Question 1

Francis Turbine Runner — Typical Number of Blades In a Francis turbine, the runner generally has how many blades?

Accepted Answer

Introduction:Runner blade count influences flow guidance, hydraulic losses, cavitation behavior, and structural integrity in reaction turbines. Francis turbines, which operate at medium heads, adopt a blade count that balances efficiency and manufacturability.Given Data / Assumptions:Francis runner with stay and guide vanes directing flow to runner blades.Standard industrial designs for medium specific speeds.Concept / Approach:Practical designs commonly employ several dozen blades for smooth guidance and to limit relative eddies. Too few blades increase incidence losses; too many raise surface friction and cost. The usual design window is approximately 16 to 24 blades for typical Francis turbines.Step-by-Step Solution:Assess flow guidance needs at part and full load.Select blade count to manage incidence and secondary flows.Adopt the widely used range 16–24 for general-purpose Francis runners.Verification / Alternative check:Manufacturer catalogues and design texts cite blade numbers in this range for standard sizes.Why Other Options Are Wrong:2–4 and 4–8: far too few for proper reaction guidance.8–16: lower bound is used in some high-specific-speed or special designs but typical mainstream range is higher.Common Pitfalls:Assuming blade counts similar to Pelton buckets or Kaplan blades; each turbine family differs.Final Answer:16 to 24

Question 2

Pelton Wheel — Recommended Bucket Depth For a Pelton wheel, the typical depth of a bucket is approximately how many times the jet diameter?

Accepted Answer

Introduction:Geometric proportions of Pelton buckets (width, depth, splitter dimensions) are standardized to capture and deflect the jet effectively while minimizing losses and avoiding jet interference.Given Data / Assumptions:Impulse turbine with free jet striking double-cupped buckets.Design guidance from classical proportions used in practice.Concept / Approach:Bucket depth affects jet containment and deflection angle. Too shallow causes spillage; too deep increases friction and weight. Empirical design ratios recommend bucket depth about 1.2 * d, where d is jet diameter, balancing capture and hydraulic efficiency.Step-by-Step Solution:Choose bucket dimensions based on jet size.Adopt depth ≈ 1.2 * d to house the jet adequately.Verify clearance to avoid interference with adjacent buckets.Verification / Alternative check:Reference proportion sets also suggest bucket width ≈ 5 * d and pitch ≈ 2.3 * d, consistent with the chosen depth.Why Other Options Are Wrong:equal to: marginal capture; prone to spillage.1.8 times and double: unnecessarily deep, increasing losses and mass.Common Pitfalls:Using nonstandard ratios that degrade hydraulic efficiency.Final Answer:1.2 times

Question 3

Jet on Fixed Plate — Normal Impact What is the force exerted by a water jet (normal impact) on a fixed flat plate in terms of specific weight w (N/m^3), jet area a (m^2), and jet velocity V (m/s)?

Accepted Answer

Introduction:The impact force of a jet on surfaces follows directly from Newton's second law applied to a control volume. Using specific weight instead of density is common in hydraulic machinery problems, necessitating careful unit conversions.Given Data / Assumptions:Jet strikes the plate normally and comes to rest tangentially (no rebound).Steady incompressible flow, neglecting air effects and splashing losses.Specific weight w = rho * g.Concept / Approach:Force equals rate of change of momentum. For normal impact on a fixed plate, the outgoing normal velocity is approximately zero, so the momentum reduction equals rho * a * V * V. Express density as rho = w / g to obtain the force in terms of w.Step-by-Step Solution:Mass flow rate: m_dot = rho * a * V = (w / g) * a * V.Momentum change per second: m_dot * V = (w / g) * a * V^2.Hence, force on plate: F = (w / g) * a * V^2.Verification / Alternative check:Dimensional check: w/g has units of kg/m^3; multiplied by a * V^2 yields N.Why Other Options Are Wrong:w * a * V: missing one factor of V and g; incorrect units.(w * a * V^2) / (2 * g): the 1/2 factor applies to energy, not momentum force in this case.(w * V) / a: dimensionally inconsistent.Common Pitfalls:Mixing energy head expressions (V^2 / 2g) with momentum-based force calculations.Final Answer:F = (w / g) * a * V^2

Question 4

Turbine Efficiencies — Relationship What is the correct relationship among hydraulic efficiency (eta_h), mechanical efficiency (eta_m), and overall efficiency (eta_o) of a turbine?

Accepted Answer

Introduction:Overall efficiency measures how effectively a turbine converts water power at inlet to shaft power at the coupling. It combines hydraulic conversion inside the runner and mechanical transmission to the shaft.Given Data / Assumptions:Definitions: eta_h = runner output / water power at inlet, eta_m = shaft output / runner output, eta_o = shaft output / water power at inlet.Steady operation; negligible measurement errors.Concept / Approach:Overall efficiency is the product of individual stage efficiencies along the energy path from fluid to shaft. Multiplying eta_h and eta_m cancels the intermediate runner output term, leaving the ratio of final to initial power.Step-by-Step Solution:eta_h = P_runner / P_water.eta_m = P_shaft / P_runner.eta_o = (P_shaft / P_runner) * (P_runner / P_water) = eta_m * eta_h.Verification / Alternative check:Energy balance across turbine confirms product form when losses in the runner and mechanical train are independent.Why Other Options Are Wrong:eta_h = eta_o * eta_m or eta_m = eta_o * eta_h: algebraically incorrect rearrangements.none of these: incorrect because the product relation is standard.Common Pitfalls:Confusing volumetric or generator efficiencies with hydraulic and mechanical efficiencies; those would extend the product further.Final Answer:eta_o = eta_h * eta_m

Question 5

Impulse Turbines — Maximum Hydraulic Efficiency What is the maximum hydraulic efficiency of an ideal impulse turbine (Pelton-type) in terms of the blade outlet angle phi (measured between relative exit velocity and the direction opposite to wheel mo…

Accepted Answer

Introduction:Hydraulic efficiency in an impulse turbine measures how much of the jet's kinetic energy is converted into useful work on the runner. The outlet blade angle controls the fraction of velocity effectively reversed, thus governing efficiency.Given Data / Assumptions:Ideal conditions: smooth buckets, no shock at entry, no losses.Jet speed V, wheel speed u, outlet relative velocity leaves the bucket at angle phi to the negative direction of motion.Single jet; standard Pelton geometry.Concept / Approach:For maximum hydraulic efficiency, the optimal speed ratio u / V is 1 / 2. Substituting this into the efficiency expression based on whirl velocity change leads to a compact formula depending only on phi. As phi decreases (sharper deflection), efficiency approaches unity, but practical values use phi around 15 degrees to avoid jet interference.Step-by-Step Solution:Work per unit mass = Vw1 * u - Vw2 * u with ideal deflection relations.Set u / V = 1 / 2 for optimal speed.Simplify hydraulic efficiency to eta_h(max) = (1 + cos phi) / 2.Verification / Alternative check:For phi = 0 (complete reversal), eta_h(max) = 1; for phi = 90 degrees (no reversal), eta_h(max) = 0, matching physical intuition.Why Other Options Are Wrong:cos phi: overestimates or underestimates depending on phi; not derived from optimal speed ratio.1 - cos phi and (1 - cos phi) / 2: correspond to different energy partitions, not the Pelton optimum.Common Pitfalls:Mixing up relative and absolute velocity angles; phi is defined at outlet relative to the bucket.Final Answer:eta_h(max) = (1 + cos phi) / 2

Question 6

Turbine Specific Speed — Definition Which expression correctly defines the specific speed (Ns) of a turbine in terms of rotational speed N (rpm), power P, and head H (m)?

Accepted Answer

Introduction:Specific speed is a dimensionless (or quasi-dimensional in rpm-based forms) index used to classify turbines and select appropriate types for a site. It reflects how speed and power scale with head for dynamically similar machines.Given Data / Assumptions:Conventional metric definition with N in rpm, P in kW (or W with consistent constants), and H in meters.Geometric similarity and dynamic similarity assumptions.Concept / Approach:By definition, specific speed is the speed at which a geometrically similar turbine would run to develop unit power under unit head. Rearranging similarity relations yields Ns ∝ N * P^0.5 / H^1.25, the widely used selection formula.Step-by-Step Solution:Start from condition: H = 1 m and P = 1 unit for the similar machine.Apply similarity: N ∝ H^0.5 and P ∝ H^1.5 for turbines.Derive Ns = N * sqrt(P) / H^(5/4).Verification / Alternative check:Type ranges: Pelton low Ns, Francis medium, Kaplan high; computed Ns sorts sites appropriately.Why Other Options Are Wrong:N * P / H and N / H^(3/2): wrong exponents; do not match similarity scaling.N * P^(3/2) / H^(1/2): incorrect power exponent; not dimensionally consistent.Common Pitfalls:Confusing pump specific speed formulas with turbine formulas; although similar, constants and exponents differ.Final Answer:Ns = N * sqrt(P) / H^(5/4)

Question 7

Hydraulic turbines — definition of hydraulic efficiency (η_h) For a reaction turbine operating under steady conditions, hydraulic efficiency is defined as the ratio of what quantity to what reference energy? Choose the most standard textbook definit…

Accepted Answer

Introduction:Hydraulic efficiency is a key performance indicator for water turbines. It compares how effectively the hydraulic energy delivered by the water is converted into useful work on the runner (wheel). For reaction turbines, both pressure and kinetic energy changes occur across the runner, so the proper definition must reference the energy actually available at the runner inlet.Given Data / Assumptions:Steady, incompressible flow of water.Losses may occur in nozzles, guide apparatus, and within the runner.We are isolating the runner to define hydraulic efficiency.Concept / Approach:Let E_in be the specific hydraulic energy available at the runner entry plane (including appropriate pressure head and velocity head components relative to exit). The useful hydraulic work rate on the runner equals the runner torque times angular speed, which is the energy transfer predicted by Euler’s turbine equation. Hydraulic efficiency is therefore η_h = (work done on runner) / (hydraulic energy supplied to runner). This definition excludes mechanical losses (bearings, seals) and volumetric slip; those are captured by other efficiencies (mechanical, volumetric, overall).Step-by-Step Solution:Identify numerator: work rate on the runner (wheel) from fluid to shaft = torque * angular speed.Identify denominator: hydraulic power at runner inlet = ρgQ*H_available at runner (or its equivalent energy flux).Form definition: η_h = (work done on the wheel) / (energy actually supplied to the turbine (runner)).Verification / Alternative check:In textbooks, “hydraulic efficiency” explicitly omits mechanical losses and leakage; hence it cannot be “shaft power / water power at penstock.” That latter ratio corresponds to overall efficiency when combined with volumetric losses.Why Other Options Are Wrong:Power produced / energy actually supplied: mixes rate with energy basis ambiguously and often implies overall efficiency.Actual work available to the turbine / energy imparted to the wheel: circular and imprecise.None of the above: incorrect because the standard definition is available among the options.Mechanical shaft power / water power: that is overall efficiency, not hydraulic efficiency.Common Pitfalls:Confusing hydraulic, mechanical, and overall efficiencies; using penstock head instead of runner-inlet available head; ignoring reaction effects through the runner.Final Answer:work done on the wheel to the energy (head of water) actually supplied to the turbine

Question 8

Unit speed of a Pelton wheel — compute using head scaling A Pelton wheel develops 1750 kW under a head of 100 m while running at 200 r.p.m. and discharging 2.5 m^3/s. Determine the unit speed (speed under 1 m head for a geometrically similar wheel).

Accepted Answer

Introduction:“Unit quantities” (unit speed, unit discharge, unit power) normalize turbine performance to a head of 1 m. They are widely used to compare and scale turbines of similar geometry. For turbines, speed scales with the square root of head, enabling quick predictions at different heads.Given Data / Assumptions:Pelton wheel (impulse turbine) running at N = 200 r.p.m.Head H = 100 m.Discharge Q = 2.5 m^3/s; Power P = 1750 kW (extraneous for unit speed).Geometric similarity and dynamic similarity for the unit conversion.Concept / Approach:For all hydraulic turbines, unit speed N_u (sometimes written N_1) is defined by N_u = N / sqrt(H). This follows from similarity laws: runner peripheral speed u ∝ sqrt(gH) and rotor diameter D fixed for a given machine during normalization. Therefore, to reduce the head from H to 1 m, divide the actual speed by sqrt(H).Step-by-Step Solution:Write definition: N_u = N / sqrt(H)Substitute: N_u = 200 / sqrt(100) = 200 / 10 = 20 r.p.m.Hence, the unit speed is 20 r.p.m.Verification / Alternative check:Dimensional check: rpm divided by sqrt(m) yields rpm for 1 m. Also, industry tables show Pelton unit speeds typically in the tens of rpm, which matches 20 r.p.m. for this head and speed combination.Why Other Options Are Wrong:10 r.p.m.: Corresponds to division by 20, not sqrt(100).40 or 80 r.p.m.: Would require multiplying by sqrt(H) or halving incorrectly.5 r.p.m.: Far too low; not consistent with scaling.Common Pitfalls:Using H instead of sqrt(H); confusing unit speed (normalize by head only) with specific speed (depends on both power and head: N_s ∝ Nsqrt(P)/H^(5/4)).Final Answer:20 r.p.m.

Question 9

Pelton wheel — classify by flow type and action Identify the correct classification of a Pelton turbine based on the nature of flow and energy conversion.

Accepted Answer

Introduction:Pelton wheels are classic high-head turbines. Understanding whether they are impulse or reaction type, and the direction of flow relative to the runner, is fundamental for selection and design in hydroelectric systems.Given Data / Assumptions:Water issued from one or more nozzles strikes spoon-shaped buckets.Jet is at atmospheric pressure at impact; pressure changes across blades are negligible.Flow is primarily tangential to the runner periphery.Concept / Approach:Impulse turbines convert the entire pressure head into jet kinetic energy before the runner. The runner then extracts energy by deflecting the jet, ideally with negligible pressure change across the buckets. Directionally, the Pelton jet strikes tangentially at the bucket splitter and is reversed to maximize momentum change and torque.Step-by-Step Solution:Identify energy conversion: pressure → velocity in nozzle (impulse)Check flow direction relative to runner: jet is tangentialThus: Pelton = tangential flow impulse turbineVerification / Alternative check:Reaction turbines (Francis, Kaplan) operate with pressure change within runner passages; Pelton does not. Radial-inflow impulse types are historical (e.g., Turgo uses oblique jet) but Pelton specifically uses tangential jets.Why Other Options Are Wrong:Inward/outward flow impulse: not the standard Pelton configuration.Inward/axial flow reaction: imply pressure change across runner, which is not the Pelton principle.Common Pitfalls:Confusing Pelton (impulse) with Francis/Kaplan (reaction); assuming “tangential” means axial—tangential is along the periphery, not the axis.Final Answer:tangential flow impulse turbine

Question 10

Pelton wheel nozzle selection by specific speed Evaluate the statement: “A Pelton wheel with one nozzle is preferred for specific speed between 35 and 60 (metric r.p.m. scale).”

Accepted Answer

Introduction:Pelton turbine selection often uses specific speed to decide the number of jets. Lower specific speeds usually require single-jet Peltons, while higher specific speeds push designers to use multiple jets for practical runner diameters and speeds.Given Data / Assumptions:Specific speed in the metric system (commonly N_s in r.p.m. with P in kW and H in m).Standard selection charts for Pelton jets.Concept / Approach:Typical guidelines: single-jet Peltons are favored at lower specific speeds (roughly up to the 30–35 range). As specific speed increases (e.g., 35–60), multiple-jet Peltons (two or more jets) are preferred to achieve feasible rotational speeds and runner sizes without exceeding jet velocity or nozzle limits.Step-by-Step Solution:Interpret the claim: single nozzle for 35–60Compare with standard ranges: single jet more typical below ~35; 35–60 more suited to 2-jet arrangementsHence the statement is falseVerification / Alternative check:Design handbooks frequently show: N_s low → 1 jet; moderate (35–60) → 2 jets; higher still → 3–6 jets. While exact cutoffs vary slightly by source, the trend is consistent.Why Other Options Are Wrong:True variants: contradict standard jet-number selection charts.Insufficient information: specific speed by itself is, in practice, adequate to suggest jet count ranges.Common Pitfalls:Assuming a single-jet Pelton suits all heads; ignoring the effect of specific speed on feasible runner diameter and speed limits.Final Answer:False

Question 11

Centrifugal pump affinity laws — delivery head vs rotational speed How does the delivery head H vary with speed N for a geometrically similar centrifugal pump operating within its normal range?

Accepted Answer

Introduction:Pump scaling and performance prediction rely on the affinity (similarity) laws. These laws relate flow rate, head, and power to impeller rotational speed for similar pumps handling the same fluid.Given Data / Assumptions:Geometrically similar pump and impeller.Same fluid, similar efficiency region.Moderate changes in N (no cavitation, adequate NPSH).Concept / Approach:The classic affinity laws state: Q ∝ N, H ∝ N^2, and P ∝ N^3. These come from dynamic similarity with velocity triangles scaling with peripheral speed u ∝ ND and energy head H ∝ u^2/g ∝ N^2.Step-by-Step Solution:Start with u = πD*N/60 ⇒ u ∝ N (D constant)Energy head contributed ∝ u^2/g ⇒ H ∝ N^2Therefore, delivery head H scales directly with the square of rotational speedVerification / Alternative check:Manufacturer curves often show that near the best-efficiency point, small speed changes follow these proportionalities closely. Deviations occur far from similarity conditions.Why Other Options Are Wrong:H ∝ N: underestimates head rise with speed.Inverse relations: contradict similarity theory.Independent of N: physically incorrect for rotodynamic machines.Common Pitfalls:Applying affinity laws outside their valid range; ignoring changes in efficiency and cavitation limits at significantly different speeds.Final Answer:directly proportional to N^2

Question 12

Specific speed of a hydraulic turbine — parameter dependence On which parameters does the specific speed (metric) of a hydraulic turbine depend?

Accepted Answer

Introduction:Specific speed is a dimensionless-like index (in a chosen unit system) used to characterize and select turbines. It indicates which turbine type (Pelton, Francis, Kaplan) is most suitable for a given head and power condition.Given Data / Assumptions:Metric definition commonly used in textbooks: N_s = Nsqrt(P)/H^(5/4).N in r.p.m., P in kW, H in m (units chosen to match metric N_s).Concept / Approach:From the definition N_s = Nsqrt(P)/H^(5/4), it is clear that specific speed depends on rotational speed N, power P, and head H simultaneously. It does not directly depend on discharge Q in this common form, although P itself is related to Q and H via P ≈ ρgQHη.Step-by-Step Solution:Write the standard expression: N_s = Nsqrt(P)/H^(5/4)Identify parameters: N, P, HConclude: depends on speed, power, and headVerification / Alternative check:Selection charts plot turbine type versus specific speed. Pelton occupies low N_s, Francis mid, Kaplan high—consistent with dependencies on N, P, H.Why Other Options Are Wrong:Speed and power only: ignores head, which is pivotal.Discharge + power: H is missing; the conventional formula uses H explicitly.Speed and head only: excludes power, which determines size and number of runners.Head and discharge only: not the textbook specific-speed form.Common Pitfalls:Confusing specific speed with unit quantities; assuming Q appears explicitly—while P encapsulates Q via P ≈ ρgQH*η.Final Answer:speed, power developed and head of water

Question 13

Power required to drive a centrifugal pump — correct formula Select the correct expression for input power P (watts) in terms of Hm (manometric head, m), w (specific weight, N/m^3), Q (discharge, m^3/s), and ηo (overall efficiency).

Accepted Answer

Introduction:To size a motor for a centrifugal pump, we must relate the hydraulic power imparted to the fluid to the electrical/mechanical input power, accounting for overall efficiency (volumetric, hydraulic, mechanical losses lumped together).Given Data / Assumptions:Incompressible fluid (water) with specific weight w (N/m^3).Flow rate Q (m^3/s) against manometric head Hm (m).Overall efficiency ηo includes all losses.Concept / Approach:Hydraulic output power (water power) equals w * Q * Hm. Since not all input power is converted to useful hydraulic power, input power must be higher by a factor 1/ηo. Thus P_input = (w * Q * Hm) / ηo.Step-by-Step Solution:Compute hydraulic power: P_water = w * Q * HmRelate to input: P_input = P_water / ηoTherefore: P = (w * Q * Hm) / ηoVerification / Alternative check:Dimensionally, w (N/m^3) * Q (m^3/s) * Hm (m) gives Nm/s = W, confirming correctness.Why Other Options Are Wrong:wQHm: omits losses; gives water power, not input.(wQHm)ηo: would be less than water power, nonphysical for ηo (ρQHm)/ηo: uses density instead of specific weight; missing factor g.(wQ)/(Hmηo): incorrect placement of head.Common Pitfalls:Mixing ρ with w; forgetting to divide by overall efficiency; confusing manometric head with static head only.Final Answer:(w * Q * Hm) / ηo

Question 14

Reciprocating pump with air vessel — percentage saving of work By fitting an air vessel to a reciprocating pump, work and power savings occur due to flow smoothing. What is the typical percentage saving for a double-acting reciprocating pump?

Accepted Answer

Introduction:Air vessels are used on suction and delivery lines of reciprocating pumps to reduce acceleration head, smoothen discharge, and lower the fluctuating power requirement. The percentage saving differs between single-acting and double-acting pumps because the flow pulsation patterns differ.Given Data / Assumptions:Double-acting reciprocating pump.Idealized analysis of acceleration head reduction with properly sized air vessels.Concept / Approach:The principal benefit is the reduction of periodic acceleration and deceleration of the fluid column. This reduces pressure fluctuations and the corresponding work over a cycle. Standard results show that for single-acting pumps the saving is higher, whereas for double-acting pumps it is approximately half that magnitude due to inherently smoother flow.Step-by-Step Solution:Recall typical results: single-acting ≈ 84.8% savingFor double-acting: ≈ 48.8% savingThus, select 48.8% for a double-acting pumpVerification / Alternative check:These percentages appear consistently in standard fluid machinery texts. Practical installations show significant reduction in pressure pulsations and motor size for the same duty.Why Other Options Are Wrong:39.2% / 24.4%: lower than the accepted value for double-acting arrangements.84.8% / 88.4%: align more with single-acting results or exceed typical theoretical values.Common Pitfalls:Mixing up single- and double-acting savings; assuming air vessels eliminate all fluctuations (they reduce, not remove, pulsations).Final Answer:48.8%

Question 15

Application domain — when to use an impulse turbine Select the head range for which impulse turbines (e.g., Pelton) are generally preferred in hydroelectric installations.

Accepted Answer

Introduction:Hydraulic turbine selection depends primarily on available head and discharge. Impulse turbines like the Pelton convert pressure head entirely to jet velocity before striking the runner, making them ideal where head is high and discharge is relatively low.Given Data / Assumptions:High elevation drop (tens to hundreds of meters).Relatively low flow rates compared to low-head sites.Stable water supply or controllable nozzles to modulate power.Concept / Approach:At high heads, nozzle velocities become large (v ≈ sqrt(2gH)), enabling efficient energy transfer by impulse action. Reaction turbines (Francis, Kaplan) are typically preferred for medium and low heads, respectively, to keep runner sizes practical and maintain efficiency.Step-by-Step Solution:Map head ranges to turbine families: Pelton (high), Francis (medium), Kaplan/Propeller (low)Consider discharge: impulse types suit lower Q at high HConclude: impulse turbines are used for high headVerification / Alternative check:Design manuals and selection charts consistently place Pelton at high head–low discharge corner of the selection map.Why Other Options Are Wrong:Low/medium head: usually reaction machines are better suited.High discharge (as an option alone): discharge magnitude alone does not define the type; high Q with low H favors Kaplan/propeller.Ultra-low head: specialized low-head turbines are used.Common Pitfalls:Assuming turbine type depends only on discharge; ignoring cavitation constraints and runner diameter practicality at given head.Final Answer:high head of water

Question 16

Flow direction in a centrifugal pump — entry vs exit Complete the statement: In a centrifugal pump, the water enters the impeller ________ and leaves the vanes axially (at design point).

Accepted Answer

Introduction:Centrifugal pumps are rotodynamic machines that add energy to the fluid by imparting tangential velocity through rotating blades, converting this to pressure in the volute or diffuser. Understanding inlet and outlet flow directions is essential for interpreting velocity triangles and avoiding inlet pre-swirl problems.Given Data / Assumptions:Proper inlet design with negligible pre-swirl (guide vanes may or may not be present depending on pump type).Design operation near the best-efficiency point.Concept / Approach:At the impeller eye, the fluid is guided to enter radially (relative velocity directed into blade passages with minimal whirl). The impeller then imparts tangential momentum, and through diffuser/volute action, flow is guided such that the absolute discharge direction can be arranged axially (for mixed/axial diffuser arrangements) or with controlled whirl depending on design. The provided statement uses the common teaching idealization: radial entry, axial leaving from vanes (in the absolute sense) at the design condition of zero whirl at outlet in certain pump designs with appropriate diffuser/guide geometry.Step-by-Step Solution:At inlet: arrange near-zero whirl ⇒ absolute velocity nearly radialThrough the impeller: tangential momentum is impartedAt outlet: diffuser/volute guide to an axial direction (idealized statement)Verification / Alternative check:Velocity triangles at design point often target zero whirl (Vw2 ≈ 0) for certain designs, leading to axial absolute exit in diffuser passages.Why Other Options Are Wrong:Axially (at inlet): contradicts the standard radial-entry assumption for typical centrifugal impellers.Tangentially opposite to rotation: describes whirl components, not the ideal entry direction.Spiral path: qualitative and ambiguous.Pre-swirl same as rotation by default: undesirable unless specifically introduced; increases NPSH requirement.Common Pitfalls:Confusing relative, absolute, and blade angles; assuming the same direction at all operating points; ignoring inlet pre-swirl control.Final Answer:radially

Question 17

Centrifugal Pump Casings and Flow Recovery In centrifugal pump design for clear liquids, the impeller can be paired with different casing arrangements to control velocity-to-pressure conversion and hydraulic losses. Identify which casing arrangement…

Accepted Answer

Introduction:Centrifugal pumps convert rotor-imparted velocity head into useful pressure head. The casing determines how efficiently this conversion happens. This question checks recognition of common casing types used with radial-flow impellers and why all of them are valid in appropriate contexts.Given Data / Assumptions:Radial-flow, single-stage centrifugal pump with closed or semi-open impeller.Newtonian liquid; steady, incompressible flow.Focus on hydraulic energy conversion and loss control within the casing.Concept / Approach:The impeller discharges high-velocity fluid. The casing must diffuse this velocity to pressure with minimal loss. Three arrangements are standard: volute (collects flow in an expanding spiral), diffuser or guide-vane casing (stationary vanes form passages to diffuse flow more gradually), and vortex/whirlpool casings (reduce eddy losses by shaping a secondary chamber). All are legitimate choices depending on duty, efficiency targets, and cost.Step-by-Step Solution:Volute: As area increases along the spiral, local velocity falls and static pressure rises; simple, robust, widely used.Diffuser (volute with guide blades): Stationary guide vanes recover pressure more efficiently at the expense of complexity and tighter clearances.Vortex casing: Introduces a vortex chamber to smooth out whirl and reduce shock/eddy losses near the tongue region.Conclusion: All three are used with centrifugal impellers depending on design priorities.Verification / Alternative check:Standard pump handbooks list these casing families and discuss their trade-offs in efficiency, head–capacity curve shape, and manufacturing cost.Why Other Options Are Wrong:Each of A, B, C singly is correct, but the most complete answer is that any of them can be used in practice.Common Pitfalls:Assuming only volutes are “real” casings. Diffusers are common in multi-stage pumps; vortex features appear in designs targeting smoother operation.Final Answer:any one of these

Question 18

Model–Prototype Similarity — Kinematic Similarity Criterion Kinematic similarity between a scale model and its prototype exists when (in addition to geometric similarity) the ratios of velocities and accelerations at corresponding points are equal a…

Accepted Answer

Introduction:Fluid model testing relies on similarity principles to extrapolate results from small-scale models to full-scale prototypes. Kinematic similarity ensures motion patterns—streamlines, velocity directions, and unsteady behavior—are dynamically consistent after applying the scale factors. This question clarifies the precise condition for kinematic similarity.Given Data / Assumptions:Incompressible or weakly compressible flow within the relevant regime.Model and prototype are geometrically similar (same shape; scaled size).We compare fields at corresponding non-dimensional positions and times.Concept / Approach:Kinematic similarity demands that the velocity and acceleration vectors at corresponding points are proportional by constant scale factors. Equivalently, dimensionless velocity and acceleration fields must match, implying equal Reynolds- or Froude-based kinematics for the chosen dominant physics, once geometry is matched. It is weaker than dynamic similarity (which equates force ratios) and stronger than geometric similarity alone.Step-by-Step Solution:Require geometric similarity so that corresponding points are well-defined.Impose velocity ratio V_m / V_p = constant at corresponding points (including direction consistency).Impose acceleration ratio a_m / a_p = constant at corresponding points and times.If the governing dimensionless groups controlling kinematics (e.g., Reynolds for viscous dominance, Froude for gravity waves) are matched, these ratio conditions are satisfied.Verification / Alternative check:In open-channel flows dominated by gravity, matching Froude number preserves free-surface kinematics; in viscous internal flows, matching Reynolds number preserves the velocity profile shapes.Why Other Options Are Wrong:Equal size and shape: geometric similarity only, insufficient.Identical absolute velocities: unnecessary and generally impossible under scaling.Equal resultant forces: dynamic similarity, a stronger requirement than kinematic.Only identical time scales: ignores geometric and velocity/acceleration requirements.Common Pitfalls:Confusing kinematic with dynamic similarity or believing absolute equality (not scaled equality) is required.Final Answer:Ratios of velocities and accelerations at corresponding points are equal (with geometric similarity satisfied).

Question 19

Cavitation Avoidance in Centrifugal Pumps — Suction Conditions To mitigate cavitation risk in a centrifugal pump, which inlet-side condition should be ensured for reliable operation at the design flow rate?

Accepted Answer

Introduction:Cavitation occurs when local static pressure in a pump falls to or below the liquid vapor pressure, forming vapor bubbles that collapse and damage components. This item checks understanding of inlet (suction) conditions required to avoid cavitation in centrifugal pumps.Given Data / Assumptions:Radial-flow centrifugal pump, steady operation.Liquid with known vapor pressure at operating temperature.Concept of Net Positive Suction Head (NPSH) is applicable.Concept / Approach:To avoid cavitation, the available NPSH at the pump suction must exceed the required NPSH of the pump by a safe margin. Practically, this means keeping the absolute suction pressure sufficiently high above the vapor pressure (reduce suction lift, minimize losses, lower temperature, or raise static head).Step-by-Step Solution:Define NPSH_available = (p_suction_absolute - p_vapor) / (rho * g) + V_suction^2 / (2 * g).For a given pump at a given flow, NPSH_required is specified by the manufacturer.Ensure NPSH_available ≥ NPSH_required + margin by raising suction head, reducing inlet losses, or lowering fluid temperature.Therefore, a “high” suction pressure (in absolute terms) is beneficial and required to suppress bubble formation.Verification / Alternative check:Field practices include placing pumps below supply tanks (flooded suction), using larger suction pipes, and avoiding throttling on the suction side—all aimed at maintaining higher inlet absolute pressure.Why Other Options Are Wrong:High delivery pressure/low delivery pressure: Discharge-side pressure does not directly fix inlet absolute pressure.Low suction pressure: Increases cavitation risk by reducing NPSH_available.Common Pitfalls:Confusing gauge and absolute pressures; cavitation depends on absolute pressure relative to vapor pressure, not on discharge head alone.Final Answer:the suction pressure should be high (sufficient NPSH available)

Question 20

Reciprocating Pumps — Suitability by Duty State whether the following is correct: “A reciprocating pump is most suitable for services requiring comparatively small discharge at relatively high heads.”

Accepted Answer

Introduction:Pumps are selected to match head–capacity requirements. Reciprocating pumps (piston/plunger) behave differently from centrifugal pumps in how head and discharge scale. This statement tests the typical application envelope for reciprocating pumps.Given Data / Assumptions:Single- or double-acting reciprocating machines.Incompressible liquids; steady average operation.Consider pulsation control via air vessels where needed.Concept / Approach:Reciprocating pumps are positive-displacement devices delivering nearly constant volume per stroke, enabling very high pressures (heads) at modest flow rates. Centrifugal pumps, in contrast, are favored for large discharges at moderate heads. Therefore, the given statement is correct for common engineering practice.Step-by-Step Solution:Identify duty: “less discharge and higher heads.”Map duty to pump families: PD reciprocating suits high head, low-to-moderate flow; centrifugal suits high flow, moderate head.Conclude statement is true for typical water and process duties.Verification / Alternative check:Manufacturer performance maps show reciprocating pumps achieving hundreds of meters of head at modest flows, while centrifugal selections for the same head often require multistage arrangements.Why Other Options Are Wrong:False: contradicts standard selection guidelines and the physics of positive displacement.Common Pitfalls:Overlooking pulsation; in practice, air vessels, dampeners, or accumulators are used to smooth discharge and reduce acceleration head losses.Final Answer:True

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