---
title: "Flood discharge in a 5-m rectangular channel"
subject: "Engineering Hydrology"
chapter: 4
key: "semester_5/engineering_hydrology:slope-area-5m-rectangular"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/slope-area-5m-rectangular
site: BCT Engineering
---

# Flood discharge in a 5-m rectangular channel

Engineering Hydrology · Chapter 4: Surface Runoff

- Asked in: 2071 Shrawan, 8 marks
- Video: [Manning's Equation](https://www.youtube.com/watch?v=JQ0FIluECP8) — HydrologyVideos
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/slope-area-5m-rectangular
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4

## Question

Estimate the flood discharge through a 5 m wide rectangular channel for the following data. The depth of water is 2 m and 1.8 m at two sections 500 m apart. The drop in water surface elevation is 0.25 m. Manning's roughness coefficient is 0.025. Assume eddy loss to be zero.

## Short answer

Manning's equation with conveyance K = (1/n)AR^(2/3) and trial on energy equation including velocity head gives friction slope; iteration yields discharge Q ≈ 8.8 m³/s.

## Answer

**Given:** rectangular channel $b=5$ m, $y_1=2.0$ m, $y_2=1.8$ m, $L=500$ m, fall $=0.25$ m, $n=0.025$, $h_e=0$, $\alpha=1$.

Section 1: $A_1=5\times2=10$ m$^2$, $P_1=5+2\times2=9$ m. Section 2: $A_2=5\times1.8=9$ m$^2$, $P_2=5+2\times1.8=8.6$ m.

**Method:** Manning's equation gives conveyance $K=\dfrac{1}{n}AR^{2/3}$ and $Q = K\sqrt{S_f}$. The friction slope is found from the energy equation between the two sections, so the discharge is found by trial.

#### Step 1: Conveyance
$$
R_1=\frac{10}{9}=1.1111\ \text{m},\qquad R_2=\frac{9}{8.6}=1.0465\ \text{m}
$$
$$
K_1=\frac{1}{0.025}\times10\times1.1111^{2/3}=429.1,\qquad K_2=\frac{1}{0.025}\times9\times1.0465^{2/3}=371.1
$$
$$
K_m=\sqrt{K_1K_2}=399.0
$$

#### Step 2: First trial (no velocity head)
$$
S_f=\frac{0.25}{500},\qquad Q_1=K_m\sqrt{S_f}=8.923\ \text{m}^3/\text{s}
$$

#### Step 3: Correct for velocity head and eddy loss
$$
h_f=(h_1-h_2)+\left(\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right)-h_e,\qquad h_e=k_e\left|\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right|
$$
With $h_e=0$: $h_f = 0.25 + (h_{v1}-h_{v2})$.

| Trial | $Q$ assumed | $v_1$ | $v_2$ | $\alpha_1 v_1^2/2g$ | $\alpha_2 v_2^2/2g$ | $h_e$ | $h_f$ | $Q=K_m\sqrt{h_f/L}$ |
|---|---|---|---|---|---|---|---|---|
| 1 | 8.923 | 0.8923 | 0.9914 | 0.04058 | 0.05010 | 0.00000 | 0.24048 | 8.751 |
| 2 | 8.751 | 0.8751 | 0.9724 | 0.03903 | 0.04819 | 0.00000 | 0.24084 | 8.758 |
| 3 | 8.758 | 0.8758 | 0.9731 | 0.03909 | 0.04826 | 0.00000 | 0.24083 | 8.758 |

The trials converge to $h_f$ and $Q$ shown in the last row.

**Answer: Discharge $Q \approx 8.8\ \text{m}^3/\text{s}$.**

## Check yourself

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

1. At section 1 (b = 5 m, y = 2 m), A and P are
   A. A = 10 m2, P = 5 m
   B. A = 10 m2, P = 7 m
   C. A = 7 m2, P = 9 m
   D. A = 10 m2, P = 9 m
   Answer: D. P = b + 2y = 9 m (wetted perimeter includes both sides). (Key point: Section properties)

2. With he = 0, hf =
   A. 0.25 + (hv1 - hv2)
   B. 0.25 - he only
   C. 0.25
   D. 0.25 x hv
   Answer: A. hf = fall + (hv1 - hv2); velocity head is larger at the shallower section 2, so hf < 0.25. (Key point: hf with no eddy loss)

3. The converged discharge is about
   A. 88 m3/s
   B. 8.8 m3/s
   C. 0.88 m3/s
   D. 20 m3/s
   Answer: B. Q converges to 8.758 m3/s. (Key point: Trial result)

## Related in Engineering Hydrology

- [Flood discharge in 10-m rectangular channel](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-rectangular-10m) — Engineering Hydrology, chapter 4
- [Slope-area flood discharge, 6 km reach](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-60-45) — Engineering Hydrology, chapter 4
- [Slope-area discharge, 10 km reach, 104.77 m](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-104-77) — Engineering Hydrology, chapter 4
- [Slope-area discharge with three sections](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-3-section-260m) — Engineering Hydrology, chapter 4
- [Slope-area flood discharge, Marin River at Kusuntar](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-marin-kusuntar) — Engineering Hydrology, chapter 4

## Related in other subjects

- [Flood discharge in rectangular channel](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aflood-discharge-slope-area) — Hydro Power, chapter 2
- [Siphon aqueduct: Q 60, Manning 0.025](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Asiphon-aqueduct-q60-bed30-fsd14) — Irrigation and Drainage Engineering, chapter 8
- [Bresse's method for 20 m wide channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Abresse-method-20m-channel-65-m3s) — Hydraulics, chapter 8
- [GVF in 5 m channel, depth 0.25 m to 0.40 m](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Agvf-5m-channel-1-85-m3s-direct-step) — Hydraulics, chapter 8, asked 2×