---
title: "Flood discharge in 10-m rectangular channel"
subject: "Engineering Hydrology"
chapter: 4
key: "semester_5/engineering_hydrology:slope-area-rectangular-10m"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/slope-area-rectangular-10m
site: BCT Engineering
---

# Flood discharge in 10-m rectangular channel

Engineering Hydrology · Chapter 4: Surface Runoff

- Asked in: 2082 Bhadra, 3+8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/slope-area-rectangular-10m
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4

## Question

During a flood flow, the depth of water in a 10 m wide rectangular channel was found to be 3.0 m and 2.7 m at two sections 200 m apart. The drop in the water surface elevation was found to be 0.12 m. Assuming Manning's coefficient to be 0.025, estimate the flood discharge through the channel. Assume no eddy loss. For the application of the hydraulic formulae for stream flow computation, what are the parameters that need to be measured in the channel?

## Short answer

The slope-area method with Manning's equation and iterative energy balance gave flood discharge Q ≈ 38.0 m^3/s. Required channel measurements: cross-sections, reach length, water-surface drop, roughness n, and velocity-head/eddy-loss coefficients.

## Answer

**Given:** rectangular channel $b=10$ m, $y_1=3.0$ m, $y_2=2.7$ m, $L=200$ m, fall $=0.12$ m, $n=0.025$, $h_e = 0$, $\alpha=1$.

Section 1: $A_1=10\times3.0=30$ m$^2$, $P_1=10+6=16$ m. Section 2: $A_2=10\times2.7=27$ m$^2$, $P_2=10+5.4=15.4$ 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
$$
R_1=\frac{30}{16}=1.8750\ \text{m},\qquad R_2=\frac{27}{15.4}=1.7532\ \text{m}
$$
$$
K_1=\frac{1}{0.025}\times30\times1.8750^{2/3}=1824.7,\qquad K_2=\frac{1}{0.025}\times27\times1.7532^{2/3}=1570.3
$$
$$
K_m=\sqrt{K_1K_2}=1692.7
$$

#### Step 2: First trial (no velocity head)
$$
S_f=\frac{0.12}{200},\qquad Q_1=K_m\sqrt{S_f}=41.463\ \text{m}^3/\text{s}
$$

#### 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 $h_e=0$: $h_f = 0.12 + (h_{v1}-h_{v2})$.

| 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 | 41.463 | 1.3821 | 1.5357 | 0.09736 | 0.12020 | 0.00000 | 0.09716 | 37.309 |
| 2 | 37.309 | 1.2436 | 1.3818 | 0.07883 | 0.09732 | 0.00000 | 0.10151 | 38.135 |
| 7 | 38.004 | 1.2668 | 1.4076 | 0.08179 | 0.10098 | 0.00000 | 0.10081 | 38.004 |

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

**Answer: Discharge $Q \approx 38.0\ \text{m}^3/\text{s}$.**

#### Parameters to be measured in the channel
To apply Manning's equation and the energy equation for stream flow computation the following are measured in the field:
1. Cross-sections at the ends (and middle) of the reach: width, depth, hence area $A$ and wetted perimeter $P$.
2. Length of the reach $L$ between the sections.
3. Elevations of the water surface at the ends of the reach (high-water marks), giving the fall $h$ and the slope.
4. Roughness coefficient $n$ from the bed and bank material and vegetation (estimated).
5. Velocity-head coefficients $\alpha$ and the eddy-loss coefficient $k_e$ (from the shape of the reach).

## Check yourself

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

1. What are A and P at section 1 (b = 10 m, y = 3.0 m)?
   A. A = 30 m^2, P = 13 m
   B. A = 30 m^2, P = 16 m
   C. A = 13 m^2, P = 30 m
   D. A = 30 m^2, P = 10 m
   Answer: B. P = b + 2y = 16 m for a rectangular channel. (Key point: Hydraulic properties)

2. With no eddy loss, how is hf obtained?
   A. hf = fall only
   B. hf = fall - 0.3(hv1-hv2)
   C. hf = fall + (hv1 - hv2)
   D. hf = hv1 + hv2
   Answer: C. hf = 0.12 + change in velocity head; here velocity head rises downstream so hf < fall. (Key point: Energy correction)

3. What is the flood discharge?
   A. About 41.5 m^3/s
   B. About 1692.7 m^3/s
   C. About 3.8 m^3/s
   D. About 38.0 m^3/s after iterating
   Answer: D. The first trial gave 41.46 m^3/s, which converges to 38.0 m^3/s. (Key point: Result)

4. Which are measured in the channel?
   A. Cross-sections, reach length, water surface elevations at ends, and roughness is estimated
   B. Only the rainfall
   C. Only the velocity with a meter
   D. Evaporation and temperature
   Answer: A. Slope-area needs geometry, length, fall and an estimated n. (Key point: Field parameters)

## Related in Engineering Hydrology

- [Flood discharge in a 5-m rectangular channel](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-5m-rectangular) — 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
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- [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

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