CHAPTER 1:TYPES OF FLOW IN OPEN CHANNEL

HYDRAULICS(BFC 2072 / BFC 21103)

Prepared by:-

MR WAN AFNIZAN BIN WAN MOHAMED

DEPT. OF WATER & ENVIRONMENTAL ENGINEERING

FAC. OF CIVIL & ENVIRONMENTAL ENGINEERING

e-mail: [email protected]

**Some part of the lecture notes are editted from Mrs

Zarina Md Ali

mailto:[email protected]




CHAPTER 1

TYPES OF FLOW IN OPEN CHANNEL

Learning Objectives

To know open channel classification and fluid characteristics.

To understand state of flow in open channel.

To understand geometric elements definition in open flow calculation.

CONTENT

☻ UNIFORM & NON-UNIFORM FLOW

☻ STEADY & NON-STEADY FLOW

☻ LAMINAR & TURBULENT FLOW

CHANNEL GEOMETRY

INTRODUCTION

STATE OF FLOW

TYPES OF FLOW

INTRODUCTION

Hydraulic is related to the principle of fluid

mechanics usage for structure of water

engineering, civil and environmental

engineering necessity especially hydraulic

structure example channel, river, weir dan

water treatment plant.

INTRODUCTION

... Con’t Properties of Fluid

Surface Tension

Viscosity

Bulk Modulus

Density

Specific Gravity

Capillarity

Compressibility

Relative Density

Properties

of Fluid

Open Channel Flow

INTRODUCTION... Con’t

Flow in open channel has a free surface which is caused by atmosphere pressure

Atmosphere Free surface

Flow

DatumZo

dDistribute

d flow

Open Channel Flow

INTRODUCTION... Con’t

Stormwater Management and Road Tunnel

(SMART), Kuala Lumpur, Malaysia

Tahan river

rapids

Siberian meandering

river

Chapter 1. Flow in Open Channel

UNIFORM

&

NON-UNIFORM

FLOW

Practical applications are the determination of:

a. flow depth in rivers, canals and other conveyance conduits,

b. changes in flow depth due to channel controls e.g. weirs,

spillways, and gates,

c. changes in river stage during floods,

d. surface runoff from rainfall over land,

e. optimal channel design, and

f. others

• Depth of flow is the same at every section

of the flow (dy/dx = 0).

TYPES OF FLOW

Uniform flow Non-uniform flow

UNIFORM FLOW

NON-UNIFORM FLOW

•Depth of flow varies at different sections of

the flow (dy/dx 0).

TYPES OF FLOW

NON-UNIFORM FLOW

Can be divided into two :-

... Con’t

1. Gradually Varied Flow (GVF)

The depth changes gradually over a long

distance along the length of a channel.

2. Rapidly Varied Flow (RVF)

The depth changes abruptly over a

comparative short distance

TYPES OF FLOW

NON-UNIFORM FLOW

Gradually Varied Flow (GVF)

... Con’t

Example: backwater at the upstream of

weir or sluice gate.

TYPES OF FLOW

NON-UNIFORM FLOW

Rapidly Varied Flow (GVF)

... Con’t

Example: hydraulic jump

STEADY

&

UNSTEADY

FLOW

Depth of flow (y) does not change and

assumed to be constant during the time

interval; dy / dt = 0.

Example: Constant flow in piping flow in

time

TYPES OF FLOW

STEADY FLOW

TYPES OF FLOW

STEADY FLOW

... Con’t

TYPES OF FLOW

UNSTEADY FLOW

... Con’t

Water surface and depth of flow fluctuate

from time to time. This means, dy / dt 0.

Unsteady uniform flow is an impossible

condition.

Example : Flood

Open Channel

Flow

Steady Flow Unsteady Flow

Uniform FlowNon-uniform

FlowUniform Flow

Non-uniform

Flow

GVUF

RVUF

GVF

RVF

In a nutshell …..

STATE OF FLOW

Can be classified according to :-

Viscosity effect

Gravity effect

STATE OF FLOW

Reynolds number depends on viscosity

Fluid viscosity is constant

Re < 500 - Laminar

500 < Re < 12500 - Transition

Re > 12500 - Turbulent

In term of viscocity effect :-

υ

vR4Re =

In open channel flow :-

... Con’t

STATE OF FLOW... Con’t

Laminar flow

Low velocity in small

cross section channel.

High viscosity.

Characteristics of laminar & turbulent flow

Transition

State of flow happen when laminar flow change to turbulent flow

before alter to fully turbulent.

Turbulent flow

Deep of flow.

Non-prismatic cross

section (i.e; river)

create the turbulent.

STATE OF FLOW

In term of gravity effect :-

... Con’t

State as inertia

force to gravity

force ratio and

known as Froude

number. In open channel flow :-

gD

vFr =

Fr = 1 - Critical

Fr < 1 - Sub critical

Fr > 1 - Super critical

STATE OF FLOW... Con’t

Combination of viscosity and gravity

effect will create 4 state of regime which

are:-

Sub critical laminar – Fr < 1.0, Re < 500

Super critical laminar – Fr > 1.0, Re < 500

Sub critical turbulent – Fr < 1.0, Re > 12500

Super critical turbulent – Fr > 1.0, Re > 12500

CHANNEL GEOMETRY

TYPES OF OPEN CHANNEL

Natural channel

Artificial channel

Prismatic channel

... Con’t

NATURAL @ EARTH CHANNEL

Natural channels are waterways that exist

naturally on earth.

The properties of natural channels are

normally very irregular and difficult to control.

Examples: hillside rivulets, brooks, streams,

rivers and tidal estuaries.

CHANNEL GEOMETRY

Three types of channel :-

A NATURAL CHANNEL

Figure 1.1 : Natural channel

CHANNEL GEOMETRY

... Con’t

ARTIFICIAL CHANNEL

Artificial channels are watercourses that are

contracted and developed by human.

The properties of artificial channels are based on

developers’ requirements and controllable.

Examples: canal, flume, chute, drop,culvert or

open channel-flow tunnel.

CHANNEL GEOMETRY

.... Cont ‘

B ARTIFICIAL CHANNEL

Figure 1.2 : Artificial channel

CHANNEL GEOMETRY

... Con’t

ARTIFICIAL CHANNEL

It is a long and mild-sloped channel built on the

ground.

Exist in earth surface or coated with concrete,

cement, wood or others.

EXAMPLE : CANAL

CHANNEL GEOMETRY

Griboyedov Canal, St. Petersburg, Russia

Terusan Wan Muhammad Saman, Kedah

EXAMPLE : CANAL

... Con’t

ARTIFICIAL CHANNEL

EXAMPLE : FLUME

It is a hard channel or passage built from

wood, metal or concrete across a depression.

It is used in laboratory for research.

CHANNEL GEOMETRY

This flume diverts water from White River,

Washington to generate electricity Bull Run Hydroelectric Project diversion flume

Open-channel flume in laboratory

EXAMPLE : FLUME

... Con’t

ARTIFICIAL CHANNEL

EXAMPLE : CHUTE & DROP

Chute is a steep-sloped channel.

Drop is similar to a chute, but the change in

elevation takes place over a very short distance.

CHANNEL GEOMETRY

Natural chute (falls) on the left and man-made logging chute on the right

on the Coulonge River, Quebec, Canada

Chute - is a channel having steep slopes.

Drop - is similar to a chute, but the change in

elevation is within a short distance. The spillway of Leasburg Diversion Dam is

a vertical hard basin drop structure

designed to dissipate energy

... Con’t

ARTIFICIAL CHANNEL

EXAMPLE : SEWERAGE

It is a closed conduit and built under soil

surface.

The water is in half-fully flow.

CHANNEL GEOMETRY

.... Cont ‘

C PRISMATIC CHANNEL

Uniform cross section & slope at whole channel

length.

Usually artificial channel.

CHANNEL GEOMETRY ELEMENT

y = Depth of water (m)T = Top width water surface (m)B = Base width water surface (m)P = Wetted perimeter (m)A = Wetted area (m2)R = Hydraulic radius R = A/P (m)D = Hydraulic depth D = A/T (m)

CHANNEL GEOMETRY

CHANNEL GEOMETRIC ELEMENT

T

B

Py

Figure 1.3 : Channel geometric element

CHANNEL GEOMETRY.... Cont ‘

.... Cont ‘


Figure 1.4 : Channel’s sides slope

I SIDES SLOPE, Z

1

z

Note : If slope, = 45 z = 1

( z at left & right side is same )

CHANNEL GEOMETRY

.... Cont ‘


Figure 1.5 : Channel’s sides slope

II CHANNEL SLOPE, So ( unit less )

CHANNEL GEOMETRY

.... Cont ‘


III RELATIONSHIP BETWEEN v & Q

AVQ …. 9.5

where ;

Q = Discharge or flow rate (m3/s)

v = Velocity (m/s)

CHANNEL GEOMETRY

.... Cont ‘


DERIVATION OF CHANNEL FORMULA

B

y

1

z

1

z

31

2

L

Figure 1.6 : Derivation of channel formula

CHANNEL GEOMETRY

.... Cont ‘

From Figure 1.6 :

T = Top width water surface

zy2BT

A = Wetted area

A = Area 1 + Area 2 + Area 3

2zyByA

y zy 2

1y By zy

2

1A

CHANNEL GEOMETRY

.... Cont ‘


Con’t …. From Figure 1.6 :

P = Wetted perimeter

2

22

222

22

z1yL

z1yL

yzyL

zyyL

CHANNEL GEOMETRY

.... Cont ‘


Con’t …. From Figure 1.6 :

P = Wetted perimeter

2z1y2BP

L2BP

therefore ;

Note :Use this trapezoidal formula (A, T & P) to find formulae for rectangular & triangular shape.

For Rectangular z = 0 for Triangular B = 0

CHANNEL GEOMETRY

.... Cont ‘

SHAPE A T P

By B B + 2y

zy2 2zy

By + zy2 B + 2zy

B

y

T

zz11 y

T

y1z

T

B

1z

d

T

y

2z+1y2

2z+1y2+B

)θsinθ(8

d2

2

θsind

2

dθ

( in radian ) ( in radian )( in angle )

To sum up .. Table 1.1 : Channel’s geometric elements

CHANNEL GEOMETRY

Based on the figure given find :-

i) Top width water surface (T), wetted area (A), wetted perimeter (P) & hydraulic radius (R).

ii) If Q = 2.4 m3/s, determine the flow state.

iii) If inclined length (L) = 50 m, find the cost to construct this channel (Given excavation cost = RM 3/m3 and lining cost = RM 5/m2)

EXAMPLE 1.1

3 m

2 m

1 m

60

.... Cont ‘

SOLUTION:

Given:-

B = 3 m

y = 2 m

t = 3 m ( channel height)

SOLUTION

Find z value first :

60z

1

5774.060tan

1z

z

160tan

Therefore ;

(a) Top width water surface, T

m 3096.5T

)2)(5774.0)(2(3T

zy2BT

(b) Wetted area, A

SOLUTION .... Cont ‘

2

2

2

m 3096.8A

)2)(5774.0()2)(3(A

zyByA

(c) Wetted perimeter, P

m 6189.7P

)5774.0(1)2)(2(3P

z1y2BP

2

2

(d) Hydraulic radius, R


(e) State of flow

m 091.1R

6159.7

3096.8R

P

AR

gD

vFr


(ii) Con’t …. State of flow

Find v & D first :

m/s 2888.03096.8

4.2v

A

Qv

m 5650.1D

3096.5

3096.8D

T

AD



Thus ;

flow critical Sub 1 074.0F

)565.1)(81.9(

2888.0F

gD

vF

r

r

r



Thus ;

flow critical Sub 1 074.0F

)565.1)(81.9(

2888.0F

gD

vF

r

r

r

(iii) Construction cost

Construction cost includes :-

(a) Excavation cost

(b) Lining cost


(iii) Con’t …. Construction cost

(a) Excavation cost

3

2

2

m 709.83 volume Excavation

(50) )3( )5774.0()3( )3( volume Excavation

(L) ztBt volume Excavation

ALvolume Excavation

Therefore ;

2129.49 RMcost Excavation

m 83.709m

3 RMcost Excavation

3

3


(iii) Con’t …. Construction cost

(b) Lining cost

Therefore ;

2

2

2

m 496.42 area Lining

)50( (0.5774)1(2)(3)3 area Lining

)L( z1t2B area Lining

PLarea Lining

2482.09 RMcost Lining

m 42.496m

5 RMcost Lining

2

2


(iii) Con’t …. Construction cost or overall cost

Hence ;

4611.58 RM cost onConstructi

2482.09 RM 2129.49 RM cost onConstructi

cost Lining cost Excavation cost onConstructi

TIME’S UP …

THANK YOU

1.6 Assignment No. 1 (due date January 10, 2011)

Q1. [Final Exam Sem. 1, Session 2010/2011]

Justify the difference between:

(a) uniform flow and nonuniform flow

(b) state of flow using Reynolds number Re and Froude number Fr


(a) What is:

(i) Wetted perimeter

(ii) Gradually varied flow

(iii) Non-uniform flow

(iv) Froude number

(b) Explain the differences between canal and sewer.


What is:

(a) Reynolds number

(b) Froude number

(c) Hydraulic radius

(d) Prismatic channel

(e) Uniform flow

Q4. A discharge of 16.0 m3/s flows with a depth of 2.0 m in a rectangular

channel of 4.0 m wide. Determine the state of flow based on (i) Froude

number, and (ii) Reynolds number. Determine the flow regime.

Q5. A triangular channel of apex angle 120 carries a discharge of 1573 L/s.

Calculate the critical depth.

Documents

CHAPTER 1:TYPES OF FLOW IN OPEN CHANNEL