---
title: "Boundary shear stress and compound channel flow"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:shear-stress-boundary-compound-channel-flow"
source: https://www.bctengineer.com/q/semester_4/hydraulics/shear-stress-boundary-compound-channel-flow
site: BCT Engineering
---

# Boundary shear stress and compound channel flow

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2077 Chaitra, 4+4 marks
- Video: [Open channel flow-6 | shear stress | Shubham sarathe #fluidmechanics](https://www.youtube.com/watch?v=DnjAme2M3JA) — EIQ
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/shear-stress-boundary-compound-channel-flow
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

Derive the equation of shear stress on the boundary of the open channel. Water flows in a channel whose bottom slope is 0.002 and whose cross section is as shown in figure below. The dimensions and the Manning's coefficients for the surfaces of different subsections are also given on the figure. Determine the flow rate through the channel and the effective Manning coefficient for the channel. [Figure: compound channel cross-section with subsections; the figure was not included in the scan]

## Short answer

Boundary shear stress τ₀=γRS₀ is derived from force equilibrium on a reach. For compound channels, divide into subsections, compute Q_i=(1/n_i)A_iR_i^{2/3}S₀^{1/2}, sum for total Q, then effective Manning n_e = (A R^{2/3}S₀^{1/2})/Q. With assumed dimensions, Q=44.67 m³/s and n_e=0.0295.

## Answer

**The figure with dimensions and Manning's coefficients was not available. The method is given in full and applied to an assumed section with stated values, so that the numbers can be replaced by those of the figure.**

#### Part 1: Boundary shear stress equation
Consider a uniform flow in a length $L$ of a prismatic channel of area $A$, wetted perimeter $P$ and bed slope $S_0$ (angle $\theta$).

```text
  ----------------------------  WSL
  \  W sin(th) ->  /         
   \ ___ control volume / tau0 on boundary
```
Forces along the flow direction (pressure forces at the two ends are equal since depth is constant):
- Gravity component: $W\sin\theta=\gamma AL\sin\theta\approx\gamma AL\,S_0$
- Boundary resistance: $\tau_0\,P\,L$

For uniform flow (no acceleration):
$$
\gamma AL\,S_0=\tau_0PL\;\Rightarrow\;\boxed{\tau_0=\gamma RS_0}
$$
where $R=A/P$. This is the average shear stress on the boundary.

#### Part 2: Compound channel (method)
1. Divide the section by vertical lines into subsections (main channel and flood plains); the dividing lines are not counted in wetted perimeters.
2. For each subsection $i$: $A_i$, $P_i$, $R_i=A_i/P_i$ and $Q_i=\dfrac{1}{n_i}A_iR_i^{2/3}S_0^{1/2}$.
3. Total discharge $Q=\sum Q_i$.
4. Effective (equivalent) Manning coefficient from the whole section, treated as a single section:
$$
n_e=\frac{A\,R^{2/3}S_0^{1/2}}{Q},\quad A=\sum A_i,\quad P=\sum P_i,\quad R=\frac AP
$$

#### Illustration with assumed data
Main channel: bed 6 m wide, 2 m deep, $n=0.025$. Flood plains: 8 m wide each side, water depth 1 m over the plain, $n=0.040$, bank vertical. $S_0=0.002$. (Assumed.)

| Subsection | $A$ (m²) | $P$ (m) | $R$ (m) | $n$ | $Q$ (m³/s) |
|---|---|---|---|---|---|
| Main channel | 12 | 8 | 1.500 | 0.025 | 28.13 |
| Flood plain (each) | 8 | 9 | 0.889 | 0.040 | 8.27 |
| Whole section | 28 | 26 | 1.077 | 0.0295 | 44.67 |


**Answer (for the assumed section): $Q=44.67$ m³/s and $n_e=0.0295$.** For the actual figure, repeat steps 1-4 with the given dimensions and $n$ values.

## Check yourself

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

1. What is the average boundary shear stress in uniform flow?
   A. tau0 = gamma R S0
   B. tau0 = gamma y
   C. tau0 = gamma A S0
   D. tau0 = rho V
   Answer: A. Balancing gamma A L S0 = tau0 P L gives tau0 = gamma R S0. (Key point: Boundary shear)

2. Which forces balance in the derivation?
   A. Pressure force at the ends against friction
   B. Weight component along the slope against boundary resistance
   C. Inertia against gravity
   D. Surface tension against weight
   Answer: B. End pressure forces cancel for constant depth, so gravity component equals tau0 P L. (Key point: Force balance)

3. How are subsections of a compound channel treated for wetted perimeter?
   A. Interface lines are included
   B. Only the bed is counted
   C. Only the flood plain is counted
   D. Interface lines are excluded
   Answer: D. Dividing vertical lines are not counted as wetted perimeter. (Key point: Subsections)

4. How is the effective Manning coefficient found?
   A. Average of the n values
   B. From whole-section A, P, R and total Q: n_e = A R^(2/3) S0^(1/2)/Q
   C. Largest n value
   D. Smallest n value
   Answer: B. Treat the whole section as one with total Q, then solve Manning's equation for n. (Key point: Effective n)

## Related in Hydraulics

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