Open Channel Flow - Cornell University
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Transcript Open Channel Flow - Cornell University
Open Channel Flow
Monroe L. Weber-Shirk
School of Civil and
Environmental Engineering
Open Channel Flow
Liquid (water) flow with a ____
________
free surface
(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 (______)
depth relationships
optimal channel design
Topics in Open Channel Flow
Uniform Flow
Channel transitions
normal depth
Discharge-Depth relationships
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
Classification of Flows
Steady and Unsteady
(Temporal)
Steady: velocity at a given point does not change with
time
Uniform, Gradually Varied, and Rapidly Varied (Spatial)
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
Momentum and Energy
Equations
Conservation of Energy
“losses” due to conversion of turbulence to heat
useful when energy losses are known or small
____________
Contractions
Must account for losses if applied over long distances
We need an equation for losses
_______________________________________________
Conservation of Momentum
“losses” due to shear at the boundaries
useful when energy losses are unknown
Expansion
____________
Open Channel Flow:
Discharge/Depth Relationship
Given a long channel of
constant slope and cross
section find the relationship
A
between discharge and depth
P
Assume
Steady Uniform Flow - ___
_____________
no acceleration
prismatic channel (no change in _________
geometry with distance)
Use Energy, Momentum, Empirical or Dimensional
dimensional
give
Analysis? What
did
givebalance)
us? us?give us?
doesenergy
momentum
(force
Compare
withequation
pipe analysis
flow
hl d
What controls depth given a discharge?
0
4l
Why doesn’t the flow accelerate? Force balance
Steady-Uniform Flow: Force
Balance
V2
oP D x
Shear force =________
2g
P
Wetted perimeter = __
Energy grade line
Hydraulic grade line
b
Dx sin
Gravitational force = A
________
Dx
c
ADx sin o PDx 0
o
A
sin
P
A
= Rh
a
d
Shear force
Hydraulic radius
W cos
W
P
S
sin
cos
Turbulence
Relationship between shear and velocity? ___________
t
o
= g Rh S
W sin
sin
Open Conduits:
Dimensional Analysis
Geometric
parameters
Hydraulic radius (Rh)
___________________
Channel length (l)
___________________
Rh
A
P
Roughness (e)
___________________
Write
the functional relationship
æl e
ö
C p = f ç , , Re, Fr , M, W÷
èRh Rh
ø
Does
No!
Fr affect shear? _________
Fr =
V
yg
Pressure Coefficient for Open
Channel Flow?
Cp
2Dp
V 2
Ch
l
2 ghl
V2
Pressure Coefficient
(Energy Loss Coefficient)
Head loss coefficient
Dp hl
hl = S f l
Friction slope
CS f =
2 gS f l
V2
Friction slope coefficient
Slope of EGL
Dimensional Analysis
CS f
æl e
ö
= f ç , , Re÷
èRh Rh
ø
CS f
ö
l æe
=
f ç , Re÷
Rh èRh
ø
CS f =
2 gS f l
V2
Head loss length of channel
Rh
æe
ö
Rh
C
=l
CS f
= f ç , Re÷ = l (like f in Darcy-Weisbach)
Sf
l
l
èRh
ø
2
LV
hl f
D 2g
2 gS f l Rh
2g
l V2
2
gS
R
f
h
V
=
S f Rh
Sf =
=l
V
=
2
l
Rh 2 g
V
l
l
Chezy Equation (1768)
Introduced
by the French engineer Antoine
Chezy in 1768 while designing a canal for
the water-supply system of Paris
V = C Rh S f
compare
2g
V=
S f Rh
l
where C = Chezy coefficient
0.0054 > l > 0.00087
m
m
60
< C < 150
0.022 > f > 0.0035
s
s
where 60 is for rough and 150 is for smooth
also a function of R (like f in Darcy-Weisbach)
For a pipe
D = 4 Rh
Darcy-Weisbach Equation (1840)
f = Darcy-Weisbach friction factor
l V2
hl = f
d 2g
l V2
Sfl = f
4 Rh 2 g
l V2
hl = f
4 Rh 2 g
V2
S f Rh = f
8g
d 2
A 4 d
Rh
P
d
4
V=
8g
S f Rh
f
æe
1
2.5 ö
Similar to Colebrook
= - 2 log ç
+
÷
è12 Rh Re f ø
f
1
For rock-bedded streams
f=
2
æ
é Rh ùö
where d84 = rock size larger than 84% of the
1.2 + 2.03log ê ú÷
ç
rocks in a random sample
è
ëd84 ûø
Manning Equation (1891)
Most popular in U.S. for open channels
1 2/3 1/2
(MKS units!)
V R h So
1/3
n
Dimensions of n? T /L
V
1.49
Is n only a function of roughness? NO!
1/2
R 2/3
S
h
o
n
Q VA
1
Q ARh2 / 3 S o1 / 2
n
(English system)
Bottom slope
very sensitive to n
Values of Manning n
Lined Canals
Cem ent plaster
Untreated gu nite
Wood , planed
Wood , u nplaned
Concrete, trow led
Concrete, w ood form s, u nfinished
Ru bble in cem ent
Asphalt, sm ooth
Asphalt, rou gh
N atural Channels
Gravel bed s, straight
Gravel bed s plu s large bou ld ers
Earth, straight, w ith som e grass
Earth, w ind ing, no vegetation
Earth , w ind ing w ith vegetation
n 0.031d 1 / 6 d in ft
n 0.038d 1 / 6 d in m
n
0.011
0.016
0.012
0.013
0.012
0.015
0.020
0.013
0.016
n = f(surface
roughness,
channel
irregularity,
stage...)
0.025
0.040
0.026
0.030
0.050
d = median size of bed material
Trapezoidal Channel
Q
1
ARh2 / 3 S o1 / 2
n
Derive
P = f(y) and A = f(y) for a
trapezoidal channel
How would you obtain y = f(Q)?
A yb y 2 z
2 1/ 2
2
ù +b
P = 2é
y
+
yz
(
)
ë
û
2 1/ 2
P = 2y é
ë1 + z ù
û +b
Use Solver!
1
y
z
b
Flow in Round Conduits
= (r sin q )(r cosq )
r y
arccos
r
radians
A r 2 sin cos
r
T 2r sin
P 2r
Maximum discharge
0.938d
when y = ______
A
y
T
Velocity Distribution
v ( y) = V +
k » 0.4
1
k
yö
æ
gdS0 1 + ln
è
dø
For channels wider than 10d
Von Kármán constant
depth [m]
V = average velocity
d = channel depth
2
At what elevation does the
velocity equal the average
velocity?
y
- 1 = ln
d
1
y= d
e
0.368d
0.8d
1.5
0.4d
0.2d
1
0.5
0
0
1
2
3
v(y) [m/s]
4
V
5
Wee Stinky Creek:
Survey on Friday
12:30
Jonathan will go out to setup stations
Need a few volunteers to help Jonathan
Each team will survey two stream cross
sections
Surveying
2:30
3:00
3:30
4:00
team
Amy, Kendra, Brenda
Dale, Justin, Thomas, Siobhan
Dan, Rick, Martin
Nami, Jason, Joanna
Open Channel Flow: Energy
Relations
velocity head
1
V12
hL = S f Dx
2g
2
V22
2g
energy
______
grade line
hydraulic
_______
grade line
y1
y2
S o Dx
Dx
Bottom slope (So) not necessarily equal to EGL slope (Sf)
Energy Relationships
p1
V12 p2
V22
+ z1 + a 1
= + z2 + a 2
+ hL
g
2g g
2g
Pipe flow
z - measured from
horizontal datum
From diagram on previous slide...
y1 + So Dx +
2
1
2
2
V
V
= y2 +
+ S f Dx
2g
2g
Turbulent flow ( 1)
y - depth of flow
Energy Equation for Open Channel Flow
2
1
2
2
V
V
y1 +
+ So Dx = y2 +
+ S f Dx
2g
2g
1
V12
hL = S f Dx
2g
2
2
2
V
2g
y1
y2
S o Dx
Dx
Specific Energy
The
sum of the depth of flow and the
velocity head is the specific energy:
E y
V2
2g
E1 So Dx E2 Sf Dx
y - _______
potential energy
V2
- _______
kinetic energy
2g
If channel bottom is horizontal and no head loss
E1 E2
For a change in bottom elevation
E1 - Dy = E2
y
Specific Energy
In a channel with constant discharge, Q
Q A1V1 A2V2
V2
Q2
E y
E y
where A=f(y)
2
2g
2gA
Consider rectangular channel (A = By) and Q = qB
q is the discharge per unit width of channel
q2
E y
y
2gy 2
A
3 roots (one is negative)
B
2
How many possible depths given a specific energy? _____
Specific Energy: Sluice Gate
10
9
y1 8
7
6
5
4
3
2
1
y2
0
sluice gate
E y
q2
y
2gy 2
1
q = 5.5 m2/s
EGL y2 = 0.45 m
V2 = 12.2 m/s
E2 = 8 m
vena contracta
2
0
1
2
3
4
5
6
7
8
9 10
E1 E2
Given downstream depth and discharge, find upstream depth.
alternate depths (same specific energy)
y1 and y2 are ___________
E
Why not use momentum conservation to find y1?
Specific Energy: Raise the Sluice
Gate
sluice gate
4
3
EGL
y
y1
2
2
1
y2 1
E1 E2
0
0
1
2
E
3
4
E y
q2
2gy 2
as sluice gate is raised y1 approaches y2 and E is minimized:
Maximum discharge for given energy.
Step Up with Subcritical Flow
Short, smooth step with rise Dy in channel
Given upstream depth and discharge find y2
4
3
y
4
3
y
Energy conserved
2
2
1
1
0
0
1
Dy
2
E
0
0
1
2
E
3
3
4
E1 = E2 + Dy
4
Is alternate depth possible? __________________________
NO! Calculate depth along step.
Max Step Up
Short, smooth step with maximum rise Dy in channel
What happens if the step is
Choked flow
increased further?___________
4
3
y
4
3
2
y
1
2
0
1
0
0
0
1
2
E
3
4
1
Dy
2
E
3
4
E1 = E2 + Dy
Step Up with Supercritical flow
Short, smooth step with rise Dy in channel
Given upstream depth and discharge find y2
4
3
y
4
y
3
2
2
1
1
0
0
1
Dy
2
E
0
0
1
2
E
3
3
4
E1 = E2 + Dy
4
What happened to the water depth?______________________________
Increased! Expansion! Energy Loss
4
y
3
yc
Critical Flow
2
1
0
0
1
2
3
E
Arbitrary cross-section
Find critical depth, yc
dE
0
dy
Q2
E y
A=f(y)
2gA2
Q 2 dA
dE
= 1=0
3
dy
gA dy
1
Q 2Tc
QT
gAc3
gA3
2
Fr
T
y
dy
dA
A
P
dA = Tdy
T=surface width
More general definition of Fr
2
V 2T
gA
Fr
2
A
=D
T
Hydraulic Depth
4
Critical Flow:
Rectangular channel
1
Q 2Tc
T
T Tc
3
c
gA
Q qT
1
q 2T 3
3
c
gy T
q
yc
g
Ac
q2
3
gyc3
1/ 3
Only for rectangular channels!
gyc3
Given the depth we can find the flow!
2
q
Ac ycT
yc
Critical Flow Relationships:
Rectangular Channels
q
yc
g
2
Vc
Vc2 yc2
yc3
g
1/ 3
1
inertial force
gravity force
Froude number
yc g
yc
Vc2
yc
g
2
E y
V2
2g
Vc2
q Vc yc
because
Kinetic energy
Potential energy
velocity head = 0.5 (depth)
2g
E yc
yc
2
yc
2
3
E
Critical Depth
Minimum
Occurs
energy for a given q
dE
dy
0
when
=___
Vc2 yc
=
2g 2
When kinetic = potential! ________
Fr=1
4
= ______critical
Super
Fr<1 = ______critical
Sub
Fr>1
3
y
q
T
Vc
=
=
Q
Fr =
3
3
gA
yc g
gyc
2
1
0
0
1
2
E
3
4
4
y
3
Critical Flow
2
1
0
0
1
2
E
Characteristics
Unstable surface
Series of standing waves
dE
0
dy
Difficult to measure depth
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
3
4
Broad-Crested Weir
q
yc
g
2
q
gy
3
c
yc
yc
1/ 3
H
P
Q = b gyc3
2
E
Hard to measure yc
3
3/ 2
Q =b g
yc
Broad-crested
weir
æ2 ö
è3 ø
E 3/ 2
3/ 2
æ2 ö
Q = Cd b g
H
è3 ø
E measured from top of weir
Cd corrects for using H rather
than E.
E
Broad-crested Weir: Example
Calculate
the flow and the depth upstream.
The channel is 3 m wide. Is H approximately
equal to E?
H
0.5
E
yc m
yc=0.3
Broad-crested
weir
How do you find flow?____________________
Critical flow relation
Energy equation
How do you find H?______________________
Solution
Hydraulic Jump
Used
for energy dissipation
Occurs when flow transitions from
supercritical to subcritical
base
of spillway
Steep slope to mild slope
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
M1 M 2 W Fp Fp Fss Conservation of Momentum
hL
EGL
M 1x M 2 x Fp Fp
1
1x
2
2x
M 1x V12 A1
y2
y1
M 2 x V A2
2
2
L
QV1 QV2 p1 A1 p2 A2
Q
2
A1
Q
2
A2
gy1 A1
2
gy2 A2
2
r gy
p=
2
V
Q
A
Hydraulic Jump:
Conjugate Depths
For a rectangular channel make the following substitutions
A By
Fr1 =
V1
gy1
Much algebra
Q By1V1
Froude number
y2
y1
1
1 8Fr12
2
y2 - 1 + 1 + 8 Fr12
=
y1
2
valid for slopes < 0.02
Hydraulic Jump:
Energy Loss and Length
Energy
Loss
E y
q2
E1 E2 hL
algebra
hL
y2 y1 3
4 y1 y2
2gy 2
significant energy loss (to turbulence) in jump
Length
of jump
No general theoretical solution
Experiments show
L 6y2 for 4.5 < Fr1 < 13
Specific Momentum
2
2
gy1 A1 Q
gy A Q
+
= 2 2+
2
A1
2
A2
y12 q 2
y22 q 2
+
= +
2 y1 g 2 y2 g
E
1:1 slope
M
4
3
y
y1 A1 Q 2 y2 A2 Q 2
+
=
+
2
A1 g
2
A2 g
5
2
1
DE
0
2
3
4
5
E or M
6
7
Gradually Varied Flow:
Find Change in Depth wrt x
V12
V22
y1 +
+ So Dx = y2 +
+ S f Dx
2g
2g
Energy equation for nonuniform, steady flow
æV22 V12 ö
So dx = ( y2 - y1 ) + ç + S f dx
÷
è2 g 2 g ø
dy y2 y1
Shrink control volume
T
æV ö
dy + d ç ÷ + S f dx = So dx
è2 g ø
dy
2
d V 2
dx
dx
Sf
So
dy dy 2 g
dy
dy
dy
y
A
P
T
Gradually Varied Flow:
Derivative of KE wrt Depth
y
dy
dA
A
P
2Q 2 dA
Q 2T
d V 2
d Q 2
Fr 2
2 gA3 dy
gA3
dy 2 g
dy 2 gA2
Change in KE
2
dA Tdy
dy d V
dx
dx
Change in PE
Sf
So
dy dy 2 g
dy
dy
1 Fr S f
2
dy
dx
dx
dy
So
So S f
1 Fr 2
dx
We are holding Q constant!
dy
The water surface slope is a function of:
bottom slope, friction slope, Froude number
Gradually Varied Flow:
Governing equation
dy
dx
So S f
1 Fr 2
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 So = S f
Surface Profiles
Mild slope (yn>yc)
Steep slope (yn<yc)
in a long channel unstable flow will occur
Horizontal slope (So=0)
in a long channel supercritical flow will occur
Critical slope (yn=yc)
in a long channel subcritical flow will occur
yn undefined
Adverse slope (So<0)
yn undefined
Note: These slopes are f(Q)!
Surface Profiles
Normal depth
Obstruction
Steep slope (S2)
Sluice gate
Steep slope
dx
So S f
1 Fr
2
S0 - Sf
1 - Fr2
dy/dx
4
+
-
+
+
+
-
yn
yc
3
y
dy
Hydraulic Jump
2
1
-
-
+
0
0
1
2
E
3
4
More Surface Profiles
dy
S0 - Sf
1 - Fr2
dy/dx
4
+
+
2 +
-
-
dx
1 Fr 2
3
y
1 +
So S f
2
yc
yn
1
3 -
-
+
0
0
1
2
E
3
4
Direct Step Method
y1
V12
So Dx y2
2g
V22
energy equation
2g
y1 y2
V12
2g
Dx
V22
2g
S f So
rectangular channel
V1
S f Dx
q
y1
V2
solve for Dx
prismatic channel
q
y2
V2
Q
A2
V1
Q
A1
Direct Step Method
Friction Slope
Manning
Darcy-Weisbach
n 2V 2
S f = 4/3
Rh
V2
Sf = f
8 gRh
n 2V 2
Sf =
2.22 Rh4 / 3
SI units
English units
Direct Step
prismatic (channel
Limitation: channel must be _________
geometry is independent of x 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 given yn and yn+1
calculate average friction slope
calculate Dx
Direct Step Method
=y*b+y^2*z
y1 y2
=2*y*(1+z^2)^0.5 +b
=A/P
V12
2g
Dx
=Q/A
=(n*V)^2/Rh^(4/3)
=y+(V^2)/(2*g)
V22
2g
S f So
=(G16-G15)/((F15+F16)/2-So)
A
B
C
D
E
F
G
H
I
y
A
P
Rh
V
Sf
E
Dx
x
0.900 1.799 4.223
0.426 0.139
0.00004 0.901
0
0.870 1.687 4.089
0.412 0.148
0.00005 0.871
0.498 0.5
J
K
L
M
T
Fr
bottom surface
3.799
0.065 0.000
0.900
3.679
0.070 0.030
0.900
Standard Step
Given a depth at one location, determine the depth at a
second given 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
Usually solved upstream for subcritical
Usually solved downstream for supercritical
Find a depth that satisfies the energy equation
y1
V12
2g
So Dx y2
V22
2g
S f Dx
What curves are available?
Steep Slope
1.4
1.2
bottom
surface
yc
yn
0.8
0.6
S3
0.4
0.2
0.0
20
15
10
5
distance upstream (m)
Is there a curve between yc and yn that increases in
depth in the downstream direction? ______
NO!
0
elevation (m)
1.0
S1
Mild Slope
If
the slope is mild, the depth is less than the
critical depth, and a hydraulic jump occurs,
what happens next?
0.6
bottom
surface
yc
yn
0.4
0.3
0.2
0.1
0.0
40
35
30
25
20
15
distance upstream (m)
10
5
0
elevation (m)
0.5
Water Surface Profiles:
Putting It All Together
reservoir
Sluice gate
2m
10 cm
1 km downstream from gate there is a broad crested
weir with P = 1 m. Draw the water surface profile.
Wave Celerity
Vw
y
V
V+V
y+y
unsteady flow
M1 + M 2 = W + Fp1 + Fp2 + Fss
Per unit width
1
1
2
Fp1 = r gy 2
Fp2 = - r g ( y + d y )
2
2
1
2
2
ù
Fp1 + Fp2 = r g é
y
y
+
d
y
(
)
ë
û
2
y
V-Vw V+V-Vw
y+y
steady flow
Fp1
Fp2
V-Vw V+V-Vw
Wave Celerity:
Momentum Conservation
M 1 V Vw y
2
M 2 = r (V + dV - Vw )(V - Vw ) y
Per unit width
M1 + M 2 = r y (V - Vw )[(V + dV - Vw ) - (V - Vw )]
1
2
2
é
F
+
F
=
r
g
y
y
+
d
y
(
) ùû
M1 + M 2 = r y (V - Vw )dV
p1
p2
ë
2
Now equate pressure and momentum
1
2
2
2
ù
r gé
y
y
2
y
d
y
d
y
= r y (V - Vw )dV
ë
û
2
gy V Vw V
y
V-Vw V+V-Vw
steady flow
y+y
Wave Celerity
yV Vw y y V V Vw
Mass conservation
yV yVw yV yV yV yV yVw yVw
V V Vw
y
y
gy V Vw V
gy V Vw
2
Momentum
y
V-Vw V+V-Vw
y
steady flow
y
gy V Vw
2
y+y
c V Vw
c
V
gy
yg
Fr
V
c
Wave Propagation
Supercritical flow
c<V
waves only propagate downstream
water doesn’t “know” what is happening downstream
upstream control
_________
Critical flow
c=V
Subcritical flow
c>V
waves propagate both upstream and downstream
Open Channel Flow Discharge
Measurements
Discharge
Weir
broad crested
sharp crested
triangular
Venturi Flume
Spillways
Sluice gates
Velocity-Area-Integration
Discharge Measurements
Sharp-Crested
V-Notch
Weir
Broad-Crested
Sluice
Weir
2
Q = Cd b 2 gH 3/ 2
3
Q=
Weir
Gate
Explain the exponents of H!
8
æq ö 5/ 2
Cd 2 g tan
H
è2 ø
15
3/ 2
æ2 ö
Q = Cd b g
H
è3 ø
Q = Cd byg 2 gy1
Summary
All
the complications of pipe flow plus
additional parameter... _________________
free surface location
Various descriptions of energy loss
Chezy,
Manning, Darcy-Weisbach
4
Fr>1
of Froude Number
3
y
Importance
2
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)
1
0
0
Methods
1
2
3
E
of calculating location of free surface
4
Broad-crested Weir: Solution
q
E
gyc3
q (9.8m / s )0.3m
2
3
0.5
q 0.5144m 2 / s
Q qL 1.54m3 / s
2
yc E
3
q2
E1 @y1
2
2 gE1
yc m
yc=0.3
Broad-crested
weir
E2
3
yc 0.45m
2
E1 = E2 + Dy = 0.95m
q2
E1 y1
2gy12
y1 0.935
H1 y1 0.5m 0.435
Summary/Overview
Energy
losses
Dimensional Analysis
Empirical
8g
V=
S f Rh
f
1 2/3 1/2
V R h So
n
Energy Equation
Specific
V12
V22
y1 +
+ So Dx = y2 +
+ S f Dx
2g
2g
V2
q2
Q2
= y+
E y
2 = y+
2
2
gy
2
gA
2g
Energy
Two depths with same energy!
do we know which depth4
is the right one?
3
Is the path to the new depth
2
possible?
y
How
1
0
0
1
2
E
3
4
What next?
Water
surface profiles
Rapidly
varied flow
A
way to move from supercritical to subcritical flow
(Hydraulic Jump)
Gradually
Surface
varied flow equations
profiles
Direct step
Standard step
Hydraulic Jump!
Open Channel Reflections
Why isn’t Froude number important for describing
the relationship between channel slope, discharge,
and depth for uniform flow?
Under what conditions are the energy and
hydraulic grade lines parallel in open channel
flow?
Give two examples of how the specific energy
could increase in the direction of flow.