Question 1

Barrage layout concept: In a river barrage, the crest level is normally kept how, relative to gate size, to control pond level and pass floods economically?

Accepted Answer

Introduction / Context: A barrage is a gated diversion structure that maintains pond level for canal offtakes and allows efficient flood passage. The choice of crest level and gate size directly influences afflux, energy dissipation, and economy. Given Data / Assumptions: River with variable discharges including floods. Gated structure (barrage), not a fixed weir. Objective: control headwater level with minimal afflux during high flows. Concept / Approach: Keeping a low crest reduces permanent obstruction; large gates provide adjustable opening, enabling low afflux during flood and controlled pond level during lean periods. This flexibility is the principal benefit of a barrage over a high fixed weir. Step-by-Step Solution: Select low crest to minimize constant head and reduce sediment deposition upstream.Use large gates to regulate pond level and pass floods efficiently.Therefore: low crest with large gates is the standard arrangement. Verification / Alternative check: Case studies of major barrages (e.g., on alluvial rivers) consistently show gated spans above a relatively low crest profile with stilling basins for energy dissipation. Why Other Options Are Wrong: High crest with large/no gates: Increases afflux and siltation; less flexible. Low crest with no gates: Behaves like a low fixed weir; poor control. Medium crest with small gates: Restricts capacity, increases upstream levels. Common Pitfalls: Confusing barrages with weirs; overlooking sediment management and afflux limits when raising crest levels. Final Answer: Low crest with large gates

Question 2

River morphology: A deficit of suspended bed-material load in flowing water will most likely cause the river to become of which type?

Accepted Answer

Introduction / Context: Alluvial rivers adjust their bed and planform to the balance between sediment supply and transport capacity. When sediment supply is insufficient relative to transport capacity, the bed tends to lower (degrade). Given Data / Assumptions: Sediment deficit relative to carrying capacity. Unarmored or weakly armored alluvial bed. Hydraulic conditions otherwise unchanged. Concept / Approach: With limited incoming bed material, the flow entrains material from the bed to satisfy transport capacity, producing net erosion. Conversely, sediment surplus leads to aggradation. Planform (e.g., meandering) is influenced by many factors, but sediment balance strongly dictates bed level trends. Step-by-Step Solution: Identify imbalance: transport capacity > sediment supply.Flow scours the bed to make up the deficit.Outcome: bed level lowers → degrading reach. Verification / Alternative check: Field evidence after sediment-trapping dams shows downstream degradation due to sediment-starved flows; grade-control structures are used to limit incision. Why Other Options Are Wrong: Aggrading type: Requires sediment surplus, opposite condition. Meandering type: Describes planform, not necessarily bed level trend caused purely by deficit. Sub-critical type: Refers to flow regime (Froude number), not sediment balance. Common Pitfalls: Confusing planform descriptors with bed level trends; ignoring that both suspended and bed loads contribute to morphodynamics. Final Answer: Degrading type (bed level lowering)

Question 3

Meander belt width definition: In river engineering, the width of a meander belt is the transverse distance between which two characteristic points of successive bends?

Accepted Answer

Introduction / Context: Meander geometry is described using wavelength, amplitude, belt width, and radius of curvature. These descriptors help to set river corridor widths and plan levee alignments. Given Data / Assumptions: Planform of a free meandering river. “Apex” refers to the outermost point of a bend. Concept / Approach: The meander belt width quantifies the lateral sweep of the river pathway. It is commonly taken as the transverse distance between the apices of two successive bends in opposite directions (a bend and its reverse bend). Step-by-Step Solution: Identify two consecutive alternate bends (left and right).Locate their outward apices (maximum lateral excursions).Measure the transverse separation between these apices → meander belt width. Verification / Alternative check: Textbook sketches show belt width as twice the meander amplitude for symmetrical planforms; apex-to-apex distance aligns with this definition. Why Other Options Are Wrong: Apex-to-crossing: Measures along-meander shift, not belt width. Two banks at one section: That is just the river width, not belt width. Midpoints of crossings: Not a standard definition. Common Pitfalls: Confusing bankfull channel width with meander belt; measuring along the thalweg instead of transverse to valley axis. Final Answer: Apex point of one bend and apex point of the reverse bend

Question 4

Canal instrumentation: Bed bars provided across a canal section are primarily intended for what operational purpose?

Accepted Answer

Introduction / Context: Operational observation features are installed on canals to monitor bed levels and silt movement. One such feature is the bed bar, a small permanent marker or bar across the bed. Given Data / Assumptions: Unlined or lined canal with potential aggradation/degradation. Need for long-term monitoring of bed elevation. Bed bars act as reference points, not flow control structures. Concept / Approach: Bed bars serve as fixed datum lines or reference features to periodically check bed profile changes. They are not meant to function as measuring weirs or to appreciably alter hydraulic grade line. Step-by-Step Solution: Install low-profile bar across bed at specified stations.Use sounding/levelling to observe bed level relative to bar elevations.Infer silting or scouring trends to guide maintenance and desilting. Verification / Alternative check: Instrumentation manuals treat bed bars as observation aids. Discharge measurement uses calibrated structures (weirs/flumes), not bed bars. Why Other Options Are Wrong: Measure discharge: Requires hydrometric structures; bed bars are unsuitable. Raise supply level: Needs control structures (regulators/weirs). Control silting by throttling: Bed bars are passive; they do not control flow. All the above: Incorrect because only observation is valid. Common Pitfalls: Assuming any cross-bed feature acts as a weir; overlooking that bed bars are primarily for monitoring. Final Answer: To watch and record the general behaviour (bed changes/silting) of the canal

Question 5

Selecting correct canal design statement: Which of the following statements is correct regarding canal levels and regime design (Lacey)?

Accepted Answer

Introduction / Context: Alluvial channel design follows empirical relations such as Lacey’s regime equations tying discharge, silt factor, hydraulic radius, velocity, and slope. Also, practical canal levels are set with regard to terrain but need not always be above ground. Given Data / Assumptions: Regime (quasi-stable) alluvial channel conveying uniform silt. Use of Lacey’s silt factor f and discharge Q. Consideration of FSL relative to natural ground. Concept / Approach: Lacey related slope S with discharge and silt factor through regime relations. Statement (b) correctly reflects that a particular slope is implied for given Q and f. FSL may lie above or below ground depending on cutting/embankment needs; it is not a universal rule. Admitting coarser silt increases f, not decreases it; hence (c) is wrong. Step-by-Step Solution: Evaluate (a): Not universally true—canals may be in cutting.Evaluate (b): True—Lacey’s equations yield a characteristic slope for given Q and f.Evaluate (c): False—coarser silt raises f; cannot reduce required slope that way.Therefore, only (b) is correct. Verification / Alternative check: From Lacey: V = k1 * f^0.5 * R^0.5 and S = k2 * f^(5/3) * Q^(−1/6) (indicative forms). These tie S to Q and f. Why Other Options Are Wrong: (a) Over-generalization. (c) Misstates effect of silt size on silt factor. (d) All-the-above fails as (a) and (c) are wrong. Common Pitfalls: Assuming canals must be in embankment; misinterpreting how sediment size affects regime parameters. Final Answer: According to Lacey, regime conditions require a particular bed slope for a given discharge and silt factor

Question 6

Kennedy’s theory – trial design step: The iterative (trial) procedure for canal design using Kennedy’s theory primarily employs which relationship between discharge and section/velocity?

Accepted Answer

Introduction / Context: Kennedy’s regime method sizes unlined canals by ensuring that mean velocity equals the “critical velocity” which does not cause silting or scouring. The trial procedure uses continuity and Kennedy’s empirical velocity-depth relation. Given Data / Assumptions: Steady, uniform flow in an alluvial canal. Critical velocity V relates to flow depth y and sediment via the critical velocity ratio m. Discharge Q must equal area A times velocity V. Concept / Approach: Kennedy proposed V = 0.55 * m * y^0.64 (SI units, with m the critical velocity ratio). The trial begins with an assumed section (hence y and A), computes V from the critical velocity relation, and checks Q = A * V. Section parameters are iterated to satisfy both hydraulics and velocity criterion. Step-by-Step Solution: Use continuity: Q = A * V.Use Kennedy: V = 0.55 * m * y^0.64.Iterate section dimensions until computed Q matches the design discharge and velocity equals the critical value. Verification / Alternative check: Some designers combine Kennedy with Chezy/Manning to ensure slope adequacy; however, the cornerstone relation in the trial is the critical velocity formula with continuity. Why Other Options Are Wrong: Q = A * y: Dimensionally incorrect; y is depth, not velocity. Chezy-only: Ignores the critical velocity criterion central to Kennedy. All the above: Incorrect because only option (a) captures Kennedy’s trial essence. Common Pitfalls: Using wrong exponent in V–y relation; omitting the m-factor; failing to reconcile depth with hydraulic slope constraints. Final Answer: Q = A * V together with V = 0.55 * m * y^0.64

Question 7

Lined canal with circular invert and straight flanks: If the straight sides of a lined canal meet a circular bottom (invert) of radius R and make an angle θ with the horizontal, the cross-sectional area A equals

Accepted Answer

Introduction / Context: Several lined canal sections use a circular invert to reduce seepage and erosion, with straight side slopes meeting tangentially. A compact formula for area is useful in hydraulic computations (discharge, velocity, and slope checks). Given Data / Assumptions: Bottom is a circular arc of radius R. Straight sides meet the invert, making angle θ with the horizontal. Geometry forms a circular segment plus two triangular side portions expressed via θ. Concept / Approach: Area is the sum of the circular segment area and the areas bounded by straight flanks up to intersection levels. For the common proportion used in design charts, the closed-form expression simplifies to A = R^2(θ + tan θ) (θ in radians). Step-by-Step Solution: Express segment area in terms of R and θ.Express triangular contributions using slope angle θ.Combine to yield A = R^2(θ + tan θ). Verification / Alternative check: Dimensional check: R^2 multiplies a dimensionless bracket, giving area units. For small θ, tan θ ≈ θ and area behaves as ~R^2(2θ), consistent with a narrow section. Why Other Options Are Wrong: R(θ + …): Wrong dimensions. cot θ variants: Do not match this standard geometry. Duplicate option with R(θ + tan θ) is dimensionally incorrect. Common Pitfalls: Using degrees instead of radians in computation; mixing geometric relations of circular segment with trapezoidal assumptions. Final Answer: R^2(θ + tan θ)

Question 8

Canal outlets classification: An outlet that maintains practically constant discharge irrespective of fluctuations in the upstream channel water level and the watercourse level is called a

Accepted Answer

Introduction / Context: Canal outlets are devices that draw water from distributaries into watercourses. Their hydraulic behaviour under upstream/downstream level fluctuations defines their type and equity of distribution among farmers. Given Data / Assumptions: Outlet required to deliver nearly constant discharge. Both distributary (supplying channel) and watercourse levels may fluctuate. Device calibrated to be insensitive to such variations. Concept / Approach: A “right modular outlet” (also called a rigid module) is designed so that its discharge is essentially independent of both upstream and downstream heads within a working range. Non-modular outlets are sensitive to both levels; semi-modular depend only on upstream head (insensitive to downstream). Step-by-Step Solution: Classify by discharge sensitivity to levels.Need: independence from both → modular (rigid).Therefore, the correct classification is right modular outlet. Verification / Alternative check: Handbooks present proportional, semi-modular, and modular types; rigid modules are used where equitable distribution regardless of backwater is essential. Why Other Options Are Wrong: Non-modular: Varies with both levels. Semi-modular: Independent of downstream only. Flexible modular: Non-standard phrasing; “rigid module” is the accepted term. Open flume outlet: Descriptive geometry, not behaviour class. Common Pitfalls: Confusing semi-modular with modular; assuming proportional outlets are constant discharge (they are not). Final Answer: Right modular outlet (rigid module)

Question 9

Canal design practice: A minimum freeboard of 90 cm is provided for which discharge range in the canal? (Select the most appropriate typical range used in irrigation design handbooks.)

Accepted Answer

Introduction / Context: Freeboard is the vertical clearance provided above the full supply level in canals to accommodate wind setup, wave action, superelevation at bends, and uncertain rises in water level. Typical values are standardized against discharge ranges for safe and economical canal sections. Given Data / Assumptions: Asked: discharge range that commonly warrants a minimum freeboard of 90 cm. Context: irrigation canal design (earthen or lined) with standard handbook recommendations. Objective: recall/recognize typical tabulated practice rather than compute. Concept / Approach: As discharge increases, wave height and velocity head effects grow, so freeboard is stepped up in ranges. Many reference tables assign about 60 cm for small distributaries, about 90 cm for medium canals in the 30–60 m^3/s range, and 120 cm or more for larger canals. Step-by-Step Solution: Link discharge range to standard freeboard tables.Identify the range that typically maps to 0.90 m.Select 30 to 60 cumecs. Verification / Alternative check: Cross-check with state PWD/Water Resources Dept. manuals that list staged freeboards (e.g., 0.6 m; 0.9 m; 1.2 m+) across discharge bands; medium canals (30–60 cumecs) usually adopt 0.9 m. Why Other Options Are Wrong: 30 to 33 cumecs: Narrow slice; handbooks group 30–60 cumecs together. Over 60 cumecs / Over 100 cumecs: Typically require ≥1.2 m, not just 0.9 m. None of these: Incorrect because 30–60 cumecs is a standard mapping to 0.9 m. Common Pitfalls: Assuming one value fits all canal classes; ignoring added allowances at bends or windy reaches; forgetting that lining type can slightly influence adopted freeboard. Final Answer: 30 to 60 cumecs

Question 10

River meanders: The meander length, meander width, and river width are empirically observed to vary roughly with which function of discharge?

Accepted Answer

Introduction / Context: Alluvial rivers adjust their planform and cross section to water and sediment supply. Empirical regime relationships relate channel width, meander wavelength, and other geometric attributes to the bankfull discharge. Given Data / Assumptions: We seek the approximate functional dependence of meander scale and channel width on discharge. Alluvial, freely meandering rivers with mobile bed and banks. Concept / Approach: Regime and hydraulic-geometry studies (e.g., Leopold–Maddock type relations) commonly show power-law trends where width and meander wavelength increase with discharge to a power less than 1. A frequently cited approximation is proportionality to Q^0.5 (square root dependence) for width and related meander metrics. Step-by-Step Solution: Recall empirical relations: width ∝ Q^b with b around 0.4–0.6.Meander wavelength and belt width scale with channel width and hence with Q^b.Approximate practical choice: square root (Q^0.5). Verification / Alternative check: Field datasets show scatter, but the square-root trend is commonly accepted for quick estimates, with constants fitted to local conditions. Why Other Options Are Wrong: Linear, square, cube powers greatly overpredict size at high discharges. None of these: incorrect because the square-root relation is widely used as a rule of thumb. Common Pitfalls: Applying a single exponent universally; ignoring sediment size, bank strength, and vegetation which modify coefficients and exponents. Final Answer: Square root of the discharge

Question 11

Regime channels (Lacey): The regime (stable) mean velocity in an alluvial channel is proportional to which combination of hydraulic radius R and bed slope S?

Accepted Answer

Introduction / Context: Velocity–slope–size relations in alluvial channels are often expressed through regime or resistance equations. Although Lacey introduced relations among velocity, silt factor, hydraulic radius, and slope for channels in regime, the resulting form reflects a square-root dependence on slope and a sub-linear dependence on hydraulic radius. Given Data / Assumptions: Mean velocity V in a stable (regime) alluvial channel. Hydraulic radius R and bed slope S as governing variables in a proportionality form. Concept / Approach: For practical purposes, regime/roughness formulations lead to V being proportional to S^(1/2) and to a fractional power of R, commonly near 2/3. This matches the general resistance behavior also echoed by Manning-type expressions for wide channels. Step-by-Step Solution: Recall proportionality: V ∝ R^m * S^n with n ≈ 1/2 for uniform open-channel flow.Empirical fits/regime theory produce m in the range 0.5–0.7; 2/3 is a practical value.Therefore choose V ∝ R^(2/3) * S^(1/2). Verification / Alternative check: Comparisons with Manning’s V = (1/n) R^(2/3) S^(1/2) corroborate the selected exponents for regime-scale proportionality, noting Lacey’s additional dependence on sediment factor in full form. Why Other Options Are Wrong: R^(1/2) * S^(3/4): Overstates slope influence and understates radius. Q^(3/4) * S^(1/3): Introduces discharge directly and a weaker slope exponent. R^(3/4) * S^(1/3): Slope exponent too small for regime resistance trends. Common Pitfalls: Confusing Lacey’s complete set (including silt factor f) with simplified proportionalities; mixing discharge-based and geometry-based forms. Final Answer: R^(2/3) * S^(1/2)

Question 12

Kennedy’s theory of stable channels: For a unique (single) design solution using Kennedy’s approach, which geometric information must be known in advance?

Accepted Answer

Introduction / Context: Kennedy’s regime theory relates non-silting, non-scouring velocity to flow depth for silt-laden channels. Because the theory does not directly yield both width and depth simultaneously, an additional geometric relation is required to arrive at a unique section. Given Data / Assumptions: Kennedy’s velocity–depth relation: V = m * y^(1/6) (m depends on silt grade). Discharge Q is known and continuity must be satisfied (Q = A * V). A shape assumption is needed to close the problem. Concept / Approach: With V tied to y, and Q known, we still have two unknowns (width and depth) for a rectangular (or nearly rectangular) section. Specifying the ratio B/D provides the necessary closure to determine both B and D uniquely. Step-by-Step Solution: Use V = m * y^(1/6) to relate velocity to depth.Apply Q = A * V with A = B * y to link width and depth.Provide B/D to close the system and solve for B, y uniquely. Verification / Alternative check: Design examples in texts specify side slopes and B/D to determine a unique section under Kennedy’s theory, followed by stability checks. Why Other Options Are Wrong: Only breadth or only depth does not close the system with the given relations. “All of the above” is false; only B/D provides the necessary closure. Common Pitfalls: Assuming Kennedy alone gives unique B and D; forgetting to adopt a shape parameter; confusing with Lacey’s set where additional parameters are available. Final Answer: Only the ratio of breadth to depth (B/D)

Question 13

Soil water forms in irrigation science: Which of the following statements is correct regarding water available to plants and forms of soil water?

Accepted Answer

Introduction / Context: Plant-available water is governed by how moisture is stored in soils. Understanding gravity, hygroscopic, and capillary water helps in irrigation scheduling and field capacity concepts. Given Data / Assumptions: We compare three forms of soil water and their relevance to plant uptake. Standard agronomy definitions are used. Concept / Approach: Gravity water drains rapidly through macropores after rainfall/irrigation and is not retained for uptake (it is not inherently “harmful,” it is simply transient). Hygroscopic water is held as thin films by adsorption with high suction and is generally unavailable to plants. Capillary water is held by surface tension in meso/micropores and constitutes the main fraction available to roots between field capacity and wilting point. Step-by-Step Solution: Identify which form plants actually use: capillary water.Recognize that hygroscopic is too tightly bound; gravity water usually drains.Select the statement that correctly reflects plant availability: option (c). Verification / Alternative check: Soil–water characteristic curves consistently show available water between field capacity and permanent wilting point—predominantly capillary-held moisture. Why Other Options Are Wrong: (a) “Harmful” is inaccurate; gravity water mainly percolates beyond root zone. (b) Hygroscopic water is not generally plant-available. (d) All of the above is false due to (a) and (b). Common Pitfalls: Equating temporary waterlogging with gravity water being intrinsically harmful; conflating adsorption (hygroscopic) with capillarity. Final Answer: Capillary moisture held in soil pores by surface tension is utilized by plants

Question 14

Sarda falls crest comparison: For identical head and geometric parameters, the ratio of discharge over a trapezoidal crest to that over a rectangular crest (Sarda type) is approximately:

Accepted Answer

Introduction / Context: Sarda fall crests are standardized profiles used in canal drops. The discharge coefficient varies slightly with crest geometry; trapezoidal crests generally pass a bit more discharge than rectangular crests under the same head and width conditions. Given Data / Assumptions: Identical head, approach conditions, and overall width; only crest shape differs. Ratio sought: Q_trapezoidal / Q_rectangular. Concept / Approach: Empirical calibration of Sarda crest shapes shows small but consistent increments for trapezoidal sections due to improved flow contraction and velocity distribution, leading to a discharge ratio modestly above unity, typically quoted near 1.07. Step-by-Step Solution: Recognize that the trapezoidal crest has a slightly higher effective coefficient.Select the tabulated ratio closest to 1.07.Choose 1.074. Verification / Alternative check: Design manuals list ratios around 1.07 for comparable Sarda geometries; the exact value may vary slightly with side slopes and approach conditions, but 1.074 is a standard choice for objective questions. Why Other Options Are Wrong: 1.084/1.064/1.054: Reasonable but less commonly cited; 1.074 aligns with standard references. 2.053: Unrealistically large. Common Pitfalls: Assuming a large difference between crest types; in reality, differences are within a few percent. Final Answer: 1.074

Question 15

Hydrostatics of plane areas: For a plane surface of area A immersed in water with its centroid at depth x below the free surface (water density ω), which statement is correct about total pressure and the depth of the centre of pressure?

Accepted Answer

Introduction / Context: The hydrostatic pressure on a plane surface varies linearly with depth. Designers need both the resultant (total) force and the location (centre of pressure) to design gates, slabs, and tank walls. Given Data / Assumptions: Plane surface of area A; centroid at depth x below the free surface. Water density (specific weight) is ω. Concept / Approach: Total hydrostatic pressure on a plane area is P = ω A x (for plane surface of uniform density fluid). The centre of pressure is deeper than the centroid; its depth y_cp from the free surface is y_cp = I_o / (A x), where I_o is the second moment of area of the surface about the free surface. Equivalently, using the parallel-axis theorem, y_cp = x + (I_G / (A x)). Step-by-Step Solution: Compute total force: P = ω A x (depends on A and x).Locate centre: y_cp = I_o / (A x) = x + I_G / (A x).Recognize the correct statement corresponds to the I/(A x) relation. Verification / Alternative check: Derivation integrates pressure p = ω y over the surface and equates first moments to obtain the I/(A x) formula; standard in fluid mechanics texts. Why Other Options Are Wrong: (a) Omits area A; incomplete. (c) 2/3 H applies only to a triangular pressure diagram on a vertical rectangle with the free surface at the top; not general. (d) Cannot be true because (a) and (c) are not universally valid. Common Pitfalls: Confusing I_o and I_G; forgetting that centre of pressure lies below centroid for vertical planes. Final Answer: Depth of the point of action (centre of pressure) equals its second moment of area about the free surface divided by A x

Question 16

Definition of waterlogging in irrigation practice: A tract is declared water-logged when which condition is met?

Accepted Answer

Introduction / Context: Waterlogging is a critical hazard in irrigated agriculture. It reduces soil aeration, impairs root respiration, and lowers yields. The operational definition focuses on the position of the water table relative to the crop root zone. Given Data / Assumptions: Root zone depth varies by crop and soil but is typically the top 0.6–1.5 m. Prolonged saturation in this zone is the key criterion. Concept / Approach: If the water table rises sufficiently to keep the root zone near-saturated, oxygen diffusion is limited, and toxic conditions develop. Hence, waterlogging is defined with respect to the root zone rather than a fixed depth such as 40 or 50 cm, which may not universally apply. Step-by-Step Solution: Identify the agronomic criterion: position of water table within root zone.Reject fixed-depth definitions as crop- and soil-dependent.Select option (c). Verification / Alternative check: Guidelines specify critical depths (e.g., < 1–1.5 m below ground) as warning thresholds, adjusted for crop rooting depth; thus the root-zone criterion is general. Why Other Options Are Wrong: (a) and (b) are arbitrary depths; not universally valid. (d) cannot be right because (a) and (b) are not general definitions. Common Pitfalls: Using a single depth threshold for all crops; neglecting seasonal variability and soil textural effects on aeration. Final Answer: The crop root zone remains saturated (water table rises into root zone)

Question 17

Rigid module (proportional outlet) in canal outlets: The discharge characteristic is best described as:

Accepted Answer

Introduction / Context: Canal outlets (modules) control delivery from distributaries to watercourses. Their hydraulic behavior—how discharge varies with upstream and downstream levels—defines whether they are rigid, semi-rigid (proportional), or flexible modules. Given Data / Assumptions: “Rigid module” is the term in question. Steady operating conditions intended. Concept / Approach: A rigid module (constant discharge module) aims to maintain nearly constant discharge irrespective of changes in head on either side, within a working range. In contrast, proportional (flexible) modules let discharge vary with upstream level, and orifices with submerged outlets may depend on both levels. Step-by-Step Solution: Recall definition: rigid module → discharge insensitive to level fluctuations.Match to statement: independence from both distributary and watercourse levels.Select option (a). Verification / Alternative check: Outlet classification tables clearly separate rigid (constant discharge), semi-rigid (proportional), and flexible (sensitive) behaviors. Why Other Options Are Wrong: (b)/(c)/(d) imply sensitivity to one or both levels, contradicting “rigid”. (e) Not applicable. Common Pitfalls: Confusing “rigid module” with “proportional module.” The term “rigid” here means hydraulically insensitive (constant discharge). Final Answer: Independent of water levels in both the distributary and the watercourse

Question 18

Inverted siphon (cross-drainage): The uplift pressure on the roof (crown) of an inverted siphon barrel is maximum under which operating condition?

Accepted Answer

Introduction / Context: Inverted siphons carry canal water or drainage under crossings under pressure. The roof (crown) may be subjected to external hydrostatic pressure from floodwater above (e.g., a drain flowing over the siphon). Uplift checks are critical for structural safety. Given Data / Assumptions: We consider external water level above the barrel and internal counter-pressure from barrel contents. Uplift is the net upward pressure acting on the roof slab. Concept / Approach: Uplift on the roof is driven by the head of water above acting downward on the surroundings while the underside pressure depends on the internal water level. The worst condition is when the external water level is highest and the internal counterbalancing pressure is least. High external level occurs at the drain H.F.L.; if the canal were empty (dry), counterpressure would be further reduced, but the key single condition indicating maximum external head is the drain at H.F.L. Step-by-Step Solution: List scenarios and compare external head over roof.Maximum external head corresponds to drain at H.F.L.Hence uplift on the roof is maximized in option (d). Verification / Alternative check: Design checks often consider the combination “drain at H.F.L., canal empty” as the absolute worst case; among the single conditions provided, H.F.L. of drain represents the controlling uplift. Why Other Options Are Wrong: (a) Drain dry → minimal external head. (b) Canal dry → reduces internal counterpressure but needs high external water to be critical. (c) Canal at F.S.L. increases internal pressure, reducing uplift risk. Common Pitfalls: Ignoring combination states; misidentifying which watercourse produces external head. Final Answer: Drain is running at highest flood level (H.F.L.)

Question 19

Canal alignment choice across a drainage: three crossings C1, C2, C3 (with bed levels C1 > C2 > C3) If a proposed canal takes off from a river at point B and three alternative alignments of approximately equal length cross the same drainage at C1, C…

Accepted Answer

Introduction / Context: Selecting the correct cross-drainage work (CDW) at a canal–drainage crossing depends on the relative elevations of the canal bed level, canal full-supply level, and the drainage bed/high flood level. When multiple crossing sites exist, the structure should be chosen to be hydraulically sound and economical. Given Data / Assumptions: Three potential crossing locations C1, C2, C3 of similar route length. Drainage bed levels: C1 > C2 > C3 (so C3 has the lowest drainage bed). Canal alignment and full-supply level are essentially the same for each alternative. Concept / Approach: At a site where the drainage bed is low relative to the canal bed, the canal can pass over the drainage on a raised waterway—this is an aqueduct. If the drainage high flood level would submerge the canal waterway, a syphon aqueduct is used. If the drainage must pass over the canal (drain bed much higher), a superpassage is adopted. A syphon (inverted syphon) carries canal water under an obstruction when canal must go below. Step-by-Step Solution: Identify that C3 has the lowest drainage bed among the options.Lower drainage bed relative to canal favors canal carried over drainage: aqueduct.Economy and head-loss considerations usually make an aqueduct preferable at such low-bed crossings. Verification / Alternative check: Check flood levels: if drainage HFL is still below the canal soffit, a simple aqueduct suffices; only if the canal waterway would be drowned should a syphon aqueduct be considered. Why Other Options Are Wrong: Syphon aqueduct: Needed when canal must pass under a high HFL; not the favored choice here.Superpassage: Used when drainage bed/HFL is above canal; opposite situation.Syphon (inverted syphon): For carrying canal below an obstacle; not indicated by the low drain bed. Common Pitfalls: Ignoring HFL and deciding solely on bed levels. Confusing aqueduct with superpassage due to naming symmetry. Final Answer: An aqueduct

Question 20

Falls in canals: which fall tends to maintain its depth (self-regulating water depth across the crest and downstream conditions)?

Accepted Answer

Introduction / Context: Falls are energy-dissipating structures on canals to negotiate ground slope while maintaining required water levels. Some fall designs are self-regulating and help maintain depth by adjusting the effective waterway as discharge changes. Given Data / Assumptions: Comparing common fall types: trapezoidal notch fall (Sarda), rectangular notch fall, and low weir fall. Focus on depth-maintaining (self-regulating) behavior for varying discharges. Concept / Approach: The Sarda (trapezoidal notch) fall uses side slopes and a notched profile so that as discharge varies, the water surface and nappe geometry adjust, helping maintain near-constant depth upstream and ensuring stable downstream energy dissipation. Rectangular notches and simple low weir falls are less self-adjusting with discharge fluctuations. Step-by-Step Solution: Identify the fall which varies its effective waterway automatically with discharge.Trapezoidal notch geometry provides changing area–depth relation.Hence it better maintains depth compared to rectangular notches and simple low weirs. Verification / Alternative check: Operational experience shows Sarda falls give more stable depth control for distributaries and minors subject to moderate discharge variations, reducing tendency for afflux or drawdown extremes. Why Other Options Are Wrong: Low weir fall: Less self-regulating; depth is more sensitive to discharge changes.Rectangular notch fall: Cross-section does not adapt as effectively, giving poorer depth maintenance.All the above: Only the trapezoidal notch fall has the desired self-regulating feature. Common Pitfalls: Assuming any fall with a notch is equally self-regulating. Confusing “measuring” falls with “depth-maintaining” falls. Final Answer: Trapezoidal notch fall (Sarda type)

Irrigation Questions