---
title: "Derive the balancing depth of a canal"
subject: "Irrigation and Drainage Engineering"
chapter: 3
key: "semester_6/irrigation_and_drainage_engineering:balancing-depth-derive"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/balancing-depth-derive
site: BCT Engineering
---

# Derive the balancing depth of a canal

Irrigation and Drainage Engineering · Chapter 3: Canal Irrigation System

- Asked in: 2077 Chaitra, 3 marks; 2062 Kartik (old course), 8 marks (asked 2 times)
- Video: [BALANCING DEPTH OF THE CANAL|WATER RESOURCE ENGG |CONCEPT & NUMERICAL](https://www.youtube.com/watch?v=JChXYF65y-U) — Tech Master Academy
- 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-derive
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-3

## Question

Derive the expression for the balancing depth for a canal section.

## Short answer

Balancing depth is where cut volume equals fill volume. Equating trapezoidal cutting area (B + n1 d)d to filling area of two banks 2h(b + n2 h) yields n1 d^2 + B d - 2h(b + n2 h) = 0. Solving gives d = [-B + sqrt(B^2 + 8 n1 h(b + n2 h))]/(2 n1).

## Answer

**Balancing depth** is the depth of excavation (depth of the bed below ground level) at which the volume of earth cut from the canal equals the volume of earth needed to form the banks, so that no earth has to be borrowed or spoiled. It gives the most economical section.

#### Notation
$B$ = bed width, $d$ = depth of the bed below ground level (balancing depth), $n_1$ = side slope in cutting ($n_1$H:1V), $n_2$ = side slope of the banks, $b$ = top width of bank, $h$ = height of bank above ground level (for a bank-on-both-sides canal).

```text
        b                      b
      |---|                  |---|
      ____                    ____
     /    \                  /    \    h
 ---/      \__    GL       __/      \---
              \            /   d
               \__________/
                    B
```

#### Derivation (per metre length of canal)
**Cutting area** (trapezoid of bottom width $B$, depth $d$, side slope $n_1$):

$$
A_c = (B + n_1 d)\,d
$$

**Filling area of one bank** (top width $b$, height $h$, slope $n_2$ on both faces):

$$
A_f = h\,(b + n_2 h)
$$

For two banks, $A_f = 2h\,(b + n_2 h)$.

For balancing, cutting = filling:

$$
(B + n_1 d)\,d = 2h\,(b + n_2 h)
$$

$$
n_1 d^2 + B d - 2h(b+n_2 h) = 0
$$

$$
d = \frac{-B + \sqrt{B^2 + 8 n_1 h\,(b + n_2 h)}}{2 n_1}
$$

#### When the bank height depends on d
For a canal of full supply depth $D$ and free board $F$, the top of the bank is at $D + F - d$ above ground level, so $h = D + F - d$. Substituting in the equation above gives a quadratic in $d$ whose smaller positive root is the balancing depth. If a service bank of top width $b_s$ is provided on one side, then fill $= h(b + n_2 h) + h(b_s + n_2 h)$.

If the actual depth is greater than $d$, earth is surplus (spoil banks are needed); if it is less, borrow pits are needed.

## Check yourself

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

1. What is balancing depth?
   A. Depth of bed below ground at which cut volume equals bank fill volume
   B. Depth at which the canal is full
   C. Depth of banks above ground
   D. Depth of free board
   Answer: A. No borrow pit or spoil bank is needed, giving most economical section. (Key point: Balancing depth meaning)

2. What is the cutting area per metre length?
   A. (B + n2 d) d
   B. (B + n1 d) d
   C. B d only
   D. (B + d) / n1
   Answer: B. Trapezoid of bed width B, depth d and side slope n1. (Key point: Cutting area)

3. What is the filling area for two banks?
   A. h(b + n2 h)
   B. 2h(B + n2 h)
   C. 2h(b + n2 h)
   D. 2b(h + n2)
   Answer: C. Each bank has top width b, height h and slope n2 on both faces. (Key point: Filling area)

4. Equating cut and fill gives which equation and root?
   A. n1 d^2 - B d + 2h(b+n2 h) = 0
   B. d = B/(2 n1)
   C. n1 d^2 + B d + 2h(b+n2 h) = 0
   D. n1 d^2 + B d - 2h(b+n2 h) = 0, positive root d = [-B + sqrt(B^2 + 8 n1 h(b+n2 h))]/(2 n1)
   Answer: D. Rearranging (B + n1 d)d = 2h(b+n2 h) and taking the positive root. (Key point: Resulting quadratic)

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