---
title: "AMC, RAM, depth and interval; headworks depth"
subject: "Irrigation and Drainage Engineering"
chapter: 2
key: "semester_6/irrigation_and_drainage_engineering:moisture-depth-interval-etc9"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/moisture-depth-interval-etc9
site: BCT Engineering
---

# AMC, RAM, depth and interval; headworks depth

Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements

- Asked in: 2072 Magh, 5 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/moisture-depth-interval-etc9
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2

## Question

With the following data: FC = 35%, PWP = 12%, root depth = 70 cm, Soil density = 1.4 gm/cc, ETc = 9 mm/day, RAM = 70% AMC, application efficiency = 85%, conveyance loss and distribution loss 20% where the abbreviations have their usual meanings. Calculate: (i) Available moisture content (ii) Readily available moisture content (iii) Depth of irrigation at the outlet of the field (iv) Irrigation interval and (v) Depth of irrigation water required at the headwork.

## Short answer

AMC = 225.40 mm, RAM = 157.78 mm. Field depth = 185.62 mm (85% application efficiency), interval = 17.5 days (17 days), headwork depth = 232.03 mm (20% conveyance loss).

## Answer

**Given:** FC = 35%, PWP = 12%, $D_r = 70$ cm $= 700$ mm, $\gamma_d = 1.4$, $ET_c = 9$ mm/day, RAM = 70% of AMC, $\eta_a = 85\%$, conveyance + distribution loss = 20%.

**(i) Available moisture content (AMC)**

$$
AMC = \frac{35-12}{100}\times 1.4\times 700 = 225.40\ \text{mm}
$$

**(ii) Readily available moisture (RAM)**

$$
RAM = 0.70\times 225.40 = 157.78\ \text{mm}
$$

**(iii) Depth of irrigation at the outlet (field)** with $\eta_a=85\%$:

$$
d_{field} = \frac{157.78}{0.85} = 185.62\ \text{mm}
$$

**(iv) Irrigation interval**

$$
I = \frac{RAM}{ET_c} = \frac{157.78}{9} = 17.53\ \text{days}\approx 17\ \text{days}
$$

**(v) Depth of water required at the headworks** (20% lost in conveyance and distribution, so 80% reaches the field):

$$
d_{head} = \frac{185.62}{0.80} = 232.03\ \text{mm}
$$

**Answer: AMC = 225.40 mm; RAM = 157.78 mm; depth at field = 185.62 mm; interval = 17.5 days (17 days); depth at headwork = 232.03 mm.**

## Check yourself

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

1. AMC = 
   A. 157.78 mm
   B. 225.4 mm
   C. 322 mm
   D. 185.6 mm
   Answer: B. 0.23 x 1.4 x 700. (Key point: AMC)

2. Depth at field outlet is RAM divided by
   A. conveyance loss
   B. 0.7
   C. application efficiency 0.85, giving 185.62 mm
   D. 0.2
   Answer: C. 157.78/0.85. (Key point: Field depth)

3. Interval is
   A. 25 days
   B. 20.7 days
   C. 157.78 days
   D. RAM/ET_c = 17.5 days
   Answer: D. 157.78/9. (Key point: Interval)

4. With 20% conveyance and distribution loss, headwork depth is
   A. field depth/0.80 = 232.03 mm
   B. field depth x 0.8
   C. field depth x 1.2 = 222.7
   D. 185.62 + 20
   Answer: A. 80% reaches the field. (Key point: Headwork depth)

## Close variants set in other exams

- [AMC, RAM and interval: FC 80%, PWP 35%](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Amoisture-fc80-pwp35-root60) — Irrigation and Drainage Engineering, chapter 2, asked 2×

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