---
title: "AMC, RAM and interval: FC 80%, PWP 35%"
subject: "Irrigation and Drainage Engineering"
chapter: 2
key: "semester_6/irrigation_and_drainage_engineering:moisture-fc80-pwp35-root60"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/moisture-fc80-pwp35-root60
site: BCT Engineering
---

# AMC, RAM and interval: FC 80%, PWP 35%

Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements

- Asked in: 2064 Jestha (old course), 6 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-fc80-pwp35-root60
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2

## Question

With following data: FC = 80%, PWP = 35%, root depth = 60 cm, soil density = 1.5 gm/cc, ETc = 5 mm/day, application efficiency = 80% and RAM = 70% AMC, where the abbreviations have usual meanings. (i) Calculate available moisture contents (ii) Calculate readily available moisture contents (iii) Calculate depth of irrigation at the outlet of the field (iv) Calculate irrigation interval.

## Short answer

Available moisture content (AMC) in root zone is 405 mm. Readily available moisture (RAM) is 283.5 mm. Depth of irrigation at field outlet is 354.4 mm, and irrigation interval is 56.7 days.

## Answer

**Given:** FC = 80%, PWP = 35%, $D_r = 60$ cm $= 600$ mm, $\gamma_d = 1.5$, $ET_c = 5$ mm/day, $\eta_a = 80\%$, RAM = 70% of AMC.

**(i) Available moisture content**

$$
AMC = \frac{80-35}{100}\times 1.5\times 600 = 405.00\ \text{mm}
$$

**(ii) Readily available moisture**

$$
RAM = 0.70\times 405.00 = 283.50\ \text{mm}
$$

**(iii) Depth of irrigation at the outlet of the field**

$$
d = \frac{RAM}{\eta_a} = \frac{283.50}{0.80} = 354.375\ \text{mm}
$$

**(iv) Irrigation interval**

$$
I = \frac{RAM}{ET_c} = \frac{283.50}{5} = 56.7\ \text{days}\approx 56\ \text{days}
$$

**Answer: AMC = 405.00 mm; RAM = 283.50 mm; depth at outlet = 354.38 mm; interval = 56.7 days (about 56 days).**

## Check yourself

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

1. In the answer, available moisture content (AMC) is found from which expression?
   A. (FC - PWP)/100 x dry density x root depth
   B. (FC + PWP)/100 x dry density x root depth
   C. (FC - PWP)/100 x root depth only, ignoring density
   D. FC/100 x dry density x root depth
   Answer: A. AMC is the water between field capacity and wilting point, converted to a depth using density and root depth: 0.45 x 1.5 x 600 = 405 mm. (Key point: Available moisture)

2. RAM here is 70% of AMC. What is RAM?
   A. 405 mm
   B. 578.6 mm
   C. 283.5 mm
   D. 170 mm
   Answer: C. RAM = 0.70 x 405 = 283.5 mm; it is the part of AMC that plants extract easily. (Key point: Readily available moisture)

3. How is the depth of irrigation at the outlet of the field obtained, and what is it?
   A. RAM x application efficiency = 226.8 mm
   B. RAM / application efficiency = 354.4 mm
   C. AMC / application efficiency = 506 mm
   D. RAM / ETc = 56.7 mm
   Answer: B. Losses in application mean more water must be delivered than is stored, so d = RAM/eta_a = 283.5/0.8 = 354.4 mm. (Key point: Depth at outlet)

4. How is the irrigation interval calculated?
   A. AMC divided by ETc = 81 days
   B. RAM x ETc = 1417 days
   C. Outlet depth divided by efficiency = 443 days
   D. RAM divided by ETc = 283.5/5 = 56.7 days
   Answer: D. The interval is the time the readily available water lasts at the crop use rate: 283.5 mm / 5 mm/day. (Key point: Irrigation interval)

## Close variants set in other exams

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