Storage capacity, depth and water at outlet

Given: FC = 27%, PWP = 13%, depth = 80 cm = 800 mm, $\gamma_d = 14.72$ kN/m$^3$, $\gamma_w=9.81$ kN/m$^3$, $\eta_a = 80\%$, watercourse and field channel loss = 15% of outlet discharge.

$$\frac{\gamma_d}{\gamma_w} = \frac{14.72}{9.81} = 1.501$$

1. Storage capacity (available moisture) in 80 cm of soil

$$d_{AM} = \frac{27-13}{100}\times 1.501\times 800 = 168.06\ \text{mm}$$

(The total water held at field capacity is $0.27\times1.501\times800 = 324.11$ mm, of which 168.06 mm is available to the crop.)

2. Depth of water to be supplied when moisture falls to 18% (net depth, to bring the soil back to FC)

$$d_{net} = \frac{27-18}{100}\times 1.501\times 800 = 108.04\ \text{mm}$$

With $\eta_a = 80\%$, the depth at the field:

$$d_{field} = \frac{108.04}{0.80} = 135.05\ \text{mm}$$

3. Water needed at the canal outlet. 15% of the outlet water is lost, so 85% reaches the field:

$$d_{outlet} = \frac{135.05}{0.85} = 158.88\ \text{mm}\ (1589\ \text{m}^3/\text{ha})$$

Answer: storage capacity = 168.06 mm; depth to be supplied = 135.05 mm at the field; water at the outlet = 158.88 mm.