Irrigation Water Requirements

Irrigation water requirement (IWR) of rice is the water that must be supplied by irrigation, after deducting effective rainfall, to meet crop evapotranspiration, field losses and land preparation needs.

Steps

1. Collect data: climate (temperature, humidity, wind, sunshine), rainfall, soil type (percolation), crop calendar (nursery, transplanting, growing stages, harvest).
2. Reference evapotranspiration ($ET_o$): compute by Penman, modified Penman or Blaney-Criddle method for each month or ten-day period.
3. Crop coefficient ($K_c$): select $K_c$ for each growth stage of rice (initial, development, mid, late) from FAO tables.
4. Crop evapotranspiration:

5. Seepage and percolation (S&P): estimate daily loss from the standing water (e.g. 2-6 mm/day depending on soil).
6. Land preparation water (puddling): estimate water for soaking and puddling, typically 150-250 mm, spread over the preparation period; also nursery requirement.
7. Water layer (WL): water needed to establish and maintain standing water, e.g. 50-100 mm at transplanting.
8. Effective rainfall ($P_e$): estimate by dependable rainfall (75% probability), USDA SCS or fixed percentage method.
9. Net irrigation requirement:

10. Gross irrigation requirement:

where $\eta_{overall}=\eta_c\times\eta_a$ (conveyance and application efficiencies). 11. Convert to discharge: the peak monthly GIR gives the canal capacity (l/s/ha, or $m^3/s$ for the command area).

Given: FC = 40%, PWP = 20%, $\gamma_d = 1.2$ g/cc (so $\gamma_d/\gamma_w=1.2$), root depth $D_r = 90$ cm = 900 mm, $ET_c = 10$ mm/day, RAM = 75% of available moisture.

Step 1: Available moisture capacity (AMC)

Step 2: Readily available moisture (RAM)

Step 3: Irrigation interval

Answer: irrigation interval = 16.2 days, say 16 days. (Round down so the crop never goes beyond RAM.)

Given: FC = 40%, PWP = 20%, $\gamma_d = 1.2$, $D_r = 900$ mm, $ET_c = 10$ mm/day, RAM = 75% of AMC.

Step 1: Available moisture capacity (AMC)

Step 2: Readily available moisture (RAM)

Step 3: Irrigation interval

Graph of available moisture against time

 Avail.
 moisture
 (mm)
 216 |= = = FC (AMC = 216 mm, full) = = = =
     |\
     | \        RAM = 162 mm used
     |  \       (75% of AMC)
  54 |- -\- - - - - - - - - - - - - - - -  limit of RAM
     |    \ slope = ET = 10 mm/day
     |
   0 |=====================================  PWP
     0                16.2          time (days)
        irrigate again at 16 days

Moisture falls from AMC (216 mm available, at FC) linearly at 10 mm/day. When 162 mm (RAM) has been used, i.e. after 16.2 days, the available moisture is 54 mm; irrigation is given to bring it back to FC. The remaining 25% of AMC (54 mm) is kept as reserve so the crop does not reach PWP.

Answer: irrigation interval = 16.2 days, say 16 days.

Given: FC = 60%, PWP = 25%, $\gamma_d = 1.2$, $D_r = 120$ cm $= 1200$ mm, $ET_c = 9$ mm/day, RAM = 85% of available moisture.

Answer: irrigation interval = 47.6 days, say 47 days.

Duty ($D$)

The area of land (in hectares) that can be irrigated by supplying one cumec ($1\ m^3/s$) of water continuously throughout the base period of the crop. Unit: ha/cumec. Duty depends on the point of measurement (field, outlet, canal head).

Delta ($\Delta$)

The total depth of water (in m or cm) required by a crop over its base period if the water is spread over the land. Unit: m.

Base period ($B$)

The time in days from the first watering to the last watering of the crop before harvest.

Relationship between duty and delta

Take 1 cumec flowing for $B$ days.

with $B$ in days and $D$ in ha/cumec. Hence a crop with a high delta has a low duty, and vice versa.

Example: wheat with $B=120$ days and $D=1800$ ha/cumec gives $\Delta = 8.64\times120/1800 = 0.576$ m.

Soil acts as a reservoir for the water used by crops, and irrigation refills it. The relationship is explained by the soil-moisture constants.

 Moisture
 content
  Saturation  ---- gravity water drains ----
  FC          ---- upper limit (irrigate up to here)
                |  available moisture
  RAM limit   ---- (allowable depletion) ----
  PWP         ---- plant wilts; lower limit
  Hygroscopic ---- not available

• Saturation: all pores are filled. Excess (gravity) water drains in 1-3 days.
• Field capacity (FC): moisture held after free drainage; the upper limit of useful soil moisture.
• Permanent wilting point (PWP): moisture at which plants wilt permanently; the lower limit.
• Available moisture = FC - PWP. Readily available moisture (RAM) is the portion (usually 50-75%) the crop can take without stress.

Link with irrigation

1. Irrigation is given when moisture falls to the RAM limit, not to PWP.
2. Depth of water per irrigation (net) = depth of moisture needed to bring the root zone back to FC:

3. Frequency (interval) of irrigation = $d/ET_c$.
4. Over-irrigation beyond FC is lost as deep percolation and may cause waterlogging; under-irrigation causes stress and yield loss.

Thus soil moisture characteristics decide how much water to apply and how often.

Crop water requirement (CWR) is the total water needed by a crop from sowing to harvest, equal to the consumptive use plus losses and special needs, supplied by rainfall, soil moisture and irrigation.

Steps of calculation

1. Collect climatic data (temperature, humidity, sunshine, wind, rainfall) and the crop calendar (sowing, growth stages, harvest).
2. Compute reference evapotranspiration $ET_o$ by Penman, Blaney-Criddle, Hargreaves, or pan evaporation method ($ET_o = K_p E_{pan}$).
3. Choose the crop coefficient $K_c$ for each growth stage from FAO-24/56 tables.
4. Compute crop evapotranspiration:

5. Add other needs: percolation and seepage (rice), land preparation, nursery, leaching requirement, and water for planting.
6. Compute effective rainfall $P_e$ (75% dependable rainfall).
7. Compute net irrigation requirement:

8. Compute gross irrigation requirement using efficiency:

9. Convert depth to discharge: $Q = \dfrac{GIR\times A}{T}$ or express as l/s/ha.
10. Find the peak demand period which governs the design capacity of canal and structures.

Given: FC = 27%, PWP = 14%, $\gamma_d = 1.5$ g/cm$^3$, $D_r = 75$ cm = 750 mm, $ET_c = 12$ mm/day. The full available moisture is taken as usable (no RAM fraction is given).

Answer: water should be supplied after about 12.2 days, i.e. every 12 days. (Round down for safety.)

Given: FC = 28%, PWP = 13%, $\gamma_d = 1.3$ g/cc, root zone depth $D_r = 70$ cm $= 700$ mm, $ET_c = 12$ mm/day.

Step 1: Available moisture depth

Step 2: Irrigation interval

Answer: irrigate after about 11.4 days, i.e. every 11 days (rounded down so the crop does not suffer stress).

Given: FC = 35%, PWP = 12%, $D_r = 70$ cm $= 700$ mm, $\gamma_d = 1.4$, $ET_c = 9$ mm/day, RAM = 70% of AMC, $\eta_a = 85\%$, conveyance + distribution loss = 20%.

(i) Available moisture content (AMC)

(ii) Readily available moisture (RAM)

(iii) Depth of irrigation at the outlet (field) with $\eta_a=85\%$:

(iv) Irrigation interval

(v) Depth of water required at the headworks (20% lost in conveyance and distribution, so 80% reaches the field):

Answer: AMC = 225.40 mm; RAM = 157.78 mm; depth at field = 185.62 mm; interval = 17.5 days (17 days); depth at headwork = 232.03 mm.

Given: FC = 80%, PWP = 35%, $D_r = 60$ cm $= 600$ mm, $\gamma_d = 1.5$, $ET_c = 5$ mm/day, $\eta_a = 80\%$, RAM = 70% of AMC.

(i) Available moisture content

(ii) Readily available moisture

(iii) Depth of irrigation at the outlet of the field

(iv) Irrigation interval

Answer: AMC = 405.00 mm; RAM = 283.50 mm; depth at outlet = 354.38 mm; interval = 56.7 days (about 56 days).

Formula: $Q = \dfrac{A\,d}{T\times 86400}$ with $A$ in m$^2$, $d$ in m and $T$ in days.

Dry-season crops: $A = 20000$ ha $= 2\times10^8$ m$^2$, $T = 5\times7 = 35$ days, $d = 0.125$ m

Wet-season crops: $A = 30000$ ha $= 3\times10^8$ m$^2$, $T = 4\times7 = 28$ days, $d = 0.095$ m

The two seasons do not overlap, so the canal is designed for the larger discharge.

Answer: canal discharge = 11.78 cumec (wet season governs; dry season needs 8.27 cumec).

Formula: $Q = \dfrac{A\,d}{T\times 86400}$ with $A$ in m$^2$, $d$ in m, $T$ in days.

Dry-season crops: $A = 30000$ ha $= 3\times10^8$ m$^2$, $T = 6\times7 = 42$ days, $d = 0.148$ m

Wet-season crops: $A = 40000$ ha $= 4\times10^8$ m$^2$, $T = 4\times7 = 28$ days, $d = 0.115$ m

The seasons are different, so the canal is designed for the larger of the two.

Answer: canal discharge = 19.01 cumec (wet season governs; dry season needs 12.24 cumec).

• Readily available moisture (RAM): the part of the available moisture (FC - PWP) which the crop can extract easily without stress, usually 50-75% of the available moisture (the upper part of the available range).
• Permanent wilting point (PWP): the soil moisture content (as % of dry weight) at which plants wilt and cannot recover even if kept in a humid atmosphere; soil moisture tension is about 15 atm (1500 kPa).
• Irrigation interval (frequency): the number of days between two successive irrigations during the growing period. It is found as

• Field capacity (FC): the amount of water held in the soil after the excess gravity water has drained away (usually 1-3 days after irrigation or rain); soil moisture tension is about 1/3 atm. It is the upper limit for irrigation.
• Available moisture (AM): the water held between field capacity and permanent wilting point, which plants can use: $AM = FC - PWP$.
• Readily available moisture (RAM): the fraction of available moisture (about 50-75%) which crops extract easily without moisture stress.
• Soil water deficiency (moisture deficiency): the amount of water required to bring the soil moisture of the root zone back to field capacity at any time; $= FC - \text{present moisture}$ (expressed as depth, mm). It is the net depth to be applied in irrigation.

• Soil moisture tension (suction): the force (negative pressure) with which water is held by the soil; the energy per unit mass that a plant has to apply to take water from soil. It is measured in atm, kPa or pF and increases as the soil dries (about 1/3 atm at FC and 15 atm at PWP).
• Osmotic pressure: pressure caused by dissolved salts in soil water, which opposes the entry of water into root cells. A saline soil has high osmotic pressure, so total water stress for the plant is the sum of soil moisture tension and osmotic pressure.
• Field capacity: moisture content retained after free gravity drainage; tension about 1/3 atm; upper limit of available moisture.
• Permanent wilting point: moisture content at which plants wilt permanently; tension about 15 atm; lower limit of available moisture.
• Available moisture: the water between FC and PWP that is usable by plants: $AM = FC - PWP$.

Soil water exists in several forms, depending on how strongly it is held by the soil particles.

  Moisture content (%)
 |  Saturation  ........ gravity water (drains out)
 |  ---------------------------------- FC (1/3 atm)
 |    capillary water    available
 |                       moisture
 |  ---------------------------------- PWP (15 atm)
 |    hygroscopic water  (not available)
 |  ---------------------------------- oven dry

1. Gravity (free) water: water in the large pores above field capacity which moves down under gravity. It drains in 1-3 days and is not stored for plant use; excess causes waterlogging.
2. Capillary water: water held in the small pores by surface tension, between field capacity and the hygroscopic range. This is the main source for plants. The part above PWP is the available moisture.
3. Hygroscopic water: a thin film held by strong adhesive forces on the soil particles from air humidity (tension above 31 atm). It is not available to plants.
4. Chemical (combined) water: water in the chemical structure of minerals; unavailable.

Soil moisture constants

• Saturation capacity: all pores filled with water.
• Field capacity: upper limit of capillary water.
• Permanent wilting point: lower limit of available water; the plant wilts permanently.
• Available moisture: $FC - PWP$; Readily available moisture: part of it extracted without stress.
• Moisture equivalent, hygroscopic coefficient are laboratory-defined related constants.

Forms of soil moisture

Soil water exists in several forms, depending on how strongly it is held by the soil particles.

  Moisture content (%)
 |  Saturation  ........ gravity water (drains out)
 |  ---------------------------------- FC (1/3 atm)
 |    capillary water    available
 |                       moisture
 |  ---------------------------------- PWP (15 atm)
 |    hygroscopic water  (not available)
 |  ---------------------------------- oven dry

1. Gravity (free) water: water in the large pores above field capacity which moves down under gravity. It drains in 1-3 days and is not stored for plant use; excess causes waterlogging.
2. Capillary water: water held in the small pores by surface tension, between field capacity and the hygroscopic range. This is the main source for plants. The part above PWP is the available moisture.
3. Hygroscopic water: a thin film held by strong adhesive forces on the soil particles from air humidity (tension above 31 atm). It is not available to plants.
4. Chemical (combined) water: water in the chemical structure of minerals; unavailable.

Soil moisture constants

• Saturation capacity: all pores filled with water.
• Field capacity: upper limit of capillary water.
• Permanent wilting point: lower limit of available water; the plant wilts permanently.
• Available moisture: $FC - PWP$; Readily available moisture: part of it extracted without stress.
• Moisture equivalent, hygroscopic coefficient are laboratory-defined related constants.

Factors affecting crop water requirement

1. Climate: temperature, humidity, wind speed and sunshine; hot, dry and windy weather increases $ET$.
2. Crop type: rice and sugarcane need much more water than wheat or pulses; leaf area and root depth matter.
3. Growth stage and crop period: mid-season stage needs the most water; the length of the crop period decides the total.
4. Soil: texture, water holding capacity, infiltration and percolation losses.
5. Water table depth: a shallow water table supplies capillary water and reduces demand.
6. Rainfall: effective rainfall reduces irrigation need.
7. Method of irrigation and efficiency: drip uses less than flooding.
8. Cultural practices: time and method of sowing, tillage, mulching, weed control, plant density.
9. Salinity: need for leaching raises water demand.
10. Land preparation and season: puddling for rice and kharif or rabi season.

Crop water requirement (CWR) is the total quantity of water, regardless of its source, needed by a crop to grow to maturity. It includes consumptive use (evapotranspiration) plus losses in application (percolation, seepage, runoff) and special needs (puddling, leaching):

and the irrigation requirement is $CWR - P_e -$ soil moisture used.

Factors affecting CWR

1. Climate: temperature, humidity, sunshine hours and wind increase evapotranspiration.
2. Crop: type, variety, leaf area, root depth and growth stage; rice and sugarcane have a high CWR.
3. Crop period: longer periods need more water.
4. Soil: texture, structure and infiltration; sandy soil has more deep percolation.
5. Water table: a shallow table contributes capillary water.
6. Rainfall: distribution and effective rainfall.
7. Method and efficiency of irrigation: surface methods lose more than drip or sprinkler.
8. Agronomic practices: tillage, mulching, planting date, spacing, weed control.
9. Salinity of soil and water: extra leaching water is needed.
10. Season and altitude of the area.

Potential evapotranspiration (PET): the amount of water evaporated and transpired per unit area per unit time from a short green grass cover (0.12 m height, well-watered, actively growing, completely shading the ground) with unlimited water supply.

Penman's combination method (steps)

Penman combines the energy-balance (radiation) and aerodynamic (wind) terms:

where $PET$ in mm/day, $\Delta$ = slope of the saturation vapour pressure curve at air temperature (mm Hg/$^\circ$C), $\gamma$ = psychrometric constant (0.49 mm Hg/$^\circ$C), $H_n$ = net radiation (mm of evaporable water/day), $E_a$ = aerodynamic term (mm/day).

1. Mean air temperature $T$ and saturation vapour pressure $e_a$ (mm Hg) at $T$; get $\Delta$ from tables.
2. Actual vapour pressure $e_d = e_a \times RH/100$.
3. Net radiation:

where $H_a$ = extraterrestrial radiation (table by latitude and month), $r$ = albedo (0.25 for grass), $n/N$ = ratio of actual to possible sunshine hours, $a = 0.18$, $b = 0.55$ (typical), $\sigma$ = Stefan-Boltzmann constant. 4. Aerodynamic term:

with $u_2$ = wind speed at 2 m height (km/day). 5. Substitute into the Penman equation to get PET (mm/day). 6. Crop evapotranspiration: $ET_c = K_c\times PET$. 7. Multiply by the number of days of the month for monthly values.

Irrigation efficiency is the ratio of the water beneficially used to the water supplied at a stage, expressed as a percentage. Types are as follows.

1. Water conveyance efficiency ($\eta_c$): the ratio of water delivered to the field (outlet) to water diverted at the canal head. It accounts for seepage and evaporation losses in canals.

2. Water application efficiency ($\eta_a$): the ratio of water stored in the root zone to water delivered to the field.

3. Water storage efficiency ($\eta_s$): the ratio of water stored in the root zone to the water needed to bring the root zone to field capacity.

4. Water distribution (uniformity) efficiency ($\eta_d$): how uniformly water is spread over the field.

where $d$ is the mean depth stored and $y$ the average numerical deviation from it. 5. Water use efficiency: crop yield per unit volume of water used (kg/m$^3$). 6. Consumptive use efficiency: consumptive use of crop water divided by water depleted from the root zone. 7. Overall (project) efficiency: $\eta_{o}=\eta_c\times\eta_a$ (also including reservoir/ distribution efficiencies); typically 25-40% in Nepal's surface systems.

Given: $Q_{diverted} = 150$ l/s, $Q_{field} = 110$ l/s, area $= 2.2$ ha $= 22000$ m$^2$, time $= 8$ h, root zone $= 1.5$ m, runoff $= 445$ m$^3$, penetration depth 1.5 m (head) to 1.1 m (tail), available moisture = 200 mm/m, irrigation started at 50% depletion of available moisture.

1. Water conveyance efficiency

2. Water applied to the field

Water stored in the root zone. The penetration depth (1.5 m at the head, 1.1 m at the tail) does not exceed the root depth of 1.5 m, so deep percolation is nil (assumed). Hence

Water application efficiency

3. Water storage efficiency. Water needed to bring the root zone back to field capacity (50% of available moisture was used):

4. Water distribution efficiency. Depth varies linearly between 1.5 m and 1.1 m, so mean depth $d = (1.5+1.1)/2 = 1.3$ m and the average numerical deviation $y = 0.2/2 = 0.1$ m.

Answer: $\eta_c = 73.33\%$, $\eta_a = 85.95\%$, $\eta_s = 82.52\%$, $\eta_d = 92.31\%$.

• Duty ($D$): the area of crop (in ha) irrigated by one cumec of water supplied continuously for the base period of the crop. Unit: ha/cumec. It decreases from the field to the canal head because of losses.
• Delta ($\Delta$): the total depth of water required by a crop during its base period (in m or cm), if it were spread over the field. Related by $\Delta = 8.64B/D$ (m).
• Base period ($B$): the time between the first watering of a crop at sowing (or nursery) and its last watering before maturity/harvest, in days.
• Crop period: the total time from sowing to harvesting of the crop (days). It is a little longer than the base period, as it includes the time before the first watering and after the last watering.

Irrigation water requirement (IWR) of rice is the depth of water that must be supplied by irrigation to rice fields during the crop period, after deducting the effective rainfall, to meet crop evapotranspiration, seepage and percolation, land preparation (puddling) and standing-water needs:

It is expressed in mm per period, or as a flow rate (l/s/ha).

The crop coefficient curve shows how $K_c$ ($ET_c/ET_o$) changes over the growing period. For rice (FAO), $K_c$ is about 1.05 at the initial stage, rises to about 1.20 in the mid-season and falls to about 0.9 at the late stage.

 Kc
1.2|              ______________
   |            /               \
1.1|          /                   \
1.0|  ______/                       \
0.9|                                  \__
   +-----+-----------+-------------+----+--> days
    Initial  Crop      Mid-season   Late
    (20-30d) development (40-50d)  (20-30d)

• Initial stage: low ground cover; $K_c$ about 1.05 (standing water in a nursery/field makes evaporation high).
• Development stage: $K_c$ rises linearly as the canopy develops.
• Mid-season: $K_c$ at the maximum, about 1.2 (full cover, flowering).
• Late-season: $K_c$ falls to about 0.9 (maturity and drying).

The values are typical (FAO Irrigation and Drainage Paper 24/56) and are used in $ET_c=K_c\,ET_o$.

Assumptions. The value 150 mm for $ET_o$ cannot be a daily rate, so it is taken as mm per month, the same time unit as the effective rainfall (30 mm/month). Crop coefficient $K_c=1$ (crop water requirement = $ET_o$), month = 30 days.

Given: CCA = 6000 ha, $ET_o = 150$ mm/month, $P_e = 30$ mm/month, overall efficiency $= 30\%$.

Step 1: Net irrigation requirement

Step 2: Gross irrigation requirement at the head

Step 3: Volume of water per month

Step 4: Discharge

Answer: irrigation demand $\approx 9.26$ cumec.

Soil is the loose upper layer of the earth's crust, formed by weathering of rocks and decay of organic matter, which supplies water, air and nutrients to plants.

Soil constituents

Mineral particles (45%), organic matter (5%), water (25%) and air (25%) by volume in a good agricultural soil.

Classification by texture

• Sand (0.05-2 mm): coarse, high infiltration, low water-holding capacity.
• Silt (0.002-0.05 mm): medium.
• Clay (<0.002 mm): fine, low infiltration, high water-holding capacity.
• Loam (mix of sand, silt and clay): the best for agriculture, with good drainage and retention.

Properties important for irrigation

Texture, structure, porosity, infiltration rate, field capacity, wilting point, available moisture, pH and salinity, and fertility.

Suitability

• Sandy loam and loam: most crops; sprinkler/drip for sandy soil.
• Clay and clay loam: rice, but poor drainage; needs careful irrigation.
• Saline-alkaline soils need leaching and gypsum.

Depth of soil, topography and drainage also decide the choice of crop and irrigation method.

Frequency of irrigation is the number of days between two successive irrigations of a crop during its growing season (also called irrigation interval).

Calculation on the basis of soil moisture

1. Find the available moisture: $AM = \dfrac{FC-PWP}{100}\times\dfrac{\gamma_d}{\gamma_w}\times D_r$ (mm).
2. Find the readily available moisture (allowable depletion): $RAM = p\times AM$, with $p$ = 0.5-0.75.
3. Divide by the daily consumptive use $ET_c$:

Alternatively, $f=\dfrac{(FC-\theta_{min})}{100}\dfrac{\gamma_d}{\gamma_w}\dfrac{D_r}{ET_c}$ if a minimum allowed moisture $\theta_{min}$ is given.

Given: FC = 32%, PWP = 13%, $\gamma_d = 1.3$ g/cc, root zone depth $D_r = 130$ cm $= 1300$ mm, $ET_c = 14$ mm/day. The depth "130 m" is a misprint and is taken as 130 cm.

Step 1: Available moisture depth

Step 2: Irrigation interval

Answer: irrigate after about 22.9 days, i.e. every 22 days (rounded down so the crop does not suffer stress).

Given: FC = 25%, PWP = 13%, $\gamma_d = 1.5$ g/cc, root zone depth $D_r = 75$ cm $= 750$ mm, $ET_c = 10$ mm/day.

Step 1: Available moisture depth

Step 2: Irrigation interval

Answer: irrigate after about 13.5 days, i.e. every 13 days (rounded down so the crop does not suffer stress).

Given: FC = 29%, PWP = 11%, $\gamma_d = 1.3$, $D_r = 65$ cm $= 650$ mm, $ET_c = 12$ mm/day, minimum allowed moisture $= 25\%$, $\eta_a = 75\%$.

Step 1: Water depleted before the next irrigation (moisture falls from FC 29% to 25%)

Step 2: Irrigation interval

Step 3: Water required at the outlet of the field. The net depth to be replaced is 33.80 mm, so with $\eta_a = 75\%$

Answer: interval $\approx$ 2 days (2.8 days computed); water needed at field outlet = 45.07 mm per irrigation. (Available moisture FC - PWP = 18% is not fully used because the limit of 25% is given.)

Given: FC = 35%, PWP = 12%, porosity $n = 0.42$, $D_r = 1.2$ m $= 1200$ mm, $ET_c = 12$ mm/day, $\eta_a = 60\%$. Moisture must stay at or above 40% of the maximum available moisture, so 60% of the available moisture may be used.

Assumption: dry density is not given, so it is found from porosity with specific gravity of solids $G = 2.65$:

Step 1: Maximum available moisture

Step 2: Depth usable between irrigations (net depth)

Step 3: Irrigation interval

Step 4: Gross irrigation water per irrigation

Answer: interval $\approx$ 21 days; net depth 254.53 mm; water to be supplied = 424.21 mm (4242 m$^3$/ha).

Given: FC = 27%, PWP = 13%, depth = 80 cm = 800 mm, $\gamma_d = 14.72$ kN/m$^3$, $\gamma_w=9.81$ kN/m$^3$, $\eta_a = 80\%$, watercourse and field channel loss = 15% of outlet discharge.

1. Storage capacity (available moisture) in 80 cm of soil

(The total water held at field capacity is $0.27\times1.501\times800 = 324.11$ mm, of which 168.06 mm is available to the crop.)

2. Depth of water to be supplied when moisture falls to 18% (net depth, to bring the soil back to FC)

With $\eta_a = 80\%$, the depth at the field:

3. Water needed at the canal outlet. 15% of the outlet water is lost, so 85% reaches the field:

Answer: storage capacity = 168.06 mm; depth to be supplied = 135.05 mm at the field; water at the outlet = 158.88 mm.

Given: FC = 80%, PWP = 30%, $D_r = 70$ cm $= 700$ mm, $\gamma_d = 1.5$, $ET_c = 4$ mm/day, $\eta_a = 80\%$, RAM $= 70\%$ of available moisture.

Step 1: Available moisture

Step 2: Readily available moisture (net depth of irrigation)

Step 3: Gross depth of irrigation at the field

Step 4: Frequency of irrigation

Answer: depth of irrigation = 367.50 mm net (459.38 mm gross); frequency $\approx$ 91 days. (The data are very high for a field soil, but the method is as shown.)

Given: $ET_c = 5$ mm/day, AM = 220 mm/m, $D_r = 1.2$ m, only 50% of AM is readily available, $\eta_a = 65\%$, CCA = 450 ha, canal runs 6 AM to 5 PM = 11 h/day.

Step 1: Net depth per irrigation

Step 2: Irrigation interval

Step 3: Gross depth and volume for the whole CCA

Step 4: Outlet (canal) discharge. All the CCA must be irrigated within one interval of 26 days, with the canal flowing 11 h each day:

Answer: interval = 26 days; required (outlet) discharge $\approx$ 0.89 cumec (888 l/s).

Frequency of irrigation on the basis of soil moisture

1. Available moisture depth: $AM = \dfrac{FC-PWP}{100}\cdot\dfrac{\gamma_d}{\gamma_w}\cdot D_r$.
2. Readily available moisture (depth that may be depleted): $RAM = p\cdot AM$, with $p\approx 0.5$-$0.75$; or use $\dfrac{FC-\theta_{min}}{100}\cdot\dfrac{\gamma_d}{\gamma_w}D_r$.
3. Frequency:

where $ET_c$ is the daily consumptive use of the crop (mm/day).

Numerical

Given: discharge at head sluice $Q = 16$ cumec, duty at the field $= 80$ ha/cumec, loss in transit $= 20\%$.

Discharge reaching the field:

Area irrigated:

Answer: area that can be irrigated = 1024 ha.

Given: water held between FC and PWP = 140 mm/m, $D_r = 0.30$ m, $CWR = 5$ mm/day, area = 60 ha, allowable depletion = 35%, $\eta_a = 40\%$.

Available moisture in the root zone

(i) Allowable depletion depth between irrigations

(ii) Frequency of irrigation

(iii) Net application depth. The soil is refilled up to field capacity by the allowable depletion:

(iv) Volume of water required per irrigation (gross, i.e. at the field inlet, for 60 ha)

(Net volume stored in the root zone = 8,820 m$^3$.)

Answer: depletion depth = 14.7 mm; frequency $\approx$ 2 days (2.94); net depth = 14.7 mm; volume = 22,050 m$^3$ per irrigation. If the interval is rounded down to exactly 2 days, only $2\times5 = 10$ mm is used between irrigations, giving gross depth 25 mm and 15,000 m$^3$ per irrigation.

Given: root zone depth $D = 1.5$ m, present moisture $= 5\%$ (by dry weight), $\gamma_d = 1600$ kg/m$^3$ so $\gamma_d/\gamma_w=1.6$, water applied $= 700$ m$^3$, evaporation loss $= 10\%$, area $= 1$ ha $= 10^4$ m$^2$. The applied water brings the soil exactly to field capacity.

Step 1: Water actually stored in the soil

Step 2: Depth of water stored

Step 3: Increase in moisture content. Using $d = \Delta w\cdot\dfrac{\gamma_d}{\gamma_w}\cdot D$:

Step 4: Field capacity

Answer: field capacity $\approx$ 7.62%.

Method. The kor (critical) watering must be given to the crop within the kor period, so the outlet discharge for a crop is

with $A$ = area of the crop (m$^2$), $d$ = kor depth (m), $T$ = kor period (days).

Areas: wheat $= 0.30\times 10000 = 3000$ ha; rice $= 0.15\times 10000 = 1500$ ha.

Wheat ($T = 4\times7 = 28$ days, $d = 0.135$ m):

Rice ($T = 3\times7 = 21$ days, $d = 0.19$ m):

Wheat (rabi) and rice (kharif) are grown in different seasons, so the kor periods do not overlap and the outlet is designed for the larger discharge.

Answer: outlet discharge = 1.67 cumec (wheat governs; if both were to be watered together the total would be 3.24 cumec).

Given: CCA = 4000 ha, intensity = 70%, crop water need = 60 cm in $B=15$ days, effective rainfall = 15 cm, total losses = 15%.

Area irrigated: $A = 0.70\times 4000 = 2800$ ha.

Net depth from irrigation (delta at the field)

(i) Duty at the head of the field

(ii) Duty at the head of the channel. Losses of 15% reduce the area served per cumec:

(iii) Head discharge of the channel

Answer: (i) 288 ha/cumec; (ii) 244.8 ha/cumec; (iii) 11.44 cumec.

Given: CCA = 40,000 ha, canal loss = 25%, reservoir loss = 15%. Area of each crop = intensity x CCA (wheat 8000 ha, sugarcane 8000 ha, cotton 4000 ha, rice 6000 ha, vegetable 6000 ha).

Method. For each crop, discharge needed at the field $Q = A/D$; volume for the base period $V = Q\times B\times 86400$ m$^3$. The losses are taken as a fraction of the water entering that part of the system, so field volume is divided by $(1-\text{loss})$.

Total volume at the field: $V_f = 403.515\times10^6$ m$^3$

Allow canal loss (25%):

Allow reservoir loss (15%):

Answer: reservoir capacity $\approx$ 632.96 million m$^3$ (0.6330 Gm$^3$).

Given: reservoir loss = 15%, canal loss = 25%; areas, base periods and duties as in the table.

Method. For each crop, discharge needed at the field $Q = A/D$; volume for the base period $V = Q\times B\times 86400$ m$^3$. The losses are taken as a fraction of the water entering that part of the system, so field volume is divided by $(1-\text{loss})$.

Total volume at the field: $V_f = 482.297\times10^6$ m$^3$

Allow canal loss (25%):