Question Bank • QB_IDE_EXP_078 NAT • 2 Marks

Section 5: Irrigation and Drainage Engineering

Irrigation Water Conveyance and Application Methods

Subtopic: Irrigation efficiencies

Question Statement:

During an irrigation event, water depths stored at four equidistant sampling locations along a furrow run are $80\text{ mm}$, $100\text{ mm}$, $120\text{ mm}$, and $100\text{ mm}$. Calculate the water distribution efficiency ($\eta_d = 100 \left[1 - \frac{\bar{y}}{\bar{d}}\right]$) in percentage, where $\bar{d}$ is the numerical average depth and $\bar{y}$ is the average absolute deviation from $\bar{d}$.

Answer Key:

90

Step-by-Step Solution & Explanation

Average depth $\bar{d} = \frac{80 + 100 + 120 + 100}{4} = \frac{400}{4} = 100\text{ mm}$. Deviations: $|80 - 100| = 20$ $|100 - 100| = 0$ $|120 - 100| = 20$ $|100 - 100| = 0$ Average deviation $\bar{y} = \frac{20 + 0 + 20 + 0}{4} = \frac{40}{4} = 10\text{ mm}$. $\eta_d = 100 \left[ 1 - \frac{10}{100} \right] = 100 [1 - 0.10] = 90.0\%$
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