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Irrigation interval: FC 40%, PWP 20%

Short answer

Available moisture capacity = (0.40–0.20) × 1.2 × 900 = 216 mm; readily available moisture = 0.75 × 216 = 162 mm; interval = 162 / 10 = 16.2 days, so irrigate every 16 days.

The field capacity of soil is 40%, permanent wilting point is 20%. Density of soil is 1.2 gm/cc, effective root depth is 90 cm, ET crop is 10 mm/day. Calculate the irrigation interval if the readily available moisture is 75% of available soil moisture capacity.

Given: FC = 40%, PWP = 20%, γd=1.2\gamma_d = 1.2 g/cc (so γd/γw=1.2\gamma_d/\gamma_w=1.2), root depth Dr=90D_r = 90 cm = 900 mm, ETc=10ET_c = 10 mm/day, RAM = 75% of available moisture.

Step 1: Available moisture capacity (AMC)

AMC=FC−PWP100×γdγw×Dr=0.20×1.2×900=216 mm\begin{aligned} AMC &= \frac{FC-PWP}{100}\times\frac{\gamma_d}{\gamma_w}\times D_r\\ &= 0.20\times 1.2\times 900 = 216\ \text{mm} \end{aligned}

Step 2: Readily available moisture (RAM)

RAM=0.75×216=162 mmRAM = 0.75\times 216 = 162\ \text{mm}

Step 3: Irrigation interval

I=RAMETc=16210=16.2 daysI = \frac{RAM}{ET_c} = \frac{162}{10} = 16.2\ \text{days}

Answer: irrigation interval = 16.2 days, say 16 days. (Round down so the crop never goes beyond RAM.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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