---
title: "Balancing depth with service bank and free board"
subject: "Irrigation and Drainage Engineering"
chapter: 3
key: "semester_6/irrigation_and_drainage_engineering:balancing-depth-bed8-service-bank"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/balancing-depth-bed8-service-bank
site: BCT Engineering
---

# Balancing depth with service bank and free board

Irrigation and Drainage Engineering · Chapter 3: Canal Irrigation System

- Asked in: 2074 Bhadra, 4 marks
- Video: [Balancing depth of canal(part-2) |Canal Irrigation System | Important numerical #irrigation ](https://www.youtube.com/watch?v=ToUJbhJGUsQ) —  The civil 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/balancing-depth-bed8-service-bank
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-3

## Question

A canal has bed width of 8 m. Full supply depth of water is 1.5 m, side slope in cutting 1:1 and filling 1.5:1. Top width of the bank is 1.8 m and service bank is 5.0 m. Free board is kept 0.6 m. Calculate balancing depth so as to get the most economical section.

## Short answer

Equate cutting area (8+d)*d to filling area h*(6.8+3h) with h=2.1-d. Solve quadratic 2d² - 27.4d + 27.51 = 0. Balancing depth = 1.09 m; bank height above ground = 1.01 m.

## Answer

**Given:** $B = 8$ m, full supply depth $D = 1.5$ m, cutting slope $n_1 = 1$ (1:1), banking slope $n_2 = 1.5$ (1.5:1), top width of ordinary bank $b_1 = 1.8$ m, top width of service bank $b_2 = 5.0$ m, free board $F = 0.6$ m.

**Set-up.** Let $d$ = depth of the bed below ground level (balancing depth). The water surface is $D - d$ above ground, and the top of the bank is at $D + F - d$ above ground:

$$
h = D + F - d = 2.1 - d
$$

**Cutting area:** $A_c = (8 + d)\,d$

**Filling area** (one ordinary bank and one service bank, slope 1.5:1 on both faces):

$$
A_f = h(b_1 + 1.5h) + h(b_2 + 1.5h) = h(6.8 + 3h)
$$

**Equate** with $h = 2.1 - d$:

$$
\begin{aligned}
d^2 + 8d &= (2.1-d)(6.8 + 6.3 - 3d)\\
d^2 + 8d &= 3d^2 - 19.4d + 27.51\\
2d^2 - 27.4d + 27.51 &= 0
\end{aligned}
$$

$$
d = \frac{27.4 - \sqrt{27.4^2 - 4\times2\times27.51}}{4} = \frac{27.4-\sqrt{530.68}}{4} = 1.091\ \text{m}
$$

(The larger root, $d>2.1$ m, is rejected because it gives a negative bank height.)

**Check:** $h = 1.009$ m; $A_c = 9.917$ m$^2$ and $A_f = 9.917$ m$^2$.

**Answer: balancing depth (bed below ground level) = 1.09 m; bank height above ground = 1.01 m.**

## Check yourself

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

1. How is the bank height related to d?
   A. h = D + F - d = 2.1 - d
   B. h = D - F + d
   C. h = D + F + d
   D. h = d
   Answer: A. The bank top is at D + F - d above ground. (Key point: Bank height)

2. What is the filling area with one ordinary and one service bank?
   A. 2h(1.8 + 1.5h)
   B. h(b1 + 1.5h) + h(b2 + 1.5h) = h(6.8 + 3h)
   C. h(6.8 + 1.5h)
   D. 2h(5 + 1.5h)
   Answer: B. b1 + b2 = 6.8 and 1.5h counted twice. (Key point: Filling area)

3. Why is the larger root rejected?
   A. It is complex
   B. It is smaller than bed width
   C. It gives d > 2.1 m, i.e. negative bank height
   D. It gives zero cut
   Answer: C. Only the root with d < 2.1 is physically valid. (Key point: Root choice)

4. What is the balancing depth?
   A. 1.01 m
   B. 2.1 m
   C. 8 m
   D. 1.09 m
   Answer: D. 2d^2 - 27.4d + 27.51 = 0 gives d = 1.091 m. (Key point: Result)

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