Chapter 9. Steady Flow in Open channels

Chapter 9

Steady Flow in O pen channels

Objectives

•

• Solve uniform open channel flow using the

Manning Equation

9.1 Uniform Flow in Open Channel

•

characterized by the presence of a liquid-gas interface called the free surface.

•

•

•

• Velocity is zero on bottom and sides of channel due to no- slip condition

• Velocity is maximum at the midplane of the free surface

• In most cases, velocity also varies in the streamwise direction

• Therefore, the flow is 3D

• Nevertheless, 1D approximation is made with good success for many practical problems.

The flow of water in a conduit may be either open channel flow or pipe flow. The two kinds of flow are similar in many ways but differ in

one important respect. Open-channel flow must have a free surface, whereas pipe flow has

none. A free surface is subject to atmospheric pressure. In Pipe flow there exist no direct

atmospheric flow but hydraulic pressure only.

differences between pipe flow and open

channel flow

Open channel flow is driven by gravity rather than by pressure work as in pipes.

9.2 Classification of Open-Channel Flows

The most common classification method is by rate of

change of free-surface depth. The classes are summarized as

1. Uniform flow (constant depth and slope) 2. Varied flow

a. Gradually varied (one-dimensional) b. Rapidly varied (multidimensional)

Classification of Open-Channel Flows

•

•

encountered in long straight sections where head loss due to friction is balanced by

elevation drop.

•

Classification of Open-Channel Flows

• Obstructions cause the flow depth to vary.

• Rapidly varied flow (RVF) occurs over a short distance near the obstacle.

• Gradually varied flow (GVF) occurs over larger distances and usually connects UF and RVF.

9.3 Properties of open channels

Artificial channels

These are channels made by man. They include irrigation canals, navigation canals, spillways, sewers, culverts and drainage

ditches. They are usually constructed in a regular cross-section shape throughout – and are thus prismatic channels (they don’t widen or get narrower along the channel.

Natural channels

Natural channels can be very different. They are not regular nor prismatic and their materials of construction can vary widely

(although they are mainly of earth this can possess many different properties.) The surface roughness will often change with time distance and even elevation.

Properties of open channels

Geometric properties necessary for analysis

Depth (y) – the vertical distance from the lowest point of the channel section to the free surface.

Stage (z) – the vertical distance from the free surface to an arbitrary datum

Area (A) – the cross-sectional area of flow, normal to the direction of flow

Wetted perimeter (P) – the length of the wetted surface measured normal to the direction of

flow.

Surface width (B) – width of the channel section at the free surface

Hydraulic radius (R) – the ratio of area to wetted perimeter (A/P) Hydraulic mean depth (Dm) – the ratio of area to surface width (A/B)

The wetted perimeter does not include the free

surface.

Multiple Choice

Consider an open rectangular channel 3m wide laid on a 1°slope. If the water depth is 2m, the hydraulic radius is:

(a)

(b)

(c)

(d)

•

•

•

9.4 Continuity and Energy Equations

g V R f L

h

2 4

=

9.5 Uniform Flow in Channels

•

•

•

•

Head loss in a Channel

„ Recall: Darcy-Weisbach Equation

¾ For circular pipes:

g V D f L h

2

=

¾ For non-circular conduits:

χ

R A g

V R f L

h

= =

2 4

2

Head loss in a Channel

g V R f L

L S

h

2 4

2

0

⋅ =

=

Solve for V:

2 / 1 0 2

/ 1 0

8

2 4

S f R

V g

f

g R

V S

⋅

⋅

=

=

Chezy Formula

„ For a given channel shape and roughness:

C Constant

f

g = =

8

„ This leads to the Chezy Formula

¾ Named after Antoine Chezy, who did experiments in the River Seine in 1760s

Ri C

RS C

V =

=

C——Chezy Coefficient

i —— the bed slope of the channel

Manning’s Analysis

Manning’s formula

„ Robert Manning (Irish engineer, 1880s)

¾ Found C increases with channel size:

6 /

1 1

n R C =

¾ Where n = a roughness coefficient

Values of Manning n

Lined Canals n

Cement plaster 0.011

Untreated gunite 0.016

Wood, planed 0.012

Wood, unplaned 0.013

Concrete, trowled 0.012

Concrete, wood forms, unfinished 0.015

Rubble in cement 0.020

Asphalt, smooth 0.013

Asphalt, rough 0.016

Natural Channels

Gravel beds, straight 0.025

Gravel beds plus large boulders 0.040 Earth, straight, with some grass 0.026 Earth, winding, no vegetation 0.030 Earth , winding with vegetation 0.050

How to get “n” ?

•

•

•

9.6 Computations in uniform flow

We can use Manning's formula for discharge to

calculate steady uniform flow. Two calculations are usually performed to solve uniform flow problems.

1. Discharge from a given depth 2. Depth for a given discharge

In steady uniform flow the flow depth is know as normal depth.

example 1

A concrete lined trapezoidal channel with uniform flow has a normal depth is 2m.The base width is 5m and the side slopes are equal at 1:2 Manning's n can be taken as 0.015

And the bed slope S0 = 0.001

What are:

a) Discharge (Q) b) Mean velocity (V)

c) Reynolds number (Re)

Discharge from depth in a trapezoidal channel

Calculate the section properties

the mean velocity

And the Reynolds number

This is very large - i.e. well into the turbulent zone - the application of the Manning's equation was therefore valid.

Solution:

Example: Manning Formula

•

1

2

3 m

1.5 m

Example

n = 0 012. S m

0 m

1 2

3000 0 0004

= . =

.

solution

2 / 1 0 3 /

1

S n AR

Q =

P m

R = A = 0 . 927

9 2

) 5 . 1 )(

3 ( ) 5 . 1 )(

3

( m m m m m

A = + =

2 2

2 (1.5 ) 9.71 )

3 ( 2 )

3

( m m m m

P = + + =

s m

Q

m m

Q

/ 3

. 14

) 0004 .

0 ( )

927 .

0 )(

9 012 ( .

0 1

3

2 / 1 3

/ 2 2

=

=

9.7 Best Hydraulic Cross Sections

•

specified cross section (or maximum hydraulic radius R_h)

• Also reflects economy of building structure with smallest perimeter

Best Hydraulic Cross Sections

• Example: Rectangular Channel

• Cross section area, A_c = yb

• Perimeter, p = b + 2y

• Solve A_c for b and substitute

• Taking derivative with respect to

• To find minimum, set derivative to zero

Best rectangular channel has a depth 1/2 of the width

Best Hydraulic Cross Sections

• Same analysis can be performed for a trapezoidal channel

• Similarly, taking the derivative of p with respect to q, shows that the optimum angle is

• For this angle, the best flow depth is

Multiple Choice

Consider an open rectangular channel 3m wide laid on a 1°slope. The most efficient water depth (best depth for a given flow and resistance ) is:

(a) 1.0m (b) 1.5m (c) 2.0m (d) 2.5m