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Open channel flow refers to liquid flow with a free surface, such as in rivers, streams, canals, and culverts. It is relevant for both natural and engineered channels. Some key topics in open channel flow include uniform flow, channel transitions, critical flow, hydraulic jumps, and gradually varied flow. The flow can be classified based on whether it is steady or unsteady, uniform or nonuniform, and laminar or turbulent. Analyzing open channel flow involves the conservation of energy and momentum equations. The Manning equation and other relationships are commonly used to relate variables like flow depth, velocity, discharge, and channel properties. Critical flow occurs when the specific energy is minimized.

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Open Channel Flow School of Civil and Monroe L. Weber-Shirk Environmental Engineering
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
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
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 - ___ 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
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? ______________
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
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
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
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)
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
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
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
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!
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
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)
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
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
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)
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?
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.
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?
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
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!
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
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 ___________________________________________
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.
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
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 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
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
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
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
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
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
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)!
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
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
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
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
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
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
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
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?
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
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
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
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 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!
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
Broad-crested Weir: Solution q = 3 gyc E yc yc=0.3 m q = (9.8m / s )( 0.3m ) 2 3 0.5 Broad-crested weir q = 0.5144m 2 / s 3 Q = qL = 1.54m 3 / s 2 E2 = yc = 0.45m yc = E 2 3 E1 = E2 + ∆y = 0.95m q2 E1 = y1 + 2gy12 q2 E1 − 2 ≅ y1 y1 = 0.935 2 gE1 H1 = y1 − 0.5m = 0.435
Sluice Gate reservoir Sluice gate 2m 10 cm S = 0.005
Summary/Overview ® Energy losses V= 8g S f Rh f ®Dimensional Analysis 1 2/3 1/2 V = R h So ®Empirical n
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
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?
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
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

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Hidro

  • 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 OpenChannel 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 forOpen 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 Manningn 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 RoundConduits = ( 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 ® Thesumof 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 achannel 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: SluiceGate 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: Raisethe 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: StepUp 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 ® Usedfor 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?
  • 31.
  • 32.
    Hydraulic Jump M1 +M2 = 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 ® Givena 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 areavailable? 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 themost 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
  • 53.
    Open Channel FlowDischarge Measurements ® Discharge ® Weir ® broad crested ® sharp crested ® triangular ® Venturi Flume ® Spillways ® Sluice gates ® Velocity-Area-Integration
  • 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 complicationsof 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
  • 56.
    Broad-crested Weir: Solution q = 3 gyc E yc yc=0.3 m q = (9.8m / s )( 0.3m ) 2 3 0.5 Broad-crested weir q = 0.5144m 2 / s 3 Q = qL = 1.54m 3 / s 2 E2 = yc = 0.45m yc = E 2 3 E1 = E2 + ∆y = 0.95m q2 E1 = y1 + 2gy12 q2 E1 − 2 ≅ y1 y1 = 0.935 2 gE1 H1 = y1 − 0.5m = 0.435
  • 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: StepUp 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