---
title: "Irrigation interval and water at outlet"
subject: "Irrigation and Drainage Engineering"
chapter: 2
key: "semester_6/irrigation_and_drainage_engineering:interval-fc29-pwp11-efficiency75"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/interval-fc29-pwp11-efficiency75
site: BCT Engineering
---

# Irrigation interval and water at outlet

Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements

- Asked in: 2075 Bhadra, 8 marks
- Video: [After how many days will you supply water to the soil | Frequency | Soil Moisture Numerical Problem](https://www.youtube.com/watch?v=nsM4OqiI51E) — SBK Concept
- 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-fc29-pwp11-efficiency75
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2

## Question

After how many days will you irrigate your field in order to ensure healthy growth of crops if: Field capacity of soil = 29%; Density of soil = 1.3 gm/cc; Daily consumptive use = 12 mm; Permanent Wilting Point = 11%; Effective root zone depth = 65 cm. For healthy growth moisture content must not fall below 25%. How much irrigation water required at outlet of the field, if the application efficiency is 75%?

## Short answer

Allowable depletion from 29% to 25%: depth = (29–25)/100 × 1.3 × 650 = 33.8 mm. Interval = 33.8 / 12 = 2.82 days (~2 days). Gross irrigation requirement at field outlet = 33.8 / 0.75 = 45.07 mm per irrigation.

## Answer

**Given:** FC = 29%, PWP = 11%, $\gamma_d = 1.3$ g/cc, $D_r = 65$ cm $= 650$ mm, $ET_c = 12$ mm/day, minimum allowed moisture $= 25\%$, $\eta_a = 75\%$.

**Step 1: Water depleted before the next irrigation** (moisture falls from FC, 29%, to 25%)

$$
d = \frac{29-25}{100}\times 1.3\times 650 = 33.80\ \text{mm}
$$

**Step 2: Irrigation interval**

$$
I = \frac{d}{ET_c} = \frac{33.80}{12} = 2.82\ \text{days, say 2 days}
$$

**Step 3: Water required at the outlet of the field.** The net depth to be replaced is 33.80 mm, so with $\eta_a = 75\%$:

$$
d_{gross} = \frac{33.80}{0.75} = 45.07\ \text{mm} = 450.7\ \text{m}^3/\text{ha}
$$

**Answer: interval $\approx$ 2 days (2.82 days computed); water needed at the field outlet = 45.07 mm per irrigation.**
(The available moisture FC - PWP = 18% is not fully used, because the moisture must not fall below 25%.)

## Check yourself

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

1. Water depleted from FC 29% to 25% is
   A. 58.5 mm
   B. 67.6 mm
   C. 33.8 mm
   D. 45.07 mm
   Answer: C. 0.04 x 1.3 x 650. (Key point: Depleted depth)

2. Interval = 
   A. 5 days
   B. 12 days
   C. 28 days
   D. 2.82 days, about 2 days
   Answer: D. 33.8/12. (Key point: Interval)

3. Gross depth at outlet is
   A. 45.07 mm
   B. 25.35 mm
   C. 33.8 mm
   D. 60 mm
   Answer: A. 33.8/0.75. (Key point: Gross depth)

4. Why is FC - PWP (18%) not fully used?
   A. Because PWP is 25%
   B. Moisture must not fall below 25% for healthy growth
   C. Because efficiency is 75%
   D. Because density is 1.3
   Answer: B. The limit is set at 25%. (Key point: Why not full AM)

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