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Flow types
CONTENT
Introduction
Types of Open Channel
Classification based on channel flow
Geometric properties of open channels
Experiment
2
INTRODUCTION
An open channel is a waterway, canal or conduit in which a liquid
flows with a free surface.
A channel is open or closed as long as its surface is exposed to
constant pressure.
In the absence of any other channel control, the flow is controlled
only by friction with the bed and the sides of the channel.
3
TYPES OF OPEN CHANNEL
Natural flows: rivers,
creeks, floods, etc.
Human-made systems:
fresh-water aqueducts,
irrigation, sewers,
drainage ditches, etc.
4
CLASSIFICATION BASED ON CHANNEL FLOW
Steady flow & Unsteady flow
Uniform flow& Non-Uniform flow
Laminar flow & turbulent flow
Subcritical flow, critical flow, Supercritical flow
5
Steady flow
Following remain
constant w.r.t time:
Depth of flow
velocity of flow
Flow rate
Unsteady flow
Following changes w.r.t
time:
Depth of flow
velocity of flow
Flow rate
Open
Channel
Steady Flow
= 0
Open Channel
Unsteady Flow
≠ 0
6
Uniform flow
Following remain
constant w.r.t Length of
channel:
Depth of flow
Velocity of flow
Slope of channel
Cross section
Non-Uniform flow
Following changes w.r.t
Length of channel:
Depth of flow
Velocity of flow
Slope of channel
Cross section
Rapidly
Varied Flow
i.e : hydraulic
jump
Non-Uniform Flow
≠ 0
Uniform Flow
= 0
Gradually
Varied flow
i.e : upstream
of obstruction7
Flow Classifications
Depending on the Reynolds number, Re
Laminar Flow (if Re < 500): very slow and shallow flowing
water in very smooth open channels.
Turbulent Flow (if Re > 1000): ordinary flow in ordinary
open channels.
Transition Flow (if 500 < Re < 1000)
V = average channel velocity
L = length of channel
v = kinematic viscosity of fluid
8
Depending on Froude number, Fr
Fr = 1 : Critical Flow
Fr < 1 : Subcritical Flow – slow flowing water
Fr > 1 : Supercritical Flow – fast flowing water
V = average channel velocity
g = gravity acceralation
D = hydraulics water depth
9
Prismatic Channel
It has constant shape & shize & is laid to a constant slope
Ex.
Some artificial channel
Laboratory flumes
Non- Prismatic channel
If its size, shape or slope changes along he length
Ex.
All natural nallas, stream, rivers, etc.,
10
GEOMETRIC PROPERTIES OF OPEN CHANNELS
Open Channel
11
GEOMETRIC PROPERTIES OF OPEN CHANNELS
The terminology of geometric elements
12
Type of channel
TOP WIDTH, T
AREA, A WETTED PERIMETER, P
RECTANGULAR B By B + 2y
TRAPEZOIDAL B+2my By + my2 B+2y √ 1+m2
Where,
GEOMETRIC PROPERTIES OF OPEN CHANNELS
13
Hydraulic Jump
The hydraulic jump is defined as the rise of water
level, which takes place due to transformation of the
unstable shooting flow (super-critical) to the stable
streaming flow (sub-critical).
When hydraulic jump occurs, a loss of energy due to
eddy formation and turbulence flow occurs.
Hydraulic Jump
Applications of Hydraulic Jump
Usually hydraulic jump reverses the flow of water. This
phenomenon can be used to mix chemicals for water
purification.
Hydraulic jump usually maintains the high water level on the
down stream side. This high water level can be used for
irrigation purposes.
Hydraulic jump can be used to remove the air from water
supply and sewage lines to prevent the air locking.
It prevents the scouring action on the down stream side of the
dam structure
Location of Hydraulic Jump
The most typical cases for the location of hydraulic
jump are:
1. Below control structures like weir, sluice are used in
the channel
2. when any obstruction is found in the channel,
3. when a sharp change in the channel slope takes
place.
4. At the toe of a spillway dam
Effect of Hydraulic Jump
Actually the hydraulic jump usually acts as the energy
dissipator. It clears the surplus energy of water.
Due to the hydraulic jump, many noticeable able disturbances
are created in the flowing water like eddies, reverse flow.
Usually when the hydraulic jump takes place, the considerable
amount of air is trapped in the water. That air can be helpful in
removing the wastes in the streams that are causing pollution.
Hydraulic jump also make the work of different hydraulic
structures, effective like weirs, notches and flumes etc.
Where,
Geometric Properties of channels
UNIFORM FLOW IN OPEN CHANNELS
Definitions
a) Open Channel: Duct through whichLiquid Flows with a Free Surface - River,Canal
b) Steady and Non- Steady Flow: InSteady Flows, all the characteristics of floware constant with time. In unsteady flows,there are variations with time.
Parameters of Open Channels
a) Wetted Perimeter, P : The Length of contactbetween Liquid and sides and base of Channel
P = B + 2 D ; D = normal depth
Hydraulic Mean Depth or Hydraulic Radius (R): Ifcross sectional area is A, then R = A/P, e.g. forrectangular channel, A = B D, P = B +2D
Area, A
Wetted Perimeter
D
B
Empirical Flow Equations for Estimating Normal Flow Velocities
a) Chezy Formula (1775):
Can be derived from basic principles. It states that: ;
Where: V is velocity; R is hydraulic radius and S isslope of the channel. C is Chezy coefficient and is afunction of hydraulic radius and channel roughness.
SRCV
Definitions
a) Freeboard: Vertical distance between
the highest water level anticipated in the
design and the top of the retaining banks. It
is a safety factor to prevent the overtopping of
structures.
b) Side Slope (Z): The ratio of the
horizontal to vertical distance of the sides of
the channel. Z = e/d = e’/D
DESIGN OF CHANNELS FOR STEADY UNIFORM FLOW
Channels are very important in Engineering projects
especially in Irrigation and, Drainage.
Channels used for irrigation are normally called canals
Channels used for drainage are normally called drains.
MOST EFFICIENT SECTION
During the design stages of an open channel, the
channel cross-section, roughness and bottom slope
are given.
The objective is to determine the flow velocity, depth
and flow rate, given any one of them. The design of
channels involves selecting the channel shape and
bed slope to convey a given flow rate with a given
flow depth. For a given discharge, slope and
roughness, the designer aims to minimize the
cross-sectional area A in order to reduceconstruction costs
The most ‘efficient’ cross-sectional shape is determined for uniform flow
conditions. Considering a given discharge Q, the velocity V is maximum
for the minimum cross-section A. According to the Manning equation the
hydraulic diameter is then maximum.
It can be shown that:
1.the wetted perimeter is also minimum,
2.the semi-circle section (semi-circle having its centre in the surface)
is the best hydraulic section
Because the hydraulic radius is equal to the water cross section area
divided by the wetted parameter, Channel section with the least wetted
parameter is the best hydraulic section
RECTANGULAR SECTIONFor a rectangular section
Q=AV, where Q=discharge through the
channel,
A=area of flow.
V=velocity with which
water is flowing in the
channel.
For Q to be maximum ,V needs to be
maximum, Since A = constant.
But V=Cmi
Where m= hydraulic mean depth.
i= bed slope.
m=A/P where P=wetted perimeter.
For m to be maximum ,P minimum
A=BDB=A/D;P=B+2D ,thenP=A/D+2D
𝑑𝑃
𝑑𝐷=
−𝐴
𝐷2+ 2
𝑑𝑃
𝑑𝐷= 0
−𝐴
𝐵2+ 2 = 0
𝐷2 =𝐴
2=
𝐵𝐷
2
D =𝐵
2
𝑚 =𝐵𝐷
𝐵 + 2𝐷=
2𝐷2
2𝐷 + 2𝐷=
𝐷
2
𝑚 =𝐷
2
TRAPEZOIDAL SECTION
For a rectangular section
Q=AV, where Q=discharge through the
channel,
A=area of flow.
V=velocity with which
water is flowing in the
channel.
For Q to be maximum ,V needs to be
maximum, Since A = constant.
But V=Cmi
Where m= hydraulic mean depth.
i= bed slope.
m=A/P where P=wetted perimeter.
For m to be maximum ,P minimum
𝐴 = 𝐵𝐷 + 𝑛𝐷2
𝑃 = 𝐵 + 2𝐷 1 + 𝑛2
𝐵 =𝐴
𝐷−𝑛𝐷2
𝐷=
𝐴
𝐷− 𝑛𝐷
𝑃 =𝐴
𝐷− 𝑛𝐷 + 2𝐷 1 + 𝑛2
𝑑𝑃
𝑑𝐷=
−𝐴
𝐷2− 𝑛 + 2 1 + 𝑛2
𝑑𝑃
𝑑𝐷= 0,⇒
−𝐴
𝐷2− 𝑛 + 2 1 + 𝑛2 = 0
2 1 + 𝑛2 = 𝑛 +𝐴
𝐷2
2 1 + 𝑛2 = 𝑛 +𝐵𝐷
𝐷2+ 𝑛
𝐵
𝐷= 2( 1 + 𝑛2 − 𝑛
𝐵 + 2𝑛𝐷
2= 𝐷 1 + 𝑛2
CONCLUSION:HALF OF THE TOP WIDTH = SIDE WALL LENGTH
m = D/2
The best side slope for
Trapezoidal section
𝑃 =𝐴
𝐷− 𝑛𝐷 + 2𝐷 1 + 𝑛2
𝑑𝑃
𝑑𝑛= 0 − 𝐷 +
2𝐷
2 1 + 𝑛2∗ 2𝑛
𝑑 𝑃 𝑑 𝑛 = 0
𝑛 = 1 √3
𝜃 = 600
open channel as shown Q=10m3/s, velocity =1.5m/s, for most
economic section. find wetted parameter, and the bed slope n=0.014.
Trapezoidal
Example 4
mD
DDDA
DkDBA
mV
QA
BD
DBD
kDBkD
78.1
667.6)2
36055.0(
667.65.1
10
6055.0
2
232
231
2
21
2
2
2
mP
kDDP
kDBP
49.72
3178.12)78.1(6055.0
126055.0
12
2
2
2
To calculate bed Slope
6.1941:1
5.189.0014.0
1
89.049.7
667.6
m 49.7
m 667.6
1
3
2
2
3
2
S
SV
P
AR
P
A
SRn
V
h
h