Question 1

In a centrifugal pump installation, where should the regulating (throttling) valve be located to control discharge safely without inducing cavitation at the inlet?

Accepted Answer

Introduction:Flow control in centrifugal pumps is commonly achieved by throttling. The location of the regulating valve is critical to avoid harmful low pressures that can cause cavitation and reduce pump life.Given Data / Assumptions:Standard end-suction centrifugal pump.Ambient water temperature; NPSH margin must be preserved.Goal: control Q while maintaining safe suction conditions.Concept / Approach:Throttling on the delivery side raises back pressure and shifts the operating point along the pump curve, reducing discharge with minimal risk to inlet pressure. Throttling on the suction side would lower inlet pressure, reduce NPSH available, and promote cavitation—therefore it must be avoided.Step-by-Step Solution:1) Install valve on delivery line downstream of pump.2) Adjust opening to meet required flow/head.3) Monitor discharge pressure and vibration to stay in stable operating region.4) Keep suction line as short, straight, and unobstructed as possible.Verification / Alternative check:NPSH analysis shows suction throttling reduces NPSH available, increasing cavitation risk; delivery throttling does not affect inlet static pressure adversely.Why Other Options Are Wrong:Casing/impeller: no such control element is placed there.Suction pipe: dangerous due to NPSH loss.Foot valve: non-return at sump, not used for regulation.Common Pitfalls:Using suction valves for flow control; neglecting the role of variable-speed drives as a superior alternative when available.Final Answer:delivery pipe

Question 2

Identify the turbine type that is not an impulse turbine from the list below (consider the primary energy conversion principle at the runner).

Accepted Answer

Introduction:Hydraulic turbines are broadly classified as impulse or reaction types. Impulse turbines extract energy solely through jet momentum at essentially atmospheric pressure in the runner, whereas reaction turbines experience a pressure drop across the runner passages.Given Data / Assumptions:Compare by operating principle, not geometry alone.Impulse examples: Pelton, Turgo, cross-flow (Banki, as commonly categorized in practice).Reaction examples: Kaplan (axial), Francis (mixed).Concept / Approach:Kaplan turbines are axial-flow reaction machines with adjustable runner blades and a draft tube, operating with substantial pressure change across the runner. In contrast, Pelton, Turgo, and Girard are classical impulse designs using free jets. Cross-flow turbines are generally treated as impulse machines because the runner operates with jets at near-atmospheric pressure, even though the flow path is through the runner interior.Step-by-Step Solution:1) Classify each listed turbine by principle.2) Identify which experiences pressure drop in the runner (reaction).3) Kaplan is reaction; others listed are impulse.Verification / Alternative check:Presence of draft tube and submerged outlet is a hallmark of reaction turbines like Kaplan.Why Other Options Are Wrong:Girad/Turgo/Pelton: classic impulse types.Banki (cross-flow): commonly treated as impulse; runner sees atmospheric pressure jets.Common Pitfalls:Confusing flow direction (axial vs radial) with impulse/reaction classification; they are related but not the same criterion.Final Answer:Kaplan turbine

Question 3

For feeding boilers (high head, reliable delivery), which pump type is most successfully employed in practice, especially for smaller to moderate capacities?

Accepted Answer

Introduction:Boiler feed service requires overcoming high pressures with dependable, often relatively modest flow rates. The pump must provide steady delivery against a steep system head with minimal slip and good controllability.Given Data / Assumptions:High discharge pressure (boiler drum/once-through systems).Flow rate small to moderate.Preference for precise metering and robust reliability.Concept / Approach:Reciprocating (positive-displacement) pumps generate high pressures efficiently at lower flow rates and maintain discharge largely independent of head. Although multistage centrifugal pumps are widely used for large utility boilers, for small-to-moderate capacities and very high heads the reciprocating type is traditionally successful and appears in many design handbooks and exam conventions as the correct choice.Step-by-Step Solution:1) Match duty: high head → positive displacement preferred.2) Evaluate options: reciprocating delivers precise, high-pressure flow.3) Conclude reciprocating pump is most suitable in the context of the question.Verification / Alternative check:Industry practice: small industrial boilers often use reciprocating or multistage centrifugal; examination standards typically key to reciprocating for “small discharge, high head.”Why Other Options Are Wrong:Centrifugal/axial flow: better for larger flows; require many stages for very high heads.Jet and air-lift pumps: inefficient for high-pressure boiler feed duty.Common Pitfalls:Assuming boiler feed is always centrifugal without noting capacity and head range.Final Answer:Reciprocating pump

Question 4

Specific speed (N_s) of a centrifugal pump helps compare geometrically similar pumps across sizes. Judge the statement: “N_s is the speed of a geometrically similar pump that would deliver one litre of water per second against a head of one metre.”

Accepted Answer

Introduction:Specific speed allows classification of pumps independent of size by tying speed, discharge, and head through similarity relationships. The correct reference discharge in the standard definition matters for dimensional consistency and practical comparison.Given Data / Assumptions:Metric/S.I. convention for centrifugal pumps.N_s relates to a pump that would deliver a standard discharge at unit head (H = 1 m).Concept / Approach:The conventional definition (metric) states: specific speed N_s is the speed of a geometrically similar pump which would deliver 1 m^3/s against a head of 1 m. Using “one litre” (1 L/s = 0.001 m^3/s) is incorrect and would change the numerical value by a factor of 1000. Hence the given statement is false as written.Step-by-Step Solution:1) Standard: N_s = N * sqrt(Q) / (H)^(3/4) (with consistent units).2) Unit head condition: H = 1 m.3) Reference discharge: Q = 1 m^3/s, not 1 L/s.4) Therefore, the statement using “one litre” is incorrect.Verification / Alternative check:Textbook tables and standards consistently cite 1 m^3/s at 1 m head for pump specific speed in the metric convention.Why Other Options Are Wrong:“True” variants contradict the standard definition; the kW-based form pertains to turbine N_s or power-specific speeds, not this definition.Common Pitfalls:Mixing litre and cubic metre bases; using turbine-style specific speed definitions for pumps.Final Answer:False

Question 5

In a reaction turbine operating under steady conditions, state whether the following statement is correct: “Water enters the runner under pressure and flows over the vanes with a pressure drop across the runner.”

Accepted Answer

Introduction:Reaction turbines (Francis, Kaplan, propeller) perform energy conversion partly through pressure changes within the runner passages. Recognizing that the runner works with a pressure gradient differentiates reaction turbines from impulse designs and explains the need for a draft tube.Given Data / Assumptions:Runner submerged, with guide (wicket) gates and a draft tube.Steady, incompressible flow.Design operation near typical efficiency conditions.Concept / Approach:In reaction turbines, a portion of the available head is converted to kinetic energy in the stay/guide passages, and an additional portion is converted across the runner via changes in relative velocity and static pressure. The outlet connects to a draft tube that recovers kinetic energy by further pressure rise, enabling sub-atmospheric pressure at the runner exit without cavitation (if properly designed).Step-by-Step Solution:1) Water enters the runner under pressure (above local vapor pressure by margin).2) As it passes over the vanes, static pressure decreases while mechanical energy is transferred to the shaft.3) Flow discharges into the draft tube where velocity head is partially recovered as pressure, raising tailrace-level efficiency.4) Thus, a pressure drop occurs across the runner, characteristic of reaction operation.Verification / Alternative check:Velocity triangles and Bernoulli between guide exit and runner exit show pressure change consistent with reaction degree > 0.Why Other Options Are Wrong:Incorrect / conditional options ignore the defining feature of reaction machines, which holds across operating points within the normal range.Common Pitfalls:Assuming turbine always works at atmospheric runner pressure (impulse behavior); neglecting the role of the draft tube in pressure recovery.Final Answer:Correct

Question 6

Hydraulic press: primary function and comparison with related devices In fluid power systems, a hydraulic press is a device used to lift or compress a much larger load by applying a comparatively small force through pressure transmission. Which stat…

Accepted Answer

Introduction / Context:A hydraulic press is a classic application of Pascal's law. It uses two cylinders with different areas connected by a fluid. A small input force on the small-area piston creates the same pressure throughout the fluid, which acts on the large-area piston to generate a much larger output force. This question checks whether you can distinguish a press from other hydraulic units such as accumulators and intensifiers.Given Data / Assumptions:Incompressible fluid and negligible leakage.Two pistons with areas A_small and A_large, hydraulically connected.Quasi-static operation where dynamic losses are minor.Concept / Approach:By Pascal's law, the pressure p applied on the small piston equals the pressure under the large piston. Therefore F_out / A_large = F_in / A_small, so F_out = F_in * (A_large / A_small). This gives a mechanical advantage equal to the area ratio, enabling a small effort to lift or press a heavy load.Step-by-Step Solution:Apply F_in on the small piston ⇒ p = F_in / A_small.The same p acts on the large piston ⇒ F_out = p * A_large.Therefore F_out = F_in * (A_large / A_small).Because A_large » A_small, a small force lifts a large load, which defines a hydraulic press.Verification / Alternative check:Energy balance (ignoring losses): F_in * s_in ≈ F_out * s_out. The larger force is achieved at the expense of reduced displacement, consistent with the area ratio.Why Other Options Are Wrong:Option (a) describes a hydraulic accumulator (energy storage). Option (b) describes a hydraulic intensifier (pressure multiplication). Option (d) groups different devices and is therefore incorrect for a press's principal use. Option (e) is a measuring device, not a press.Common Pitfalls:Confusing energy storage (accumulator) and pressure multiplication (intensifier) with force multiplication via area ratio (press). Mixing up force advantage with pressure increase.Final Answer:to lift a larger load by applying a comparatively much smaller force (hydraulic press)

Question 7

Reciprocating pump operation: condition for negative slip Slip of a reciprocating pump becomes negative (i.e., actual discharge exceeds theoretical discharge) under which practical condition?

Accepted Answer

Introduction / Context:Slip in reciprocating pumps is defined as theoretical discharge minus actual discharge. It is usually positive because of leakage and valve losses. However, under certain dynamic conditions the actual discharge momentarily exceeds the theoretical value, producing negative slip. Recognizing when this occurs is important for diagnosing performance anomalies.Given Data / Assumptions:Single- or double-acting reciprocating pump with delivery and suction valves.Significant inertia and frictional effects in the delivery line.High-speed operation relative to system time constants.Concept / Approach:At high speeds with a long delivery line, the moving water column possesses substantial momentum. During the suction stroke, the inertia of this column may continue pushing water through the delivery valve, so flow does not reverse as quickly as the piston motion would suggest. This makes the measured actual discharge exceed the kinematic theoretical discharge based on piston motion alone, yielding negative slip.Step-by-Step Solution:Define slip S = Q_theoretical − Q_actual.At high speed and with long delivery pipe, inertia sustains delivery flow even during part of the suction stroke.Therefore Q_actual > Q_theoretical ⇒ S Hence the condition is long delivery pipe combined with high pump speed.Verification / Alternative check:Indicator diagrams and instantaneous flow measurements show continued positive delivery flow beyond the theoretical cut-off in such cases, confirming negative slip behavior.Why Other Options Are Wrong:Short lines and low speeds minimize inertia effects, so slip stays positive. Long suction with high speed mainly aggravates suction problems (cavitation), not negative slip in delivery.Common Pitfalls:Assuming leakage always makes S positive; ignoring water hammer and fluid inertia in long pipelines.Final Answer:delivery pipe is long and the pump is running at high speeds

Question 8

Centrifugal pump terminology: manometric head Statement: The manometric head is the actual head of water against which a centrifugal pump has to work (including suction and delivery static heads and relevant losses).

Accepted Answer

Introduction / Context:For centrifugal pumps, various heads are defined to analyze performance: suction head, delivery head, static head, and manometric head. Understanding the manometric head is essential for comparing pump curves with system curves and selecting the right pump for an installation.Given Data / Assumptions:Steady operation with water as the working fluid.Heads measured relative to the same datum.Losses in suction and delivery lines may be present.Concept / Approach:Manometric head H_m is effectively the head against which the pump must deliver water. It is commonly expressed as the difference between the total head at the pump discharge flange and the total head at the pump suction flange, accounting for velocity heads and losses. Practical definitions in textbooks also relate H_m to the static head plus pipeline losses up to the discharge point.Step-by-Step Solution:Identify suction-side energy level (pressure head + velocity head + elevation).Identify delivery-side energy level at the point of delivery.Compute H_m = (Total head at delivery) − (Total head at suction).Recognize that this is the head the pump has to work against.Verification / Alternative check:When plotted on a system head curve, H_m corresponds to the ordinate at the operating flow, matching the pump’s head–capacity performance point.Why Other Options Are Wrong:Limiting the statement to reciprocating pumps or to specific inlet conditions is incorrect; the concept applies generally to centrifugal pumps. Zero suction head or flooded eye are operating states, not definitional requirements.Common Pitfalls:Confusing manometric head with static head only; omitting velocity head or line losses.Final Answer:Correct

Question 9

Jet impact on a moving plate (normal incidence) A water jet of velocity V and area a strikes a flat plate normally. The plate moves in the jet direction with constant velocity v. The force exerted by the jet on the plate equals:

Accepted Answer

Introduction / Context:Force due to a jet on a surface derives from momentum change. When the surface moves in the same direction as the jet, the effective mass flow striking the plate changes because the relative velocity is reduced. This problem is a standard result used in jet propulsion and impulse turbine analyses.Given Data / Assumptions:Jet cross-sectional area = a, jet velocity relative to ground = V.Flat plate moves colinearly with the jet at speed v (0 ≤ v Jet sticks momentarily and leaves tangentially with zero normal component (idealized), so normal velocity component is brought to rest.Concept / Approach:The mass flow actually intercepted by the moving plate is ρ * a * (V − v) because fluid parcels approach with relative speed (V − v). The normal momentum lost per unit time equals mass flow times change in normal velocity, which is (V − v) to 0 in the plate frame. Transforming back to the ground frame gives the same force magnitude in steady conditions.Step-by-Step Solution:Relative speed at impact = V_rel = V − v.Mass flow striking plate = ρ * a * V_rel.Normal velocity reduction = V_rel to 0 ⇒ ΔV = V_rel.Force = mass flow * ΔV = (ρ * a * V_rel) * V_rel = ρ * a * (V − v)^2.Verification / Alternative check:Limiting cases: v = 0 ⇒ F = ρ a V^2 (stationary plate). v → V ⇒ F → 0 (plate runs with the jet). Both match physical intuition.Why Other Options Are Wrong:Options (b) and (c) ignore the reduced intercepted mass flow or double-count terms. Option (d) is unphysical for colinear motion. Option (e) inappropriately halves the correct result.Common Pitfalls:Using V instead of V − v for intercepted flow; forgetting that momentum change uses the relative velocity squared in this setup.Final Answer:ρ * a * (V - v)^2

Question 10

Pump selection for supplying water to hydraulic turbines Which pump is commonly and successfully used for lifting and supplying water to turbines (e.g., for hydro-testing or auxiliary supply lines)?

Accepted Answer

Introduction / Context:Where water must be delivered continuously to a turbine system—such as in auxiliary feeds, testing loops, or small-head transfers—the pump should be capable of steady flow, simple control, and reliable operation over a range of discharges. Centrifugal pumps are the industry workhorse in such applications.Given Data / Assumptions:Clean or lightly silted water service.Moderate heads and continuous duty.Preference for smooth, non-pulsating flow to prevent hydraulic transients.Concept / Approach:Centrifugal pumps impart kinetic energy to water with a rotating impeller and then convert it to pressure head in the volute or diffuser. They provide a nearly steady discharge, are easy to throttle, and can be configured in series/parallel to meet head and flow requirements typical of turbine feed or test circuits.Step-by-Step Solution:Identify the duty: continuous supply with controllable flow.Match pump type: centrifugal pumps excel at continuous, moderate-head duties.Verify suitability: wide availability, straightforward maintenance, and low pulsation.Conclude centrifugal pump is the preferred, commonly used solution.Verification / Alternative check:Plant standards and vendor catalogs list centrifugal units for turbine auxiliary services; reciprocating pumps are used when very high heads at low flows are required, but they introduce pulsation and complexity.Why Other Options Are Wrong:Reciprocating pumps deliver pulsating flow and are maintenance intensive. Jet and air-lift pumps are specialized; peristaltic pumps handle slurries/chemicals at low flow and head, not turbine feeds.Common Pitfalls:Assuming any pump that moves water is suitable; ignoring flow steadiness and controllability requirements.Final Answer:Centrifugal pump

Question 11

Specific speed range and outlet flow type for centrifugal pumps Which pump configuration is preferred for a specific speed between 80 and 160 rpm (metric specific speed class)?

Accepted Answer

Introduction / Context:Specific speed groups centrifugal pumps by how head and discharge scale with speed and impeller diameter. As specific speed increases, the outlet flow pattern transitions from radial to mixed and then to axial. Selecting the right outlet type within a specific speed band ensures good efficiency and cavitation performance.Given Data / Assumptions:Specific speed Ns in the range 80–160 (metric convention).Water service and standard pump designs.Concept / Approach:Low Ns (≈10–80) favors radial-flow impellers (high head, low flow). Mid-to-high Ns (≈80–160) corresponds to mixed-flow designs that combine radial and axial components for improved capacity at moderate heads. Very high Ns (>160–300+) trends toward axial-flow (propeller) pumps.Step-by-Step Solution:Recognize Ns = 80–160 lies above the classic radial-flow band.Map Ns to outlet pattern: this band aligns with mixed-flow impellers.Therefore the preferred configuration is a high-speed, mixed-flow pump.Verification / Alternative check:Manufacturer selection charts plot efficiency islands vs. Ns, showing mixed-flow optima in this range.Why Other Options Are Wrong:Options (a) and (b) indicate radial-flow suited to lower Ns. Option (c) keeps radial discharge at high Ns, which is inefficient. Option (e) corresponds to yet higher Ns typical of axial pumps.Common Pitfalls:Confusing rotational speed in rpm with specific speed; ignoring that Ns is a non-dimensionalized grouping that dictates geometry trends.Final Answer:high-speed pump with mixed flow at outlet

Question 12

Francis turbine speed ratio: typical range The speed ratio (runner peripheral speed / √(2 g H)) for Francis turbines generally lies in which range?

Accepted Answer

Introduction / Context:The speed ratio is a non-dimensional parameter comparing the blade tip speed of a turbine runner with the theoretical jet speed derived from head. For reaction machines like Francis turbines, this ratio is higher than for impulse machines because part of the energy conversion occurs under pressure inside the passages.Given Data / Assumptions:Speed ratio φ = u / √(2 g H).Francis turbine operating near best efficiency with typical design coefficients.Concept / Approach:Impulse turbines (Pelton) have φ around 0.43–0.5. Francis turbines, with reaction and submerged runners, have larger φ typically in the 0.6–0.9 band, depending on specific speed and blade design. This reflects higher runner peripheral speeds relative to head-based velocity.Step-by-Step Solution:Define φ = u / √(2 g H).Consult typical design ranges: Pelton φ ≈ 0.46; Kaplan higher; Francis in-between but closer to reaction values.Select the range 0.6–0.9 as characteristic for Francis turbines.Verification / Alternative check:Textbook design charts and manufacturer data show φ in this range for conventional Francis runners across moderate specific speeds.Why Other Options Are Wrong:0.15–0.3 is too low (would suit very different machines). 0.4–0.5 suits Pelton. Values ≥1 imply blade tips faster than theoretical jet speed, not typical for Francis.Common Pitfalls:Confusing speed ratio with specific speed; mixing Francis and Pelton characteristic ratios.Final Answer:0.6 to 0.9

Question 13

Centrifugal pump behavior with speed changes Statement: For a centrifugal pump of fixed geometry operating under constant head and constant discharge, the power requirement increases with rotational speed.

Accepted Answer

Introduction / Context:Pump affinity laws relate flow, head, and power to rotational speed for a given impeller and casing. However, those relations apply when the system allows these variables to change with speed. This statement locks head and discharge as constants, so it tests whether you can separate operating point constraints from free-scaling behavior.Given Data / Assumptions:Fixed pump geometry.Head H and discharge Q are held constant (for example by throttling and system control).Efficiency changes with speed are not specified.Concept / Approach:Hydraulic power delivered to the fluid is P_h = ρ * g * Q * H. If Q and H are both fixed, the hydraulic power requirement is constant regardless of speed. Shaft power equals P_h / η. If efficiency remains similar, input power stays approximately constant. Therefore, a generic claim that power increases solely because speed is higher is false under the stated constraints.Step-by-Step Solution:Given: Q = constant, H = constant.Hydraulic power P_h = ρ g Q H = constant.Shaft power P_in = P_h / η; if η does not systematically decrease, P_in does not inherently increase with speed.Hence the blanket statement is false.Verification / Alternative check:If the system let Q and H vary per affinity laws, then P ∝ N^3. But that is explicitly not the case here, confirming that speed alone does not dictate higher power at fixed Q and H.Why Other Options Are Wrong:Options tying truth to positive displacement or zero efficiency are irrelevant; the governing relation is P_h = ρ g Q H.Common Pitfalls:Blindly applying affinity P ∝ N^3 without verifying whether Q and H are fixed or allowed to vary.Final Answer:False

Question 14

Classification check: which is not a reaction turbine? Identify the turbine from the list that is not a reaction type.

Accepted Answer

Introduction / Context:Turbines are broadly divided into impulse and reaction types based on how they convert hydraulic energy. Reaction turbines exhibit a pressure drop across the runner passages; impulse turbines rely on high-velocity jets at essentially constant runner pressure. Knowing classic examples prevents misclassification.Given Data / Assumptions:Furneyron and Jonval are early reaction designs (radial- and axial-flow respectively).Thomson's (Fourneyron/Thomson variants) are also reaction-type machines.Pelton is a pure impulse turbine.Concept / Approach:Pelton wheels operate with one or more free jets at atmospheric pressure at the runner; the jet momentum is redirected by buckets to produce torque, with negligible static pressure change through the runner. Therefore, Pelton is not a reaction turbine.Step-by-Step Solution:List types: Furneyron (reaction), Jonval (reaction), Thomson's (reaction), Pelton (impulse).Select the non-reaction option: Pelton wheel.Verification / Alternative check:Any turbine using a draft tube is reaction; Pelton does not use a draft tube and is installed above tail race, typical of impulse machines.Why Other Options Are Wrong:Furneyron, Jonval, and Thomson's all involve pressure changes across the runner and are classic reaction examples; Francis (in option list extension) is also reaction.Common Pitfalls:Confusing historical names; assuming only Francis and Kaplan are reaction while older names are impulse.Final Answer:Pelton wheel

Question 15

Effects of cavitation in hydraulic machines What are the principal adverse effects when cavitation occurs in pumps or turbines?

Accepted Answer

Introduction / Context:Cavitation occurs when local static pressure falls to or below the liquid's vapor pressure, leading to the formation of vapor bubbles that subsequently collapse in higher-pressure regions. The implosion of these bubbles generates shock waves and micro-jets that damage surfaces and degrade hydraulic performance.Given Data / Assumptions:Water as the working fluid.Operation near cavitation inception conditions (low NPSH, off-design flow, or high local velocities).Material surfaces susceptible to pitting under repeated bubble collapse.Concept / Approach:Cavitation affects machines mechanically and hydraulically. Mechanically, bubble collapse near solid boundaries produces pitting, erosion, and induces vibration and noise. Hydraulically, vapor cavities displace liquid, altering velocity triangles and limiting effective head transfer, which reduces discharge, head, and overall efficiency. Severe cavitation causes unstable operation and sudden power loss.Step-by-Step Solution:As pressure drops below vapor pressure, vapor bubbles form.Bubbles convect to higher pressure zones and collapse violently.Surface damage and unsteady forces arise ⇒ noise and vibration increase.Flow passages partially filled with vapor reduce effective flow area and energy transfer ⇒ discharge, head, and efficiency decrease.Verification / Alternative check:Performance curves under cavitating conditions show reduced head-flow capability; inspection reveals pitted blades or volute walls. Vibration spectra exhibit characteristic broadband and discrete components associated with cavity dynamics.Why Other Options Are Wrong:Since cavitation simultaneously causes noise/vibration, discharge reduction, and efficiency/power loss, selecting any single effect alone is incomplete; “all of the above” is the correct comprehensive choice. “None of the above” is clearly incorrect.Common Pitfalls:Assuming cavitation only causes noise without performance loss; overlooking NPSH requirements and off-design incidence as root causes.Final Answer:all of the above

Question 16

Centrifugal Pumps — Selection by Specific Speed (Ns = 20 rpm-based) For a very low specific speed of approximately 20 (rpm-based units), which type of centrifugal pump is generally preferred at the outlet in terms of flow direction and speed categor…

Accepted Answer

Introduction:Pump specific speed is a powerful index for choosing impeller geometry. Very low specific speed (around 20 in rpm-based form) points to pumps designed for high head and low discharge, which naturally aligns with radial-flow (centrifugal) impellers operating at relatively slow rotational speeds.Given Data / Assumptions:Specific speed Ns ≈ 20 (rpm-based).Standard water service; geometric similarity assumptions.Goal: choose the suitable pump outlet-flow type and speed category.Concept / Approach:Specific speed for pumps correlates with outlet flow pattern: low Ns → radial-flow impellers (slow speed, high head); medium Ns → mixed-flow; high Ns → axial-flow (propeller) pumps. Thus, Ns ≈ 20 clearly indicates a radial-flow design at slow speed.Step-by-Step Solution:Interpret Ns banding: low Ns values favor radial-flow pumps.Relate hydraulic duty: high head, small Q → slow rotational speed for given impeller diameters.Conclude appropriate choice: slow speed, radial outlet flow.Verification / Alternative check:Manufacturer selection charts map Ns ≈ 20 squarely into the radial and slow-speed region.Why Other Options Are Wrong:medium speed radial: Ns is too low for this category.high speed radial/axial: correspond to higher Ns bands, not ~20.Common Pitfalls:Confusing pump Ns (uses discharge) with turbine Ns (uses power).Final Answer:slow speed pump with radial flow at outlet

Question 17

Jet Propulsion of Ships — Efficiency with Lateral (Right-Angle) Inlets For a ship using water-jet propulsion with inlet orifices arranged at right angles to the direction of motion, which expression gives the propulsive efficiency (η) in terms of sh…

Accepted Answer

Introduction:Jet propulsion of ships accelerates an incoming stream of water and ejects it astern to produce thrust. Propulsive efficiency compares useful power delivered to the hull (thrust * ship speed) with the rate of kinetic-energy input to the jet stream. When inlets are at right angles to the motion, the inflow has negligible streamwise momentum before acceleration, simplifying the efficiency relation.Given Data / Assumptions:Ship speed relative to water: V.Jet exit speed relative to the ship: V_j (directed astern).Lateral (right-angle) inlets so the incoming water has near-zero axial momentum.Idealized momentum exchange and steady incompressible flow.Concept / Approach:The thrust T equals mass flow rate times the change in axial velocity. With right-angle inlets, the incoming axial component is approximately zero, and the outgoing axial velocity relative ground is V_j − V (backward). The propulsive efficiency is the ratio of useful power T * V to the jet power input in the slipstream. Under standard jet-propulsion analysis, this simplifies to η = 2V / (V + V_j), which shows that efficiency is highest when V is close to V_j (small velocity slip), but thrust then diminishes.Step-by-Step Solution:Let m_dot = rho * A * V_j (relative discharge through the pump and nozzle).Axial speed change (ground frame): 0 → (V_j − V) backward; thrust magnitude T = m_dot * (V_j − V).Useful power: P_use = T * V = m_dot * (V_j − V) * V.Jet power added: P_jet = (1/2) * m_dot * (V_j − V)^2 + (1/2) * m_dot * V^2 − approximately reduces to m_dot * (V_j^2 − V^2) / 2 under idealization.Propulsive efficiency reduces to η = 2V / (V + V_j).Verification / Alternative check:Limits: if V = V_j → η → 1 (perfect, but zero thrust). If V ≪ V_j → η ≈ 2V / V_j → low efficiency, matching practical experience.Why Other Options Are Wrong:V / V_j: ignores kinetic-energy balance; only a thrust ratio proxy.(V_j − V) / V_j: describes slip fraction, not efficiency.1 − (V / V_j)^2: energy ratio form that does not match momentum–energy analysis for jet propulsion.Common Pitfalls:Confusing air-jet and water-jet forms; the same ideal relation holds when inlet axial momentum is negligible.Using power-based slip factors without relating them to V and V_j.Final Answer:η = 2V / (V + V_j)

Question 18

Hydraulic Machines — Optimum Vane Speed for Maximum Efficiency A jet of water strikes the center of a smooth curved vane that moves in the jet direction at uniform speed. For maximum efficiency of energy transfer to the moving vane, the vane (bucket…

Accepted Answer

Introduction:When a free jet impinges on a moving vane, useful work is done through the change of whirl (streamwise) velocity. There exists an optimum vane speed that maximizes the fraction of jet kinetic energy converted into work. This classical result is widely applied in Pelton buckets and other jet–vane systems.Given Data / Assumptions:Jet speed V, vane speed u along jet direction.Jet hits the center of a smooth, lossless curved vane and exits without shock.Relative exit speed equals relative inlet speed (ideal case).Concept / Approach:Power delivered is proportional to the product of vane speed and change in whirl component. For an ideal deflection, optimizing with respect to u yields the condition u = V / 2 for maximum hydraulic efficiency of the moving surface.Step-by-Step Solution:Work per unit mass = u * (V_w1 − V_w2).Ideal turning with no losses and no shock gives V_w2 proportional to (V − u).Differentiate work with respect to u and set derivative to zero.The optimum occurs at u = V / 2.Verification / Alternative check:Substitute u = V / 2 into the hydraulic efficiency expression to obtain the known maximum value for an ideal jet–vane pair.Why Other Options Are Wrong:one-third, two-third, three-fourth: do not satisfy the optimum from the derivative of the power expression; they either underutilize or overrun the jet.Common Pitfalls:Confusing fixed-vane formulas (where u = 0) with moving-vane optimization.Final Answer:one-half

Question 19

Centrifugal Pump vs Reaction Turbine — Reversed Action State whether the following statement is correct: “The action of a centrifugal pump is that of a reversed reaction turbine.”

Accepted Answer

Introduction:Pumps add energy to a fluid, while turbines extract energy from a fluid. A reaction turbine and a centrifugal pump share similar flow passages and velocity triangles but operate in opposite directions of energy transfer. Recognizing this duality clarifies many performance relations and similarity laws.Given Data / Assumptions:Incompressible flow; similar impeller/runner geometry.Loss mechanisms present but not altering the fundamental action.Concept / Approach:In a reaction turbine, water enters the runner with high pressure/velocity and leaves with lower energy as the runner extracts work. In a centrifugal pump, the impeller does work on the water, raising its head; the velocity triangles are effectively reversed. Thus, a centrifugal pump can be regarded as a reaction turbine operating in reverse.Step-by-Step Solution:Compare inlet/outlet triangles: turbine reduces whirl at outlet; pump increases whirl at outlet.Energy direction: turbine converts fluid energy → shaft; pump converts shaft energy → fluid.Conclude reversed action holds conceptually and kinematically.Verification / Alternative check:Some small turbines can be driven as pumps (and vice versa) under suitable conditions, illustrating the reversibility of flow passages.Why Other Options Are Wrong:Incorrect: contradicts standard turbomachinery theory.Shut-off or axial-only caveats: reversibility is not limited to those cases.Common Pitfalls:Assuming impulse turbines share the same reverse equivalence; the statement is specific to reaction-type machines.Final Answer:Correct

Question 20

Turbine Selection by Head — 450 m Head of Water For an available head of approximately 450 m at a hydropower site, which turbine type is generally preferred?

Accepted Answer

Introduction:Choosing the turbine type depends primarily on site head and discharge, summarized by specific speed. High heads with relatively small discharges suit impulse machines, while low heads with large discharges suit axial-flow reaction machines.Given Data / Assumptions:Gross head ≈ 450 m (very high head).Standard water properties; feasible penstock designs.Goal: select an appropriate turbine family.Concept / Approach:Pelton turbines (impulse type) are the go-to choice for high heads (typically above ~250–300 m), converting the entire head into jet kinetic energy before striking buckets. Francis turbines suit medium heads, whereas Kaplan (propeller) turbines serve low heads. At 450 m, Pelton provides excellent efficiency and manageable runner speeds via multiple jets if needed.Step-by-Step Solution:Classify head: 450 m → high-head regime.Match to turbine: Pelton wheel designed for high head, low to moderate discharge.Reject others: Francis (medium head), Kaplan (low head).Verification / Alternative check:Specific-speed charts place 450 m sites in the Pelton domain; many alpine stations use Pelton runners at similar heads.Why Other Options Are Wrong:Kaplan: optimized for very low head and high flow.Francis: best for medium heads and mixed-flow conditions.none of these: incorrect since Pelton fits.Common Pitfalls:Overlooking cavitation limits and runner speed; Pelton mitigates these at high heads.Final Answer:Pelton wheel

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