1

Chapter 3Chapter 3

Uniform FlowUniform Flow

2

3.1 INTRODUCTION3.1 INTRODUCTION A flow is said to be uniform if its properties remain constant

with respect to distance. As mentioned earlier, the term uniform flow in open channels is understood to mean steady uniform flow. The depth of flow remains constant at all sections in a uniform flow (Fig. 3.1). Considering two sections 1 and 2, the depths

and hence Since , it follows that in uniform flow

. Thus in a uniform flow, the depth of flow, area of cross-section and velocity of flow remain constant along the channel. The trace of the water surface and channel bottom slope are parallel in uniform flow (Fig.3.1)

021 yyy

021 AAA constantAVQ

VVV 21

3

3.2 CHEZY EQUATION3.2 CHEZY EQUATION By definition there is no acceleration in uniform flow. By ap

plying the momentum equation to a control volume encompassing sections 1 and 2, distance L apart, as shown in Fig. 3.1,

(3.1) 1221 sin MMPFWP f

4

where and are the pressure forces and and are the momentum fluxes at section 1 and 2 respectively

= weight of fluid in the control volume and = shear force at the boundary.

Since the flow is uniform,

Also, where = average shear stress on the wetted perimeter of

length and = unit weight of water. Replacing by (= bottom slope), Eq. (3.1) can be written as

or (3.2)

1P 2P 1M2M

WfF

2121 MMandPP PLFandALW f 0

0P

sin 0SPLALS 00

000 RSSP

A

5

where is defined as the hydraulic radius. is a length parameter accounting for the shape of the chann

el. It plays a very important role in developing flow equations which are common to all shapes of channels.

Expressing the average shear stress as , where =a coefficient which depends on the nature of the

surface and flow parameters, Eq. (3.2) is written as

leading to (3.3)

PAR R

0 20 Vk

02 RSVk

k

0RSCV

6

where = a coefficient which depends on the nature of the surface and the flow. Equation (3.3) is known

as the Chezy formula after the French engineer Antoine Chezy, who is credited with developing this basic simple relationship in 1769. The dimensions of are and it can be made

dimensionless by dividing it by . The coefficient is known as the Chezy coefficient.

kC

1

C 121 TL

g

7

3.3 DARCY-WEISBACH FRICTION3.3 DARCY-WEISBACH FRICTION FACTOR f FACTOR f Incompressible, turbulent flow over plates, in pipes and du

cts has been extensively studied in the fluid mechanics discipline. From the time of Prandtl (1875- 1953) and Von karman (1881 一 1963) research by numerous eminent investigators has enabled considerable understanding of turbulent flow and associated useful practical applications. The basics of velocity distribution and shear resistance in a turbulent flow are available in any good text on fluid mechanics .

Only relevant information necessary for our study is summed up in this section.

2,1

8

Pipe Flow A surface can be termed hydraulically smooth, rough or in t

ransition depending on the relative thickness of the roughness magnitude to the thickness of the laminar sub-layer. The classification is as follows:

where =sand grain roughness, = shear velocity and = kinematic viscosity.

lsmooth wal-llyhydraulica4* v

vs

regime altransition604 * v

vs

flowrough full60* v

vs

s 00* gRSv v

9

For pipe flow, the Darcy-Weisbach equation is

(3.4)

where = head loss due to friction in a pipe of diameter and length ; = Darcy-Weisbach friction factor. For smooth pipes, is found to be a

function of the Reynolds number only. For rough turbulent flows, is a function of the relati

ve roughness and type of roughness and is independent of the Reynolds number. In the transition regime, both the Reynolds number and relative roughness play important roles. The roughness magnitudes for commercial pipes are expressed as equivalent sand-grain roughness .

g

V

D

Lfh f 2

2

fhD L f

f

v

VDRe

f Ds

)( s

10

The extensive experimental investigations of pipe flow have yielded the following generally accepted relations for the variation of in various regimes of flow:

1. For smooth walls and

(Blasius formula) (3.5)

2. For smooth walls and

(karman-Prandtl equation) (3.6)

f

510Re

41Re

316.0f

510Re

8.0Relog0.21

ff

11

3.For rough boundaries and

(Karman-Prandtl equation) (3.7) 4. For the transition zone

(Colebrook-White equation) (3.8) It is usual to show the variation of with and by a three-parameter graph known as the Moody chart.

14.1log21

Df

s

510Re

f

D

Dfss

Re35.91log214.1log2

1

ReD

s

12

Studies on non-circular conduits, such as rectangular, oval and triangular shapes have shown that by introducing the hydraulic radius ,the formulae developed for pipes are applicable for non-circular ducts also. Since for a circular shape

, by replacing by , Eqs. (3.5) through (3.8) can be used for any duct shape provided the conduit areas are close enough to the area of a circumscribing circle or semicircle.

Open channels For purposes of flow resistance which essentially takes plac

e in a thin layer adjacent to the wall, an open channel can be considered to be a conduit cut into two.

D R4

R

4DR

13

The appropriate hydraulic radius would then be a length parameter and a prediction of the friction factor can be done by using Eqs. (3.5) through (3.8). It should be remembered that and the relative roughness is .

Equation (3.4) can then be written for an open channel flow as

which on rearranging gives (3.9)

Noting that for uniform flow in an open channel = slope of the energy line = = , it may be

f

v

RV4Re Rs 4

g

V

R

Lfh f 24

2

LhRf

gV f

8

L

h f

fS0S

14

seen that Eq. (3.9) is the same as Eq. (3.3) with (3.10) For convenience of use, Eq (3.10) along with Eqs (3.5) through (3.8) can be used to prepare a modified Moody cha

rt showing the variation of C with

If is to be calculated by using one of the Eqs (3.5) through (3.8), Eqs (3.6) and (3.8) are inconvenient to use as is involved on both sides of the equations. Simplified empirical forms of Eqs (3.6) and (3.8), which are accurate enough for all practical purposes, are given by Jain as follows:

(3.6a)

fgC 8

s

Rand

v

RV

44

Re

f

f

1.5146-Relog80.11

f

15

and (3.8a) Equation (3.8a) is valid for

These two equations are very useful for obtaining explicit solutions of many flow-resistance problems.

Generally, the open channels that are encountered in the field are very large in size and also in the magnitude of roughness elements.

0.9s

Re

25.21

42.0log-14.1

1

Rf

268 104

1010Re5000 R

and s

16

3.4 MANNING’S FORMULA3.4 MANNING’S FORMULA A resistance formula proposed by Robert Manning, an Irish

engineer, for uniform flow in open channels, is

(3.11) where = a roughness coefficient known as Manning’s

. This coefficient is essentially a function of the nature of boundary surface. It may be noted that the dimensions of dimensions of are

. Equation (3.11) is popularly known as the Manning's formula. Owing to its simplicity and acceptable degree of accuracy in a variety of practical applications, the Manning’s formula is probably the most widely used uniform-flow formula in the world. Comparing Eq. (3.11) with the Chezy formula, Eq. (3.3), we have

210

321SR

nV

nn

n TL 31

17

(3.12)

From Eq. (3.10)

i.e. (3.13) since Eq. (3.13) does not contain any velocity term (and he

nce the Reynolds number), we can compare Eq. (3.13) with Eq. (3.7), i.e. the Pranal-Karman relationship for rough turbulent flow. If Eq. (3.7) is

plotted as vs. on a log-log paper, a smooth

611R

nC

6118R

nf

gC

gR

nf 8

31

2

fs

R

4

18

curve that can be approximated to a straight line with a slope of is obtained (Fig. 3.2). From this the term can be expressed as

3

1

f3131

4

Rfor

Rf s

s

19

Since from Eq. (3.13), , it follow that .

Conversely, if , the Manning’s formula and Dracy-Weisbach formula both represent rough

turbulent flow

31

2

R

nf 61

sn

61sn

60*

v

vs

20

3.5 OTHER RESISTANCE3.5 OTHER RESISTANCE FORMULAE FORMULAE Several forms of expressions for the Chezy coefficient have been proposed by different investigators in the past.

Many of these are archaic and are of historic interest only. A few selected ones are listed below:

1. Pavlovski Formula

(3.14) in which and = Manning’s coefficient. This formula appears to be in use in Russia.

C

XRn

C1

10.075.013.05.2 nRnxn

21

2. Ganguillet and Kutter Formula

(3.15)

in which = Manning’s coefficient 3. Bazin’s Formula

in which = a coefficient dependent on the surface roughness.

R

nS

SnC

0

0

00155.0231

00155.0123

n

RMC

1

0.87

M

22

3.6 VELOCITY DISTRIBUTION3.6 VELOCITY DISTRIBUTION(a)Wide Channels (i) Velocity-defect Law: In channels with large aspect

ratio , as for example in rivers and very large canals, the flow can be considered to be essentially two dimensional. The fully developed velocity distributions are similar to the logarithmic form of velocity-defect law found in turbulent flow in pipes. The maximum velocity occurs essentially at the water surface, (Fig.3.3). The velocity at a height

above the bed in a channel having uniform flow at a depth is given by the velocity-defect law for

as

(3.17)

0yB

muu

0y15.00 yy

0*

ln1

y

y

ku

uum

010log3.2

yyk

23

where = shear velocity = = , = hydraulic radius, = longitudinal slope, and = Karman constant = 0.41 for open channel flow .

*u 0 0gRSR

0Sk 5

24

This equation is applicable to both rough and smooth boundaries alike. Assuming the velocity distribution of Eq. (3.17) is applicable to the entire depth , the velocity can be expressed in terms of the average velocity

(3.18) From Eq (3.18), it follows that

(3.19)

u0y

0

*

00

ln1

as1 0

y

y

k

uVu

udyy

Vy

k

uuV m

*

25

(ii) Law of the wall: For smooth boundaries, the flow of the wall as

(3.20) is found applicable in the inner wall region ( < 0.20). Th

e values of the constants are found to be = 0.41 and = 5.29 regardless of the Froude number and Re

ynolds number of the flow . Further, there is an overlap zone between the law of the wall region and the velocity-defect law region.

For completely rough turbulent flows, the velocity distribution in the wall region ( < 2.0) is given by

(3.21)

sAv

yu

ku

u *

*

ln1

0yy

0yy

ksA

rs

Ay

ku

u

ln

1

*

5

26

where = equivalent sand grain roughness. It has been found that is a universal constant irrespective of the roughness size . Values of = 0.41 and = 8.5 are appropriate.

For further details of the velocity distributions Ref. [5] can be consulted.

(b) Channels with Small Aspect Ratio In channels which are not wide enough to have two dimensi

onal flow, the resistance of the sides will be significant to alter the two-dimensional nature of the velocity distribution given by Eq.(3.17). The most important feature of the velocity distributions in such channels is the occurrence of velocity-dip, where the maximum velocity occurs not at the free surface

sk

k5

rA

27

but rather some distance below it, (Fig. 3.4).

Typical velocity distributions in rectangular channels with = 1.0 and 6.0 are shown in Fig. 3.5(a) and (b) respectively.

0yB

28

29

30

3.7 SHEAR STRESS3.7 SHEAR STRESS DISTRIBUTION DISTRIBUTION The average shear stress on the boundary of a channel is,

by Eq. (3.2), given as . However, this shear stress is not uniformly distributed over

the boundary. It is zero at tile intersection of the water surface with the boundary and also at the corner in the boundary. As such, the boundary shear stress will have certain local maxima on the side as well as on the bed. The turbulence of the flow and the presence of secondary currents in the channel also contribute to the non-uniformity of the shear stress distribution. A knowledge of the shear stress distribution in a channel is of interest not only in the understanding of the mechanics of flow but also in certain problems involving sediment transport and design of stable channels in non-cohesive material, (Chapter 11).

000 RS

0

31

Preston tube is a very convenient device for the boundary shear stress measurements in a laboratory channel. Distributions of boundary shear stress by using Preston tube in rectangular , trapezoidal and compound channels have been reported. Is sacs and Macintosh report the use of a modified Preston tube to measure shear stresses in open channels.

Lane obtained the shear stress distributions on the sides and bed of trapezoidal and rectangular channels by the use of membrane analogy. A typical distribution of the boundary shear stress on the side

and bed in a trapezoidal channel of =4.0 and side slope =1.5 obtained by Lane is shown in Fig.

(3.6).

6

8,7 8

10,9

11

12

s b 0yBm

32

The variation of the maximum shear stress on the bed and on the sides in rectangular and trapezoidal channels is shown in Fig. (3.7). It is noted from this figure that for trapezoidal sections approximately and

when .

bm sm

0076.0~ Sysm 00~ Sybm 0.6

~0 yB

33

3.8 RESISTANCE FORMULA FOR3.8 RESISTANCE FORMULA FOR PRACTICAL USE PRACTICAL USE Since a majority of the open channel flows are in the rough

turbulent range, the Manning's formula (Eq. 3.11) is the most convenient one for practical use. Since it is simple in form and is also backed by considerable amount of experience, it is the most preferred choice of hydraulic engineers. However, it has a limitation in that it cannot adequately represent the resistance in situations where the Reynolds number effect is predominant and this must be borne in mind. In this book, the Manning's formula is used as the resistance equation.

The Darcy-Weisbach coefficient used with the Chezy formula is also an equally effective way of representing the resistance in uniform flow.

f

34

However, field-engineers generally do not prefer this approach, partly because of the inadequate information to assist in the estimation of and partly because it is not sufficiently backed by experimental or field observational data. It should be realised that for open channel flows with hydrodynamically smooth boundaries, it is perhaps the only approach available to estimate the resistance.

s

35

36

3.9 MANNING’S ROUGHNESS 3.9 MANNING’S ROUGHNESS COEFFICIENT COEFFICIENT In the Manning's formula, all the terms except are capabl

e of direct measurement. The roughness coefficient, being a parameter representing the integrated effects of the channel cross-sectional resistance, is to be estimated. The selection of a value for n is subjective, based on one's own experience and engineering judgement. However, a few aids are available which reduce to a certain extent the subjectiveness in the selection of an appropriate value of n for a given channel. These include:

1. Photographs of selected typical reaches of canals, their description and measured values of .

n

14,13n

n

37

These act as type values and by comparing the channel under question with a figure and description set that resembles it most, one can estimate the value of fairly well. Movies, sterioscopic colour photographs and video recordings of selected typical reaches are other possible effective aids under this category. 2. A comprehensive list of various types of channels, their descriptions with the associated range of values of . Some typical values of for various normally encountered channel surfaces prepared from information gathered from various sources are presented in Table 3.2.

n

n n

17,16,15,13

38

39

40

EXAMPLE 3.1 A rectangular channel 2.0m wide carries water at at a depth of 0.5m.The channel is laid on a slope of 0.0004. Find the hydrody- namic nature of the surface if the channel is made of (a) very smooth concrete and (b) rough concrete.

Solution Hydraulic radius

C20

mR 333.05.022

5.02

0004.0333.01081.9 300 RS

m308.1

smv 03617.010

308.1cityshear velo

30

*

41

(a) For a Smooth Concrete Surface Form Table 3.1,

Since this value is slightly greater than 4.0, the boundary is hydrodynamically in the early transition from smooth to rough surface.

(b) For a Rough Concrete Surface From Table 3.1,

Since this value is greater than 60, the boundary is hydrodynamically rough.

mmms 00025.025.0 smv 2610C20at

04.910

03617.000025.06

*

v

vs

0035.05.3 mms6.126*

v

vs

42

EXAMPLE 3.2 For the two cases in Example 3.1, estimate the discharge in the channel using (i) the Chezy formula with Darcr-Weisbach and (ii) the Manning's formula.

Solution Case (a) : Smooth Concrete Channel (i)

Since the boundary is in the transitional stage, Eq. (3.8a) would be used.

Here Re is not known to start with and hence a trial and err

or method has to be adopted. By trial

f

43

10894.11033.04

25.0

425.0

Randmm s

s

0.9Re

25.21

4log0.214.1

1

Rfs

43

(ii) Referring to Table 3.2, the value of for smooth trowel-finished concrete can be taken as 0.012. By the Manning’s formula (Eq. 3.11),

Case (b): Rough Concrete Channel

0145.0f

6.738 fgC

smRSCV 850.00004.0333.06.730 smAVQ 3850.0

2132 0004.0333.0012.0

1V

sm801.0smAVQ 3801.0

n

44

(i) Since the flow is in the rough-turbulent state, by Eq. (3.7),

(ii) By the Manning’s Formula Form Table 3.2, for rough concrete, = 0.015 is appropriate.

310625.24

5.3 R

andmm ss

310625.2log214.11 f

025.0f

0.56025.0

81.98

C

smV 647.00004.0333.056 smAVQ 3647.0

n

45

Empirical Formulae for n Many empirical formulae have been presented for estimati

ng Manning's coefficient in natural streams. These relate to the bed-particle size. The most popular form under this type is the Strickler formula:

(3.22) Where is in meters and represents the particle

2132 0004.0333.0015.0

1V

sm641.0

smQ 3641.0

nn

1.21

6150d

n

50d

46

size in which 50 per cent of the bed material is her. For mixtures of bed materials with considerable coarse-grained sizes, Eq. (3.17) has been modified by Meyer . As

(3.23) where = size in metres and in which 90 per cent of the pa

rticles are finer than .This equation is reported to be useful in predicting in mountain streams paved with coarse gravel and cobbles.

Factors Affecting n The Manning's is essentially a coefficient representing th

e integrated effect of a large number of factors contributing to the energy loss in a reach.

18alet

26

6190d

n

90d

90dn

n

47

Some important factors are: (a) surface roughness, (b) vegetation, (c) cross-section irregularity and (d) irregularity alignment of channel. The chief among these are the characteristics of the surface. The dependence of the value of n on the surface roughness in indicated in Tables 3.1 and 3.2. Since n is proportional to ,a large variation in the absolute roughness magnitude of a surface causes correspondingly a small change in the value of n.

The vegetation on the channel perimeter acts as a flexible roughness element. At low velocities and small depths vegetation, such as grass and weeds, can act as a rigid roughness element which bends and deforms at higher velocities and depths of flow to yield lower resistance.

61s

48

For grass-covered channels, the value of n is known to decrease as the product VR increases. The type of grass and density of coverage also influence the value of n. For other types of vegetation, such as brush, trees in Rood plains, etc. the only recourse is to account for their presence by suitably increasing the values of n given in Table 3.2, which of course is highly subjective.

Channel irregularities and curvature, especially in natural streams, produce energy losses which are difficult to evaluate separately. As such, they are combined with the boundary resistance by suitably increasing the value of n. The procedure is sometimes also applied to account for other types of form losses, such as obstructions that may occur in a reach of channel.

49

3.10 EQUIVALENT3.10 EQUIVALENT ROUGHNESS ROUGHNESS In some channels different parts of the channel perimeter

may have different roughnesses. Canals in which only the sides are lined, laboratory flumes with glass walls and rough beds, rivers with a sand bed in deepwater portion and flood plains covered with vegetation, are some typical examples. This equivalent roughness, also called the composite roughness, represents a weighted average value for the roughness coefficient. Several formulae exist for calculating the equivalent roughness. All are based on certain assumptions and are approximately effective to the same degree. One such method of calculation of equivalent roughness is given below.

50

Consider a channel having its perimeter composed of types of roughnesses. are the lengths of these parts and

are the respective roughness coefficients (Fig. 3.8). Let each port be associated with a partial area such that

N Ni PPPP ,...,,...,, 21

N Ni nnnn ,...,...,, 21

iP iA

N

iNi AAAAA

121 area total...

51

It is assumed that the mean velocity in each partial area is the mean velocity for the entire area of flow, i.e.

By the Mannning’s formula

(3.24) where = equivalent roughness

From Eq. (3.24)

V

VVVVV Ni ......21

3232322

2232

1

11210 ......

N

NN

i

ii

R

nV

R

nV

R

nV

R

nVS

32R

Vn

n

32

3232

nP

Pn

A

A iii

52

(3.25)

i.e. (3.26)

This equation affords a means of estimating the equivalent roughness of a channel having multiple roughness types in its perimeter.

If the Darcy-Weisbach friction formula is used under the same assumption of (i) velocity being equal in all the partial areas and (ii) slope is common to all partial areas, then

Pn

PnAA ii

i 23

23

Pn

PnAAA ii

i 23

23

32

2323

P

Pnn ii

0S

53

Hence

Thus and on summation

i.e.

or (3.27)

gA

PfV

gR

fVSLh f 88

22

0

ii

i

fP

A

Pf

A

gS

V

0

2

8

pf

fPAA ii

i 11

Pf

fPAA

N

iiN

ii

PffP ii

P

fPf ii

54

EXAMPLE 3.3 An earthen trapezoidal channel (n = 0.025) has a bottom width of 5.0 m, side slopes of 1.5 horizontal:1 vertical and a uniform flow depth of 1.1m. In an economic study to remedy excessive seepage from the canal two proposals, viz. (a) to line the sides only and (b) to line the bed only are considered. If the lining is of smooth concrete (n=0.012), determine the equivalent roughness in the above two cases.

Solution Case (a) : Lining on the side only Here for the bed

For the sides:

mPandn 0.5025.0 11

222 5.1110.12012.0 Pandn

m966.3

55

Equivalent roughness, by Eq. (3.26)

Case b: Lining on the bottom only

mPPP 966.8966.30.521

32

325.15.1

966.8

012.0966.3025.05 n

020.031585.4

085447.0

012.00.5 11 nmP

025.0966.3 22 nmP

966.8P

56

Equivalent roughness

32

325.15.1

966.8

025.0966.3012.05 n

018.031585.4

079107.0

57

3.11 UNIFORM FLOW 3.11 UNIFORM FLOW COMPUTATIONS COMPUTATIONS The Manning's formula (Eq. 3.11) and the continuity equati

on, Q =AV form the basic equations for uniform-flow computations. The discharge Q is then given by

(3.28)

(3.28a) where, is called the conveyance if the

channel and expresses the discharge capacity of the channel per unit longitudinal slope. The term

is sometimes called the section factor for uniform-flow computations.

210

321SAR

nQ

0SK321

ARn

K

32ARnK

58

For a given channel, is a function of the depth of flow. For example, consider a trapezoidal section of bottom width =B and side slope m horizontal: 1 vertical. Then,

(3.29)

32AR

ymyBA

12 2 myBP

12 2

myB

ymyBR

ymBf

myB

ymyBAR ,,

1232

2

353532

59

For a given channel, and are fixed and = . Figure 3.9 shows the relationship of Eq (3.29)

32AR)(yf

B m

60

in a non-dimensional manner by plotting

for different values of .

It may be seen that for , there is only one value for each value of , indicating that for

, is a single-valued function of . This is also true for any other shape of channel provided that the top width is either constant or increases with depth. we shall denote these channels as channels of the first kind.

Since and if and are fixed for a

ByB

ARvs

38

32

m

0mBy

0m32AR y

0

32

S

QnAR n

0S

61

channel, the channels of the first kind have a unique depth in uniform flow associated with each discharge. This depth is called the normal depth. Thus the normal depth is defined as the depth of flow at which a given discharge flows as uniform flow in a given channel. The normal depth is designated as , the suffix ‘0’, being usually used to indicate uniform-flow conditions. The channels of the first kind thus have one normal depth only.

While a majority of the channels belong to the first kind, sometimes one encounters channels with closing top width. Circular and ovoid sewers are typical examples of this category. Channels with a closing top-width can be designated as channels of the second kind.

0y

62

63

The variation of = with depth of flow for few channels of this second kind is shown in Fig. 3.10. It may be seen that in some ranges of depth, is not a single-valued function of depth. For example, the following regions of depth have two values of for a given value of : (i) y/D>0.82 in circular channels, (ii) y/B>0.71 in trapezoidal channels with m = -0.5, (iii) y/B>1.30 in trapezoidal channels with m = -0.25. Thus in these regions for any particular discharge, two normal depths are possible. As can be seen from Fig. 3.10, the channels of the second kind will have a finite depth of flow at which ,and hence the discharge for a given channel, is maximum.

32AR

32AR

32AR

32AR

64

Types of Problems Uniform flow computation problems are relatively simple.

The available relations are: 1.Manning's formula 2.Continuity equation 3.Geometry of the cross-section. The basic variables in uniform flow situations can be the

discharge , velocity of flow , normal depth ,roughness coefficient , channel slope

and the geometric elements (e.g. and for a trapezoidal channel). There can be many other derived variables accompanied by corresponding relationships. From among the above, the following five types of basic problems are recognised.

Q V0y n

0S B m

65

Problems of the types 1, 2 and 3 normally have explicit sol

utions and hence do not present any difficulty in their calculations. Problems of the types 4 and 5 usually do not have explicit solutions and as such may involve trial-and-error solution procedures. A typical example for each type of problem is given below.

66

EXAMPLE 3.4 A trapezoidal channel is 10.0 m wide and has a side slope of 1.5 horizontal: 1 vertical. The bed slope is 0.0003. The channel is lined with smooth concrete of n =0.012. Compute the mean velocity and discharge for a depth of flow of 3.0 m.

67

Solution Let Here Area

Wetted perimeter

Hydraulic radius

Mean velocity

depth flow uniform0 y5.1 slope side and m 10.0 mB

ymyBA

250.430.30.35.10.10 m

ymBP 12 2

m817.200.3125.220.10

mP

AR 090.2

210

321SR

nV

68

Discharge

213.2 0003.009.2012.0

1

sm36.2

smAVQ 36.102

69

EXAMPLE 3.5 In the channel of Example 3.4 find the bottom slope necessary to carry only 50 of the discharge at a depth of 3.0 m.

Solution

sm3

250.43 mAmP 817.20

mR 09.2

342

22

342

22

009.25.43

012.00.50

RA

nQS

0000712.0

70

EXAMPLE 3.6 A triangular channel with an apex angle of 75 carries a flow of at a depth of 0.80 m. If the bed slope is 0.009, find the roughness coefficient of the channel.

Solution

sm32.1

my 80.0depth normal0

71

Referring to Fig. 3.12

Area

Wetted perimeter

2

75tan8.0280.0

2

1A

2491.0 mmP 0168.25.37sec8.02

mPAR 243.0

20.1

009.0243.0491.0 2132210

32

Q

SARn

0151.0n

72

EXAMPLE 3.7 A trapezoidal channel 5.0 m wide and having a side slope of 1.5 horizontal: 1 vertical is laid on a slope of 0.00035.The roughness coefficient n=0.015. Find the normal depth for a discharge of 20

through this channel.Solution Let Area

Wetted perimeter

sm3

depth normal0 y 005.10.5 yyA

025.320.5 yP

0606.30.5 y 0

00

606.30.5

5.10.5

y

yyPAR

73

The section factor

Algebraically, can be found from the above equation by the trial-and-error method. The normal depth is found to be 1.820 m.

0

32

S

QnAR

036.1600035.0

015.020

606.30.5

5.10.52132

0

350

350

y

yy

0y

74

EXAMPLE 3.8 A concrete-lined trapezoidal channel (n=0.0155) is to have a side slope of 1.0 horizontal: 1 vertical. Find the bottom slope is to be 0.0004. Find the bottom width of the channel necessary to carry 100 of discharge at a normal depth of 2.50 m.

Solution Let = bottom width. Here = normal depth=

2.20 m Area

Wetted perimeter

sm3

B 0y

5.25.2 BA

071.75.222 BBP

32

0

750004.0

015.0100AR

S

Qn

75

By trial-and-error = 16.33 m.

0.75071.7

5.25.232

35

B

B

B

76

Computation of Normal Depth It is evident from Example 3.7 that the calculation of

normal depth for a trapezoidal channel involves a trial-and-error solution. This is true for many other channel shapes also. Since practically all open channel problems involve normal depth, special attention towards providing aids for quicker calculations of normal depth is warranted. A few aids for computing normal depth in some common channel sections are given below.

Rectangular Channel(a) Wide Rectangular Channel

77

For a rectangular channel, (Fig. 3.13) Area Wetted perimeter

Hydraulic radius

As , the aspect ratio of the channel decreases, . Such channels with large bed-widths as compared to their respective depths are known as wide rectangular channels. In these channels, the hydraulic radius approximates to the depth of flow.

Considering a unit width of a wide rectangular channel,

0ByA

02yBP

By

y

yB

ByR

0

0

0

0

212

By0

0yR

78

(3.31)

This approximation of a wide rectangular channel is found applicable for rectangular channels with

< 0.02.(b) Rectangular Channels with

For these channels

0.1, 00 BandyRyA

210

350

1 unit widthper discharge Sy

nq

B

Q

53

0

0

S

qny

By0

32

0

ARS

Qn

02.00 By

79

(3.31)

where

Equation (3.25) when plotted as vs will provide a non-dimensional graphical solution aid for

general application. Since , one can

easily find from this plot for any combination of

3832

0

350

320

35032

212B

By

By

yB

ByAR

0320

350

38

32

380 21

B

AR

BS

Qn

B

y00

0 0

380 BS

Qn

By0

80

, , and in a rectangular channel.Trapezoidal Channel Following a procedure similar to the above, for a trapezoid

al section of side slope : 1, (Fig. 3.14)

Q n0S B

m

81

Area

Wetted perimeter

Hydraulic radius

Non-dimensionalising the variables,

(3.32)

00 ymyBA

02 12 ymBP

0

2

00

12 ymB

ymyBPAR

32

02

350

35032

0 12 ymB

ymyBAR

S

Qn

m

m

m

BS

Qn

B

AR,

121

1032

02

350

350

380

38

32

82

where A curve of vs with as the third parameter will provi

de a general normal depth solution aid. It may be noted that =0 is the case of a rectangular channel. Table 3A.1 given in Appendix 3A at the end of this chapter gives values of for in the range 0.01 to 4.0 and in the range 0 to 3.0. The values of have been calculated to several decimal places so that they can be truncated to any desired level. Values of are close enough for linear interpolation between successive values. This table will be very useful in quick solution of a variety of uniform-now problems.

0 m

m

0 m

0

83

EXAMPLE 3.9 Solve the problem of Example 3.7 by using Table 3A.1.

Solution For example 3.7

Looking at Table 3A.1 under = 1.5

Bt interpolation, for Hence

38

32

38380

21936.05

036.16

B

AR

BS

Qn

m360.021485.0 0 Byfor

370.022619.0 0 Byfor364.0,21936.0 0 By

my 820.10.5364.00

84

Circular Channel Let be the diameter of a circular channel (Fig.

3.15) and be the, angle in radians subtended by the water surface at the centre.

= area of the flow section =area of the sector-area of the triangular portion

D2

A

85

(3.33)

= wetted perimeter (3.34)

Also

Hence

cossin22

12

2

100

20 rrr

2sin22

1 20

20 rr

2sin28

2

D

P

Dr 02

D

y

r

yr 0

0

00 21cos

Dyf 021

0321SAR

nQ

86

Assuming = constant for all depths

Non-dimensionalising both sides

(3.35) The functional relationship of Eq. (3.35) has been evaluate

d for various values of and is given in Table 2A.1 in Appendix 2A.Besides , other geometric elements of a circular channel are also given in the table which is very handy in solving problems related to circular channels.

n

32

35

35

310

32

35

0

2sin2

8

D

D

P

A

S

Qn

32

35

38

32

380

2sin

32

1

D

AR

DS

Qn

Dy0

Dy03832 DAR

87

Using this table, with linear interpolations wherever necessary, the normal depth for a given , ,

and in a circular channel can be determined easily. The graphical plot of Eq. (3.35)is also shown in Fig. 3.10.

As noted earlier, for depths of flow greater than 0.82 , there will be two normal depths in a circular channel. In practice, it is usual to restrict the depth of flow to a value of 0.8 to avoid the region of two normal depths. In the region y/D>0.82, a small disturbance in the water surface may lead the water surface to seek alternate normal depths, thus contributing to the instability of the water surface.

D

D Q n0S

D

88

EXAMPLE 3.10 A trunk sewer pipe of 2.0 m diameter is laid on a slope of 0.0004. Find the depth of flow when the discharge is 2.0 .(Assurnp n=0.014.)

Solution

From Table 2A.2

By interpolation, for The normal depth of flow

sm3

38380

38

32

0.20004.0

014.00.2

DS

Qn

D

AR

22049.0

62.022004.0 038

32

D

yat

D

AR

63.022532.0 0 Dyat

621.0,22049.0 03832 DyDAR

my 242.10

89

3.12 STANDARD LINED CANAL3.12 STANDARD LINED CANAL SECTIONS SECTIONS Canals are very often lined to reduce seepage losses and rel

ated problems. Exposed hard surface lining using materials such as cement concrete, brick tiles, asphaltic concrete and stone masonry form one of the important category of canal lining and especially SO for canals with large discharges.

90

Standard Lined Trapezoidal section Referring to Fig. 3.16, the full supply depth = normal depth

at design discharge = . At normal depth Area (3.36)

where (3.37)

0y

20200 ymyByA

00 yyB

mmm

1tan 1

91

Wetted perimeter (3.38)

Hydraulic radius

By Mannning’s formula

Non-dimensionalising the variables,

(3.39)

where

000 222 yBymyBP

yB

yyBPAR

200

32

0

350

350

0138210

35

21

1

BS

Qn

21032

0

350

350

2

1S

yB

yyB

nQ

B

y 00

92

From Eq. (3.39) the function can be easily evaluated for various values of . A table of vs

or a curve of vs affords a quick method for the solution of many types of problems associated with lined trapezoidal channels.

Standard Lined Triangular Section Referring to Fig. 3.17, at normal depth ,

Area (3.40)

where as before

10 1

0 1 0

0y

20

20

20

22 yy

myA

mmm

1tan 1

93

Wetted perimeter (3.41) and hydraulic radius (3.42)

By Manning’s formula

or (3.43)

Bt using Eq. (3.43), elements of standard lined triangular channels in uniform flow can be easily determined.

02yP 20yPAR

210

320

20 2

1Syy

nQ

63.038

021

0

yS

QnT

94

EXAMPLE 3.11 A standard lined trapezoidal canal section is to be designed to convey 100 of flow. The side slopes are to be 1.5 horizontal: 1 vertical and Manning's n =0.016. The longitudinal slope of tile bed is 1 in 5000.If a bed width of 10.0 m is preferred what would be the normal depth?

Solution Referring to Fig. 3.16, = side slope= 1.5

Further, here = 100.0 , =0.016

m

sm3

sm3

088.25.11tan5.11

tan 11

mm

Q nmBS 0.100002.00

8314.00.100002.0

088.2016.0100382/1

35

38210

35

1

BS

Qn

95

By Eq. (3.39)

On simplifying,

On solving by trial and error

The normal depth

8314.021

132

0

350

350

1

8951.021

152

0

350

350

74.000

B

y

my 544.3088.2

0.1074.00

96

EXAMPLE 3.12 Show that for a standard lined trapezoidal canal section with side slopes of m horizontal: 1 vertical, and carrying a discharge of Q with a velocity ,

where ; ; and is Manning’s coefficient. Also examine the situation when (i) (ii)

sV

4

411

2

10 M

B

y 00

mm

1tan 1

34

230

nV

QSM

s

n4M

4M

97

Solution For a standard lined trapezoidal canal section (Fig.

3.16) Area (i) Perimeter

Hydraulic radius (ii)

From Manning’s formula

i.e. (iii)

sVQyyBA 00 02yBP

PV

QPAR

s

210

321SR

nvs

230

332

S

nVR s

98

Substituting for in Eq. (ii)

Hence (iv)

Putting

from Eq. (i)

Substituting for in Eq. (iv)

230

33

22

2

S

nV

PV

Q s

s

R

2

0235

230

22 2

1

B

yB

nV

SQP

s

B

y 00

00

2

1

1

sV

QB

2B

Q

V

nV

SQ s

s

35

230

2

020

2021

99

Hence

On solving

(i) When , . Since and

are finite values this corresponds to . Thus , corresponds to the case of standard lined triangular channel section.

MnV

QS

s

34

230

0200

20 441 MM

0144 020 MM

4

411

2

10 M

4M B

y 00 0y

0B4M

100

(ii) when , is imaginary and hence this is not a physically realisable propsition

4M 0

101

3.13 MAXIMUM DISCHARGE OF A3.13 MAXIMUM DISCHARGE OF A CHANNEL OF THE SECOND CHANNEL OF THE SECOND KIND KIND It was shown in Section 3.9 that the channels of the second

kind have two normal depths in a certain range and there exists a finite depth at which these sections carry maximum discharge. The condition for maximum discharge can be expressed as

(3.44)

Assuming = constant at all depths, for a constant , Eq. (3.44) can be rewritten as

(3.45)

0dy

dQ

n0S

032 ARdy

d

102

i.e. (3.45a)

Knowing for a given channel, Eq. (3.45) can be use to evaluate the depth for maximum discharge.

EXAMPLE 3.13 Analyse the maximum discharge in a circular channel.

Solution Referring to Fig. 3.15, from Eq. (3.33)

and from Eq. (3.34)

For the maximum discharge, from Eq.(3.45a)

025 PAdy

d

yfAR 32

2sin28

2

D

A

DP

103

i.e.

The solution of flow for maximum discharge

025 PAd

d

025 d

dPA

d

dAP

02sin28

22cos228

522

DDD

D

02sin2cos53

'11151

938.02

cos210

Dy

104

Hence the depth of flow for maximum discharge

At

Also when

Hence if = discharge with , i.e. the pipe running just full, and = maximum discharge then

thus the maximum discharge will be 7.6 per cent more than the pipe full discharge.

Dy 938.00

3353.0938.038

32

0

D

ARDy

3117.0,0.138

32

0

D

ARDy

FQ Dy 0

mQ

0757.13117.0

3353.0

F

m

Q

Q

105

3.14 HYDRAILICALLY-EFFICIENT3.14 HYDRAILICALLY-EFFICIENT CHANNEL SECTION CHANNEL SECTION The conveyance of a channel section of a given area increas

es with a decrease in its perimeter. Hence a channel section having the minimum perimeter for a given area of flow provides the maximum value of the conveyance. With the slope, roughness coefficient and area of flow fixed, a minimum perimeter section will represent the hydraulically-efficient section as it conveys the maximum discharge. This channel section is also called the best section.

Of all the various possible open channel sections, the semicircular shape has the least amount of perimeter for a given area.

106

(a) Rectangular Section Bottom width = and depth of flow = Area of flow Wetted perimeter

If is to be minimum with = constant

Which gives

i.e. (3.46)

B yconstantByA

yBP 2y

y

A2

P A

022

y

A

dy

dP

22 eyA

22,2 e

eeeee

yRandyBBy

107

the suffix ‘e‘ denotes the geometric elements of a hydraulically-efficient section. Thus it is seen that for a rectangular channel when the depth of flow is equal to half the bottom width i.e. when the channel section is a half-square, a hydraulically-efficient section is obtained, (Fig. 3.18).

108

(b) Trapezoidal Section Bottom width = , side slope = horizontal: 1

vertical Area

(3.47)

Wetted perimeter

(3.48)

Keeping and as fixed, for a hydraulically-efficient section,

B m

constant ymyBA

myy

AB

ymBP 12 2

ymmyy

A12 2

A m

109

i.e. (3.49)

Substituting in Eqs (3.47) and (3.48),

(3.50)

(3.51)

(3.52)

A hydraulically-efficient trapezoidal section having the

2212 eymmA

mmyB ee 212

mmyP ee 212

2

122

122

22

e

e

ee y

ymm

ymmR

110

proportions given by Eqs (3.49) through (3.52) is indicated in Fig. 3.19. Let be the centre of the water surface. And arc perpendiculars drawn to the bed and sides respectively.

OOS OT

eyOS

1sin

2

m

OROROT

111

Substituting for form Eq. (3.50),

Thus the proportions of a hydraulically-efficient trapezoidal section will be such that a semicircle can be inscribed in it

In the above analysis, the side dope was held constant. However, if m is allowed to vary, the optimum value of to make most efficient is

obtained by putting = 0. Form Eqs (3.51) and (3.49)

21 myOR e

ee myBOR 2

1

eB

eyOSOT

m

ePm

dm

dPe

112

(3.53)

Setting = 0 in Eq. (3.53) gives

where the suffix ’em’ denotes the most efficient section. Further,

(3.54a)

(3.54b)

dm

dPe

mmAPe 2122

cot3

1emm

60em

ememem yyP 323

131122

ememem yyB3

2

3

131122

113

(3.54c)

If = length of the inclined side of the canal, it is easily seen that

Thus the hydraulically most efficient trapezoidal section is one-half of a regular hexagon.

Using the above approach, the relationship between the various geometrical elements to make different channel shapes hydraulically efficient can be determined. Table 3.3 contains the geometrical relation of some most efficient sections.

22 33

13112 emem yyA

L

ememem ByL 3

2

114

115

EXAMPLE 3.14 A slightly rough brick-lined trapezoidal channel carrying a discharge of 25.0 is to have a longitudinal slope of 0.0004. Analyse the proportions of (a) an efficient trapezoidal channel section having a side of 1.5 horizontal: 1 vertical, (b) the most efficient-channel section of trapezoidal shape.

Solution From Table 3.2, = 0.017 Case (a): = 1.5 For an efficient trapezoidal channel section, by Eq. (3.49)

3m

nm

2212 ee ymmA

116

mye 830.2

smQyR ee30.25,2

21

32

2 )0004.0(2

)1056.2()017.0(

1250

e

e

yy

222 1056.25.15.112 ee yy

117

(by Eq 3.50)

Case (b): For the most-efficient trapezoidal channel section

5.15.11830.22 2

eB

m714.1

57735.03

1emm

2,732.1 2 em

ememem

yRyA

4753.1938 emy

myem 045.3

mBem 516.3045.33

2

118

3.15 THE SECOND HYDRAULIC 3.15 THE SECOND HYDRAULIC EXPONENT EXPONENT NN The conveyance of a channel is in general a function of the

depth of flow. In calculations involving gradually-varied flow, for purposes of integration, Bakhmeteff introduced the following assumption

(3.55) where = a coefficient and = an exponent called here as

the second hydraulic exponent to distinguish it from the first hydraulic exponent associated with the critical depth. It is found that the second hydraulic exponent is essentially constant for a channel over a wide range of depth. Alternatively, is usually a slowly varying function of the aspect ratio of the channel.

NyCK 22

2C N

N

N

119

To determine for any channel, a plot of log vs log is prepared. If is constant between two point and in this plot, it is determined as

(3.56)

For a trapezoidal channel, if given in

Table 3A.1 is plotted against on a log-log paper, from the slope of the curve at any , the value of at that point can be estimated. Figure 3.20 shows the variation of for trapezoidal channels.

By

N Ky N 11, yK 22 , yK

21

21

log

log2

yy

KKN

38

32

B

AR

N

N

120

121

The values of in this curve have been generated based on the slope of the log -log relation using a computer. Figure 320 is useful in the quick estimation of . It is seen from this figure that is a slowly-varying function of . For a trapezoidal section, the minimum value of = 2.0 is obtained for a deep rectangular channel and a maximum value of = 5.33 is obtained for a triangular channel. It may be noted that if the Chezy formula with = constant is used, values of different from the above would result.

NK y

NBy

N

N

C N

122

EXAMPLE 3.15 Obtain the value of for (a) a wide rectangular channel and (b) a triangular channel.

Solution (a) For a Wide Rectangular Channel Considering unit width,

By equating the exponents of on both sides, (b) For a Triangular Channel of Side Slope Horizontal: 1 Vertical

N

yAyR

NyCyyn

K 2342

22 1

33.3Ny

ymPmyA 12, 22

ym

mR

12 2

123

By equating the exponents of on both sides, = 5.33.Ny

NyCym

mmy

nK 2

34

2

222

2

12

1

124

3.16 COMPOUND SECTIONS3.16 COMPOUND SECTIONS Some channel sections may be formed as a combination of

elementary sections. Typically natural channels, such as rivers, have flood plains which are wide and shallow compared to the deep main channel. Figure 3.21 represents a simplified section of a stream with Hood banks. Channels of this kind are known as compound sections.

125

Consider the compound section to be divided into subsections by arbitrary lines. These can be either extensions of the deep channel boundaries as in Fig. 3.21 or vertical lines drawn at the edge of the deep channels. Assuming the logitudinal dope to be same for all subsections, it is easy to see that the subsections will have different mean velocities depeding upon the depth and roughness of the boundaries. Generally, overbanks have larger size roughness than the deeper main channel.

If the depth of flow is confined to the deep channel only , calculation of discharge by using the Manning's formula is very simple. ) i.e.( hy

126

However, when the flow spills over into the flood plain , the problem of the discharge calculation is complicated as the calculation may give a smaller hydraulic radius for the whole stream section and hence the discharge may be underestimated. This underestimation of the discharge happens in a small range of , say

, where = maximum value of beyond which the underestimation of the discharge as abo

ve does not occur. For a value of , the calculation of the discharge by considering the whole s

ection as one unit would be adequate. For values of in the range , the channel has to be considered to be made up of sub-areas and the discharge in each sub-area determined separately.

) i.e.( hy

y

myyh my y

myy

y myyh

127

The total discharge is obtained as a sum of discharges through all such sub- areas. The value of

would depend upon the channel geometry. However, for practical purposes the following method of discharge estimation can be adopted .

(i) The discharge is calculated as the sum of the partial discharges in the sub-areas; for e.g. units 1.2 and 3 in Fig.321.

(ii) The discharge is also calculated by considering the whole section as one unit, (portion ABCDEFGH in Fig. 3.21), say .

my

15,14

iiip AVQQ

mQ

128

(iii) The larger of the above two discharges, and ,is adopted as the discharge at the depth . For determining the partial discharges and hence in

step (i) above, two methods are available.Posey’s method In this method, while calculating the wetted perimeter for t

he sub-areas, the imaginary divisions (FJ and CK in Fig. 3.21) are considered as boundaries for the deeper portion only and neglected completely in the calculation relating to the shallower portion. This way the shear stress that occurs at the interface of the deeper and shallower parts is empirically accounted for.

PQ

WQ y

iQPQ

23,22

129

Zero shear method Some investigators mostly in computational work,

treat the interface as purely a hypothetical interface with zero shear stress. As such, the interfaces are not counted as perimeter cither for the deep portion or for the shallow portion. The procedure can be better understood through Examples 3.16 and 3.17. Further aspects of compound channel sections are discussed in Section 5.7.2 in Chapter 5.

24

130

EXAMPLE 3.16 For the compound channel shown in Fig. 3.22 determine the discharge for a depth of flow of (a) 1.20 m and (b) 1.60 m. Use Posey's method for computing partial discharges.

131

Solution Case (a): (i) Partial area Discharge by Posey’s Method Sub-area 1:

Similarly Sub-area 2:

21 1.23.00.7 mA

mP 3.70.73.01 mR 288.03.71.2

21321 0002.0288.01.2

02.0

1PQ

sm3647.0

smQP3

3 647.02

2 6.32.10.3 mA

132

(ii) By the Total-Section Method

mP 4.52.12.10.32

mR 667.04.56.32

21322 0002.0667.06.3

02.0

1PQ

sm3943.1areas partialby discharge totalPQ

sm3237.3943.1647.0647.0

28.76.31.21.2 mA

mP 4.193.00.79.00.39.00.73.0

mR 402.04.198.7

133

Since , the discharge in the channel is taken as .

Case (b): (i) Partial Area Discharge by Posey’s Method Sub-area 1:

2132 0002.0402.08.702.0

1WQ

sm3005.3

PW QQ smQQ P

3237.3my 6.10

21 9.47.07 mA

mP 7.70.77.01 mR 636.07.79.41

21321 0002.0636.09.4

02.0

1PQ

134

Similarly, Sub-area 2:

(ii) By the Total –Section Method

sm3563.2

smQP3

2 563.22

2 8.46.13 mA mP 8.40.36.12

mR 636.02.68.42

21322 0002.0774.08.4

02.0

1PQ

sm3862.2smQP

2988.7862.2563.2563.2

26.148.49.49.4 mA

135

Since , the discharge in the channel is taken as .

mP 2.207.07.09.00.39.07.07.0

mR 723.02.206.14

2132 0002.0723.06.1402.0

1WQ

sm3315.8

PW QQ smQQ W

3315.8

136

EXAMPLE 3.17 Calculate the discharge for Case (a) of Example 3.16 by using zero shear method for the partial areas.

Solution (i) By Partial Areas Using Zero Shear Method Here = 1.2 m. By using the zero shear method Sub-area 1: Area Perimeter

Partial discharge

Similarly

0y2

1 10.23.00.7 mA mP 3.70.73.01 mR 288.03.71.21

21321 0002.0288.06.3

02.0

1PQ

sm3647.0

smQP3

3 647.0

137

Sub-area 2: Area Perimeter

Partial discharge

Total discharge by partial areas

(ii) By the Total-Section Method: Area Perimeter

22 60.32.10.3 mA

mP 8.49.09.00.32 mR 8.48.46.32

21322 0002.0750.06.3

02.0

1PQ

sm310.2321 PPPP QQQQ

smQP3395.3647.0100.2647.0

28.76.31.21.2 mA

mP 4.193.07.09.00.39.07.03.0

138

Discharge

Since , the discharge in the channel is taken as

mR 402.04.198.7

2132 0002.0402.08.702.0

1wQ

sm3005.3wp QQ

smQQ p3395.3

139

3.17 GENERALISED-FLOW3.17 GENERALISED-FLOW RELATION RELATION Since the Froude number of the flow in a channel is

(3.57) If the discharge occurs as a uniform flow, the slope req

uired to sustain this discharge is, by the Manning’s formula,

(3.58)

Substituting Eq. (3.57) in Eq. (3.58) and simplifying

TgA

VF

T

AF

g

Q 322

Q0S

342

22

0 RA

nQS

140

or (3.59) For a trapezoidal channel of side slope ,

(3.60a)

Non-dimensionalising both sides, through multiplication by ,

(3.60)

31

3422

0 TA

PgnFS

yfTA

P

gnF

S

31

34

220

m

31

00

342

220

2

12

ymyBmyB

ymB

gnF

S

31B

31

342

22

310

121

121

mm

m

gnF

BS

141

in which . Designating

= generalised slope (3.61) Equation (3.60) represents the relationship between the v

arious elements of uniform flow in a trapezoidal channel in a generalised manner. The functional relationship of Eq. (3.60) is plotted in Fig. 3.23. This figure can be used to find, for a given trapezoidal channel, (a) the bed slope required to carry a uniform flow at a known depth and Froude number and (b) the depth of flow necessary for generating a uniform flow of a given Froude number in a channel of known bed slope.

By0*22

310 SgnF

BS

,* mfS

142

For a rectangular channel, m =0 and hence Eq. (3.60) becomes

(3.62a)

For a triangular channel, B = 0 and hence Eq. (3.60) cannot be used. However, by redefining the generalised slope for triangular channels, by Eq. (3.60a).

(3.63)

Roots and Limit Values of S. for TrapezoidalChannels

31

34

*

21

S

2

231

*22

310 1

2m

mS

gnF

ySt

143

Equation (3.60) can be written as

(3.64)

This is a fifth-degree equation in , except for m = 0 when it reduces to a fourth, degree equation. Out of its five roots it can be shown that (a) at least one root shall be real and positive and (b) two roots are always imaginary. Thus depending upon the value Of m and , there may be one, two or three roots.

The limiting values of are obtained by putting, , which results in

mm

mS

121

1213

42

3*

*S

*S

ddS*

144

145

(3.65) Solving Eq. (3.65) the following significant results are obtai

ned 1. Rectangular channels (m=0), a single limiting value with = 8/3 and = l/6 is obtained. 2. Between m = 0 and m = 0.46635 there are two limiting values. 3. At m = 0.46635, the two limit values merge into one at = 2.l545 and =0.7849. 4. For m > 0.46635, there are no limiting points. These features are easily discernible from Fig. 3.23.

22 12121118 mmmm

010101 22 mm

*S

*S

146

3.17.1 CRITICAL SLOPE AND3.17.1 CRITICAL SLOPE AND LIMIT SLOPE LIMIT SLOPE The slope of a channel which carries a given discharge as a

uniform flow at the critical depth is called the critical slope, . The condition governing the critical slope in any channel can be easily obtained from Eq. (3.60) by putting F=1.0. For trapezoidal channels, by denoting the generalised

critical slope, and , the

behaviour of can be studied using Fig 3.23. All the conclusions derived in the previous section for will also apply to relationship.

cS

cc Sgn

BS*2

31

cc

B

y

cS*

*SccS *

147

For a channel of given shape and roughness will have a least value under coditions corresponding to a limit value of . The least value of is called the limit slope, Keeping the critical slope and limit slope in mind, Fig. 3.23 can be studied to yield the following points:

1. For a trapezoidal channel of given geometry and roughness, a given depth of critical flow can be maintained by one and only one critical slope. However, for a given critical slope there can be more than one critical depth. 2. For channels of the second kind ( m is negative) and for rectangular channels (m=0), only one limit slope exists.

cS

cS* cSLcS

148

Slopes flatter than this cannot be critical and the dopes steeper than this can be critical at two different depths. For a rectangular channel, the limit value of is 8/3 at = 1/6. 3. when , any slope can be critical and for each slope there will be only one critical depth. There are no limit slopes in this range. For m = 0.46635, the limit value of is 2.15446 at = 0.7849. 4. For , there are two values of limit slopes, and with . (a) For , there are three critical depth for each value of ;

cS* c46635.0m

cS*

c46635.00 m

1LcS 2LcS 21 LcLc SS 12 LccLc SSS

cS

149

the largest of these, however, may be impracticably large. (b) For or there are two critical depths. (c) For or , there is only one critical depth for each value of the slope.

EXAMPLE 3.18 A rectangular channel is 4.0 m wide and has n = 0.015. Find the bed-slope required to maintain a uniform flow in this channel with a depth of 1.25m and a Froude number, (a) F = 2.0, (b) F = 1.0 and (c) F = 0.5. Also End the limit slope and the corresponding critical depth.

Solution

1Lcc SS 2Lcc SS 2Lcc SS

1Lcc SS

150

Recalling Eq. (3.62)

Substituting in the right-hand side of the above equation,

Thus (a) For (b) (c)

3122

310

*

21

nFgn

BSS

3125.00.4

25.1

22

310

*015.081.9

0.481528.2

F

SS

015658.0,0.2 0 SF015658.0,0.1 0 cSSF

015658.0,5.0 0 SF

151

At the limit slope, = 1.0, and limit and

F 38* cS61c

003708.04

015.081.9667.231

2

LcS

myLc 667.06

0.4

152

EXAMPLE 3.19 A trapezoidal channel section with m = 0.25, B = 3.0, and n = 0.015,has to carry a uniform flow with a Froude number of 0.5.

(a) If the bed slope of = 0.001052 is to be used, at what depths would this flow be possible? (b) within what range of would the above feature of three possible depths be feasible?Solution

(a)

From Fig. (3.23), for = 0.25. given given

0S

0S

75.25.0015.081.9

0.3001052.022

31

22

310

* Fgn

BSS

m75.01 00.12

my 25.21 my 00.32

153

and from Eq. (3.64) by trial and error, giving . (b) From Fig. (3.23), the limit values of are 2.40 and 3.25.

As such, the slope has to lie between and , i.e.

70.183 10.563 y

*S

0S31240.2 Bgn

310243.1

154

3.18 DESIGN OF IRRIGATION3.18 DESIGN OF IRRIGATION CANNALS CANNALS For a uniform flow in a

where and are in general, functions of the geometric elements of the canal. If the canal is of trapezoidal cross-section,

(3.66) Equation (3.66) has six variables out of which one is

a dependent variable and the rest five are independent ones. Similarly, for other channel shapes, the number of variables depend upon the channel geometry. In a channel design problem, the independent variables are known either explicitly or implicitly, or as inequalities, mostly in terms of empirical relationships.

210

321SAR

nQ

A R

mBSynfQ ,,,, 00

155

In this section the canal-design practice adopted by the Irrigation Engineering profession in India is given. This practice may have application in other fields also. The guidelines given below are meant only for rigid-boundary channels, i.e. for lined and unlined non-erodible channels. The design considerations for unlined alluvial channels follow different principles governed by sediment transport and related aspects. The wide variety of soil and topographical features of the country led different states and agencies, in the past, to adopt their own design practices. Reference 26 indicates the effort of the Central water Commissions (CWC), India, towards standardisation and general guidelines applicable to the whole country. Relevant Indian standards for irrigation canal design are found in IS : 4745-1968, IS :7112-1973 .

16,15

156

Canal Section Normally a trapezoidal section is adopted.

Rectangular cross-sections are also in use in special situations, such as in rock cuts, steep chutes and in cross-drainage works.

The side slope, expressed as m horizontal: 1 vertical, depends on the type of canal, i.e. lined or unlined, nature and type of soil through which the canal is laid. The slopes are designed to withstand seepage forces under critical conditions, such as (i) a canal running full with banks saturated due to rainfall and (ii) the sudden drawdown of canal supply. Usually the slopes are steeper in cutting than in filling.

157

For lined canals, the slopes roughly correspond to the angle of repose of the natural soil and the values of m range from 1.0 to 1.5 and rarely up to 2.0. The slopes recommended by CWC for unlined canal in cutting are given in Table 3.4.26

158

Longitudinal Slope The longitudinal slope is fixed on the basis of topography to

command as much area as possible with the limiting velocities acting as constraints. Usually the slopes are of the order of 0.0001. For lined canals a velocity of about 2.0 m/s is usually recommended.

Roughness Since the cost for a given length of canal depends upon its s

ize, if the available slope permits, it is economical to use highest safe velocities. High velocities may cause scour and erosion of the boundaries. As such, in unlined channels the maximum permissible velocities refer to the velocities that can be safely allowed in the channel without causing scour or erosion of the channel material.

159

In lined canals, where the material of lining can withstand very high velocities, the maximum permissible velocity is determined by the stability and durability of the lining and also on the erosive action of any abrasive material that may be carried in the stream. The permissible maximum velocities normally adopted for a few soil types and lining materials are indicated in Table 3.5.

160

In addition to the maximum velocities mentioned above, a minimum velocity in the channel is also an important constraint in the canal design.

Too low a velocity would cause deposition of suspended matter, like silt, which cannot only impair the carrying capacity but also increase the maintenance costs. Also, in unlined canals, too low a velocity may encourage weed growth. The minimum velocity in irrigation channels is of the order of

0.30 .Free Board Free board for lined canals is the vertical distance between

the full supply level to the top of the lining (Fig.3.24). For unlined canals, it is the vertical distance from the full supply level to the level of the top of the bank.

sm

161

This distance should be sufficient to prevent overtopping of the canal lining or banks due to waves. The amount of free board provided depends on the Cana1 size, location, velocity and depth of flow. The relevant Indian standards suggest the minimum free board to be as below: 16,15

162

Width to Depth Ratio The relationship between width and depth varies widely de

pending upon the design practice. If the hydraulically most-efficient channel section is

adopted (Sec6tion 3.14), ,3

2,

3

10

0 yy

Bm

163

i.e. . If any other value of is used,

the corresponding value of for the efficient section would be, from Eq. (3.50)

However, in practice it is usual to adopt a shallower section, i.e. a value of larger than that suggested by Eq. (3.50). The CWC recommendation for as a function of discharge is as follows:

1547.10

y

Bm

0yB

mmy

B 2

0

12

0yB

0yB

26

164

In large canals it is necessary to limit the depth to avoid dangers of bank failure. Usually depths higher than about 4.0 m are adopted only when it is absolutely necessary.

For selection of width and depth, the usual procedure is to adopt a recommended value of

and to find the corresponding using Table 3A.1. Knowing Q, n and A, the values of B and are found.

The bottom width is usually adopted to the nearest 25 cm or l0 cm and the depth adjusted accordingly. The resulting velocity is then checked to see that permissible velocity constraints are not exceeded.

The typical cross-section of a lined irrigation canal is shown in Fig. 3.24.

165

EXAMPLE 3.20 A trapezoidal channel is to carry a discharge of 50 . The maximum slope that can be used is 0.004. The soil is hard. Design the channel as (a) a lined canal with concrete lining and (b) an unlined non- erodible channel.

Solution (a) Lined Canal Adopt side slope of 1 : 1, i.e. = 1.0 (from Table 3.4) for c

oncrete = 0.013 (from Table 3.2) Recommended for = 50 is about 8.0 For = 8.0 (i.e. = 0.125), from Table 3A.1

Substituting = 50.0, = 0.013, = 0.0004

sm3

mn

0yB Q sm3

0yB 0yB

03108.038

0

BS

Qn

Q n0S

166

in the above = 13.5605 m. Adopt = 13.50 m. Then actual

Corresponding = 0.12588 giving = 1.700 m

This value is greater than the minimum velocity of 0.3 ; is of the order of 2.0 m hand further is less than the maximum permissible velocity of 6.0 m/s for concrete. Hence the selection of B and are all right. The recommended geometric parameters of the canal are therefore

B B

03108.0

5.310004.0

013.05038

0yB 0y 840.25700.1700.15.13 A

smV 935.1

sm

0y

0004.0,0.1,50.13 0 SmmB

167

Adopt a free board of 0.75 m. The normal depth for = 0.013 will be 1.70 m.

(b) Unlined Canal From Table 3.4, a side slope of 1 :1 is adopted. From Table

3.2, take n for hard soil surface as 0.020. Recommended for Q = 50 is about 8.0. From Tabl

e 3A.1.

For

Substituting = 50.0, = 0.020 and = 0.0004 in the above, = 15.988 m, hence adopt = 16.00 m. Actual = 0.030760 and the corresponding = 0.12422.

n

0yB sm3

03108.0,0.838

00

BS

Qn

y

B

Q n 0SB

0yB

168

Then

But this velocity is larger than the permissible velocity of 0.90-1.10 for hard soil (Table 3.5). In this case, therefore, the maximum permissible velocity will control the channel dimensions.

Adopt

For Adopt

my 988.11612422.00 276.35988.1988.100.16 mA

smV 398.176.3550

sm

smV 10.1

2002 1455.4510.1

0.50B

B

y

B

mymA

mByB 978.17,0.80 mB 0.18

169

From , substituting

Substituting in the general discharge equation

Hence the recommended parameters of the canal are =18.0 m, =1.0 and = 0.0002106. Adopt a free board of

0.75 m. The normal depth for =0.020 will be 2.245 m.

00 ymyBA 455.45A

mymB 245.2,0.1,0.18 0

mP 35.24245.21120.18

mPAR 867.1

210

32867.1455.4502.0

150 S

0002106.00 S

B m 0Sn

170

習題習題3-1 A trapezoidal channel has a bottom width of 2.50 m and a depth of flow 0.80 m. The side slopes are 1.5 horizontal: 1 vertical. The channel is lined with bricks ( = 3.0 mm). If the longitudinal slope of the channel is 0.0003, estimate (a) the average shear stress, (b) the hydrodynamic nature of the surface, (c) Chezy C by using f, (d) Manning’s n, (e) the uniform-flow discharge for cases (c) and (d).3-6 A trapezoidal channel of bed-width 4.0 m and side slopes 1.5 horizontal: 1 vertical has sand bed ( = 0.025). At a certain reach he sides are lined by smooth concrete ( = 0.012). Calculate the equivalent roughness of this reach if the depth of flow is 1.50 m.

s

1n

2n

171

3-9 Find the discharge in the follow channels with bed slope of 0.0006 and n = 0.016: (a) Rectangular, B = 3.0 m, =1.20 m (b) Trapezoidal, B = 3.0 m, m = 1.5 and = 1.10 m (c) Triangular, m = 1.5, = 1.50 m.3-12 A circular channel 2.50 m in diameter is made of concrete (n = 0.014) and is laid on a slope of 1 in 200. (a) Calculate the discharge if the normal depth is 1.50 m. (b) Calculate the depth of flow for a discharge of 15.0 .

0y

0y

0y

sm3

172

3-17 A trapezoidal channels of bed width 3.0 m and side slope 1.5 horizontal: 1 vertical carries a full supply of 10.0 at a depth 1.50 m. What would be the discharge at half of full supply depth (i.e. at 0.75 m)? What would be the depth at half of full supply discharge ?3.20 A concrete storm water drain (n = 0.012) is 0.75 m in diameter and is to discharge 0.10 . What is the minimum slope that has to be employed if the depth of flow should not exceed 0.8 diameter ?

sm3

sm3

173

3-23 A flow of 10.0 is to be passed in a rectangular channel with the depth of flow equal to one third the width. The channel is lined with smooth concrete (n = 0.014). Calculate the channel dimensions and its longitudinal slope necessary to carry the above discharge with a mean velocity of 2.5 .3-25 The specific energy in a 2.0 m wide rectangular channel is not to exceed 1.2 m. What maximum discharge can be carried in such a channel? What longitudinal slope is required to sustain such a flow? Assume Manning’s n=0.015.

sm3

sm

174

3-28 Determine the bottom width and full supply depth of a standard lined trapezoidal section (Fig 3.25) to carry 180 of flow with a velocity of 2.0 when laid on a slope if 1 in 4500. The side slopes are to be 1.25 horizontal: 1 vertical Manning’s n can be assumed to be 0.014.

sm3 sm

175

3-32 A standard lined triangular canal section (Fig 3.26) is to carry a discharge of 25 when laid on a slope of 1 in 1000. The side slopes are 1.25 H : 1 V. Calculate the depth of flow needed. What is the average boundary shear stress in this channel? (Assume n= 0.015)

sm3

176

3-33 A standard lined triangular channel is designed to carry the full supply discharge at a depth of 2.5 m when laid on a slope of 0.0004. The side of the channel is 1.25 H: 1 V and Manning’s n = 0.015. Determine the full supply discharge in the canal.

177

3-37 A triangular duct (Fug.3.27) resting on a side is carrying water with a free surface. Obtain the condition for maximum discharge when (a) m = 0.5, (b) m = 0.25 and (c) m = 0.10.

178

3-38 Water flows in a channel of the shape of an isosceles triangle of bed width a and sides making an angle of 45 with the bed. Determine the relation between the depth of flow d and the bed with a for maximum velocity condition an for maximum discharge condition. Use Manning’s formula and note that d is less than 0.5 a.

3-40 A trapezoidal channel is 5.0 m wide and has a side slope of 0.5 horizontal: 1 vertical. Find the depth of flow which can make the channel an efficient section, If = 0.0002 and n = 0.02, find the corresponding discharge.

0S

179

3-44 A trapezoidal channel of efficient section is to have an area of 60.0 . The side slope is 1.5 horizontal: 1 vertical. Find the bottom width and depth of flow.

3-48 A lined channel (n = 0.014) is of a trapezoidal section with one side vertical and other side on a slope of 1 H : 1V. If the canal has to deliver 5

when laid on a slope of 0.0001. Calculate the dimensions of the efficient section which requires minimum of lining.

2m

180

3-54 Using Fig 3.20, estimate the value of the second hydraulic exponent N for the following cases:

0.2,0.1,5.0,0.1 Bym0.2,0.1,5.0,0.2 Bym

181

3-55 For the compound section shown in Fig. 3.28, find the discharge when (a) h =0.2 m and (b) h = 1.0 m. Assume n = 0.02 and = 0.0009 for all parts of the perimeter. Use Posey’s method for computing partial discharges.

0S