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
normal depth
 Discharge-Depth relationships

Uniform Flow

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

Classification of Flows

Steady and Unsteady
(Temporal)
Steady: velocity at a given point does not change with
time
 Uniform, Gradually Varied, and Nonuniform (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 and Momentum, Empirical or
Dimensional Analysis?
 What controls depth given a discharge?
 Why doesn’t the flow accelerate? Force balance

Steady-Uniform Flow: Force
Balance
oP D x
Shear force =________
V2
P
Wetted perimeter = __
b
2g
Energy grade line
Hydraulic grade line
Dx sin
Gravitational force = A
________
ADx sin    o PDx  0
o  
A
P
A
sin 
Dx
c
a
d
W cos  
Shear force
= R h Hydraulic radius

W
W sin 
S 
sin 
 sin 
cos 
Turbulence
Relationship between shear and velocity? ______________
P
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) S f
l
l
èRh
ø
2 gS f l Rh
=l
2
V
l
V=
2 gS f Rh
l
V=
2g
S f Rh
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
m
m
60
< C < 150
s
s
where 60 is for rough and 150 is for smooth
also a function of R (like f in Darcy-Weisbach)
Darcy-Weisbach equation (1840)
 d 2 



A  4  d
Rh  

P
d
4
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
For rock-bedded streams
8g
V=
S f Rh
f
f =
1
2
æ
é Rh ùö
1.2 + 2.03log ê ú÷
ç
ëd84 ûø
where d84 = rock size larger than 84% of è
the rocks in a random sample
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
n  0.038d 1 / 6 dftin 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

P  2 y   yz 

b

b
2 1/ 2
2

P  2 y 1 z
1
2 1/ 2
Use Solver!
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
Open Channel Flow: Energy
Relations
velocity head
1
V12
hL = S f Dx
2g
V22
2
2g
energy
grade line
hydraulic
grade line
y1
y2
water surface
S o Dx
Dx
Bottom slope (So) not necessarily equal to surface slope (Sf)
Energy relationships
Pipe flow
p1
V12 p2
V22
+ z1 + a 1
= + z2 + a 2
+ hL
g
2g g
2g
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
V12
V22
y1 +
+ So Dx = y2 +
+ S f Dx
2g
2g
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
y
For a change in bottom elevation
E1 - Dy = E2
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
B
3 roots (one is negative)
Specific Energy: Sluice Gate
sluice gate
10
9
y1 8
7
6
5
4
3
2
1
y2
0
E  y 
q2
1
y
2gy 2
q = 5.5 m2/s
EGL y2 = 0.45 m
V2 = 12.2 m/s
E2 = 8 m
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.
Specific Energy: Step Up
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
3
4
E
0
0
1
2
E
3
4
E1 = E2 + Dy
Increase step height?
4
y
3
yc
Critical Flow
2
1
0
0
1
2
3
E
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
Arbitrary cross-section
T
y
2
gA
A
P
dA  Tdy
V 2T
dy
dA
 Fr
2
T=surface width
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
yc
q2

3
gyc3
1/ 3




Only for rectangular channels!
gyc3
Given the depth we can find the flow!
2
q 
Ac  ycT
Critical Flow Relationships:
Rectangular Channels
q
yc  
g

2
Vc
 Vc2 yc2 

yc3  
 g 


1/ 3




 1
Froude number
yc g
yc 
Vc2
yc
g
2
E  y 
V2
2g

inertial force
gravity force
Vc2
q  Vc yc
because
Kinetic energy
Potential energy
velocity head = 0.5 (depth)
2g
E  yc 
yc
2
yc 
2
3
E
4
y
3
Critical Flow
2
1
0
0
1
2
E

Characteristics
Unstable surface
 Series of standing waves


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
relationship between depth and discharge
 Unique
___________________________________________
3
4
Broad-crested Weir
q
yc  
g

2
q 
gy
3
c
yc 
1/ 3




Q = b gyc3
2
E
P
yc
Broad-crested
weir
Hard to measure yc
3
3/ 2
Q =b g
E
H
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.
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
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
 We
of spillway
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
M1  M 2  W  Fp  Fp  Fss Conservation of Momentum
hL
EGL
M 1x  M 2 x  Fp  Fp
1
1x
2
2x
y2
M 1x   V12 A1
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
p
gy
2
V 
Q
A
Hydraulic Jump:
Conjugate Depths
For a rectangular channel make the following substitutions
A  By
V1
Fr1 =
gy1
Much algebra
Q  By1V1
Froude number
y2 
y1
2
valid for slopes < 0.02
 1 
1  8Fr12

Hydraulic Jump:
Energy Loss and Length
•Energy Loss E1  E2  hL
E  y 
q2
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  Fr1  20
Gradually Varied Flow
V12
V22
y1 +
+ So Dx = y2 +
+ S f Dx
2g
2g
æV22 V12 ö
So dx = ( y2 - y1 ) + ç + S f dx
÷
è2 g 2 g ø
æV 2 ö
dy + d ç ÷ + S f dx = So dx
è2 g ø
d  V 2 
dx
dx

Sf
 So


dy dy  2 g 
dy
dy
dy
Energy equation for nonuniform, steady flow
dy  y2  y1
T
dy
y
A
P
Gradually Varied Flow
  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

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
1  Fr  
2
 So
dx
We are holding Q constant!
dy
So  S f
dy
dx

So  S f
1  Fr 2
Gradually Varied Flow
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 
V22
2g
2g
y1  y2 
V12

2g
Dx 
V22
2g
S f  So
rectangular channel
V1 
energy equation
 S f Dx
q
y1
V2 
q
y2
solve for Dx
prismatic channel
V2 
Q
A2
V1 
Q
A1
Direct Step Method
Friction Slope
Manning
Darcy-Weisbach
n 2V 2
S f = 4/3
Rh
fV 2
Sf =
8 gRh
n 2V 2
Sf =
2.22 Rh4 / 3
SI units
English units
Direct Step
Limitation: channel must be _________
prismatic (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

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 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
y1 
V12
2g
 So Dx  y2 
V22
2g
 S f Dx
What curves are available?
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
decreases in depth in the upstream direction?
0
elevation (m)
1.0
S1
Wave Celerity
Vw
y
V
y
y+y
V+V
unsteady flow
F1 
1
gy
2
F2 
2
Fr  F1  F2 
steady flow
1
g  y  y 2
F1
2
1
2
V-Vw V+V-Vw y+y

g y   y  y 
2
2

F2
V-Vw V+V-Vw
Wave Celerity:
Momentum Conservation
Fr  M 2  M 1
M 1    V  Vw  y
Per unit width
2
Fr  yV  Vw V  V  Vw   V  Vw 
Fr  yV  Vw V
1
2
1


g y 2   y  y 2  yV  Vw V
g  2 yy   y V  Vw V
2
 gy  V  Vw V
y
V-Vw V+V-Vw y+y
steady flow
Wave Celerity
yV  Vw    y  y V  V  Vw 
Mass conservation
yV  yVw  yV  yV  yV  yV  yVw  yVw
V  V  Vw 
y
y
 gy  V  Vw V
gy  V  Vw 
2
Momentum
y
V-Vw V+V-Vw y+y
y
steady flow
y
gy  V  Vw 
2
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

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
 Triangular
Weir
 Broad-Crested
 Sluice
Weir
Weir
Gate
Explain the exponents of H!
2
Q = Cd b 2 gH 3/ 2
3
Q=
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 head loss term
 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
Sluice Gate
reservoir
2m
10 cm
Sluice gate
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
Specific Energy: Step Up
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
3
4
E
0
0
1
2
E
3
4
E1 = E2 + Dy
Increase step height?
Critical Depth
When
When
energy for q
dE
 0
dy
kinetic
= potential!
Fr=1
q
T
Vc
=
=
Q
Fr =
3
3
gA
yc g
gyc
Fr>1
= Supercritical
Fr<1 = Subcritical
yc

2
Vc2
2g
4
3
y
 Minimum
2
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