Question Bank • QB_IDE_ADV_045 NAT • 2 Marks

Section 5: Irrigation and Drainage Engineering

Soil-Water-Plant Relationship

Subtopic: Consumptive use and evapotranspiration

Question Statement:

Daily reference evapotranspiration ($ET_0$) is estimated using the Hargreaves-Samani temperature-radiation equation: $ET_0 = 0.0023 \cdot R_a \cdot (T_{\text{mean}} + 17.8) \cdot (T_{\text{max}} - T_{\text{min}})^{0.5}$, where $R_a$ is extraterrestrial solar radiation expressed in equivalent water evaporation depth of $15.0\text{ mm/day}$. For a sunny summer day with recorded temperatures of $T_{\text{max}} = 34.0^\circ\text{C}$, $T_{\text{min}} = 18.0^\circ\text{C}$, and $T_{\text{mean}} = 26.0^\circ\text{C}$, the estimated reference evapotranspiration $ET_0$ in $\text{mm/day}$ is ________ (round off to two decimal places).

Answer Key:

6.04

Step-by-Step Solution & Explanation

**Method 1: Hargreaves-Samani Direct Evaluation** 1. Diurnal temperature range parameter: $\Delta T = T_{\text{max}} - T_{\text{min}} = 34.0 - 18.0 = 16.0^\circ\text{C}$ $(\Delta T)^{0.5} = \sqrt{16.0} = 4.0$ 2. Mean temperature factor: $T_{\text{mean}} + 17.8 = 26.0 + 17.8 = 43.8$ 3. Product of components: $ET_0 = 0.0023 \times R_a \times (T_{\text{mean}} + 17.8) \times (\Delta T)^{0.5}$ $ET_0 = 0.0023 \times 15.0 \times 43.8 \times 4.0$ $0.0023 \times 15.0 = 0.0345$ $43.8 \times 4.0 = 175.2$ $ET_0 = 0.0345 \times 175.2 = 6.0444\text{ mm/day} \approx 6.04\text{ mm/day}$ **Method 2: Grouped Coefficient Check** $\text{Factor} = 0.0023 \times 4.0 = 0.0092$ $ET_0 = 0.0092 \times 15.0 \times 43.8 = 0.138 \times 43.8 = 6.0444\text{ mm/day} \approx 6.04\text{ mm/day}$
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