---
title: "Depth and frequency of irrigation"
subject: "Irrigation and Drainage Engineering"
chapter: 2
key: "semester_6/irrigation_and_drainage_engineering:depth-frequency-fc80-pwp30"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/depth-frequency-fc80-pwp30
site: BCT Engineering
---

# Depth and frequency of irrigation

Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements

- Asked in: 2076 Bhadra, 6 marks
- Video: [Depth & Frequency of Irrigation | Irrigation Engineering | Available Moisture | Consumptive Use](https://www.youtube.com/watch?v=cw2gkeVc72U) — Civil Engineering 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/depth-frequency-fc80-pwp30
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2

## Question

Following data are given below, calculate the depth and frequency of irrigation. Field capacity = 80%, Permanent Wilting Point (PWP) = 30%, Root depth = 70 cm, soil density = 1.5 gm/cc, Consumptive use = 4 mm/day, Application efficiency = 80%, Readily available moisture (RAM) = 70%.

## Short answer

Readily available moisture (net depth) is 367.50 mm, gross depth is 459.38 mm. Irrigation frequency is about 92 days (91.9 days).

## Answer

**Given:** FC = 80%, PWP = 30%, $D_r = 70$ cm $= 700$ mm, $\gamma_d = 1.5$, $ET_c = 4$ mm/day, $\eta_a = 80\%$, RAM $= 70\%$ of available moisture.

**Step 1: Available moisture**

$$
AM = \frac{80-30}{100}\times 1.5\times 700 = 525.00\ \text{mm}
$$

**Step 2: Readily available moisture (net depth of irrigation)**

$$
RAM = 0.70\times 525.00 = 367.50\ \text{mm}
$$

**Step 3: Gross depth of irrigation at the field**

$$
d_{gross} = \frac{RAM}{\eta_a} = \frac{367.50}{0.80} = 459.38\ \text{mm}
$$

**Step 4: Frequency of irrigation**

$$
f = \frac{RAM}{ET_c} = \frac{367.50}{4} = 91.88\ \text{days}
$$

**Answer: depth of irrigation = 367.50 mm net (459.38 mm gross); frequency $\approx$ 91 days.** (The data are very high for a field soil, but the method is as shown.)

## Check yourself

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

1. AM = 
   A. 367.5 mm
   B. 750 mm
   C. 525 mm
   D. 350 mm
   Answer: C. 0.5 x 1.5 x 700. (Key point: Available moisture)

2. Net depth equals
   A. 525 mm
   B. 459.38 mm
   C. 1.5 x 525
   D. RAM = 0.7 x 525 = 367.5 mm
   Answer: D. RAM is the net depth. (Key point: Net depth)

3. Gross depth is
   A. 367.5/0.8 = 459.38 mm
   B. 367.5 x 0.8 = 294 mm
   C. 525/0.8
   D. 367.5/4
   Answer: A. Divide by efficiency. (Key point: Gross depth)

4. Frequency is
   A. gross/ET_c, 115 days
   B. RAM/ET_c, about 91 days
   C. AM x ET_c
   D. ET_c/RAM
   Answer: B. 367.5/4 = 91.88. (Key point: Frequency)

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