written 19 months ago by
Chandan15
• 300

•
modified 19 months ago

(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,
$$
\begin{array}{l}
n_{d}=\left(1\frac{d}{D}\right) \times 100 \\
D=\frac{1.5+1.1}{2}=1.3
\end{array}
$$
$$
\begin{array}{l}
d=\operatorname{area~of~hatched~portion~} \\
d=\left(\frac{1}{2} \times \frac{1}{2} \times 0.2\right) \times 2=0.1 \\
{ n_{d}}=\left(1\frac{0.1}{1.3}\right) \times 100=92.31 \%
\end{array}
$$