Irrigation Engineering covers the artificial application of water to agricultural land — from the necessity and types of irrigation, through duty/delta/base period water-demand calculations, canal design (Kennedy's and Lacey's theories) and distribution systems, hydraulic structures (weirs, barrages) and gravity dams, to waterlogging control and command area development. Every formula, design procedure, diagram, solved example and exam-pattern table is included.
After studying this chapter you will be able to:
Prerequisite: Engineering Hydrology (runoff estimation and groundwater hydraulics developed there directly feed canal discharge design and conjunctive-use planning here). Leads to: Environmental Engineering, which covers water treatment and supply systems that share many hydraulic design principles with irrigation infrastructure.
Irrigation is the artificial application of water to land to assist the growing of agricultural crops. In India, about 60% of net sown area is rain-fed. Irrigation is necessary due to: (1) uneven spatial distribution of rainfall across the country; (2) uneven temporal distribution — the monsoon pattern concentrates rainfall into a few months, leaving long dry spells; (3) soil moisture deficits that occur during critical crop growth periods even in generally wet regions.
| Type | Method | Suitability |
|---|---|---|
| Flow / Canal irrigation | Water flows by gravity from river/reservoir through canal network | Flat terrain; large areas; perennial rivers |
| Lift irrigation | Water pumped from river/well/tank to field | Where gravity flow not possible; wells, tanks |
| Tank / Reservoir irrigation | Small impoundments store runoff; gravity supply | South India; hilly and undulating terrain |
| Drip (trickle) irrigation | Water delivered directly to root zone through emitters | High-value crops; water-scarce areas; 40–60% water saving |
| Sprinkler irrigation | Water sprayed over crop canopy like rainfall | Undulating land; sandy soils; 30–50% saving over surface |
| Sub-surface irrigation | Water applied below soil surface through buried pipes | Limited use; high-water-table areas |
Field Capacity (FC) is the maximum water held by soil after excess drains away by gravity — the upper limit of readily available soil moisture. The Permanent Wilting Point (PWP) is the minimum soil moisture at which plants can no longer extract water and wilt permanently. The difference between these two — the Available Water Capacity (AWC) — represents the moisture reservoir that irrigation scheduling aims to keep replenished, since allowing soil moisture to approach PWP causes crop stress well before wilting actually occurs.
Available Water Capacity: \( AWC = FC - PWP \)
Depth of water required: \( d = (FC-\theta_{actual})\times\dfrac{\gamma_b}{\gamma_w}\times(\text{depth of root zone}) \)
\(\gamma_b\)=bulk density of soil; \(\gamma_w\)=density of water
Given: A soil has FC = 28% (by weight), current moisture \(\theta_{actual}=18\%\), bulk density \(\gamma_b=1.4\ \text{g/cm}^3\), root zone depth 1.0 m. Find the depth of water to be applied.
Solution:
\( d = (0.28-0.18)\times\dfrac{1.4}{1.0}\times1.0 = 0.10\times1.4\times1.0 = 0.14\ \text{m} \)
Answer: \(d = 0.14\ \text{m} = 140\ \text{mm}\) of water must be applied.
Given: A high-value horticulture crop is grown in a water-scarce region with undulating terrain. Recommend a suitable irrigation method.
Solution:
Water scarcity favours a high-efficiency method; undulating terrain rules out simple flow/canal irrigation. Drip irrigation delivers water directly to the root zone with 40–60% savings, ideal for high-value crops where the higher capital cost per hectare is justified by crop value.
Answer: Drip (trickle) irrigation is the most suitable choice.
Fig. 1.1 — Soil moisture zones: from saturation, through field capacity (upper limit of available water) and the readily-available range, down to the permanent wilting point where plants can no longer extract water.
| Term | Definition | Units |
|---|---|---|
| Duty (D) | Area of land that can be irrigated by unit discharge (1 cumec) flowing continuously for the base period | hectares/cumec |
| Delta (Δ) | Total depth of water required by a crop for its full growth (entire base period) | metres |
| Base period (B) | Total time from first to last watering of a crop | days |
Duty, delta and base period are three ways of expressing the same underlying quantity — the total water demand of a crop — related by a simple unit-conversion identity. A high duty (large command area per cumec) implies either a small delta (crop needs little water) or a long base period (the same flow rate is spread over more time); conversely, water-intensive crops like rice have low duty and high delta.
The Gross Command Area is the entire area within the irrigation project's physical boundary; the Culturable Command Area (CCA) excludes land unfit for cultivation (waterbodies, forests, habitation) from the GCA; and the Net Command Area (NCA) is the area actually irrigated in a given season, which is typically well below the CCA since not every eligible field grows an irrigated crop every season. The ratio of NCA to CCA — the intensity of irrigation — is a key planning parameter distinguishing kharif (monsoon) and rabi (winter) cropping seasons.
\( \Delta = \dfrac{8.64B}{D} \) or equivalently \( D = \dfrac{8.64B}{\Delta} \)
Derivation: 1 cumec for B days delivers \(86400B\ \text{m}^3\), covering \(D\times10^4\ \text{m}^2\) to depth \(\Delta\): \(86400B=D\times10^4\times\Delta\).
Rice (Kharif): \(\Delta\approx1.0\text{–}1.2\ \text{m}\); base period ≈120–150 days
Wheat (Rabi): \(\Delta\approx0.35\text{–}0.40\ \text{m}\); base period ≈120–130 days
Sugarcane: \(\Delta\approx0.9\text{–}1.2\ \text{m}\); base period ≈360 days
\( GCA \ge CCA \ge NCA \)
Intensity of irrigation \( = \dfrac{NCA}{CCA} \) (typically 40–60% Kharif; 30–40% Rabi)
Canal discharge: \( Q = \dfrac{A_{ir}\times\Delta}{B\times86400} \) (\(A_{ir}\)=area irrigated, m²)
Given: A crop has duty \(D=1000\ \text{ha/cumec}\) and base period \(B=120\) days. Find the delta.
Solution:
\( \Delta = 8.64\dfrac{\times120}{1000} = \dfrac{1036.8}{1000} = 1.0368\ \text{m} \)
Answer: \(\Delta\approx1.037\ \text{m}\) (consistent with a rice-like crop).
Given: A wheat crop has \(\Delta=0.38\ \text{m}\), \(B=125\) days. Find the duty.
Solution:
\( D = 8.64\dfrac{\times125}{0.38} = \dfrac{1080}{0.38} \approx 2842\ \text{ha/cumec} \)
Answer: \(D\approx2842\ \text{ha/cumec}\) — much higher than rice, consistent with wheat's lower water demand.
Given: A canal is to irrigate \(A_{ir}=5000\ \text{ha}=5\times10^7\ \text{m}^2\), with \(\Delta=1.0\ \text{m}\), \(B=120\) days. Find the required canal discharge.
Solution:
\( Q = \dfrac{5\times10^7\times1.0}{120\times86400} = \dfrac{5\times10^7}{1.0368\times10^7} \approx 4.82\ \dfrac{\text{m}^3}{\text{s}} \)
Answer: \(Q\approx4.82\ \text{cumecs}\).
Fig. 2.1 — Nested relationship: GCA (entire project area) ⊇ CCA (land fit for cultivation) ⊇ NCA (area actually irrigated in a given season).
R.G. Kennedy studied canals in Punjab and concluded that a canal is in regime (no silting or scouring) when the critical velocity equals the mean velocity of flow. Critical velocity is the velocity that keeps silt in suspension without either depositing it (silting, if velocity is too low) or eroding the bed (scouring, if velocity is too high).
Kennedy's critical velocity depends only on the depth of flow (through an empirical power law) and a Critical Velocity Ratio (CVR) that adjusts the formula for silt grades other than the standard silt Kennedy originally studied — finer silt needs a lower critical velocity (CVR < 1) to stay in regime, while coarser silt needs a higher one (CVR > 1). Since Kennedy's formula alone has two unknowns (width and depth) for one equation, a design also requires checking against Manning's equation (or Kutter's formula), iterating until both the critical-velocity criterion and the flow-resistance equation are simultaneously satisfied.
Kennedy's theory has several recognised shortcomings: it does not account for bed load (coarser) sediment transport separately from suspended load; it does not define a unique cross-section (infinitely many width/depth combinations can satisfy the same velocity criterion, requiring an assumed side slope and trial-and-error); it treats bed slope as an external design input rather than deriving it; and it provides only a velocity criterion, giving no direct guidance on the channel's overall shape.
\( V_0 = C\times m\times y^{0.64} \)
\(C=0.55\) for Indian channels (original value); \(y\)=depth of flow (m)
\(m\)=Critical Velocity Ratio (CVR): \(m=1.0\) standard silt; \(m<1\) fine silt; \(m>1\) coarse silt
1. Assume depth \(y\); compute \(V_0=0.55\,m\,y^{0.64}\)
2. Compute area: \(A=\dfrac{Q}{V_0}\)
3. For trapezoidal section (side slope 1H:2V standard): solve for \(B\) and \(y\)
4. Check with Manning: \(V=\tfrac1nR^{2/3}S^{1/2}\)
5. Iterate until \(V=V_0\) and Manning's \(V\) agree
Given: A canal has a flow depth of 1.5 m, standard silt (\(m=1.0\)). Find the critical velocity.
Solution:
\( V_0 = 0.55\times1.0\times1.5^{0.64} \)
\( 1.5^{0.64} \approx 1.298 \)
\( V_0 \approx 0.55\times1.298 \approx 0.714\ \text{m/s} \)
Answer: \(V_0\approx0.71\ \text{m/s}\).
Given: The same canal as Example 1 (\(y=1.5\ \text{m}\)) carries fine silt with \(m=0.8\). Find the critical velocity.
Solution:
\( V_0 = 0.55\times0.8\times1.298 \approx 0.571\ \text{m/s} \)
Answer: \(V_0\approx0.57\ \text{m/s}\) — lower than the standard-silt case, since fine silt requires less velocity to stay in suspension.
Given: A canal must carry \(Q=15\ \dfrac{\text{m}^3}{\text{s}}\) with critical velocity \(V_0=0.75\ \text{m/s}\). Find the required cross-sectional area.
Solution:
\( A = \dfrac{Q}{V_0} = \dfrac{15}{0.75} = 20\ \text{m}^2 \)
Answer: \(A=20\ \text{m}^2\) — the starting point for solving the trapezoidal section dimensions.
Fig. 3.1 — Kennedy's critical velocity concept: a canal is "in regime" when the mean velocity equals the critical velocity that keeps silt in suspension without silting or scouring the bed.
Gerald Lacey recognised that a channel is in true regime when it carries a particular discharge with a particular silt load without silting or scouring — a more complete formulation than Kennedy's, since Lacey additionally introduced the silt factor \(f\), a quantitative measure derived directly from the mean particle size of the bed material, rather than a qualitative ratio like Kennedy's CVR.
Lacey's method provides a complete, self-consistent set of equations for velocity, flow area, hydraulic radius, wetted perimeter and bed slope — all expressed purely in terms of the design discharge \(Q\) and the silt factor \(f\), requiring no separately-assumed side slope or additional resistance equation the way Kennedy's method does. The theoretical regime channel shape is semicircular, though practical canals use a trapezoidal approximation with a standard 1H:2V side slope, and the width-to-depth ratio for typical regime channels falls in a fairly narrow, well-established range.
| Feature | Kennedy's Theory | Lacey's Theory |
|---|---|---|
| Silt parameter | CVR (m) — qualitative | Silt factor f — quantified from particle size |
| Slope | External input (design assumed) | Determined from regime equations |
| Section shape | Not uniquely defined | Semi-circular (regime); unique section |
| Wetted perimeter | Not addressed | \(P=4.75\sqrt Q\) |
| Accuracy | Less accurate | More accurate; widely used in India |
\( f = 1.76\sqrt{m_r} \) (\(m_r\)=mean particle diameter in mm)
\(f=1.0\) for standard Kennedy silt (\(d=0.323\ \text{mm}\))
\( V = 0.6385\times(fQ)^{1/6} \)
Flow area: \( A = \dfrac{Q}{V} \)
\( R = 0.4715\times(\dfrac{Q}{f^2})^{1/3} \) or \( V^2=\dfrac{fR}{2.5} \Rightarrow R=\dfrac{2.5V^2}{f} \)
Wetted perimeter: \( P=4.75\sqrt Q \)
Bed slope: \( S = \dfrac{f^{5/3}}{3340\,Q^{1/6}} \)
Theoretical: semicircular; practical: trapezoidal, side slope 1H:2V
Width-to-depth ratio \(\dfrac{B}{y}\approx3.8\text{–}4.5\) for typical channels
Given: Bed material has a mean particle size of 0.5 mm. Find the silt factor.
Solution:
\( f = 1.76\sqrt{0.5} = 1.76\times0.707 \approx 1.245 \)
Answer: \(f\approx1.25\).
Given: A canal carries \(Q=50\ \dfrac{\text{m}^3}{\text{s}}\) with silt factor \(f=1.0\). Find the regime velocity and wetted perimeter.
Solution:
\( V = 0.6385\times(1.0\times50)^{1/6} = 0.6385\times50^{1/6} \)
\( 50^{1/6} \approx 1.919 \); \( V \approx 0.6385\times1.919 \approx 1.225\ \text{m/s} \)
\( P = 4.75\sqrt{50} = 4.75\times7.071 \approx 33.6\ \text{m} \)
Answer: \(V\approx1.23\ \text{m/s}\), \(P\approx33.6\ \text{m}\).
Given: Using \(Q=50\ \dfrac{\text{m}^3}{\text{s}}\), \(f=1.0\) from Example 2, find the hydraulic radius and bed slope.
Solution:
\( R = 0.4715\times(\dfrac{50}{1.0^2})^{1/3} = 0.4715\times50^{1/3} = 0.4715\times3.684 \approx 1.737\ \text{m} \)
\( S = \dfrac{1.0^{5/3}}{3340\times50^{1/6}} = \dfrac{1.0}{3340\times1.919} \approx \dfrac{1.0}{6410.5} \approx 1.56\times10^{-4} \)
Answer: \(R\approx1.74\ \text{m}\), \(S\dfrac{\approx1}{6410}\) (or \(\approx0.000156\)).
Fig. 4.1 — Lacey's theoretical regime channel is semicircular; practical design uses a trapezoidal approximation with 1H:2V side slopes and a width-to-depth ratio of about 3.8–4.5.
| Canal Level | Takes off from | Function |
|---|---|---|
| Main canal | River headworks / reservoir | Carries bulk discharge; no direct irrigation |
| Branch canal | Main canal | Feeds distributaries; may have direct irrigation |
| Distributary | Branch canal | Carries water to watercourses; \(Q\) typically 0.03–0.3 m³/s |
| Minor | Distributary | Smaller channel for a group of villages |
| Watercourse | Minor / distributary | Last channel before farm; \(Q<0.03\) m³/s |
| Field channel | Watercourse | On-farm; directly irrigates plots |
Lining a canal — with concrete, boulder/stone masonry, brick, shotcrete, or plastic film — dramatically reduces seepage (roughly a 10-fold reduction compared to an unlined earthen canal), increases the permissible velocity (since erosion resistance no longer limits flow speed), reduces maintenance (no re-desilting or bank repair from erosion), and suppresses weed growth that would otherwise obstruct flow. Concrete lining is the most durable and widely used option for major and medium canals.
Water is lost between the headworks and the farm through seepage (by far the dominant loss, especially in old unlined systems) and evaporation (a comparatively minor loss). These transit losses, combined with losses during actual field application, are captured by three efficiency terms: water conveyance efficiency (how much water reaches the farm relative to what was released at the headworks), water application efficiency (how much of the water delivered to the farm is actually stored in the crop's root zone), and the product of the two — the overall project efficiency.
Seepage loss, unlined canal: 1–3 m³/s per million m² of wetted perimeter (typical India)
Seepage loss, lined canal: 0.1–0.3 m³/s per million m² (≈10× reduction)
Transit losses = Seepage + Evaporation
Evaporation loss ≈ 0.25–1.0% of flow
Seepage loss: 15–25% of water at head; 50–60% in old unlined systems
Water application efficiency: \( \eta_a = \dfrac{\text{water stored in root zone}}{\text{water delivered at farm}} \)
Water conveyance efficiency: \( \eta_c = \dfrac{\text{water delivered at farm}}{\text{water released at headworks}} \)
Overall project efficiency: \( \eta = \eta_a\times\eta_c \) (typically 35–50%)
Given: A project has water conveyance efficiency \(\eta_c=70\%\) and water application efficiency \(\eta_a=65\%\). Find the overall project efficiency.
Solution:
\( \eta = 0.70\times0.65 = 0.455 \)
Answer: \(\eta\approx45.5\%\) — within the typical 35–50% range.
Given: A crop requires 800 mm of water stored in the root zone. Using \(\eta_a=60\%\) and \(\eta_c=75\%\), find the volume that must be released at the headworks per hectare.
Solution:
Water delivered at farm \(=\dfrac{800}{0.60}=1333\ \text{mm}\)
Water released at headworks \(=\dfrac{1333}{0.75}\approx1778\ \text{mm}\)
Answer: Approximately 1778 mm (1.778 m) per hectare must be released at the headworks — more than double the crop's actual need, illustrating the cumulative impact of conveyance and application losses.
Given: An unlined canal has a wetted perimeter of 2 million m² and loses water at 2 m³/s per million m². Estimate the seepage saving if the canal is lined (using 0.2 m³/s per million m² for the lined case).
Solution:
Unlined seepage \(= 2\times2 = 4\ \dfrac{\text{m}^3}{\text{s}}\)
Lined seepage \(= 0.2\times2 = 0.4\ \dfrac{\text{m}^3}{\text{s}}\)
Saving \(= 4-0.4 = 3.6\ \dfrac{\text{m}^3}{\text{s}}\)
Answer: Lining saves approximately 3.6 m³/s of seepage loss — a 90% reduction.
Fig. 5.1 — Canal distribution hierarchy: water flows from the main canal through progressively smaller branch, distributary, minor, watercourse and field channels before reaching individual farm plots.
A weir is a low-head dam built across a river to raise the water level for diversion, with no gates (or a fixed crest). A barrage has adjustable gates spanning the full width, allowing discharge control at all river stages — barrages are preferred for modern irrigation headworks precisely because gate operation lets the pond level be maintained even as river flow varies through the year, unlike a fixed weir crest.
Bligh (1910) assumed that seepage beneath a weir follows the base profile of the structure, losing head uniformly along this "creep length" — a simple linear model that only checks total creep length against total head, without distinguishing horizontal from vertical seepage paths and without addressing the exit gradient (where seepage water re-emerges at the downstream ground surface, the location where piping failure actually initiates).
Lane refined Bligh's simple creep-length concept by observing that vertical (or steeply-inclined) seepage paths are more effective at dissipating head than horizontal paths, so vertical creep is given three times the weight of horizontal creep in the total "weighted creep length" — a design with plenty of vertical cutoffs (sheet piles) is therefore safer than an equivalent length of horizontal apron, a refinement Bligh's method misses entirely.
Khosla (1936) provided an exact mathematical solution (via conformal transformation) for seepage flow beneath a hydraulic structure, directly yielding both the pressure distribution along the base and — critically — the exit gradient, which neither Bligh's nor Lane's method computes directly. Since the exit gradient governs the actual onset of piping (soil particles being carried away by seepage force at the downstream end), Khosla's theory superseded the older creep theories as the basis for safe design, with permissible exit gradient values tabulated by soil type (finer soils require a smaller, safer exit gradient than coarser ones).
Safe hydraulic gradient: \( i_{safe} = \dfrac{1}{C_B} \)
Required creep length: \( L_c = C_B\times H \)
\(C_B\) values: fine micaceous sand 15; fine sand 12; coarse sand 9; boulders/gravel 4–6
\( L_w = \tfrac13L_H+L_V \)
\(L_H\)=total horizontal creep; \(L_V\)=total vertical creep
Required: \( L_w/H\ge C_L \) (weighted creep coefficient)
\( G_E = \dfrac{H}{d\,\lambda} \) (sheet pile at downstream end)
\( \lambda = \dfrac{1+\sqrt{1+\alpha^2}}{2} \), \(\alpha=\dfrac{b}{d}\)
\(H\)=total head; \(d\)=depth of downstream pile; \(b\)=base width of structure
Fine sand: 1/6; Medium sand: 1/5; Coarse sand: 1/4; Gravel: 1/3
Given: A weir on fine sand (\(C_B=12\)) has a total head of 5 m. Find the required creep length.
Solution:
\( L_c = 12\times5 = 60\ \text{m} \)
Answer: \(L_c=60\ \text{m}\).
Given: A structure has base width \(b=40\ \text{m}\), downstream pile depth \(d=5\ \text{m}\), total head \(H=4\ \text{m}\). Find the exit gradient.
Solution:
\( \alpha = \dfrac{b}{d} = \dfrac{40}{5} = 8 \)
\( \lambda = \dfrac{1+\sqrt{1+64}}{2} = \dfrac{1+8.062}{2} = 4.531 \)
\( G_E = \dfrac{H}{d\lambda} = \dfrac{4}{5\times4.531} = \dfrac{4}{22.66} \approx 0.1766 \)
Answer: \(G_E\approx0.177 \approx \dfrac{1}{5.66}\) — should be checked against the safe value for the actual foundation soil.
Given: A weir has total horizontal creep \(L_H=45\ \text{m}\) and total vertical creep \(L_V=12\ \text{m}\), with head \(H=6\ \text{m}\). Find the weighted creep ratio.
Solution:
\( L_w = \dfrac{45}{3}+12 = 15+12 = 27\ \text{m} \)
\( L_w/H = \dfrac{27}{6} = 4.5 \)
Answer: Weighted creep ratio \(=4.5\) — compare against the safe \(C_L\) for the foundation material to check adequacy.
Fig. 6.1 — Seepage beneath a weir: Bligh's theory follows the creep path along the base profile; Khosla's theory additionally computes the exit gradient at the downstream toe, where piping failure actually initiates.
| Force | Direction | Effect | Point of Action |
|---|---|---|---|
| Water pressure (\(F_{WS}\)) | Horizontal | Destabilising (overturning) | Triangular distribution, acts at \(\dfrac{H}{3}\) from base |
| Self-weight (\(W\)) | Vertical | Stabilising | Centre of gravity of dam section |
| Uplift (\(U\)) | Vertical (upward) | Destabilising | Varies from full reservoir head at heel to tailwater head at toe |
| Silt pressure (\(F_s\)) | Horizontal | Destabilising | Acts at \(\dfrac{H_s}{3}\) from base (\(H_s\)=silt depth) |
| Earthquake force | Horizontal + vertical | Dynamic, destabilising | Acts at centre of gravity (inertia force) |
| Wave pressure | Horizontal | Destabilising (minor) | Near reservoir full level (FRL) |
| Ice pressure | Horizontal | Destabilising (minor, cold climates) | Near FRL |
The dominant forces for design are water pressure, self-weight, and uplift — the remaining forces (silt, earthquake, wave, ice) are typically smaller but must still be checked in the load combinations relevant to the site.
The theoretical elementary profile of a gravity dam is a right-angled triangle with a vertical upstream face, zero width at the top (apex at reservoir full level), and base width \(B\) — this shape gives the minimum practical cross-section that satisfies both the no-overturning and no-tension stability requirements simultaneously for the full-reservoir condition.
For no tension to develop at the heel under full reservoir condition (uplift considered), the required base width works out to \( B = \dfrac{H}{\sqrt{G-1}} \), where \(G\) is the specific gravity of the dam material (concrete/masonry, typically \(G=2.4\)–\(2.5\)). A commonly used approximate (no-uplift) form is \( B \approx \dfrac{H}{\sqrt{G}} \) for the no-flotation check, but the no-tension condition including uplift governs practical design and gives the larger, controlling base width.
In practice, the actual dam section is built wider than this theoretical triangle (practical profile) to accommodate a road/gallery at the top and provide a margin against the idealised assumptions.
A gravity dam section must be checked against four independent modes of failure:
A spillway is the safety valve of a dam — it passes surplus flood water from the reservoir to the downstream channel, preventing overtopping of the dam. The most common type on a gravity dam is the ogee (overflow) spillway, whose crest profile follows the lower nappe of a freely falling sheet of water so that the flow just touches the surface at the design head (avoiding negative pressure and cavitation).
River training guides the flow of a river along a desired course to protect banks, hydraulic structures (weirs, barrages, bridges) and adjoining land from erosion and flooding. The main works are:
\( B = \dfrac{H}{\sqrt{G-1}} \) (no tension, with uplift)
\(G\)=specific gravity of dam material (≈2.4–2.5)
\( FOS_O = \dfrac{\Sigma M_R}{\Sigma M_O}\ge1.5 \)
\( FOS_S = \dfrac{\mu\Sigma V}{\Sigma H}\ge1.0 \)
\(\mu\)=0.7–0.75 typical (concrete on rock)
\( p = \dfrac{\Sigma V}{B}\left(1\pm\dfrac{6e}{B}\right) \)
No tension: \( e\le \dfrac{B}{6} \) (resultant within middle third)
Given: A gravity dam of height \(H=60\ \text{m}\) uses concrete with \(G=2.4\). Find the theoretical base width for no tension.
Solution:
\( B = \dfrac{H}{\sqrt{G-1}} = \dfrac{60}{\sqrt{1.4}} = \dfrac{60}{1.183}\approx50.7\ \text{m} \)
Answer: \(B\approx50.7\ \text{m}\).
Given: At the base of a dam section, \(\Sigma V=4500\ \text{kN/m}\), base width \(B=15\ \text{m}\), and the resultant acts at \(2\ \text{m}\) from the base centre (\(e=2\ \text{m}\)). Check whether tension develops.
Solution:
Middle-third limit: \( \dfrac{B}{6} = \dfrac{15}{6} = 2.5\ \text{m} \)
Since \(e=2\ \text{m} < 2.5\ \text{m}\), the resultant lies within the middle third — no tension develops.
\( p_{max,min} = \dfrac{4500}{15}\left(1\pm\dfrac{6\times2}{15}\right) = 300(1\pm0.8) = 540\ \text{and}\ 60\ \text{kN/m}^2 \)
Answer: Both \(p_{max}=540\ \text{kN/m}^2\) and \(p_{min}=60\ \text{kN/m}^2\) are compressive (positive) — no tension, section is safe on this criterion.
Given: \(\Sigma V=5000\ \text{kN/m}\), \(\Sigma H=2600\ \text{kN/m}\), \(\mu=0.75\). Check sliding safety.
Solution:
\( FOS_S = \dfrac{\mu\Sigma V}{\Sigma H} = \dfrac{0.75\times5000}{2600} = \dfrac{3750}{2600}\approx1.44 \)
Answer: \(FOS_S\approx1.44 \ge 1.0\) — safe against sliding (though below the more conservative 1.5 threshold sometimes used when shear-friction is not separately credited).
Fig. 7.1 — Elementary triangular profile of a gravity dam showing the three dominant forces: triangular water pressure (destabilising, acting at H/3 from base), self-weight (stabilising, at the section's centre of gravity), and uplift (destabilising, varying from full head at the heel to tailwater head at the toe).
Waterlogging occurs when the water table rises so close to the ground surface that it enters the plant root zone, cutting off aeration and reducing crop yield or killing the crop entirely. It is a widespread problem in intensively irrigated command areas in India, affecting well over 8.4 million hectares of cultivable land, and is a direct consequence of the same canal irrigation that raised agricultural productivity — making it a critical design and management concern, not merely an incidental side-effect.
| Cause | Mechanism |
|---|---|
| Over-irrigation | Applying water in excess of crop requirement percolates below the root zone, gradually raising the water table |
| Canal seepage | Unlined canals typically lose 15–30% of their discharge to seepage, which directly recharges and raises the surrounding water table |
| Poor natural drainage | Flat topography and/or impermeable sub-surface layers prevent percolated water from moving away, so it accumulates |
| Intensive cropping | Continuous multiple cropping keeps the soil saturated for longer periods, sustaining a permanently high water table |
Preventive measures (avoid the problem before it starts): canal lining (directly cuts seepage recharge — the single most effective preventive measure), correct assessment and application of duty/delta (avoids over-irrigation), preferring high-consumptive-use crops in areas already prone to waterlogging, and conjunctive use of surface and groundwater (pumping groundwater for irrigation lowers the water table while also supplying additional water).
Curative measures (remedy an already-waterlogged area): open surface drains to intercept and remove excess surface/near-surface water, sub-surface tile or pipe drainage systems buried below the root zone to intercept and carry away the shallow water table, and tube-well pumping to actively draw down the water table while the pumped water is reused for irrigation.
In poorly drained irrigated soils, dissolved salts carried by irrigation water accumulate in the root zone as water evaporates or is transpired, leaving salts behind (salinization). To keep the root-zone salt concentration in balance, some extra water beyond crop requirement must be applied to leach accumulated salts below the root zone — this extra fraction is the leaching requirement (LR), and both waterlogging and salinity control ultimately depend on adequate drainage to carry this leached water away.
\( LR = \dfrac{EC_w}{5\,EC_e - EC_w} \)
\(EC_w\)=electrical conductivity of irrigation water; \(EC_e\)=permissible EC of saturation extract for the crop
\( NIR = ET - P_e \)
\(ET\)=crop evapotranspiration; \(P_e\)=effective rainfall
\( GIR = \dfrac{NIR}{(1-LR)\,\eta_a} \)
\(\eta_a\)=field application efficiency
Given: Irrigation water has \(EC_w=1.2\ \text{dS/m}\); the crop's permissible saturation-extract EC is \(EC_e=4\ \text{dS/m}\). Find the leaching requirement.
Solution:
\( LR = \dfrac{1.2}{5\times4-1.2} = \dfrac{1.2}{18.8}\approx0.064 \)
Answer: \(LR\approx6.4\%\) — about 6.4% extra water beyond crop need must be applied to control salt build-up.
Given: Crop \(ET=6\ \text{mm/day}\), effective rainfall \(P_e=1\ \text{mm/day}\), \(LR=0.10\), field application efficiency \(\eta_a=0.70\). Find GIR.
Solution:
\( NIR = 6-1 = 5\ \text{mm/day} \)
\( GIR = \dfrac{5}{(1-0.10)\times0.70} = \dfrac{5}{0.63}\approx7.94\ \text{mm/day} \)
Answer: \(GIR\approx7.94\ \text{mm/day}\).
Given: A command area has water table rising steadily by 0.4 m/year and unlined distributaries losing an estimated 25% of discharge to seepage. What preventive measure would most directly address the dominant cause?
Solution: With 25% conveyance loss (well within the typical 15–30% range for unlined canals), canal seepage is very likely the dominant contributor to the rising water table.
Answer: Lining the distributaries is the most direct and effective preventive measure, since it attacks the seepage recharge at its source rather than only treating the symptom after the water table has already risen.
Fig. 8.1 — Waterlogging cross-section: canal seepage raises the water table into the crop root zone; a sub-surface tile drain intercepts and removes the shallow water table as a curative remedy.
A persistent problem in Indian irrigation is the large gap between irrigation potential created (the CCA that canal infrastructure could theoretically serve) and irrigation potential utilised (the CCA actually receiving water) — the CADP was launched to bridge this gap through on-farm and distribution-level interventions. Its main components are: construction of field channels (linking outlets to individual fields), land levelling and shaping (so water spreads evenly instead of pooling), on-farm development works generally, and promotion of conjunctive use of surface and groundwater to make the most of the available supply.
Warabandi is a rotational water-distribution system, historically associated with North-West India (Punjab, Haryana, and parts of Rajasthan), under which each irrigator on a watercourse receives water for a fixed duration each week, proportional to the area of land they hold — rather than continuous or on-demand supply, everyone gets a guaranteed, equitable turn regardless of their position along the watercourse.
The weekly duration allotted to a given holding is computed as:
| Category | Culturable Command Area (CCA) |
|---|---|
| Major project | CCA > 10,000 ha |
| Medium project | 2,000 ha ≤ CCA ≤ 10,000 ha |
| Minor project | CCA < 2,000 ha |
\( t = 168\times\dfrac{A_i}{A_{total}} \) (hours/week)
\(A_i\)=area of individual holding; \(A_{total}\)=total area served by the watercourse; \(168\)=hours in a week
Major: \(CCA>10{,}000\ \text{ha}\)
Medium: \(2{,}000\le CCA\le10{,}000\ \text{ha}\)
Minor: \(CCA<2{,}000\ \text{ha}\)
Given: A watercourse serves a total command area of \(A_{total}=80\ \text{ha}\). A farmer's holding is \(A_i=5\ \text{ha}\). Find the weekly Warabandi turn duration.
Solution:
\( t = 168\times\dfrac{5}{80} = 168\times0.0625 = 10.5\ \text{hours/week} \)
Answer: The farmer receives \(10.5\) hours of water per week, proportional to their share of the total command area.
Given: A proposed irrigation project has a culturable command area of \(6{,}500\ \text{ha}\). Classify the project.
Solution: Since \(2{,}000\ \text{ha} \le 6{,}500\ \text{ha} \le 10{,}000\ \text{ha}\), the project falls in the medium category.
Answer: Medium irrigation project.
Given: A watercourse commands \(120\ \text{ha}\) in total, divided among 6 farmers each holding \(20\ \text{ha}\). Verify that the sum of individual turn durations equals one full week.
Solution:
Each farmer's share: \( t_i = 168\times\dfrac{20}{120} = 28\ \text{hours} \)
Total for 6 equal farmers: \( 6\times28 = 168\ \text{hours} = 1\ \text{week} \)
Answer: The sum exactly equals \(168\) hours, confirming the rotation covers the full week with no gaps or overlaps, as the Warabandi system requires.
Fig. 9.1 — Warabandi rotational schedule: the 168-hour week is divided among farmers on a watercourse, each sector's angular size proportional to that farmer's share of the total command area.
A cross-drainage (CD) work is a structure built where an irrigation canal crosses a natural drain (stream) so the two can pass without interfering. The choice of type depends on the relative bed levels of the canal and the drain and on their discharges.
| Type | Relative levels & arrangement |
|---|---|
| Aqueduct | Canal carried over the drain in a trough; drain flows at atmospheric pressure below. Used when the canal bed is well above the drain's high flood level. |
| Syphon aqueduct | Canal over the drain, but the drain is forced through a barrel under pressure (syphon) because its flood level is above the canal trough bottom. |
| Super-passage | The drain is carried over the canal in a trough; used when the drain bed is above the canal's full-supply level. |
| Canal syphon | The canal passes under the drain, running under pressure through a barrel. |
| Level crossing | Canal and drain water mix in a common pool at the same level, regulated by cross-regulators; used for large drains. |
| Inlet & outlet | A small drain is simply admitted into the canal (inlet) and let out further down (outlet) — the cheapest option for very small drains. |
The decisive factor is the relative bed and flood levels of the canal and the drain. When the canal is comfortably above the drain's flood level → aqueduct; when the drain floods above the canal trough → syphon aqueduct; when the drain is above the canal → super-passage or canal syphon. Discharge magnitudes, available head loss and cost then refine the choice.
\( \Delta = \dfrac{8.64\,B}{D} \)
\( V_0 = 0.55\,m\,y^{0.64} \)
\( f = 1.76\sqrt{d_{mm}} \)
\( V^2 = \dfrac{fR}{2.5} \)
\( P = 4.75\sqrt{Q} \)
\( S = \dfrac{f^{5/3}}{3340\,Q^{1/6}} \)
\( L_c = C_B\times H \)
\( G_E = \dfrac{H}{d\,\lambda} \), \( \lambda=\dfrac{1+\sqrt{1+\alpha^2}}{2} \)
\( e \le \dfrac{B}{6} \)
\( \dfrac{\Sigma M_R}{\Sigma M_O}\ge1.5 \)
\( \dfrac{\mu\Sigma V}{\Sigma H}\ge1.0 \)
Major \(CCA>10{,}000\ \text{ha}\); Minor \(CCA<2{,}000\ \text{ha}\)
| Chapter | GATE | ESE | SSC-JE / State PSC |
|---|---|---|---|
| Necessity & Types | Low | Moderate — conceptual | High — direct definitions |
| Duty, Delta & Base Period | High — numerical | High — numerical | High — numerical & conceptual |
| Kennedy's Theory | High — numerical | High — numerical | High |
| Lacey's Theory | High — numerical | High — numerical | Moderate |
| Canal Distribution | Low | Moderate | High — hierarchy & efficiency |
| Weirs & Barrages | High — Khosla numerical | High — Khosla numerical | Moderate |
| Gravity Dams | High — stability numerical | High — stability numerical | Moderate |
| Waterlogging & Drainage | Moderate | Moderate | High — causes/remedies |
| Command Area Development | Low | Moderate — Warabandi | High — classification |
Q1. A crop has a base period of 100 days and requires a duty of 1200 ha/cumec at the field. Find the delta in metres.
Q2. For a channel with normal depth \(y=2.5\ \text{m}\) and Kennedy's critical velocity ratio \(m=1.1\), find the critical velocity.
Q3. A channel carries \(Q=50\ \text{cumecs}\) with Lacey's silt factor \(f=1.0\). Find the regime wetted perimeter and bed slope.
Q4. A weir on coarse sand (\(C_B=9\)) has a total head of \(6\ \text{m}\). Find the required Bligh creep length, and compare with a weir on fine micaceous sand under the same head.
Q5. A gravity dam section has \(\Sigma V=6000\ \text{kN/m}\), base width \(B=18\ \text{m}\), and eccentricity \(e=3.5\ \text{m}\). Does tension develop at the heel?