BE Civil Engineering (IOE, TU) Irrigation and Drainage Engineering (IOE, CE 705) Question Paper 2078 Nepal

Definitions

Duty (D): The area of land (in hectares) that can be irrigated by a unit discharge (1 m$^3$/s = 1 cumec) of water flowing continuously throughout the base period. Units: ha/cumec.

Delta ($\Delta$): The total depth of water (in cm or m) required by a crop over its entire base period to bring it to maturity. Units: cm or m.

Base period (B): The total time (in days) between the first watering at sowing and the last watering before harvesting.

Relation between Duty and Delta

A discharge of 1 cumec running for the base period $B$ days delivers a volume:

$$V = 1 \times (B \times 24 \times 60 \times 60) = 86400\,B \ \text{m}^3$$

This volume spreads over $D$ hectares ($D \times 10^4$ m$^2$) to a depth $\Delta$ (m):

$$86400\,B = D \times 10^4 \times \Delta$$

With $\Delta$ in metres:

$$\boxed{\Delta = \frac{8.64\,B}{D}} \quad\text{(}\Delta\text{ in m, }B\text{ in days, }D\text{ in ha/cumec)}$$

(a) Duty and area on the field

Given $B = 120$ days, $\Delta = 40\,\text{cm} = 0.40\,\text{m}$.

$$D = \frac{8.64\,B}{\Delta} = \frac{8.64 \times 120}{0.40} = \frac{1036.8}{0.40} = 2592\ \text{ha/cumec}$$

Area irrigated by $Q = 2.5$ cumec:

$$A = D \times Q = 2592 \times 2.5 = \mathbf{6480\ ha}$$

Field duty $= \mathbf{2592\ ha/cumec}$.

(b) Duty at canal head

Conveyance efficiency $\eta_c = 75\% = 0.75$. The duty at the head accounts for transit losses, so:

$$D_{head} = \eta_c \times D_{field} = 0.75 \times 2592 = \mathbf{1944\ ha/cumec}$$

Less area is served per cumec at the head because part of the water is lost in conveyance.

(c) Delta for rice

Given $B = 90$ days, $D = 800$ ha/cumec.

$$\Delta = \frac{8.64 \times 90}{800} = \frac{777.6}{800} = 0.972\,\text{m} = \mathbf{97.2\ cm}$$

The large delta is consistent with rice being a heavy water-consuming crop.

Assumptions of Kennedy's Silt Theory

1. The silt is kept in suspension solely by the vertical eddies generated from the channel bed (not the sides).
2. The quantity of silt carried is a function of the mean velocity of flow.
3. A channel is in a state of non-silting and non-scouring when its mean velocity equals the critical velocity $V_0$, which depends only on the depth of flow.
4. The eddies supporting silt are proportional to the bed width.

Limitations: No equation for bed slope is given (a $B/D$ ratio must be assumed), so the solution is not unique; the silt grade factor $m$ is arbitrary; trial-and-error is required; the effect of silt charge and side eddies is ignored.

Design Procedure (trial-and-error on D)

Given $Q=12$ m$^3$/s, $m=1.0$, side slope $0.5{:}1$, $S=1/5000$, $N=0.0225$, $B/D=5$.

Geometry (trapezoidal, side slope 0.5H:1V):

• Area $A = BD + 0.5D^2 = 5D\cdot D + 0.5D^2 = 5.5D^2$
• Wetted perimeter $P = B + 2D\sqrt{1+0.5^2} = 5D + 2D(1.118) = 7.236D$
• Hydraulic radius $R = A/P = 5.5D^2 / 7.236D = 0.7601\,D$

Step 1 — assume D, compute V0: Try $D = 1.7$ m.

$$V_0 = 0.55 \times 1.0 \times 1.7^{0.64}$$

$1.7^{0.64} = e^{0.64\ln 1.7} = e^{0.64\times 0.5306} = e^{0.3396} = 1.4044$

$$V_0 = 0.55 \times 1.4044 = 0.7724\ \text{m/s}$$

Step 2 — discharge from this V0 and area: $A = 5.5 \times 1.7^2 = 5.5 \times 2.89 = 15.895\ \text{m}^2$ $Q_{calc} = A V_0 = 15.895 \times 0.7724 = 12.28\ \text{m}^3/\text{s}$ — close to 12, slightly high.

Try $D = 1.68$ m: $1.68^{0.64} = e^{0.64\times 0.5188} = e^{0.3320} = 1.3937$; $V_0 = 0.55\times1.3937 = 0.7665$ m/s. $A = 5.5\times1.68^2 = 5.5\times2.8224 = 15.523\ \text{m}^2$; $Q = 15.523\times0.7665 = 11.90\ \text{m}^3/\text{s}$.

Interpolating, take $D = 1.685$ m, giving $Q \approx 12.0$ m$^3$/s. Then $B = 5D = \mathbf{8.43\ m}$, $D \approx \mathbf{1.685\ m}$.

At $D=1.685$: $V_0 = 0.55\times 1.685^{0.64}$; $1.685^{0.64}=e^{0.64\times0.5217}=e^{0.3339}=1.3964$, $V_0=0.7680$ m/s. $A = 5.5\times1.685^2 = 15.616\ \text{m}^2$, $R = 0.7601\times1.685 = 1.281\ \text{m}$.

Step 3 — Verify mean velocity by Kutter–Chezy:

Kutter's $C = \dfrac{23 + \frac{1}{N} + \frac{0.00155}{S}}{1 + \left(23 + \frac{0.00155}{S}\right)\frac{N}{\sqrt{R}}}$

$\frac{1}{N}=44.44$, $\frac{0.00155}{S}=0.00155\times5000=7.75$. Numerator $= 23 + 44.44 + 7.75 = 75.19$. Denominator $= 1 + (23+7.75)\frac{0.0225}{\sqrt{1.281}} = 1 + 30.75 \times \frac{0.0225}{1.1318} = 1 + 30.75\times0.01988 = 1 + 0.6113 = 1.6113$. $C = 75.19/1.6113 = 46.66$.

Mean velocity $V = C\sqrt{RS} = 46.66\sqrt{1.281\times \tfrac{1}{5000}} = 46.66\sqrt{0.0002562} = 46.66\times0.016006 = 0.7468\ \text{m/s}$.

Check: $V = 0.747$ m/s vs $V_0 = 0.768$ m/s. $V/V_0 = 0.97 \approx 1$ — very close. The channel is essentially in regime; a marginal steepening of the slope (to about $1/4700$) would make $V = V_0$ exactly.

Final Design

Bligh's Creep Theory

Bligh assumed water creeps along the contact of the base profile of the structure with the subsoil, losing head uniformly along the total creep length $L$ (sum of horizontal + vertical paths, treated as equivalent). Safety against piping requires:

$$L \ge C\,H \qquad\text{i.e.}\qquad \frac{H}{L} \le \frac{1}{C}$$

Shortcomings of Bligh's theory (overcome by Khosla):

1. No distinction between horizontal and vertical creep (Khosla showed vertical creep is more effective).
2. The actual uplift pressure distribution is not linear, as Bligh assumed.
3. The position and importance of sheet piles (especially the outer faces) is not properly accounted for.
4. Bligh overestimated the value of intermediate sheet piles and ignored the exit gradient concept (Khosla introduced the exit gradient criterion $G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}}$).

Khosla's Theory of Independent Variables

Khosla solved the Laplace seepage equation treating the composite floor as a set of simple standard forms (a straight horizontal floor; a floor with a pile at one end; a floor with an intermediate pile; a depressed floor). The actual uplift pressures at key points are obtained from these standard forms and then corrected for (i) mutual interference of piles, (ii) the thickness of the floor, and (iii) the slope of the floor.

(a) Bligh check for piping

Creep length by Bligh = horizontal floor + 2 × (depth of d/s pile):

$$L = 30 + 2(6) = 42\ \text{m}$$

Required creep length $= C H = 12 \times 5 = 60\ \text{m}$.

Since $L = 42\ \text{m} < 60\ \text{m}$, the floor is NOT safe against piping by Bligh's theory for the given head. The percentage shortfall is $(60-42)/60 = 30\%$; either the floor must be lengthened or the head reduced (safe head by Bligh $= L/C = 42/12 = 3.5$ m).

(b) Residual head and uplift at base of d/s pile (Khosla)

For a floor with a pile at the downstream end, the relevant Khosla key points at the pile are $E$ (floor level, u/s side of pile) and $D$ (bottom of pile). Given:

• $\phi_E = 71\%$ (pressure head as % of $H$ at point E)
• $\phi_D = 80\%$ (pressure head as % of $H$ at the base D of the pile)

These are the residual seepage pressures expressed as a percentage of the seepage head $H = 5$ m.

Uplift pressure at base of downstream pile (point D):

$$P_D = \phi_D \times H = 0.80 \times 5 = \mathbf{4.0\ m \ of \ water}$$

Uplift pressure at floor level just u/s of pile (point E):

$$P_E = \phi_E \times H = 0.71 \times 5 = \mathbf{3.55\ m \ of \ water}$$

Residual seepage head dissipated between E and D (along the pile): $(\phi_D - \phi_E)H = (0.80-0.71)\times5 = 0.45$ m.

Floor thickness required at d/s end (to balance the uplift $P_D$ with submerged weight of concrete, $G=2.24$, so floor specific submerged weight $\approx G-1 = 1.24$):

$$t = \frac{P_D}{G-1} = \frac{4.0}{1.24} = 3.23\ \text{m}$$

A practical provision (with factor of safety ~4/3) $t \approx \frac{4}{3}\times3.23 \approx \mathbf{4.3\ m}$ of concrete at the downstream end, demonstrating why the downstream pile location experiences the most severe uplift.

Note: Khosla's distribution gives higher, non-linear pressures concentrated near the downstream pile compared with Bligh's uniform assumption, which is precisely why Bligh's design underestimated the danger here.

Cross-Drainage Work

A cross-drainage work (CD work) is a hydraulic structure constructed at the crossing of an irrigation canal and a natural drainage (stream/river), to allow both to function without interference.

Classification

Type I — Canal over drainage: Aqueduct, Siphon-aqueduct. Type II — Drainage over canal: Super-passage, Canal-siphon (siphon super-passage). Type III — Canal and drainage at same level: Level crossing, Inlet–outlet.

Distinctions

• Aqueduct: Canal is carried over the drain in a trough; the drainage water passes below with its water surface below the trough bottom (free flow). Sketch: canal trough on piers, drain flowing as open channel beneath.
• Super-passage: The reverse — the drain is carried over the canal; canal runs below with free surface flow.
• Siphon-aqueduct: Canal over the drain, but the HFL of the drain is above the trough bottom, so the drain flows under pressure (siphonic) flow through a depressed/contracted barrel below the canal trough. Sketch: trough above, drain bed depressed and the drainage runs full under pressure.

Selection criteria

Relative bed levels and water levels of canal and drain; relative discharges; availability of head/drop; cost and foundation conditions; afflux permissible on the drain; ease of construction and maintenance.

Numerical

(a) Lacey's regime waterway of the drain

$$P = 4.75\sqrt{Q_d} = 4.75\sqrt{250} = 4.75 \times 15.811 = 75.10\ \text{m}$$

Lacey waterway $\approx 75.1$ m.

(b) Number of spans and piers

Clear span = 8 m, pier thickness = 1.5 m. One "module" = one span + one pier = $8 + 1.5 = 9.5$ m, but the end abutments make total waterway = $n\times(\text{clear span}) + (n-1)\times(\text{pier})$ for the clear waterway.

Use clear waterway = Lacey waterway = 75.1 m (clear opening between piers governs the regime perimeter):

$$n \times 8 = 75.1 \Rightarrow n = 9.39$$

Provide $n = \mathbf{10\ spans}$ → clear waterway $= 10\times8 = 80$ m ($>75.1$ m, safe). Number of intermediate piers $= n - 1 = \mathbf{9\ piers}$. Total structural length $= 10\times8 + 9\times1.5 = 80 + 13.5 = 93.5$ m between abutments.

(c) Width of canal trough

Canal discharge in trough $Q_c = 30$ m$^3$/s, permissible velocity $V = 2.0$ m/s, depth $y = 2.5$ m (rectangular trough).

Required flow area:

$$A = \frac{Q_c}{V} = \frac{30}{2.0} = 15\ \text{m}^2$$

Width:

$$B = \frac{A}{y} = \frac{15}{2.5} = \mathbf{6.0\ m}$$

So a 6.0 m wide rectangular trough (plus freeboard above the 2.5 m depth) carries the canal across the aqueduct at the permissible 2.0 m/s velocity.

Waterlogging

Waterlogging is the condition in which the water table rises so close to the ground surface that the soil pores in the root zone become saturated, depriving plant roots of air and harming crop growth. An area is generally considered waterlogged when the water table is within about 1.5–2 m of the surface.

Causes

• Over-irrigation / excessive seepage from unlined canals and reservoirs.
• Inadequate natural drainage / obstruction of natural drainage by roads, embankments.
• Flat topography and impervious subsoil.
• Heavy rainfall and floods; seepage from higher adjacent lands.
• Rise in regional water table from canal irrigation without drainage.

Ill-effects

• Lack of soil aeration → roots cannot respire; crop yield falls.
• Salinity / alkalinity: capillary rise brings salts to the surface (salt efflorescence).
• Difficulty in tillage and field operations; growth of weeds and water-loving plants.
• Lower soil temperature delaying germination.

Preventive and curative measures

Preventive: lining of canals to cut seepage; controlled/economical use of irrigation water; conjunctive use of groundwater (pumping lowers the table); preventing over-irrigation; proper crop rotation. Curative (drainage): surface drains (open ditches); subsurface drainage by tile/perforated pipe drains; pumping from tube wells (vertical drainage); leaching of salts with extra water plus drains; improving outfall/natural drainage.

Numerical — Hooghoudt drain spacing

Given $K = 0.8$ m/day, $R = 4$ mm/day $= 0.004$ m/day, $d = 3.0$ m, $h = 1.2$ m.

Hooghoudt ellipse equation:

$$S^2 = \frac{4K h (2d + h)}{R}$$

Compute the bracket: $2d + h = 2(3.0) + 1.2 = 6.0 + 1.2 = 7.2\ \text{m}$.

Numerator: $4 K h (2d+h) = 4 \times 0.8 \times 1.2 \times 7.2$ $= 4 \times 0.8 = 3.2;\quad 3.2 \times 1.2 = 3.84;\quad 3.84 \times 7.2 = 27.648\ \text{m}^3/\text{day}$.

$$S^2 = \frac{27.648}{0.004} = 6912\ \text{m}^2$$ $$S = \sqrt{6912} = \mathbf{83.1\ m}$$

Required drain spacing $S \approx 83\ \text{m}$. Adopt 83 m (or a slightly smaller practical value such as 80 m to provide a margin of safety against a higher recharge).

Comparison of irrigation methods

Drip irrigation

Advantages:

1. Very high water-use efficiency (water applied directly to root zone, minimal evaporation/percolation losses) — vital for water-scarce areas.
2. Reduced weed growth and possibility of applying soluble fertilizers with water (fertigation); suitable for saline water and uneven land.

Limitation:

• High initial cost and the risk of emitter clogging by silt, salts, or algae, requiring good filtration and regular maintenance.

Canal alignment types

Ridge (watershed) canal: Aligned along the ridge line / watershed of the area. Irrigation is possible on both sides by gravity, and since the watershed separates drainage basins, no natural drain crosses the canal. Used in plains.

Contour canal: Aligned roughly along a contour (with a slight slope for flow), used in hilly/sloping terrain. It irrigates only the area on the lower (valley) side. Because it runs across the slope, it intercepts the natural drains flowing down the slope, requiring many cross-drainage works.

Side-slope canal: Aligned along the side slope, roughly at right angles to the contours (i.e. more or less along the direction of the slope / parallel to natural drainage). Since it runs parallel to the drains, it is nearly perpendicular to the contours and generally does not cross drains, so few CD works are needed; however it cannot command much area and water flows fast.

Which minimizes cross-drainage works

The ridge (watershed) canal minimizes cross-drainage works. Because it follows the watershed, the natural drains run away from the canal on both sides (the ridge is the highest line separating drainage), so practically no drain has to cross the canal — eliminating costly aqueducts/super-passages. (Side-slope canals also avoid drains but command little area; contour canals require the most CD works.)

Given

Field capacity $FC = 28\%$, wilting point $PWP = 12\%$, dry unit weight $\gamma_d = 14\ \text{kN/m}^3$, root depth $D = 1.0$ m, allowable depletion $= 50\%$, consumptive use $C_u = 5$ mm/day. Unit weight of water $\gamma_w = 9.81\ \text{kN/m}^3$.

(a) Total available water (depth) in root zone

Depth of water held between FC and PWP:

$$d_{avail} = \frac{\gamma_d}{\gamma_w}\,(FC - PWP)\,D$$ $$= \frac{14}{9.81}(0.28 - 0.12)(1.0) = 1.4271 \times 0.16 \times 1.0 = 0.2283\ \text{m}$$ $$= \mathbf{228.3\ mm}$$

(b) Readily available moisture (50% depletion)

$$d_{readily} = 0.50 \times d_{avail} = 0.50 \times 228.3 = \mathbf{114.2\ mm}$$

(c) Frequency of irrigation

The readily available moisture is consumed at $C_u = 5$ mm/day:

$$F = \frac{d_{readily}}{C_u} = \frac{114.2}{5} = 22.8\ \text{days}$$

Frequency of irrigation $\approx 22$ days (round down to ensure moisture does not fall below the allowable depletion level).

Diversion headworks

A diversion headworks is a structure built across a river to raise the water level and divert the required supply into an off-taking canal, while passing the surplus flood water safely downstream. It does not store water (unlike a storage dam); it only heads up and regulates.

Main components

Weir or barrage, divide wall, under-sluices (scouring sluices), canal head regulator, fish ladder, silt-excluder/silt-prevention devices, marginal/guide banks, river training works, and the upstream/downstream impervious floor with sheet piles.

Functions

(i) Weir / Barrage: The main obstruction across the river that raises (heads up) the water level to enable diversion and passes floods. A weir holds up water mainly by a raised crest (with shutters); a barrage uses low crest with gates, giving better control over levels and afflux.

(ii) Divide wall: A masonry/concrete wall built parallel to the river flow, separating the under-sluice (scouring) pocket from the rest of the weir. It keeps the flow in front of the canal head regulator calm and silt-free, promotes still water for silt deposition, and prevents cross-currents/oblique flow.

(iii) Under-sluices (scouring sluices): Gated openings in the weir adjacent to the canal head regulator, kept at a lower crest level. They maintain a deep channel (pocket) in front of the regulator, scour and flush out silt deposited before the canal head, and pass low/dry-season flows and part of floods.

(iv) Fish ladder: A stepped/baffled passage provided in the weir body that allows fish to migrate upstream and downstream past the obstruction, by dissipating energy and keeping velocities low enough for fish to swim through.

Irrigation development in Nepal

Agriculture is the backbone of Nepal's economy, and irrigation is central to it. Of the total cultivable land (~2.6 million ha), a large share has irrigation infrastructure, but year-round (round-the-year) irrigation is available on only a limited fraction; most systems are monsoon/seasonal and depend on rainfall. Irrigation is provided through Farmer-Managed Irrigation Systems (FMIS), agency-managed systems, and increasingly groundwater (shallow/deep tube wells) in the Terai. Government policy (e.g. the Irrigation Policy and the Irrigation Master Plan) emphasizes year-round irrigation, participatory/joint management, and inter-basin transfers.

Two major irrigation projects

1. Sunsari–Morang Irrigation Project (eastern Terai) — one of the largest agency-managed surface systems.
2. Babai Irrigation Project / Bagmati Irrigation Project, and the multipurpose Sikta Irrigation Project; the Bheri–Babai Diversion Multipurpose Project is a flagship inter-basin transfer (tunnel) scheme for year-round irrigation plus hydropower.

Two key challenges

1. Seasonal water availability and lack of storage — most rivers are rain-fed; dry-season flows are low, so reliable year-round irrigation needs storage and inter-basin transfer, which are costly and slow.
2. Difficult topography, sediment and poor O&M — steep hills make canal construction and cross-drainage works expensive; heavy sediment damages headworks; weak operation, maintenance and water-user management reduce system efficiency. (Other challenges: fragmented land holdings, funding constraints, and climate-change-induced flood/drought variability.)

Given

$Q = 40\ \text{m}^3/\text{s}$, silt factor $f = 1.0$.

(a) Regime velocity

$$V = \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{40 \times 1^2}{140}\right)^{1/6} = \left(0.28571\right)^{1/6}$$

$\ln(0.28571) = -1.2528$; divide by 6 $= -0.20880$; $e^{-0.20880} = 0.8116$.

$$V = \mathbf{0.812\ m/s}$$

(b) Hydraulic mean radius

$$R = \frac{5}{2}\frac{V^2}{f} = 2.5 \times \frac{(0.8116)^2}{1.0} = 2.5 \times 0.6587 = \mathbf{1.647\ m}$$

(c) Regime bed slope

$$S = \frac{f^{5/3}}{3340\,Q^{1/6}}$$

$f^{5/3} = 1^{5/3} = 1$. $Q^{1/6} = 40^{1/6}$: $\ln 40 = 3.6889$, $/6 = 0.61482$, $e^{0.61482} = 1.8493$.

$$S = \frac{1}{3340 \times 1.8493} = \frac{1}{6176.7} = 1.619\times10^{-4}$$ $$\boxed{S \approx \frac{1}{6177} \approx 1.62\times10^{-4}}$$

Summary: $V = 0.812$ m/s, $R = 1.647$ m, $S \approx 1/6177$.