Transcript Slide 1

Hydraulic Engineering
Open Channel Flow
(Part 1)
Open Channel
• Open channel hydraulics, a subject of great
importance to civil engineers, deals with flows
having a free surface in channels constructed for
water supply, irrigation, drainage, and
hydroelectric power generation; in sewers,
culverts, and tunnels flowing partially full; and in
natural streams and rivers.
Pipe system
Open Channel
Classification
• Steady Flow
• Unsteady Flow
– Not time dependent
– Is time dependent
Typical situations
 Uniform flow
 Gradually Varied Flow
 Rapidly Varied Flow
Open channel section types
Uniform Flow
Chezy equation
V C Rh S
C is the Chezy C, a dimensional factor which characterizes the resistance to flow
wettedA
Rh  hydraulicRadius 
wettedP
S  bed slope
Manning equation
1 2 / 3 1/ 2
V  Rh S
n
Rh  hydraulicRadius 
S  bed slope
n  ManningCoefficient
wettedA
wettedP
Example 1
1 2
V  Rh 3 S
n
A  0.5  3  9  1.5  9 m 2


P  2 3  1.5  3  9.708
2
2
A
9

 0.927
P 9.708
2
1
V
 0.9273 1
 0.538 m/s
5000
0.025
Q  VA  0.538 9  4.84 m 3 / s
Rh 
1.5m
open channel of width = 3m as shown, bed slope = 1:5000, d=1.5m
find the flow rate using Manning equation, n=0.025.
1
2
3.0m
Example 2
open channel as shown, bed slope = 69:1584, find the flow rate using
Chezy equation, C=35.
V  C Rh S
A
2.52  5.04
0.72  2.52
 2.52  16.8 
 3.6  0.72  150  162.52 m 2
2
2
P  0.72  150 
1.8
2

 3.6 2  16.8 
A 162.52

 0.917
P 177.18
0.69
V  35 0.917 
 0.7 m/s
1584
Q  VA  0.7  162.52  113.84 m 3 / s
Rh 
2.52
2

 5.042  177.18m
Most Efficient Sections
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 reduce
construction costs
Most Efficient Sections
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
Most Efficient Sections
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 section
A  B D
P  2D  B
A
P2D 
D
dP
0
dD
dP
A BD
 A 
2 2 0 2 2  2
dD
D
D
D 
B
D
2
B
2 
D
Trapezoidal section
A(Bk D )D
or
k
B
A
 kD
D
PB2 D 1k 2
k
dP
0
dD
A
P (
 kD )  2 D 1k 2
D
dP
A
A
  2  k  2 1k 2  0 2 1k 2  2 k
dD
D
D
(BkD) D
B2k D
2 1k 
k
2
D
D
2
B2kD
D 1k 
2
2
Other criteria for economic Trapezoidal section
OFD
k
The best side slope for Trapezoidal section
dP
0
dk
k
1
3
   60

Circular section
d 2
d2
A
 sin2
4
8
P  2 r   d
Maximum Flow using Manning
 154  D  0.95d
Maximum Flow using Chezy
 151  D  0.94d
Maximum Velocity using Manning or Chezy
 128.75  D  0.81d
Example 3
Circular open channel as shown d=1.68m, bed slope = 1:5000, find
the Max. flow rate & the Max. velocity using Chezy equation, C=70.
Max. flow rate
 154
V  C Rh S
d2
d2
1.682

1.682
A

sin 2 
 154

sin 2  154  2.17 m 2
4
8
4
180
8
P   d  154

180
 1.68  4.5 m
A 2.17

 0.485m
P 4 .5
1
V  70 0.485
 0.69 m/s
5000
Q  VA  0.69  2.17  1.496 m 3 / s
Rh 
Max. Velocity
 128.75
V  C Rh S
d2
d2
1.682

1.682
A

sin 2 
 128.75 

sin 2  128.75  1.93 m 2
4
8
4
180
8
P   d  128.75 

180
 1.68  3.378 m
A
1.93

 0.57m
P 3.3775
1
V  70 0.57 
 0.748 m/s
5000
Rh 
Variation of flow and velocity with depth in circular pipes
Example 4
Trapezoidal 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.
B  2kD
D 1 k 2 
2
B  2 3 D
2
2
D 1 3 
2
2
0.6055D  B
Q 10
A 
 6.667m 2
V 1 .5
A  B  kDD
3
A  (0.6055D  D)  D  6.667
2
D  1.78m
P  B  2D 1  k 2
P  0.6055D  2 D 1  k 2
2
3
P  0.6055(1.78)  2 1.78 1     7.49m
2
To calculate bed Slope
1 2
V  Rh 3 S
n
A  6.667 m 2
P  7.49 m
A 6.667
Rh  
 0.89
P 7.49
2
1
V
 0.89 3 S  1.5
0.014
S  1 : 1941.6