Question 1

Define the gross (total) head for turbines Gross head between headrace and tailrace is obtained by what relation between their water surface levels?

Accepted Answer

Introduction: Before accounting for conduit and entry losses, engineers begin with the gross or total head available from a site. This head reflects the vertical energy difference driving the hydropower installation. Given Data / Assumptions: Steady water levels at headrace and tailrace. Neglecting velocity heads for a basic definition (or considered small). Concept / Approach: Gross head H_gross is the elevation (energy) difference between the upstream headrace water level and the downstream tailrace water level. It is a site attribute, independent of penstock routing. The net head used for power calculations is obtained after subtracting hydraulic losses. Step-by-Step Solution: Measure z_headrace and z_tailrace relative to the same datumForm H_gross = z_headrace − z_tailraceLater compute H_net = H_gross − h_losses for turbine power Verification / Alternative check: Bernoulli’s equation between headrace and tailrace simplifies to the difference in piezometric heads when velocities are small and pressures are hydrostatic. Why Other Options Are Wrong: Sum/product/ratio/average: have no physical meaning for head definition. Common Pitfalls: Confusing gross head with rated head at the turbine inlet; forgetting that tailrace levels rise during floods, reducing gross head. Final Answer: difference

Question 2

Impulse vs reaction turbines — identify the correct statement Choose the statement that correctly describes pressure/energy conditions across the runner for impulse and reaction turbines.

Accepted Answer

Introduction: Understanding where pressure changes occur differentiates impulse and reaction machines. Correctly identifying these conditions is vital for selecting casings, draft tubes, and governing systems. Given Data / Assumptions: Impulse (e.g., Pelton) uses nozzles to convert pressure head fully to jet velocity before the runner. Reaction (e.g., Francis/Kaplan) allows pressure changes across runner passages. Concept / Approach: In impulse turbines, jets strike buckets at atmospheric pressure; the runner primarily redirects momentum with negligible pressure change. In reaction turbines, both pressure and velocity change across the runner blades, producing lift-like forces while the casing and draft tube contain sub-atmospheric regions. Step-by-Step Solution: Impulse: p_runner_in ≈ p_atm, p_runner_out ≈ p_atm; energy change is kinetic → mechanical via momentum exchangeReaction: p_runner_in ≠ p_runner_out; part of pressure head converts to mechanical work within runnerTherefore, the correct statement is constant (atmospheric) pressure across an impulse runner Verification / Alternative check: Pelton installations show free jets and open buckets exposed to atmosphere; reaction turbines require sealed casings and draft tubes due to pressure variations. Why Other Options Are Wrong: Pressure energy at impingement (impulse): nozzle already converted it to velocity head. “Kinetic only” for reaction: reaction also involves pressure changes. Pressure increases over reaction vanes: it typically decreases as energy is extracted. Both constant: false for reaction machines. Common Pitfalls: Confusing nozzle exit conditions with runner passage conditions; assuming enclosed casings imply constant pressure. Final Answer: In an impulse turbine, the pressure of the flowing water remains essentially atmospheric across the runner.

Question 3

Impulse turbine operation — purpose of a braking jet In an impulse turbine (e.g., Pelton), what is the function of the braking jet during shutdown or speed reduction?

Accepted Answer

Introduction: High-inertia runners in impulse turbines can spin for long periods after the main jet is deflected off the buckets. A braking jet provides controlled deceleration to reduce stopping time and improve operational safety and responsiveness. Given Data / Assumptions: Pelton-type runner with one or more nozzles and deflector/gate control. Shutdown procedures aiming to avoid overspeed and long coast-downs. Concept / Approach: The braking jet is a small jet directed against the back of the buckets or in a direction that produces a torque opposite to the current rotation. This negative torque extracts kinetic energy from the rotating runner and dissipates it in the water, shortening the stopping time without mechanical friction brakes on the shaft. Step-by-Step Solution: Main jet closed or deflected to avoid overspeedBraking jet opened to apply reverse torque on bucketsRunner decelerates rapidly to rest or to a lower safe speed Verification / Alternative check: Operator manuals and control sequences show coordinated jet deflection plus braking jet application during trips and load rejections to avoid excessive rotational coast-down. Why Other Options Are Wrong: Breaking the jet into droplets: not the objective; atomization wastes water without controlled torque. Changing direction of runner: not feasible by a braking jet. Increasing nozzle coefficient: unrelated to shutdown behavior. “None”: the function is specific and well established. Common Pitfalls: Confusing the braking jet with deflectors; assuming mechanical brakes alone handle all stopping (they are often supplemented or avoided to reduce wear). Final Answer: to bring the runner to rest in a short time

Question 4

Hydraulic accumulator — form of energy stored A hydraulic accumulator is a device used to store which form of energy for later delivery to a hydraulic machine?

Accepted Answer

Introduction: Hydraulic systems use accumulators to smooth transients, supply peak demands, and maintain pressure during pump stoppage. Knowing the energy form stored clarifies accumulator function and sizing. Given Data / Assumptions: Accumulator types: weight-loaded, spring-loaded, and gas-charged (bladder or piston). Working fluid is typically oil under pressure. Concept / Approach: Accumulators store potential energy of pressurized fluid. In gas-charged accumulators, compressed gas (e.g., nitrogen) provides the elastic element; in weight-loaded designs, potential energy arises from a heavy weight acting on the fluid; in spring-loaded designs, the spring stores elastic energy. In all cases, the usable output is hydraulic pressure/flow supplied later to the circuit. Step-by-Step Solution: Charge the accumulator to a preset pressure (gas precharge or weight/spring force)Fluid enters, compressing gas or deflecting spring/raising weightOn demand, stored pressure energy discharges fluid to the system Verification / Alternative check: Power backup and pulsation dampening in presses and machines are achieved by releasing the stored hydraulic energy to maintain pressure and flow. Why Other Options Are Wrong: Strain: generic term; the device delivers pressure/flow, not strain as an output. Kinetic: flywheels store kinetic energy; accumulators do not spin. Thermal/electrical: unrelated to hydraulic accumulators. Common Pitfalls: Confusing accumulator storage medium (compressed gas, weight, spring) with the delivered energy form—what the system receives is hydraulic pressure energy. Final Answer: pressure

Question 5

Reciprocating pump — classification by displacement A reciprocating pump is also called a ________ displacement pump. Choose the correct term used in fluid machinery.

Accepted Answer

Introduction: Pumps are broadly divided into positive displacement (PD) and rotodynamic categories. Recognizing which class a reciprocating pump belongs to is essential for understanding its flow characteristics, control methods, and protections. Given Data / Assumptions: Piston/ plunger mechanism with suction and delivery valves. Finite swept volume per stroke; relatively low speed. Concept / Approach: Positive displacement machines transfer a nearly fixed volume of fluid each cycle irrespective of discharge pressure (within mechanical limits). Reciprocating pumps trap fluid in a chamber and displace it by piston motion, making them quintessential PD pumps. Consequently they require relief valves to prevent overpressure and often benefit from air vessels to smooth pulsating flow. Step-by-Step Solution: Identify operating principle: trapping and forcing fixed volumesRelate to definition: flow nearly proportional to speed, not headConclude: “positive displacement” is the correct classification Verification / Alternative check: Manufacturer curves for PD pumps show near-vertical head–flow characteristics: flow rate changes little with discharge pressure compared to centrifugal pumps. Why Other Options Are Wrong: Negative/non-displacement/centrifugal: do not describe reciprocating mechanisms. Variable-inertia: not a standard category. Common Pitfalls: Operating PD pumps with closed discharge without relief valves; confusing their control logic with that of centrifugal pumps. Final Answer: positive

Question 6

Turbines — Definition of Unit Speed (Speed at Unit Head) In turbine performance scaling, the unit speed of a runner is defined as the rotational speed the turbine would have if it operated under a head of 1 m. Select the correct expression for unit …

Accepted Answer

Introduction: Unit quantities help compare turbines independent of head by normalizing performance to a head of 1 m. Unit speed indicates how fast the runner would turn at unit head, enabling fair comparison across different sites and tests. Given Data / Assumptions: Actual rotational speed = N (rpm). Operating head = H (m). Geometric similarity and dynamically similar operation are implied when comparing across heads. Concept / Approach: Under dynamically similar operation, characteristic velocities scale with sqrt(H). Therefore rotational speed, which is proportional to characteristic velocity divided by a length scale that is held fixed for a given machine, also scales with sqrt(H). To reduce from head H to unit head, divide by sqrt(H). Step-by-Step Solution: Velocity scale: V ∝ sqrt(g * H).Speed scale: N ∝ V / D with fixed D ⇒ N ∝ sqrt(H).Unit speed is the speed at H = 1 m: N_u = N / sqrt(H). Verification / Alternative check: The related unit relations are N_u = N / sqrt(H), Q_u = Q / sqrt(H), and P_u = P / H^(3/2). All consistently remove the head dependence for similar operation. Why Other Options Are Wrong: N * sqrt(H), N * H: These increase with head; unit speed must be smaller at higher H. N / H, N / H^2: Over-correct; rotational speed does not scale linearly or quadratically with head. Common Pitfalls: Mixing unit speed with specific speed, or using linear head scaling instead of the square-root law. Final Answer: N_u = N / sqrt(H)

Question 7

Model–Prototype Similarity — When All Linear Dimension Ratios Match If the ratios of all corresponding linear dimensions between a model and its prototype are equal (constant scale factor) and shapes are identical, what type of similarity is satisfi…

Accepted Answer

Introduction: Engineering models are built to predict prototype behavior. The first level of similarity required is geometric similarity, which ensures that the model is a scaled replica of the prototype so that corresponding points and angles are well-defined. Given Data / Assumptions: All linear dimensions in the model maintain a constant ratio to the prototype (same scale factor in x, y, z). Angles and shape proportions are preserved. Concept / Approach: Similarity is categorized as geometric, kinematic, and dynamic. Geometric similarity concerns only size and shape. Kinematic adds equality of non-dimensional velocities/accelerations; dynamic requires equality of non-dimensional forces (e.g., Reynolds, Froude, Euler numbers) in addition to the first two. Step-by-Step Solution: Check dimension ratios: all constant ⇒ shapes are identical up to scale.No statement about velocity/acceleration fields ⇒ kinematics not guaranteed.No statement about force ratios or dimensionless numbers ⇒ dynamics not guaranteed.Therefore, only geometric similarity is assured. Verification / Alternative check: Wind-tunnel and hydraulic-lab practices explicitly distinguish geometric similarity before targeting kinematic/dynamic similarity via matched dimensionless groups. Why Other Options Are Wrong: Kinematic similarity needs velocity/acceleration ratios equal. Dynamic similarity needs force ratios (dimensionless groups) equal. None of these: incorrect because geometric similarity is indeed satisfied. Common Pitfalls: Assuming geometric similarity automatically implies correct forces or flows; it does not. Final Answer: geometric similarity

Question 8

Centrifugal Pump Affinity Laws — Head Scaling with Speed and Impeller Diameter For dynamically similar operation of a centrifugal pump, how does the developed head H scale with impeller speed N and impeller diameter D?

Accepted Answer

Introduction: Selection and scaling of centrifugal pumps use the affinity laws relating flow rate, head, and power to speed and impeller size. Knowing the correct proportionality prevents mis-sizing during variable-speed operation or trim changes. Given Data / Assumptions: Dynamically similar operation (same efficiency region and flow coefficient). Impeller diameter = D, rotational speed = N. Concept / Approach: The pump affinity laws state: Q ∝ N * D^3H ∝ N^2 * D^2P ∝ N^3 * D^5 Thus head varies with the square of both speed and diameter, not linearly with either one or their simple product. Step-by-Step Solution: Compare each option to H ∝ N^2 * D^2.Option A (∝ D) and B (∝ N) omit the square dependence.Option C (∝ N * D) is still linear overall and incorrect.Therefore, none of the listed linear relations match; choose “none of the above”. Verification / Alternative check: Dimensionless analysis using head coefficient psi = g * H / (N^2 * D^2) confirms psi remains approximately constant under similarity, leading to H ∝ N^2 D^2. Why Other Options Are Wrong: Linear statements underestimate/overestimate head change when speed or diameter changes. Common Pitfalls: Confusing linear speed scaling with the quadratic dependence inherent to head. Final Answer: none of the above

Question 9

Reciprocating Pump Components — Definition of an Air Vessel Consider the statement: “The air vessel in a reciprocating pump is a cast iron closed chamber having an opening at its base (connected to the pipeline).” Choose the correct response.

Accepted Answer

Introduction: Air vessels are classic accessories used with reciprocating pumps to smoothen discharge, reduce acceleration head, minimize power fluctuations, and prevent separation. Understanding their construction and connection helps in correct placement and operation. Given Data / Assumptions: Reciprocating pump with suction and/or delivery air vessel. Traditional construction uses cast iron; modern variants may use steel pressure vessels. Connection is near the pump cylinder to be effective. Concept / Approach: An air vessel is a closed chamber partially filled with air. It is connected to the pipeline at its base so that water can enter/leave below, compressing/expanding the air cushion above. This arrangement absorbs flow pulsations during stroke reversals and maintains near-uniform pipe velocity. Step-by-Step Solution: Identify construction: closed pressure vessel with trapped air.Connection detail: opening at the base into the suction/delivery line near the cylinder.Function: air compressibility stores/releases energy, smoothing the flow and reducing acceleration head. Verification / Alternative check: Standard pump texts depict bell-shaped cast-iron vessels mounted vertically with flange at the base connecting to the pipeline tee near the pump body. Why Other Options Are Wrong: Disagree: contradicts the conventional design description. Side-specific agreements are unnecessary; vessels can be provided on suction, delivery, or both. Common Pitfalls: Assuming the vessel must be at the top or confusing it with surge tanks used on long mains. Final Answer: Agree

Question 10

Reaction Turbines — Role of Guide Vanes What is the principal function of the stationary guide vanes (wicket gates) in a reaction turbine such as a Francis machine?

Accepted Answer

Introduction: Guide vanes in reaction turbines condition the inflow to the runner. They set both the flow rate and the flow angle to maximize efficiency and avoid hydraulic losses from incidence or separation. Given Data / Assumptions: Steady operation; incompressible flow. Adjustable wicket gates used to regulate discharge and swirl. Concept / Approach: The guide apparatus provides two key controls: quantity control (opening changes flow area) and quality control (exit angle sets whirl component), thereby minimizing shock on entry and suppressing eddies. Step-by-Step Solution: Set proper incidence: vane exit angle matches runner inlet velocity triangle to avoid shock.Regulate discharge: gate opening sets the required flow rate.Shape flow: smooth passages reduce eddy formation and pressure losses. Verification / Alternative check: Runner performance maps show sharp efficiency penalties when incidence is off-design, highlighting the importance of guide vane angle control. Why Other Options Are Wrong: Each single function is true, but the complete and correct choice is “all of the above”. Common Pitfalls: Assuming guide vanes only throttle flow; their primary role also includes direction control to match velocity triangles. Final Answer: all of the above

Question 11

Pelton Wheel Design — Empirical Rule for Number of Buckets For a Pelton wheel of pitch-circle diameter D supplied by a jet of diameter d, the empirical relation used to select the number of buckets around the periphery is:

Accepted Answer

Introduction: Bucket count influences jet interference, discharge smoothness, and mechanical balance in Pelton turbines. An empirical guideline is commonly used at preliminary design to ensure proper spacing and acceptance of the jet by successive buckets. Given Data / Assumptions: Single- or multi-jet Pelton wheel. Pitch-circle diameter = D, jet diameter = d. Standard splitter geometry and deflection angles. Concept / Approach: Empirical practice suggests keeping roughly one bucket pitch between the jet and the bucket leaving the jet to avoid interference, leading to the well-known rule: z ≈ 15 + D / (2 d) This gives adequate bucket count for smooth torque with minimal overlap. Step-by-Step Solution: Compute D / (2 d) from geometry.Add baseline count (≈15) to account for spacing and starting buckets.Select nearest integer to obtain the final bucket number. Verification / Alternative check: Design handbooks and many worked examples apply z = 15 + D/(2 d), yielding practical bucket counts for common jet ratios (D/d) between 10 and 16. Why Other Options Are Wrong: Other formulas either overshoot or undershoot the count and are not standard. Common Pitfalls: Confusing pitch diameter with outer diameter or using jet nozzle diameter after contraction instead of the effective jet diameter. Final Answer: z = 15 + D / (2 d)

Question 12

Water Wheels — Undershot Type and Driving Mechanism Undershot water wheels are those in which the wheel is driven primarily by which action of the water?

Accepted Answer

Introduction: Traditional water wheels are categorized as overshot, breastshot, and undershot depending on where water strikes the wheel and how energy is transferred. Understanding the dominant driving mechanism clarifies where each type is suitable. Given Data / Assumptions: Low head, high flow rivers suitable for undershot wheels. Undershot: water meets paddles near the bottom of the wheel. Concept / Approach: Undershot wheels are impulse devices. Flowing water strikes the paddles at low head, imparting momentum change to produce torque. Overshot relies mainly on weight (potential energy), while breastshot is mixed (weight + impulse). Step-by-Step Solution: Identify strike position: near bottom indicates undershot.Dominant energy transfer: impulse from water velocity acting on paddles.Therefore, wheel runs by the impulse of water. Verification / Alternative check: Classic classifications: undershot (impulse), overshot (weight), breastshot (combination) are widely documented in hydraulic machinery texts. Why Other Options Are Wrong: Entirely by weight: describes overshot wheels. Partly weight and partly impulse: describes breastshot wheels. None of the above: incorrect since one option fits precisely. Common Pitfalls: Confusing geographical installation style with energy mechanism; the key is where and how water meets the wheel. Final Answer: the wheel runs entirely by the impulse of water

Question 13

Turbine Selection by Specific Speed — 1500 kW at 300 rpm under 150 m Head A turbine must develop 1500 kW while running at 300 rpm under a net head of 150 m. Which turbine is most appropriate?

Accepted Answer

Introduction: Selecting a turbine type depends primarily on head and specific speed. High heads with relatively low specific speed favor impulse (Pelton) machines, while medium heads suit Francis and low heads suit Kaplan/propeller types. Given Data / Assumptions: Power P = 1500 kW (≈ 1.5 MW). Speed N = 300 rpm. Head H = 150 m. Standard metric specific speed definition. Concept / Approach: Compute specific speed Ns ≈ N * sqrt(P) / H^(5/4). For the given data, Ns falls within the Pelton range (typically ~8–35). To achieve the required power at this speed, multi-jet arrangements are commonly used to increase discharge without oversizing the wheel, hence a Pelton with two or more nozzles is preferred. Step-by-Step Solution: Estimate Ns qualitatively: high head (150 m) drives the choice toward Pelton.Given the moderate speed (300 rpm) and MW-scale power, multiple jets are used to deliver higher flow at the same runner diameter and speed.Therefore choose a Pelton wheel with two or more nozzles. Verification / Alternative check: Manufacturer charts show Francis/Kaplan specific speeds far larger than the computed Ns for this duty, reinforcing the Pelton choice; multi-jet designs extend Pelton capacity at a given speed. Why Other Options Are Wrong: Single-nozzle Pelton may be insufficient to deliver the needed flow at 300 rpm. Kaplan: suited to low head, high flow. Francis: suited to medium head and higher Ns than indicated. Common Pitfalls: Ignoring speed constraints and assuming head alone determines type; both head and specific speed matter. Final Answer: Pelton wheel with two or more nozzles

Question 14

Model Testing — Is Equal Velocity Ratio Dynamic or Kinematic Similarity? Consider the statement: “If the ratios of corresponding velocities at corresponding points are equal, then the model and prototype have dynamic similarity.” Choose the correct …

Accepted Answer

Introduction: It is important to distinguish between kinematic and dynamic similarity in fluid modeling. Mislabeling these leads to wrong scaling laws and erroneous extrapolation from model to prototype. Given Data / Assumptions: Velocity ratios at corresponding points are equal (with geometric similarity). No information given about forces or dimensionless numbers. Concept / Approach: Kinematic similarity means equality of the ratios of velocities and accelerations at corresponding points (and time correspondence), given geometric similarity. Dynamic similarity is stricter: it requires equality of force ratios, typically enforced by matching appropriate dimensionless groups (e.g., Reynolds, Froude, Weber, Mach) depending on dominant physics. Step-by-Step Solution: Check what is stated: equal velocity ratios ⇒ kinematic similarity.Dynamic similarity requires forces to scale; velocity equality alone does not ensure this.Therefore the statement is incorrect. Verification / Alternative check: Examples: Matching Reynolds number ensures dynamic similarity for viscous flows; matching Froude number ensures dynamic similarity where gravity waves dominate. Both go beyond mere velocity scaling. Why Other Options Are Wrong: Correct: would conflate kinematic with dynamic similarity. Common Pitfalls: Assuming that if velocity fields look similar, then force fields do too; not necessarily true without matched dimensionless numbers. Final Answer: Incorrect

Question 15

Pelton Terminology — Definition of Jet Ratio In Pelton wheel design, the jet ratio is defined as the ratio of which two diameters?

Accepted Answer

Introduction: Jet ratio is a key geometric parameter in Pelton design influencing bucket sizing, bucket count, and efficiency. It links the runner diameter to the jet diameter provided by the nozzle(s). Given Data / Assumptions: Pelton wheel with one or more jets. Runner pitch-circle diameter = D, jet diameter = d. Concept / Approach: By definition, jet ratio JR is the diameter ratio JR = D / d. Practical ranges (often around 10–16) balance jet coherence, bucket acceptance, and mechanical size. Step-by-Step Solution: Identify diameters: D = runner diameter; d = jet diameter.Form the ratio: JR = D / d.Conclude: jet ratio is diameter of Pelton wheel to diameter of jet. Verification / Alternative check: This definition appears alongside empirical relations like bucket number z ≈ 15 + D/(2 d) and sizing rules for bucket width and depth. Why Other Options Are Wrong: Jet-to-wheel diameter reverses the definition. Velocity ratios are unrelated to the geometric jet ratio. Common Pitfalls: Confusing pitch-circle diameter with overall outside diameter; design commonly uses the pitch-circle for D in ratio definitions. Final Answer: diameter of Pelton wheel to the diameter of jet

Question 16

In an impulse turbine used in hydraulic machinery, the nozzle converts the entire available head into kinetic energy before the jet strikes the runner. State whether the following assertion is correct: “The water pressure at the entry to the runner …

Accepted Answer

Introduction: Impulse turbines (e.g., Pelton, Turgo) extract energy almost entirely from the jet's kinetic head. Understanding the pressure conditions at runner inlet and outlet is fundamental to distinguishing impulse from reaction turbines and explains why draft tubes are unnecessary for impulse machines. Given Data / Assumptions: Nozzle expands to atmosphere and delivers a free jet. Jet strikes runner buckets in an open casing (not submerged). Negligible pressure rise within the runner passage; losses limited to friction and splash. Concept / Approach: In an impulse turbine, the entire pressure drop occurs in the nozzle. The runner operates in air. Consequently, static pressure at the runner entry (just as the jet leaves the nozzle) is atmospheric, and after the jet is deflected by the buckets and discharged, it again mixes with the surrounding atmosphere, remaining at atmospheric pressure. Step-by-Step Solution: 1) Convert head to velocity in the nozzle: V ≈ (2 * g * H)^(1/2).2) At nozzle exit, static pressure = atmospheric (free jet condition).3) Runner changes only the jet momentum (direction and slightly magnitude); static pressure stays ≈ atmospheric.4) Discharged water leaves to atmosphere; no draft tube or submergence is used. Verification / Alternative check: Pressure taps placed just upstream of the nozzle show high pressure; immediately in the free jet and around the runner, static pressure readings are essentially atmospheric. Why Other Options Are Wrong: Disagree/part-load/small jets: the impulse principle holds over the operating range; pressure in the runner region remains ≈ atmospheric. Draft tube condition: draft tubes belong to reaction turbines, not impulse wheels. Common Pitfalls: Confusing impulse with reaction operation and assuming a pressure drop across the runner, which does not occur for ideal impulse machines. Final Answer: Agree

Question 17

A jet-propelled boat draws water through orifices that face the boat’s direction of motion and discharges it astern to generate thrust. Express the propulsion efficiency η in terms of the boat speed V and the jet speed relative to still water V_j (m…

Accepted Answer

Introduction: Jet propulsion in watercraft relies on momentum exchange: water is ingested and expelled at a different velocity to create thrust. Propulsive efficiency compares useful power (thrust * boat speed) to the rate of kinetic energy imparted to the water stream. Knowing the correct efficiency expression is key to optimizing jet speed for best economy. Given Data / Assumptions: Boat speed relative to water = V. Jet absolute discharge speed (opposite to motion) has magnitude V_j relative to still water. Steady operation; intake faces the direction of travel (scooping), discharge astern. Neglect hull drag, viscous losses, and inlet/discharge losses for the ideal relation. Concept / Approach: For an ideal momentum jet, thrust T equals mass flow rate m_dot times change in axial velocity between intake (≈ V forward) and discharge (≈ −V_j aft), giving T = m_dot * (V + V_j). Useful power is T * V. The rate of kinetic energy imparted to the stream is (1/2) m_dot * [(V + V_j)^2 − V^2] = (1/2) m_dot * (V_j^2 + 2 V V_j). The well-known simplification yields η = 2V / (V + V_j). Step-by-Step Solution: 1) Axial velocity change: Δu = V + V_j.2) Thrust: T = m_dot * Δu.3) Useful power: P_use = T * V = m_dot * (V + V_j) * V.4) Jet power increase (ideal): P_jet = (1/2) m_dot * [(V + V_j)^2 − V^2].5) Efficiency: η = P_use / P_jet = 2V / (V + V_j). Verification / Alternative check: Check limits: if V_j → V (small slip), η → 1; if V_j » V, η drops, matching intuition that excessively fast jets waste energy. Why Other Options Are Wrong: V/V_j and (V_j − V)/V_j ignore momentum-power balance. 2V_j/(V + V_j) inverts the dependence. (V + V_j)/(2V) is the reciprocal of the correct form. Common Pitfalls: Mixing absolute and relative velocities; omitting the inflow kinetic energy carried on board with the incoming stream. Final Answer: η = 2V / (V + V_j)

Question 18

For an outward-flow reaction turbine used in hydraulic power plants, identify the correct description of the water's path through the runner.

Accepted Answer

Introduction: Reaction turbines are categorized by the direction of flow relative to the runner: radial (inward or outward), axial, and mixed. Understanding the flow path is essential to interpret velocity triangles, specific speed, and efficiency ranges. Given Data / Assumptions: Outward-flow reaction turbine (historical designs; modern practice mostly uses inward/mixed). Steady incompressible flow; pressure drop occurs partly in the runner. Concept / Approach: In an outward-flow machine, water enters near the center (small radius) and moves to larger radii, gaining tangential momentum while experiencing pressure changes through guide and runner passages. This contrasts with inward-flow (e.g., Francis) where entry is near the periphery and exit near the center. Step-by-Step Solution: 1) Guide vanes admit water toward the runner center.2) Runner blades turn the flow outward to larger radii.3) Energy conversion happens via change of absolute velocity components and pressure (reaction).4) Flow leaves near the outer periphery to the casing/draft path. Verification / Alternative check: Velocity triangles for outward-flow runners show increasing radius across the blade passage, consistent with the description. Why Other Options Are Wrong: Parallel to axis: that is axial flow (e.g., Kaplan). Outer to center: describes inward-flow types. Partly radial and partly axial: a mixed-flow description (e.g., Francis modern). Tangential in/out only: an impulse idealization, not a reaction radial machine. Common Pitfalls: Confusing outward- with inward-flow terminology used in many standard textbooks. Final Answer: the water enters at the centre of the wheel and then flows towards the outer periphery of the wheel

Question 19

A jet of water strikes and leaves a fixed, smooth curved vane tangentially at inlet and outlet. What is the magnitude of the force component exerted by the jet on the vane along the normal to the vane surface? (α = inlet jet angle with tangent, β = …

Accepted Answer

Introduction: Hydraulic forces on curved vanes are evaluated by resolving momentum change along convenient axes. When a jet meets and leaves a vane tangentially, the tangential component relates to whirl, while the normal component depends on the normal projections of the jet velocity at entry and exit. Given Data / Assumptions: Fixed, smooth curved vane. Jet speed magnitude V; area a; density ρ. Inlet and outlet are tangential to the vane surface. Angles α (inlet) and β (outlet) measured from the tangent to the jet line. Concept / Approach: By linear momentum, force on the vane equals mass flow rate times the vector change in jet velocity. Resolve along the normal to the vane. For tangential approach/exit, the normal components at inlet and outlet are V sin α and V sin β (both directed into the vane normal on their respective sides). Because the vane is fixed, the normal force component sums these magnitudes. Step-by-Step Solution: 1) Mass flow rate: m_dot = ρ a V.2) Normal component of inlet velocity: V_n,in = V sin α.3) Normal component of outlet velocity: V_n,out = V sin β (same sense into the vane).4) Force along normal: F_n = m_dot * (V_n,in + V_n,out) = ρ a V^2 (sin α + sin β). Verification / Alternative check: Special case β = 0 (perfectly tangential exit) gives F_n = ρ a V^2 sin α, matching flat-plate normal result. Why Other Options Are Wrong: Cosine forms correspond to tangential components, not normal. Difference forms imply opposing normal components, which is incorrect for tangential approach/exit on a fixed vane. 2 cos α is unrelated to normal summation. Common Pitfalls: Confusing angle definitions (with tangent vs with vane chord) and mixing tangential/normal projections. Final Answer: ρ * a * V^2 * (sin α + sin β)

Question 20

Select the pump type best suited for duties requiring very high heads with relatively small discharge (typical of boiler feed to high-pressure systems and precise dosing).

Accepted Answer

Introduction: Pumps are matched to head–discharge requirements. Applications demanding very high delivery head but comparatively small flow rates (e.g., boiler feed to high-pressure drums, hydraulic presses, metering) call for machines that can generate large pressures efficiently at low capacity. Given Data / Assumptions: Required head is high; discharge is small-to-moderate. Steady, incompressible liquids (e.g., water). Preference for positive, reliable delivery against high system resistance. Concept / Approach: Reciprocating pumps are positive-displacement machines: head rise is largely independent of flow (within mechanical limits). They easily achieve very high pressures at small discharges. Centrifugal, axial, and mixed-flow pumps are roto-dynamic; very high heads typically require multi-staging or very high specific speed compromises and are less efficient at tiny flows. Step-by-Step Solution: 1) Map duty: high head, small Q → positive displacement preferred.2) Evaluate pump families: reciprocating achieves target without excessive staging.3) Conclude: reciprocating pump is most suitable. Verification / Alternative check: Industry practice: boiler feed at very high pressures often uses reciprocating or multistage centrifugal; at very small flows, reciprocating dominates. Why Other Options Are Wrong: Centrifugal/axial/mixed-flow: better for medium-to-large flows and moderate heads unless multistaged. Regenerative pumps can produce moderate high heads but at low efficiencies and not to the extent of robust reciprocating units for very high heads. Common Pitfalls: Assuming “high head” always implies centrifugal multistage without considering discharge magnitude and efficiency at small Q. Final Answer: Reciprocating pump

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