Water conveyance and application efficiencies

Given: $Q_{diverted} = 150$ l/s, $Q_{field} = 110$ l/s, area $= 2.2$ ha $= 22000$ m$^2$, time $= 8$ h, root zone $= 1.5$ m, runoff $= 445$ m$^3$, penetration depth 1.5 m (head) to 1.1 m (tail), available moisture = 200 mm/m, irrigation started at 50% depletion of available moisture.

1. Water conveyance efficiency

$$\eta_c = \frac{110}{150}\times 100 = 73.33\%$$

2. Water applied to the field

$$W_f = 0.110\times 8\times 3600 = 3168\ \text{m}^3$$

Water stored in the root zone. The penetration depth (1.5 m at the head, 1.1 m at the tail) does not exceed the root depth of 1.5 m, so deep percolation is nil (assumed). Hence

$$W_s = W_f - \text{runoff} = 3168 - 445 = 2723\ \text{m}^3$$

Water application efficiency

$$\eta_a = \frac{W_s}{W_f}\times 100 = \frac{2723}{3168}\times 100 = 85.95\%$$

3. Water storage efficiency. Water needed to bring the root zone back to field capacity (50% of available moisture was used):

$$W_n = 0.5\times 0.200\times 1.5\times 22000 = 3300\ \text{m}^3$$ $$\eta_s = \frac{W_s}{W_n}\times 100 = \frac{2723}{3300}\times 100 = 82.52\%$$

4. Water distribution efficiency. Depth varies linearly between 1.5 m and 1.1 m, so mean depth $d = (1.5+1.1)/2 = 1.3$ m and the average numerical deviation $y = 0.2/2 = 0.1$ m.

$$\eta_d = \left(1-\frac{y}{d}\right)\times 100 = \left(1-\frac{0.1}{1.3}\right)\times 100 = 92.31\%$$

Answer: $\eta_c = 73.33\%$, $\eta_a = 85.95\%$, $\eta_s = 82.52\%$, $\eta_d = 92.31\%$.