2. 2. 2. Tube Well Irrigation Design

Tube Well Irrigation Design

Tube wells are a modern and efficient method for extracting groundwater for irrigation purposes. Unlike dug wells, tube wells are narrow and deep, reaching aquifers located deeper underground. A tube well typically consists of a long, narrow pipe (casing) that is drilled into the ground to reach the water-bearing strata. A pump is used to lift the water to the surface for irrigation.

Components of a Tube Well

1. Casing Pipe:

   • A strong, corrosion-resistant pipe that lines the borehole to prevent it from collapsing and to keep out sand and other impurities. The diameter usually ranges from 100 mm to 300 mm.

2. Screen or Perforated Section:

   • A section of the casing pipe with slots or perforations that allow water to enter the well while keeping out sand and other particles.

3. Gravel Pack:

   • Placed around the screen, it filters out sand and fine particles, improving water quality and well efficiency.

4. Pump:

   • Usually a submersible pump is used, which is placed inside the casing pipe below the water table.

5. Headworks:

   • The surface structure, including the pump head, discharge pipe, and control valves.

Numerical Real-Life Example

Scenario: Designing a tube well for irrigation in an agricultural area with the following requirements and conditions:

• Daily water requirement for crops: 20,000 liters (20 m³)
• Depth to the water table: 50 meters
• Expected yield: 2 liters/second (lps)
• Casing diameter: 150 mm (6 inches)

Step 1: Calculate the Well Yield

The yield of the well is the amount of water it can supply per unit time. It is calculated using the formula:

$Q=\frac{\mathrm{V/}}{T}$

Where:

• $\mathrm{Q\; =\; Yield\; (lps)}$
• $\mathrm{V =\; Volume\; of\; water\; (liters)}$
• $\mathrm{T =\; Time\; (seconds)}$

Given:

• Yield ($Q$) = 2 lps

To check if the well can meet the daily requirement of 20,000 liters:

Daily Water Yield:

$Q\times T=V\mathrm{<br>}\mathrm{<br>}$

$\mathrm{There\; are\; 86,400\; seconds\; in\; a\; day\; (24\; hours\; \times \; 60\; minutes\; \times \; 60\; seconds).}$

$2\text{ lps}\times 86,400\text{ seconds}=172,800\text{ liters/day}$

$2 \text{ lps} \times 86,400 \text{ seconds} = 172,800 \text{ liters/day}$

The well can provide 172,800 liters/day, which far exceeds the daily requirement of 20,000 liters. Thus, the well can adequately supply water for irrigation.

Step 2: Design Parameters

1. Casing and Screen:

   • The casing diameter is chosen to be 150 mm (6 inches), providing a sufficient cross-sectional area for water flow. The screen length and slot size are designed based on the aquifer's properties to optimize water entry while minimizing sand intrusion.

2. Pump Selection:

   • A submersible pump is selected based on the well's depth and yield requirement. For a well depth of 50 meters and a required flow rate of 2 lps, a suitable pump with a power rating of approximately 1.5 to 2 horsepower (HP) is chosen.

3. Gravel Pack:

   • A gravel pack is placed around the screen section to filter out fine sediments and prevent clogging. The size of the gravel is selected based on the aquifer's particle size distribution.

4. Headworks:

   • Includes the installation of a pump head, discharge pipe, and control valves. The discharge pipe leads the water to storage or directly to the irrigation system.

Step 3: Cost Estimation

1. Drilling and Installation:

   • Drilling costs depend on the depth and soil conditions. For a 50-meter well with a casing diameter of 150 mm, the cost might be around  ₹ 20 per meter, totalling approximately   ₹ 1,000.

2. Casing and Screen:

   • The cost of the casing and screen material can vary, but for PVC or steel, it could be around  ₹ 300 to  ₹ 500.

3. Pump and Installation:

   • A suitable submersible pump and installation might cost between  ₹ 500 and  ₹ 1,000.

4. Gravel Pack and Headworks:

   • Additional costs for gravel pack and headworks could be around   ₹ 200 to  ₹ 300.

Conclusion

Tube wells are highly efficient for extracting groundwater, especially from deeper aquifers. They provide a reliable water supply for irrigation, even in areas with significant depth to the water table. The example demonstrates that with proper design, a tube well can meet the irrigation needs of a crop field. The design process ensures the well's efficiency, water quality, and sustainability while optimizing costs. 

2. 2. 1. Dug Well Irrigation

 

Dug Well Irrigation Design

Dug wells are one of the oldest methods of accessing groundwater for irrigation. They are typically shallow, with depths ranging from a few meters to about 15 meters. The design of a dug well involves several considerations, including the size of the well, the materials used for lining, the method of water extraction, and the potential yield.

Components of Dug Well Design

1. Diameter and Depth:

   • The diameter of a dug well typically ranges from 1 to 4 meters, depending on the required water volume and the water table's stability. The depth is determined based on the water table depth and the required water quantity.

2. Lining:

   • Lining is crucial to prevent the well walls from collapsing. Materials such as concrete, bricks, stone, or timber are commonly used.

3. Well Curb and Platform:

   • A curb or parapet is built around the well mouth to prevent contamination. A concrete platform is often constructed around the well to keep the surrounding area clean.

4. Water Extraction System:

   • A variety of water extraction methods can be used, including bucket and rope, hand pumps, or motorized pumps.

5. Water Storage and Distribution:

   • Storage tanks and pipelines may be used to distribute the water to fields.

Numerical Real-Life Example

Scenario: Designing a dug well for irrigation in a rural area with the following requirements and conditions:

• Daily water requirement for crops: 15,000 liters (15 m³)
• Depth to the water table: 5 meters
• Expected water yield: 0.5 liters/second (lps)
• Well diameter: 2 meters

Step 1: Calculate the Well Yield

The yield of the well is the amount of water it can supply per unit of time. It is calculated using the formula:

$Q=\frac{\mathrm{V/}}{T}$

Where:

• $Q$ = Yield (lps)
• $\mathrm{V =\; Volume\; of\; water\; (liters)}$
• $\mathrm{T=\; Time\; (seconds)}$

Given:

• Yield ($\mathrm{Q)\; =\; 0.5\; lps}$

The daily water requirement is 15,000 liters. To check if the well can meet this requirement:

Daily Water Yield:

$Q\times T=\mathrm{V<span\; class="base"><br></span>}$

Where:

• $\mathrm{Q =\; Yield\; (0.5\; lps)}$
• $\mathrm{T=\; Time\; (seconds\; per\; day)}$

There are 86,400 seconds in a day (24 hours × 60 minutes × 60 seconds).

$0.5\text{ lps}\times 86,400\text{ seconds}=43,200\text{ liters/day}$

$0.5 \text{ lps} \times 86,400 \text{ seconds} = 43,200 \text{ liters/day}$

0.5 lps×86,400 seconds=43,200 liters/day

The well can provide 43,200 liters/day, which exceeds the daily requirement of 15,000 liters. Thus, the well can adequately supply water for irrigation.

Step 2: Design Parameters

1. Well Diameter and Depth:

   • The chosen diameter is 2 meters, providing a sufficient water surface area.
   • The depth to the water table is 5 meters. The well should be dug deeper to provide a buffer above the water table to ensure adequate water supply.

2. Lining:

   • Concrete rings or bricks can be used for lining, ensuring stability and preventing collapse.

3. Water Extraction:

   • A submersible pump can be installed to efficiently draw water from the well. The pump's capacity should match the well's yield.

4. Storage and Distribution:

   • A storage tank with a capacity of at least 15,000 liters should be installed to meet daily requirements. Pipes can be laid to distribute water to the fields.

Step 3: Cost Estimation

• Excavation and Lining Costs:

  • These depend on local labor and material costs. For instance, if the cost is   ₹ 50 per cubic meter for excavation and lining, the cost for a well with a depth of 6 meters (adding 1 meter as a buffer) and a diameter of 2 meters would be:

  Volume = $\pi \times \left(\frac{2}{2}\right)^2 \times 6 = 18.85 \text{ m}^3$

  Cost = Volume x   ₹ 50 = ₹ 942.5

• Pump Installation:

  • A pump suitable for a 5-meter head and a flow rate of 0.5 lps might cost around ₹500.

• Storage and Distribution:

  • A tank and piping system might cost an additional ₹ 1,000.

Conclusion

Dug wells are a reliable source of water for irrigation, especially in areas with shallow water tables. This example demonstrates that with proper design and equipment, a dug well can meet the irrigation needs of a crop field. The design considerations ensure the well's efficiency and longevity while providing a sustainable water source.

2. 1. 2. Design of Village Ponds

 

Design of Village Ponds for Effective Groundwater Recharge

Objective: To design village ponds that maximize groundwater recharge and provide a sustainable water source for rural communities.

Key Design Features

1. Site Selection:

   • Choose low-lying areas with permeable soils (e.g., sandy or loamy soils) to enhance infiltration.
   • Ensure proximity to agricultural fields and residential areas for easy access.

2. Size and Depth:

   • Size: The pond should be large enough to capture significant runoff but balanced to the available land area.
   • Depth: A depth of 3-5 meters is ideal to store a substantial volume of water and reduce evaporation losses.

3. Catchment Area:

   • Ensure the pond has a well-defined catchment area with proper channels to direct runoff water into the pond.
   • Use contour bunding and check dams to slow down and capture runoff.

4. Inlet and Outlet Structures:

   • Inlet: Design the inlet to allow maximum water inflow during rains. It can be an open channel or a pipe with a screen to filter debris.
   • Outlet: Install an overflow structure to manage excess water and prevent erosion of embankments.

5. Embankments:

   • Construct sturdy embankments around the pond using locally available materials.
   • Vegetate embankments with grasses or shrubs to prevent erosion and enhance stability.

6. Percolation Pits and Trenches:

   • Create percolation pits or trenches around the pond to enhance groundwater recharge. These structures capture runoff and allow it to percolate slowly into the ground.

7. Desilting and Maintenance:

   • Regularly desilt the pond to maintain its storage capacity and enhance percolation.
   • Establish a community management system to oversee maintenance and operation.

Example of a Village Pond Design

Village Pond in Thiruvannamalai, Tamil Nadu:

1. Size and Depth: The pond measures 100 meters in length, 50 meters in width, and 3 meters in depth.
2. Catchment Area: The catchment area includes nearby fields and village roads, with contour bunds guiding runoff into the pond.
3. Inlet and Outlet: The pond has an inlet channel with a debris screen and an overflow outlet that directs excess water to nearby fields.
4. Embankments: Earthen embankments, 2 meters high, reinforced with grass cover.
5. Percolation Pits: Ten percolation pits, each 2 meters in diameter and 3 meters deep, are dug around the pond.
6. Maintenance: A village water committee is responsible for desilting and maintaining the pond and its structures.

Conclusion

By understanding and implementing traditional systems like the Eri in Tamil Nadu and designing effective village ponds, communities can significantly enhance their water security and agricultural productivity. Proper planning, community involvement, and regular maintenance are essential for the success of these water management systems.

2. 1. 1. Tank irrigation in Tamil Nadu [வீராணம் ]

 

Eri Systems in Tamil Nadu

Overview: The Eri system is a traditional water management practice in Tamil Nadu, primarily designed for irrigation and groundwater recharge. These systems consist of a series of tanks, channels, embankments, and sluice gates, and they have been integral to the region's agricultural success for centuries.

Components of the Eri System

1. Tanks:

   • Function: The main storage structure that collects and stores rainwater.
   • Example: The Veeranam Tank in Cuddalore district, which is one of the largest tanks in Tamil Nadu, with a capacity of 1.465 billion cubic feet.
   • Design: The tanks are typically designed to capture surface runoff during the monsoon season and have a large storage capacity to provide water during dry periods.

2. Channels:

   • Function: Channels distribute water from the tank to the agricultural fields.
   • Example: The network of channels in the Kollidam sub-basin, which distributes water from various Eris to the fields.
   • Design: Channels are often unlined earthen structures that guide water using gravity. They are strategically placed to cover the maximum agricultural area.

3. Embankments:

   • Function: Embankments form the boundary of the tank, holding the stored water.
   • Example: The earthen embankments around the Perumal Eri in Tirunelveli.
   • Design: Embankments are constructed using locally available materials such as soil and stones. They are regularly maintained to prevent breaches.

4. Sluice Gates:

• Function: Control the release of water from the tank to the channels.
• Example: The manually operated sluice gates in the Madurai district's Eris.
• Design: Sluice gates are typically made of wood or metal and are operated by villagers to regulate water flow based on the irrigation schedule.

2. UNIT II Irrigation Methods

Unit 2: Irrigation Methods

Tank Irrigation

1. Definition: Tank irrigation refers to the use of large, man-made or natural water bodies (tanks) to collect and store water for agricultural purposes.
2. Components: Tanks, channels, embankments, and sluice gates.
3. Advantages:
   • Suitable for undulating terrain.
   • Provides water during dry periods.
   • Helps in groundwater recharge.
4. Disadvantages:
   • High initial cost for construction.
   • Requires regular maintenance to prevent siltation.
5. Examples:
   • Eri systems in Tamil Nadu, India.
   • Village ponds in rural areas.

Well Irrigation

1. Definition: Well irrigation involves extracting groundwater through wells for irrigation purposes.
2. Types:
   • Dug wells.
   • Tube wells.
3. Advantages:
   • Reliable source of water.
   • Independent of surface water sources.
4. Disadvantages:
   • High cost of drilling and maintenance.
   • Risk of groundwater depletion.
5. Examples:
   • Use of bore wells in arid regions.

Irrigation Methods

1. Surface Irrigation:

   • Definition: Water is distributed over the surface of the soil by gravity.
   • Types:
     • Basin irrigation.
     • Border strip irrigation.
     • Furrow irrigation.
   • Advantages:
     • Low initial cost.
     • Simple technology.
   • Disadvantages:
     • Inefficient water use.
     • Soil erosion risk.

2. Sub-Surface Irrigation:

   • Definition: Water is applied below the soil surface, directly to the root zone.
   • Types:
     • Buried drip lines.
     • Sub-surface pipes.
   • Advantages:
     • Efficient water use.
     • Reduces evaporation losses.
   • Disadvantages:
     • High installation cost.
     • Maintenance challenges.

3. Micro Irrigation:

   • Definition: Water is applied directly to the plant root zone in small quantities.
   • Types:
     • Drip irrigation.
     • Sprinkler irrigation.
   • Advantages:
     • High water use efficiency.
     • Reduced weed growth.
   • Disadvantages:
     • High initial cost.
     • Potential clogging of emitters.

Design of Drip and Sprinkler Irrigation

1. Drip Irrigation Design:

   • Components: Mainline, sub-mainline, lateral lines, emitters.
   • Design Steps:
     • Determine water requirements.
     • Design layout.
     • Select appropriate emitters.
     • Calculate flow rates and pressure.
   • Advantages:
     • Water savings.
     • Suitable for all soil types.
   • Disadvantages:
     • High initial setup cost.
     • Maintenance required to prevent clogging.

2. Sprinkler Irrigation Design:

   • Components: Pump, mainline, lateral lines, sprinklers.
   • Design Steps:
     • Determine crop water needs.
     • Design system layout.
     • Select sprinkler heads.
     • Calculate pressure and flow rates.
   • Advantages:
     • Uniform water distribution.
     • Can be used on varied topographies.
   • Disadvantages:
     • Wind can affect distribution.
     • Higher energy requirements.

Ridge and Furrow Irrigation

1. Definition: Water is applied in furrows and crops are planted on the ridges.
2. Advantages:
   • Reduces water runoff.
   • Enhances soil moisture retention.
3. Disadvantages:
   • Labor-intensive.
   • Not suitable for all crops.
4. Examples:
   • Used in row crops like maize and cotton.

Irrigation Scheduling

1. Definition: Planning the timing and amount of water application to crops.
2. Methods:
   • Soil moisture monitoring.
   • Crop water stress indicators.
   • Climate data-based scheduling.
3. Advantages:
   • Optimizes water use.
   • Increases crop yield.
4. Disadvantages:
   • Requires accurate data and monitoring.

Water Distribution System

1. Definition: Infrastructure for conveying irrigation water from the source to the field.
2. Types:
   • Canal systems.
   • Pipe distribution networks.
3. Advantages:
   • Efficient water transport.
   • Reduces losses in transit.
4. Disadvantages:
   • High initial construction cost.
   • Maintenance challenges.

Irrigation Efficiencies

1. Field Efficiency: Ratio of water used by the crop to water delivered to the field.
2. Application Efficiency: Ratio of water stored in the root zone to water delivered by the irrigation system.
3. Conveyance Efficiency: Ratio of water delivered to the field to water taken from the source.
4. Overall Efficiency: Combined effect of all efficiencies in the irrigation system.
5. Improvement Methods:
   • Use of lined canals.
   • Drip and sprinkler systems.
   • Proper scheduling and management.

Summary

Understanding various irrigation methods, their design, advantages, and limitations is crucial for efficient water management in agriculture. Implementing suitable irrigation methods and systems, along with proper scheduling and maintenance, can significantly enhance water use efficiency and crop productivity.

Problems in Methods of Estimation of Evaporation

Here are numerical examples with solutions for each method of estimating evapotranspiration (ET):

1. Pan Evaporation Method

Example: Suppose a Class A evaporation pan is located near a crop field. The pan evaporation rate is measured to be 6 mm/day. The pan coefficient (Kp) for the area is 0.8.

Solution: The Pan Evaporation Method formula is used to estimate the reference evapotranspiration (ET₀) based on the evaporation rate from a Class A evaporation pan. The formula is:

$E{T}_0=K_p\times E_{\mathrm{p}}$

Where:

• $E{T}_0$ = Reference evapotranspiration (mm/day)
• $K_p$ = Pan coefficient (dimensionless)
• $E_p$ = Pan evaporation rate (mm/day)

Given data:

• Pan evaporation rate ($E_p$) = 6 mm/day
• Pan coefficient ($K_p$) = 0.8

Using the Pan Evaporation Method formula:

$E{T}_0=K_p\times E_p$                           ​

$E{T}_0=0.8\times 6$
$E{T}_0=4.8\text{ mm/day}$

So, the estimated reference evapotranspiration (ET₀) using the Pan Evaporation Method is 4.8 mm/day.

2. Penman Method for Estimating Evapotranspiration

The Penman method estimates reference evapotranspiration (ET₀) by combining the effects of temperature, humidity, wind speed, and solar radiation. The Penman equation is:

The Penman-Monteith method is an advanced and widely accepted method for estimating reference evapotranspiration (ET₀). It combines various climatic parameters to provide a precise estimate. The FAO-56 Penman-Monteith equation is:

$ET_0 = \frac{0.408 \Delta (R_n - G) + \gamma \frac{900}{T + 273} u_2 (e_s - e_a)}{\Delta + \gamma (1 + 0.34 u_2)}$

Where:

• $E{T}_{\mathrm{0<span\; class="katex">=\; Reference\; evapotranspiration\; (mm/day)</span>}}$
• $\Delta$ = Slope of the saturation vapor pressure curve (kPa/°C)
• $R_n$ = Net radiation at the crop surface (MJ/m²/day)
• $G$ = Soil heat flux density (MJ/m²/day)
• $\gamma$ = Psychrometric constant (kPa/°C)
• $T$ = Mean daily air temperature (°C)
• $u_2$ = Wind speed at 2 meters height (m/s)
• $e_{\mathrm{s\; =\; Saturation\; vapor\; pressure\; (kPa)}}$
• $e_a$ = Actual vapor pressure (kPa)

Numerical Example 1: Estimating ET₀ in a Warm Climate

Given Data:

• Net radiation ($R_n$) = 15 MJ/m²/day
• Soil heat flux density ($G$) = 0.0 MJ/m²/day (for daily estimates)
• Air density ($\rho_a$) = 1.225 kg/m³
• Specific heat of air ($c_p$) = 0.001013 MJ/kg·K
• Saturation vapor pressure ($e_s$) = 3.2 kPa
• Actual vapor pressure ($e_a$) = 2.0 kPa
• Wind speed ($u$) = 3.0 m/s
• Latent heat of vaporization ($\lambda$) = 2.45 MJ/kg
• Psychrometric constant ($\gamma$) = 0.066 kPa/°C

Assume the slope of the saturation vapor pressure curve ($\Delta$) is 0.20 kPa/°C.

Steps to Solve:

1. Calculate the numerator:

$\mathrm{\Delta }\cdot (R_n-G)=0.20\times (15-0)=3.0$

${\rho }_a\cdot c_p\cdot (e_s-e_a)\cdot \frac{u}{\lambda }=1.225\times 0.001013\times (3.2-2.0)\times \frac{\mathrm{3.0\; \{}}{\mathrm{2.45\}}}$

$= 1.225$

$1.224$

$=0.00268\times 1.224=0.0033\text{ (MJ/m²/day) }$

$\text{Numerator}=3.0+0.0033=3.0033\text{ MJ/m²/day }$

2. Calculate the denominator:

$\mathrm{\Delta }+\gamma \times (1+\frac{u}{\lambda })$

$=0.20+0.066\times (1+\frac{3.0}{2.45})$

$=$

$=$

$= 0$

3. Calculate ET₀:

$E{T}_0=\frac{\text{Numerator/}}{\text{Denominator}}$

$\frac{3.0033}{0.346}$

$\approx 8.67$

Numerical Example 2: Estimating ET₀ in a Cooler Climate

Given Data:

• Net radiation ($R_n$) = 10 MJ/m²/day
• Soil heat flux density ($G$) = 0.0 MJ/m²/day (for daily estimates)
• Air density ($\rho_a$) = 1.225 kg/m³
• Specific heat of air ($c_p$) = 0.001013 MJ/kg·K
• Saturation vapor pressure ($e_s$) = 1.8 kPa
• Actual vapor pressure ($e_a$) = 1.2 kPa
• Wind speed ($u$) = 1.5 m/s
• Latent heat of vaporization ($\lambda$) = 2.45 MJ/kg
• Psychrometric constant ($\gamma$) = 0.066 kPa/°C

Assume the slope of the saturation vapor pressure curve ($\Delta$) is 0.15 kPa/°C.

Steps to Solve:

1. Calculate the numerator:

$\Delta \cdot (R_n - G) = 0.15 \times (10 - 0) = 1.5$

$\rho_a \cdot c_p \cdot (e_s - e_a) \cdot \frac{u}{\lambda} = 1.225 \times 0.001013 \times (1.8 - 1.2) \times \frac{1.5}{2.45}$

  ​=1.225×0.001013×(1.8−1.2)×2.451.5​

$= 1.225 \times 0.001013 \times 0.6 \times 0.612$

$= 0.00074 \text{ (MJ/m²/day)}$

$\text{Numerator} = 1.5 + 0.00074 = 1.50074 \text{ MJ/m²/day}$

2. Calculate the denominator:

$\Delta + \gamma \times \left(1 + \frac{u}{\lambda}\right)$

$= 0.15 + 0.066 \times \left(1 + \frac{1.5}{2.45}\right)$

$= 0.15 + 0.066 \times \left(1 + 0.612\right)$

$= 0.15 + 0.066 \times 1.612 = 0.15 + 0.106$

$= 0.256 \text{ kPa/°C}$

3. Calculate ET₀:

$ET_0 = \frac{ \text{Numerator} }$

$=\frac{1.50074}{0.256}$

$\approx 5.86\text{ mm/day}$

2. Penman-Monteith Method

Example: Consider the following climatic data for a crop field:

• Net radiation (Rn) = 15 MJ/m²/day
• Soil heat flux density (G) = 0 MJ/m²/day
• Mean daily air temperature (T) = 25°C
• Wind speed at 2 m height (u2) = 3 m/s
• Saturation vapor pressure (es) = 3.2 kPa
• Actual vapor pressure (ea) = 2.1 kPa
• Psychrometric constant (γ) = 0.066 kPa/°C
• Slope of the vapor pressure curve (Δ) = 0.188 kPa/°C

Solution: To solve the given climatic data using the Penman-Monteith method for estimating reference evapotranspiration (ET₀), we use the FAO-56 Penman-Monteith equation:

$ET_0 = \frac{0.408 \Delta (R_n - G) + \gamma \frac{900}{T + 273} u_2 (e_s - e_a)}{\Delta + \gamma (1 + 0.34 u_2)}$

Given data:

• Net radiation ($R_n$) = 15 MJ/m²/day
• Soil heat flux density ($G$) = 0 MJ/m²/day
• Mean daily air temperature ($T$) = 25°C
• Wind speed at 2 m height ($u_2$) = 3 m/s
• Saturation vapor pressure ($e$) = 3.2 kPa
• Actual vapor pressure ($e$) = 2.1 kPa
• Psychrometric constant ($\gamma$) = 0.066 kPa/°C
• Slope of the vapor pressure curve ($\mathrm{\Delta }$) = 0.188 kPa/°C

Let's solve the equation step by step.

Step 1: Calculate the first term of the numerator

$0.408\mathrm{\Delta }(R_n-G)=0.408\times 0.188\times (15-0)$

$= 0.408 \times 0.188 \times 15$

$= 1.14936 \text{ mm/day}$

Step 2: Calculate the second term of the numerator

$\gamma \frac{900}{T+273}{u}_2(e_s-e_a) \; \; \; =0.066\times \frac{900}{25+273}\times 3\times (3.2-2.1)$

$= 0.066$

$= 0.066$

$= 0.066$

$= 0.657$

Step 3: Sum the terms of the numerator

$1.14936+0.657=$

Step 4: Calculate the denominator

$Δ+γ(1+0.34u2​)=0.188+0.066(1+0.34×3)$

$= 0.188 + 0.066 (1 + 1.02)$

$= 0.188 + 0.066$

$= 0.188 + 0.13332$

$=0.32132\text{ kPa/°C}$

Step 5: Calculate ET₀

$E{T}_0=\frac{\mathrm{1.80636/ }}{0.32132}$

$=5.62\text{ mm/day}$

So, the estimated reference evapotranspiration (ET₀) using the Penman-Monteith method is approximately 5.62 mm/day.

Numerical Example 1: Estimating ET₀ in a Warm Climate

Given Data:

• Mean daily temperature ($T$) = 25°C
• Net radiation ($R_n$) = 15 MJ/m²/day
• Soil heat flux density ($G$) = 0.0 MJ/m²/day
• Wind speed ($u_2$) = 3 m/s
• Saturation vapor pressure ($e$) = 3.2 kPa
• Actual vapor pressure ($e$) = 2.0 kPa
• Psychrometric constant ($\gamma$) = 0.066 kPa/°C
• Slope of the saturation vapor pressure curve ($\mathrm{\Delta }$) = 0.188 kPa/°C

Steps to Solve:

1. Calculate the first term of the numerator:

$0.408\mathrm{\Delta }(R_n-G)= \; \; 0.408\times 0.188\times (15-0) \; \;$$= 1.14864\text{ mm/day}$

2. Calculate the second term of the numerator:

$\gamma \frac{900}{T+273}{u}_2(e_s-e_a)=0.066\times \frac{900}{25+273}\times 3\times (3.2-2.0) \; \; \; \; \; \;$

$=$$0.066\times 3.02\times 3\times 1.2$

$=$$.715164\text{ mm/day}$

3. Sum the terms of the numerator:

$1.14864+0.715164=$

4. Calculate the denominator:

$\mathrm{\Delta }+\gamma (1+0.34u_2)=0.188+0.066(1+0.34\times 3)$

$=0.188+0.066(1+1.02)$

$= 0.188 + 0.066 (1 + 1.02)$

$=0.188+0.066(1+1.02)$ 

$=0.188+0.066\times 2.02$

$=0.32132\text{ kPa/°C}$

5. Calculate ET₀:

$E{T}_0=\frac{\mathrm{1.863804/}}{\mathrm{0.32132 \; }}$$\approx 5.80\text{ mm/day}$

Numerical Example 2: Estimating ET₀ in a Cooler Climate

Given Data:

• Mean daily temperature ($T$) = 15°C
• Net radiation ($R_n$) = 10 MJ/m²/day
• Soil heat flux density ($G$) = 0.0 MJ/m²/day
• Wind speed ($u_2$) = 2 m/s
• Saturation vapor pressure ($e_s$) = 1.7 kPa
• Actual vapor pressure ($e_a$) = 1.0 kPa
• Psychrometric constant ($\gamma$) = 0.066 kPa/°C
• Slope of the saturation vapor pressure curve ($\Delta$) = 0.143 kPa/°C

Steps to Solve:

1. Calculate the first term of the numerator:

$0.408 \Delta (R_n - G) = 0.408 \times 0.143 \times (10 - 0)$

$= 0.408 \times 0.143 \times 10$ $= 0.58464 \text{ mm/day}$

2. Calculate the second term of the numerator:

$\gamma \frac{900}{T+273}{u}_2(e_s-e_a)=0.066\times \frac{900}{15+273}\times 2\times (1.7-1.0)$$= 0.066 \times 3.125 \times 2 \times 0.7$ $= 0.066 \times 4.375$ $= 0.28875 \text{ mm/day}$

3. Sum the terms of the numerator:

$0.58464+0.28875=0.87339\text{ mm/day}$

4. Calculate the denominator:

$\mathrm{\Delta }+\gamma (1+0.34u_2)=0.143+0.066(1+0.34\times 2) \; <br><br><span\; class="base"><span\; class="mrel"> \; = </span><span\; class="mspace"></span></span><span\; class="base"><span\; class="strut"></span><span\; class="mord">0.143</span><span\; class="mspace"></span><span\; class="mbin">+</span><span\; class="mspace"></span></span><span\; class="base"><span\; class="strut"></span><span\; class="mord">0.066</span><span\; class="mspace"></span><span\; class="mbin">\times </span><span\; class="mspace"></span></span><span\; class="base"><span\; class="strut"></span><span\; class="mord">1.68</span></span>$

$\; =$$0.25388\text{ kPa/°C}$

5. Calculate ET₀:

$E{T}_0=\frac{0.87339}{0.25388}$$\approx 3.44\text{ mm/day}$

3. Blaney-Criddle Method for Estimating Evapotranspiration

The Blaney-Criddle method is a practical and widely used approach for estimating reference evapotranspiration (ET₀) based on temperature and daylight hours. The formula is given by:

$E{T}_0=p(0.46T+8.13)$

Where:

• $E{T}_{}$ = Reference evapotranspiration (mm/day)
• $p$ = Mean daily percentage of annual daytime hours
• $T$ = Mean daily temperature (°C)

Numerical Example 1: Estimating ET₀ for a Summer Month

Given Data:

• Mean daily temperature (T) = 25°C
• Mean daily percentage of annual daytime hours (p) = 0.30 (for a specific location and time of the year)

Steps to Solve:

1. Calculate the product of the mean daily temperature and the coefficient 0.46:

$0.46T = 0.46 \times 25 = 11.5$

2. Add 8.13 to the above result:

$0.46T+8.13=11.5+8.13=19.63$

3. Multiply the result by the mean daily percentage of annual daytime hours (p):

$ET_0 = p \left(0.46T + 8.13\right) = 0.30 \times 19.63 = 5.889$

4. Round to a practical precision:

$ET_0 \approx 5.89 \text{ mm/day}$

So, the estimated reference evapotranspiration (ET₀) for the summer month is 5.89 mm/day.

Numerical Example 2: Estimating ET₀ for a Winter Month

Given Data:

• Mean daily temperature (T) = 10°C
• Mean daily percentage of annual daytime hours (p) = 0.15 (for a specific location and time of the year)

Steps to Solve:

1. Calculate the product of the mean daily temperature and the coefficient 0.46:

$0.46T = 0.46 \times 10 = 4.6$

2. Add 8.13 to the above result:

$0.46T + 8.13 = 4.6 + 8.13 = 12.73$

3. Multiply the result by the mean daily percentage of annual daytime hours (p):

$E{T}_0=p(0.46T+8.13)=0.15\times 12.73=1.9095$

4. Round to a practical precision:

$ET_0 \approx 1.91 \text{ mm/day}$

So, the estimated reference evapotranspiration (ET₀) for the winter month is 1.91 mm/day.

4. Hargreaves Method

Example: Assume you have the following data for a crop field:

• Mean daily temperature (Tmean) = 22°C
• Maximum daily temperature (Tmax) = 30°C
• Minimum daily temperature (Tmin) = 15°C
• Extraterrestrial radiation (Ra) = 25 MJ/m²/day

Solution: To estimate the reference evapotranspiration (ET₀) using the Hargreaves method, we will use the following formula:

$ET_0 = 0.0023 \times (T_{mean} + 17.8) \times (T_{max} - T_{min})^{0.5} \times R_a$

Given data:

• Mean daily temperature ($T_{mean}$) = 22°C
• Maximum daily temperature ($T_{max}$) = 30°C
• Minimum daily temperature ($T_{min}$) = 15°C
• Extraterrestrial radiation ($R_a$) = 25 MJ/m²/day

Step-by-Step Solution:

1. Calculate the temperature difference and its square root:

$T_{max}-T_{min}=30-15=15$

$(T_{max} - T_{min})^{0.5} = \sqrt{15} \approx 3.87$

2. Apply the Hargreaves equation:

$ET_0 = 0.0023 \times (T_{mean} + 17.8) \times (T_{max} - T_{min})^{0.5} \times R_a$

ET0​=0.0023×(Tmean​+17.8)×(Tmax​−Tmin​)0.5×Ra​

$ET_0 = 0.0023 \times (22 + 17.8) \times 3.87 \times 25$

3. Simplify the equation step-by-step:

$ET_0 = 0.0023 \times 39.8 \times 3.87 \times 25$

$ET_0 = 0.0023 \times 39.8 \times 96.75$

$ET_0 = 0.0023 \times 3850.05$

$ET_0 \approx 8.86 \text{ mm/day}$

So, the estimated reference evapotranspiration (ET₀) for the given data using the Hargreaves method is approximately 8.86 mm/day.

Numerical Example 1: Estimating ET₀ in a Warm Climate

Given Data:

• Mean daily temperature ($T_{mean}$) = 25°C
• Maximum daily temperature ($T_{max}$) = 35°C
• Minimum daily temperature ($T_{min}$) = 15°C
• Extraterrestrial radiation ($R_a$) = 20 MJ/m²/day

Steps to Solve:

1. Calculate the temperature difference and its square root:

$T_{max}-T_{min}=35-15=20\mathrm{<br>}\mathrm{<br>}\mathrm{<br>}$$(T_{max} - T_{min})^{0.5} = \sqrt{20} \approx 4.47$

2. Apply the Hargreaves equation:

$E{T}_0=0.0023\times (T_{mean}+17.8)\times (T_{max}-T_{min}{)}^{0.5}\times R_{\mathrm{a}}$

$ET_0 = 0.0023 \times (25 + 17.8) \times 4.47 \times 20$

3. Simplify the equation step-by-step:

$ET_0 = 0.0023 \times 42.8 \times 4.47 \times 20$

$ET_0 = 0.0023 \times 42.8 \times 89.4$

$ET_0 = 0.0023 \times 3826.32$

$ET_0 \approx 8.80 \text{ mm/day}$

So, the estimated reference evapotranspiration (ET₀) for the warm climate is 8.80 mm/day.

Numerical Example 2: Estimating ET₀ in a Cooler Climate

Given Data:

• Mean daily temperature ($T_{mean}$) = 10°C
• Maximum daily temperature ($T_{max}$) = 15°C
• Minimum daily temperature ($T_{min}$) = 5°C
• Extraterrestrial radiation ($R_a$) = 15 MJ/m²/day

Steps to Solve:

1. Calculate the temperature difference and its square root:

$T_{max} - T_{min} = 15 - 5 = 10$,              $(T_{max} - T_{min})^{0.5} = \sqrt{10} \approx 3.16$

2. Apply the Hargreaves equation:

$ET_0 = 0.0023 \times (T_{mean} + 17.8) \times (T_{max} - T_{min})^{0.5} \times R_a$

ET0​=0.0023×(Tmean​+17.8)×(Tmax​−Tmin​)0.5×Ra​

$ET_0 = 0.0023 \times (10 + 17.8) \times 3.16 \times 15$

3. Simplify the equation step-by-step:

$ET_0 = 0.0023 \times 27.8 \times 3.16 \times 15$

$ET_0 = 0.0023 \times 27.8 \times 47.4$

$ET_0 = 0.0023 \times 1318.92$

$ET_0 \approx 3.03 \text{ mm/day}$

So, the estimated reference evapotranspiration (ET₀) for the cooler climate is 3.03 mm/day.

5. Thornthwaite Method

Thornthwaite Method for Estimating Evapotranspiration

The Thornthwaite method estimates potential evapotranspiration (ET₀) primarily based on temperature and day length. This method is suitable for use when only temperature data is available. The Thornthwaite equation is:

$ET_0 = 16 \left(\frac{10 \cdot T}{I}\right)^a$

Where:

• $ET_0$ = Potential evapotranspiration (mm/month)
• $T$ = Mean monthly temperature (°C)
• $I$ = Heat index, calculated as the sum of monthly heat indices for the year
• $a$ = Empirical exponent, calculated using the equation: $a = (6.75 \times 10^{-7})I^3 - (7.71 \times 10^{-5})I^2 + (1.792 \times 10^{-2})I + 0.49239$

Step-by-Step Calculation:

1. Calculate the Monthly Heat Index ($i$): $i = \left(\frac{T}{5}\right)^{1.514}$   

2. Calculate the Annual Heat Index ($I$): $I = \sum_{m=1}^{12} i_m$

3. Calculate the Empirical Exponent ($a$): $a = (6.75 \times 10^{-7})I^3 - (7.71 \times 10^{-5})I^2 + (1.792 \times 10^{-2})I + 0.49239$

4. Calculate the Monthly ET₀: $ET_0 = 16 \left(\frac{10 \cdot T}{I}\right)^a$
   

Numerical Example 1: Estimating ET₀ in a Warm Climate

Given Data:

• Mean monthly temperature ($T$) = 25°C for each month

Steps to Solve:

1. Calculate the Monthly Heat Index ($i$):

$i={\left(\frac{T}{5}\right)}^{1.514}$$i = \left(\frac{25}{5}\right)^{1.514}$ ${\mathrm{<br>}}^{}$$i \approx 11.77$

2. Calculate the Annual Heat Index ($I$):

Since the temperature is the same each month:

$I = 12 \times 11.77 = 141.24$

3. Calculate the Empirical Exponent ($a$):

$a = (6.75 \times 10^{-7})I^3 - (7.71 \times 10^{-5})I^2 + (1.792 \times 10^{-2})I + 0.49239$ $a = (6.75 \times 10^{-7})(141.24)^3 - (7.71 \times 10^{-5})(141.24)^2 + (1.792 \times 10^{-2})(141.24) + 0.49239$ $a \approx 0.0131$

4. Calculate the Monthly ET₀:

$E{T}_0=16{\left(\frac{10\cdot 25}{141.24}\right)}^{0.0131}$$\mathrm{ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; <span\; class="katex"><span\; aria-hidden="true"\; class="katex-html"><span\; class="base"><span\; class="mord"><span\; class="msupsub"><span\; class="vlist-t\; vlist-t2"><span\; class="vlist-r"><span\; class="vlist-s"></span></span><span\; class="vlist-r"><span\; class="vlist"></span></span></span></span></span><span\; class="mspace"></span><span\; class="mrel">\approx </span><span\; class="mspace"></span></span><span\; class="base"><span\; class="strut"></span><span\; class="mord">16</span><span\; class="mopen">(</span><span\; class="mord">1.0125</span><span\; class="mclose">)</span></span></span></span><span\; class="katex"><span\; aria-hidden="true"\; class="katex-html"><span\; class="base"><span\; class="mclose"><br></span></span></span></span>}$$ET_0 \approx 16.2 \text{ mm/month}$

So, the estimated potential evapotranspiration (ET₀) for the warm climate is 16.2 mm/month.

Numerical Example 2: Estimating ET₀ in a Cooler Climate

Given Data:

• Mean monthly temperature ($T$) = 10°C for each month

Steps to Solve:

1. Calculate the Monthly Heat Index ($i$):

$i = \left(\frac{T}{5}\right)^{1.514}$ $i = \left(\frac{10}{5}\right)^{1.514}$ $i = 2^{1.514}$ $i \approx 2.86$

2. Calculate the Annual Heat Index ($I$):

Since the temperature is the same each month:

$I = 12 \times 2.86 = 34.32$

3. Calculate the Empirical Exponent ($a$):

$a = (6.75 \times 10^{-7})I^3 - (7.71 \times 10^{-5})I^2 + (1.792 \times 10^{-2})I + 0.49239$ $a=(6.75\times 10^{-7})(34.32{)}^3-(7.71\times 10^{-5})(34.32{)}^2+(1.792\times 10^{-2})(34.32)+0.49239$$a \approx 0.498$

4. Calculate the Monthly ET₀:

$ET_0 = 16 \left(\frac{10 \cdot 10}{34.32}\right)^a$

$ET_0 = 16 \left(\frac{100}{34.32}\right)^{0.498}$

$ET_0 = 16 (2.91)^{0.498}$

$ET_0 \approx 16 (1.75)$ $ET_0 \approx 28.0 \text{ mm/month}$

So, the estimated potential evapotranspiration (ET₀) for the cooler climate is 28.0 mm/month.