---
title: "Irrigation demand from ETo and efficiency"
subject: "Irrigation and Drainage Engineering"
chapter: 2
key: "semester_6/irrigation_and_drainage_engineering:project-irrigation-demand-eto"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/project-irrigation-demand-eto
site: BCT Engineering
---

# Irrigation demand from ETo and efficiency

Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements

- Asked in: 2066 Bhadra (old course), 5 marks
- Video: [Consumptive use of water and Irrigation efficiencies : WRE](https://www.youtube.com/watch?v=Cr8DAqXyp8o) — The Classroom
- 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/project-irrigation-demand-eto
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2

## Question

An irrigation project has 6000 ha of CCA and ETo is 150 mm/day, effective rainfall is 30 mm/month and the overall efficiency of the project is 30%. Calculate the irrigation demand in cumec.

## Short answer

Net irrigation requirement = 150–30 = 120 mm/month. Gross requirement = 120 / 0.30 = 400 mm/month. Volume = 0.40 m × 6000×10⁴ m² = 2.4×10⁷ m³/month; discharge = 2.4×10⁷ / (30×86400) = 9.26 cumec.

## Answer

**Assumptions.** The value of 150 mm for $ET_o$ cannot be a daily rate, so it is taken as **mm per month**, the same time unit as the effective rainfall (30 mm/month). Crop coefficient $K_c=1$ (crop water requirement = $ET_o$), and a month = 30 days.

**Given:** CCA = 6000 ha, $ET_o = 150$ mm/month, $P_e = 30$ mm/month, overall efficiency $= 30\%$.

**Step 1: Net irrigation requirement**

$$
NIR = ET_o - P_e = 150 - 30 = 120\ \text{mm/month}
$$

**Step 2: Gross irrigation requirement at the head**

$$
GIR = \frac{NIR}{\eta} = \frac{120}{0.30} = 400\ \text{mm/month}
$$

**Step 3: Volume of water per month**

$$
V = 0.40\ \text{m}\times 6000\times 10^4\ \text{m}^2 = 2.4\times 10^7\ \text{m}^3
$$

**Step 4: Discharge**

$$
Q = \frac{V}{30\times 86400} = \frac{2.4\times 10^7}{2\,592\,000} = 9.26\ \text{m}^3/\text{s}
$$

**Answer: irrigation demand $\approx 9.26$ cumec.**

## Check yourself

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

1. Why is ET_o = 150 taken as mm/month?
   A. Because CCA is in ha
   B. Because efficiency is 30%
   C. It is the standard FAO value
   D. 150 mm/day is unrealistic and must match effective rainfall units
   Answer: D. Consistent units are needed. (Key point: Time unit assumption)

2. NIR is
   A. 150 - 30 = 120 mm/month
   B. 150 + 30
   C. 150 / 0.3
   D. 30 - 150
   Answer: A. ET_o minus Pe. (Key point: Net requirement)

3. Gross requirement is NIR divided by
   A. 0.30 giving 36 mm
   B. overall efficiency 0.30 giving 400 mm/month
   C. 3.0
   D. 1.30
   Answer: B. 120/0.30 = 400. (Key point: Gross requirement)

4. Final demand is about
   A. 0.926 cumec
   B. 92.6 cumec
   C. 9.26 cumec
   D. 24 cumec
   Answer: C. 24e6 m3 / 2,592,000 s. (Key point: Discharge)

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