---
title: "Slope-area flood discharge, 7 km reach"
subject: "Engineering Hydrology"
chapter: 4
key: "semester_5/engineering_hydrology:slope-area-45-50-7km"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/slope-area-45-50-7km
site: BCT Engineering
---

# Slope-area flood discharge, 7 km reach

Engineering Hydrology · Chapter 4: Surface Runoff

- Asked in: 2076 Chaitra, 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-45-50-7km
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4

## Question

Calculate the flood discharge of a stream by the slope area method given as below:

| Upstream | Downstream |
|---|---|
| Upstream flow area = 45 m$^2$ | Downstream flow area = 50 m$^2$ |
| Upstream wetted perimeter = 26 m | Downstream wetted perimeter = 28 m |
| Upstream velocity head coefficient = 1.18 | Downstream velocity head coefficient = 1.23 |

Falling difference = 1.15 m, Reach length = 7 km, Manning's coefficient n = 0.03. The eddy loss coefficient of 0.3 for gradual expansion and 0.1 for gradual contraction.

## Short answer

Compute conveyance K1, K2, Km from Manning's n. Trial with energy equation including velocity head and eddy loss for expansion yields friction slope and discharge. Q ≈ 29.6 m³/s.

## Answer

**Given:** $A_1=45$ m$^2$, $P_1=26$ m, $\alpha_1=1.18$; $A_2=50$ m$^2$, $P_2=28$ m, $\alpha_2=1.23$; fall $=1.15$ m; $L=7000$ m; $n=0.03$. The area increases downstream (expansion), so $k_e=0.3$.

**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
$$
R_1=\frac{45}{26}=1.7308\ \text{m},\qquad R_2=\frac{50}{28}=1.7857\ \text{m}
$$
$$
K_1=\frac{1}{0.03}\times45\times1.7308^{2/3}=2162.3,\qquad K_2=\frac{1}{0.03}\times50\times1.7857^{2/3}=2453.1
$$
$$
K_m=\sqrt{K_1K_2}=2303.1
$$

#### Step 2: First trial (no velocity head)
$$
S_f=\frac{1.15}{7000},\qquad Q_1=K_m\sqrt{S_f}=29.520\ \text{m}^3/\text{s}
$$

#### Step 3: Correct for velocity head and eddy loss
$$
h_f=(h_1-h_2)+\left(\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right)-h_e,\qquad h_e=k_e\left|\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right|
$$
Here $h_{v1}>h_{v2}$ (expansion), so $h_e = 0.3(h_{v1}-h_{v2})$.

| 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 | 29.520 | 0.6560 | 0.5904 | 0.02588 | 0.02185 | 0.00121 | 1.15282 | 29.556 |
| 2 | 29.556 | 0.6568 | 0.5911 | 0.02595 | 0.02191 | 0.00121 | 1.15283 | 29.557 |

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

**Answer: Discharge $Q \approx 29.6\ \text{m}^3/\text{s}$.**

## Check yourself

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

1. Why is ke = 0.3?
   A. The reach is contracting
   B. n is 0.03
   C. Area increases downstream, so it is an expansion
   D. alpha1 < alpha2
   Answer: C. Area goes from 45 to 50 m^2. (Key point: Eddy loss)

2. How is velocity head computed with coefficients?
   A. v^2/(2g) without alpha
   B. alpha v/(2g)
   C. alpha v^2 g
   D. alpha v^2/(2g) at each section
   Answer: D. Coefficients 1.18 and 1.23 multiply v^2/2g. (Key point: Velocity head)

3. What is the discharge?
   A. About 29.6 m^3/s
   B. About 2303 m^3/s
   C. About 1.15 m^3/s
   D. About 295 m^3/s
   Answer: A. First trial 29.52 converges to 29.557 m^3/s. (Key point: Result)

## Close variants set in other exams

- [Slope-area discharge, 1300 m reach](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-3500-3250) — Engineering Hydrology, chapter 4, asked 2×

## Related in Engineering Hydrology

- [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](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-306-920) — 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
- [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
- [Stream flow by slope-area method](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-method-explain) — Engineering Hydrology, chapter 4, asked 3×

## Related in other subjects

- [DO deficit 50 km downstream, 1.9 m³/s stream](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Astream-do-deficit-50km) — Sanitary Engineering, chapter 6, asked 2×
- [Design: 230 m head plant, 45 km line](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2071-230m-45km) — Power Plant Design, chapter 4, asked 12×
- [DO deficit 60 km downstream, 2 m³/s stream](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Astream-do-deficit-60km) — Sanitary Engineering, chapter 6, asked 2×
- [Non-linear kinematic wave discharge at 1 km](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Akinematic-wave-implicit-numerical) — Computational Techniques in Civil Engineering, chapter 5