AMC, RAM, depth and interval; headworks depth

Given: FC = 35%, PWP = 12%, $D_r = 70$ cm $= 700$ mm, $\gamma_d = 1.4$, $ET_c = 9$ mm/day, RAM = 70% of AMC, $\eta_a = 85\%$, conveyance + distribution loss = 20%.

(i) Available moisture content (AMC)

$$AMC = \frac{35-12}{100}\times 1.4\times 700 = 225.40\ \text{mm}$$

(ii) Readily available moisture (RAM)

$$RAM = 0.70\times 225.40 = 157.78\ \text{mm}$$

(iii) Depth of irrigation at the outlet (field) with $\eta_a=85\%$:

$$d_{field} = \frac{157.78}{0.85} = 185.62\ \text{mm}$$

(iv) Irrigation interval

$$I = \frac{RAM}{ET_c} = \frac{157.78}{9} = 17.53\ \text{days}\approx 17\ \text{days}$$

(v) Depth of water required at the headworks (20% lost in conveyance and distribution, so 80% reaches the field):

$$d_{head} = \frac{185.62}{0.80} = 232.03\ \text{mm}$$

Answer: AMC = 225.40 mm; RAM = 157.78 mm; depth at field = 185.62 mm; interval = 17.5 days (17 days); depth at headwork = 232.03 mm.