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Irrigation interval and water at outlet

Short answer

Allowable depletion from 29% to 25%: depth = (29–25)/100 × 1.3 × 650 = 33.8 mm. Interval = 33.8 / 12 = 2.82 days (~2 days). Gross irrigation requirement at field outlet = 33.8 / 0.75 = 45.07 mm per irrigation.

After how many days will you irrigate your field in order to ensure healthy growth of crops if: Field capacity of soil = 29%; Density of soil = 1.3 gm/cc; Daily consumptive use = 12 mm; Permanent Wilting Point = 11%; Effective root zone depth = 65 cm. For healthy growth moisture content must not fall below 25%. How much irrigation water required at outlet of the field, if the application efficiency is 75%?

Given: FC = 29%, PWP = 11%, γd=1.3\gamma_d = 1.3 g/cc, Dr=65D_r = 65 cm =650= 650 mm, ETc=12ET_c = 12 mm/day, minimum allowed moisture =25%= 25\%, ηa=75%\eta_a = 75\%.

Step 1: Water depleted before the next irrigation (moisture falls from FC, 29%, to 25%)

d=29−25100×1.3×650=33.80 mmd = \frac{29-25}{100}\times 1.3\times 650 = 33.80\ \text{mm}

Step 2: Irrigation interval

I=dETc=33.8012=2.82 days, say 2 daysI = \frac{d}{ET_c} = \frac{33.80}{12} = 2.82\ \text{days, say 2 days}

Step 3: Water required at the outlet of the field. The net depth to be replaced is 33.80 mm, so with ηa=75%\eta_a = 75\%:

dgross=33.800.75=45.07 mm=450.7 m3/had_{gross} = \frac{33.80}{0.75} = 45.07\ \text{mm} = 450.7\ \text{m}^3/\text{ha}

Answer: interval ≈\approx 2 days (2.82 days computed); water needed at the field outlet = 45.07 mm per irrigation. (The available moisture FC - PWP = 18% is not fully used, because the moisture must not fall below 25%.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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