Frequency of irrigation; area irrigable

Frequency of irrigation on the basis of soil moisture

1. Available moisture depth: $AM = \dfrac{FC-PWP}{100}\cdot\dfrac{\gamma_d}{\gamma_w}\cdot D_r$.
2. Readily available moisture (depth that may be depleted): $RAM = p\cdot AM$, with $p\approx 0.5$-$0.75$; or use $\dfrac{FC-\theta_{min}}{100}\cdot\dfrac{\gamma_d}{\gamma_w}D_r$.
3. Frequency:

$$f = \frac{RAM}{ET_c}\ \text{days}$$

where $ET_c$ is the daily consumptive use of the crop (mm/day).

Numerical

Given: discharge at head sluice $Q = 16$ cumec, duty at the field $= 80$ ha/cumec, loss in transit $= 20\%$.

Discharge reaching the field:

$$Q_{field} = 16\times(1-0.20) = 12.80\ \text{cumec}$$

Area irrigated:

$$A = Q_{field}\times D = 12.80\times 80 = 1024\ \text{ha}$$

Answer: area that can be irrigated = 1024 ha.