1. Open Channel Flow
School of Civil and
Monroe L. Weber-Shirk Environmental Engineering
2. Open Channel Flow
® Liquid (water) flow with a ____surface
free ________
(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
depth
® optimal channel design
3. Topics in Open Channel Flow
® Uniform Flow normal depth
® 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
4. 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
5. 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
® ____________
6. 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 - ___ acceleration
no _____________
geometry
® 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? Force balance
7. Steady-Uniform Flow: Force
Balance
V2
τo P ∆ x
Shear force =________ 2g Energy grade line
P Hydraulic grade line
Wetted perimeter = __ b
Gravitational force = γA ∆x sinθ
________ ∆x c
γA∆ sin θ −τo P∆ = 0
x x
a
A
τo =γ sinθ d
P W cos θ θ
θ
Shear force
W
A sin θ
= R h Hydraulic radius S = ≅ sin θ
P cos θ
W sin θ
Turbulence
Relationship between shear and velocity? ______________
8. Open Conduits:
Dimensional Analysis
® Geometric parameters A
Hydraulic radius (Rh)
® ___________________ Rh =
P
Channel length (l)
® ___________________
® ___________________
Roughness (ε)
® Write the functional relationship
l ε
C p = f , , Re, Fr , M, W
Rh Rh
V
® Does No! Fr =
Fr affect shear? _________ yg
9. Pressure Coefficient for Open
Channel Flow?
− 2 ∆p Pressure Coefficient
Cp = − ∆p = γhl
ρV 2 (Energy Loss Coefficient)
2 ghl hl = S f l
Ch = Head loss coefficient
V2
l
Friction slope
2 gS f l Friction slope coefficient Slope of EGL
CS f =
V2
10. Dimensional Analysis
l ε 2 gS f l
CS f = f , , Re CS f =
Rh Rh V2
l ε
CS f = f , Re Head loss ∝ length of channel
Rh Rh
Rh ε R
CS f = f , Re = λ (like f in Darcy-Weisbach) CS f h = λ
l Rh l
2 gS f l Rh 2 gS f Rh 2g
=λ V= V= S f Rh
V 2
l λ λ
11. Chezy equation (1768)
® Introducedby the French engineer Antoine
Chezy in 1768 while designing a canal for
the water-supply system of Paris
2g
V = C Rh S f compare V= S f Rh
λ
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)
12. Darcy-Weisbach equation (1840)
πd 2
f = Darcy-Weisbach friction factor
A 4 d
=
l V2 l V2 Rh = =
hl = f hl = f P πd 4
d 2g 4 Rh 2 g
l V2 V2 8g
Sfl = f S f Rh = f V= S f Rh
4 Rh 2 g 8g f
For rock-bedded streams 1
f = 2
Rh
1.2 + 2.03log d
where d84 = rock size larger than 84% of 84
the rocks in a random sample
13. Manning Equation (1891)
®Most popular in U.S. for open channels
1 2/3 1/2 (MKS units!)
V = R h So
n Dimensions of n? T /L1/3
Is n only a function of roughness? NO!
1.49
V = R 2/3S1/2
h o (English system)
n
Bottom slope
Q = VA
1
Q = ARh / 3 S o / 2
2 1
very sensitive to n
n
14. Values of Manning n
Lined Canals n
Cement plaster 0.011
Untreated gunite 0.016
Wood, planed 0.012 n = f(surface
Wood, unplaned 0.013
Concrete, trowled 0.012 roughness,
Concrete, wood forms, unfinished
Rubble in cement
0.015
0.020
channel
Asphalt, smooth 0.013 irregularity,
Asphalt, rough 0.016
Natural Channels
stage...)
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
n = 0.031d 1 / 6 d in ft d = median size of bed material
n = 0.038d 1 / 6 d in m
15. 1
Trapezoidal Channel Q = ARh / 3 S o / 2
2 1
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
1 y
z
[
P = 2 y 2 + ( yz ) ]
2 1/ 2
+b b
[
P = 2 y 1+ z ]
2 1/ 2
+b
Use Solver!
16. Flow in Round Conduits
= ( r sin θ ) ( r cos θ )
r − y
θ = arccos
r
radians
A = r 2 (θ − sin θ cosθ )
r
T = 2r sin θ
θ
P = 2 rθ A y
Maximum discharge
when y = ______
0.938d T
17. Open Channel Flow: Energy
Relations
V12 hL = S f ∆x
velocity head α1
2g V22
energy
α2 grade line
2g
hydraulic
y1 grade line
y2 water surface
S o ∆x
∆x
Bottom slope (So) not necessarily equal to surface slope (Sf)
18. Energy relationships
p1 V12 p2 V22 Pipe flow
+ z1 + α1 = + z2 + α 2 + hL
γ 2g γ 2g z - measured from
horizontal datum
From diagram on previous slide...
2 2 Turbulent flow (α ≅ 1)
V V
y1 + So ∆x + 1
= y2 + 2
+ S f ∆x y - depth of flow
2g 2g
Energy Equation for Open Channel Flow
V12 V22
y1 + + So ∆x = y2 + + S f ∆x
2g 2g
19. Specific Energy
® Thesum of the depth of flow and the
velocity head is the specific energy:
V2 y - potential energy
E = y +
2g V2
- kinetic energy
2g
E1 + S o ∆x = E2 + S f ∆x
If channel bottom is horizontal and no head loss
E1 = E2
y
For a change in bottom elevation
E1 − ∆y = E2
20. 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
q2 q is the discharge per unit width of channel
E = y +
2gy 2 y A
B
3 roots (one is negative)
21. Specific Energy: Sluice Gate
10
9
sluice gate q = 5.5 m2/s
y1 8 EGL y2 = 0.45 m
7 q2
6
E = y + V2 = 12.2 m/s
2gy 2 1
5
E2 = 8 m
y
4
3
2
2
y2 1
0
0 1 2 3 4 5 6 7 8 9 10
E1 = E2
E
Given downstream depth and discharge, find upstream depth.
alternate
y1 and y2 are ___________ depths (same specific energy)
Why not use momentum conservation to find y1?
22. Specific Energy: Raise the Sluice
Gate
4
sluice gate
3
y1 EGL
2
y
1 2
y2 1
E1 = E2
0
0 1 2 3 4 q2
E E = y +
2gy 2
as sluice gate is raised y1 approaches y2 and E is
minimized: Maximum discharge for given energy.
23. Specific Energy: Step Up
Short, smooth step with rise ∆y in channel
Given upstream depth and 4
discharge find y2
4 3
3 2
y
2 1
y
1 0
0 1 ∆y 2 3 4
0 E
0 1 2 3 4
E E1 = E2 + ∆y Increase step height?
24. 4
3
2
yc
y
Critical Flow 1
0
0 1 2 3 4
E
Find critical depth, yc Arbitrary cross-section
dE
= 0 T
dy dy
Q2
E = y + A=f(y) y A dA
2gA2
P
dE Q 2 dA
= 1− 3 =0 dA = Tdy T=surface width
dy gA dy
Q 2Tc 2
QT V 2T A
1 = = Fr 2
= Fr 2
=D Hydraulic Depth
gAc3 gA 3
gA T
25. Critical Flow:
Rectangular channel
Q 2Tc T
1 = T = Tc
gAc3
Q = qT Ac = ycT Ac yc
q 2T 3 q2
1= =
3 3 3
gy T
c gyc
1/ 3
q 2
yc = Only for rectangular channels!
g
q = 3
gyc Given the depth we can find the flow!
26. Critical Flow Relationships:
Rectangular Channels
q 2
1/ 3
Vc2 yc
2
yc = yc =
3 because q = Vc yc
g g
Vc inertial force Kinetic energy
= 1 Froude number
yc g gravity force Potential energy
Vc2 yc Vc2
yc = = velocity head = 0.5 (depth)
g 2 2g
V2 yc 2
E = y + E = yc + yc = E
2g 2 3
27. 4
3
2
y
Critical Flow 1
0
0 1 2 3 4
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
® Unique relationship between depth and discharge
___________________________________________
28. Broad-crested Weir
1/ 3
q 2 E
yc = H
g yc
P Broad-crested
q = gy 3
c Q = b gyc
3
weir
2
yc = E Hard to measure yc
3
3/ 2
2 E measured from top of weir
Q=b g E 3/ 2
3
3/ 2
2 Cd corrects for using H rather
Q = Cd b g H
3 than E.
29. Broad-crested Weir: Example
® Calculate the flow and the depth upstream.
The channel is 3 m wide. Is H approximately
equal to E? E
yc
yc=0.3 m
0.5 Broad-crested
weir
How do you find flow?____________________
Critical flow relation
Energy equation
How do you find H?______________________
Solution
30. 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?
32. Hydraulic Jump
M1 +M 2 = W +Fp1 +Fp2 +Fss Conservation of Momentum
hL
EGL
M 1 x + M 2 x = F p + Fp 1x 2x
M 1 x = −ρ 12 A1
V y2
y1
M 2 x = ρ A2
V 2
2
L
− ρQV1 + ρQV2 = p1 A1 − p2 A2
2 2 ρgy Q
Q Q gy1 A1 gy2 A2 p= V =
− + = − 2
A1 A2 2 2 A
33. Hydraulic Jump:
Conjugate Depths
For a rectangular channel make the following substitutions
A = By Q = By1V1
V1
Fr1 = Froude number
gy1
Much algebra y2 =
y1
(− 1 + 1 + 8 Fr12 )
2
valid for slopes < 0.02
34. Hydraulic Jump:
Energy Loss and Length
•Energy Loss E1 = E2 + hL
q2 ( y2 − y1 ) 3
E = y + algebra hL =
2gy 2 4 y1 y2
significant energy loss (to turbulence) in jump
•Length of jump
No general theoretical solution
Experiments show
L = 6 y2 for 4 < Fr1 < 20
35. Gradually Varied Flow
V12 V22 Energy equation for non-
y1 + + So ∆x = y2 + + S f ∆x
2g 2g uniform, steady flow
V22 V12 dy = y2 − y1
So dx = ( y2 − y1 ) + − + S f dx
2g 2g T
V 2 dy
dy + d + S f dx = So dx
2g y A
P
dy d V 2
+ S f dx = S o dx
+
dy dy 2 g
dy dy
36. Gradually Varied Flow
d V 2
= d Q2
− 2Q 2 dA
= ⋅
− Q 2T
= = − Fr 2
dy 2 g
dy 2 gA2
2 gA3 dy
gA3
Change in KE
dy d V 2 dx dx
+ +Sf = So Change in PE
dy dy 2 g
dy dy
We are holding Q constant!
dx dx
1 − Fr + S f
2
= So
dy dy
dy dy So − S f
[1 − Fr ]
2
= So − S f =
dx dx 1 − Fr 2
37. Gradually Varied Flow
dy So − S f
= Governing equation for
dx 1 − Fr 2 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
38. 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)!
39. Surface Profiles
Normal depth Obstruction
Sluice gate Steep slope (S2)
Steep slope Hydraulic Jump
dy So − S f
= S0 - S f 1 - Fr2 dy/dx
dx 1 − Fr 2
4
+ + +
yn 3
- + - yc
2
y
1
- - + 0
0 1 2 3 4
E
40. More Surface Profiles
dy So − S f
S0 - S f 1 - Fr2 dy/dx =
4
dx 1 − Fr 2
1 + + + 3
yc
2
y
2 + - -
yn 1
3 - - + 0
0 1 2 3 4
E
41. Direct Step Method
V12 V22
y1 + + S o ∆x = y2 + + S f ∆x energy equation
2g 2g
V12 V22
y1 − y2 + −
2g 2g
∆x = solve for ∆x
S f − So
rectangular channel prismatic channel
q q Q Q
V1 = V2 = V2 = V1 =
y1 y2 A2 A1
42. Direct Step Method
Friction Slope
Manning Darcy-Weisbach
n 2V 2 fV 2
S f = 4/3 SI units Sf =
Rh 8 gRh
n 2V 2
Sf = English units
2.22 Rh / 3
4
43. Direct Step
® Limitation: channel must be _________ (so that
prismatic
velocity is a function of depth only and not a
function of x)
® Method
® identify type of profile (determines whether ∆y is + or -)
® choose ∆y 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 ∆x
44. Direct Step Method
=y*b+y^2*z
V12 V22
=2*y*(1+z^2)^0.5 +b y1 − y2 + −
2g 2g
=A/P ∆x =
S f − So
=Q/A
=(n*V)^2/Rh^(4/3)
=y+(V^2)/(2*g)
=(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
45. Standard Step
® Given a depth at one location, determine the depth at a
second location
® Step size (∆x) 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
V12 V22
y1 + + S o ∆x = y2 + + S f ∆x
2g 2g
46. What curves are available?
1.4
1.2
1.0
S1
elevation (m)
0.8
bottom
S3 0.6
0.4
surface
yc 0.2
yn
0.0
20 15 10 5 0
distance upstream (m)
Is there a curve between yc and yn that
decreases in depth in the upstream direction?
47. Wave Celerity
Vw
y y+δy y
V V+δV V-Vw V+δV-Vw y+δy
unsteady flow steady flow
1 1
F1 = ρgy 2
F2 = ρg ( y + δy ) 2
2 2 F1 F2
Fr = F1 − F2 =
1
[
ρg y − ( y + δy )
2 2
] V-Vw V+δV-Vw
2
48. Wave Celerity:
Momentum Conservation
Fr = M 2 − M 1 M 1 = − ρ (V − Vw ) y
2
Per unit width
Fr = ρy ( V − Vw ) [ ( V + δV − Vw ) − ( V − Vw ) ]
Fr = ρy ( V − Vw ) δV
y V-Vw V+δV-Vw y+δy
1
[ ]
ρg y 2 − ( y + δy ) 2 = ρy ( V − Vw ) δV
2
1 steady flow
g [ − 2 yδy ] = y ( V − Vw ) δV
2
− gδy = ( V − Vw ) δV
49. Wave Celerity
y ( V − Vw ) = ( y + δy )( V + δV − Vw ) Mass conservation
yV − yVw = yV + δyV + yδV + δyδV − yVw − δyVw
δy
δV = −( V − Vw )
y
y
− gδy = ( V − Vw ) δV Momentum V-Vw V+δV-Vw y+δy
δy
gδy = ( V − Vw )
2
y steady flow
V V
gy = ( V − Vw )
2
c = V − Vw c = gy = Fr =
yg c
50. 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
51. 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
52. 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
54. Discharge Measurements
2
® Sharp-Crested Q = Cd b 2 gH 3/ 2
Weir 3
8 θ 5/ 2
® Triangular Weir Q= Cd 2 g tan H
15 2
3/ 2
® Broad-Crested 2
Weir Q = Cd b g H
3
® Sluice Gate Q = Cd by g 2 gy1
Explain the exponents of H!
55. Summary
® Allthe complications of pipe flow plus
additional parameter... _________________
free surface location
® Various descriptions of head loss term
® Chezy, Manning, Darcy-Weisbach
4
® Importance of Froude Number 3
2
® Fr>1 decrease in E gives increase in y
y
1
® Fr<1 decrease in E gives decrease in y 0
0 1 2 3 4
E
® Fr=1 standing waves (also min E given Q)
® Methods of calculating location of free surface
57. Sluice Gate
reservoir Sluice gate
2m
10 cm
S = 0.005
58. Summary/Overview
® Energy losses V=
8g
S f Rh
f
®Dimensional Analysis 1 2/3 1/2
V = R h So
®Empirical n
59. V12 V22
Energy Equation y1 +
2g
+ So ∆x = y2 +
2g
+ S f ∆x
V2 q2 Q2
® Specific
Energy E = y + = y+ 2 = y+
2g 2 gy 2 gA2
® Two depths with same energy!
®How do we know which depth4
is the right one?
3
®Is the path to the new depth
2
y
possible?
1
0
0 1 2 3 4
E
60. Specific Energy: Step Up
Short, smooth step with rise ∆y in channel
Given upstream depth and 4
discharge find y2
4 3
3 2
y
2 1
y
1 0
0 1 ∆y 2 3 4
0 E
0 1 2 3 4
E E1 = E2 + ∆y Increase step height?
61. Critical Depth
® Minimum energy for q
®When dE
= 0
dy yc Vc2
®When kinetic = potential! =
2 2g
®Fr=1
4
V q T
Fr = c = =Q 3
yc g 3
gyc gA3
2
y
®Fr>1 = Supercritical 1
®Fr<1 = Subcritical 0
0 1 2 3 4
E
62. 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
