---
title: "Slope-area flood discharge, 6 km reach"
subject: "Engineering Hydrology"
chapter: 4
key: "semester_5/engineering_hydrology:slope-area-60-45"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/slope-area-60-45
site: BCT Engineering
---

# Slope-area flood discharge, 6 km reach

Engineering Hydrology · Chapter 4: Surface Runoff

- Asked in: 2078 Bhadra, 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-60-45
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4

## Question

During a flood flow the cross section area of a river were measured as 60 m$^2$ for u/s and 45 m$^2$ for d/s at two sections 6 km apart. Wetted perimeters of these sections were 18 m and 14 m respectively. Elevation difference between u/s and d/s bed of the river was 0.45 m. Estimate the flood discharge of the river. Take Manning's rugosity coefficient = 0.025 and eddy loss coefficient = 0.15.

## Short answer

Slope-area method with Manning’s n=0.025 and contraction loss coefficient 0.15, corrected for velocity head, yields flood discharge ≈ 38.8 m³/s.

## Answer

**Given:** $A_1=60$ m$^2$, $P_1=18$ m, $A_2=45$ m$^2$, $P_2=14$ m, $L=6$ km, $n=0.025$, $k_e=0.15$. The bed levels differ by 0.45 m. Assuming the flow depth is the same at both sections (uniform depth reach), the water-surface fall equals the bed fall $=0.45$ m. $\alpha=1$.

**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
$$
R_1=\frac{60}{18}=3.3333\ \text{m},\qquad R_2=\frac{45}{14}=3.2143\ \text{m}
$$
$$
K_1=\frac{1}{0.025}\times60\times3.3333^{2/3}=5355.5,\qquad K_2=\frac{1}{0.025}\times45\times3.2143^{2/3}=3920.4
$$
$$
K_m=\sqrt{K_1K_2}=4582.1
$$

#### Step 2: First trial (no velocity head)
$$
S_f=\frac{0.45}{6000},\qquad Q_1=K_m\sqrt{S_f}=39.682\ \text{m}^3/\text{s}
$$

#### Step 3: Correct for velocity head and eddy loss
$$
h_f=(h_1-h_2)+\left(\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right)-h_e,\qquad h_e=k_e\left|\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right|
$$
The area decreases downstream (contraction), so $h_e = 0.15(h_{v2}-h_{v1})$.

| 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 | 39.682 | 0.6614 | 0.8818 | 0.02229 | 0.03963 | 0.00260 | 0.43006 | 38.793 |
| 2 | 38.793 | 0.6465 | 0.8621 | 0.02131 | 0.03788 | 0.00249 | 0.43094 | 38.833 |
| 3 | 38.833 | 0.6472 | 0.8629 | 0.02135 | 0.03796 | 0.00249 | 0.43090 | 38.831 |
| 4 | 38.831 | 0.6472 | 0.8629 | 0.02135 | 0.03795 | 0.00249 | 0.43091 | 38.831 |

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

**Answer: Discharge $Q \approx 38.8\ \text{m}^3/\text{s}$.**

## Check yourself

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

1. What is used as the water-surface fall when only bed levels are given?
   A. Zero
   B. The average depth
   C. The difference of hydraulic radii
   D. Bed elevation difference, assuming equal flow depth at both sections (0.45 m)
   Answer: D. With the same flow depth at both sections the water-surface fall equals the bed fall, 0.45 m. (Key point: Water-surface fall)

2. The area decreases downstream (60 to 45 m2). How is the eddy loss computed?
   A. he = 0.15 (hv2 - hv1)
   B. he = 0.15 x fall
   C. he = 0
   D. he = 0.15 (hv1 - hv2)
   Answer: A. In a contraction hv2 > hv1, and he = ke |hv1 - hv2| = 0.15 (hv2 - hv1). (Key point: Contraction eddy loss)

3. How is the first trial discharge Q1 obtained?
   A. Q1 = Km x 0.45
   B. Q1 = Km sqrt(0.45/6000), ignoring velocity head
   C. Q1 = A1 x V where V is assumed 1 m/s
   D. Q1 = Km sqrt(6000/0.45)
   Answer: B. The first trial uses Sf = fall/L with no velocity-head correction. (Key point: First trial discharge)

## Related in Engineering Hydrology

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