BE Civil Engineering (IOE, TU) Irrigation and Drainage Engineering (IOE, CE 705) Question Paper 2080 Nepal

(a) Definitions and derivation

Duty (D): The area (in hectares) that a unit discharge (1 cumec) of water flowing continuously can irrigate during the entire base period of a crop. Units: ha/cumec.

Delta (\u0394): The total depth of water (in cm or m) required by a crop over its entire base period, expressed as a depth spread uniformly over the irrigated area.

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

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

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

This volume spread over the duty area $D$ (ha $= D\times10^4$ m$^2$) gives depth

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

Thus $\boxed{\Delta = \dfrac{8.64\,B}{D}}$ with $\Delta$ in m, $B$ in days, $D$ in ha/cumec.

(b) Duty at the field

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

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

Duty at the field = 2592 ha/cumec.

(c) Duty at the head and discharge

Conveyance efficiency $\eta_c = 0.65$. Duty at head $=$ field duty $\times \eta_c$ (less area can be served per cumec because of losses):

$$D_{head} = 2592 \times 0.65 = 1684.8\ \text{ha/cumec}.$$

Discharge at head for 4000 ha:

$$Q = \frac{A}{D_{head}} = \frac{4000}{1684.8} = 2.374\ \text{cumec}.$$

Duty at head = 1684.8 ha/cumec; required head discharge \u2248 2.37 cumec.

(Check: at the field, $Q_{field}=4000/2592=1.543$ cumec; $Q_{head}=Q_{field}/0.65=2.374$ cumec. Consistent.)

(d) Rice comparison

For rice, $\Delta=1.20$ m, $B=120$ days:

$$D = \frac{8.64\times120}{1.20} = 864\ \text{ha/cumec}.$$

Duty falls from 2592 to 864 ha/cumec — because rice needs 3× the depth, the same cumec irrigates only one-third the area. Duty is inversely proportional to delta for a fixed base period.

(a) Lacey's regime concept

A channel is in regime when it flows in unlimited incoherent alluvium of the same character as that transported, carrying a steady discharge and silt charge, so that there is neither net scour nor net silting over a cycle. Lacey's equations:

$$V = 0.6395\,f^{1/2}R^{1/2}\ \text{(SI form via } V=\sqrt{\tfrac{2}{5}fR}),\quad P = 4.75\sqrt{Q},$$ $$A = \frac{Q}{V},\quad f = 1.76\sqrt{d_{mm}},\quad S = \frac{f^{5/3}}{3340\,Q^{1/6}}.$$

A convenient set: $V = \left(\dfrac{Q f^2}{140}\right)^{1/6}$.

(b) Regime parameters ($Q=30$, $f=1.0$)

Velocity:

$$V = \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{30 \times 1}{140}\right)^{1/6} = (0.21429)^{1/6}.$$

$\ln(0.21429) = -1.5404$; $\div 6 = -0.25674$; $e^{-0.25674}=0.7736$.

$$V = 0.774\ \text{m/s}.$$

Area:

$$A = \frac{Q}{V} = \frac{30}{0.774} = 38.76\ \text{m}^2.$$

Wetted perimeter:

$$P = 4.75\sqrt{Q} = 4.75\sqrt{30} = 4.75 \times 5.4772 = 26.02\ \text{m}.$$

Hydraulic radius: Lacey $R = \dfrac{5}{2}\dfrac{V^2}{f} = 2.5 \times \dfrac{0.774^2}{1.0} = 2.5 \times 0.5990 = 1.498\ \text{m}.$ (Check $A/P = 38.76/26.02 = 1.49$ m — consistent.)

Regime slope:

$$S = \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1}{3340 \times 30^{1/6}}.$$

$30^{1/6}=e^{(\ln 30)/6}=e^{3.4012/6}=e^{0.5669}=1.7628$.

$$S = \frac{1}{3340 \times 1.7628} = \frac{1}{5887.8} = 1.698\times10^{-4} \approx \frac{1}{5888}.$$

Summary: $V=0.774$ m/s, $A=38.76$ m², $P=26.02$ m, $R=1.498$ m, $S \approx 1/5888$.

(c) Trapezoidal proportioning (side slope $m=0.5$)

For a trapezoid: $A = By + my^2$, $P = B + 2y\sqrt{1+m^2}$. With $m=0.5$, $\sqrt{1+m^2}=\sqrt{1.25}=1.1180$, so $P = B + 2.236\,y$.

From $P$: $B = 26.02 - 2.236\,y$. Substitute in $A$:

$$(26.02 - 2.236y)y + 0.5y^2 = 38.76$$ $$26.02y - 2.236y^2 + 0.5y^2 = 38.76 \Rightarrow -1.736y^2 + 26.02y - 38.76 = 0.$$ $$1.736y^2 - 26.02y + 38.76 = 0.$$ $$y = \frac{26.02 \pm \sqrt{26.02^2 - 4(1.736)(38.76)}}{2(1.736)} = \frac{26.02 \pm \sqrt{677.0 - 269.2}}{3.472} = \frac{26.02 \pm 20.19}{3.472}.$$

Practical (shallow regime) root: $y = (26.02-20.19)/3.472 = 1.679\ \text{m}.$ Then $B = 26.02 - 2.236(1.679) = 26.02 - 3.75 = 22.27\ \text{m}.$

Design: bed width $B \approx 22.3$ m, depth $y \approx 1.68$ m (slope $1/5888$).

(a) Theories

Bligh's creep theory: Water creeps along the contact of the base profile of the structure with the subsoil, losing head uniformly along the creep length $L =$ (sum of horizontal + vertical creep paths, vertical counted twice). Residual head at any point $= H\,(1 - l/L)$. Safety against piping uses the coefficient of creep $C$: $L \ge C\,H$. It treats horizontal and vertical creep as equally effective — a known defect.

Khosla's theory: Treats seepage as 2-D potential (Laplacian) flow. The composite profile is split into simple elementary forms (straight floor with end pile, intermediate pile, depressed floor) for which closed-form pressure distributions exist (the method of independent variables). Corrections are applied for floor thickness, mutual interference of piles, and floor slope. Exit gradient is computed rigorously rather than empirically.

Why Khosla superseded Bligh: Bligh wrongly assumed uniform head loss and equal weight to horizontal/vertical creep, over-estimated outer floor effectiveness, and gave no rational exit-gradient (piping) criterion. Khosla's potential-flow solution matched observed uplift and provided the exit-gradient concept.

(b) Khosla pressures — d/s end pile

For a floor of length $b$ with a pile of depth $d$ at the downstream end, define

$$\alpha = \frac{b}{d} = \frac{40}{6} = 6.667,\qquad \lambda = \frac{1 + \sqrt{1+\alpha^2}}{2}.$$ $$\sqrt{1+\alpha^2} = \sqrt{1 + 44.44} = \sqrt{45.44} = 6.741.$$ $$\lambda = \frac{1 + 6.741}{2} = 3.871.$$

Pressure at E (floor junction, d/s pile):

$$\phi_E = \frac{1}{\pi}\cos^{-1}\!\left(\frac{\lambda - 2}{\lambda}\right) = \frac{1}{\pi}\cos^{-1}\!\left(\frac{3.871 - 2}{3.871}\right) = \frac{1}{\pi}\cos^{-1}(0.4834).$$

$\cos^{-1}(0.4834) = 61.08^\circ = 1.0661\ \text{rad}$.

$$\phi_E = \frac{1.0661}{\pi} = 0.3393 = 33.9\%.$$

Uplift head at E $= 0.339 \times 5 = 1.70\ \text{m}$ (before thickness correction).

Exit gradient:

$$G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}}.$$ $$\pi\sqrt{\lambda} = \pi\sqrt{3.871} = \pi \times 1.9675 = 6.181.$$ $$G_E = \frac{5}{6} \times \frac{1}{6.181} = 0.8333 \times 0.16179 = 0.1348 \approx \frac{1}{7.4}.$$

(c) Safety against piping

Computed exit gradient $G_E = 0.135 = 1/7.4$. The safe value for fine sand is $1/6 = 0.167$. Since

$$G_E = 0.135 < 0.167\ \text{(safe)},$$

the structure is safe against piping at the downstream end. The factor of safety $= 0.167/0.135 \approx 1.24$.

(a) Cross-drainage works (CDW)

Structures built where an irrigation canal crosses a natural drainage (stream/river) so both can pass without interference. Classification by relative levels:

Selection criteria: governed by relative bed/HFL levels of canal and drainage, discharge of each, available head, cost, and afflux permissible upstream.

(b) Aqueduct hydraulics

Velocity in normal canal section (trapezoidal assumed rectangular bed portion; here treat as rectangular with $B=12$ m, $y=1.6$ m):

$$A_1 = B\,y = 12 \times 1.6 = 19.2\ \text{m}^2,\qquad V_1 = \frac{Q}{A_1} = \frac{25}{19.2} = 1.302\ \text{m/s}.$$

Velocity in flumed trough ($B_2 = 8$ m). Assuming the same depth is maintained (lined trough, $y_2 \approx 1.6$ m as first estimate):

$$A_2 = 8 \times 1.6 = 12.8\ \text{m}^2,\qquad V_2 = \frac{25}{12.8} = 1.953\ \text{m/s}.$$

Acceptability: $V_2 = 1.95\ \text{m/s} < 3\ \text{m/s}$ and well above non-scouring/silting limits, so the fluming ratio $8/12 = 0.67$ is acceptable; no choking (flow stays subcritical: $Fr_2 = V_2/\sqrt{g y_2} = 1.953/\sqrt{9.81\times1.6} = 1.953/3.962 = 0.49 < 1$).

Afflux / head drop at entry (energy + continuity): The rise in velocity head between canal and trough is

$$\Delta h_v = \frac{V_2^2 - V_1^2}{2g} = \frac{1.953^2 - 1.302^2}{2\times9.81} = \frac{3.814 - 1.695}{19.62} = \frac{2.119}{19.62} = 0.108\ \text{m}.$$

Allowing an entry (contraction) loss coefficient of about 0.2 of this velocity-head change at transitions, the upstream water surface must rise by roughly

$$h_{afflux} \approx \Delta h_v (1 + 0.2) \approx 0.108 \times 1.2 = 0.130\ \text{m}.$$

Afflux ≈ 0.13 m, which is small and acceptable. (A gradual splayed transition, e.g. 2:1 contraction and 3:1 expansion, keeps losses low.)

(a) Waterlogging

Definition: A land is said to be waterlogged when the water table rises so high that the crop root zone becomes saturated, cutting off air to the roots and harming plant growth (typically when the water table is within about 1.5–2 m of the surface for sensitive crops).

Causes: over-irrigation; seepage from unlined canals/reservoirs; inadequate natural/surface drainage; impervious subsoil strata; flat topography; high rainfall; rise in regional groundwater; obstruction of drainage by embankments/roads.

Effects: reduced soil aeration (root suffocation); accumulation of harmful salts at surface (salinity/alkalinity); lowering of soil temperature; growth of water-loving weeds; reduced crop yield; difficulty in tillage; structural damage to foundations.

Measures: (1) line canals and control over-irrigation to cut seepage and recharge; (2) provide surface drains and subsurface tile/pipe drains or tube-well (vertical) drainage to lower the water table; (3) bio-drainage (tree plantations).

(b) Drain spacing (Hooghoudt ellipse)

Given $K = 1.0$ m/day, $R = 0.004$ m/day, $h = 1.0$ m, $D = 3.0$ m.

$$L^2 = \frac{4K h\,(2D + h)}{R}.$$

Compute the bracket: $2D + h = 2(3.0) + 1.0 = 7.0\ \text{m}$. Numerator: $4 \times 1.0 \times 1.0 \times 7.0 = 28.0$.

$$L^2 = \frac{28.0}{0.004} = 7000\ \text{m}^2.$$ $$L = \sqrt{7000} = 83.67\ \text{m}.$$

Required drain spacing $L \approx 83.7$ m.

Check sensitivity: if the recharge doubles to $R=0.008$ m/day, $L^2 = 28/0.008 = 3500$, $L = 59.2$ m — higher recharge demands closer drains, as expected.

Comparison of irrigation methods

Sub-types of surface irrigation:

• Border: land divided into strips bounded by low ridges; a sheet of water advances down the strip.
• Check basin: level basins enclosed by ridges, flooded and ponded; best for paddy and orchards.
• Furrow: water runs in furrows between ridged rows; only part of the surface is wetted — good for row crops (maize, potato, sugarcane).

Two situations where drip is clearly preferable

1. Water-scarce regions / widely spaced orchard or vegetable crops (e.g. citrus, tomato), where the high application efficiency (85–95%) maximises yield per drop and only the root zone is wetted.
2. Saline water or saline soils, and steep/undulating terrain, where drip keeps salts pushed to the wetting front away from roots, avoids runoff/erosion, and enables fertigation — conditions where surface flooding would waste water or worsen salinity.

(a) Crop evapotranspiration

$$ET_c = K_c \times ET_0 = 1.15 \times 5.5 = 6.325\ \text{mm/day}.$$

$ET_c = 6.33\ \text{mm/day}$.

(b) Net irrigation depth (readily available moisture)

Total available water (TAW) per unit depth = (FC − PWP) = $0.22 - 0.10 = 0.12$ (volume fraction). Over root depth $d_r = 0.9$ m:

$$TAW = 0.12 \times 0.9 = 0.108\ \text{m} = 108\ \text{mm}.$$

Readily available water (allowable depletion, $p=0.5$):

$$d_{net} = p \times TAW = 0.5 \times 108 = 54\ \text{mm}.$$

Net irrigation depth = 54 mm per application.

(c) Irrigation interval

$$\text{Interval} = \frac{d_{net}}{ET_c} = \frac{54\ \text{mm}}{6.325\ \text{mm/day}} = 8.54\ \text{days}.$$

Rounded down to a safe whole number, irrigate every 8 days (applying ≈ 54 mm net each time; gross depth would be larger after dividing by application efficiency).

Principles of canal alignment

An ideal alignment should: (1) command the maximum culturable area by gravity; (2) minimise cross-drainage works and earthwork (balanced cut/fill); (3) avoid sharp curves, villages, valuable land, religious/forest land; (4) follow as straight and short a route as practicable; (5) be economical in construction and maintenance; (6) avoid waterlogging-prone reaches and bad foundation soils.

Three alignment types

1. Watershed / Ridge canal: Aligned along the watershed (ridge line) dividing two drainages. Water can be supplied to both sides by gravity, and since drainage lines fall away on either side, no cross-drainage works are needed on the main canal. Preferred for main canals in the upper reaches.

        canal on ridge
   \        ___/\___        /
    \      /        \      /   <- ground falls away both sides
     \____/          \____/
   drainage          drainage

2. Contour canal: Aligned roughly along a contour with a small longitudinal bed slope (less than ground slope), so it gradually departs from the contour. It can irrigate land only on the lower (valley) side, and it crosses natural drainage, requiring cross-drainage works. Common in hilly terrain (much of Nepal's hill irrigation).

3. Side-slope canal: Aligned roughly at right angles to the contours, i.e. down the side slope, more or less parallel to the drainage. Its bed slope nearly equals the ground slope, so it generally needs no cross-drainage works, but it commands little area (land on both sides is higher than or close to the canal level on a steep slope only on one limited side).

Advantages of watershed alignment

• Irrigation possible on both banks → larger command area.
• No cross-drainage works on the main canal → large cost saving.
• Surface drainage flows away from the canal → less risk of breach, silting, and waterlogging.

Components of a diversion headworks

A diversion headworks raises the river water level and diverts a regulated supply into the off-taking canal. Main components:

1. Weir / Barrage (with crest and gates): Raises (ponds) the water level across the river to command the canal. A barrage uses gates over a low crest (better flood/silt control); a weir relies mainly on raised crest.
2. Divide wall: A masonry/concrete wall parallel to flow, separating the weir bay from the undersluice pocket; creates still pond in front of the canal head, controls cross-currents and helps silt exclusion.
3. Undersluices (scouring sluices): Gated openings adjacent to the canal head regulator with a lower crest; they maintain a deep clear channel toward the canal, scour deposited silt, and pass low/initial floods.
4. Canal head regulator (head sluice): Gated structure controlling the discharge entering the canal, regulating supply, excluding silt and floods, and providing pondage control. Its crest is kept higher than the undersluice crest.
5. Fish ladder: Provides a path of moderate velocity for fish migration past the weir.
6. Silt excluder / silt control devices: (see below).
7. River training works — guide banks (Bell's bunds), marginal/afflux bunds, launching aprons — guide the flow axially through the structure and protect approaches from scour and flanking.
8. Impervious & protective floor with cutoffs (sheet piles), upstream/downstream aprons and energy dissipators (cistern/blocks) — control seepage/uplift and dissipate energy.

Two silt-control devices

1. Silt excluder: A set of horizontal tunnels constructed in the river bed in front of the canal head regulator. The bottom (silt-laden) layer of water is intercepted by the tunnels and ejected downstream through the undersluices, while the upper, relatively clear water enters the canal. It works before silt enters the canal.

2. Silt ejector (silt extractor): Located in the canal a short distance downstream of the head regulator. Tunnels in the canal bed collect the bottom silt-laden layer that has already entered and eject it back to the river/escape, leaving cleaner water to continue in the canal. (Other measures: still-pond regulation, vortex-tube ejectors.)

Irrigation development in Nepal — status and challenges

Agriculture is the backbone of Nepal's economy and most arable land lies in the Terai and river valleys. Of the total irrigable area (around 2.1 million ha in the Terai, hills and mountains), a large share has some irrigation infrastructure, but year-round (reliable monsoon + winter) irrigation remains limited — a substantial portion is monsoon-dependent surface irrigation. Both agency-managed and farmer-managed systems exist.

Key challenges:

• Steep, fragile Himalayan topography making large gravity command areas costly; many small hill schemes.
• Strong seasonality — abundant monsoon flow but low dry-season river discharge; storage is scarce.
• High sediment load in rivers damaging headworks and canals.
• Inadequate canal lining → seepage, conveyance loss, waterlogging in patches.
• Weak operation/maintenance, financing, and need for inter-basin transfer/storage.
• Climate change altering snowmelt and rainfall patterns.

Two major irrigation projects

1. Sikta Irrigation Project (Banke) — a large Terai project drawing from the Rapti River to irrigate tens of thousands of hectares in the western Terai.
2. Babai Irrigation Project (Bardiya) / Bagmati Irrigation Project (Rautahat-Sarlahi) — major surface schemes diverting river flow for Terai command areas. (The Sunsari-Morang and Narayani lift/gravity systems and the inter-basin Bheri-Babai Diversion Multipurpose Project are other landmark schemes.)

Role of Farmer-Managed Irrigation Systems (FMIS)

FMIS are traditional, community-built and community-operated schemes (e.g. raj kulo, hill kulos) that irrigate a very large share of Nepal's irrigated land. Farmers themselves handle water allocation, distribution rules, fee collection, and routine maintenance through water-user associations. They embody strong local institutions and equitable rotational water sharing; government policy promotes participatory irrigation management (PIM) and joint management to strengthen and modernise these systems while preserving local ownership.

Kennedy's method (trial depth)

Kennedy's critical (non-silting, non-scouring) velocity: $V_0 = 0.55\,m\,y^{0.64}$ (m, with $y$ in m).

Trial: assume $y = 1.7$ m. Critical velocity:

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

$1.7^{0.64} = e^{0.64\ln 1.7} = e^{0.64\times0.5306} = e^{0.3396} = 1.4044.$

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

Required area: $A = Q/V_0 = 15/0.7724 = 19.42\ \text{m}^2.$

For a trapezoidal section with side slope $0.5H{:}1V$: $A = By + 0.5y^2$.

$$19.42 = B(1.7) + 0.5(1.7)^2 = 1.7B + 1.445 \Rightarrow B = \frac{19.42 - 1.445}{1.7} = \frac{17.975}{1.7} = 10.57\ \text{m}.$$

Check velocity by Kutter/Manning. Wetted perimeter:

$$P = B + 2y\sqrt{1+0.5^2} = 10.57 + 2(1.7)(1.1180) = 10.57 + 3.80 = 14.37\ \text{m}.$$ $$R = A/P = 19.42/14.37 = 1.351\ \text{m}.$$

Using Manning ($n = N = 0.0225$):

$$V = \frac{1}{n}R^{2/3}S^{1/2} = \frac{1}{0.0225}(1.351)^{2/3}(1/5000)^{1/2}.$$

$1.351^{2/3} = e^{(2/3)\ln1.351} = e^{0.667\times0.3009} = e^{0.2006} = 1.2222.$ $(1/5000)^{1/2} = (0.0002)^{0.5} = 0.014142.$

$$V = 44.44 \times 1.2222 \times 0.014142 = 0.7682\ \text{m/s}.$$

Comparison: Kennedy $V_0 = 0.772$ m/s vs computed mean $V = 0.768$ m/s — agreement within ~0.5%, so the trial depth is acceptable.

Design (one trial): depth $y = 1.7$ m, bed width $B \approx 10.6$ m, velocity ≈ 0.77 m/s, $R = 1.35$ m, on slope $1/5000$. (If the two velocities differed more, repeat with a revised $y$.)