Chapter 3: Steady uniform flow in open channels. Learning outcomes By the end of this lesson,...

Chapter 3:Steady uniform flow in open

channels

Learning outcomes

• By the end of this lesson, students should be able to:– Understand the concepts and equations used in

open channel flow

– Determine the velocity and discharge using Chezy’s & Manning’s equation

– Able to solve problems related to optimum cross section in both conduits and open channel

2UiTMSarawak/ FCE/ BCBidaun/

ECW301

Introduction • Comparison between full flow in closed conduit and

flow in open channel

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Full flow in closed conduit Open channel flow

No free surface and pressure in the pipe is not constant

Existence of free water surface through out the length of flow in the channel. Pressure at the free surface remains constant, with value equal to atmospheric pressure.

Flow cross sectional area remains constant and it is equal to the cross sectional area of the conduit (pipe).

Flow cross sectional area may change throughout the length depending on the depth of flow.

UiTMSarawak/ FCE/ BCBidaun/ ECW301


ECW301

Flow classification

• Turbulent flow:– Characterized by the random and irregular

movement of fluid particles.

– Movement of fluid particles in turbulent flow is accompanied by small fluctuations in pressure.

– Flows in open channel are mainly turbulent.

– E.g. Hydraulic jump from spillway, Flow in fast flowing river


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Flow classification• Laminar flow:

– Flow characterized by orderly movement of fluid particles in well defined paths.

– Tends to move in layers.

– May be found close to the boundaries of open channel.


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For flow in pipes


7

Flow in open channel


8


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Steady uniform flow

•Flow parameters do not change wrt space (position) or time.

•Velocity and cross-sectional area of the stream of fluid are the same at each cross-section.

•E.g. flow of liquid through a pipe of uniform bore running completely full at constant velocity.


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Steady non-uniform flow

• Flow parameters change with respect to space but remain constant with time.

• Velocity and cross-sectional area of the stream may vary from cross-section to cross-section, but, for each cross-section, they will not vary with time.

• E.g. flow of a liquid at a constant rate through a tapering pipe running completely full.


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Unsteady uniform flow• Flow parameters remain constant wrt space but

change with time.

• At a given instant of time the velocity at every point is the same, but this velocity will change with time.

• E.g. accelerating flow of a liquid through a pipe of uniform bore running full, such as would occur when a pump is started.


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Unsteady non-uniform flow

• Flow parameters change wrt to both time & space.

• The cross-sectional area and velocity vary from point to point and also change with time.

• E.g. a wave travelling along a channel.


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Flow classification

• Normal depth – depth of flow under steady uniform condition.

• Steady uniform condition – long channels with constant cross-sectional area & constant channel slope.– Constant Q

– Constant terminal v

– Therefore depth of flow is constant (yn or Dn)


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Total energy line for flow in open channel


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• From the figure :– Water depth is constant

• slope of total energy line = slope of channel

• When velocity and depth of flow in an open channel change, then non-uniform flow will occur.


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Flow classification

• Non uniform flow in open channels can be divided into two types:

– Gradually varied flow, GVF • Where changes in velocity and depth of flow take place

over a long distance of the channel

– Rapidly varied flow, RVF• Where changes in velocity and depth of flow occur over

short distance in the channel


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Non-uniform flow


19


20


21


22

Total energy line for flow in open channel


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Continuity equation:

• For rectangular channel:

• Express as flow per unit width, q:

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2211

21

vAvA

QQ

222111

2211

21

vDBvDB

vAvA

QQ

vDB

vBD

B

Qq


Momentum equation

• Produced by the difference in hydrostatic forces at section 1 and 2:

• Resultant force,

25

)( 12 vvQF

22

11

gA2section at motion ofdirection opposingin Force

gA1section at motion ofdirection in Force

x

x

g

vvQxx

vvQxx

122211

122211

AA

AAg

momentum of change of Rate Force


Energy equation

• But hydrostatic pressure at a depth x below free surface,

• Therefore,

• Energy equation rewritten as,


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)(2

2

xdzg

v

g

pH

g

px

dzg

vH

2

2

Lhdzg

vdz

g

v 22

22

11

21

22

Energy equation

• For steady uniform flow,

• Therefore the head loss is,

• And energy equation reduces to,

• Known as specific energy, E, (total energy per unit weight measured above bed level),

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21

21

vv

dd

21 zzhL

2

22

1

21

22d

g

vd

g

v

Dg

vE

2

2


Geometrical properties of open channels

• Geometrical properties of open channels:– Flow cross-sectional area, A

– Wetted perimeter, P

– Hydraulic mean depth, m


ECW301

• A – covers the area where fluid takes place.

• P – total length of sides of the channel cross-section which is in contact with the flow.

BDA

29

D

B

DBP 2

DB

BD

P

Am

2


Example 3.1

Determine the hydraulic mean depth, m, for the trapezoidal channel shown below.


ECW301


31

miCv 21

321im

nv

Chezy’s coefficient, C• From chapter 1,

• Rearranging to fit for open channel, the velocity:

• Where Chezy roughness coefficient,

• For open channel i can be taken as equal to the gradient of the channel bed slope s. Therefore,

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g

v

m

fLhL 2

2

miCv

f

gC

2

msCv


Manning’s n• Introduced roughness coefficient n of the

channel boundaries.


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21

321im

nv

Example 3.2

Calculate the flow rate, Q in the channel shown in Figure 3.5, if the roughness coefficient n = 0.025 and the slope of the channel is 1:1600.


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Example 3.3• Determine the flow velocity, v and the flow rate for

the flow in open channel shown in the figure. The channel has a Manning’s roughness n = 0.013 and a bed slope of 1:2000.


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B

2.75 m

900

Example 3.4 (Douglas, 2006)

• An open channel has a cross section in the form of trapezium with the bottom width B of 4 m and side slopes of 1 vertical to 11/2 horizontal. Assuming that the roughness coefficient n is 0.025, the bed slope is 1/1800 and the depth of the water is 1.2 m, find the volume rate of flow Q using

a. Chezy formula (C=38.6)

b. Manning formula36


Example 3.5 (Munson, 2010)

• Water flows along the drainage canal having the properties shown in figure. The bottom slope so = 0.002. Estimate the flow rate when the depth is 0.42 m .


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Example 3.6 (Bansal, 2003)

Find the discharge of water through the channel shown in figure. Take the value of Chezy’s constant = 60 and slope of the bed as 1 in 2000.

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C

A D

B

E

2.7 m

1.2 m

1.5 m


Example 3.8 (Bansal, 2003)Find the diameter of a circular sewer pipe which is laid at a slope of 1 in 8000 and carries a discharge of 800 L/s when flowing half full. Take the value of Manning’s n = 0.020.

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d

D


Optimum cross-sections for open channels

• Optimum cross section – producing Qmax for a given area, bed slope and surface roughness, which would be that with Pmin and Amin therefore tend to be the cheapest.

• Qmax : Amin, Pmin


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Example 3.9Given that the flow in the channel shown in figure is a maximum, determine the dimensions of the channel.


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Optimum depth for non-full flow in closed conduits

• Partially full in pipes can be treated same as flow in an open channel due to presence of a free water surface.


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• Flow cross sectional area,

• Wetted perimeter,

43

2sin2

1

cossin22

1

OMP triangle-OMNPsector

2

22

r

rr

A

rP 2



• Under optimum condition, vmax,

• Substituting & simplifying,


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d

dPA

d

dAP

d

dPA

d

dAP

PdPAd

0

12

05.257

2tan2


• Hence depth, D, at vmax,

• Using Chezy equation,


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d

rD

81.0

62.1

212

13

21

21

iP

AC

iACmQ


• Qmax occurs when (A3/P) is maximum,

• Substituting & simplifying,

• Therefore depth at Qmax,


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d

dPA

d

dAP

d

dPA

d

dAPA

PdP

Ad

3

031 32

2

3

0

0

154

3082

d

rD

95.0

9.1

Review of past semesters’ questions

OCT 2010

• Analysis of flow in open channels is based on equations established in the study of fluid mechanics. State the equations.


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OCT 2010

• Determine the discharge in the channel (n = 0.013) as shown in Figure Q3(b). The channel has side slopes of 2 : 3 (vertical to horizontal) and a slope of 1 : 1000. Determine also the discharge if the depth increases by 0.1 m by using Manning's equation.


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OCT 2010

• State the differences between :

i) Steady and unsteady flow

ii) Uniform and non-uniform flow


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OCT 2010

• Figure Q4(b) shows the channel. Prove that for a channel, the optimum cross-section occurs when the width is 4 times its depth (B = 4D). (Hint: A = 4/3BD and P = B + 4D)


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Chapter 3: Steady uniform flow in open channels. Learning outcomes By the end of this lesson,...

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