28
CHE 493: FLUID MECHANICS Chapter 9: Flow in Open Channels 1

Chapter 9 Flow in Open Channel

Embed Size (px)

DESCRIPTION

fluid mechanic

Citation preview

Page 1: Chapter 9 Flow in Open Channel

CHE 493: FLUID MECHANICS

Chapter 9: Flow in Open Channels

1

Page 2: Chapter 9 Flow in Open Channel

Learning outcome

1. Explain the concept of uniform flow

2. Describe velocity transmission of a wave

3. Describe and calculate hydraulic jump phenomenon

2

Page 3: Chapter 9 Flow in Open Channel

Introduction

• Open channel flow implies flow of liquids inchannels open to the atmosphere or in partiallyfilled conduits

• Characterized by the presence of a liquid-gasinterface called the free surface

• Most of natural flows encountered in practiceare open-channels flow

• Eg: Rivers, floods, draining of rainwater throughroofs, highways

3

Page 4: Chapter 9 Flow in Open Channel

Uniform/varied flow• Uniform flow - if the flow depth (average velocity)

remains constant

• Encountered in long straight sections of channels withconstant slope and cross section – the liquidaccelerates until the head loss due equals theelevation drop – reaches terminal velocity – uniformflow is established

• Remains uniform as long as the slope, cross sectionand surface roughness of the channel remainunchanged

4

Page 5: Chapter 9 Flow in Open Channel

Uniform/varied flow

• Flow depth is called the normal depth –important characteristic for open-channelflows

• Non-uniform/varied flow - Flow depth varieswith distance in the flow direction

5

Page 6: Chapter 9 Flow in Open Channel

Why are open-channel flows important?

• Many natural systems responsible for the transport of sediment are channelized, in both sub- aerial and subaqueous environments.

• Nearly all of the modeling performed on the entrainment and transport of sediment is either in open channels or in 1-D boundary layers.

6

Page 7: Chapter 9 Flow in Open Channel

Uniform Flow in Channel

7

Page 8: Chapter 9 Flow in Open Channel

Head loss = Elevation Loss

Flow depth = y

Average flow velocity = V

Bottom slope = S0 = tan α

During open channel, Head Loss = Elevation Drop

--------------- (1)

Since hL = S0L and Dh = 4Rh ------------------------- (2)

Sub (2) in (1): ------------------------ (3)

8

Page 9: Chapter 9 Flow in Open Channel

• Rearrange (3), uniform flow velocity:

where

• Flow rate:

Chezy CoefficientAntoine Chezy (1718-1798)

hoc RSCAQ

Note: Determine using Moody chart,

open channel typically is

turbulent flow and fully develop.

9

Page 10: Chapter 9 Flow in Open Channel

Gauckler and Manning made recommendations:

Where: a = dimensional constant = 1 m1/3/s

n = Manning coefficient (depends on roughness of the channel surface)

For uniform flow velocity & flow rate:

and 2/13/2

oc SRAn

aQ

h

10

Page 11: Chapter 9 Flow in Open Channel

Mean value for Manning coefficient

11

Page 12: Chapter 9 Flow in Open Channel

Types of Channels

Hydraulic radius

Circular channel

Rectangular channel

Trapezoidal channel

12

Page 13: Chapter 9 Flow in Open Channel

Example 1

Water is flowing in a weedy excavated earthchannel of trapezoidal cross-section with a bottomwidth of 0.8m, trapezoid angle of 60˚ and abottom slope angle of 0.3˚. If the flow depth ismeasured to be 0.52 m, determine the flow rate ofwater through the channel. (Given n = 0.030)

13

Page 14: Chapter 9 Flow in Open Channel

Specific energy

Consider flow of a liquid in a channel

Where:

y - flow depth

V - average velocity

Z – elevation of the bottom of channel at that location relative to some reference datum

14

Page 15: Chapter 9 Flow in Open Channel

Total mechanical energy in terms of head:

• Not realistic representing true energy

• It can be realistic if the reference datum is taken to be the bottom of the channel so Z = 0

• Then, the total mechanical energy = Pressure + Dynamic Head

• This term is called specific energy, Es

- ------------------------ (1)

15

Page 16: Chapter 9 Flow in Open Channel

• Consider flow in an open channel of constant width, b.

• Volume flowrate: .

• So, the average flow velocity

--------------- (2)

• Sub (2) into (1)

ybVVAQ c

yb

Q

A

QV

c

22

2

2 bgy

QyEs

16

Page 17: Chapter 9 Flow in Open Channel

There is minimum specific energy Es,min required to support specific flow rate, Q

Therefore, Es cannot be below Es,min for a given Q

So,

17

Page 18: Chapter 9 Flow in Open Channel

Critical flow depth

Critical velocity

To find character and flow, using Froude Number

Lc = Critical LengthFr < 1 = Subcritical or tranquil flowFr = 1 = Critical flowFr > 1 = Supercritical or rapid flow

3/1

2

2

gb

Qyc

18

Page 19: Chapter 9 Flow in Open Channel

Example 2

Water is flowing steadily in a 0.65 m widerectangular open channel at a rate of 0.25 m3/s.If the flow depth is 0.15 m, determine

(a) The flow velocity and type of flow

(b) The alternate flow depth (Es1=Es2 ) if thecharacter of flow were to change

19

Page 20: Chapter 9 Flow in Open Channel

Hydraulic jump

• It called rapidly varied flow (RVF) if the flow depthchanges markedly over a relatively short distance inthe flow direction.

• Occur when there is a sudden change in flow, such asan abrupt change in cross section.

• RVF is complicated—since there will be affect ofbackflow and flow separation.

20

Page 21: Chapter 9 Flow in Open Channel

Hydraulic jump• In compressible flow, a liquid can accelerate from

subcritical to supercritical flow

• It can also decelerate from supercritical tosubcritical flow by undergoing a shock which isknown as hydraulic jump

• Hydraulic jump involves considerable mixing andagitation and thus significant amount of mechanicalenergy dissipation

21

Page 22: Chapter 9 Flow in Open Channel

22

Hydraulic jump formed on a spillway model

for the Karnafuli Dam in East Pakistan.

Classification of hydraulic jumps: (a) Fr =

1.0 to 1.7:undular jumps; (b) Fr= 1.7 to

2.5:weak jump; (c) Fr= 2.5 to 4.5:

oscillating jump; (d) Fr=4.5 to 9.0:

steady jump; (e) Fr= 9.0: strong

jump.

Page 23: Chapter 9 Flow in Open Channel

Assumption from figure:

• Velocity is nearly constant across the channel at section 1 & 2 –therefore the momentum flux correction factors β1 = β2

• Pressure in the liquid varies hydrostatically, we consider gage pressure only since atmospheric pressure acts on all surfaces and its effect cancel out.

• The wall shear stress and associated losses negligible relative to the losses that occur during the hydraulic jump due to intense agitation.

• The channel is wide and horizontal

• No external or body forces23

y2

ρgy2ρgy1

y1

(1) (2)

hL

Energy line

Control

volume

v1 v2

Consider steady

flow through a

control volume

that encloses

the hydraulic

jump

x

Page 24: Chapter 9 Flow in Open Channel

• From momentum equation

• For channel width b

24

Page 25: Chapter 9 Flow in Open Channel

• Substituting and simplifying:

• Eliminating V2 by from the continuity equation gives:

• Canceling factor y1 – y2 from both side and rearranging gives:

where

25

Page 26: Chapter 9 Flow in Open Channel

• Therefore, depth ratio:

• The energy equation for this horizontal flow section can be expressed as:

• Noting that;

and

• The head loss associated with hydraulic jump is expressed as:

26

Page 27: Chapter 9 Flow in Open Channel

• The specific energy of the liquid before the hydraulic jump is

• Then , the energy dissipation ratio:

27

Page 28: Chapter 9 Flow in Open Channel

Example 3

Water is discharged into a 8 m wide rectangularhorizontal channel from a sluice gate is observed tohave undergone a hydraulic jump. The flow depth andvelocity before the jump are 0.8 m and 7 m/srespectively. Determine:

(a) The flow depth and the Froude number after thejump

(b) The head loss and the dissipation ratio

(c) The wasted power production potential due to thehydraulic jump

28