---
title: "Flood discharge in rectangular channel"
subject: "Hydro Power"
chapter: 2
key: "semester_6/hydro_power:flood-discharge-slope-area"
source: https://www.bctengineer.com/q/semester_6/hydro_power/flood-discharge-slope-area
site: BCT Engineering
---

# Flood discharge in rectangular channel

Hydro Power · Chapter 2: Introduction to Hydrology

- Asked in: 2079 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/flood-discharge-slope-area
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-2

## Question

During a flood flow the depth of water in a 10 m wide rectangular channel was found to be 3 m and 2.9 m at two sections 200 m apart. The drop in the water surface elevation was found to be 0.12 m. Estimate the flood discharge through the channel. Assume Manning's n to be 0.025.

## Short answer

Applying the slope-area method with Manning’s n=0.025, average conveyance K=√(K1K2), and solving by trial for velocity head correction, the flood discharge converges to approximately 42.3 m³/s.

## Answer

The **slope-area method** is used: discharge from Manning's equation with friction slope found by the energy equation, solved by trial.

#### Data
$b = 10$ m, $y_1 = 3.0$ m, $y_2 = 2.9$ m, $L = 200$ m, fall $\Delta Z = 0.12$ m, $n = 0.025$.

#### Section properties and conveyance
$$
\begin{aligned}
A_1 &= 30\ \text{m}^2,\; P_1 = 16\ \text{m},\; R_1 = 1.875\ \text{m} \\
A_2 &= 29\ \text{m}^2,\; P_2 = 15.8\ \text{m},\; R_2 = 1.835\ \text{m} \\
K_1 &= \frac{1}{n}A_1R_1^{2/3} = \frac{30 \times 1.875^{2/3}}{0.025} = 1824.7 \\
K_2 &= \frac{29 \times 1.835^{2/3}}{0.025} = 1738.9 \\
K &= \sqrt{K_1K_2} = 1781.3
\end{aligned}
$$

#### Trial solution
Area decreases downstream (contracting reach), so eddy loss is taken as zero. Then
$$
h_f = \Delta Z + \frac{V_1^2 - V_2^2}{2g}, \qquad Q = K\sqrt{h_f/L}
$$

| Trial | $h_f$ (m) | $S_f$ | Q (m³/s) | $V_1$ | $V_2$ | $\frac{V_1^2 - V_2^2}{2g}$ (m) |
|---|---|---|---|---|---|---|
| 1 | 0.1200 | 0.000600 | 43.63 | 1.454 | 1.505 | −0.00756 |
| 2 | 0.1124 | 0.000562 | 42.23 | 1.408 | 1.456 | −0.00709 |
| 3 | 0.1129 | 0.000565 | 42.32 | 1.411 | 1.459 | −0.00712 |
| 4 | 0.1129 | 0.000564 | 42.32 | — | — | converged |

Sample (trial 1): $Q = 1781.3\sqrt{0.12/200} = 1781.3 \times 0.02449 = 43.63$ m³/s.

**Answer:** Flood discharge ≈ 42.3 m³/s (taking $K_e = 0$ for the contracting reach).

## Check yourself

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

1. How is conveyance K computed at each section?
   A. K = n A R^(2/3)
   B. K = (1/n) A R^(2/3)
   C. K = (1/n) A R^2
   D. K = A / (nR)
   Answer: B. Manning's equation gives conveyance K = (1/n) A R^(2/3). (Key point: Conveyance)

2. How are the two sectional conveyances combined (1824.7 and 1738.9)?
   A. Arithmetic mean = 1781.8
   B. Sum = 3563.6
   C. Geometric mean sqrt(K1 K2) = 1781.3
   D. Smaller value = 1738.9
   Answer: C. The method uses K = sqrt(K1 K2), which is 1781.3. (Key point: Mean conveyance)

3. Why is the eddy loss taken as zero?
   A. The channel is rectangular
   B. The reach is contracting since the area decreases downstream
   C. n is small
   D. The fall is 0.12 m
   Answer: B. The area falls from 30 to 29 m2, a contracting reach, so Ke is taken as about 0. (Key point: Eddy loss)

4. What is the converged discharge after the trials?
   A. 43.6 m3/s
   B. 44.5 m3/s
   C. 41.0 m3/s
   D. 42.3 m3/s
   Answer: D. The first trial gives 43.63, correcting hf for velocity head gives hf = 0.1129 and Q about 42.3 m3/s. (Key point: Final discharge)

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