11 open_channel.ppt

Monroe L. Weber-Shirk
School of Civil and
Environmental Engineering
Open Channel Flow

Sensitivity: LNT Construction Internal Use
Open Channel Flow
 Liquid (water) flow with a ____ ________
(interface between water and air)
 relevant for
 natural channels: rivers, streams
 engineered channels: canals, sewer
lines or culverts (partially full), storm drains
 of interest to hydraulic engineers
 location of free surface
 velocity distribution
 discharge - stage (______) relationships
 optimal channel design
free surface
depth

Topics in Open Channel Flow
 Uniform Flow
 Discharge-Depth relationships
 Channel transitions
 Control structures (sluice gates, weirs…)
 Rapid changes in bottom elevation or cross section
 Critical, Subcritical and Supercritical Flow
 Hydraulic Jump
 Gradually Varied Flow
 Classification of flows
 Surface profiles
normal depth

Classification of Flows
 Steady and Unsteady
 Steady: velocity at a given point does not change with
time
 Uniform, Gradually Varied, and Nonuniform
 Uniform: velocity at a given time does not change
within a given length of a channel
 Gradually varied: gradual changes in velocity with
distance
 Laminar and Turbulent
 Laminar: flow appears to be as a movement of thin
layers on top of each other
 Turbulent: packets of liquid move in irregular paths
(Temporal)
(Spatial)

Momentum and Energy
Equations
 Conservation of Energy
 “losses” due to conversion of turbulence to heat
 useful when energy losses are known or small
 ____________
 Must account for losses if applied over long distances
 _______________________________________________
 Conservation of Momentum
 “losses” due to shear at the boundaries
 useful when energy losses are unknown
 ____________
Contractions
Expansion
We need an equation for losses

 Given a long channel of
constant slope and cross
section find the relationship
between discharge and depth
 Assume
 Steady Uniform Flow - ___ _____________
 prismatic channel (no change in _________ with distance)
 Use Energy and Momentum, Empirical or
Dimensional Analysis?
 What controls depth given a discharge?
 Why doesn’t the flow accelerate?
Open Channel Flow:
Discharge/Depth Relationship
P
no acceleration
geometry
Force balance
A

Steady-Uniform Flow: Force
Balance

W

W sin 
Dx
a
b
c
d
Shear force
Energy grade line
Hydraulic grade line
Shear force =________
0
sin 
D

D x
P
x
A o





 sin
P
A
o 
h
R
=
P
A



sin
cos
sin


S
W cos 
g
V
2
2
Wetted perimeter = __
Gravitational force = ________
Hydraulic radius
Relationship between shear and velocity? ______________
oP D x
P
A Dx sin
Turbulence

 Geometric parameters
 ___________________
 ___________________
 ___________________
 Write the functional relationship
 Does Fr affect shear? _________
P
A
Rh 
Hydraulic radius (Rh)
Channel length (l)
Roughness (e)
Open Conduits:
Dimensional Analysis
f , ,Re,
p
h h
l
C r
R R
e
æ ö
= ç ÷
è ø
F ,M,W
V
Fr
yg
=
No!

Pressure Coefficient for Open
Channel Flow?
2
2
C
V
p
p

D


2
2
C
V
ghl
hl

2
2
C f
f
S
gS l
V
=
l f
h S l
=
l
h
p 

D

Pressure Coefficient
Head loss coefficient
Friction slope coefficient
(Energy Loss Coefficient)
Friction slope
Slope of EGL

Dimensional Analysis
f , ,Re
f
S
h h
l
C
R R
e
æ ö
= ç ÷
è ø
f
h
S
R
C
l
l
=
2
2 f h
gS l R
V l
l
=
2 f h
gS R
V
l
=
2
f h
g
V S R
l
=
f ,Re
f
S
h h
l
C
R R
e
æ ö
= ç ÷
è ø
Head loss  length of channel
f ,Re
f
h
S
h
R
C
l R
e
l
æ ö
= =
ç ÷
è ø
2
2
f
f
S
gS l
C
V
=
(like f in Darcy-Weisbach)

Chezy equation (1768)
 Introduced by the French engineer Antoine
Chezy in 1768 while designing a canal for
the water-supply system of Paris
h f
V C R S
=
150
<
C
<
60
s
m
s
m
where C = Chezy coefficient
where 60 is for rough and 150 is for smooth
also a function of R (like f in Darcy-Weisbach)
2
f h
g
V S R
l
=
compare

Darcy-Weisbach equation (1840)
where d84 = rock size larger than 84% of
the rocks in a random sample
For rock-bedded streams
f = Darcy-Weisbach friction factor
2
84
1
1.2 2.03log h
f
R
d
=
æ ö
é ù
+ ê ú
ç ÷
è ø
ë û
4
4
P
2
d
d
d
A
Rh 












2
2
l
l V
h f
d g
=
2
4 2
l
h
l V
h f
R g
=
2
4 2
f
h
l V
S l f
R g
=
2
8
f h
V
S R f
g
=
8
f h
g
V S R
f
=

Manning Equation (1891)
 Most popular in U.S. for open channels
(English system)
1/2
o
2/3
h S
R
1
n
V 
1/2
o
2/3
h S
R
49
.
1
n
V 
VA
Q 
2
/
1
3
/
2
1
o
h S
AR
n
Q  very sensitive to n
Dimensions of n?
Is n only a function of roughness?
(MKS units!)
NO!
T /L1/3
Bottom slope

Values of Manning n
Lined Canals n
Cement plaster 0.011
Untreated gunite 0.016
Wood, planed 0.012
Wood, unplaned 0.013
Concrete, trowled 0.012
Concrete, wood forms, unfinished 0.015
Rubble in cement 0.020
Asphalt, smooth 0.013
Asphalt, rough 0.016
Natural Channels
Gravel beds, straight 0.025
Gravel beds plus large boulders 0.040
Earth, straight, with some grass 0.026
Earth, winding, no vegetation 0.030
Earth , winding with vegetation 0.050
d = median size of bed material
n = f(surface
roughness,
channel
irregularity,
stage...)
6
/
1
038
.
0 d
n 
6
/
1
031
.
0 d
n  d in
ft
d in m

Trapezoidal Channel
 Derive P = f(y) and A = f(y) for a
trapezoidal channel
 How would you obtain y = f(Q)?
z
1
b
y
z
y
yb
A 2


 
  b
yz
y
P 


2
/
1
2
2
2
  b
z
y
P 


2
/
1
2
1
2
2
/
1
3
/
2
1
o
h S
AR
n
Q 
Use Solver!

Flow in Round Conduits







 

r
y
r
arccos

 


 cos
sin
2

 r
A

sin
2r
T 

y
T
A
r

r
P 2

radians
Maximum discharge
when y = ______
0.938d
( )( )
sin cos
r r
q q
=

Open Channel Flow: Energy
Relations
2g
V2
1
1

2g
V2
2
2

x
SoD
2
y
1
y
x
D
L f
h S x
= D
energy
grade line
hydraulic
grade line
velocity head
Bottom slope (So) not necessarily equal to surface slope (Sf)
water surface

Energy relationships
2 2
1 1 2 2
1 1 2 2
2 2 L
p V p V
z z h
g g
a a
g g
+ + = + + +
2 2
1 2
1 2
2 2
o f
V V
y S x y S x
g g
+ D + = + + D
Turbulent flow (  1)
z - measured from
horizontal datum
y - depth of flow
Pipe flow
Energy Equation for Open Channel Flow
2 2
1 2
1 2
2 2
o f
V V
y S x y S x
g g
+ + D = + + D
From diagram on previous slide...

Specific Energy
 The sum of the depth of flow and the
velocity head is the specific energy:
g
V
y
E
2
2


If channel bottom is horizontal and no head loss
2
1 E
E 
y - potential energy
g
V
2
2
- kinetic energy
For a change in bottom elevation
1 2
E y E
- D =
x
S
E
x
S
E o D


D
 f
2
1
y

Specific Energy
In a channel with constant discharge, Q
2
2
1
1 V
A
V
A
Q 

2
2
2gA
Q
y
E 

g
V
y
E
2
2

 where A=f(y)
Consider rectangular channel (A=By) and Q=qB
2
2
2gy
q
y
E 

A
B
y
3 roots (one is negative)
q is the discharge per unit width of channel

0
1
2
3
4
5
6
7
8
9
10
0 1 2 3 4 5 6 7 8 9 10
E
y
Specific Energy: Sluice Gate
2
2
2gy
q
y
E 

1
2
2
1 E
E 
sluice gate
y1
y2
EGL
y1 and y2 are ___________ depths (same specific energy)
Why not use momentum conservation to find y1?
q = 5.5 m2/s
y2 = 0.45 m
V2 = 12.2 m/s
E2 = 8 m
alternate
Given downstream depth and discharge, find upstream depth.

0
1
2
3
4
0 1 2 3 4
E
y Specific Energy: Raise the Sluice
Gate
2
2
2gy
q
y
E 

1 2
E1  E2
sluice gate
y1
y2
EGL
as sluice gate is raised y1 approaches y2 and E is
minimized: Maximum discharge for given energy.

0
1
2
3
4
0 1 2 3 4
E
y
Specific Energy: Step Up
Short, smooth step with rise Dy in channel
Dy
1 2
E E y
= + D
Given upstream depth and
discharge find y2
0
1
2
3
4
0 1 2 3 4
E
y
Increase step height?

P
A
Critical Flow
T
dy
y
T=surface width
Find critical depth, yc
2
2
2gA
Q
y
E 

0

dy
dE
Tdy
dA 
0
dE
dy
= =
3
2
1
c
c
gA
T
Q

Arbitrary cross-section
A=f(y)
2
3
2
Fr
gA
T
Q

2
2
Fr
gA
T
V

dA
A
D
T
= Hydraulic Depth
2
3
1
Q dA
gA dy
-
0
1
2
3
4
0 1 2 3 4
E
y
yc

Critical Flow:
Rectangular channel
yc
T
Ac
3
2
1
c
c
gA
T
Q

qT
Q  T
y
A c
c 
3
2
3
3
3
2
1
c
c gy
q
T
gy
T
q


3
/
1
2









g
q
yc
3
c
gy
q 
Only for rectangular channels!
c
T
T 
Given the depth we can find the flow!

Critical Flow Relationships:
Rectangular Channels
3
/
1
2









g
q
yc c
c y
V
q 









g
y
V
y
c
c
c
2
2
3
g
V
y
c
c
2

1

g
y
V
c
c
Froude number
velocity head =
because
g
V
y c
c
2
2
2

2
c
c
y
y
E 
 E
yc
3
2

force
gravity
force
inertial
0.5 (depth)
g
V
y
E
2
2


Kinetic energy
Potential energy

Critical Flow
 Characteristics
 Unstable surface
 Series of standing waves
 Occurrence
 Broad crested weir (and other weirs)
 Channel Controls (rapid changes in cross-section)
 Over falls
 Changes in channel slope from mild to steep
 Used for flow measurements
 ___________________________________________
Unique relationship between depth and discharge
Difficult to measure depth
0
1
2
3
4
0 1 2 3 4
E
y

Broad-crested Weir
H
P
yc
E
3
c
gy
q  3
c
Q b gy
=
E
yc
3
2

3/ 2
3/ 2
2
3
Q b g E
æ ö
=
è ø
Cd corrects for using H rather
than E.
3
/
1
2









g
q
yc
Broad-crested
weir
E measured from top of weir
3/ 2
2
3
d
Q C b g H
æ ö
=
è ø
Hard to measure yc

Broad-crested Weir: Example
 Calculate the flow and the depth upstream.
The channel is 3 m wide. Is H approximately
equal to E?
0.5
yc
E
Broad-crested
weir
yc=0.3 m
Solution
How do you find flow?____________________
How do you find H?______________________
Critical flow relation
Energy equation

Hydraulic Jump
 Used for energy dissipation
 Occurs when flow transitions from
supercritical to subcritical
 base of spillway
 We would like to know depth of water
downstream from jump as well as the
location of the jump
 Which equation, Energy or Momentum?

Hydraulic Jump!

Hydraulic Jump
y1
y2
L
EGL
hL
Conservation of Momentum
2
2
1
1
2
1 A
p
A
p
QV
QV 


 

2
gy
p


A
Q
V 
2
2
2
2
1
1
2
2
1
2
A
gy
A
gy
A
Q
A
Q




x
x p
p
x
x F
F
M
M 2
1
2
1 


1
2
1
1 A
V
M x 


2
2
2
2 A
V
M x 

ss
p
p F
F
F
W
M
M 



 2
1
2
1

Hydraulic Jump:
Conjugate Depths
Much algebra
For a rectangular channel make the following substitutions
By
A  1
1V
By
Q 
1
1
1
V
Fr
gy
= Froude number
 
2
1
1
2 8
1
1
2
Fr
y
y 



valid for slopes < 0.02

Hydraulic Jump:
Energy Loss and Length
No general theoretical solution
Experiments show
2
6y
L  20
4 1 
 Fr
•Length of jump
•Energy Loss
2
2
2gy
q
y
E 

L
h
E
E 
 2
1
significant energy loss (to turbulence) in jump
 
2
1
3
1
2
4 y
y
y
y
hL


algebra
for

Gradually Varied Flow
2 2
1 2
1 2
2 2
o f
V V
y S x y S x
g g
+ + D = + + D
Energy equation for non-
uniform, steady flow
1
2 y
y
dy 

2
2 f o
V
dy d S dx S dx
g
æ ö
+ + =
ç ÷
è ø
P
A
T
dy
y
dy
dx
S
dy
dx
S
g
V
dy
d
dy
dy
o
f 










2
2
( )
2 2
2 1
2 1
2 2
o f
V V
S dx y y S dx
g g
æ ö
= - + - +
ç ÷
è ø

2
3
2
3
2
2
2
2
2
2
2
2
Fr
gA
T
Q
dy
dA
gA
Q
gA
Q
dy
d
g
V
dy
d









 









 


















  f
o S
S
Fr
dx
dy


 2
1
dy
dx
S
dy
dx
S
g
V
dy
d
dy
dy
o
f 










2
2
dy
dx
S
dy
dx
S
Fr o
f 

 2
1
Change in KE
Change in PE
We are holding Q constant!
2
1 Fr
S
S
dx
dy f
o




2
1 Fr
S
S
dx
dy f
o


 Governing equation for
gradually varied flow
 Gives change of water depth with distance along channel
 Note
 So and Sf are positive when sloping down in
direction of flow
 y is measured from channel bottom
 dy/dx =0 means water depth is constant
yn is when o f
S S
=

Surface Profiles
 Mild slope (yn>yc)
 in a long channel subcritical flow will occur
 Steep slope (yn<yc)
 in a long channel supercritical flow will occur
 Critical slope (yn=yc)
 in a long channel unstable flow will occur
 Horizontal slope (So=0)
 yn undefined
 Adverse slope (So<0)
 yn undefined
Note: These slopes are f(Q)!

Normal depth
Steep slope (S2)
Hydraulic Jump
Sluice gate
Steep slope
Obstruction
Surface Profiles
2
1 Fr
S
S
dx
dy f
o


 S0 - Sf 1 - Fr2 dy/dx
+ + +
- + -
- - + 0
1
2
3
4
0 1 2 3 4
E
y
yn
yc

More Surface Profiles
S0 - Sf 1 - Fr2 dy/dx
1 + + +
2 + - -
3 - - + 0
1
2
3
4
0 1 2 3 4
E
y
yn
yc
2
1 Fr
S
S
dx
dy f
o




Direct Step Method
x
S
g
V
y
x
S
g
V
y f
o D



D


2
2
2
2
2
2
1
1
o
f S
S
g
V
g
V
y
y
x





D
2
2
2
2
2
1
2
1
energy equation
solve for Dx
1
1
y
q
V 
2
2
y
q
V 
2
2
A
Q
V 
1
1
A
Q
V 
rectangular channel prismatic channel

Direct Step Method
Friction Slope
2 2
4/3
f
h
n V
S
R
=
2 2
4/3
2.22
f
h
n V
S
R
=
2
8
f
h
fV
S
gR
=
Manning Darcy-Weisbach
SI units
English units

Direct Step
 Limitation: channel must be _________ (so that
velocity is a function of depth only and not a
function of x)
 Method
 identify type of profile (determines whether Dy is + or -)
 choose Dy and thus yn+1
 calculate hydraulic radius and velocity at yn and yn+1
 calculate friction slope yn and yn+1
 calculate average friction slope
 calculate Dx
prismatic

Direct Step Method
=(G16-G15)/((F15+F16)/2-So)
A B C D E F G H I J K L M
y A P Rh V Sf E Dx x T Fr bottom surface
0.900 1.799 4.223 0.426 0.139 0.00004 0.901 0 3.799 0.065 0.000 0.900
0.870 1.687 4.089 0.412 0.148 0.00005 0.871 0.498 0.5 3.679 0.070 0.030 0.900
=y*b+y^2*z
=2*y*(1+z^2)^0.5 +b
=A/P
=Q/A
=(n*V)^2/Rh^(4/3)
=y+(V^2)/(2*g)
o
f S
S
g
V
g
V
y
y
x





D
2
2
2
2
2
1
2
1

Standard Step
 Given a depth at one location, determine the depth at a
second location
 Step size (Dx) must be small enough so that changes in
water depth aren’t very large. Otherwise estimates of the
friction slope and the velocity head are inaccurate
 Can solve in upstream or downstream direction
 upstream for subcritical
 downstream for supercritical
 Find a depth that satisfies the energy equation
x
S
g
V
y
x
S
g
V
y f
o D



D


2
2
2
2
2
2
1
1

What curves are available?
S1
S3
Is there a curve between yc and yn that
decreases in depth in the upstream direction?
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
0
5
10
15
20
distance upstream (m)
elevation
(m)
bottom
surface
yc
yn

Wave Celerity
2
1
2
1
gy
F 
  2
2
2
1
y
y
g
F 
 

 
 
2
2
2
1
2
1
y
y
y
g
F
F
Fr 
 




F1
y+y
V+V
V
Vw
unsteady flow
y y y+y
V+V-Vw
V-Vw
steady flow
V+V-Vw
V-Vw
F2

Wave Celerity:
Momentum Conservation
     
 
w
w
w
r V
V
V
V
V
V
V
y
F 




 

  V
V
V
y
F w
r 
 

 
    V
V
V
y
y
y
y
g w 


 



2
2
2
1
    V
V
V
y
y
y
g w 
 

 2
2
1
  V
V
V
y
g w 
 


y y+y
V+V-Vw
V-Vw
steady flow
1
2 M
M
Fr 
   y
V
V
M w
2
1 

  Per unit width

Wave Celerity
    
w
w V
V
V
y
y
V
V
y 



 

w
w
w yV
yV
V
y
V
y
yV
yV
yV
yV 



 






 
y
y
V
V
V w

 


  V
V
V
y
g w 
 


 
y
y
V
V
y
g w

 2


 2
w
V
V
gy 
 w
V
V
c 
 gy
c 
Mass conservation
y y+y
V+V-Vw
V-Vw
steady flow
Momentum
c
V
Fr
yg
V



Wave Propagation
 Supercritical flow
 c<V
 waves only propagate downstream
 water doesn’t “know” what is happening downstream
 _________ control
 Critical flow
 c=V
 Subcritical flow
 c>V
 waves propagate both upstream and downstream
upstream

Most Efficient Hydraulic
Sections
 A section that gives maximum discharge for a
specified flow area
 Minimum perimeter per area
 No frictional losses on the free surface
 Analogy to pipe flow
 Best shapes
 best
 best with 2 sides
 best with 3 sides

Why isn’t the most efficient
hydraulic section the best design?
Minimum area = least excavation only if top of
channel is at grade
Cost of liner
Complexity of form work
Erosion constraint - stability of side walls

Open Channel Flow Discharge
Measurements
 Discharge
 Weir
 broad crested
 sharp crested
 triangular
 Venturi Flume
 Spillways
 Sluice gates
 Velocity-Area-Integration

Discharge Measurements
Sharp-Crested Weir
Triangular Weir
Broad-Crested Weir
Sluice Gate
5/ 2
8
2 tan
15 2
d
Q C g H
q
æ ö
=
è ø
3/ 2
2
3
d
Q C b g H
æ ö
=
è ø
3/ 2
2
2
3 d
Q C b gH
=
1
2
d g
Q C by gy
=
Explain the exponents of H!

Summary
 All the complications of pipe flow plus
additional parameter... _________________
 Various descriptions of head loss term
 Chezy, Manning, Darcy-Weisbach
 Importance of Froude Number
 Fr>1 decrease in E gives increase in y
 Fr<1 decrease in E gives decrease in y
 Fr=1 standing waves (also min E given Q)
 Methods of calculating location of free surface
free surface location
0
1
2
3
4
0 1 2 3 4
E
y

Broad-crested Weir: Solution
0.5
yc
E
Broad-crested
weir
yc=0.3 m
3
c
gy
q 
 3
2
3
.
0
)
/
8
.
9
( m
s
m
q 
s
m
q /
5144
.
0 2

s
m
qL
Q /
54
.
1 3


E
yc
3
2

m
y
E c 45
.
0
2
3
2 

1 2 0.95
E E y m
= + D =
2
1
2
1
1
2gy
q
y
E 

435
.
0
5
.
0
1
1 

 m
y
H
935
.
0
1 
y
2
1 1
2
1
2
q
E y
gE
- @

Sluice Gate
2 m
10 cm
Sluice gate
reservoir

Summary/Overview
Energy losses
Dimensional Analysis
Empirical
8
f h
g
V S R
f
=
1/2
o
2/3
h S
R
1
n
V 

Energy Equation
Specific Energy
Two depths with same energy!
How do we know which depth
is the right one?
Is the path to the new depth
possible?
2 2
1 2
1 2
2 2
o f
V V
y S x y S x
g g
+ + D = + + D
2
2
2
q
y
gy
= +
g
V
y
E
2
2


2
2
2
Q
y
gA
= +
0
1
2
3
4
0 1 2 3 4
E
y

Critical Depth
Minimum energy for q
When
When kinetic = potential!
Fr=1
Fr>1 = Supercritical
Fr<1 = Subcritical
0

dy
dE
0
1
2
3
4
0 1 2 3 4
E
y
g
V
y c
c
2
2
2

3
T
Q
gA
=
3
c
q
gy
=
c
c
V
Fr
y g
=

What next?
 Water surface profiles
 Rapidly varied flow
 A way to move from supercritical to subcritical flow
(Hydraulic Jump)
 Gradually varied flow equations
 Surface profiles
 Direct step
 Standard step

11 open_channel.ppt

Recommended

Recommended

More Related Content

Similar to 11 open_channel.ppt

Similar to 11 open_channel.ppt (20)

Recently uploaded

Recently uploaded (20)

11 open_channel.ppt