---
title: "Slope-area discharge with three sections"
subject: "Engineering Hydrology"
chapter: 4
key: "semester_5/engineering_hydrology:slope-area-3-section-260m"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/slope-area-3-section-260m
site: BCT Engineering
---

# Slope-area discharge with three sections

Engineering Hydrology · Chapter 4: Surface Runoff

- Asked in: 2076 Asoj, 8 marks
- Video: [Slope Area Method of Stream Flow Measurement by Dr. Upain Kumar Bhatia](https://www.youtube.com/watch?v=NdtMFAQxuY0) — ACET CIVIL TECH
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/slope-area-3-section-260m
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4

## Question

For the purpose of discharge measurement in a stream by Slope-Area method the following data has been obtained.

| | U/S Section | Middle Section | D/S Section |
|---|---|---|---|
| Area (m$^2$) | 105.75 | 102.63 | 96.63 |
| Wetted perimeter (m) | 64.25 | 60.20 | 58.00 |
| Gauge reading (m) | 315.5 | - | 315.15 |
| Manning's roughness | 0.025 | 0.027 | 0.029 |

Determine the stream discharge for length between U/S and D/S sections as 260 m assuming coefficient of contraction $K_c$ as 0.1.

## Short answer

With fall 0.35 m over 260 m, geometric mean conveyance K_m=5310.8. First Q=194.9 m³/s; after iterative correction for velocity heads and eddy loss (k_c=0.1), discharge converges to 185.2 m³/s.

## Answer

**Given:** reach length $L=260$ m between the u/s and d/s sections; fall $=315.50-315.15=0.35$ m; contraction coefficient $k_c=0.1$ (area decreasing downstream); $\alpha=1$. Roughness differs at each section ($n = 0.025, 0.027, 0.029$).

**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
With three sections of different roughness, the conveyance of each section is computed with its own $n$:

| Section | $A$ (m$^2$) | $P$ (m) | $R$ (m) | $n$ | $K$ |
|---|---|---|---|---|---|
| U/S | 105.75 | 64.25 | 1.646 | 0.025 | 5896.7 |
| Middle | 102.63 | 60.20 | 1.705 | 0.027 | 5424.5 |
| D/S | 96.63 | 58.00 | 1.666 | 0.029 | 4682.8 |

The mean conveyance of the reach is taken as the geometric mean of the three sections:
$$
K_m=(K_1K_2K_3)^{1/3}=5310.8
$$
The middle section has no gauge reading, so the fall is that between the end sections, and the eddy loss is applied between the end sections.

#### Step 2: First trial (no velocity head)
$$
S_f=\frac{0.35}{260},\qquad Q_1=K_m\sqrt{S_f}=194.853\ \text{m}^3/\text{s}
$$

#### Step 3: Correct for velocity head and eddy loss
$$
h_f=(h_1-h_2)+\left(\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right)-h_e,\qquad h_e=k_e\left|\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right|
$$
Velocity heads are taken at the u/s and d/s sections, and $h_e=k_c|h_{v1}-h_{v2}|$ with $k_c = 0.1$.

| 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 | 194.853 | 1.8426 | 2.0165 | 0.17304 | 0.20725 | 0.00342 | 0.31237 | 184.082 |
| 2 | 184.082 | 1.7407 | 1.9050 | 0.15444 | 0.18497 | 0.00305 | 0.31642 | 185.269 |
| 6 | 185.154 | 1.7509 | 1.9161 | 0.15625 | 0.18713 | 0.00309 | 0.31603 | 185.155 |

The trials converge to $h_f$ and $Q$ shown in the last row (trials 3 to 5 are not shown).

**Answer: Discharge $Q \approx 185.2\ \text{m}^3/\text{s}$.**

## Check yourself

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

1. The middle section has no gauge reading. What fall is used?
   A. Half of 0.35 m
   B. Zero
   C. 315.5 - 315.0
   D. 315.50 - 315.15 = 0.35 m between end sections
   Answer: D. Only the end sections have gauge readings so the fall is between them. (Key point: Fall used)

2. How is the mean conveyance of the three sections found?
   A. Geometric mean (K1 K2 K3)^(1/3)
   B. Sum of K
   C. Arithmetic mean
   D. Smallest K
   Answer: A. Km = (K1 K2 K3)^(1/3) = 5310.8. (Key point: Mean conveyance)

3. Since n differs at each section (0.025, 0.027, 0.029), how is K computed?
   A. n is ignored
   B. Each section's K uses its own n
   C. Using an average n for all sections
   D. Using the largest n only
   Answer: B. Each K = (1/n) A R^(2/3) with that section's own n. (Key point: Different roughness)

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