---
title: "Canal design discharge at 0, 1, 2, 3 km"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:canal-design-discharge-along-reach"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/canal-design-discharge-along-reach
site: BCT Engineering
---

# Canal design discharge at 0, 1, 2, 3 km

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2068 Baisakh (old course), 9 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-design-discharge-along-reach
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4

## Question

Calculate the design discharge of canal at 0, 1, 2 and 3 km from headworks. The GCA at the head of the canal is 40000 ha and after each km it is reduced by 6000 ha. Of this command, the CCA is only 70%. The intensity of irrigation for Wheat and Rice are 40% and 25% respectively. Assume total loss below 3 km = 0.4 m³/s; channel losses per km = 4% of discharge at beginning of each km; Kor period for wheat and Rice is 4 and 3 weeks respectively. Kor depth for wheat and rice are 14 and 20 cm respectively.

## Short answer

Rice demand (20 cm, 3 weeks) governs. Outlet demand per km is 1.157 m³/s. Working upstream: Q at 3 km = 4.64, 2 km = 6.04, 1 km = 7.50, 0 km = 9.02 m³/s, accounting for 4% channel loss per km and tail loss.

## Answer

#### Assumptions

- Wheat and rice are grown in different seasons, so the canal is designed for the crop that needs the larger discharge (not their sum).
- Intensity is on CCA. The GCA at a km is the area still commanded below that point. Total loss below 3 km $= 0.4$ m³/s is a fixed loss added at 3 km; the channel loss in each km is 4% of the discharge at the beginning of that km.

#### Discharge per hectare of crop

$$
\begin{aligned}
\text{Wheat}: q_w &= \frac{0.14\times 10^4}{28\times 86400} = 0.5787\times 10^{-3}\ \text{m}^3/\text{s per ha} \\
\text{Rice}: q_r &= \frac{0.20\times 10^4}{21\times 86400} = 1.1023\times 10^{-3}\ \text{m}^3/\text{s per ha}
\end{aligned}
$$

Demand per hectare of GCA: wheat $= 0.7\times 0.40\times q_w = 0.1620\times 10^{-3}$; rice $= 0.7\times 0.25\times q_r = 0.1929\times 10^{-3}$ m³/s. Rice governs at every section.

#### Demand of outlets (rice) and discharge

| Section | GCA below (ha) | CCA (ha) | Rice area (ha) | Wheat area (ha) | Rice demand (m³/s) | Wheat demand (m³/s) |
|---|---|---|---|---|---|---|
| 0 km | 40,000 | 28,000 | 7,000 | 11,200 | 7.716 | 6.481 |
| 1 km | 34,000 | 23,800 | 5,950 | 9,520 | 6.559 | 5.509 |
| 2 km | 28,000 | 19,600 | 4,900 | 7,840 | 5.401 | 4.537 |
| 3 km | 22,000 | 15,400 | 3,850 | 6,160 | 4.244 | 3.565 |

#### Working from the tail end

Area served between two consecutive km marks $= 6000$ ha GCA, with outlet demand $6000\times 0.1929\times 10^{-3} = 1.157$ m³/s.

$$
\begin{aligned}
Q_{3} &= 4.244 + 0.4 = 4.644\ \text{m}^3/\text{s} \\
Q_{begin} &= Q_{end} + \text{outlet demand} + 0.04\,Q_{begin} \Rightarrow Q_{begin} = \frac{Q_{end} + 1.157}{0.96} \\
Q_2 &= \frac{4.644 + 1.157}{0.96} = 6.043\ \text{m}^3/\text{s} \\
Q_1 &= \frac{6.043 + 1.157}{0.96} = 7.500\ \text{m}^3/\text{s} \\
Q_0 &= \frac{7.500 + 1.157}{0.96} = 9.019\ \text{m}^3/\text{s}
\end{aligned}
$$

**Answer:** Design discharge at 0, 1, 2, 3 km = **9.02, 7.50, 6.04, 4.64 m³/s** (rice season governs).

## Check yourself

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

1. Why is the canal designed for rice discharge and not the sum of wheat and rice?
   A. Rice has a higher intensity than wheat
   B. Wheat demand is zero
   C. Rice area is larger than wheat area
   D. They are grown in different seasons and rice needs the larger discharge
   Answer: D. Wheat and rice seasons differ, so the larger of the two (rice, 0.1929 x 10^-3 per ha of GCA) governs. (Key point: Governing crop)

2. What is rice discharge per hectare of crop, q_r?
   A. 0.20 x 10^4 / (21 x 86400) = 1.102 x 10^-3 m3/s
   B. 0.14 x 10^4 / (28 x 86400) = 0.579 x 10^-3 m3/s
   C. 0.20 x 10^4 / (3 x 86400) = 7.7 x 10^-3 m3/s
   D. 0.20 / (21 x 86400) = 1.1 x 10^-7 m3/s
   Answer: A. Kor period of 3 weeks = 21 days, depth 0.20 m, so q = 2000/(21 x 86400) = 1.102 x 10^-3. (Key point: Rice discharge per ha)

3. How is the discharge at the start of a km found from the end of that km?
   A. Q_begin = Q_end + 0.04 Q_end
   B. Q_begin = Q_end + 1.157 x 0.04
   C. Q_begin = (Q_end + 1.157)/0.96
   D. Q_begin = (Q_end + 1.157) x 0.96
   Answer: C. Q_begin = Q_end + outlet demand + 0.04 Q_begin, so Q_begin = (Q_end + 1.157)/0.96. (Key point: Working from tail end)

4. What is the design discharge at 3 km?
   A. 4.244 m3/s
   B. 0.4 m3/s
   C. 1.157 m3/s
   D. 4.644 m3/s
   Answer: D. Q3 = rice demand 4.244 + fixed loss 0.4 = 4.644 m3/s. (Key point: Result at 3 km)

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