BE Civil Engineering (IOE, TU) Irrigation and Drainage Engineering (IOE, CE 705) Question Paper 2077 Nepal

(a) Definitions and the duty–delta relation

Delta ($\Delta$): the total depth of water (in metres or cm) required by a crop over its entire base period to bring it to maturity, applied at the field.

Duty ($D$): the area of land (in hectares) that can be irrigated by a continuous flow of 1 cumec ($1\ \text{m}^3/\text{s}$) of water supplied throughout the base period $B$ (in days). Units: ha/cumec.

Derivation. A discharge of 1 cumec flowing continuously for $B$ days delivers a volume:

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

This volume spreads over an area $D$ hectares ($= 10^4 D\ \text{m}^2$) to a depth $\Delta$ (m):

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

With $\Delta$ expressed in metres:

$$\boxed{\;\Delta = \dfrac{8.64\,B}{D}\ \text{(m)}\quad\Longleftrightarrow\quad D = \dfrac{8.64\,B}{\Delta}\ \text{(ha/cumec)}\;}$$

(b) Duty at the field

Here $B = 120$ days and $\Delta = 48\ \text{cm} = 0.48\ \text{m}$.

$$D = \frac{8.64 \times 120}{0.48} = \frac{1036.8}{0.48} = 2160\ \text{ha/cumec}.$$

Duty at the field = 2160 ha/cumec.

(c) Discharge to be diverted at canal head

Field (net) discharge to irrigate $A = 1500$ ha at field duty:

$$Q_{field} = \frac{A}{D} = \frac{1500}{2160} = 0.6944\ \text{cumec}.$$

Overall efficiency $\eta = 60\% = 0.60$, so the gross discharge at the head:

$$Q_{head} = \frac{Q_{field}}{\eta} = \frac{0.6944}{0.60} = 1.157\ \text{cumec}.$$

Discharge to be diverted at the canal head $\approx$ 1.16 cumec.

Check: equivalently the duty at the head $= 0.60 \times 2160 = 1296$ ha/cumec, giving $Q_{head} = 1500/1296 = 1.157$ cumec. Consistent.

(a) Assumptions and limitations of Kennedy's theory

Assumptions: (i) silt is kept in suspension by the vertical eddies generated from the channel bed only (not the sides); (ii) the quantity of silt transported is a function of a critical velocity $V_0$ that depends on depth; (iii) the channel is in regime (non-silting, non-scouring) when the mean velocity equals $V_0$ for that depth.

Limitations: no equation for bed slope is provided (a separate flow formula such as Kutter's is needed); the $B/D$ ratio must be assumed/trial; it does not account for silt grade/charge explicitly; ignores side-eddy contribution; trial-and-error solution; applicable mainly to channels in similar alluvium to Kennedy's data (Upper Bari Doab).

(b) Design

Geometry with side slope $0.5\!:\!1$ (i.e. $z = 0.5$) and $B = 6.5\,D$:

• Area: $A = (B + zD)D = (6.5D + 0.5D)D = 7.0\,D^2$
• Wetted perimeter: $P = B + 2D\sqrt{1+z^2} = 6.5D + 2D\sqrt{1.25} = 6.5D + 2.2361D = 8.7361\,D$

Step 1 — Trial on D using $Q = A V_0$. Critical velocity $V_0 = 0.55\,(1.0)\,D^{0.64} = 0.55\,D^{0.64}$.

$$Q = A V_0 = 7.0D^2 \times 0.55 D^{0.64} = 3.85\,D^{2.64} = 30.$$ $$D^{2.64} = \frac{30}{3.85} = 7.792 \implies D = 7.792^{1/2.64}.$$

$\ln 7.792 = 2.0530$; $\;2.0530/2.64 = 0.7777$; $\;e^{0.7777} = 2.176$.

$$\boxed{D \approx 2.18\ \text{m}}$$

Step 2 — Bed width. $B = 6.5 \times 2.18 = 14.17\ \text{m} \approx \boxed{14.2\ \text{m}}.$

Step 3 — Velocity and hydraulic radius. $V_0 = 0.55\,(2.18)^{0.64}$. $\;(2.18)^{0.64}$: $\ln 2.18 = 0.7793$, $\times0.64 = 0.4988$, $e^{0.4988}=1.647$.

$$V_0 = 0.55 \times 1.647 = 0.906\ \text{m/s}.$$

$A = 7.0(2.18)^2 = 7.0 \times 4.7524 = 33.27\ \text{m}^2;\quad P = 8.7361 \times 2.18 = 19.04\ \text{m}.$

$$R = A/P = 33.27/19.04 = 1.747\ \text{m}.$$

(Discharge check: $A V_0 = 33.27 \times 0.906 = 30.1\ \text{m}^3/\text{s} \approx 30$. OK.)

Step 4 — Bed slope from Kutter's formula.

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

With $N=0.0225$, $1/N = 44.44$, $\sqrt{R}=1.322$. Solve by trial for $S$ so that $C\sqrt{RS}=V=0.906$.

Try $S = 1/5000 = 0.0002$: $0.00155/S = 7.75$. Numerator $= 23 + 44.44 + 7.75 = 75.19$. Denominator $= 1 + (23+7.75)(0.0225/1.322) = 1 + 30.75\times0.01702 = 1 + 0.5234 = 1.5234$. $C = 75.19/1.5234 = 49.36$. Then $V = 49.36\sqrt{1.747\times0.0002} = 49.36\sqrt{0.0003494} = 49.36\times0.01869 = 0.923\ \text{m/s}$ (slightly high).

Try $S = 1/5300 = 0.0001887$: $0.00155/S = 8.213$. Num $= 23+44.44+8.213 = 75.65$. Den $= 1 + (31.213)(0.01702) = 1.5314$. $C = 49.40$. $V = 49.40\sqrt{1.747\times0.0001887} = 49.40\sqrt{0.0003297}=49.40\times0.01816 = 0.897\ \text{m/s} \approx 0.906$. Close.

Adopt $S \approx 1/5250$ (about $0.00019$).

Summary

(a) Bligh's creep theory and its limitations

Bligh's creep theory assumes that water seeping below a hydraulic structure creeps along the contact (the bottom profile of the floor and cutoffs) and that the head is lost uniformly per unit length of creep, treating vertical and horizontal creep as equally effective. The total creep length $L =$ (sum of horizontal lengths) $+ 2\times$(sum of vertical cut-off depths). Safety against piping is judged by Bligh's coefficient $C = L/H_s$.

Why superseded by Khosla: Bligh wrongly weighted vertical and horizontal creep equally and could not predict the actual uplift distribution; observed failures of works designed by Bligh's method showed it was unsafe. Khosla's analysis (potential-flow / electrical analogy) showed that the exit gradient at the downstream end governs undermining (piping) — particles are dislodged when the upward seepage force exceeds the submerged weight of soil. Khosla introduced independent variables (treating each cut-off and floor as a standard form), correction factors (for floor thickness, interference of piles, slope), and a quantitative exit gradient criterion, giving safer, more accurate designs.

(b) Khosla pressures and exit gradient

Geometry: $b = 36\ \text{m}$, $d = 6\ \text{m}$, $\alpha = b/d = 6$, seepage head $H_s = 5.0\ \text{m}$.

Uplift pressure heads at key points (pressure as % of $H_s$, measured above downstream floor):

• Point $E$ (upstream face of d/s pile): $\phi_E = 38\%$

$$h_E = 0.38 \times 5.0 = 1.90\ \text{m of water}.$$

• Point $D$ (bottom of d/s pile): $\phi_D = 26\%$

$$h_D = 0.26 \times 5.0 = 1.30\ \text{m of water}.$$

(Point $C$ at d/s floor level has $\phi_C = 0\%$, head $= 0$, the exit.)

These uplift heads act upward on the floor; the floor thickness must be designed so its submerged weight balances the net uplift (e.g. at $E$, net uplift head above floor $= 1.90\ \text{m}$).

Exit gradient. Khosla's exit-gradient formula for a floor of length $b$ with a d/s vertical cut-off of depth $d$:

$$G_E = \frac{H_s}{d}\cdot\frac{1}{\pi\sqrt{\lambda}},\qquad \lambda = \frac{1+\sqrt{1+\alpha^2}}{2}.$$

With $\alpha = 6$: $\sqrt{1+\alpha^2} = \sqrt{37} = 6.0828$, so

$$\lambda = \frac{1 + 6.0828}{2} = 3.5414,\qquad \sqrt{\lambda} = 1.8819.$$ $$G_E = \frac{5.0}{6}\times\frac{1}{\pi \times 1.8819} = 0.8333 \times \frac{1}{5.9123} = 0.8333 \times 0.16914 = 0.1410.$$ $$\boxed{G_E \approx 0.141 \;\;(\text{i.e. } 1\,\text{in}\,7.1)}$$

Safety check: for typical alluvial sand the safe (permissible) exit gradient is about $1/5$ to $1/7$ ($G \approx 0.20$ to $0.14$). With a factor of safety of $\sim 5$ the safe value is $\approx 0.14$–$0.17$; here $G_E = 0.141$ is at the safe limit, so the section is acceptable (provide adequate d/s floor thickness for the computed uplift).

(a) Classification of cross-drainage (CD) works

A cross-drainage work carries a canal across a natural drainage (stream) where the two cross. Three families depending on relative bed levels:

1. Canal over drainage (canal on top):

• Aqueduct — canal is carried in a trough/bridge above the drain; the drainage flows freely (at atmospheric pressure) underneath with its HFL below the canal trough bottom.

        ====== Canal trough ======
   ~~~~~~ drain HFL ~~~~~~  (free flow, below trough)

Selected when drainage HFL is sufficiently below the canal bed and the drain is small relative to the canal.

• Siphon aqueduct — same arrangement, but the drainage HFL is above the trough bottom, so the drain is forced to flow under pressure through barrels (runs full). Selected when the drain HFL is high / drain discharge large.

2. Drainage over canal (drain on top):

• Super-passage — the drain is carried in a trough above the canal; canal flows freely below at atmospheric pressure. Selected when the canal FSL is below the drainage bed.
• Canal siphon (siphon) — drain on top, but the canal is forced to flow under pressure through barrels below the drain trough. Selected when canal FSL is above the drain trough bottom.

3. Level crossing / inlet–outlet — drain and canal water are allowed to mix at the crossing (controlled by regulators); used when both are at nearly the same level and the drain flood is small or intermittent.

(b) Siphon-aqueduct barrel hydraulics

Barrel area (one rectangular vent):

$$A = b\times h = 10 \times 3 = 30\ \text{m}^2.$$

Velocity through the barrel (running full):

$$V = \frac{Q_d}{A} = \frac{200}{30} = 6.667\ \text{m/s}.$$

(This is high; in practice multiple vents reduce $V$, but we proceed with the given single vent.)

Velocity head:

$$\frac{V^2}{2g} = \frac{(6.667)^2}{2\times 9.81} = \frac{44.45}{19.62} = 2.266\ \text{m}.$$

Afflux / loss of head across the barrel (entry + friction + exit lumped, coefficient $1.5$):

$$h_L = 1.5\times\frac{V^2}{2g} = 1.5\times 2.266 = 3.40\ \text{m}.$$ $$\boxed{V = 6.67\ \text{m/s},\qquad \text{afflux } h_L \approx 3.40\ \text{m}}$$

Remark: an afflux of 3.4 m and $V\approx6.7$ m/s are excessive (scour risk, large upstream heading-up). The design should be revised by providing additional barrel vents to bring $V$ down to about 2–3 m/s. For example with three such vents $A = 90\ \text{m}^2$, $V = 2.22\ \text{m/s}$, $V^2/2g = 0.252\ \text{m}$ and $h_L = 0.38\ \text{m}$, which is acceptable.

(a) Waterlogging

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

Causes: over-irrigation and excessive percolation; seepage from unlined canals and reservoirs; inadequate natural/artificial drainage; impervious sub-strata; flat topography and obstruction of natural drainage by embankments/roads; heavy rainfall and floods; rising regional groundwater.

Harmful effects: lack of soil aeration (root suffocation); accumulation of harmful salts at the surface as water evaporates (salinity/alkalinity); fall in soil temperature retarding bacterial activity; difficulty in tillage; reduction or total loss of crop yield; growth of weeds and water-loving plants.

Reclamation / preventive measures: lining of canals to cut seepage; controlled/optimum irrigation (correct duty); provision of surface and subsurface (tile) drains; intercepting drains; pumping (vertical tube-well drainage to lower the water table); leaching of salts with good drainage; afforestation/biological drainage; improving natural drainage outfalls.

(b) Tile-drain spacing (Hooghoudt steady-state)

Given: $d = 2.0\ \text{m}$, $h = 0.4\ \text{m}$, $K = 1.0\ \text{m/day}$, $R = 8\ \text{mm/day} = 0.008\ \text{m/day}$.

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

Numerator: $4 \times 1.0 \times 0.4 \times (2\times2.0 + 0.4) = 1.6 \times (4.0 + 0.4) = 1.6 \times 4.4 = 7.04\ \text{m}^3/\text{day}.$

$$S^2 = \frac{7.04}{0.008} = 880\ \text{m}^2.$$ $$S = \sqrt{880} = 29.66\ \text{m}.$$ $$\boxed{S \approx 29.7\ \text{m (adopt about 29 m)}}$$

Interpretation: placing the tile drains about 29–30 m apart keeps the mid-spacing water table no higher than $0.4\ \text{m}$ above drain level under the design recharge of $8\ \text{mm/day}$, thereby keeping the root zone adequately drained.

Surface vs pressurized irrigation

Two Nepalese situations favouring drip:

1. Mid-hill / Terai vegetable and high-value cash-crop farming (tomato, cucumber, coffee, off-season vegetables) on sloping terraces where surface irrigation causes erosion and water is limited — drip gives high efficiency and fertigation.
2. Water-scarce, lift/groundwater-dependent areas (e.g. parts of the western Terai and inner-Madhesh, or hill areas served by small reservoirs/jars) where pumped/stored water is costly, so the high efficiency of drip conserves scarce, expensive water.

(a) Net irrigation requirement

For paddy, the field water requirement comprises evapotranspiration plus percolation (and any saturation/puddling demand). Net irrigation requirement after deducting effective rainfall:

$$NIR = (ET_c + \text{percolation}) - R_e = (18.0 + 9.0) - 12.0 = 27.0 - 12.0 = 15.0\ \text{cm}.$$ $$\boxed{NIR = 15.0\ \text{cm over 30 days}}$$

(b) Field (gross) irrigation requirement

With field application efficiency $\eta_a = 70\% = 0.70$:

$$FIR = \frac{NIR}{\eta_a} = \frac{15.0}{0.70} = 21.43\ \text{cm}.$$ $$\boxed{FIR \approx 21.4\ \text{cm over 30 days}}$$

Equivalent continuous flow rate (per hectare)

Depth $= 0.2143\ \text{m}$ over $1\ \text{ha} = 10^4\ \text{m}^2$ in $30$ days.

$$\text{Volume} = 0.2143 \times 10^4 = 2143\ \text{m}^3.$$

Time $= 30 \times 24 \times 3600 = 2.592\times10^6\ \text{s}.$

$$Q = \frac{2143}{2.592\times10^6} = 8.27\times10^{-4}\ \text{m}^3/\text{s} = 0.827\ \text{L/s}.$$ $$\boxed{\approx 0.83\ \text{L/s per hectare (gross)}}$$

(a) Types of canal alignment

Watershed (ridge) canal: aligned along the ridge line (watershed) separating two drainage valleys. Because the highest ground is followed, the canal can irrigate both sides by gravity and no drainage crosses it, so cross-drainage works are largely eliminated. Main canals in plains are kept on the watershed.

Contour canal: aligned roughly along a contour (with a small longitudinal slope) on hill/sloping country. Ground rises on the upper side and falls on the lower side, so it can irrigate only the lower side, and natural drains coming from above must be crossed by cross-drainage works.

Side-slope canal: aligned roughly at right angles to the contours, running down the side slope (parallel to the natural drainage). Drainage runs parallel to it, so no cross-drainage works are needed, but it commands very little area and is used as a feeder/branch.

Advantages of watershed alignment: irrigation on both banks by gravity; no/minimal cross-drainage works (economical); the canal is safe from flooding by adjacent drains; commands the maximum area.

(b) Lining of canals

Advantages: drastically reduces seepage losses (saves water, prevents waterlogging/salinity); permits higher velocity (smaller section, lower cost, less land); prevents scour and silting, weed growth, and bank erosion; lower maintenance; safer against breaching.

Disadvantages: high initial (capital) cost; difficulty of repair if lining cracks/fails and of taking out new outlets; requires skilled construction and curing; possible uplift/drainage problems behind the lining; longer construction time.

Layout of a diversion headworks

           RIVER (flow direction  ↓)
   Guide bank )                       ( Guide bank
      |   ----------- Weir / Barrage bays -----------   |
      |   | (with crest gates / shutters)            |  |
      |   |                                           |  |
      |   |======= DIVIDE WALL =======|               |  |
      |   | Undersluice bays (lower   |  Fish ladder  |  |
      |   |  crest, near canal head)  |  (side bay)   |  |
   Canal HEAD REGULATOR  →→→  Main Canal (off-take)
         (silt-excluder tunnels in pocket upstream)

Functions of components

• Weir / Barrage: the obstruction across the river that raises (heads up) the water level so it can flow into the canal by gravity. A weir raises level mainly by a solid raised crest; a barrage uses gates over a low crest to control the level (better silt control, less afflux).
• Undersluices (scouring sluices): gated bays of lower crest adjacent to the canal head regulator; they keep a deep, silt-free pocket in front of the regulator, scour out deposited silt, and pass low flows/floods downstream.
• Divide wall: a wall (parallel to flow) separating the undersluice pocket from the weir proper; it provides a still silt pocket, prevents cross-currents, and separates the scouring action from the main weir flow.
• Fish ladder: a stepped/baffled channel that lets fish migrate up- and downstream past the weir by dissipating energy into small drops with low velocity.
• Canal head regulator: the gated entrance to the off-taking canal; it controls the discharge entering the canal, regulates the supply, and excludes silt and excess flood water.
• Guide banks: earthen banks (with stone pitching) on either approach that guide the river flow axially through the works, prevent outflanking, and protect the structure and approach during floods.
• Silt-excluder: a system of tunnels in the floor of the canal head-regulator pocket that carries the bottom (silt-laden) layer of water back to the river through the undersluices, so that comparatively clear top water enters the canal.

Irrigation development in Nepal

Classification of systems:

• By management: Farmer-Managed Irrigation Systems (FMIS) — built, operated and maintained by farmers themselves (e.g. traditional raj kulo, kulo systems); these irrigate a large share of Nepal's irrigated area. Agency-Managed Irrigation Systems (AMIS) — built and run by the government / Department of Irrigation (now under provincial/federal agencies). Jointly managed systems also exist (turnover/management transfer).
• By source: Surface irrigation — diversion of rivers/streams via canals (dominant in Terai and hills). Groundwater irrigation — shallow and deep tube wells, important in the Terai. Lift irrigation — pumping from rivers/canals for higher land. Non-conventional — sprinkler/drip, ponds, rainwater-harvesting jars in hills.

Major projects (examples): Sunsari-Morang Irrigation Project, Bagmati Irrigation Project, Babai Irrigation Project, Mahakali (Sarada/Tanakpur), Kankai, Narayani Lift, Praganna/Chhatis Mauja (FMIS), and the multipurpose Sikta and Bheri-Babai Diversion Multipurpose Project. (Naming any two is sufficient, e.g. Sunsari-Morang and Bheri-Babai.)

Key challenges:

• Seasonal / unreliable water supply — many systems irrigate only in the monsoon (year-round/round-the-year irrigation is limited); dependence on monsoon-fed rivers.
• Aging and poorly maintained infrastructure, low conveyance efficiency, siltation and sediment in Himalayan rivers.
• Difficult topography in the hills/mountains (steep terraces, small command areas, high cost of canals and cross-drainage).
• Floods, landslides and changing river morphology damaging headworks; climate-change variability.
• Institutional/governance issues — weak Water User Associations in some AMIS, inadequate cost recovery, fragmented federal/provincial/local mandates.
• Limited storage (mostly run-of-river diversion) restricting dry-season supply; under-utilised groundwater in the Terai.

Lacey regime design

Given: $Q = 45\ \text{m}^3/\text{s}$, silt factor $f = 1.1$.

1. Regime velocity.

$$V = \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{45\times 1.1^2}{140}\right)^{1/6} = \left(\frac{45\times 1.21}{140}\right)^{1/6} = \left(\frac{54.45}{140}\right)^{1/6} = (0.38893)^{1/6}.$$

$\ln 0.38893 = -0.9444$; $\div 6 = -0.15740$; $e^{-0.15740} = 0.8543$.

$$\boxed{V = 0.854\ \text{m/s}}$$

2. Hydraulic mean radius.

$$R = \frac{5}{2}\frac{V^2}{f} = 2.5\times\frac{(0.8543)^2}{1.1} = 2.5\times\frac{0.72983}{1.1} = 2.5\times 0.66348 = 1.659\ \text{m}.$$ $$\boxed{R = 1.66\ \text{m}}$$

3. Wetted perimeter.

$$P = 4.75\sqrt{Q} = 4.75\sqrt{45} = 4.75\times 6.7082 = 31.86\ \text{m}.$$ $$\boxed{P = 31.9\ \text{m}}$$

4. Bed slope.

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

$f^{5/3}$: $\ln 1.1 = 0.09531$, $\times 5/3 = 0.15885$, $e^{0.15885} = 1.1722$. $Q^{1/6} = 45^{1/6}$: $\ln 45 = 3.8067$, $\div 6 = 0.63445$, $e^{0.63445} = 1.8860$.

$$S = \frac{1.1722}{3340\times 1.8860} = \frac{1.1722}{6299.2} = 1.861\times10^{-4} \approx \frac{1}{5373}.$$ $$\boxed{S \approx 1.86\times10^{-4}\ \;(\approx 1\ \text{in}\ 5370)}$$

Summary table

(Consistency check: $A = P\,R = 31.86\times1.659 = 52.9\ \text{m}^2 \approx Q/V = 52.7\ \text{m}^2$. OK.)