Question

A pump delivering 80 L/s irrigates 2 ha in 10 h. Soil moisture increases from 18% to 30% in a 50 cm root zone. Soil bulk density is 1500 kg/m\(^3\). Compute water application efficiency (round off to two decimals).

Hint:

Efficiency = (water stored in root zone) / (water applied).

Answer 1

Correct Answer: 62.25

Volume of water applied:
\[ Q = 80\ \text{L/s} = 0.08\ \text{m}^3/\text{s} \] \[ V = Q (10 \times 3600) = 0.08 \times 36000 = 2880\ \text{m}^3 \] Water stored in root zone:
\[ \Delta \theta = 0.30 - 0.18 = 0.12 \] Root zone depth = 0.5 m
Area = 2 ha = \(20000\ \text{m}^2\) Mass of soil:
\[ M_s = \rho_b A Z = 1500(20000)(0.5) \] \[ M_s = 15\times10^9\ \text{kg} \] Water stored:
\[ W_s = M_s (0.12) = 1.8\times10^9\ \text{kg} \] Convert to volume:
\[ V_s = \frac{W_s}{1000} = 1800\ \text{m}^3 \] Application efficiency:
\[ \eta = \frac{1800}{2880} = 0.625 \] \[ \eta = 62.5% \]