---
title: "Sandy loam irrigation frequency and volume"
subject: "Irrigation and Drainage Engineering"
chapter: 2
key: "semester_6/irrigation_and_drainage_engineering:interval-sandy-loam-140mm"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/interval-sandy-loam-140mm
site: BCT Engineering
---

# Sandy loam irrigation frequency and volume

Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements

- Asked in: 2064 Kartik (old course), 8 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-sandy-loam-140mm
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2

## Question

A sandy loam soil holds water at 140 mm/m depth between FC and PWP. The crop has a root depth of 30 cm and CWR (Crop Water Requirement) equal to 5 mm/day. The cropping area is equal to 60 ha in which allowable depletion of soil is 35% and irrigation application efficiency equal to 40%. Determine: (i) Allowable depletion depth between irrigations (ii) Frequency of irrigation (iii) Net application depth of water (iv) Volume of water required.

## Short answer

Available moisture in root zone is 42 mm. Allowable depletion depth is 14.7 mm, giving irrigation frequency 2.94 days (≈2 days). Net application depth is 14.7 mm; gross depth 36.75 mm yields required volume 22,050 m³ per irrigation (or 15,000 m³ for a 2-day interval).

## Answer

**Given:** water held between FC and PWP = 140 mm/m, $D_r = 0.30$ m, $CWR = 5$ mm/day, area = 60 ha, allowable depletion = 35%, $\eta_a = 40\%$.

**Available moisture in the root zone**

$$
AM = 140\times 0.30 = 42\ \text{mm}
$$

**(i) Allowable depletion depth between irrigations**

$$
d = 0.35\times 42 = 14.7\ \text{mm}
$$

**(ii) Frequency of irrigation**

$$
f = \frac{14.7}{5} = 2.94\ \text{days}\approx 2\ \text{days}
$$

**(iii) Net application depth.** The soil is refilled up to field capacity by the allowable depletion:

$$
d_{net} = 14.7\ \text{mm}\quad\left(d_{gross} = \frac{14.7}{0.40} = 36.75\ \text{mm}\right)
$$

**(iv) Volume of water required per irrigation** (gross, i.e. at the field inlet, for 60 ha)

$$
V = \frac{36.75}{1000}\times 60\times 10^4 = 22,050\ \text{m}^3
$$

(Net volume stored in the root zone = 8,820 m$^3$.)

**Answer: depletion depth = 14.7 mm; frequency $\approx$ 2 days (2.94); net depth = 14.7 mm; volume = 22,050 m$^3$ per irrigation.**
If the interval is rounded down to exactly 2 days, only $2\times5 = 10$ mm is used between irrigations, giving gross depth 25 mm and 15,000 m$^3$ per irrigation.

## Check yourself

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

1. AM in 30 cm is
   A. 42 mm
   B. 140 mm
   C. 14.7 mm
   D. 420 mm
   Answer: A. 140 x 0.30. (Key point: Available moisture)

2. Allowable depletion depth is
   A. 42 mm
   B. 0.35 x 42 = 14.7 mm
   C. 0.65 x 42
   D. 14.7/0.4
   Answer: B. 35% of AM. (Key point: Allowable depletion)

3. Frequency is
   A. 8.4 days
   B. 29 days
   C. 14.7/5 = 2.94, about 2 days
   D. 0.34 days
   Answer: C. Depth/CWR. (Key point: Frequency)

4. Gross volume for 60 ha is
   A. 14.7 mm over 60 ha = 8,820 m3
   B. 36.75 m3
   C. 147,000 m3
   D. 36.75 mm over 60 ha = 22,050 m3
   Answer: D. Gross depth = net/0.40. (Key point: Volume)

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