Unit 5 Open Channel flow.pdf Unit 5 Open Channel flow

Unit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flow

1. CE 205 : ENGG. FLUID MECHANICS Unit 5
Open Channel Flow

2. CE 205 : ENGG. FLUID MECHANICS Unit 5
Open Channel Flow
Open Channel Flow ORGravity Flow
 The flow is under gravity.
 There will be a free surface for the flow open to atmosphere
 The position of the free surface can change in space and time
Properties of Open Channels
Many different types
River, stream, canal, flume, ditch, culverts
Many different Cross sectional types
Rectangular, Trapezoidal, Circular, Triangular

3. Definition of Geometric Elements of a Channel
R = hydraulic radius = A/P
D = hydraulic depth = A/T
y = depth of flow
T = top width
P = wetted perimeter
A = flow area
CE 205 : ENGG. FLUID MECHANICS Unit 5

4. Flow Classification
•Uniform (normal) flow:
Depth is constant at every section along length of
channel
•Nonuniform(varied) flow:
Depth changes along channel
–Rapidly-varied flow: Depth changes suddenly
–Gradually-varied flow: Depth changes gradually
CE 205 : ENGG. FLUID MECHANICS Unit 5
Two Types

5. Equations of Motion
•There are three general principles used in solving
problems of flow in open channels:
–Continuity (conservation of mass)
–Energy
–Momentum
•For problems involving steady uniform flow,
continuity and energy principles are sufficient
CE 205 : ENGG. FLUID MECHANICS Unit 5

6. Continuity Equation (Conservation of Mass)
•Since water is essentially incompressible,
conservation of mass (continuity) reduces to the
following:
Discharge in = Discharge out
•Stated in terms of velocity and area:
Q = V
1A
1= V
2A
2
CE 205 : ENGG. FLUID MECHANICS Unit 5

7. Control Volume for Open Channels
CE 205 : ENGG. FLUID MECHANICS Unit 5

8. Conservation of Energy
•Conservation of energy applied to control volume
results in the following:Z
1y
1
1
V
1
2
2g
Z
2y
2
2
V
2
2
2g
h
f
whereZ
1,Z
2are elevations of the bed,
y
1, y
2are depths of flow,
V
1, V
2are velocities,
a
1, a
2are kinetic energy corrections, and
h
fis the frictional loss.
CE 205 : ENGG. FLUID MECHANICS Unit 5

9. Energy Coefficient
•The term associated with each velocity head () is
the energy coefficient
•This term is needed because we are using the
average velocity over the depth to compute the total
kinetic energy
CE 205 : ENGG. FLUID MECHANICS Unit 5

10. Uniform Flow Computations
•Equations are developed for steady-state conditions
–Depth, discharge, area, velocity all constant along
channel length
•Rarely occurs in natural channels (even for constant
geometry) since it implies a perfect balance of all
forces
•Two general equations are in use:
-Chezyand Manning formulas
CE 205 : ENGG. FLUID MECHANICS Unit 5

11. ChezyEquation
V is mean velocity
R is hydraulic radius (area/wetted perimeter)
S is the slope of energy gradeline, and
C is the ChezycoefficientRSCV
CE 205 : ENGG. FLUID MECHANICS Unit 5
C is a function of the roughness of the channel bottom

12. Manning Equation
•The Manning equation is given as:
V is mean velocity (m/s)
R is hydraulic radius (m)
S is the slope of the energy gradeline(m/m)
n is the Manning’s roughness coefficientV
1
n
R
2/3
S
1/2
CE 205 : ENGG. FLUID MECHANICS Unit 5

13. Manning’s Roughness Coefficient(n)
•Roughness coefficient (n) is a function of:
–Channel material
–Surface irregularities
–Variation in shape
–Vegetation
–Flow conditions
–Channel obstructions
–Degree of meandering
CE 205 : ENGG. FLUID MECHANICS Unit 5

14. Manning’s n
CE 205 : ENGG. FLUID MECHANICS Unit 5

15. Discharge EquationQ
1
n
AR
2/3
S
1/2
CE 205 : ENGG. FLUID MECHANICS Unit 5
For uniform flow, A, R and n are constants, thusQKS
1/2 3/21
AR
n
K
The term K is conveyance, given as

18. CE 205 : ENGG. FLUID MECHANICS Unit 5
Exercise 5.1 :
A rectangular channel is 2.5 m wide and has a uniform bed
slope of 1 in 500. If the depth of flow is constant at 1.7 m,
calculate (a) the hydraulic mean depth, (b) the velocity of
flow, ( c ) the volume rate of flow. Assume that the value of
the coefficient C in Chezy’sformula is 50 in SI units.

19. CE 205 : ENGG. FLUID MECHANICS Unit 5
Exercise 5. 2 :
An open channel has a Veeshaped cross section with sides
inclined at an angle of 60
0
to the vertical. If the rate of flow is
80 dm
3
/s when the depth at the centre is 0.25 m, calculate the
slope of the channel assuming Chezy’sconstant is 45 in SI units.
Exercise 5. 3 :
A rectangular channel is 6 m wide and will carry a discharge of
22.5 m
3
s
-1
of water. Determine the necessary slopes to achieve
uniform flow at (a) a depth of 3 m (b) a depth of 0.6 m.
Assume manning’s n = 0.02

20. CE 205 : ENGG. FLUID MECHANICS Unit 5
Exercise 5. 4 :
A channel is 5 m wide at the top and 2 m deep has sides
sloping 2 vertically in 1 horizontally. The slope of the channel
is 1 in 1000. Find the volume rate of flow when the depth of
water is constant at 1 m. Take Chezy’sconstant as 53 in SI
units.

21. Measurement of Discharge in Open Channel
CE 205 : ENGG. FLUID MECHANICS Unit 5
Notches or weirs are used to measure discharge in an open channel
A notch is an opening in the vertical side of a tank or in a
channel, such that the free surface of the liquid in the tank or
channel is below the top edge of the opening
A weir is a concrete or masonry dam built across an open
channel (river, canal etc.) over which water overflows

22. Measurement of Discharge in Open Channel
CE 205 : ENGG. FLUID MECHANICS Unit 5
A Triangular Notch
(Also Called as V Notch)

23. Discharge Equations
CE 205 : ENGG. FLUID MECHANICS Unit 5
1. Rectangular Notch2/3
2..
3
2
hgbQ
th 
2. V -Notch2/5
2
tan2.
15
8
hgQ
th 







Note : Q
act= C
dx Q
th

24. CE 205 : ENGG. FLUID MECHANICS Unit 5
Exercise 5. 5 :
The width of a rectangular Notch is 2 m and the height of water surface
above the crest is 30 cm. Determine the discharge through the Notch. Take
Cd= 0.62
Exercise 5. 6 :
In a rectangular notch, the discharge of water is 0.25 m
3
/s. If width of the
Notch is 3m, determine the head above the sill. Assume Cd= 0.6
Exercise 5. 7 :
Determine the height of water surface above the crest of a V-Notch when
the flow rate is 0.135 m
3
/s. Take Cd= 0.6. Also determine the width of
water surface. The angle of the Notch is 60
0

25. CE 205 : ENGG. FLUID MECHANICS Unit 5
Exercise 5. 8 :
Exercise 5. 9 :
More Exercise In Class :