Question 1

Runaway Speed of Hydraulic Turbines — Definition What does the term “runaway speed” of a hydraulic turbine signify?

Accepted Answer

Introduction:Runaway speed is a critical safety parameter in hydropower engineering. It defines the highest speed a turbine would attain when the electrical or mechanical load is suddenly removed yet the waterway continues to supply head and flow before the governor throttles the gates.Given Data / Assumptions:Steady head at the turbine inlet.Load is removed; inlet guide vanes may still be open momentarily.No external braking; negligible mechanical losses for the short transient.Concept / Approach:With no load, the resisting torque drops dramatically, so the hydraulic torque accelerates the runner until hydraulic and mechanical losses balance. The resulting equilibrium speed (often 1.6–2.0 times rated, depending on type) is called the runaway speed. Design of shafts, bearings, and generators must account for this contingency.Step-by-Step Solution:Assume sudden load rejection → torque imbalance.Runner accelerates due to excess hydraulic torque.Speed rises until losses equal input torque; that speed is the runaway speed under zero load.Verification / Alternative check:Manufacturer data specify runaway multipliers for Pelton, Francis, Kaplan; protection systems are sized accordingly.Why Other Options Are Wrong:full load/maximum overload: these involve positive shaft power absorption, not free running.no damage: not a definition; survivability is a separate design requirement.Common Pitfalls:Confusing synchronous/generator overspeed with the hydraulic runaway condition; concepts are related but not identical.Final Answer:at which the turbine will run freely without load

Question 2

Pelton Wheel Geometry — Bucket Width Proportion For Pelton wheels, the recommended bucket width is typically how many times the jet diameter?

Accepted Answer

Introduction:Pelton turbines are impulse machines in which a free jet is divided and turned by double-cupped buckets. Proper geometric proportions of the bucket relative to jet diameter are critical for capture efficiency and to prevent jet interference.Given Data / Assumptions:Jet diameter = d; design follows standard empirical ratios.Smooth buckets; negligible leakage and splash for proportioning logic.Concept / Approach:Design handbooks recommend bucket width B ≈ 5d, bucket depth ≈ 1.2d, and bucket pitch ≈ 2.3d. The width allows the jet to expand and split around the splitter without spillage while maintaining a compact wheel diameter and acceptable bucket spacing.Step-by-Step Solution:Start from empirical proportion set for Pelton buckets.Select width B to be about 5 times the jet diameter d.Check with other ratios (depth, pitch) to avoid interference with adjacent buckets.Verification / Alternative check:Physical model tests and CFD confirm that B ≈ 5d offers high hydraulic efficiency with manageable bucket weight.Why Other Options Are Wrong:double/three times: too narrow, risking spillage and reduced energy capture.four times: still narrow for typical high-efficiency designs; 5d is the common practice.Common Pitfalls:Confusing width with pitch or depth; each has different recommended multiples of d.Final Answer:five times

Question 3

Pump Suction — Consequences of Insufficient NPSH If the pump’s Net Positive Suction Head (NPSH) requirement is not satisfied in operation, what will occur?

Accepted Answer

Introduction:NPSH ensures that the absolute pressure at a pump’s suction stays above the vapor pressure of the liquid, preventing vapor bubble formation. Failing to meet NPSH leads directly to cavitation, which damages components and degrades performance.Given Data / Assumptions:Liquid: water at operating temperature.Known pump NPSH required (NPSHr) from manufacturer curves.Available NPSH (NPSHa) falls below NPSHr.Concept / Approach:When NPSHa Step-by-Step Solution:Assess suction conditions to compute NPSHa.Compare to NPSHr: if NPSHa Expect performance issues: head breakdown and noise.Verification / Alternative check:Observation: a drop in head–capacity curve at high flows and “gravel” sound indicates cavitation.Why Other Options Are Wrong:no flow: pumps may still deliver some flow; the primary issue is vapor formation.efficiency will be low / excessive power: these can occur secondary to cavitation; the root cause/outcome is cavitation itself.Common Pitfalls:Confusing suction recirculation or air entrainment with cavitation; NPSH specifically addresses vapor pressure margin.Final Answer:cavitation will be formed

Question 4

Centrifugal Pumps — General Characteristics Which of the following statements about centrifugal pumps is correct?

Accepted Answer

Introduction:Centrifugal pumps dominate water supply and process services because they handle large flows smoothly and compactly. Comparing them with reciprocating pumps clarifies typical application domains and performance differences.Given Data / Assumptions:Standard single-stage centrifugal pump at its best efficiency region.Conventional single-acting reciprocating pump for comparison.Water-like liquids; steady-state operation assumptions.Concept / Approach:Centrifugal pumps produce head through dynamic action and are best for moderate to low head with large discharge. They are compact, need simpler foundations, and deliver near-pulsation-free flow. Reciprocating pumps excel at high heads and low flows with higher mechanical efficiency but require heavier foundations and produce pulsating discharge.Step-by-Step Solution:Relate flow/head domain: centrifugal → large Q, small–moderate H.Assess installation: centrifugal → smaller footprint, simpler baseplate.Compare efficiencies: reciprocating often higher peak efficiency; centrifugal lower but adequate and more practical for big Q.Verification / Alternative check:Industry selection charts and textbooks state these comparative traits consistently.Why Other Options Are Wrong:Each individual statement a–c is true; therefore “all of the above” is the most complete correct choice.Common Pitfalls:Assuming centrifugal pumps are always more efficient; at very low flow/high head, reciprocating machines may win.Final Answer:all of the above

Question 5

Turbine Specific Speed — Physical Meaning The specific speed of a turbine is defined as the speed of an imaginary turbine, geometrically similar to the given turbine, which operates under unit head and what output condition?

Accepted Answer

Introduction:Specific speed allows turbines of different sizes and heads to be compared on a common basis. It specifies the speed at which a geometrically similar turbine would run to produce a standard output at a standard head, guiding type selection and speed matching to generators.Given Data / Assumptions:Unit head means H = 1 m (in conventional metric usage).Output condition used in the definition is unit power.Geometric and dynamic similarity are implied.Concept / Approach:The accepted definition for turbines is: “speed of an imaginary geometrically similar turbine which develops unit power under unit head.” From similarity, the common formula is Ns = N * sqrt(P) / H^(5/4) (rpm-based), consistent with the physical definition stated.Step-by-Step Solution:Fix H = 1 and P = 1 (unit values).Determine the rotational speed N that a similar turbine would need at those unit conditions.That N is, by definition, the specific speed of the actual turbine.Verification / Alternative check:Type ranges: Pelton has low Ns, Francis medium, Kaplan high; the computed Ns places a site within these families.Why Other Options Are Wrong:unit discharge phrases: apply to pump specific speed, not turbine definition.unit speed: not part of the turbine Ns definition.Common Pitfalls:Mixing turbine and pump specific speed formulas; turbines use power, pumps use discharge.Final Answer:develops unit power under unit head

Question 6

Kaplan turbine selection by specific speed (metric system) A Kaplan (axial-flow reaction) turbine is generally preferred for installations having a specific speed in the range of approximately 300 to 1000 r.p.m. (metric specific speed scale). State …

Accepted Answer

Introduction:Specific speed is a widely used index to match hydraulic turbines to site conditions of head and discharge. Kaplan turbines are axial-flow reaction machines suited to low heads and high discharges, which correspond to higher specific speed values. This question checks recognition of the typical specific speed band for Kaplan selection in the metric system.Given Data / Assumptions:Specific speed in metric form, customarily expressed in r.p.m. with power in kW and head in metres.Standard turbine families: Pelton (impulse, high head, low specific speed), Francis (mixed flow, medium specific speed), Kaplan/propeller (axial flow, high specific speed).Typical guideline ranges are approximate and may vary slightly by source and manufacturer.Concept / Approach:Metric specific speed for turbines is commonly defined as N_s = N*sqrt(P) / H^(5/4). For lower heads H with comparable power P, N_s rises; thus axial-flow turbines occupy the high N_s region. Practical design charts place Kaplan around N_s ≈ 300 to 1000, Francis around ≈ 60 to 300 (overlapping), and Pelton typically ≈ 10 to 60 (single-jet lower, multi-jet higher but still well below Kaplan).Step-by-Step Solution:Relate head classes to turbine types: low head → axial-flow (Kaplan), medium head → mixed-flow (Francis), high head → impulse (Pelton).Recall typical metric specific speed bands from selection charts.Identify that 300–1000 r.p.m. aligns with Kaplan practice; therefore the statement is correct.Verification / Alternative check:Manufacturer catalogues and classic charts consistently show Kaplan at high N_s due to the axial-flow geometry and blade pitch control designed for low-head sites with large discharges. While exact cutoffs differ slightly by text, the band 300–1000 is a standard teaching guideline.Why Other Options Are Wrong:No / below 200 / above 1000: Do not match accepted ranges for Kaplan; below 200 is closer to Francis, and above 1000 is beyond common Kaplan guidelines.Applicable to Pelton: Pelton is a low N_s impulse turbine, not axial-flow.Common Pitfalls:Confusing specific speed (selection index) with unit speed; mixing metric with imperial definitions; assuming the ranges are exact rather than practical guidelines with overlap between families.Final Answer:Yes

Question 7

Compute the specific speed of a turbine (metric) A hydraulic turbine develops 10,000 kW under a head of 25 m and runs at 135 r.p.m. Determine the specific speed N_s (metric), choosing the nearest value.

Accepted Answer

Introduction:Specific speed is a similarity parameter that helps select the turbine type suited to the site. In the metric definition most used in textbooks, it combines rotational speed, power, and head into a single index. This problem tests correct application of the formula and basic arithmetic handling of exponents.Given Data / Assumptions:Power P = 10,000 kW.Head H = 25 m.Speed N = 135 r.p.m.Metric definition: N_s = N * sqrt(P) / H^(5/4).Steady operation and geometric similarity implied by the definition.Concept / Approach:The metric specific speed relates families of turbines at different scales but similar flow patterns. The exponent 5/4 on head arises from dimensional analysis using power scaling with head and discharge for dynamically similar turbines. Careful evaluation of H^(5/4) is essential to avoid exponent mistakes.Step-by-Step Solution:Write the formula: N_s = N * sqrt(P) / H^(5/4)Compute sqrt(P): sqrt(10,000) = 100Compute H^(5/4): H^(5/4) = 25^(1.25) = 25^(5/4) = (25^(1/4))^5; numerical value ≈ 55.9017Substitute: N_s ≈ 135 * 100 / 55.9017 ≈ 241.5 r.p.m.Verification / Alternative check:Sanity check by order of magnitude: for H = 25 m and moderate speed, a value in the low hundreds is expected (Francis range). The computed ≈ 241.5 r.p.m. fits selection charts for medium-head turbines.Why Other Options Are Wrong:175.4 or 215.5: result from incorrect exponent handling or rounding.275.4 or 305.0: typically arise from using H^(1/2) instead of H^(5/4), or from arithmetic slips.Common Pitfalls:Using discharge instead of power in the formula; replacing H^(5/4) with H or sqrt(H); forgetting consistent units (kW and metres) that the metric N_s convention assumes.Final Answer:241.5 r.p.m.

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