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.