Uniform Flow: Basic concepts of free surface flows, velocity and pressure distribution, Mass, energy and momentum principle for prismatic and non-prismatic channels, Review of Uniform flow: Standard equations, hydraulically efficient channel sections, compound sections, Energy-depth relations: Concept of specific energy, specific force, critical flow, critical depth, hydraulic exponents, and Channel transitions. An open channel is a conduit in which a liquid flows with a free surface. The free surface is actually an interface between the moving liquid and an overlying fluid medium and will have constant pressure. In civil engineering applications; water is the most common liquid with air at atmospheric pressure as the overlying fluid. The prime motivating force for open channel flow is gravity. An open channel is a conduit in which a liquid flows with a free surface. The free surface is actually an interface between the moving liquid and an overlying fluid medium and will have constant pressure. In civil engineering applications; water is the most common liquid with air at atmospheric pressure as the overlying fluid. The prime motivating force for open channel flow is gravity. An open channel is a conduit in which a liquid flows with a free surface. The free surface is actually an interface between the moving liquid and an overlying fluid medium and will have constant pressure. In civil engineering applications; water is the most common liquid with air at atmospheric pressure as the overlying fluid. The prime motivating force for open channel flow is gravity.

1. Open Channel Flow
LEARNING OBJECTIVES
 Uniform Flow: Basic concepts of free surface flows,
 velocity and pressure distribution,
 Mass, energy and momentum principle for prismatic and non-prismatic
channels,
 Review of Uniform flow: Standard equations,
 hydraulically efficient channel sections,
 compound sections,
 Energy-depth relations:
 Concept of specific energy, specific force,
 critical flow, critical depth,
 hydraulic exponents, and
 Channel transitions.
1

2. Open Channel Flow
An open channel is a conduit in which a liquid flows with a free
surface.
The free surface is actually an interface between the moving liquid
and an overlying fluid medium and will have constant pressure.
In civil engineering applications; water is the most common liquid
with air at atmospheric pressure as the overlying fluid.
The prime motivating force for open channel flow is gravity.
OPEN CHANNEL FLOW
2

3. Open Channel Flow
3

4. Open Channel Flow
TYPES OF CHANNELS
Rigid channels are those in which the boundary is not deformable in the sense
that the shape, platform and roughness magnitudes are not functions of the
flow parameters.
Typical examples include lined canals, sewers and non-erodible unlined canals.
The flow velocity and shear-stress distribution will be such that no major
scour, erosion or deposition takes place in the channel and the channel
geometry and roughness are essentially constant with respect to time.
We have many unlined channels in alluvium—both man- made channels and
natural rivers—in which the boundaries undergo deformation due to the
continuous process of erosion and deposition due to the flow.
The boundary of the channel is mobile in such cases and the flow carries
considerable amounts of sediment through suspension and in contact with the
bed. Such channels are classified as mobile-boundary channels.
On the basis of the nature of the boundary open channel
Rigid and Mobile Boundary Channels
4

5. Open Channel Flow
5

6. Open Channel Flow
Prismatic Channel-
A channel in which the cross-sectional shape and size and also the
bottom slope are constant is termed as a prismatic channel.
Most of the man-made (artificial) channels are prismatic channels over
long stretches.
he rectangle, trapezoid, triangle and circle are some of the commonly
used shapes in manmade channels.
Non- Prismatic Channel-
All natural channels generally have varying cross-sections and
consequently are non-prismatic
Prismatic & Non Prismatic Channel
6

7. Open Channel Flow
1. Steady and Unsteady Flow
2. Uniform and Non-uniform Flow
3. Laminar and Turbulent Flow
4. Sub-critical, Critical and Super-critical Flow
CLASSIFICATION OF FLOWS
7

8. Open Channel Flow
CLASSIFICATION OF FLOWS
Steady and Unsteady Flows
 A steady flow occurs when the flow properties, such as the depth
or discharge at a section do not change with time.
 And if the depth or discharge changes with time the flow is termed
unsteady.
 Flood flows in rivers and rapidly
varying surges in canals are
some examples of unsteady
flows. Unsteady flows are
considerably more difficult to
analyse than steady flows.
8

9. Open Channel Flow
 If the flow properties, say the depth of flow, in an open channel
remain constant along the length of the channel, the flow is said to
be uniform.
 As a corollary of this, a flow in which the flow properties vary along
the channel is termed as non-uniform flow or varied flow.
 A prismatic channel carrying a certain discharge with a constant
velocity is an example of uniform flow [Fig. 1.1(a)].
 In this case the depth of flow will be constant along the channel
length and hence the free surface will be parallel to the bed.
 It is easy to see that an unsteady uniform flow is practically
impossible, and hence the term uniform flow is used for steady
uniform flow.
 Flow in a non-prismatic channel and flow with varying velocities in
a prismatic channel are examples of varied flow. Varied flow can be
either steady or unsteady.
Uniform and Non-uniform Flows
9

10. Open Channel Flow
Typesof Non-uniform Flow
1. Gradually Varied Flow (GVF)
If the depth of the flow in a channelchangesgradually over a
length of the channel.
2. Rapidly Varied Flow (RVF)
If the depth of the flow in a channelchangesabruptly over a
smalllength of channel
10

11. Open Channel Flow
Laminar and Turbulent Flow
Bothlaminar and turbulent flow canoccurin open channels depending
onthe Reynoldsnumber(Re)
Re = ρVR/µ Where,
ρ = density of water = 1000 kg/m3
µ = dynamicviscosity
R= Hydraulic Mean Depth = Area / WettedPerimeter
11

12. Open Channel Flow
Sub-critical, Critical and Super-critical Flow
12

13. Open Channel Flow
13

14. Open Channel Flow
14

15. Open Channel Flow
15

16. Open Channel Flow
Velocity Distribution
 Velocity is always vary across channel; because of friction along the boundary
 The maximum velocity usually found just below the surface.
 the velocity vectors of the flow to have components only in the longitudinal and
lateral direction but also in normal direction to the flow.
 In a macro-analysis, one is concerned only with the major component, viz., the
longitudinal component, vx. The other two components being small are ignored.
 The distribution of v in a channel is
dependent on the geometry of the
channel.
 Figure show isovels (contours of
equal velocity) of v for a natural and
rectangular channel respectively.
 The influence of the channel
geometry is apparent. The velocity v
is zero at the solid boundaries and
gradually increases with distance
from the boundary
16

17. Open Channel Flow
17

18. Open Channel Flow
a. Natural Channel , b. Rectangular Channel
18

19. Open Channel Flow
 A typical velocity profile at a section in a plane normal to the direction of flow is
presented in figure.
 Max velocity occurs at a distance 0.0yo to 0.25yo from the free surface.
 Deeper and narrower channel, more deep will be the max velocity from the surface
 Field observations in rivers and canals have shown that the average velocity at any
vertical Vav occurs at a level of 0.6 yo from the free surface, where yo = depth of
flow.
Further, it is found that
Vav = (V0.2 + V0.8) /2
which V0.2 = velocity at a depth of 0.2 yo from
the free surface, and V0.8 = velocity at a depth
of 0.8 yo surface.
The surface velocity Vs is related to the average
velocity Vav as
Vav = k Vav
where, k = a coefficient with a value between
0.8 and 0.95.
Depends on Channel section. 19

20. Open Channel Flow
VELOCITY DISTRIBUTION – 1 D APPROACH
Velocity profile :
Discharge through elemental area dA having velocity v
dQ = vdA
 So, Total discharge Q = ʃ vdA
 Considering Avg velocity V and area A
Q= VA
VA = ʃ vdA
Therefore Avg velocity V = 1/A ʃ vdA
dA
Velocity -
Depth
20

21. Open Channel Flow
21

22. Open Channel Flow
22

23. Open Channel Flow
PRESSURE DISTRIBUTION IN OPEN CHANNEL
 The intensity of pressure for a liquid at its free surface is equal to that of
the surrounding atmosphere.
 Since the atmospheric pressure is commonly taken as a reference and of
value equal to zero, the free surface of the liquid is thus a surface of zero
pressure.
 This linear variation of pressure with depth having the constant of
proportionality equal to the unit weight of the liquid is known as
hydrostatic-pressure distribution.
Pressure distribution in still water:
W = PA – Patm
W = PA -0
P =
𝑊
𝐴
=
𝑦 𝐴 Υ
𝐴
= y. ϒ = ρw g .y
P = y. ϒ = ρw g .y 23

24. Open Channel Flow
 Let usconsider a channel with a very small value of the longitudinal
slope θ.
 Let θ~ sin θ ~ 1/1000. For such channels the vertical section is
practically the same as the normal section.
 If a flow takes place in this channel with the water surface parallel to
the bed, i.e. uniform flow, the streamlines will be straight lines and as
such in a vertical direction the normal acceleration an = 0.
P = y. ϒ
𝑃
ϒ
=y and
𝑃
ϒ
+ Z = Z1
 Thus the piezometric head at any point in the channel will be equal to the water
surface elevation.
 The hydraulic grade line will therefore lie essentially on the water surface.
Channels with Small Slope
24

25. Open Channel Flow
Figure shows a uniform free-surface flow in achannel with a large value of
inclination θ.
The flow is uniform, i.e. the water surfaceis parallel to the bed. An element
of length Δ L and unit width is considered at the
Channels with Large Slope
At any point A at a depth y measured
normal to the water surface, the weight
of column A1 1′A′ = γΔLy and acts
vertically downwards. The pressure at
AA′ supports the normal component of
the column A1 1′A′.
Thus
pAΔL = γ y ΔL cosθ
pA = γ y cosθ
pA / γ = γ cosθ
25

26. Open Channel Flow
 The pressure pA varies linearly with the depth y but the constant of
proportionality is γ cos θ.
 If h = normal depth of flow, the pressure on the bed at point 0, p0= γ h cos
θ.
 If d = vertical depth to water surface measured at the point O,
then h = d cos θ and the pressure head at point O, on the bed is given by
p0 = h cos θ = d cos2 θ
26

27. Open Channel Flow
1. For the channels given below – Calculate the discharge.
5mX3m
Depth of flow = 2.30 m
Velocity i) = 2.8m/s ( 32.2 )
ii) = 3.15 m/s (36.5)
Depth of flow = 4 m
Velocity i) = 2.25m/s (99)
ii) = 5.12 m/s (225.8)
5m
1v:1.5H
(b)
(a)
27

28. Open Channel Flow
In the case of steady-uniform flow in an open channel, the
following main features must be satisfied:
 The water depth, water area, discharge, and the velocity
distribution at all sections throughout the entire channel length
must remain constant, i.e.; Q , A , y , V remain constant through
the channel length.
The slope of the energy gradient line (S), the water surface
slope (Sws), and the channel bed slope (S0) are equal i.e.; S = Sws
= S0
Flow Formulas in Open Channels (steady-uniform flow):
28

29. Open Channel Flow
The Chezy Formula: (1769)
Empirical formulas are used to describe the flow in
open channels.
The Chezy formula is probably the first formula derived for uniform flow. It may
be expressed in the following form:
𝐕 = 𝐂 𝐑𝐒
where C = Chezy coefficient (Chezy’s resistance factor), m1/2/s.
29

30. Open Channel Flow
The Manning Formula: (1895)
Using the analysis performed on his own experimental data and
on those of others, Manning derived the following empirical
relation:
𝐶 =
1
𝑛
𝑅
1
6
where n = Manning’s coefficient for the channel roughness, m-
1/3/s.
𝑉 =
1
𝑛
𝑅2/3 𝑆
30

31. Open Channel Flow
31

32. Open Channel Flow
GEOMETRIC ELEMENTS OF OPEN CHANNELS
A channel section is defined as the cross-section taken perpendicular to the
main flow direction.
Referring to Figure, the geometric elements of an open channel are defined as
follows;
Flow depth, y: Vertical distance from the channel bottom to the free
surface.
Depth of flow section, d : Flow depth measured perpendicular to the
channel bottom.
The relationship between d and y is d= yCosϴ .
For most manmade and natural channels Cosϴ = 1.0,
and therefore y = d.
The two terms are used interchangeably. 32

33. Open Channel Flow
Top width, T: Width of the channel section at free surface.
Wetted perimeter, P: Length of the interface between the water
and the channel boundary.
Flow area, A: Cross-sectional area of the flow.
Hydraulic depth, D: Flow area divided by top width,
D = A/T.
Hydraulic radius, R: Flow area divided by wetted perimeter,
R = A/P.
Bottom slope,S0: Longitudinal slope of the channel bottom,
S0=tanϴ = sinϴ .
33

34. Open Channel Flow
Geometric elements of channel sections
34

35. Open Channel Flow
35

36. Open Channel Flow
1. The cross section of an open channel is a trapezoid with a bottom width of 4
m and side slopes 1:2, calculate
i) the discharge if the depth of water is 1.5 m and bed slope = 1/1600. Take Chezy
constant C = 50.
Ii) Measure the discharge using Manning’s formula if the channel lining is of smooth
concrete.
36

37. Open Channel Flow
 A section of a channel is said to be most economical when the cost of construction
of the channel is minimum.
 But the cost of construction of a channel depends on excavation and the lining.
 To keep the cost down or minimum, the wetted perimeter, for a given discharge,
should be minimum.
 This condition is utilized for determining the dimensions of economical sections of
different forms of channels.
 Most economical section is also called the best section or most efficient section as
the discharge, passing through a most economical section of channel for a given
cross-sectional area A, slope of the bed S0 and a resistance coefficient, is
maximum.
 But the discharge Q = AV
𝑄 = 𝐴𝐶 𝑅𝑆
𝑄 = 𝐴𝐶
𝐴
𝑃
𝑆 = Const.
1
𝑃
 Hence the discharge Q will be maximum when the wetted perimeter P is minimum
MOST ECONOMICAL SECTION OF CHANNELS
37

38. Open Channel Flow
Most Economical Rectangular Channel:
Consider a rectangular section of channel as shown.
Let B = width of channel, D = depth of flow.
Area of flow,
A = B x D ……………………………….(7.4a)
Wetted perimeter,
P = 2D + B……………….……(7.4b)
from Eq. (7.4a) we have B = A/D
P = 2D+ A/D ………………………….…….. (7.4c)
For most economical cross section, P should be
minimum for a given area;
𝑑𝑃
𝑑𝐷
= 0
𝑑
𝑑𝐷
2𝐷 +
𝐴
𝐷
= 2 −
𝐷
𝐴2
= 0
𝐷
(𝐵𝐷)2
= 2
D =
𝑩
38

39. Open Channel Flow
From equations (7.5) and (7.6), it is clear that rectangular channel
will be most economical when either:
(a) The depth of the flow is half the width or
(b) The hydraulic radius is half the depth of flow
39

40. Open Channel Flow
1. For a rectangular channel with width = 4m and bed
slope = 1/1500, Calculate the max discharge for the
most economical condition when Chezy’s constant C =
50.
D = B/2
= 4/2 = 2m
A = 2X4 = 8m2
Q = AV
= 8 50 (R (1/1500))1/2 = 10.328 m3/s
R = D/2 = 1
40

41. Open Channel Flow
Consider a trapezoidal section of channel as shown.
Most Economical Trapezoidal Channel
41

42. Open Channel Flow
42

43. Open Channel Flow
43

44. Open Channel Flow
 Best side slope for most economical trapezoidal section can be shown to
be when n = 1/ 𝟑
 So far we assumed that the side slopes are constant. Let us now consider
the case when the side slopes can also vary.
 The most economical side slopes of a most economical trapezoidal
section can be obtained as follows:
44

45. Open Channel Flow
45

46. Open Channel Flow
 Therefore, best side slope is at 60o to the horizontal, i.e.;
of all trapezoidal sections a half hexagon is most
economical.
 However, because of constructional difficulties, it may
not be practical to adopt the most economical side
slopes.
46

47. Open Channel Flow
 For a most economical section the discharge, for a constant cross-sectional
area, slope of bed and resistance coefficient, is maximum (or P is minimum).
 But in the case of circular channels, the area of the flow cannot be maintained
constant.
 Indeed, the cross-sectional area A and the wetted perimeter P both do not
depend on D but they depend on the angle a.
Most Economical Circular Channel
 Referring to the figure shown, we can
determine the wetted perimeter P and the
area of flow A as follows:
Let
D = depth of flow
d = diameter of pipe
r = radius of pipe
2a = angle subtended by the free surface at
the centre (in radians)
47

48. Open Channel Flow
A1
a b
O
A2
D
α
M
48

49. Open Channel Flow
Consequently, the cross-sectional area A and the wetted perimeter
P both depend on the angle α which is the most suitable variable.
Thus in case of circular channels, for most economical section, two
separate conditions are obtained:
1) condition for maximum discharge, and
2) condition for maximum velocity
49

50. Open Channel Flow
Condition for Maximum Discharge for Circular Section
50

51. Open Channel Flow
51

52. Open Channel Flow
 This means that the maximum discharge (minimum P) in a
circular channel occurs when the of the pipe.
 The above results holds good when the Chezy formula is
used.
 If Manning’s formula is used, results will be:
α = 151o and D = 0.94 d
52

53. Open Channel Flow
Condition for Maximum Velocity for Circular Section
Since the cross-sectional area A varies with α, the condition for the
maximum velocity is different from the condition for the maximum
discharge.
53

54. Open Channel Flow
54

55. Open Channel Flow
1. For a most economical open channel of Trapezoidal section with side slope
of 2V:3H, Evaluate the wetted perimeter if discharge flowing through it is
10m3/s and the velocity of flow is 1.5 m/s and bed slope is 0.014.
2. A nearly horizontal channel has a bottom width of 3 ft, and it carries a
discharge of 60 cfs at a depth of 4 ft. Determine the magnitude and direction of
the hydrostatic pressure force exerted on each of the sidewalls per unit length of
the channel if -
(a) the channel is rectangular with vertical sidewalls
(b) the channel is trapezoidal with each sidewall sloping outward at a slope 2
horizontal over 1 vertical, that is m= 2.
3. The base of the most economical trapezoidal channel section is 6m and the
side slope is 1H:2V, Calculate the maximum discharge through the channel if
the bed slope is 1 in 1000 and C = 50.
NUMERICALS
55

56. Open Channel Flow
4. The top width of a most economical trapezoidal channel section is 8m,
determine the hydraulic radius of the channel if the side slope is 1H:3V.
(1.9m)
5. The top width of a most economical trapezoidal channel section is 7m
and the side slope of the channel is 1H:2V, determine the depth of the
channel section. (3.13m)
6. A circular channel is proposed to lay on a slope of 1 in 2000 and is
required to carry 1.5cumec. What size of circular channel should be used
if it has to flow half-full take n=0.015. (2.1m)
56

57. Open Channel Flow
Energy Principles in Open Channel Flow
Referring to the figure shown, the total energy of a flowing liquid per unit weight is
given by,
Total Energy
E =Z +y +
𝑽𝟐
𝟐𝒈
Where
Z = height of the bottom of
channel above datum,
y = depth of liquid,
V = mean velocity of flow.
If the channel bed is taken as the datum (as shown), then the total energy per unit
weight will be,
Es= y +
𝑽𝟐
𝟐𝒈
This energy is known as specific energy, Es, Specific energy of a flowing liquid in
a channel is defined as energy per unit weight of the liquid measured from the
channel bed as datum. It is a very useful concept in the study of open channel flow.
57

58. Open Channel Flow
which is valid for any cross section.
58

59. Open Channel Flow
It is defined as the curve which shows the variation of specific energy (Es ) with depth
of flow y. Let us consider a rectangular channel in which a constant discharge is taking
place.
If q = discharge per unit width =
𝑄
𝐵
= constant ( since Q and B are constants),
Then, Velocity of flow,
V =
𝑄
𝐴
=
𝑄
𝐵𝑋𝑦
=
𝑞 𝐵
𝐵 𝑦
=
𝑞
𝑦
Substituting for V into above equations, we get
Es = y +
𝑞2
2𝑔𝑦2 = Ep +Ek
Specific Energy Curve (rectangular channel)
59

60. Open Channel Flow
θ
y
Z
V2/2g
60

61. Open Channel Flow
SECTION FACTOR Z
61

62. Open Channel Flow
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 Z with y in an exponential form.
The (Z – y) relationship
Z2 = C1yM
In this equation C1 = a coefficient and M= an exponent called the
first hydraulic exponent. It is found that generally M is a slowly-
varying function of the aspect ratio for most of the channel shapes.
FIRST HYDRAULIC EXPONENT
62

63. Open Channel Flow
 The concepts of specific energy and critical energy are useful in
the analysis of transition problems.
 Transitions in rectangular channels are presented here. The
principles are equally applicable to channels of any shape and
other types of transitions
CHANNEL TRANSITIONS
1.Channel with a Hump
1.Transition with a Change in Width
2.General Transition
63

64. Open Channel Flow
a)Subcritical Flow
Channel with a Hump
 Consider a horizontal, frictionless rectangular channel of width B carrying
discharge Q at depth y1.
 Let the flow be subcritical. At a section 2, smooth hump of height ΔZ is built on
the floor.
 Since there are no energy losses between sections 1 and 2, construction of a
hump causes the specific energy at section to decrease by ΔZ. Thus the specific
energies at sections 1 and 2 are,
E1 =y1 +
𝑽1
𝟐
𝟐𝒈
E2 =E1 - ΔZ
64

65. Open Channel Flow
 The flow is subcritical, the water surface will drop due to a decrease in the
specific energy.
 In Figure the water surface which was at P at section 1 will come down to
point R at section 2. The depth y2 will be given by-
E2 = y2 +
𝑽2
𝟐
𝟐𝒈
= 𝒚𝟐 +
Q𝟐
𝟐𝒈𝑩𝟐
𝒚𝟐
𝟐
 As the value of ΔZ is increased, the
depth at section 2, y2, will decrease.
The minimum depth is reached when
the point R coincides with C, the
critical depth.
 At this point the hump height will be
maximum, ΔZmax, y2 = yc = critical
depth, and E2 = Ec = minimum energy
for the flowing discharge Q.
 The condition at ΔZmax is given by the
relation,
E1 – ΔZ =E2 = Ec = 𝒚𝒄 +
Q𝟐
𝟐𝒈𝑩𝟐
𝒚𝒄
𝟐
65

66. Open Channel Flow
when ΔZ > Δzmax
 The flow is not possible with the given conditions (given discharge).
 The upstream depth has to increase to cause and increase in the specific
energy at section 1.
 If this modified depth is represented by y1`,
E`1 = 𝒚𝟏
, +
Q𝟐
𝟐𝒈𝑩𝟐
𝒚𝟏,𝟐
with E`1>E1 and y`1>y1)
 At section 2 the flow will continue at the minimum specific energy level, i.e. at the
 critical condition. At this condition, y2 = yc, and,
66

67. Open Channel Flow
 when 0 < ΔZ < ΔZmax the upstream water level remains
stationary at y1 while the depth of flow at section 2
decreases with ΔZ reaching a minimum value of yc at ΔZ =
ΔZmax.
 With further increase in the value of ΔZ,
i.e. for ΔZ > ΔZmax , y1 will change to y1` while y2 will
continue to remain yc.
67

68. Open Channel Flow
b) Supercritical Flow
If y1 is in the supercritical flow regime, Fig. (5.13) shows that the depth of flow
increases due to the reduction of specific energy.
In Fig. point P` corresponds to y1 and point R` to depth at the section 2. Up to the
critical depth, y2 increases to reach yc at ΔZ = ΔZmax. For ΔZ > ΔZmax , the depth
over the hump y2 = yc will remain constant and the upstream depth y1 will change.
It will decrease to have a higher specific energy E1`by increasing velocity V1. The
variation of the depths y1 and y2 with ΔZ in the supercritical flow is shown in Fig.
68

69. Open Channel Flow
Q. 1. A rectangular channel has a width of 2.0 m and carries a
discharge of 4.80 m3/sec with a depth of 1.60 m. At a certain cross-
section a small, smooth hump with a flat top and a height 0.10 m is
proposed to be built. Calculate the likely change in the water
surface. Neglect the energy loss. (E2 = y2 = 1.48m, )
1. a) If the height of the hump is 0.50 m, estimate the water surface
elevation on the hump and at a section upstream of the hump.
69

70. Open Channel Flow
2. Water flow in a wide channel approaches a 10 cm high hump
at 1.50 m/sec velocity and a depth of 1 m. Estimate a) The
water depth y2 over the hump and b) The hump height that will
cause the crest flow to be critical.
70

71. Open Channel Flow
Transition with a Change in Width
Subcritical Flow in a Width Constriction
 Consider a frictionless horizontal channel of width B1 carrying a discharge Q at a
depth y1 as in Fig.
 At a section 2 channel width has been constricted to B2 by a smooth transition.
 Since there are no losses involved and since the bed elevations at sections 1 and 2 are
the same, the specific energy at section is equal to the specific energy at section 2.
 It is convenient to analyze the flow
in terms of the discharge intensity
q = Q/B.
 At section 1, q1 = Q/B1 and
at section 2, q2 = Q/B2.
71

72. Open Channel Flow
 Since B2 < B1, q2 > q1.
 In the specific energy diagram drawn with the discharge intensity, point P on the curve
q1 corresponds to depth y1 and specific energy E1
 Since at section 2, E2 = E1 and q = q2, point P will move vertically downward to point R
on the curve q2 to reach the depth y2.
 Thus, in subcritical flow the depth is y2 < y1. If B2 is made smaller, then q2 will increase
and y2 will decrease.
 The limit of the contracted width B2 = B2min is reached when corresponding to E1, the
discharge intensity q2 = q2max, i.e. the maximum discharge intensity for a given specific
energy (critical flow condition) will prevail.
72

73. Open Channel Flow
73

74. Open Channel Flow
𝑩𝟐𝐦𝐢𝐧 =
𝟐𝟕 𝑸𝟐
𝟖𝒈𝑬𝟏
𝟐
 If B2 < B2min, the discharge intensity q2 will be larger than qmax, the maximum
discharge intensity consistent E1. The flow will not, therefore, be possible with
the given upstream conditions.
 The upstream depth will have to increase to y1`. The new specific energy will
be formed which will 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,
 Since B2 < B2min , yc2 will be larger that ycm, yc2 > ycm.
 Thus even though critical flow prevails for all B2 <
B2min , the depth section 2 is not constant as in the
hump case but increases as y1` and hence E1`
rises.
74

75. Open Channel Flow
 If the upstream depth y1 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 y2.
 Point P` corresponds to y1 and point R` to y2. As the width B2 is decreased,
R` moves up till it becomes critical at B2 = B2min .
 Any further reduction in B2 causes the upstream depth to decrease to y1` so
that E1 rises to E1`.
 At section2, critical depth yc` corresponding to the new specific energy E1`
will prevail.
 The variation of y1, y2 and E with B2/B1 in supercritical flow regime is
indicated in Fig.
Supercritical Flow in a Width Constriction
75

76. Open Channel Flow
General Transition
A transition in general form may have a change of channel shape, provision of
a hump or a depression, contraction or expansion of channel width, in any
combination. In addition, there may be various degrees of loss of energy at
various components. However, the basic dependence of the depths of flow on
the channel geometry and specific energy of flow will remain the same. Many
complicated transition situations can be analyzed by using the principles of
specific energy and critical depth. In subcritical flow transitions the emphasis is
essentially to provide smooth and gradual changes in the boundary to prevent
flow separation and consequent energy losses. The transitions in supercritical
flow are different and involve suppression of shock waves related disturbances.
76

77. Open Channel Flow
77

78. Open Channel Flow
78

79. Open Channel Flow
 A compound channel is a channel section composed of a main deep portion and
one or two flood plains that carry high-water flows.
 The main channel carries the dry weather flow and during wet season, the flow
may spillover the banks of the main channel to the adjacent flood plains.
 A majority of natural rivers have compound sections. A compound section is
also known as two-stage channel.
 The hydraulic conditions of the main channel and the flood plain differ
considerably, especially in the channel geometry and in its roughness.
 The flood plains generally have considerably larger and varied roughness
elements.
COMPOUND CHANNEL SECTION
79

80. Open Channel Flow
 In one-dimensional analysis, Manning’s formula is applied to the
compound channel by considering a common conveyance K and a
common energy slope Sf for the entire section to obtain the discharge as
𝑄 = 𝐾 𝑆𝑓.
 However, to account for the different hydraulic conditions of the main
and flood plain sections, the channel is considered to be divided into
subsections with each subsection having its own conveyance, Ki.
 The sum of the conveyances will give the total channel conveyance
(ΣKi = K) for use in discharge computation.
 Various methods for defining the boundaries of the sub-sections are
proposed by different researchers leading to a host of proposed
methods.
 However, the overall method of considering the channel as a composite
of sub sections is well accepted and the method is known as Divided
Channel method (DCM).
 Currently DCM is widely used and many well-known software
packages, including HEC-RAS (2006), adopt this method in dealing
with compound channels 80

81. Open Channel Flow
Evaluation of compound section –
1. Vertical Interface Method
2. Diagonal Interface Method
81

82. Open Channel Flow
1. For the compound channel shown below (Fig. 3.32) estimate the discharge for a depth of
flow of (i) 1.20 m, and (ii) 1.6 m, by using DCM with vertical interface procedure.
2. A compound channel is symmetrical in cross section and has the following geometric
properties. Main channel: Trapezoidal cross section, Bottom width = 15.0 m, Side slopes
= 1.5 H : 1V, Bank full depth = 3.0 m, Manning’s coefficient = 0.03, Longitudinal slope
= 0.0009. Flood plains: Width = 75 m, Side slope = 1.5 H : 1V, Manning’s coefficient =
0.05, Longitudinal slope = 0.0009. Compute the uniform fl ow discharge for a fl ow with
total depth of 4.2 m by using DCM with (i) diagonal interface, and (ii) vertical interface
procedures. 82

83. Open Channel Flow
3. The base of the most economical trapezoidal channel section is 6m
and the side slope is 1H:2V, Calculate the maximum discharge through
the channel if the bed slope is 1 in 1000 and C = 50.
83

84. Open Channel Flow
1. The top width of a most economical trapezoidal channel section is 8m,
determine the hydraulic radius of the channel if the side slope is 1H:3V.
The top width of a most economical trapezoidal channel section is 7m and
the side slope of the channel is 1H:2V, determine the depth of the channel
section.
84

85. Open Channel Flow
85

86. Open Channel Flow
86