Ce154 lecture 3 reservoirs, spillways, & energy dissipators

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  • 60 ft diameter outlet tunnel
  • Ce154 lecture 3 reservoirs, spillways, & energy dissipators

    1. 1. Reservoirs, Spillways, & Energy Dissipators CE154 – Hydraulic Design Lecture 3 Fall 2009 1CE154
    2. 2. Fall 2009 2 Lecture 3 – Reservoir, Spillway, Etc. • Purposes of a Dam - Irrigation - Flood control - Water supply - Hydropower - Navigation - Recreation • Pertinent structures – dam, spillway, intake, outlet, powerhouse CE154
    3. 3. Fall 2009 3 Hoover Dam – downstream face CE154
    4. 4. Fall 2009 4 Hoover Dam – Lake Mead CE154
    5. 5. Fall 2009 5 Hoover Dam – Spillway Crest CE154
    6. 6. Fall 2009 6 Hoover dam – Outflow Channel CE154
    7. 7. Fall 2009 7 Hoover Dam – Outlet Tunnel CE154
    8. 8. Fall 2009 8 Hoover Dam – Spillway CE154
    9. 9. Fall 2009 9 Dam Building Project • Planning - Reconnaissance Study - Feasibility Study - Environmental Document (CEQA in California) • Design - Preliminary (Conceptual) Design - Detailed Design - Construction Documents (plans & specifications) • Construction • Startup and testing • Operation CE154
    10. 10. Fall 2009 10 Necessary Data • Location and site map • Hydrologic data • Climatic data • Geological data • Water demand data • Dam site data (foundation, material, tailwater) CE154
    11. 11. Dam Components • Dam - dam structure and embankment • Outlet structure - inlet tower or inlet structure, tunnels, channels and outlet structure • Spillway - service spillway - auxiliary spillway - emergency spillway Fall 2009 11CE154
    12. 12. Spillway Design Data • Inflow Design Flood (IDF) hydrograph - developed from probable maximum precipitation or storms of certain occurrence frequency - life loss ⇒ use PMP - if failure is tolerated, engineering judgment ⇒ cost-benefit analysis ⇒ use certain return-period flood Fall 2009 12CE154
    13. 13. Spillway Design Data (cont’d) • Reservoir storage curve - storage volume vs. elevation - developed from topographic maps - requires reservoir operation rules for modeling • Spillway discharge rating curve Fall 2009 13CE154
    14. 14. Reservoir Capacity Curve Fall 2009 14CE154
    15. 15. Spillway Discharge Rating Fall 2009 15CE154
    16. 16. Spillway Design Procedure • Route the flood through the reservoir to determine the required spillway size ∆S = (Qi – Qo) ∆t Qi determined from IDF hydrograph Qo determined from outflow rating curve ∆S determined from storage rating curve - trial and error process Fall 2009 16CE154
    17. 17. Spillway Capacity vs. Surcharge Fall 2009 17CE154
    18. 18. Spillway Cost Analysis Fall 2009 18CE154
    19. 19. Spillway Design Procedure (cont’d) • Select spillway type and control structure - service, auxiliary and emergency spillways to operate at increasingly higher reservoir levels - whether to include control structure or equipment – a question of regulated or unregulated discharge Fall 2009 19CE154
    20. 20. Spillway Design Procedure (cont’d) • Perform hydraulic design of spillway structures - Control structure - Discharge channel - Terminal structure - Entrance and outlet channels Fall 2009 20CE154
    21. 21. Types of Spillway • Overflow type – integral part of the dam -Straight drop spillway, H<25’, vibration -Ogee spillway, low height • Channel type – isolated from the dam -Side channel spillway, for long crest -Chute spillway – earth or rock fill dam - Drop inlet or morning glory spillway -Culvert spillway Fall 2009 21CE154
    22. 22. Sabo Dam, Japan – Drop Chute Fall 2009 22CE154
    23. 23. New Cronton Dam NY – Stepped Chute Spillway Fall 2009 23CE154
    24. 24. Sippel Weir, Australia – Drop Spillway Fall 2009 24CE154
    25. 25. Four Mile Dam, Australia – Ogee Spillway Fall 2009 25CE154
    26. 26. Upper South Dam, Australia – Ogee Spillway Fall 2009 26CE154
    27. 27. Winnipeg Floodway - Ogee Fall 2009 27CE154
    28. 28. Hoover Dam – Gated Side Channel Spillway Fall 2009 28CE154
    29. 29. Valentine Mill Dam - Labyrinth Fall 2009 29CE154
    30. 30. Ute Dam – Labyrinth Spillway Fall 2009 30CE154
    31. 31. Matthews Canyon Dam - Chute Fall 2009 31CE154
    32. 32. Itaipu Dam, Uruguay – Chute Spillway Fall 2009 32CE154
    33. 33. Itaipu Dam – flip bucket Fall 2009 33CE154
    34. 34. Pleasant Hill Lake – Drop Inlet (Morning Glory) Spillway Fall 2009 34CE154
    35. 35. Monticello Dam – Morning Glory Fall 2009 35CE154
    36. 36. Monticello Dam – Outlet - bikers heaven Fall 2009 36CE154
    37. 37. Grand Coulee Dam, Washington – Outlet pipe gate valve chamber Fall 2009 37CE154
    38. 38. Control structure – Radial Gate Fall 2009 38CE154
    39. 39. Free Overfall Spillway • Control - Sharp crested - Broad crested - many other shapes and forms • Caution - Adequate ventilation under the nappe - Inadequate ventilation – vacuum – nappe drawdown – rapture – oscillation – erratic discharge Fall 2009 39CE154
    40. 40. Overflow Spillway • Uncontrolled Ogee Crest - Shaped to follow the lower nappe of a horizontal jet issuing from a sharp crested weir - At design head, the pressure remains atmospheric on the ogee crest - At lower head, pressure on the crest is positive, causing backwater effect to reduce the discharge - At higher head, the opposite happensFall 2009 40CE154
    41. 41. Overflow Spillway Fall 2009 41CE154
    42. 42. Overflow Spillway Geometry • Upstream Crest – earlier practice used 2 circular curves that produced a discontinuity at the sharp crested weir to cause flow separation, rapid development of boundary layer, more air entrainment, and higher side walls - new design – see US Corps of Engineers’ Hydraulic Design Criteria III-2/1 Fall 2009 42CE154
    43. 43. Overflow Spillway overcrestheadenergydesign crestoverheadenergytotal spillwayofwidtheffectiveL esubmergencdownstreamPfC CLQ H H H H H o e o e e = = = = = ),,,( 2/3 θ Fall 2009 43CE154
    44. 44. Overflow Spillway • Effective width of spillway defined below, where L = effective width of crest L’ = net width of crest N = number of piers Kp = pier contraction coefficient, p. 368 Ka = abutment contraction coefficient, pp. 368-369 HKKL eap NL )(2 ' +−= Fall 2009 44CE154
    45. 45. Overflow Spillway • Discharge coefficient C C = f( P, He/Ho, θ, downstream submergence) • Why is C increasing with He/Ho? He>Ho → pcrest<patmospheric → C>Co • Designing using Ho=0.75He will increase C by 4% and reduce crest length by 4% Fall 2009 45CE154
    46. 46. Overflow Spillway • Why is C increasing with P? - P=0, broad crested weir, C=3.087 - P increasing, approach flow velocity decreases, and flow starts to contract toward the crest, C increasing - P increasing still, C attains asymptotically a maximum Fall 2009 46CE154
    47. 47. C vs. P/Ho Fall 2009 47CE154
    48. 48. C vs. He/Ho Fall 2009 48CE154
    49. 49. C. vs. θ Fall 2009 49CE154
    50. 50. Downstream Apron Effect on C Fall 2009 50CE154
    51. 51. Tailwater Effect on C Fall 2009 51CE154
    52. 52. Overflow Spillway Example • Ho = 16’ • P = 5’ • Design an overflow spillway that’s not impacted by downstream apron • To have no effect from the d/s apron, (hd+d)/Ho = 1.7 from Figure 9-27 hd+d = 1.7×16 = 27.2’ P/Ho = 5/16 = 0.31 Co = 3.69 from Figure 9-23 Fall 2009 52CE154
    53. 53. Example (cont’d) • q = 3.69×163/2 = 236 cfs/ft • hd = velocity head on the apron • hd+d = d+(236/d)2 /2g = 27.2 d = 6.5 ft hd = 20.7 ft • Allowing 10% reduction in Co, hd+d/He = 1.2 hd+d = 1.2×16 = 19.2 Saving in excavation = 27.2 – 19.2 = 8 ft Economic considerations for apronFall 2009 53CE154
    54. 54. Energy Dissipators • Hydraulic Jump type – induce a hydraulic jump at the end of spillway to dissipate energy • Bureau of Reclamation did extensive experimental studies to determine structure size and arrangements – empirical charts and data as design basis Fall 2009 54CE154
    55. 55. Hydraulic Jump energy dissipator • Froude number Fr = V/(gy)1/2 • Fr > 1 – supercritical flow Fr < 1 – subcritical flow • Transition from supercritical to subcritical on a mild slope – hydraulic jumpFall 2009 55CE154
    56. 56. Hydraulic Jump Fall 2009 56CE154
    57. 57. Hydraulic Jump yy11 VV11 VV22 yy22 LLjj Fall 2009 57CE154
    58. 58. Hydraulic Jump • Jump in horizontal rectangular channel y2/y1 = ½ ((1+8Fr1 2 )1/2 -1) - see figure y1/y2 = ½ ((1+8Fr2 2 )1/2 -1) • Loss of energy ∆E = E1 – E2 = (y2 – y1)3 / (4y1y2) • Length of jump Lj ≅ 6y2 Fall 2009 58CE154
    59. 59. Hydraulic Jump • Design guidelines - Provide a basin to contain the jump - Stabilize the jump in the basin: tailwater control - Minimize the length of the basin • to increase performance of the basin - Add chute blocks, baffle piers and end sills to increase energy loss – Bureau of Reclamation types of stilling basin Fall 2009 59CE154
    60. 60. Type IV Stilling Basin – 2.5<Fr<4.5 Fall 2009 60CE154
    61. 61. Stilling Basin – 2.5<Fr<4.5 Fall 2009 61CE154
    62. 62. Stilling Basin – 2.5<Fr<4.5 Fall 2009 62CE154
    63. 63. Type IV Stilling Basin – 2.5<Fr<4.5 • Energy loss in this Froude number range is less than 50% • To increase energy loss and shorten the basin length, an alternative design may be used to drop the basin level and increase tailwater depth Fall 2009 63CE154
    64. 64. Stilling Basin – Fr>4.5 • When Fr > 4.5, but V < 60 ft/sec, use Type III basin • Type III – chute blocks, baffle blocks and end sill • Reason for requiring V<60 fps – to avoid cavitation damage to the concrete surface and limit impact force to the blocks Fall 2009 64CE154
    65. 65. Type III Stilling Basin – Fr>4.5 Fall 2009 65CE154
    66. 66. Type III Stilling Basin – Fr>4.5 Fall 2009 66CE154
    67. 67. Type III Stilling Basin – Fr>4.5 • Calculate impact force on baffle blocks: F = 2 γ A (d1 + hv1) where F = force in lbs γ = unit weight of water in lb/ft3 A = area of upstream face of blocks in ft2 (d1+hv1) = specific energy of flow entering the basin in ft. Fall 2009 67CE154
    68. 68. Type II Stilling Basin – Fr>4.5 • When Fr > 4.5 and V > 60 ft/sec, use Type II stilling basin • Because baffle blocks are not used, maintain a tailwater depth 5% higher than required as safety factor to stabilize the jump Fall 2009 68CE154
    69. 69. Type II Stilling Basin – Fr>4.5 Fall 2009 69CE154
    70. 70. Type II Stilling Basin – Fr>4.5 Fall 2009 70CE154
    71. 71. Example • A rectangular concrete channel 20 ft wide, on a 2.5% slope, is discharging 400 cfs into a stilling basin. The basin, also 20 ft wide, has a water depth of 8 ft determined from the downstream channel condition. Design the stilling basin (determine width and type of structure). Fall 2009 CE154 71
    72. 72. Example 1. Use Manning’s equation to determine the normal flow condition in the upstream channel. V = 1.486R2/3 S1/2 /n Q = 1.486 R2/3 S1/2 A/n A = 20y R = A/P = 20y/(2y+20) = 10y/(y+10) Q= 400 = 1.486(10y/(y+10))2/3 S1/2 20y/n Fall 2009 CE154 72
    73. 73. Example • Solve the equation by trial and error y = 1.11 ft check ⇒ A=22.2 ft2, P=22.2, R=1.0 1.486R2/3S1/2/n = 18.07 V=Q/A = 400/22.2 = 18.02 • Fr1 = V/(gy)1/2 = 3.01 ⇒ a type IV basin may be appropriate, but first let’s check the tailwater level Fall 2009 CE154 73
    74. 74. Example 2. For a simple hydraulic jump basin, y2/y1 = ½ ((1+8Fr1 2 )1/2 -1) Now that y1=1.11, Fr1=3.01 ⇒ y2 = 4.2 ft This is the required water depth to cause the jump to occur. We have a depth of 8 ft now, much higher than the required depth. This will push the jump to the upstream 3. A simple basin with an end sill may work well.Fall 2009 CE154 74
    75. 75. Example • Length of basin Use chart on Slide #62, for Fr1 = 3.0, L/y2 = 5.25 L = 42 ft. • Height of end sill Use design on Slide #60, Height = 1.25Y1 = 1.4 ft • Transition to the tailwater depth or optimization of basin depth needs to be worked outFall 2009 CE154 75

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