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

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

Open Channel Flow
Open Channel Flow OR Gravity 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

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

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
Two Types

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

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 = V1A1 = V2A2

Control Volume for Open Channels

Conservation of Energy
• Conservation of energy applied to control volume
results in the following:
Z1 y11
V
1
2
2g
Z2 y2 2
V
2
2
2g
hf
where Z1,Z2 are elevations of the bed,
y1, y2 are depths of flow,
V1, V2 are velocities,
a1, a2 are kinetic energy corrections, and
hf is the frictional loss.

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

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:
- Chezy and Manning formulas

Chezy Equation
V is mean velocity
R is hydraulic radius (area/wetted perimeter)
S is the slope of energy gradeline, and
C is the Chezy coefficient
RS
C
V
C is a function of the roughness of the channel bottom

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 coefficient
V1
n
R
2/3
S
1/2

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

Manning’s n

Discharge Equation
Q1
n
AR
2/3
S1/2
For uniform flow, A, R and n are constants, thus
QKS
1/2
3
/
2
1AR
n
K
The term K is conveyance, given as

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

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’s formula is 50 in SI units.

Exercise 5. 2 :
An open channel has a Vee shaped cross section with sides
inclined at an angle of 600 to the vertical. If the rate of flow is
80 dm3/s when the depth at the centre is 0.25 m, calculate the
slope of the channel assuming Chezy’s constant is 45 in SI units.
Exercise 5. 3 :
A rectangular channel is 6 m wide and will carry a discharge of
22.5 m3 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

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’s constant as 53 in SI
units.

Measurement of Discharge in Open Channel
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

Measurement of Discharge in Open Channel
A Triangular Notch
(Also Called as V Notch)

Discharge Equations
1. Rectangular Notch
2
/
3
2
.
.
3
2 h
g
b
Q
th 

2. V - Notch
2
/
5
2
tan
2
.
15
8 h
g
Q
th 






 
Note : Qact = Cd x Qth

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 m3/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 m3/s. Take Cd = 0.6. Also determine the width of
water surface. The angle of the Notch is 600

Exercise 5. 8 :
Exercise 5. 9 :
More Exercise In Class :

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

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