1 chapter 2 energy depth relationships. 2 2.1 specific energy the total energy of a channel flow...
TRANSCRIPT
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Chapter Chapter 22
Energy Depth Energy Depth RelationshipsRelationships
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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
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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,
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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
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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
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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
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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
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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
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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
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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
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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'
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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
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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
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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
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(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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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10
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
![Page 46: 1 Chapter 2 Energy Depth Relationships. 2 2.1 SPECIFIC ENERGY The total energy of a channel flow referred to a datum is given by Eq. (1.39) as If the](https://reader038.vdocuments.site/reader038/viewer/2022110320/56649c905503460f9494a61d/html5/thumbnails/46.jpg)
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
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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
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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
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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
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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.
![Page 51: 1 Chapter 2 Energy Depth Relationships. 2 2.1 SPECIFIC ENERGY The total energy of a channel flow referred to a datum is given by Eq. (1.39) as If the](https://reader038.vdocuments.site/reader038/viewer/2022110320/56649c905503460f9494a61d/html5/thumbnails/51.jpg)
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
![Page 52: 1 Chapter 2 Energy Depth Relationships. 2 2.1 SPECIFIC ENERGY The total energy of a channel flow referred to a datum is given by Eq. (1.39) as If the](https://reader038.vdocuments.site/reader038/viewer/2022110320/56649c905503460f9494a61d/html5/thumbnails/52.jpg)
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
![Page 53: 1 Chapter 2 Energy Depth Relationships. 2 2.1 SPECIFIC ENERGY The total energy of a channel flow referred to a datum is given by Eq. (1.39) as If the](https://reader038.vdocuments.site/reader038/viewer/2022110320/56649c905503460f9494a61d/html5/thumbnails/53.jpg)
53
![Page 54: 1 Chapter 2 Energy Depth Relationships. 2 2.1 SPECIFIC ENERGY The total energy of a channel flow referred to a datum is given by Eq. (1.39) as If the](https://reader038.vdocuments.site/reader038/viewer/2022110320/56649c905503460f9494a61d/html5/thumbnails/54.jpg)
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
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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
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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
![Page 57: 1 Chapter 2 Energy Depth Relationships. 2 2.1 SPECIFIC ENERGY The total energy of a channel flow referred to a datum is given by Eq. (1.39) as If the](https://reader038.vdocuments.site/reader038/viewer/2022110320/56649c905503460f9494a61d/html5/thumbnails/57.jpg)
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
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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
![Page 59: 1 Chapter 2 Energy Depth Relationships. 2 2.1 SPECIFIC ENERGY The total energy of a channel flow referred to a datum is given by Eq. (1.39) as If the](https://reader038.vdocuments.site/reader038/viewer/2022110320/56649c905503460f9494a61d/html5/thumbnails/59.jpg)
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
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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
![Page 61: 1 Chapter 2 Energy Depth Relationships. 2 2.1 SPECIFIC ENERGY The total energy of a channel flow referred to a datum is given by Eq. (1.39) as If the](https://reader038.vdocuments.site/reader038/viewer/2022110320/56649c905503460f9494a61d/html5/thumbnails/61.jpg)
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
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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
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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
![Page 64: 1 Chapter 2 Energy Depth Relationships. 2 2.1 SPECIFIC ENERGY The total energy of a channel flow referred to a datum is given by Eq. (1.39) as If the](https://reader038.vdocuments.site/reader038/viewer/2022110320/56649c905503460f9494a61d/html5/thumbnails/64.jpg)
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
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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
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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
![Page 67: 1 Chapter 2 Energy Depth Relationships. 2 2.1 SPECIFIC ENERGY The total energy of a channel flow referred to a datum is given by Eq. (1.39) as If the](https://reader038.vdocuments.site/reader038/viewer/2022110320/56649c905503460f9494a61d/html5/thumbnails/67.jpg)
67
22
22
2
2
22
22 22 ygB
Qy
g
VyE
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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
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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
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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
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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
![Page 72: 1 Chapter 2 Energy Depth Relationships. 2 2.1 SPECIFIC ENERGY The total energy of a channel flow referred to a datum is given by Eq. (1.39) as If the](https://reader038.vdocuments.site/reader038/viewer/2022110320/56649c905503460f9494a61d/html5/thumbnails/72.jpg)
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
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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
![Page 74: 1 Chapter 2 Energy Depth Relationships. 2 2.1 SPECIFIC ENERGY The total energy of a channel flow referred to a datum is given by Eq. (1.39) as If the](https://reader038.vdocuments.site/reader038/viewer/2022110320/56649c905503460f9494a61d/html5/thumbnails/74.jpg)
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
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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
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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
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(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''
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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
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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
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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.
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Z
Z Z
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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
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(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
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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
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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
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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.)
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Bm
m
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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?
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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: