1

Chapter Chapter 22

Energy Depth Energy Depth RelationshipsRelationships

2

2.1 SPECIFIC ENERGY2.1 SPECIFIC ENERGY The total energy of a channel flow referred to a datum is giv

en by Eq. (1.39) as

If the datum coincides with the channel bed at the section, the resulting expression is known as specific energy and is denoted by . Thus

(2.1) When and ,

(2.2)

g

VyZH

2cos

2

E

g

VyE

2cos

2

0.1cos 0.1

g

VyE

2

2

3

The concept of specific energy, introduced by Bakhmeteff, is very useful in defining critical depth and in the analysis of flow problems. It may be noted that while the total energy in a real fluid flow always decreases in the downstream direction, the specific energy is constant for a uniform flow and can either decrease or increase in a varied flow, since the elevation of the bed of the channel relative to the elevation of the total energy line, determines the specific energy. If the frictional resistance of the flow can be neglected, the total energy in non-uniform flow will be constant at all sections while the specific energy for such flows, however,

4

will be constant only for a horizontal bed channel and in all other cases the specific energy will vary.

To simplify the expressions it will be assumed, for use in all further analysis, that the specific energy is given by Eq. (2.2) i.e. and

.This is with the knowledge that and can be appended to and terms respectively, without difficulty if warranted.

0.1cos 0.1cos

y gV 22

5

2.2 CRITICAL DEPTH2.2 CRITICAL DEPTHConstant Discharge Situation Since the specific energy

(2.2a) for a channel of known geometry, . Keeping =constant = , the variation of with

is represented by a cubic parabola (Fig. 2.1). It is seen that there are two positive roots for the equation of indicating that any particular discharge can be passed in a given channel at two depths and still maintain the same specific energy

. In Fig. 2.1 the ordinate represents

2

22

22 gA

Qy

g

VyE

QyfE ,Q

1Q Ey

E 'PP

1QE

6

the condition for a specific energy of E1. The depths of flow can be either or .

These two possible depths having the same specific energy are known as alternate depths.

1yPR 1'' yPR

7

In Fig.2.1, a line (OS) drawn such that (i.e. at to the abscissa) is the asymptote of the upper lim

b of the specific-energy curve. It may be noticed that the intercept or represents the velocity head. Of the two alternate depths, one ( )is smaller and has a large velocity head while the other ( ) has a larger depth and consequently a smaller velocity head. For a given

, as the specific energy is increased the difference between the two alternate depths increases. On the other hand, if is decreased, the difference ( )

will decrease and at a certain value ,

yE 45

''RP RP'

1yPR 1'' yPR

Q

11' yy E

cEE

8

the two depths will merge with each other (point in Fig. 21). No value for can be obtained when

,denoting that the flow under the given conditions is not possible in this region. The condition of minimum specific energy is known as the critical-flow condition and the corresponding depth is known as the critical depth.

At critical depth, the specific energy is minimum. Thus differentiating Eq.(2.2a) with respect to

(keeping constant) and equating to zero,

(2.3)

cEE

Cy

cy

yQ

013

2

dy

dA

gA

Q

dy

dE

9

But = top width, i.e. width of the channel at

the water surface. Designating the critical-flow conditions by the suffix , (2.4) or

(2.4a) If an value other than unity is to be used, Eq. (2.4) will b

ecome (2.5)

Tdy

dA

''c 13

2

c

c

gA

TQ

c

c

T

A

g

Q 32

0.13

2

c

c

gA

TQ

10

Equation (2.4) or (2.5) is the basic equation governing the critical-flow conditions in a channel. It may be noted that the critical-flow condition is governed solely by the channel geometry and discharge (and ). Other channel properties such as the bed slope and roughness do not influence the critical flow condition for any given . If the Froude number of the flow is denned as

(2.6) it is easy to see that by using in Eq. (2.4), at the critical flo

w and . we thus get an important result that the critical flow corres

ponds to the minimum specific energy and at this condition the Froude number of the flow is unity.

Q

TgAVF

cyy 0.1 cFFF

11

For a channel with large longitudinal slope and having a flow with an energy correction factor of

, the Froude number will be defined as

(2.6a)

Referring to Fig. 2.1, considering any specific energy other than , (say ordinate at

)the Froude number of the flow corresponding to both the alternate depths will be different from unity as

or .

F

cos

1TA

gVF

cE 'PP 1EE

1y cyy 1'

12

At the lower limb, CR of the specific-energy curve, the depth . As such, and . This region is called the supercritical flow region. In the upper limb , . As such and

. This denotes the subcritical flow region.Discharge as a Variable In the above section the critical-flow condition was derived

by keeping the discharge constant. The specific-energy diagram can be plotted for different discharges ( =1,2,3...), as in Fig. 2.2. In this figure, and is constant along the respective vs plots.

cyy 1 cVV 1' 0.11 F

'CR cyy 1' cVV 1'0.1'1 F

constant1 QQ i 321 QQQ

E y

13

Consider a section in this plot. It is seen that for the ordinate , . Different curves give different intercepts. The difference between the alternate depths decreases as the value increases.

'PPconstant1 EE Q'PP

Q

14

It is possible to imagine a value of at a point at which the corresponding specific-energy curve would

be just tangential to the ordinate . The dotted line in Fig. 2.2 indicating represents the maximum value of discharge that can be passed in the channel while maintaining the specific energy at a constant value ( ). Any specific-energy curve of higher value (i.e. ) will have no intercept with the ordinate and hence there will be no depth at which such a discharge can be passed in the channel with the given specific energy. Since by Eq. (2.2a)

'PP

mQQ C

mQQ

1EmQQ Q'PP

2

2

2gA

QyE

15

(2.7) The condition for maximum-discharge can be obtained by

differentiating Eq. (2.7) with respect to and equating it to zero while keeping =constant.

Thus

Putting and

(2.8) This is same as Eq.(2.4)and hence represent the critical-flo

w conditions.

yEgAQ 2

Ey

02

2

yEg

gA

dy

dAyEg

dy

dQ

Tdy

dA yields2 yEg

A

Q

0.13

2

gA

TQ

16

EXAMPLE 2.1 A rectangular channel 2.5m wide has a specific energy of 1.5m when carrying a discharge of 6.48 . Calculate the alternate depths and corresponding Froude numbers.

Solution From Eq. (2.2a)

sm3

22

22

22 ygB

Qy

g

VyE

22

2

5.281.92

48.65.1

yy

2

34243.0

yy

17

Solving this equation by trial and error, the alternate depths and are obtained as and

. Froude number

At , ;and at , The depth is in the subcritical flow region and th

e depth is the supercritical flow region.

1y 2y my 296.11 my 625.02

my 296.11 561.01 F

my 296.11 my 625.02 675.12 F

my 625.0

23

82756.0

81.95.2

48.6

yyygy

VF

18

EXAMPLE 2.2 A flow of is passing at a depth of 1.5 m through a rectangular channel of width 2.5 m. The kinetic energy correction factor α is found to be 1.20. what is the specific energy of the flow? what is the value of the depth alternate to the existing depth if α =1.0 is assumed for the alternate flow ?

Solution

sm30.5

smA

QV 33.1

5.15.2

0.51

m

g

V1087.0

81.92

33.120.1

2

221

1

19

Specific energy

For the alternate depth ,

i.e.

By trial and error

1087.05.12

21

111 g

VyE

m6087.12y

0.1sin6087.15.281.92

0.522

2

2

2

cey

y

6087.12039.0

22

2 y

y

my 413.02

20

2.3 CALCULATION OF THE2.3 CALCULATION OF THE CRITICAL DEPTH CRITICAL DEPTH Using Eq. (2.4), expressions for the critical depth in channel

s of various geometric shapes can be obtained as follows:Rectangular Section For a rectangular section, and (Fig. 2.4). Hence by

Eq. (2.4)

or

(2.9)

ByA BT

12

3

2

c

c

c

c

gy

V

gA

TQ

cc yg

V

2

1

2

2

21

Specific energy at critical depth

(2.10) Note that Eq. (2.10) is independent of the width of the chan

nel. Also, if = discharge per unit width = ,

i.e. (2.11) Since , from Eq. (2.6) the Froude number for a recta

ngular channel will be defined as (2.12)

cc

cc yg

VyE

2

3

2

2

q BQ

32

cyg

q 312

g

qyc

gy

VF

yTA

22

Triangular Channel For a triangular channel having a side slope of horizontal:

1 vertical (Fig. 2.4), and . By Eq. (2.4a),

(2.13) Hence (2.14)

m2myA

myT 2

22

526332c

c

c

c

c ym

my

ym

T

A

g

Q

51

2

22

gm

Qyc

23

The specific energy at critical

i.e. (2.15) It is noted that Eq. (2.15) is independent of the side slope

m of the channel. Since the Froude number for a triangular channel is denned by using Eq. (2.6) as

(2.16)

g

VyE ccc 2

2

42

52

2

2

42 c

cc

cc ym

ymy

gA

Qy

cc yE 25.1

2yTA

gy

VF

2

24

Circular Channel Let be the diameter of a circular channel (Fig. 25)

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

D2

25

Top width and

section flow theof areaAportionr triangula theof area section theof area

2sin22

1

cossin22

12

2

1

20

20

002

0

rr

rrr

2sin28

2

D

A

sinDT

DyfD

y

21cos22 1

26

Substituting these in Eq. (2.4a) yields

(2.17)

Since explicit solution for cannot be obtained from Eq. (2.17), a non-dimensional representation of Eq. (2.17) is obtained as

(2.18) This function is evaluated and is given in Table 2A.1

c

cc

D

D

g

Q

sin

2sin28

32

2

cy

DyfgD

Qc

c

cc

21

23

5 sin

2sin2044194.0

27

of Appendix 2A at the end of this chapter as an aid for the estimation of .

Since , the Froude number for a given at any depth will be

cy

)(D

yfnTA

Q y

DyfnTAg

Q

TAg

VF

3

28

Trapezoidal Channel For a trapezoidal channel having a bottom width of

and side slopes of horizontal : 1 vertical (Fig.(2.6)) Area and Top width

Bm ymyBA myBT 2

29

At the critical flow

Here also an explicit expression for the critical depth is not possible. The non-dimensional representation of Eq. (2.19) facilitates the solution of

by the aid of tables or graphs. Rewriting the right-hand side of Eq. (2.19) as

c

cc

c

c

myB

ymyB

T

A

g

Q

2

3332

cy

cy

B

myB

yB

myB

myB

ymyB

c

cc

c

cc

21

1

2

33

333

30

where gives

(2.20) or (2.20a) Equation 2.20(a) can easily be evaluated for various value o

f and plotted as vs . It may be noted that if , can be defined as

c

cc

m

B

21

1 33

3

5

c

cc

gB

mQ

21

1 33

5

32

B

mycc

21

2323

25

23

21

1

c

cc

Bg

Qm

c c1

21

5

32

gB

mQ

31

One such plot, shown in Fig. 2.7, is very helpful in quick estimation of critical depth and other parameters related to the critical-flow condition in trapezoidal channels. Table 2A.2 which gives values of for various values is useful for constructing a plot of vs as in Fig. 2.7 on a lager scale.

Since the Froude

number at any depth is

cc

Bmy

yBmy

myB

ymyBTA

21

1

2

y

32

10

33

for a given discharge . Further the specific energy at critical depth, is a functio

n of ( ) and it can be shown that (Problem 2.7)

where

BmyfnTgA

AQ

TgA

VF

Q

cEBym c

c

c

c

c

y

E

21

53

2

1

B

mycc

34

2.4 SECTION FACTOR Z2.4 SECTION FACTOR Z The expression is a function of the depth for a given channel geometry and is known as the section

factor . Thus (2.22) At the critical-flow condition, and (2.23) As a corollary of Eq. (2.23), if is the section factor for an

y depth of flow , then (2.24) where represents the discharge that would make the dep

th critical and is known as the critical discharge.

TAAy

Z

cyy gQTAAZ cccc Z

y

cQy

TAAZ

ZgQc

35

2.5 FIRST HYDRAULIC 2.5 FIRST HYDRAULIC EXPONENT M EXPONENT M In many computations involving a wide range of depths in a

channel, such as in the GVF computations, It is convenient to express the variation of with

in an exponential form. The ( )relationship (2.25) Is found to be very advantageous. In this equation = a coefficient and =an exponent called the first hydraulic

exponent. It is found that generally is a slowly-varying function of the aspect ratio for most of the channel shapes. The variation of and

for a trapezoidal channel is indicated in Fig. 2.8.

Zy

yZ MyCZ 1

2 1C

MM

MB

my

36

37

The value of for a given channel can be determined by preparing a plot of vs on a log-log scale. If is constant between two points ( , ) and ( , )in this plot, the value of

is determined as

(2.26) In Eq.(2.26), instead of Z, a non-dimensionalised Z value c

an also be used. For a trapezoidalchannel,Eq.(220a)represents a non-dimen

sionalised value of Z if the suffix ‘c‘ is removed.

Z yM

1Z 1y 2Z 2y

12

12

log

log2

yy

ZZM

M

38

Hence the slope of vs on a log-log plot, such as in Fig. 2.7, can be used to obtain the value of

at any value of . It may be noted that for a trapezoidal channel is a unitque function of

and will have a value in the range 3.0 to5.0. An estimate of can also be obtained by the relation

(2.27)

Bmy

M MBmy

M

dy

dT

T

AT

A

yM 3

39

EXAMPLE 2.3 Obtain the value of for (a) a rectangular channel and (b) a triangular channel.

Solution For a rectangular channel,

and by Eq. (2.25)

By equating the exponent of on both sides, For a triangular channel of side slope horizontal : 1 verti

cal, , and

by By equating the exponent of on both sides,

M

BTByA ,MyCyB

T

AZ 1

323

2 y 0.3M

m2myA myT 2

MyCmy

ym

T

AZ 1

6332

2

0.5My

40

2.6 COMPUTATIONS2.6 COMPUTATIONS The problems concerning critical depth involve the

following parameters: geometry of the channel, or or . EXAMPLE 2.4 Calculate the critical depth and the

corresponding specific energy for a discharge of in the following channels: (a) Rectangular channel, = 2.0 m (b) Triangular channel, = 0.5 (c) Trapezoidal channel, = 2.0 m, = 1.5 (d) Circular channel, = 2.0 m

Q

cE cy

sm30.5B

Bm

mD

41

Solution (a) Rectangular Channel

Since for a rectangular channel ,

(b) Triangular Channel

From Eq. (2.14)

msmBQq 35.20.2

0.5

mgqyc 860.0

81.9

5.2312

312

5.1c

c

y

EmEc 290.1

51

2

22

gm

Qyc

42

Since for a triangular channel

(c) Trapezoidal Channel

Using Table 2A.2 the corresponding value of

m828.15.081.9

525/1

2

2

mEy

Ec

c

c 284.2,25.1

51843.00.281.9

5.10.525

23

25

23

Bg

Qm

536.0B

mycc

myc 715.0 2197.2715.0715.05.10.2 mAc

43

(d) Circular Channel

From Table 2A.1 showing vs , the corresponding value of by suitable linear interpolation is

smVc 276.2197.20.5

mgVc 265.022

mg

VyE ccc 979.0264.0715.0

2

2

5964.181.9

0.5

g

QZc

2822.05.2 DZc5.2DZc Dy

Dycmy

D

yc

c 074.1,537.0

44

Also, from Table 2A.1 for

Hence

4297.0,537.02

D

A

D

y cc

22 7187.14297.00.2 mAc smVc 909.27187.10.5

mgVc 431.022

mg

VyE ccc 505.1431.0074.1

2

2

45

EXAMPLE 2.5 Calculate the bottom width of a channel required to carry a discharge of as a critical flow at a depth of 1.2m, if the channel section is (a) rectangular and (b) trapezoidal with side slope of 1.5 horizontal : 1 vertical.

Solution (a) Rectangular Section The solution here is straightforward

sm30.15

3

312

i.e. cc gyqg

qy

msmq 33 117.42.181.9

mB 643.34.117

15.0 widthbottom

46

(b) Trapezoidal Channel The solution in this case id by trial-and-error

By trial-and-error

2.18.12.12.15.1 BBAc 6.32.15.12 BBTc

c

c

T

A

g

Q 32

81.9

15

6.3

2.18.1 233

B

B

273.13

6.3

8.1 3

B

B

mB 535.2

47

EXAMLE 2.6 Find the critical depth for a specific energy head of 1.5 m in the following channels:

(a) Rectangular channel, = 2.0 m (b) Triangular channel, = 1.5 (c) Trapezoidal channel, = 2.0 m, = 1.0 (d) Circular channel, = 1.5 mSolution (a) Rectangular channel By Eq. (2.10)

m

B

Bm

D

myE cc 50.12

3

myc 00.13

250.1

48

(b) Triangular channel By Eq. (2.15)

(c) Trapezoidal channel

Since by Eq. (2.4a)

Solving by trial and error,

myE cc 50.125.1

myc 20.125.1

50.1

2

22

22 cc

ccc gA

Qy

g

VyE

c

ccccc T

AyETA

g

Q

2,3

2

c

ccc y

yyy

20.22

0.25.1

myc 095.1

49

(d) Circular channel

By non-dimension with respect to the diameter .

From Table2A.1, values of and for a chosen are read and a trial and error procedure is adopted to solve for . It is found that

c

ccc T

AyE

2

D 0.1

5.1

5.1

2

2

D

E

DT

DA

D

y c

c

cc

2DAc DTc DycDyc

myD

yc

c 035.150.169.0and69.0

50

2.7 TRANSITIONS2.7 TRANSITIONS The concepts of specific energy and critical depth are

extremely useful in the analysis of problems connected with transitions. To illustrate the various aspects, a few simple transitions in rectangular channels are presented here. The principles are nevertheless equally applicable to channels of any shape and other types of transitions.

51

2.7.1 Channel with a Hump2.7.1 Channel with a Hump(a)Subcritical Flow Consider a horizontal, frictionless rectangular channel of width carrying at a depth . Let the flow be subcritical. At a section 2 (Fig. 2.9) a smoot

h hump of height is built on the floor. Since there are no energy losses between sections 1 and 2, and construction of a hump causes the specific energy at section 2 to decrease by . Thus the specific energies at sections 1 and 2 are given by

(2.28)

B Q 1y

Z

Z

g

VyE

2

21

11

ZEE 12

52

Since the flow is subcritical, the water surface will drop due to a decrease in the specific energy. In Fig. 2.10, the water surface which was at at section 1 will come down to point at section 2. The depth

will be given by

(2.29)

PR

2y

22

2

2

2

22

22 22 ygB

Qy

g

VyE

53

54

It is easy to see from Fig. 2.10 that as the value of is increased, the depth at section 2, i.e. , will decrease.

The minimum depth is reached when the point coincides with , the critical-depth point. At this point the hump height will be maximum, say=

, =critical depth and . The condition at is given by the relation

(2.30) The question naturally arises as to what happens when

. The upstream depth has to increase to cause an increase in the specific energy at section 1. If this modified depth is represented by

,then

2yZ

R C

mZ cyy 2 cEE 2

mZ

22

2

21 2 cccm ygB

QyEEZE

mZZ

'1y

55

(2.31) Recollecting the various sequences, when the upstream water level remains stationary at while the depth of flow at section 2 decreases with reaching

a minimum value of at (Fig. 2.11). With further increase in the value of, i.e. for , will change to while y2

will continue to remain at . The variation of and with in the subcritical regime

can be clearly noticed in Fig. 2.11.

21

2

2

11 '2''

ygB

QyE 1111 ' and 'with yyEE

mZZ 0

cy1y

cy mZZ Z

mZZ 1y 1'y

cy

1y 2y Z

56

(b) Supercritical Flow If is in the supercritical flow regime, Fig. 2.10 shows that t

he depth of flow increases due to the reduction of specific energy. In Fig. 2.10 point ,

corresponds to and point to depth at the section 2. Up to the critical depth, increases to reach at .

cy mZZ

1y

1y2y

'R'P

57

the depth over the hump will remain constant and the upstream depth will change. It will decrease to have a higher specific energy .

The variation of the depths and with in the supercritical flow is shown in Fig. 2.12.

12 yy 1y

1'E

1y 2y Z

58

EXAMPLE 2.7 A rectangular channel has a width of 2.0 m and carries a discharge of 4.80 with a depth of 1.60 m. At a certain section a small, smooth hump with a flat top and of height 0.10 m is proposed to be built. Calculate the likely change in the water surface. Neglect the energy loss.

Solution Let the suffixes 1 and 2 refer to the upstream and

downstream sections respectively as in Fig. 2.9.

msmq 340.20.2

8.4

mg

VsmV 115.0

2,50.1

6.1

40.2 21

1

sm3

59

, hence the upstream flow is subcritical and the hump will cause a drop in the water-surface elevation.

At section 2,

391.0111 gyVF

mE 7515.1115.060.11

mZEE 615.110.0715.112

myc 837.0

81.9

4.2312

myE cc 256.15.1

60

The minimum specific energy at section 2, is less than , the available specific energy at that section. Hence and the upstream depth will remain unchanged. The depth is calculated by solving the specific-energy relation

i.e.

Solving by trial and error,

2cE2E

cyy 2 1y2y

2

22

2 2E

g

Vy

615.1

81.92

4.222

2

2

y

y

my 481.12

61

EXAMPLE 2.8 In Example 2.7, if the height of the hump is 0.5 m, estimate the water surface elevation on the hump and at a section upstream of the hump.

Solution Form Example 2.7 : =0.391, =1.715 m and . Available specific energy at section

The minimum specific energy at section 2 is greater than ,the available specific energy at that section. Hence, the depth at section 2 will be at the critical depth. Thus = = 1.256 m. The upstream depth will increase to a depth such that the new specific energy at the upstream section 1 is

1F 1Emyy cc 837.02

ZEE 122mE 215.1500.0715.12

myE cc 256.15.1 22

2E

2E 2cE1y 1'y

62

Thus

ZEE c 21'

ZEg

VyE c 2

21

11 2

'''

756.1500.0256.1'2

'21

2

1 gy

qy

756.1

'81.92

4.2'

21

2

1

y

y

756.1'

2936.0'

21

1 y

y

63

Solving by trial and error and selecting the positive root which gives , = 1.648 m.

The nature of the water surface is shown in Fig. 2.1321' yy 1'y

64

EXAMPLE 2.9 A rectangular channel 2.5 m wide carries 6.0 of flow at a depth of 0.5 m. Calculate the height of a flat topped hump required to be placed at a section to cause critical flow. The energy loss due to the obstruction by the hump can be taken as 0.1 times the upstream velocity head.

Solution

hence the flow is supercritical

sm3

smVmsmq 8.45.0

4.2,4.2

5.2

0.61

3

mgV 174.1221

,167.25.081.9

80.41

F

mE 674.1174.150.01

65

Since the critical flow is desired at section 2

By the energy equation between section 1 and 2,

Where Hence

2

312

837.081.9

4.2yyc

g

Vy

g

V cc

2419.0

22

22

2

Zg

VyEE L

2

22

21

mgVEL 117.021.0lossenergy 21

hump theofheight ZZ 419.0837.0117.0674.1

mZ 301.0

66

2.7.2 Transition with a Change in 2.7.2 Transition with a Change in WidthWidth(a) Subcritical Flow in a Width Constriction Consider a frictionless horizontal channel of width carrying a discharge at a depth as in Fig. 2.14. At sec

tion 2 the channel width has been constricted to by a smooth transition. Since there are no losses involved and since the bed elevations at sections 1 and 2 are same, the specific energy at section 1 is equal to the specific energy at section 2 .

1B Q 1y

2B

21

21

2

1

21

11 22 ygB

Qy

g

VyE

67

22

22

2

2

22

22 22 ygB

Qy

g

VyE

68

It is convenient to analysis the flow in terms of the discharge intensity . At section 1,

and at section 2, . Since , . In the specific energy diagram (Fig. 2.15) drawn with

the discharge intensity as the third parameter, point

BQq 11 BQq 22 BQq 12 BB 12 qq

69

on the curve corresponds to depth and specific energy . Since at section 2, and

, point will move vertically downward to point on the curve to reach the depth .

Thus, in subcritical flow the depth .If is made smaller, then will increase and will decrease. The limit of the contracted width

is obviously reached when corresponding to , the discharge intensity , i.e. the maximum discharge intensity for a given specific energy (critical-flow condition) will prevail 1. At this minimum width, =critical depth at Section 2,

and (2.33)

1qP 1y

1E 12 EE 2qq PR 2q 2y

12 yy 2B2q 2y

mBB 22

1Emqq 2

2ycmy

222

2

1 )(2 cmmcmcm yBg

QyEE

70

For a rectangular channel, at critical flow Since (2.34)

and

i.e. (2.35) If , the discharge intensity will be larger than the

maximum discharge intensity consistent with .The flow will not, therefore, be possible with the given upstream conditions.

cc Ey3

2

cmEE 1

12 3

2

3

2EEyy cmcm

3

2

2

31

22

2

cmm

mc gy

QBor

gB

Qy

31

2

2 8

27

gE

QB m

mBB 22 2qmq

1E

71

The upstream depth will have to increase to so that a new specific energy is formed which will just be sufficient to cause critical

flow at section 2. It may be noted that the new critical depth at section 2 for a rectangular channel is

'1y

21

21

2

11 '2''

yBg

QyE

72

and

Since , will be larger than . Further, . Thus even though critical flow prevails for all

, the depth at section 2 is not constant as in the hump case but increases as

,and hence , rises. The variation of , and with shown schematically in Fig.216.

3122

31

22

2

2 gqgB

Qyc

2

22

22 5.12

' cc

cc yg

VyE

mBB 22 2cy cmy

221 5.1' cc yEE mBB 22

1'y 1'E 1y 2yE 12 / BB

73

(b) Supercritical Flow in a Width Constriction If the upstream depth is in the supercritical flow

regime, a reduction of the flow width and hence an increase in the discharge intensity cause a rise in depth . In Fig.2.15, point , corresponds to

and point to .

1y

2y 'P 1y'R 2y

74

Choking In the case of a channel with a hump, and also in the

case of a width constriction, it is observed that the upstream water-surface elevation is not affected by the conditions at section 2 till a critical stage is first achieved.

EXAMPLE 2.10 A rectangular channel is 3.5 m wide and conveys a discharge of 15.0 at a depth of 2.0 m. It is proposed to reduce the width of the channel at a hydraulic structure. Assuming the transition to be horizontal and the flow to be frictionless determine the water surface elevations upstream and downstream of the constriction when the constricted width is (a) 2.50 m and (b) 2.20 m

sm3

75

Solution Let suffixes l and 2 denote sections upstream and downstre

am of the transition respectively. Discharge

The upstream flow is subcritical and the transition will cause a drop in the water surface.

111 VyBQ

smV 143.20.25.3

0.151

484.00.281.9

143.2number Froude

1

11

gy

VF

m

g

VyE 234.2

81.92

143.20.2

2

221

11

76

Let = minimum width at section 2 which does not cause choking.

Then

Since

mB2

mEEcm 234.21

mEy cmcm 489.1234.23

2

3

2

22

23

mcm gB

Qy

mgy

QB

cm 636.2

489.181.9

0.1521

3

221

32

2

2

77

(a) When =2.50 m and hence choking conditions prevail. The depth at

the section .The upstream depth will increase to .

Actual

At the upstream section 1: with new upstream depth of such that

.

2BmBB 22

222 cyy 1y 1'y

msmq 32 0.6

5.2

0.15

m

g

qyc 542.1

81.9

0.6312312

22

myE cc 3136.2542.15.15.1 22

3136.2' 21 cEE 1'ymsmVyq 3

111 2857.45.315''

78

Hence

Solving by trial and error and selecting a root that gives subcritical flow,

(b) when =2.20 m As choking condition prevail. Depth at section

3136.22

''

21

1 g

Vy

3136.2

'81.92

2857.4'

21

21

1

y

y

3136.2'

9362.0'

21

1 y

y

my 102.2'1 2B

mBB 22

222 cyy

msmq 22 8182.6

20.2

0.15

79

At upstream section 1 : New upstream depth= and

Hence

6797.1

81.9

8182.6312

2

cy

5195.25.1 22 cc yE

mEE c 5195.2' 21

msmyVq 3111 2857.45.315''

5195.2'2

'' 2

1

21

1 gy

qy

5195.2

'81.92

2857.4'

21

2

1

y

y

5195.2'

9362.0'

21

1 y

y my 350.2'1

1'y

80

2.7.3 General Transition2.7.3 General Transition A transition in its general form may have a change of chann

el shape, provision of a hump or a depression, and contraction or expansion of channel width, in any combination. In addition, there may be various degrees of loss of energy at various components.

EXAMPLE 2.11 A discharge of 16.0 flows with a depth of 2.0 m in a rectangular channel 4.0 m wide. At a downstream section the width is reduced to 3.5 m and the channel bed is raised by . Analysis the water-surface elevations in the transitions when (a) =0.20m and (b) =0.35m.

sm3

Z

Z Z

81

Solution Let the suffixes 1 and 2 refer to the upstream and downstream sections respectively. At the upstream section,

The upstream flow is subcritical and the transition will cause a drop in the water surface elevation.

smV 0.224

161

452.00.281.9

0.2number Froude

1

11

gy

VF

mgV 204.0221

mE 204.2204.00.21 section downstream at theintensity dischange2 q

msmB

Q 3571.45.3

0.16

82

(a) When = 0.2 m

Hence the depth and the upstream depth will remain unchanged at .

22 toingcorresponddepth critical qyc

m

g

q287.1

81.9

571.4312312

2

myE cc 930.12

322

Z2section at energy specific available2 E

21 004.220.0204.2 cEmZE 22 cyy 1y

1

22

2 2EZ

g

Vy

83

Solving by trial and error, = 1.575 m. Hence when = 0.20 m, = 2.00 m and = 1.575 m (b) When = 0.35,

Hence the contraction will be working under choked conditions. The upstream depth must rise to create a higher total head. The depth of flow at section 2 will be critical with .

If the new upstream depth is

20.0204.2

81.92

571.422

2

2

y

y

004.2065.1

22

2 y

y

2yZ

2y1y

Z2section at energy specific available2 E

2854.1350.0204.2 cEm

myy c 287.122 1'y

84

i.e.

By trial-and-error, The upstream depth will therefore rise by 0.094 m due to t

he choked condition at the constriction. Hence, when = 0.35 m

350.0930.1'2

' 221

21

2

1 ZEygB

Qy c

28.2'0.481.92

16'

21

2

2

1

y

y

280.2'

8155.0'

21

1 y

y

Z

myy

y

c 287.1

094.2'

22

1

85

PROBLEMSPROBLEMS2.3 A rectangular channel 5.0 m wide carries 20 of discharge at depth of 2.0 m. The width beyond a certain section is to be charged to 3.5 m. If it is desired to keep the water-surface elevation unaffected by this change, what modifications are needed to the bottom elevation?2.4 Find the alternate depths corresponding to a specific head of 2.0 m and a discharge of 6.0 in (a) trapezoidal channel, = 0.9 m, = 1.0, (b) triangular channel, = 1.5, (c) circular channel, = 2.50m. (Use the trial and error method. For Part (c) use Table 2A.1.)

sm3

sm3

Bm

m

D

86

2.7 A trapezoidal channel has a bottom width of 6.0 m and side slope of 1:1. The depth of flow is 1.5 m at a discharge of 15 . Determine the specific energy and alternate depth.2.18 A triangular channel has an apex angle of 60 and carries a flow with a velocity of 2.0 and depth of 1.25 m. (a) Is the flow subcritical or super-critical? (b) What is the critical depth? (c) what is the specific energy? (d) What is the alternate depth possible for this specific energy?

sm3

sm

87

2.19 Fill the missing data in the following table connected with critical depth computations in rectangular channels:

2.20 Fill the missing data in the following table connected with critical depth computations in triangular channels: