptions: Steady flow of water from headrace to the turbine runner. Hydraulic losses occur in penstock, fittings, nozzles, and intake structures. Head is measured as energy per unit weight (m of water). Concept / Approach: Gross head equals the elevation/energy difference between headrace and tailrace water levels. As water travels through conduits, a portion of the energy is lost to friction and local losses. The head actually available at the runner inlet, after subtracting these losses from gross head, is called net or effective head. This is the head that produces torque and power on the turbine. Step-by-Step Solution: Define gross head: H_gross = z_headrace − z_tailrace (plus velocity corrections, if used)Compute losses: h_losses = h_friction + Σh_localCompute net head: H_net = H_gross − h_losses Verification / Alternative check: Power available to the runner (water power) is ρgQ*H_net. Using H_gross in power without subtracting losses overestimates turbine output, demonstrating why H_net is the correct performance basis. Why Other Options Are Wrong: “Incorrect” or intake-limited variants: The definition is standard across impulse and reaction turbines. “Only for impulse turbines” or “only when friction is zero”: Net head applies universally; frictionless flow is an idealization where H_net ≈ H_gross. “Applies to pumps”: Net head terminology is also used in pumps but the definition here is for turbines and is correct. Common Pitfalls: Mixing up net head with effective head at a nozzle exit; neglecting seasonal changes in headrace/tailrace levels; ignoring air vents and valve losses. Final Answer: Correct

Question 17

Francis turbine classification — inward or outward flow? State whether a Francis turbine is an outward flow reaction turbine or not. Choose the correct classification for standard Francis machines used in hydropower.

Accepted Answer

Introduction: Turbines are categorized by how water flows through the runner (radial, axial, mixed) and the type of energy conversion (impulse vs reaction). The Francis turbine is a widely used reaction machine for medium heads and is frequently misdescribed regarding the direction of flow. Given Data / Assumptions: Conventional Francis design with spiral casing and guide vanes. Runner passages shaped for combined pressure and velocity changes (reaction action). Concept / Approach: Francis turbines are predominantly inward radial flow at entry, with the flow turning and leaving in an axial direction—thus often described as a mixed-flow reaction turbine. They are not outward radial flow machines. Inward flow reduces runner diameter toward the exit and suits the conversion of pressure head to mechanical work across the blades. Step-by-Step Solution: Identify energy mode: reaction (pressure changes within runner)Identify flow path: radially inward at inlet, turning to axial at outletConclude: Francis is inward (mixed) flow reaction, not outward flow Verification / Alternative check: Velocity triangles at inlet show significant whirl and radial components pointing inward; at exit, the design often targets reduced whirl and axial discharge into the draft tube. Why Other Options Are Wrong: “True” (outward flow): contradicts the standard Francis configuration. “Axial flow”: describes Kaplan/propeller class, not Francis. “Tangential impulse”: describes Pelton-type impulse turbines. Common Pitfalls: Assuming “radial flow” means outward; confusing the inlet/outlet directions; overlooking that many Francis runners are explicitly described as mixed-flow. Final Answer: False

Question 18

Hydraulic power transmission — device for varying torque Which hydraulic unit is specifically used to transmit and vary (increase/decrease) torque to the driven shaft in fluid couplings and automotive drivetrains?

Accepted Answer

Introduction: Many hydraulic devices sound similar but perform very different functions—pumping water uphill by water hammer, boosting pressure, storing energy, or transmitting torque. This question targets the component that manipulates torque between a prime mover and a driven shaft using fluid dynamics. Given Data / Assumptions: Closed-loop hydraulic device using impeller, turbine, and stator. Operates with circulating working fluid (typically oil). Designed to multiply torque at low output speeds. Concept / Approach: A hydraulic torque converter is a hydrodynamic device with three main elements: pump (impeller) connected to the engine, turbine connected to the load, and a stator that redirects flow to provide reaction and torque multiplication at stall or low-speed conditions. As output speed approaches input speed, the torque ratio drops toward unity, functioning like a fluid coupling but with torque multiplication capability when needed. Step-by-Step Solution: Recognize target function: vary and multiply torque hydrodynamicallyMatch to device: torque converter with impeller–turbine–statorSelect “Hydraulic torque converter” as correct Verification / Alternative check: Automotive automatic transmissions historically use torque converters for launch and low-speed torque boost before lock-up clutches engage. Why Other Options Are Wrong: Hydraulic ram: lifts water using water hammer; no torque transmission. Hydraulic intensifier: increases static pressure, not rotational torque transmission. Hydraulic accumulator: stores pressure energy. Hydraulic press: converts hydraulic pressure into linear force for forming. Common Pitfalls: Confusing torque converters with simple fluid couplings (which do not multiply torque) or with mechanical gear torque multipliers. Final Answer: Hydraulic torque converter

Question 19

Axial-flow reaction turbine — flow direction relative to shaft Complete the statement: In an axial flow reaction turbine, the water flows ________ to the axis of the wheel (runner).

Accepted Answer

Introduction: Turbine classification by flow direction guides selection for head and discharge. Axial-flow turbines, such as Kaplan and propeller types, are preferred for low heads and high discharges, where the fluid ideally moves along the shaft axis through the runner passages. Given Data / Assumptions: Reaction turbine with guide vanes and adjustable or fixed-pitch runner blades. Design near best-efficiency point with minimal swirl at exit. Concept / Approach: In axial-flow turbines, the absolute velocity of water is substantially along the axis at both inlet and outlet (with small swirl components adjusted by blade pitch). This reduces secondary losses and allows a compact draft tube to recover kinetic energy efficiently at low heads. Step-by-Step Solution: Identify turbine type: axial-flow reaction (e.g., Kaplan)Relate to geometry: runner blades aligned for near-axial absolute velocityConclude the flow direction: parallel to the shaft axis Verification / Alternative check: Velocity triangles for Kaplan turbines show small whirl components targeted so that discharge is nearly axial into the draft tube, maximizing pressure recovery. Why Other Options Are Wrong: Perpendicular: describes radial flow, not axial. Radially inward then axial: describes Francis (mixed flow). Tangential: impulse action along periphery (Pelton). Spiral with large pre-swirl: increases losses; not a defining feature. Common Pitfalls: Confusing axial with radial-mixed configurations; ignoring that small swirl may still exist but the dominant direction is axial. Final Answer: parallel

Question 20

Pelton wheel performance — typical overall efficiency range Select the most appropriate overall efficiency (ratio of shaft power to water power at runner inlet) usually achieved by well-designed Pelton turbines.

Accepted Answer

Introduction: Overall efficiency incorporates hydraulic, mechanical, and volumetric effects. Pelton turbines, used at high heads and modest flows, are renowned for high efficiencies due to clean jet formation and limited cavitation risk when designed correctly. Given Data / Assumptions: Well-designed runner and nozzles with proper jet deflection. Good bearings and seals; minimal leakage. Operation near best-efficiency point. Concept / Approach: Overall efficiency η_o = P_shaft / (ρgQ*H_net). Empirical data and manufacturer curves commonly show Pelton turbines achieving mid-to-high 80% efficiency in practical installations. Achieving above ~90% consistently is rare and typically reflects idealized or test conditions. Step-by-Step Solution: Identify typical range from industry practiceExclude unrealistically low or overly optimistic rangesSelect 0.85–0.90 as the most representative Verification / Alternative check: Comparative data: Francis ~0.90 at mid-heads, Kaplan ~0.90 at low heads under ideal settings; Pelton generally 0.85–0.90 depending on specific speed and jet quality. Why Other Options Are Wrong: 0.50–0.75: Undervalues modern machines. 0.75–0.85: Possible but below the usual best range. 0.90–0.96: Overly optimistic for routine operation. Common Pitfalls: Confusing hydraulic efficiency with overall; ignoring part-load operation and nozzle throttling effects which reduce efficiency. Final Answer: 0.85 to 0.90

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