---
title: "Interval and amount with 40% moisture limit"
subject: "Irrigation and Drainage Engineering"
chapter: 2
key: "semester_6/irrigation_and_drainage_engineering:interval-fc35-porosity-40pct"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/interval-fc35-porosity-40pct
site: BCT Engineering
---

# Interval and amount with 40% moisture limit

Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements

- Asked in: 2081 Chaitra, 8 marks
- Video: [Soil - Moisture Relationship | Irrigation Engineering | Irrigation depth,Irrigation interval](https://www.youtube.com/watch?v=pNjYV5scQOA) — Sudip Dhakal
- 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-fc35-porosity-40pct
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2

## Question

For the following data pertaining to a cultivated land, determine irrigation interval and amount of irrigation water needed at each irrigation so that the moisture content at any stage does not fall below 40% of the maximum available moistures. Data: field capacity of soil = 35%; permanent wilting point = 12%; porosity of soil = 0.42; depth of root-zone soil = 1.20 m; consumptive use = 12 mm per day; application efficiency = 60%.

## Short answer

Irrigation interval is 13.8 days (about 13 days). Net irrigation depth needed is 165.6 mm, and gross application at 60% efficiency is 276 mm (2760 m³/ha).

## Answer

**Given:** FC = 35%, PWP = 12%, porosity $n = 0.42$, $D_r = 1.2$ m $= 1200$ mm, $ET_c = 12$ mm/day, $\eta_a = 60\%$. Moisture must stay at or above 40% of the maximum available moisture, so 60% of the available moisture may be used.

**Interpretation:** no dry density is given, only porosity, so the moisture contents are taken **by volume**. This is the only consistent reading: water held at FC (0.35 of the soil volume) is less than the pore volume (0.42). If FC were by weight, any realistic dry density (for example $(1-0.42)\times 2.65 = 1.54$) would give a volumetric FC of $0.35\times1.54 = 0.54 > 0.42$, which is impossible.

**Step 1: Maximum available moisture**

$$
AM = \frac{35-12}{100}\times D_r = 0.23\times 1200 = 276\ \text{mm}
$$

**Step 2: Depth usable between irrigations (net depth)**

$$
d_{net} = 0.60\times 276 = 165.6\ \text{mm}
$$

**Step 3: Irrigation interval**

$$
I = \frac{165.6}{12} = 13.8\ \text{days}\approx 13\ \text{days}
$$

**Step 4: Gross irrigation water per irrigation**

$$
d_{gross} = \frac{d_{net}}{\eta_a} = \frac{165.6}{0.60} = 276\ \text{mm}
$$

**Answer: interval $\approx$ 13.8 days (irrigate every 13 days); net depth 165.6 mm; water to be supplied = 276 mm (2760 m$^3$/ha) per irrigation.**

## Check yourself

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

1. Dry density from porosity is
   A. n x G = 1.11
   B. 1 + n
   C. (1-n)G = 1.537
   D. G/n
   Answer: C. 0.58 x 2.65. (Key point: Dry density)

2. If moisture must not fall below 40% of max available, usable fraction is
   A. 40%
   B. 100%
   C. 80%
   D. 60%
   Answer: D. Only 40% remains. (Key point: Usable fraction)

3. Interval = 
   A. about 21 days
   B. about 35 days
   C. about 12 days
   D. about 42 days
   Answer: A. 254.53/12. (Key point: Interval)

4. Water supplied is net depth divided by
   A. multiplied by efficiency, 152 mm
   B. application efficiency, giving 424.21 mm
   C. consumptive use
   D. porosity
   Answer: B. 254.53/0.60. (Key point: Gross water)

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