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Problem on Irrigation Efficiency

A Discharge of 150 lit/sec was delieved from canal and 110 Lit /sec reached the field.In 8 hr, 2.2 Ha area was irrigated.

• Runoff Loss in the field = 445m³
• Depth of water Penetration varies Linearly from 1.5m at the head End of the field to 1.1m at the tail end.
• Effective Root Zone Depth = 1.5m Available moisture holding capacity of soil is 200mm per m depth of soil.
• Note that irrigation was started at a moisture extracted level of 50%.

Then, Find a) Nc b) Na c) Ns d) Nd

(a) $n_{c}=\frac{\text { water delivered to the Field }}{\text { (water Extracted From the source) }} \times 100$ $$n_{c}=\frac{110 lit/ \mathrm{s}}{150 lit/ \mathrm{s}} \times 100=73.33 \%$$ (B) Application efficiency, $n_{a}=\frac{ \text { water actually stored in the Root }}{\text { water delivered to the Field }} \times 100$ - Water actually stored in the Root zone \begin{aligned} &=110lit/s-\frac{445 \times 10^{3}}{8 \times 3600} \mathrm{lit/s} \ &=94.548 \mathrm{lit/s} \ n_{a} &=\frac{94.548}{110} \times 100=85.95 \% \end{aligned} (c) Storage Efficiency, $n_{S}=\frac{\text { water actually stored in the Root } 20 x e}{\text { water Required to be stored in the Root zone}} \times 100$ - Water Required to be stored in the Root zone - \begin{aligned} &=\frac{\left(0.5 \times 1.5 \times 0.2 \times 2.2 \times 10^{4}\right) \times 10^{3}}{8 \times 3600}\\ &=114.58 \mathrm{l} / \mathrm{s} \\ n_{s}=\frac{94.548}{114.58} \times 100=82.52 \% \end{aligned} (d) Water distribution efficiency, $$\begin{array}{l} n_{d}=\left(1-\frac{d}{D}\right) \times 100 \\ D=\frac{1.5+1.1}{2}=1.3 \end{array}$$ $$\begin{array}{l} d=\operatorname{area~of~hatched~portion~} \\ d=\left(\frac{1}{2} \times \frac{1}{2} \times 0.2\right) \times 2=0.1 \\ { n_{d}}=\left(1-\frac{0.1}{1.3}\right) \times 100=92.31 \% \end{array}$$