Question 1

Impulse Turbine Performance — Definition of Overall Efficiency Evaluate the statement: “The overall efficiency of an impulse turbine is the ratio of the actual power produced by the turbine to the energy actually supplied by the turbine.”

Accepted Answer

Introduction: Efficiency definitions must clearly distinguish between energy supplied to the machine and energy produced by it. This question tests precise terminology for overall efficiency in impulse turbines such as Pelton wheels. Given Data / Assumptions: Steady operation at rated head and discharge. Neglecting small leakage and windage for the definition. Hydraulic input comes from the water jet or jets impinging on the buckets. Concept / Approach: The overall efficiency η_overall is defined as: useful shaft power output divided by the net hydraulic power supplied to the turbine by the water at inlet. The phrase “energy actually supplied by the turbine” is incorrect because the turbine does not supply energy to itself; it receives energy from the fluid. Therefore, the statement must be marked incorrect. Step-by-Step Solution: Define P_out = actual shaft power measured at the runner shaft.Define P_in = rho * g * Q * H_net (net head at turbine inlet).Then η_overall = P_out / P_in.The quoted statement misidentifies the denominator as “energy supplied by the turbine,” which is a wording error. Verification / Alternative check: Texts present overall efficiency as the product of hydraulic and mechanical efficiencies, both referenced to energy supplied by the water (not by the machine). Why Other Options Are Wrong: Yes: Accepting the statement would propagate the wording error and confuse input vs output energy roles. Common Pitfalls: Mixing “to the turbine” with “by the turbine.” Precision in prepositions is crucial in efficiency definitions. Final Answer: No

Question 2

Inward Flow Reaction Turbine — Flow Direction Through Runner In an inward flow reaction turbine, how does the water move through the runner relative to the wheel radius?

Accepted Answer

Introduction: Reaction turbines exchange energy with the fluid across the runner as pressure changes occur. Their classification by flow direction clarifies the geometry of inlet and outlet. This question focuses on the classical inward-flow reaction arrangement. Given Data / Assumptions: Steady operation; water is incompressible. Runner passages guide flow radially inward. Neglect minor 3D effects; consider idealized runner. Concept / Approach: In inward-flow reaction turbines, water enters at a larger radius (outer periphery) with higher absolute velocity and static pressure, and discharges at a smaller radius (towards the centre), transferring energy to the runner through a drop in both pressure and absolute velocity components as guided by the blades. Step-by-Step Solution: Identify turbine type: inward radial flow.At inlet (outer radius), absolute velocity has tangential and radial components designed to match blade speed and incidence.Through the runner, radius decreases, static pressure and whirl adjust, doing work on blades; discharge approaches inner radius.Hence, motion is from outer periphery towards the centre. Verification / Alternative check: Velocity triangle sketches for inward reaction turbines show positive work extraction with decreasing radius and appropriate whirl reduction. Why Other Options Are Wrong: Parallel to axis: describes axial-flow machines (e.g., Kaplan). Centre to periphery: outward-flow type, not inward. Partly radial and partly axial: mixed/diagonal flow machines, not purely inward radial. Common Pitfalls: Confusing inward radial turbines with axial-flow turbines due to blade appearance in schematics. Final Answer: the water enters at the outer periphery and then flows towards the centre

Question 3

Pelton Wheel Scaling — Unit Discharge Calculation A Pelton wheel develops 1750 kW under a head H = 100 m while running at N = 200 rpm and discharging Q = 2500 litres per second. Compute the unit discharge of the wheel (i.e., discharge at 1 m head, Q…

Accepted Answer

Introduction: Unit quantities allow comparison of turbines independent of the operating head by normalizing speed, discharge, and power to a 1 m head. This problem asks for unit discharge, a commonly used parameter for model–prototype scaling and performance catalogs. Given Data / Assumptions: Pelton wheel (impulse turbine), head H = 100 m. Actual discharge Q = 2500 L/s = 2.5 m^3/s. Standard definition Q_unit = Q / sqrt(H). Concept / Approach: Unit discharge is defined as the discharge at unit head (1 m) for dynamic similarity. Since discharge scales with sqrt(H) for dynamically similar operation of impulse turbines, we divide the actual discharge by sqrt(H). Step-by-Step Solution: Convert units: Q = 2500 L/s = 2.5 m^3/s.Compute sqrt(H) = sqrt(100) = 10.Q_unit = Q / sqrt(H) = 2.5 / 10 = 0.25 m^3/s.Therefore, the unit discharge is 0.25 m^3/s. Verification / Alternative check: Dimensional analysis yields Q ∝ D^2 * V and V ∝ sqrt(g * H); with D scaled to preserve similarity at unit head, Q scales with sqrt(H), confirming the formula. Why Other Options Are Wrong: 0.5 m^3/s: Implies division by sqrt(25) or an arithmetic slip. 1.5 m^3/s and 2.5 m^3/s: Do not apply the head normalization; 2.5 m^3/s is the given actual discharge. Common Pitfalls: Forgetting to convert litres to cubic metres and neglecting the square-root head scaling. Final Answer: 0.25 m^3/s

Question 4

Reciprocating Pump with Air Vessel — Percentage Saving in Work (Single-Acting) An air vessel is fitted to a single-acting reciprocating pump to smoothen flow and reduce acceleration head. The percentage saving in work (and thus power) due to the air…

Accepted Answer

Introduction: Reciprocating pumps experience periodic acceleration and deceleration of the liquid column in the suction and delivery pipes, creating acceleration head and extra work per cycle. Air vessels reduce this unsteadiness, cutting energy losses. This question asks for the approximate saving for a single-acting pump. Given Data / Assumptions: Single-acting reciprocating pump with sinusoidal piston motion. Long pipes where acceleration head is significant. Air vessel properly located near cylinder on suction and/or delivery side. Concept / Approach: With an air vessel, the flow in the pipe becomes nearly uniform, and the fluctuating component of kinetic energy is largely eliminated. Classical analysis shows the reduction in work per cycle is about 39.2% for a single-acting pump (and about 48.8% for double-acting), assuming ideal placement and negligible secondary losses. Step-by-Step Solution: Without air vessel: periodic velocity in pipe causes added acceleration head h_as ∝ L * A_piston / (g * a_pipe) * r * ω^2 * cos(ωt), leading to extra work.With air vessel: the vessel supplies/absorbs flow during stroke reversals, flattening the pipe velocity waveform.Theoretical derivation yields percentage saving ≈ 39.2% for single-acting geometry and motion. Verification / Alternative check: Design texts cite ~39.2% for single-acting and ~48.8% for double-acting pumps; field results vary slightly depending on pipe inertia and vessel sizing. Why Other Options Are Wrong: 48.8%: Corresponds to double-acting pumps, not single-acting. 84.8% and 88.4%: Unrealistically high; would imply near-total elimination of dynamic effects. Common Pitfalls: Confusing single- vs double-acting values or misplacing the air vessel too far from the cylinder, which reduces the actual benefit. Final Answer: 39.2%

Question 5

Francis Turbine — Relative Pressure at Runner Inlet and Outlet In a reaction (Francis) turbine operating under design conditions, the static pressure at the runner inlet is typically __________ the static pressure at the runner outlet (draft tube en…

Accepted Answer

Introduction: Francis turbines are mixed-flow reaction machines where a portion of the total head is converted to pressure rise/drop within stationary and rotating passages. Understanding how pressure changes across the runner is key to draft tube design and cavitation control. Given Data / Assumptions: Properly designed guide vanes and runner; steady operation. Draft tube recovers kinetic energy by diffusion, so outlet pressure is near tailrace pressure. Total head drop occurs partly in guide vanes, runner, and draft tube. Concept / Approach: In a Francis turbine, water enters the runner at a higher static pressure and leaves at a lower static pressure, having done work on the blades. The draft tube then gradually converts some kinetic energy to static pressure before discharge to the tailrace. Therefore, p_inlet > p_outlet across the runner. Step-by-Step Solution: Across runner: energy is extracted, so pressure plus kinetic head decreases.At outlet: lower static pressure region exists before the draft tube restores part of the velocity head.Hence, static pressure at runner inlet is more than at runner outlet. Verification / Alternative check: Velocity triangles and Euler’s turbine equation confirm energy transfer from fluid to runner and associated pressure drop. Why Other Options Are Wrong: Less than: contradicts the reaction nature across the runner. Common Pitfalls: Confusing conditions inside the runner with those at the draft tube exit; the draft tube increases static pressure relative to runner outlet but not above runner inlet. Final Answer: more than

Question 6

In reciprocating pumps used in fluid mechanics and hydraulic machines, air vessels are fitted on the suction and delivery sides to improve performance under unsteady flow. Which primary effect best explains why air vessels are provided on reciprocat…

Accepted Answer

Introduction: Reciprocating pumps create pulsating flow because the piston accelerates and decelerates water each stroke. Air vessels are classic attachments that mitigate the adverse effects of this unsteadiness. Understanding their principal function is essential for sizing, power estimation, and cavitation avoidance. Given Data / Assumptions: Single- or double-acting reciprocating pump with suction and delivery pipes. Water is essentially incompressible; trapped air in the vessel is compressible. Objective: reduce unsteady inertial effects in the pipeline. Concept / Approach: During piston motion, the water column in pipes undergoes acceleration and deceleration, generating an acceleration head h_a proportional to (l/gd) * (dv/dt). An air vessel provides a compressible cushion that absorbs and supplies flow during transients, flattening velocity fluctuations and thereby reducing the required acceleration of the water column and the corresponding acceleration head. Step-by-Step Solution: 1) Without an air vessel: high dv/dt → large h_a → high instantaneous power and pressure swings.2) With an air vessel: part of the stroke volume is briefly stored/released in the vessel → lower dv/dt in long pipes.3) Reduced dv/dt ⇒ reduced h_a in both suction and delivery lines.4) Consequences: smoother discharge, reduced risk of separation/cavitation on suction, and power savings. Verification / Alternative check: Power input peaks diminish when h_a is reduced; measured pressure traces show attenuated pulsations with an air vessel installed. Why Other Options Are Wrong: Smoothen the flow: true as an outcome but not the fundamental targeted head term; the more precise primary effect is reducing acceleration head. Reduce suction head: static suction head is geometric; air vessels do not change sump level. Increase delivery head: they do not increase head; they reduce pulsations. Increase pump speed without cavitation risk: speed limits still depend on NPSH and valve dynamics. Common Pitfalls: Confusing acceleration head with friction head; assuming air vessels change static levels rather than transient components. Final Answer: reduce acceleration head

Question 7

In turbine similarity analysis, “unit power” P_u is defined as the power developed by a turbine when operating under unit head (H = 1). Given actual power P at head H, express the unit power in terms of P and H.

Accepted Answer

Introduction: Unit quantities (unit speed, unit discharge, unit power) allow comparison of hydraulically similar turbines at different heads. By reducing head effects via dimensional analysis, designers can scale performance between model and prototype efficiently. Given Data / Assumptions: Geometric and dynamic similarity between compared operating points. Head H is the effective net head across the turbine. Define unit head condition as H = 1. Concept / Approach: From similarity, discharge Q ∝ H^(1/2), rotational speed N ∝ H^(1/2) (for constant specific speed form), and therefore power P ∝ ρ g Q H ∝ H^(3/2). Setting H = 1 defines unit power P_u that removes head scaling, yielding P_u = constant for dynamically similar operation. Thus, P_u relates to actual P via division by H^(3/2). Step-by-Step Solution: 1) Use Q ∝ H^(1/2) from velocity head scaling.2) Power P = ρ * g * Q * H ⇒ P ∝ H^(1/2) * H = H^(3/2).3) Define unit power as power at H = 1: P_u = P / H^(3/2).4) Therefore, for any measured P at head H, compute unit power with the above relation. Verification / Alternative check: Dimensionless groups: P / (ρ g) scales like Q H, and with Q ∝ H^(1/2), the H^(3/2) law is recovered, confirming the formula. Why Other Options Are Wrong: P_u = P * H^(3/2): incorrectly amplifies head effect. P_u = P / H or P * H or P / H^2: do not follow similarity exponents derived from velocity scaling. Common Pitfalls: Mixing unit quantities: note N_u = N / H^(1/2), Q_u = Q / H^(1/2), but P_u = P / H^(3/2). Final Answer: P_u = P / H^(3/2)

Question 8

In hydraulic turbines classified as mixed-flow reaction turbines, the water path through the runner is a combination of two components. State whether the following is correct: “The flow is partly radial and partly axial.”

Accepted Answer

Introduction: Turbines are categorized by the direction of flow through the runner: radial, axial, or mixed. Mixed-flow reaction turbines blend radial and axial components to balance head utilization and discharge capacity across a range of sites and heads. Given Data / Assumptions: Reaction turbine with pressure change across the runner. Runner passages guide flow with both radial and axial velocity components. Operation near design point. Concept / Approach: In mixed-flow turbines (e.g., modern Francis types), water enters the runner radially inward (or outward in some historical designs) and exits with a significant axial component. This combination enables efficient energy conversion over medium heads by controlling whirl and flow angles to maintain favorable incidence and reaction. Step-by-Step Solution: 1) Guide vanes impart whirl; runner blades turn the flow, extracting energy.2) Velocity triangles show both radial and axial components at inlet/outlet.3) Outlet arranged to reduce residual swirl and minimize draft losses.4) Hence, the qualitative statement “partly radial and partly axial” holds true for mixed-flow reaction turbines. Verification / Alternative check: Performance maps and runner geometry confirm non-zero radial and axial velocity components under design conditions. Why Other Options Are Wrong: Incorrect/part-load only: mixed nature persists across loads, though angles vary. Pelton wheel: impulse, tangential flow; not mixed reaction. Kaplan: primarily axial flow, not mixed. Common Pitfalls: Confusing Francis (mixed) with Kaplan (axial) or Pelton (tangential impulse). Final Answer: Correct

Question 9

For a centrifugal pump handling water under steady conditions, define the overall efficiency in practical engineering terms. Select the correct ratio representing overall efficiency.

Accepted Answer

Introduction: Overall efficiency of a centrifugal pump quantifies how effectively input shaft power from the prime mover is converted into useful hydraulic power delivered to the fluid at the delivery outlet. It captures all internal losses collectively. Given Data / Assumptions: Steady operation at a given duty point. Definition of useful output as water horsepower at delivery: ρ * g * Q * H_d. Input power measured at the pump shaft (after mechanical transmission losses). Concept / Approach: Overall efficiency η_o = (useful hydraulic output) / (input shaft power) = (ρ * g * Q * H_d) / P_in. It equals the product of volumetric, hydraulic, and mechanical efficiencies, but the simplest definition is output-over-input referred to the prime mover power supplied to the pump shaft. Step-by-Step Solution: 1) Compute hydraulic output: P_out = ρ * g * Q * H_d.2) Measure input: P_in at shaft.3) Overall efficiency: η_o = P_out / P_in.4) Recognize that η_o = η_v * η_h * η_m, but the ratio definition remains output/input. Verification / Alternative check: Laboratory tests plotting pump curves routinely derive η_o from measured head–discharge and shaft power. Why Other Options Are Wrong: Energy at impeller vs energy supplied: describes manometric or hydraulic sub-efficiencies, not overall. Energy supplied to pump divided by energy at impeller: inverted meaning. Manometric head ratio per kN: partial metric, not overall definition. Product of only two sub-efficiencies: omits mechanical losses. Common Pitfalls: Confusing overall efficiency with manometric or volumetric efficiency; ignoring mechanical losses in bearings and seals. Final Answer: actual workdone by the pump to the energy supplied to the pump by the prime mover

Question 10

A hydraulic ram (hydram) is a water-lifting device that uses the momentum of a large flow available at low head to deliver a small flow at a high head. Identify the most accurate description of its function among the choices.

Accepted Answer

Introduction: The hydraulic ram is a century-old, robust pump that requires no external power. It exploits water hammer from an intermittent waste flow to raise a fraction of that water to a much higher elevation, serving remote communities and hilly terrains. Given Data / Assumptions: Continuous supply of water at low head (drive head). Hydram with waste (impulse) valve and delivery (check) valve. Short, rigid drive pipe for effective pressure rise. Concept / Approach: During each cycle, the waste valve shuts suddenly, generating a pressure surge (water hammer) in the drive pipe. This surge opens the delivery valve momentarily, pushing a small parcel of water into the air chamber and up the delivery pipe to a higher head. Over many cycles, a modest but continuous high-head discharge is obtained, sacrificing a larger low-head flow to waste. Step-by-Step Solution: 1) Water flows through the waste valve until velocity builds.2) Waste valve slams shut → pressure rise.3) Delivery valve opens → water enters air chamber and delivery line.4) Pressures equalize; cycle repeats, achieving net elevation gain. Verification / Alternative check: Energy balance shows output head greatly exceeds drive head, with efficiency typically 50–70% for the lifted fraction. Why Other Options Are Wrong: Store energy: an air chamber smooths flow but storage is not the primary function. Increase pressure of water: true locally but incomplete and nonspecific. Lift from deep wells: hydram needs a flowing source, not deep suction. Convert alternating to steady flow: incidental smoothing via air chamber only. Common Pitfalls: Underestimating the need for proper drive pipe design; confusing hydram with conventional reciprocating pumps. Final Answer: to lift small quantity of water to a greater height when a large quantity of water is available at a smaller height

Question 11

Classify a Kaplan turbine with respect to flow direction through the runner. State whether the assertion “A Kaplan turbine is an axial flow reaction turbine” is correct.

Accepted Answer

Introduction: Kaplan turbines are widely used for low-head, high-discharge hydro sites. Their hallmark is adjustable runner blades combined with wicket gates to maintain high efficiency across varying loads, and an essentially axial flow path through the runner. Given Data / Assumptions: Kaplan runner with variable-pitch blades. Reaction turbine with pressure drop through the runner. Draft tube present for pressure recovery. Concept / Approach: In axial flow turbines, the bulk direction of water relative to the runner shaft is largely parallel to the axis at inlet and outlet. Kaplan geometry guides flow so that tangential components are minimized at exit, limiting swirl and improving draft tube recovery. The machine operates with significant static pressure change (reaction), distinguishing it from impulse wheels. Step-by-Step Solution: 1) Wicket gates regulate swirl and discharge.2) Runner blades adjust pitch to keep incidence optimal as load changes.3) Flow remains predominantly axial through the runner passages.4) Draft tube converts kinetic energy to pressure for head recovery. Verification / Alternative check: Velocity triangles at best efficiency show small whirl at outlet and axial dominance, confirming classification. Why Other Options Are Wrong: Incorrect/part-load/fixed-blade-only: Kaplan retains axial classification across loads and with variable pitch. Impulse operation: Kaplan is a reaction turbine; pressure drop occurs within the runner. Common Pitfalls: Confusing Kaplan (axial reaction) with Francis (mixed flow) or Pelton (impulse). Final Answer: Correct

Question 12

In a mixed flow centrifugal pump, the through-flow within the impeller passages combines two components. State whether the statement “the flow through the impeller is a combination of radial and axial flows” is correct.

Accepted Answer

Introduction: Centrifugal pumps are categorized by the direction of flow leaving the impeller: radial, axial, or mixed. Mixed-flow pumps are preferred when higher specific speeds are needed than purely radial impellers can provide, while still maintaining good head characteristics. Given Data / Assumptions: Mixed-flow impeller with blade geometry imparting both radial and axial velocity components. Steady, incompressible flow at design conditions. Concept / Approach: In mixed-flow impellers, the velocity triangle at outlet shows significant meridional (axial) component alongside radial ejection, unlike purely radial pumps where the exit velocity is largely radial. This blend increases discharge for a given head and diameter, aligning with higher specific speed requirements. Step-by-Step Solution: 1) Impeller blades are swept to direct part of the flow axially.2) Outlet velocity components: V_r (radial) and V_a (axial) both non-zero.3) Diffuser/volute converts kinetic energy to pressure for both components.4) Performance curves reflect mixed nature: greater Q at moderate head compared to radial-only designs. Verification / Alternative check: Specific speed trends place mixed-flow pumps between radial and axial types, substantiating the combined-flow description. Why Other Options Are Wrong: Incorrect/only at design/diffuser-only/low specific speed: the mixed character is inherent to impeller geometry and persists across typical operating ranges. Common Pitfalls: Assuming all centrifugal pumps are strictly radial; ignoring axial component in mixed-flow designs. Final Answer: Correct

Question 13

Cavitation risk in reaction turbines can be mitigated by appropriate design and layout measures. Which of the following collectively help in avoiding cavitation to a great extent?

Accepted Answer

Introduction: Cavitation occurs when local pressure drops below vapor pressure, forming vapor bubbles that collapse and damage surfaces. Reaction turbines, especially high-speed runners, are susceptible near low-pressure zones. Design measures must target higher local pressures and better surface resistance. Given Data / Assumptions: Reaction turbine with draft tube. Tail race level known; runner elevation selectable. Material and surface finish influence erosion resistance. Concept / Approach: Submerging the runner (placing it below tail race level) increases static pressure margin against vapor pressure. Stainless steels exhibit better cavitation erosion resistance than plain carbon steels. Polished, smooth runner blades reduce micro-crevice nucleation and suppress boundary layer separation, elevating local static pressure and delaying bubble inception. Step-by-Step Solution: 1) Increase static head at runner by lowering installation relative to tail water.2) Select cavitation-resistant materials (e.g., stainless steel).3) Achieve smooth surface finish by polishing to reduce nucleation sites and losses.4) Combined, these measures reduce sigma (Thoma cavitation factor) demands. Verification / Alternative check: Manufacturer data correlates improved surface finish and material selection with longer cavitation life; submergence calculations show higher absolute pressure at runner exit. Why Other Options Are Wrong: Any single measure alone helps but the most comprehensive answer is the combination, reflected in “all of the above”. Installing above tail water (option e) worsens cavitation by lowering pressure at the runner. Common Pitfalls: Focusing only on material while neglecting installation elevation and hydraulic design of the draft tube. Final Answer: all of the above

Question 14

In model–prototype testing of hydraulic machines, dynamic similarity is a stronger requirement than geometric similarity. Under which condition do we say that model and prototype have dynamic similarity?

Accepted Answer

Introduction: Similarity laws enable prediction of prototype performance from model tests. Three levels exist: geometric (shape), kinematic (motion), and dynamic (forces). Dynamic similarity requires that the ratios of all types of forces match between model and prototype so that the dimensionless parameters governing the flow are equal. Given Data / Assumptions: Hydraulic machine or flow system with inertia, viscous, pressure, gravity, and surface tension forces potentially relevant. Objective: accurate force and performance prediction from model to prototype. Concept / Approach: Dynamic similarity is achieved when the corresponding dimensionless numbers (e.g., Reynolds, Froude, Weber, Euler numbers) are equal for model and prototype. The options presented do not explicitly state this condition. Identical velocities, equal size/shape, or merely geometric similarity are insufficient; they do not ensure matched force ratios. Step-by-Step Solution: 1) Identify relevant forces → derive dimensionless groups.2) Impose equality of these groups between model and prototype.3) Recognize that options (a), (b), (c) fail to guarantee force ratio equality.4) Therefore, among given choices, the correct selection is “none of the above”. Verification / Alternative check: Example: ship resistance requires Froude similarity; spillway aeration may require Weber too; pumps/turbines often target Reynolds and specific speed similarity. Why Other Options Are Wrong: Identical velocities: does not scale forces properly across sizes. Equal in size and shape: that is identity, not similarity. Identical shape but different size: only geometric similarity. Common Pitfalls: Assuming geometric similarity implies dynamic similarity; ignoring which dimensionless numbers dominate for a given problem. Final Answer: none of the above

Question 15

When the ratios of all corresponding forces acting at corresponding points in a model and its prototype are equal, what type of similarity exists between them? Choose the most appropriate classification.

Accepted Answer

Introduction: Engineering similarity is categorized into geometric (shape), kinematic (motion), and dynamic (forces). For accurate force and performance prediction, matching the force ratios or, equivalently, the relevant dimensionless numbers is essential. Given Data / Assumptions: Model and prototype are compared under conditions where multiple forces act (inertia, pressure, viscous, gravity, surface tension). Ratios of corresponding forces are stated equal. Concept / Approach: Equal ratios of corresponding forces define dynamic similarity. This implies the equality of the relevant dimensionless parameters (Reynolds, Froude, Euler, Weber, Mach as applicable). Geometric or kinematic similarity alone cannot guarantee correct scaling of loads or performance. Step-by-Step Solution: 1) Identify that the condition given is “equal force ratios”.2) Map this to dynamic similarity by definition.3) Note that geometric similarity concerns shapes; kinematic similarity concerns velocity and acceleration fields.4) Therefore, the correct classification is dynamic similarity. Verification / Alternative check: In turbine model testing, matching Re and specific speed ensures force fields scale correctly; this is a dynamic similarity requirement. Why Other Options Are Wrong: Geometric: needs similar shapes but says nothing about forces. Kinematic: matches motion patterns but not force magnitudes. None/geometric+kinematic only: incomplete without force equivalence. Common Pitfalls: Equating geometric similarity with complete similarity; overlooking dominant force regimes (e.g., Froude vs Reynolds) for the application. Final Answer: dynamic similarity

Question 16

Francis turbine operating significantly below design head Statement: If a Francis turbine is run for long periods at less than 50% of its design/net head, the efficiency drops and the danger of cavitation becomes more serious.

Accepted Answer

Introduction / Context: Francis turbines are reaction machines designed to operate around a specific head and discharge. Running far away from this design point—especially at much lower head—changes pressure distribution and velocities inside the runner and draft tube. This not only reduces efficiency but can also aggravate cavitation risk in low-pressure zones. Given Data / Assumptions: Francis (reaction) turbine with spiral casing, stay/guide vanes, runner, and draft tube. Operation at < 50% of design/net head for extended periods. Speed and guide vane settings adjusted to keep unit on line. Concept / Approach: At reduced head, velocities and pressure recovery patterns change. Local static pressures can fall below vapor pressure near the runner exit and within the draft tube, promoting cavitation inception. Off-design incidence on blades increases losses (eddy, separation), so hydraulic efficiency falls. Draft tube swirl and vortex ropes are stronger off-design, further worsening performance and cavitation susceptibility. Step-by-Step Solution: Lower head ⇒ lower available specific energy and altered velocity triangles.Blade incidence deviates from optimum ⇒ higher profile and secondary losses.Pressure at runner exit may dip below vapor pressure ⇒ cavitation bubbles form/collapse.Cavitation damages surfaces and introduces unsteady forces ⇒ additional efficiency loss and reliability issues. Verification / Alternative check: Unit characteristic curves show peak efficiency near design head; as head is reduced substantially, efficiency drops. Cavitation Thoma coefficient requirements become harder to satisfy at off-design, confirming higher risk. Why Other Options Are Wrong: “Disagree” ignores well-known reaction turbine off-design behavior. Pelton (impulse) turbines are different machines. Draft tube presence does not negate low-pressure pockets; even with a draft tube, off-design operation can cavitate. Saying cavitation reduces at low head is contrary to reaction turbine practice. Common Pitfalls: Confusing best-efficiency-point shifts with universally high efficiency; assuming lower head always means safer pressures (local pockets can still drop below vapor pressure). Final Answer: Agree

Question 17

Draft tube performance definition If V1 and V2 are the absolute water velocities at the inlet and outlet of a draft tube, respectively, the efficiency of the draft tube is:

Accepted Answer

Introduction / Context: A draft tube is fitted at the outlet of reaction turbines (e.g., Francis, Kaplan) to convert kinetic energy at runner exit into static pressure, allowing recovery of part of the velocity head and enabling the turbine to be installed above tail water level without excessive loss of head. Given Data / Assumptions: Inlet absolute velocity to draft tube = V1; outlet absolute velocity = V2. Same datum for assessing kinetic head; density and gravity constant. Neglecting elevation changes inside the short tube for the definition (focus on kinetic head recovery). Concept / Approach: The draft tube efficiency quantifies the fraction of the inlet kinetic head recovered as useful static head. Kinetic head is proportional to V^2. Therefore, recovered fraction equals (V1^2 − V2^2)/V1^2, a non-dimensional number between 0 and 1 in ideal cases (losses make it less than 1). Step-by-Step Solution: Kinetic head at inlet = V1^2 / (2 g).Kinetic head at outlet = V2^2 / (2 g).Recovered head = (V1^2 − V2^2) / (2 g).Efficiency η_d = (Recovered head) / (Inlet kinetic head) = (V1^2 − V2^2)/V1^2. Verification / Alternative check: If V2 ≈ 0 (excellent diffusion), η_d → 1. If no recovery (V2 = V1), η_d = 0. These limits match physical expectations. Why Other Options Are Wrong: Reversing numerator sign makes η_d negative. Velocity ratios (c) and (d) ignore the squared relation of kinetic energy. Option (e) uses the wrong reference in the denominator. Common Pitfalls: Mixing velocity head with velocity; forgetting that energy terms scale with the square of velocity. Final Answer: η_d = (V1^2 - V2^2) / V1^2

Question 18

Impulse turbine optimum velocity ratio The hydraulic efficiency of an impulse turbine is maximum when the runner (wheel) peripheral speed is what fraction of the jet velocity?

Accepted Answer

Introduction / Context: Impulse turbines (e.g., Pelton) extract energy by changing the momentum of a high-speed jet. The wheel speed relative to jet speed controls how much of the jet's kinetic energy is converted to shaft work. There is a well-known optimum speed ratio that maximizes hydraulic efficiency under typical assumptions. Given Data / Assumptions: Ideal impulse action with no shaft work from pressure changes (atmospheric pressure at runner). Jet deflection approximates 180° minus losses; relative exit speed nearly equals relative inlet speed. Neglecting mechanical losses for the hydraulic-optimum reasoning. Concept / Approach: The hydraulic power depends on the change in whirl component of velocity across the runner. Treating losses modestly, differentiating power with respect to runner speed yields a maximum when the wheel speed equals approximately half the jet velocity. Step-by-Step Solution: Let jet speed be V and wheel peripheral speed be u.Hydraulic efficiency depends on the product of mass flow and change in whirl velocity (V − u) terms.Optimization of the idealized expression gives u/V ≈ 0.5.Therefore, the best hydraulic efficiency is achieved when u is one-half of V. Verification / Alternative check: Pelton design guides and textbook derivations converge on the 0.46–0.5 practical range; ideal analysis gives 0.5, with real losses shifting slightly lower. Why Other Options Are Wrong: One-fourth or three-fourth deviate from the optimum, reducing momentum change effectiveness. “Double” or “equal to jet” produce very low relative jet deflection or zero energy extraction. Common Pitfalls: Confusing mechanical (overall) efficiency with hydraulic efficiency; ignoring the impact of bucket losses that slightly reduce the optimal ratio in practice. Final Answer: one-half

Question 19

Hydraulic intensifier principle Statement: A hydraulic intensifier increases the pressure (intensity) of water using the energy available from a large flow at low pressure.

Accepted Answer

Introduction / Context: Hydraulic intensifiers are used when a system needs a short-duration, high-pressure fluid supply but only a low-pressure source is available. They convert energy from a larger-volume, low-pressure stream into a smaller-volume, high-pressure output by force balance across areas. Given Data / Assumptions: Incompressible fluid behavior and negligible leakage. Two interconnected pistons or rams with different effective areas. Quasi-static (hydrostatic) operation governed by Pascal's law. Concept / Approach: By applying a low pressure to a large-area piston, a larger force is generated that acts on a smaller-area piston, producing a higher pressure on the outlet side. Energy is conserved neglecting losses: low-pressure side provides high flow at low pressure; high-pressure side delivers low flow at high pressure. Step-by-Step Solution: Input pressure p1 acts over area A1 ⇒ force F1 = p1 * A1.This force drives a smaller area A2 piston ⇒ output pressure p2 = F1 / A2.Thus p2 = p1 * (A1 / A2) > p1 when A1 > A2, intensifying pressure.Flow is reduced proportionally to satisfy energy balance (neglecting losses). Verification / Alternative check: Industrial intensifiers (ram or differential piston types) follow the same area-ratio principle; measured efficiencies < 100% due to friction and leakage but pressure multiplication remains. Why Other Options Are Wrong: “Incorrect” contradicts the device's purpose. The working fluid can be water or oil. Draft tubes relate to turbines, not intensifiers. Intensifiers do not reduce pressure; they multiply it. Common Pitfalls: Confusing intensifiers with accumulators (which store energy) or boosters (which may be gas-driven but share a similar area-ratio idea). Final Answer: Correct

Question 20

General characteristics of water wheels Which of the following statements about traditional water wheels is/are correct?

Accepted Answer

Introduction / Context: Before modern turbines, water wheels (overshot, breastshot, undershot) were widely used to convert hydraulic energy into mechanical work. Understanding their operating traits clarifies why turbines replaced them in many applications and where water wheels still make sense (e.g., heritage mills, micro-hydro with very low heads). Given Data / Assumptions: Low to moderate heads and variable streams. Simple construction with buckets/paddles; open-channel approach flow. Speed governed by wheel diameter and flow; no high-speed gearing assumed. Concept / Approach: Water wheels rely on weight (overshot), impulse (undershot), or a combination (breastshot). Their large diameters and direct mechanical linkage lead to low rotational speeds. Overshot/breastshot types can use small heads effectively. Because of their broad buckets and inertia, efficiency can remain reasonably steady across moderate discharge changes, though not as high or as flat as modern turbines. Step-by-Step Solution: Low head suitability: overshot/breastshot can operate on a few meters or less.Slow speeds: large wheels with direct drive rotate slowly compared to turbines.Efficiency variation: within typical operating ranges, efficiency is fairly stable though absolute peak is lower than Francis/Kaplan. Verification / Alternative check: Historical performance records and modern reconstructions show efficiencies in the 60–80% range depending on type, relatively constant over moderate flow fluctuations. Why Other Options Are Wrong: Each of (a), (b), and (c) is broadly correct in context; “none of the above” is therefore wrong. Common Pitfalls: Comparing water wheels directly to high-specific-speed turbines; ignoring that severe flow variations still reduce efficiency. Final Answer: all of the above

Hydraulic Machines Questions