BE Civil Engineering (IOE, TU) Irrigation and Drainage Engineering (IOE, CE 705) Question Paper 2076 Nepal

Definitions

Duty (D): The area of land (in hectares) that can be irrigated by a unit discharge of water (1 cumec) flowing continuously throughout the base period of the crop. Unit: ha/cumec.

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

Relationship between Duty and Delta

Consider a discharge of $D$ ha irrigated by 1 cumec ($1\ \text{m}^3/\text{s}$) flowing for the base period $B$ days.

Volume of water supplied in base period:

$$V = 1\ \text{m}^3/\text{s} \times (B \times 24 \times 3600)\ \text{s} = 86400\,B\ \text{m}^3$$

This water spreads over $D$ hectares $= D \times 10^4\ \text{m}^2$.

Depth (delta) in metres:

$$\Delta = \frac{86400\,B}{D \times 10^4} = 8.64\frac{B}{D}\ \text{m}$$

Expressing in cm:

$$\boxed{\Delta = \frac{864\,B}{D}\ \text{cm}}\qquad (B\text{ in days},\ D\text{ in ha/cumec})$$

(a) Delta of wheat

Given $B = 120$ days, $D = 1800$ ha/cumec:

$$\Delta = \frac{864 \times 120}{1800} = \frac{103680}{1800} = 57.6\ \text{cm}$$

Delta of wheat = 57.6 cm

(b) Discharge at the head of the distributary

Area under wheat $= \text{CCA} \times \text{intensity} = 8000 \times 0.40 = 3200\ \text{ha}$

Discharge required at the field:

$$Q_{field} = \frac{\text{Area}}{\text{Duty}} = \frac{3200}{1800} = 1.778\ \text{cumec}$$

The duty 1800 ha/cumec is measured at the field. With 12% conveyance loss, only 88% of the head discharge reaches the field:

$$Q_{head} = \frac{Q_{field}}{1 - 0.12} = \frac{1.778}{0.88} = 2.020\ \text{cumec}$$

Discharge required at the head of the distributary = 2.02 cumec (≈ 2.0 m³/s)

Kennedy's Theory

Kennedy's critical velocity (non-silting, non-scouring) for depth $D$:

$$V_0 = 0.55\,m\,D^{0.64}\ \text{m/s}$$

Step 1 — Assume depth and compute geometry

Take $B/D = 5.7 \Rightarrow B = 5.7D$. Side slope $\tfrac{1}{2}:1$ means horizontal $= 0.5D$ per side.

Area of trapezoidal section:

$$A = BD + \tfrac{1}{2}D^2 = 5.7D^2 + 0.5D^2 = 6.2D^2$$

Kennedy velocity:

$$V = 0.55(1.0)D^{0.64} = 0.55\,D^{0.64}$$

Discharge continuity:

$$Q = A\,V \Rightarrow 15 = 6.2D^2 \times 0.55\,D^{0.64} = 3.41\,D^{2.64}$$ $$D^{2.64} = \frac{15}{3.41} = 4.399$$ $$D = 4.399^{1/2.64} = e^{(\ln 4.399)/2.64} = e^{1.4815/2.64} = e^{0.5612} = 1.753\ \text{m}$$

Take $D = 1.75\ \text{m}$.

Step 2 — Compute B, A, V

$$B = 5.7 \times 1.75 = 9.975 \approx 10.0\ \text{m}$$ $$A = BD + 0.5D^2 = 10.0(1.75) + 0.5(1.75)^2 = 17.5 + 1.531 = 19.03\ \text{m}^2$$ $$V = \frac{Q}{A} = \frac{15}{19.03} = 0.788\ \text{m/s}$$

Kennedy critical velocity at $D=1.75$:

$$V_0 = 0.55(1.75)^{0.64} = 0.55 \times 1.426 = 0.784\ \text{m/s}$$

$V \approx V_0$ (0.788 vs 0.784) — design is consistent with Kennedy's theory.

Step 3 — Verify by Kutter's formula

Wetted perimeter:

$$P = B + 2D\sqrt{1 + 0.5^2} = 10.0 + 2(1.75)(1.118) = 10.0 + 3.913 = 13.91\ \text{m}$$ $$R = \frac{A}{P} = \frac{19.03}{13.91} = 1.368\ \text{m}$$

Kutter's $C$ with $N = 0.0225$, $S = 1/5000 = 0.0002$:

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

$\dfrac{0.00155}{0.0002} = 7.75$, $\dfrac{1}{N} = 44.44$

Numerator $= 23 + 7.75 + 44.44 = 75.19$

Denominator $= 1 + (23 + 7.75)\dfrac{0.0225}{\sqrt{1.368}} = 1 + 30.75 \times \dfrac{0.0225}{1.170} = 1 + 30.75(0.01923) = 1 + 0.5913 = 1.5913$

$$C = \frac{75.19}{1.5913} = 47.25$$

Velocity by Chezy/Kutter:

$$V = C\sqrt{RS} = 47.25\sqrt{1.368 \times 0.0002} = 47.25\sqrt{0.0002736} = 47.25 \times 0.01654 = 0.782\ \text{m/s}$$

This matches the Kennedy velocity (0.784 m/s) closely.

Final Design

Channel: B = 10.0 m, D = 1.75 m, side slope 1/2:1, V ≈ 0.78 m/s — satisfies both Kennedy and Kutter.

Bligh's Creep Theory

Bligh assumed that water from the upstream creeps along the contact (creep) path of the base profile of the impervious floor and the underside of the foundation, losing head uniformly in proportion to the length of travel. Total creep length:

$$L = L_{horizontal} + 2\,(\text{vertical depths of piles/cutoffs})$$

Safety against piping (Bligh): $L \geq C\,H$, i.e. the creep length must be at least the coefficient of creep times the seepage head.

Hydraulic gradient (uplift): head loss is linear, so residual head at any point at creep distance $l$ from upstream:

$$h = H\left(1 - \frac{l}{L}\right)$$

Limitations of Bligh's theory

1. No distinction between horizontal and vertical creep — both treated equally, which is incorrect.
2. Loss of head is assumed proportional to length, but the actual exit gradient and the position of cutoffs matter greatly.
3. The importance of the downstream (end) cutoff/pile for preventing undermining is not recognized.
4. No mention of the exit gradient concept; piping is governed by exit gradient, not just average creep.

How Khosla's theory overcomes them

Khosla used the theory of potential flow (flow nets / Laplace's equation) to obtain seepage pressures. Key contributions:

• Outer faces of the end sheet piles are far more effective than inner faces and horizontal floor; vertical cutoffs are most effective.
• Introduced the concept of exit gradient $G_E$ for piping safety, with permissible values for each soil.
• Provided independent-variable method with correction factors (for floor thickness, mutual interference of piles, and floor slope) using Khosla's charts.

(a) Safety against piping (Bligh)

Safe creep length required:

$$L_{req} = C\,H = 12 \times 4.5 = 54\ \text{m}$$

Provided $L = 30\ \text{m} < 54\ \text{m}$.

The structure is UNSAFE against piping by Bligh's criterion; creep length must be increased (longer floor or deeper cutoffs).

(b) Uplift pressure and floor thickness at l = 18 m

Residual seepage head at 18 m:

$$h = H\left(1 - \frac{l}{L}\right) = 4.5\left(1 - \frac{18}{30}\right) = 4.5 \times 0.4 = 1.8\ \text{m of water}$$

Uplift pressure head at that point = 1.8 m of water.

Required floor thickness to balance uplift (floor weight must resist net uplift). The net uplift head above the floor underside is $h$, and the floor of specific gravity $G$ provides a buoyant resisting head $t(G-1)$:

$$t(G - 1) = \frac{4}{3}\,h \quad\text{(with usual factor of safety } \tfrac{4}{3})$$ $$t = \frac{4}{3}\cdot\frac{h}{G - 1} = \frac{4}{3}\cdot\frac{1.8}{2.24 - 1} = \frac{4}{3}\cdot\frac{1.8}{1.24} = \frac{4}{3}(1.452) = 1.935\ \text{m}$$

Required floor thickness ≈ 1.94 m at the point 18 m from upstream.

(Note: the large required thickness again confirms the section is heavily under-designed by Bligh's creep length.)

Cross-Drainage Work (CDW)

A cross-drainage work is a hydraulic structure constructed where an irrigation canal and a natural drainage (stream/river) cross each other, to convey one across the other without interference, intermixing, or damage.

Classification (with descriptions)

1. Canal over drainage

• Aqueduct: Canal is carried over the drain in a trough/superpassage; drainage water passes underneath with free water surface (HFL well below canal trough). Canal banks continue as the trough.
• Siphon aqueduct: Same as aqueduct but the drainage HFL is above the canal trough bottom, so the drain flows under pressure (siphonic) beneath the canal. Used when drain HFL is high.

Aqueduct (canal above):
   ====[ canal trough ]====   <- canal FSL
        |||  drain  |||        <- drainage flows below (free surface)

2. Drainage over canal

• Super-passage: Drain is carried over the canal in a trough; canal flows underneath with free surface.
• Canal siphon (siphon super-passage): Drain over canal, but canal runs full/pressurized (siphon) beneath the drainage trough.

3. Drainage through/at canal level

• Level crossing: Both meet at the same level; flows are mixed and regulated by cross-regulators and head regulators. Used when both discharges are large.
• Inlet–outlet: Small drain admitted into canal at an inlet and let out downstream at an outlet; only for very small drains.

Criteria for selecting the type

• Relative bed levels and water levels of canal and drain (FSL of canal vs HFL of drain).
• Relative discharges of canal and drainage.
• Available difference of levels and economy.
• Nature of foundation, importance of canal, and likely afflux/heading-up.

Recommendation for the given data

Canal FSL = +102.0 m; Drain HFL = +103.5 m. The drainage HFL (103.5) is above the canal FSL (102.0). Carrying the large flood (250 m³/s) over the comparatively small canal would need a high trough; instead the canal (smaller, controllable) is taken below the drain, but since drain HFL exceeds canal levels, the drain must pass over while canal runs under pressure — however the conventional and most economical choice when canal is taken under a drain whose HFL is high is to keep the canal in a barrel running under pressure.

Given canal FSL (102.0) < drain HFL (103.5), the suitable work is a SIPHON AQUEDUCT is not applicable (that is canal-above). Here drainage is higher and larger, so the canal should pass below the drainage as a pressurized barrel: this is a CANAL SIPHON (siphon super-passage) — drainage trough above, canal siphons underneath.

Recommended: Canal Siphon (drainage over canal, canal under pressure), because the drain discharge (250 m³/s) is large and its HFL (+103.5) is above the canal FSL (+102.0), making it uneconomical to lift the drain or the canal above the other with free flow.

Waterway of the drainage trough (Lacey)

Lacey's regime wetted perimeter for the drain:

$$P = 4.75\sqrt{Q_d} = 4.75\sqrt{250} = 4.75 \times 15.811 = 75.1\ \text{m}$$

For a wide waterway the surface width ≈ $P$, so provide a clear drainage waterway of about 75 m, split into a suitable number of bays/spans (e.g., 75 m / 5 m ≈ 15 bays of 5 m clear span, plus piers).

Waterway of drainage trough ≈ 75 m (Lacey regime perimeter, f = 1.0).

Waterlogging

Waterlogging is the condition of agricultural land in which the soil pores in the root zone become saturated and the water table rises to or near the ground surface, restricting aeration of the root zone (commonly when the water table is within about 1.5–2.0 m of the surface).

Causes

1. Over-irrigation and excessive seepage from unlined canals.
2. Inadequate natural or artificial drainage / obstruction of natural drainage by embankments, roads, railways.
3. Heavy rainfall and floods / submergence.
4. Impervious subsoil strata and flat topography preventing percolation/outflow.
5. Seepage from reservoirs and rise of regional water table.

Ill-effects

1. Lack of soil aeration — roots cannot breathe; crop yield falls.
2. Salinity and alkalinity (salt efflorescence) as capillary rise brings salts to the surface.
3. Growth of weeds and water-loving plants; difficulty in tillage operations.
4. Spread of water-borne diseases (e.g. malaria) and unhealthy living conditions.
5. Lowering of soil temperature, delaying crop maturity.

Preventive / remedial measures

1. Lining of canals and distributaries to cut seepage.
2. Controlled, optimum irrigation (avoid over-watering); proper water management and crop rotation.
3. Surface drainage (open drains) and subsurface drainage (tile/perforated pipe drains).
4. Tube wells / vertical drainage to lower the water table (also gives conjunctive use).
5. Provision of adequate cross-drainage and intercepting drains; raising cropping intensity wisely.

Drain spacing computation

Given: $K = 0.6\ \text{m/day}$, $b = 1.8\ \text{m}$, $a = 0.8\ \text{m}$, $q = 0.004\ \text{m/day}$.

$$S^2 = \frac{4K(b^2 - a^2)}{q}$$

Compute $b^2 - a^2 = 1.8^2 - 0.8^2 = 3.24 - 0.64 = 2.60\ \text{m}^2$.

$$S^2 = \frac{4 \times 0.6 \times 2.60}{0.004} = \frac{6.24}{0.004} = 1560\ \text{m}^2$$ $$S = \sqrt{1560} = 39.5\ \text{m}$$

Spacing of subsurface (tile) drains ≈ 39.5 m (provide ~39 m c/c).

(For the 60 ha field, with parallel drains at 39.5 m spacing, the number and length of laterals can then be laid out to discharge into a collector/main drain.)

Given

FC = 28%, PWP = 13%, $\gamma_d = 14.5\ \text{kN/m}^3$, root depth $d = 0.90\ \text{m}$, readily available = 75% of available, $C_u = 5\ \text{mm/day}$, $\eta = 65\%$.

Unit weight of water $\gamma_w = 9.81\ \text{kN/m}^3$.

(a) Available moisture depth (FC to PWP)

Depth of water held between field capacity and wilting point:

$$d_w = \frac{\gamma_d}{\gamma_w}\,(FC - PWP)\,d$$ $$d_w = \frac{14.5}{9.81}\,(0.28 - 0.13)(0.90) = 1.4781 \times 0.15 \times 0.90$$ $$d_w = 1.4781 \times 0.135 = 0.1995\ \text{m} = 19.95\ \text{cm}$$

Total available moisture in root zone ≈ 19.95 cm (≈ 199.5 mm).

(b) Frequency of irrigation

Readily available moisture (allowed depletion) = 75% of available:

$$d_{readily} = 0.75 \times 199.5 = 149.6\ \text{mm}$$

Frequency (interval between irrigations):

$$F = \frac{d_{readily}}{C_u} = \frac{149.6}{5} = 29.9\ \text{days}$$

Frequency of irrigation ≈ 30 days.

(c) Depth of water at canal head per irrigation

Net depth to be replenished in field = readily available moisture = 149.6 mm.

Accounting for irrigation (application + conveyance) efficiency of 65%:

$$d_{gross} = \frac{d_{net}}{\eta} = \frac{149.6}{0.65} = 230.2\ \text{mm} = 23.0\ \text{cm}$$

Gross depth of water to be supplied at the canal head per irrigation ≈ 23.0 cm (230 mm).

Comparison of irrigation methods

Numerical

Area $= 1.5\ \text{ha} = 15000\ \text{m}^2$; spacing $4 \times 4 = 16\ \text{m}^2$ per plant.

(a) Number of plants

$$N = \frac{15000}{16} = 937.5 \approx 937\ \text{plants}$$

Number of plants ≈ 937.

(b) Daily water requirement per plant

Wetted (canopy) area per plant = 60% of ground area allotted:

$$A_w = 0.60 \times 16 = 9.6\ \text{m}^2$$

Depth required = 6 mm/day = 0.006 m/day. Volume per plant per day:

$$V = A_w \times \text{depth} = 9.6 \times 0.006 = 0.0576\ \text{m}^3 = 57.6\ \text{litres/day}$$

Water requirement per plant ≈ 57.6 litres/day.

(c) Discharge per emitter

4 emitters per plant, running 3 h/day:

$$q_{plant} = \frac{57.6\ \text{L}}{3\ \text{h}} = 19.2\ \text{L/h per plant}$$ $$q_{emitter} = \frac{19.2}{4} = 4.8\ \text{L/h}$$

Discharge of each emitter ≈ 4.8 litres/hour.

Types of canal alignment

1. Watershed (ridge) canal Aligned along the ridge line / watershed dividing two drainage basins. Irrigation is possible on both sides by gravity, and no cross-drainage works are required since the canal does not cross any drainage (drains start below the ridge and flow away from it). Most preferred alignment for main and branch canals.

        ___ridge (watershed)___
      /   <-- canal on top -->  \
  drain                         drain   (water commands both sides)

2. Contour canal Aligned (almost) along a contour of the ground, with a slight downward slope for flow. Used in hilly / steep terrain where the ridge cannot be followed. Irrigation is possible only on the lower (valley) side; the canal crosses natural drains, so several cross-drainage works (aqueducts, etc.) are needed.

  hill side (no command)
  ~~~~~~~~~ canal along contour ~~~~~~~~~
  valley side (commanded area)

3. Side-slope canal Aligned roughly at right angles to the contours, i.e. down the side slope between two drains, running more or less parallel to the natural drainage. Since it runs along the slope and parallel to drains, no cross-drainage works are required, but it commands a comparatively small area and is generally not used for irrigation distribution.

General principles of canal alignment

1. Align on the watershed (ridge) as far as possible to command maximum area on both sides and avoid cross-drainage works.
2. Keep the alignment as short and straight as practicable to reduce cost and head loss; avoid sharp curves.
3. Avoid deep cutting and high filling; balance earthwork (cut ≈ fill) for economy.
4. Avoid valuable land, villages, religious/important structures, and difficult/marshy or rocky ground.
5. Cross drainage at the most economical and suitable site when crossing cannot be avoided; cross drains as nearly at right angles as possible.
6. Maintain the required bed slope and ensure the full supply level commands the irrigable area.

Components of a diversion headworks

1. Weir / Barrage: The main barrier across the river that raises (heads up) the water level to divert it into the canal; passes surplus flood downstream.
2. Under-sluices (scouring sluices): Openings adjacent to the canal head regulator with crest lower than the weir crest; maintain a deep clear channel in front of the regulator and scour out silt.
3. Canal head regulator: Regulates and controls supply entering the off-taking canal and excludes silt and excess flood water.
4. Divide wall: A masonry wall separating the weir bay from the under-sluice pocket; creates still pond conditions for silt control and separates the two flows.
5. Fish ladder: Provides a passage for migratory fish across the weir by dissipating head in steps.
6. Silt excluder / silt vanes: Structures in the pocket to keep silt away from the canal entry.
7. River training works (guide banks, marginal bunds, spurs): Guide the flow centrally through the works and protect against outflanking.
8. Impervious floor, cutoffs and protection (launching apron): Resist seepage, uplift, piping and downstream scour.

Numerical

Head over crest

$$Q = C\,L\,H^{3/2} \Rightarrow H^{3/2} = \frac{Q}{C\,L} = \frac{1100}{1.7 \times 80} = \frac{1100}{136} = 8.088$$ $$H = 8.088^{2/3} = e^{(2/3)\ln 8.088} = e^{(2/3)(2.0904)} = e^{1.3936} = 4.030\ \text{m}$$

Head over the crest $H \approx 4.03$ m.

Afflux

Afflux = rise in upstream water level (above crest) due to the weir, over the normal (pre-weir) upstream level (above crest):

$$\text{Afflux} = H_{u/s} - h_{normal} = 4.03 - 3.6 = 0.43\ \text{m}$$

Afflux ≈ 0.43 m.

(This afflux determines the heading-up against which the marginal bunds and guide banks free-board must be designed.)

Step 1 — Required area from permissible velocity

Limiting velocity $V = 2.0\ \text{m/s}$.

$$A = \frac{Q}{V} = \frac{40}{2.0} = 20\ \text{m}^2$$

Step 2 — Required hydraulic radius from Manning

Manning: $V = \dfrac{1}{n}R^{2/3}S^{1/2}$, $S = 1/4000 = 0.00025$, $\sqrt{S} = 0.015811$.

$$R^{2/3} = \frac{V\,n}{\sqrt{S}} = \frac{2.0 \times 0.016}{0.015811} = \frac{0.032}{0.015811} = 2.024$$ $$R = 2.024^{3/2} = e^{1.5\ln 2.024} = e^{1.5(0.7050)} = e^{1.0575} = 2.880\ \text{m}$$

Required: $A = 20\ \text{m}^2$, $R = 2.88\ \text{m}$.

Step 3 — Solve for B and D (trapezoid, side slope $z = 1.25$)

Area: $A = BD + zD^2 = BD + 1.25D^2 = 20$ Perimeter: $P = B + 2D\sqrt{1+z^2} = B + 2D\sqrt{1+1.5625} = B + 2D(1.6008) = B + 3.2016D$ $R = A/P = 2.88 \Rightarrow P = A/R = 20/2.88 = 6.944\ \text{m}$

From area: $B = \dfrac{20 - 1.25D^2}{D}$. Substitute into $P$:

$$\frac{20 - 1.25D^2}{D} + 3.2016D = 6.944$$ $$\frac{20}{D} - 1.25D + 3.2016D = 6.944$$ $$\frac{20}{D} + 1.9516D = 6.944$$

Multiply by $D$:

$$1.9516D^2 - 6.944D + 20 = 0$$

Discriminant $= 6.944^2 - 4(1.9516)(20) = 48.22 - 156.1 = -107.9 < 0$.

No real solution — meaning the section that gives $A=20$ and $R=2.88$ simultaneously is not geometrically attainable for this $z$ (the required $R$ is too large for that area). The required $R = 2.88$ m would need an area larger than 20 m². Hence the velocity is governed by the section, not the limit: adopt a practical $B/D$ and let velocity fall below the permissible 2.0 m/s (safe).

Step 4 — Practical design (adopt B/D)

Take $B/D = 3$ (typical for lined canals). With $A = BD + 1.25D^2 = 3D^2 + 1.25D^2 = 4.25D^2$.

We size from Manning instead of forcing $V=2$. Try $D$ such that discharge = 40. $P = B + 3.2016D = 3D + 3.2016D = 6.2016D$, $R = A/P = 4.25D^2/(6.2016D) = 0.6853D$.

$$Q = \frac{1}{n}A R^{2/3} S^{1/2} = \frac{1}{0.016}(4.25D^2)(0.6853D)^{2/3}(0.015811)$$

$(0.6853)^{2/3} = e^{(2/3)\ln0.6853}=e^{(2/3)(-0.3779)}=e^{-0.2519}=0.7773$, so $(0.6853D)^{2/3}=0.7773\,D^{2/3}$.

$$Q = 62.5 \times 4.25D^2 \times 0.7773\,D^{2/3} \times 0.015811 = 3.264\,D^{8/3}$$

Set $=40$: $D^{8/3} = 40/3.264 = 12.26 \Rightarrow D = 12.26^{3/8} = e^{0.375\ln12.26}=e^{0.375(2.506)}=e^{0.9398}=2.560\ \text{m}$.

$B = 3D = 7.68\ \text{m}$. Adopt $D = 2.55\ \text{m}$, $B = 7.65\ \text{m}$.

Check: $A = 4.25(2.55)^2 = 27.6\ \text{m}^2$, $V = Q/A = 40/27.6 = 1.45\ \text{m/s} < 2.0\ \text{m/s}$ — safe (non-eroding).

Final Design

Adopt B = 7.65 m, D = 2.55 m, side slope 1.25:1; velocity 1.45 m/s is within the permissible limit.

Irrigation development in Nepal

Present status and potential

• Nepal is predominantly agrarian; agriculture depends heavily on irrigation, yet a large share of farmland still relies on monsoon rain.
• Of the total cultivable land (roughly 2.6–3 million ha), only a part has assured year-round irrigation. The terai holds the greatest irrigable potential because of fertile land and groundwater availability; hills and mountains rely largely on small surface and FMIS schemes.
• Potential sources: perennial snow-fed rivers (Koshi, Gandaki, Karnali, Mahakali systems), medium rivers, and groundwater (shallow & deep tube wells in the terai).

Classification of irrigation systems

• Farmer-Managed Irrigation Systems (FMIS): Built, operated, and maintained by farmers/communities themselves; traditional, locally governed, cover a very large fraction of irrigated area in Nepal (especially hills and small schemes). Strong community ownership.
• Agency-Managed Irrigation Systems (AMIS): Designed, constructed, and managed by government agencies (e.g. Department of Water Resources and Irrigation); generally larger surface schemes. Increasingly handed over to Water User Associations under joint/participatory management.
• Also classified by source/scale: surface (canal/diversion), groundwater (tube wells), and lift / non-conventional irrigation; and by size into large, medium and small/micro schemes.

Major challenges

1. Seasonal and unreliable river flows; lack of storage to provide year-round (round-the-year) irrigation.
2. High sediment loads and flooding damaging headworks and canals.
3. Difficult Himalayan topography raising costs and limiting hill irrigation.
4. Weak operation & maintenance, deteriorating systems, and inadequate cost recovery.
5. Inter-basin/inter-state water sharing issues and need for diversion/storage (e.g. trans-basin transfers).
6. Financing, institutional capacity, and climate-change variability.

Two important irrigation projects (with sources)

1. Sunsari–Morang Irrigation Project — utilises the Koshi (Sapta Koshi) River in eastern terai.
2. Bagmati Irrigation Project — utilises the Bagmati River (central/eastern terai).

(Other notable schemes: Babai Irrigation Project on the Babai River, and the Sikta Irrigation Project on the Rapti River; the Bheri–Babai Diversion Multipurpose Project transfers water from the Bheri to the Babai for year-round irrigation.)

Summary: Nepal has large irrigation potential (especially in the terai), a mix of community-run FMIS and agency-run AMIS, but is constrained by seasonal flows, sediment/flood damage, terrain, and O&M/financing limitations; the policy thrust is toward storage, multipurpose diversion projects, and participatory management for round-the-year irrigation.