---
title: "Frequency of irrigation; area irrigable"
subject: "Irrigation and Drainage Engineering"
chapter: 2
key: "semester_6/irrigation_and_drainage_engineering:frequency-and-area-irrigated-16-cumec"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/frequency-and-area-irrigated-16-cumec
site: BCT Engineering
---

# Frequency of irrigation; area irrigable

Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements

- Asked in: 2073 Magh, 8 marks
- Video: [Discharge Rate, Irrigatated Area and Depth of Irrigation](https://www.youtube.com/watch?v=CdJPxj-rKFI) — Amit Bhatnagar Classes
- 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/frequency-and-area-irrigated-16-cumec
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2

## Question

How do you calculate the frequency of irrigation on the basis of soil moisture? Water is released at the rate of 16 cumecs at the head sluice. If the duty at the field is 80 ha/cumecs and loss of water in transit 20%. Find the area of the land that can be irrigated.

## Short answer

Frequency = readily available moisture (mm) divided by daily ET (mm/day). With 16 cumec head discharge, 20% loss gives 12.80 cumec at field; at duty 80 ha/cumec, area = 1024 ha.

## Answer

#### Frequency of irrigation on the basis of soil moisture
1. Available moisture depth: $AM = \dfrac{FC-PWP}{100}\cdot\dfrac{\gamma_d}{\gamma_w}\cdot D_r$.
2. Readily available moisture (depth that may be depleted): $RAM = p\cdot AM$, with $p\approx 0.5$-$0.75$; or use $\dfrac{FC-\theta_{min}}{100}\cdot\dfrac{\gamma_d}{\gamma_w}D_r$.
3. Frequency:

$$
f = \frac{RAM}{ET_c}\ \text{days}
$$

where $ET_c$ is the daily consumptive use of the crop (mm/day).

#### Numerical
**Given:** discharge at head sluice $Q = 16$ cumec, duty at the field $= 80$ ha/cumec, loss in transit $= 20\%$.

Discharge reaching the field:

$$
Q_{field} = 16\times(1-0.20) = 12.80\ \text{cumec}
$$

Area irrigated:

$$
A = Q_{field}\times D = 12.80\times 80 = 1024\ \text{ha}
$$

**Answer: area that can be irrigated = 1024 ha.**

## Check yourself

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

1. Frequency based on soil moisture is
   A. AM x ET_c
   B. RAM/ET_c
   C. FC/PWP
   D. RAM x ET_c
   Answer: B. Days between irrigations. (Key point: Frequency formula)

2. Discharge reaching the field with 20% loss is
   A. 16 x 1.2
   B. 16 - 0.2
   C. 16 x 0.8 = 12.8 cumec
   D. 16/0.8 = 20
   Answer: C. 20% lost. (Key point: Discharge at field)

3. Area irrigated is
   A. head discharge x duty / 0.8
   B. 16/80
   C. 12.8/80
   D. discharge at field x duty at field
   Answer: D. 12.8 x 80 = 1024 ha. (Key point: Area)

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