Question 1

Reciprocating pump: theoretical power requirement (watts) Given: specific weight of liquid w (N/m³), discharge Q (m³/s), suction head H_s (m), delivery head H_d (m). Write the theoretical power to drive the pump (neglecting loss…

Accepted Answer

Concept Pump power (ideal) = weight flow rate × total static head raised. Calculation Weight flow rate = w Q (N/s). Total head = Hs + Hd. Ptheoretical = w Q (Hs + Hd) (watts). Final Answerw Q (Hs + Hd)

Question 2

Reciprocating pump slip: definition based on discharges  Define the slip of a reciprocating pump using theoretical discharge and actual discharge. Pick the correct expression.  Choose the correct option.

Accepted Answer

Definition Slip = Qtheoretical − Qactual. Notes Slip fraction or percentage divides this difference by Qtheoretical and expresses it as a fraction or percent. Final AnswerDifference of theoretical discharge and actual discharge

Question 3

Pelton wheel unit power: compute from given head and output A Pelton wheel develops 1750 kW under a head of 100 m, runs at 200 r.p.m., and discharges 2500 L/s. Find the unit power of the wheel (power at unit head for a similar turbine). C…

Accepted Answer

Given data P = 1750 kW H = 100 m Unit power definition: Pu = P / H3/2 Step-by-step calculation H3/2 = 1003/2 = 100 × √100 = 100 × 10 = 1000 Pu = 1750 / 1000 = 1.75 kW Verification Unit quantities remove head effect; values of order 1–5 kW are typical for such data. Final Answer1.75 kW

Question 4

Turbine runner speed–head relation  For dynamically similar turbines, how does runner speed N vary with available head H? Assume unit-speed similarity criteria apply.  Choose the correct proportionality.

Accepted Answer

Concept Unit speed Nu = N / √H (constant for similar operation). Result Therefore N ∝ √H = H1/2. Final AnswerN ∝ H1/2

Question 5

Maximum propulsive efficiency for jet-propelled ship with side inlets  A ship uses jet propulsion with inlet orifices at right angles to its motion. What is the maximum possible propulsive efficiency for this configuration?  Choose the correct perce…

Accepted Answer

Concept Propulsive efficiency ηp = (useful power to ship) / (jet power), maximized when jet and ship speeds are optimally related. Result for side inlets For inlets at right angles to motion, analysis of momentum and energy shows a theoretical maximum ηp of 50%. Common pitfalls Confusing this with axial inlets or aircraft jets where the optimal values differ. Final Answer50%

Question 6

Centrifugal pump similarity: discharge vs. rotational speed  How does the discharge Q of a centrifugal pump vary with impeller rotational speed N (all else equal)?  Choose the correct relation.

Accepted Answer

Concept For a given geometry and diameter, the flow (via Vf) scales with tip speed ∝ N, so Q ∝ area × Vf ∝ N. Affinity law With constant diameter: Q ∝ N; Head H ∝ N²; Power P ∝ N³. Final AnswerQ ∝ N

Question 7

Centrifugal pump similarity: discharge vs. impeller diameter  For a given speed and inlet width, how does discharge Q vary with impeller diameter D in a centrifugal pump?  Choose the correct dependence.

Accepted Answer

Concept At inlet, Q = π D b Vf. For fixed b and similar Vf (same N), Q scales with D. Notes Under full similarity (b ∝ D and Vf ∝ D N), Q would scale as D³ N, but with b and N fixed as in many textbook MCQs, the direct proportionality to D is appropriate. Final AnswerDirectly proportional to D

Question 8

Fluid mechanics: definition of geometric similarity between model and prototype  Compare model and prototype for geometric similarity in hydraulic/thermal machinery modeling. Choose the correct statement that defines geometric similarity.

Accepted Answer

ConceptGeometric similarity means the model and prototype are the same shape; all corresponding linear dimensions are in a constant ratio (scale). Size may differ, but shape must be identical.Dynamic or kinematic similarity adds matching force/velocity ratios, but those are separate requirements.Why the other options are incorrectIdentical velocities or forces are aspects of kinematic or dynamic similarity, not strictly geometric.Equal size and shape implies the same object, not a scale model.Final AnswerIdentical in shape but different in size according to a constant scale.

Question 9

Turbomachinery performance parameter: unit discharge  For a turbine operating under head H with discharge Q, express the unit discharge (i.e., the discharge referred to a head of 1 m). Choose the correct formula.

Accepted Answer

GivenActual discharge through the turbine = Q.Operating head = H.ConceptUnit quantities are values reduced to a reference head of 1 m. For discharge, the scaling with head is proportional to √H (from continuity with velocity ∝ √H), hence:FormulaUnit discharge, Qu = Q / √HChecksIf H increases while geometry is similar, Q increases roughly with √H, so Q/√H stays about constant.Final AnswerQ / √H.

Question 10

Starting condition for a centrifugal pump to deliver liquid  Identify the correct head rise condition inside the impeller that must be met before flow is delivered to the discharge line. Choose the correct head term.

Accepted Answer

ConceptA centrifugal pump begins to deliver when the pressure rise produced by the impeller equals or exceeds the system's manometric head.Manometric headHmano includes static elevation difference (delivery minus suction) plus frictional and minor losses between suction and delivery gauge points.Why others are incorrectKinetic/velocity head are components inside the impeller/volute, but delivery starts when the full manometric requirement is met.Final AnswerManometric head.

Question 11

Hydrostatics principle in devices Which unit operates primarily on Pascal's law (pressure applied to a confined fluid is transmitted undiminished in all directions)?

Accepted Answer

Introduction / Context:Pascal’s law underpins many hydraulic machines where force multiplication is achieved through static pressure transmission in incompressible fluids. Recognizing which devices rely on hydrostatics versus hydrodynamics is essential for correct selection and analysis.Given Data / Assumptions:Incompressible fluid and sealed system for Pascal’s principle.Quasi-static operation where dynamic effects are negligible.Concept / Approach:Pascal’s law: a pressure increment applied to a confined fluid is transmitted equally to every point and to the walls of the container. A hydraulic press uses two pistons of different areas connected by fluid; equal pressure gives F_out / F_in = A_out / A_in, yielding mechanical advantage.Step-by-Step Solution:Apply an input force on the small piston creating pressure p = F_in / A_in.The same pressure acts on the large piston: F_out = p * A_out.Therefore F_out / F_in = A_out / A_in, demonstrating force amplification.Verification / Alternative check:Energy balance (ignoring losses): F_in * s_in ≈ F_out * s_out, consistent with the area ratio trade-off.Why Other Options Are Wrong:Air lift pump and jet pump exploit fluid momentum and entrainment (hydrodynamics). Hydraulic coupling transmits torque via circulating oil and relative motion, not static pressure multiplication. A venturimeter measures flow using Bernoulli’s principle, not Pascal’s law.Common Pitfalls:Confusing pressure transmission (static) with momentum transfer (dynamic). Assuming any “hydraulic” device uses Pascal’s law, which is not always true.Final Answer:Hydraulic press

Question 12

Head–power relationship for turbines If H is the net head on a turbine of fixed geometry operating at its best efficiency point, the developed power varies how with H?

Accepted Answer

Introduction / Context:Turbine similarity laws link discharge, runner speed, and power to the operating head. For a given turbine design running near its best efficiency point, these relations provide quick scaling rules useful for preliminary sizing and performance prediction.Given Data / Assumptions:Turbine geometry fixed; operating near best efficiency.Fluid is water, density approximately constant.Losses scale comparably so similarity holds.Concept / Approach:From similarity: specific speed and unit quantities give Q ∝ H^(1/2) and power P ∝ ρ g Q H (times efficiency). With efficiency roughly constant at BEP, this yields P ∝ H * H^(1/2) = H^(3/2).Step-by-Step Solution:Discharge scaling: Q ∝ H^(1/2).Hydraulic power: P_h = ρ * g * Q * H.Assume η ≈ constant near BEP; then P ≈ η * P_h ∝ Q * H.Hence P ∝ H^(1/2) * H = H^(3/2).Verification / Alternative check:Unit power P_u = P / H^(3/2) is approximately constant for dynamically similar operation, commonly used to compare turbines of different sizes.Why Other Options Are Wrong:H^(1/2) scaling applies to discharge or speed, not power. Inverse relations contradict energy scaling. Independence from H is physically untenable for hydraulic machines.Common Pitfalls:Mixing up unit speed/discharge with power; assuming efficiency changes negate the fundamental scaling (they modify constants but not the basic exponent near BEP).Final Answer:Directly proportional to H^(3/2)

Question 13

Pumps for viscous liquids Which pump type is most suitable for pumping highly viscous fluids in industrial service?

Accepted Answer

Introduction / Context:Fluid viscosity strongly influences pump selection. Dynamic (centrifugal) pumps lose efficiency and capacity as viscosity rises, while positive displacement pumps maintain flow with less slip, making them preferable for viscous services like oils, syrups, and slurries.Given Data / Assumptions:High viscosity fluids where Reynolds number is low.Steady flow requirements without excessive shear or heating.Concept / Approach:Screw and progressive-cavity pumps are positive displacement machines generating flow by trapping fluid in cavities and transporting it axially. Their performance is relatively insensitive to viscosity (within limits), unlike centrifugal pumps that rely on imparting kinetic energy and suffer large head losses at high viscosity.Step-by-Step Solution:Identify viscosity range: as viscosity increases, centrifugal efficiency drops and NPSH requirements rise.Positive displacement options (reciprocating, gear, screw) become favorable.Among them, screw/progressive-cavity pumps provide smooth, low-pulsation flow suitable for continuous processes.Therefore, the best general choice is a screw (or progressive cavity) pump.Verification / Alternative check:Manufacturer selection charts map acceptable viscosity ranges, consistently favoring PD pumps for high-viscosity applications.Why Other Options Are Wrong:Centrifugal and jet pumps are poor at high viscosity; air-lift depends on buoyancy with gas injection, not suitable for thick liquids; reciprocating pumps can handle viscosity but introduce pulsation and complex valving—screw pumps are generally preferred for steady viscous service.Common Pitfalls:Overlooking shear sensitivity of the product; ignoring suction limitations that demand careful NPSH margin for viscous feeds.Final Answer:Screw pump (progressive cavity/rotary screw)

Question 14

Pelton wheel: optimal speed ratio for maximum efficiency For a Pelton turbine operating at design head and discharge, the maximum efficiency occurs at which speed ratio (runner peripheral speed / jet velocity)?

Accepted Answer

Introduction / Context:Impulse turbines like the Pelton wheel extract energy from a high-velocity jet striking double-cupped buckets. The runner speed relative to jet speed—the speed ratio—controls the exit relative velocity and thus the work extracted per unit jet momentum.Given Data / Assumptions:Single-jet Pelton at design condition (nozzle coefficient and bucket losses typical).Jet deflection near 165–170 degrees with some relative speed loss in the buckets.Concept / Approach:The ideal analysis (neglecting losses) suggests maximum efficiency near a speed ratio of 0.5. Accounting for practical losses and incomplete flow deflection, empirical practice places the optimum slightly lower, about 0.46.Step-by-Step Solution:Let jet velocity be V_jet and runner peripheral speed u.Power ∝ mass flow * (V_jet - u) * (velocity change along jet direction).Differentiate efficiency with respect to u and set derivative to zero (including loss coefficients).Solution yields u / V_jet ≈ 0.46 for typical Pelton buckets.Verification / Alternative check:Design manuals commonly recommend 0.43–0.48 as the practical optimum; 0.46 is a standard textbook value.Why Other Options Are Wrong:0.26 and 0.36 under-speed the runner and waste jet kinetic energy; 0.56 and 0.66 over-speed the wheel, reducing relative jet deflection and work.Common Pitfalls:Using the ideal 0.5 without accounting for bucket losses; confusing specific speed trends of reaction turbines with Pelton settings.Final Answer:0.46

Question 15

Jet propulsion efficiency for ships with transverse intakes Statement: For a ship whose water intakes are at right angles to the direction of motion, the propulsive efficiency will be maximum when the relative velocity between the discharged jet and…

Accepted Answer

Introduction / Context:Water-jet propulsion produces thrust by accelerating a mass of water astern. Propulsive efficiency compares useful power (thrust * ship speed) to the rate of kinetic energy imparted to the jet plus losses. Maximum efficiency is achieved when the velocity increase of the water over the ship speed is minimized for a given thrust (large mass flow, small velocity rise).Given Data / Assumptions:Steady motion; intake orifices transverse to ship path.Incompressible flow, neglecting detailed intake and nozzle losses for the trend argument.Concept / Approach:Propulsive efficiency for a momentum jet can be expressed (in simplified form) as η_p = 2 * V_s / (V_s + V_j,rel), where V_s is ship speed and V_j,rel is jet speed relative to the ship. This monotonically increases as V_j,rel approaches V_s from above; it does not peak at 2 V_s. The closer the jet exit speed is to the ship speed (for sufficient mass flow to produce required thrust), the higher the efficiency.Step-by-Step Solution:Let relative jet speed be V_j,rel.Compute η_p trend: as V_j,rel decreases toward V_s, η_p increases toward a maximum (approaching 1 in the ideal limit).If V_j,rel = 2 V_s, η_p = 2 V_s / (3 V_s) = 2/3, which is below the maximum achievable trend.Therefore the statement claiming the maximum at twice the ship speed is false.Verification / Alternative check:Momentum theory of propulsors yields the same conclusion: for fixed thrust, raising jet velocity rather than mass flow reduces efficiency due to higher kinetic energy losses in the slipstream.Why Other Options Are Wrong:Answers asserting conditional yes are incorrect; maximum does not occur at 2 V_s under typical idealizations, regardless of depth or Froude number.Common Pitfalls:Confusing intake angle effects (mainly affecting losses) with the optimal jet/ship velocity ratio; assuming “faster jet is better,” which reduces efficiency.Final Answer:No

Question 16

Francis turbine: definition of flow ratio The flow ratio for a Francis turbine is defined as the ratio of which velocities?

Accepted Answer

Introduction / Context:The flow ratio is a non-dimensional parameter used in the design and analysis of reaction turbines, particularly the Francis turbine. It links the meridional (axial/radial) component of velocity at entry to a head-based reference speed and helps set guide vane openings and runner passage areas.Given Data / Assumptions:Inlet head H after accounting for losses relevant to √(2 g H).Velocity of flow (meridional component) at inlet denoted V_f1.Concept / Approach:By definition, flow ratio φ_f = V_f1 / √(2 g H). It differs from speed ratio (runner peripheral ratio) and from the whirl coefficient that involves the tangential component of velocity.Step-by-Step Solution:Compute theoretical jet speed from head: V_th = √(2 g H).Measure or design for inlet flow component: V_f1.Form ratio: φ_f = V_f1 / V_th.Use φ_f in preliminary sizing of guide vane settings and blade heights.Verification / Alternative check:Typical φ_f values for Francis designs fall in standard ranges used by manufacturers; comparing with case studies confirms the definition.Why Other Options Are Wrong:Inverting the ratio changes the non-dimensional group meaning. The runner peripheral ratio concerns blade speed, not meridional flow. “None of the above” is incorrect because option (a) is the standard definition.Common Pitfalls:Confusing flow ratio with speed ratio or with whirl velocity fraction; mixing inlet and outlet symbols.Final Answer:Velocity of flow at inlet to the theoretical jet velocity (√(2 g H))

Question 17

Hydraulic Turbines — Draft Tube Usage Is a draft tube used with impulse turbines such as the Pelton wheel, or is it associated with reaction turbines only?

Accepted Answer

Introduction:A draft tube is a gradually expanding conduit fitted at the outlet of reaction turbines. Its purpose is to convert part of the kinetic energy at runner exit into useful pressure energy and to allow installation of the turbine above the tailrace without losing head. The question tests whether this device is relevant to impulse turbines like Pelton wheels.Given Data / Assumptions:Impulse turbine (e.g., Pelton) exhausts water to atmosphere after striking buckets.Reaction turbines (e.g., Francis, Kaplan) discharge with significant velocity head at runner exit.Steady incompressible flow; standard textbook definitions of turbine components.Concept / Approach:Impulse turbines operate at essentially atmospheric pressure throughout the runner and extract energy solely from the change in jet momentum. There is no useful pressure recovery possible downstream; water simply falls into the tailrace. Reaction turbines, however, discharge at sub-atmospheric pressure with high exit velocity; a draft tube recovers kinetic energy by diffusion and raises the static pressure closer to tailrace level, improving overall efficiency.Step-by-Step Solution:Identify turbine type: impulse vs reaction.For impulse: runner exit is atmospheric; no pressure rise is feasible via a diffuser.For reaction: runner exit velocity is high; draft tube converts velocity head to pressure head.Conclusion: draft tube is not used with impulse turbines.Verification / Alternative check:Plant layouts show Pelton tailraces without draft tubes, while Francis/Kaplan runners connect to elbow or conical draft tubes.Why Other Options Are Wrong:Yes: contradicts operating principle of impulse machines.Only in high-head plants: head magnitude does not change the impulse principle.Only for speed control: draft tubes are not speed-control devices.Common Pitfalls:Assuming all water turbines require draft tubes; only reaction types do.Final Answer:No

Question 18

Reaction Turbines — Effect of Unit Speed on Discharge How does the discharge through a reaction turbine vary with an increase in unit speed (speed at unit head)?

Accepted Answer

Introduction:Unit quantities normalize turbine performance to a head of 1 m, enabling comparisons across heads and scales. Understanding how unit speed relates to flow helps in model testing, selection, and regulation of reaction turbines.Given Data / Assumptions:Reaction turbine operating under dynamic similarity.Definitions: unit speed Nu = N / H^0.5 and unit discharge Qu = Q / H^0.5.Geometrically similar settings and comparable guide-vane openings.Concept / Approach:From similarity laws, peripheral velocity u ∝ N * D and flow rate Q ∝ V_f * A. Under similitude at fixed geometry, characteristic velocities scale with H^0.5 and with shaft speed. Consequently, Qu correlates with Nu; increasing unit speed implies a proportionate tendency for unit discharge to rise.Step-by-Step Solution:Define Nu = N / H^0.5.Define Qu = Q / H^0.5.For similar operation, Qu ∝ Nu.Therefore, as unit speed increases, discharge increases correspondingly.Verification / Alternative check:Performance curves (Qu vs Nu) for reaction turbines typically show a rising trend within the stable operating zone.Why Other Options Are Wrong:remains same: contradicts similarity relations.decreases: opposite to observed and theoretical trend.depends only on gate opening: gate affects flow, but unit speed also influences it under similitude.Common Pitfalls:Confusing absolute speed changes with unit-normalized quantities; unit variables remove the head effect but keep speed influence.Final Answer:increases

Question 19

Centrifugal Pumps — Affinity Laws The power required to drive a centrifugal pump varies how with rotational speed N (other factors geometrically similar)?

Accepted Answer

Introduction:Pump affinity laws predict how head, flow, and power scale with speed for a given impeller and fluid. These relations are essential for estimating motor requirements and evaluating speed-control strategies.Given Data / Assumptions:Geometrically similar operation; same impeller and casing geometry.Same fluid, negligible change in efficiency over the considered range.Rotational speed N is varied.Concept / Approach:The affinity laws state: Q ∝ N, H ∝ N^2, and P ∝ N^3 for similar conditions. Intuitively, power combines flow (N) and head (N^2), giving N^3 overall. Thus, modest changes in speed can have large power implications.Step-by-Step Solution:Flow scaling: Q2 / Q1 = N2 / N1.Head scaling: H2 / H1 = (N2 / N1)^2.Power scaling: P2 / P1 = (Q2 / Q1) * (H2 / H1) ≈ (N2 / N1) * (N2 / N1)^2 = (N2 / N1)^3.Verification / Alternative check:Manufacturer performance curves shifted by speed align when scaled by the affinity laws.Why Other Options Are Wrong:inversely proportional to N^3: physically unreasonable; higher speed increases power draw.proportional to N or N^2: underestimates the cubic effect.Common Pitfalls:Ignoring efficiency changes at off-design, which can slightly alter the exact scaling but not the cubic trend.Final Answer:directly proportional to N^3

Question 20

Fixed Curved Vane — Jet Tangential Entry and Exit A jet of water enters and leaves a fixed curved vane tangentially. What is the force component of the jet perpendicular to the vane surface?

Accepted Answer

Introduction:Force on a vane depends on the change in momentum of the jet along and normal to the vane surface. If the jet meets and leaves tangentially, its velocity has no normal component at entry and exit, which has direct implications for the normal force component.Given Data / Assumptions:Fixed curved vane; steady, incompressible jet.Jet velocity direction coincides with vane tangent at entry and at exit.Losses neglected for the directional argument.Concept / Approach:By momentum principle, the force on the vane in any direction equals the rate of change of momentum of the jet in that direction. If the normal component of velocity is zero both at inlet and outlet, the change of normal momentum is zero; therefore, the normal (perpendicular) force is zero.Step-by-Step Solution:Resolve jet velocity into tangential and normal components relative to vane surface.At tangential entry: normal component = 0.At tangential exit: normal component = 0.Change of normal momentum = 0 thus normal force = 0.Verification / Alternative check:Free-body of control volume aligned with local normal shows no flux of normal momentum in or out.Why Other Options Are Wrong:False: contradicts the momentum change reasoning.Perpendicular force is maximum / equals density * V^2 * area: these apply when the jet strikes normally, not tangentially.Common Pitfalls:Confusing tangential force (which can be non-zero) with normal force (zero here).Final Answer:True

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