---
title: "Irrigation interval with AMC, RAM graph"
subject: "Irrigation and Drainage Engineering"
chapter: 2
key: "semester_6/irrigation_and_drainage_engineering:interval-ram75-graph"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/interval-ram75-graph
site: BCT Engineering
---

# Irrigation interval with AMC, RAM graph

Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements

- Asked in: 2075 Baisakh, 5+2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/interval-ram75-graph
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2

## Question

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.

## Short answer

Available moisture capacity is 216 mm. With 75% RAM (162 mm), irrigation interval is 16.2 days (about 16 days).

## Answer

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

**Step 1: Available moisture capacity (AMC)**

$$
AMC = \frac{40-20}{100}\times 1.2\times 900 = 216\ \text{mm}
$$

**Step 2: Readily available moisture (RAM)**

$$
RAM = 0.75\times 216 = 162\ \text{mm}
$$

**Step 3: Irrigation interval**

$$
IR = \frac{RAM}{ET_c} = \frac{162}{10} = 16.2\ \text{days}\approx 16\ \text{days}
$$

**Graph of available moisture against time**

```text
 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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. AMC = 
   A. 162 mm
   B. 54 mm
   C. 216 mm
   D. 180 mm
   Answer: C. 0.2 x 1.2 x 900. (Key point: AMC)

2. On the graph, what remains when RAM has been used?
   A. 0
   B. 216 mm
   C. 162 mm
   D. 54 mm reserve (25% of AMC)
   Answer: D. 216-162. (Key point: Reserve)

3. The slope of the depletion line is
   A. ET_c = 10 mm/day
   B. RAM
   C. 1.2
   D. 16.2 mm/day
   Answer: A. Moisture falls linearly. (Key point: Slope)

4. The interval is
   A. 21.6 days
   B. 16.2 days, about 16
   C. 5.4 days
   D. 162 days
   Answer: B. 162/10. (Key point: Interval)

## Close variants set in other exams

- [Irrigation interval: FC 40%, PWP 20%](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Ainterval-fc40-pwp20-root90) — Irrigation and Drainage Engineering, chapter 2, asked 4×
- [Irrigation interval: FC 60%, PWP 25%](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Ainterval-fc60-pwp25-ram85) — Irrigation and Drainage Engineering, chapter 2, asked 4×

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