---
title: "Interval and outlet discharge, 11 hour canal"
subject: "Irrigation and Drainage Engineering"
chapter: 2
key: "semester_6/irrigation_and_drainage_engineering:canal-operation-hours-interval-outlet"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/canal-operation-hours-interval-outlet
site: BCT Engineering
---

# Interval and outlet discharge, 11 hour canal

Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements

- Asked in: 2074 Bhadra, 6 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/canal-operation-hours-interval-outlet
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2

## Question

If daily consumptive use of the crop is 5 mm and the canal may operate from 6 AM to 5 PM only. Available moisture for the given soil is 220 mm per m and maximum depth of root zone for the crop is 1.2 m. Assume that only 50% of soil moisture is available to the crop. Application efficiency is 65%. Calculate the required discharge if CCA is 450 ha. Calculate irrigation interval and outlet discharge.

## Short answer

Irrigation interval is 26.4 days (about 26 days). Required outlet discharge is 0.89 m³/s (888 l/s) for 450 ha with canal running 11 h/day.

## Answer

**Given:** $ET_c = 5$ mm/day, AM = 220 mm/m, $D_r = 1.2$ m, only 50% of AM is readily available, $\eta_a = 65\%$, CCA = 450 ha, canal runs 6 AM to 5 PM = 11 h/day.

**Step 1: Net depth per irrigation**

$$
d_{net} = 0.50\times 220\times 1.2 = 132\ \text{mm}
$$

**Step 2: Irrigation interval**

$$
I = \frac{132}{5} = 26.4\ \text{days}\approx 26\ \text{days}
$$

**Step 3: Gross depth and volume for the whole CCA**

$$
\begin{aligned}
d_{gross} &= \frac{132}{0.65} = 203.08\ \text{mm}\\
V &= 0.203077\times 450\times 10^4 \approx 913,846\ \text{m}^3
\end{aligned}
$$

**Step 4: Outlet (canal) discharge.** All the CCA must be irrigated within one interval of 26 days, with the canal flowing 11 h each day:

$$
T = 26\times 11\times 3600 = 1,029,600\ \text{s},\qquad Q = \frac{V}{T} = \frac{913,846}{1,029,600} = 0.888\ \text{m}^3/\text{s}
$$

**Answer: interval = 26 days; required (outlet) discharge $\approx$ 0.89 cumec (888 l/s).**

## Check yourself

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

1. Net depth per irrigation is
   A. 264 mm
   B. 132 mm
   C. 220 mm
   D. 203 mm
   Answer: B. 0.5 x 220 x 1.2. (Key point: Net depth)

2. Interval is
   A. 44 days
   B. 13 days
   C. 132/5 = 26.4 days, 26 days
   D. 40.6 days
   Answer: C. Net depth / ET_c. (Key point: Interval)

3. Gross volume is based on
   A. 132 mm x 450 ha
   B. 132 x 0.65
   C. 220 mm x 450 ha
   D. 132/0.65 = 203.08 mm over 450 ha
   Answer: D. Divide by efficiency. (Key point: Gross volume)

4. The canal operating time used is
   A. 26 days x 11 h/day
   B. 26 x 24 h
   C. 11 days
   D. 26 x 5 h
   Answer: A. 6 AM to 5 PM is 11 h. (Key point: Operating time)

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