Question

A sprinkler irrigation system with an irrigation efficiency of 70% is used to irrigate 16 ha of maize crop. The crop evapotranspiration of 6 mm day$^{-1}$ is used for estimating the irrigation depth. If the irrigation system is operated 20 hours per day for 10 days, the system capacity in L s$^{-1}$ is _____. \(\textit{[Round off to two decimal places]}\)

Hint:

To calculate the system capacity, divide the irrigation volume by the total operation time, considering the area and irrigation efficiency.

Answer 1

Correct Answer: 18.9

The irrigation depth \( D_{\text{irrigation}} \) can be calculated using the formula: \[ D_{\text{irrigation}} = \frac{ET \times A}{\eta} \] where:
- \( ET = 6 \, \text{mm/day} = 0.006 \, \text{m/day} \) is the evapotranspiration,
- \( A = 16 \, \text{ha} = 160000 \, \text{m}^2 \) is the area to be irrigated,
- \( \eta = 0.70 \) is the irrigation efficiency.
The irrigation depth over 10 days is: \[ D_{\text{irrigation}} = \frac{0.006 \times 160000}{0.70} = 1371.43 \, \text{m}^3. \] The system capacity is given by: \[ Q = \frac{D_{\text{irrigation}}}{\text{time}} = \frac{1371.43}{10 \times 20 \times 3600} = 19.1 \, \text{L/s}. \] Thus, the system capacity is approximately \( \boxed{19.10} \, \text{L/s} \) (rounded to two decimal places).