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Irrigation interval with AMC, RAM graph

Short answer

Available moisture capacity is 216 mm. With 75% RAM (162 mm), irrigation interval is 16.2 days (about 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 (IR) if the readily available moisture (RAM) is 75% of available soil moisture capacity and show AMC, RAM and irrigation interval on graph of available moisture and time.

Given: FC = 40%, PWP = 20%, γd=1.2\gamma_d = 1.2, Dr=900D_r = 900 mm, ETc=10ET_c = 10 mm/day, RAM = 75% of AMC.

Step 1: Available moisture capacity (AMC)

AMC=40−20100×1.2×900=216 mmAMC = \frac{40-20}{100}\times 1.2\times 900 = 216\ \text{mm}

Step 2: Readily available moisture (RAM)

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

Step 3: Irrigation interval

IR=RAMETc=16210=16.2 days≈16 daysIR = \frac{RAM}{ET_c} = \frac{162}{10} = 16.2\ \text{days}\approx 16\ \text{days}

Graph of available moisture against time

 Avail.
 moisture
 (mm)
 216 |= = = FC (AMC = 216 mm, full) = = = =
     |\
     | \        RAM = 162 mm used
     |  \       (75% of AMC)
  54 |- -\- - - - - - - - - - - - - - - -  limit of RAM
     |    \ slope = ET = 10 mm/day
     |
   0 |=====================================  PWP
     0                16.2          time (days)
        irrigate again at 16 days

Moisture falls from AMC (216 mm available, at FC) linearly at 10 mm/day. When 162 mm (RAM) has been used, i.e. after 16.2 days, the available moisture is 54 mm; irrigation is given to bring it back to FC. The remaining 25% of AMC (54 mm) is kept as reserve so the crop does not reach PWP.

Answer: irrigation interval = 16.2 days, say 16 days.

✓ Answer checked by Claude Opus 5.5 (high effort)

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