Reservoirs, Spillways, & Energy Dissipators CE154 – Hydraulic Design Lecture 3 Fall 2009 CE154

Lecture 3 – Reservoir, Spillway, Etc. Purposes of a Dam - Irrigation - Flood control - Water supply - Hydropower - Navigation - Recreation Pertinent structures – dam, spillway, intake, outlet, powerhouse Fall 2009 CE154

Hoover Dam – downstream face Fall 2009 CE154

Hoover Dam – Lake Mead Fall 2009 CE154

Hoover Dam – Spillway Crest Fall 2009 CE154

Hoover dam – Outflow Channel Fall 2009 CE154

Hoover Dam – Outlet Tunnel 60 ft diameter outlet tunnel Fall 2009 CE154

Hoover Dam – Spillway Fall 2009 CE154

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 Fall 2009 CE154

Necessary Data Location and site map Hydrologic data Climatic data Geological data Water demand data Dam site data (foundation, material, tailwater) Fall 2009 CE154

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 CE154

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 CE154

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 CE154

Reservoir Capacity Curve Fall 2009 CE154

Spillway Discharge Rating Fall 2009 CE154

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 CE154

Spillway Capacity vs. Surcharge Fall 2009 CE154

Spillway Cost Analysis Fall 2009 CE154

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 CE154

Spillway Design Procedure (cont’d) Perform hydraulic design of spillway structures - Control structure - Discharge channel - Terminal structure - Entrance and outlet channels Fall 2009 CE154

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 CE154

Sabo Dam, Japan – Drop Chute Fall 2009 CE154

New Cronton Dam NY – Stepped Chute Spillway Fall 2009 CE154

Sippel Weir, Australia – Drop Spillway Fall 2009 CE154

Four Mile Dam, Australia – Ogee Spillway Fall 2009 CE154

Upper South Dam, Australia – Ogee Spillway Fall 2009 CE154

Winnipeg Floodway - Ogee Fall 2009 CE154

Hoover Dam – Gated Side Channel Spillway Fall 2009 CE154

Valentine Mill Dam - Labyrinth Fall 2009 CE154

Ute Dam – Labyrinth Spillway Fall 2009 CE154

Matthews Canyon Dam - Chute Fall 2009 CE154

Itaipu Dam, Uruguay – Chute Spillway Fall 2009 CE154

Itaipu Dam – flip bucket Fall 2009 CE154

Pleasant Hill Lake – Drop Inlet (Morning Glory) Spillway Fall 2009 CE154

Monticello Dam – Morning Glory Fall 2009 CE154

Monticello Dam – Outlet - bikers heaven Fall 2009 CE154

Grand Coulee Dam, Washington – Outlet pipe gate valve chamber Fall 2009 CE154

Control structure – Radial Gate Fall 2009 CE154

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 CE154

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 happens Fall 2009 CE154

Overflow Spillway Fall 2009 CE154

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 CE154

Overflow Spillway Fall 2009 CE154

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 Fall 2009 CE154

Overflow Spillway Discharge coefficient C C = f( P, He/Ho, , 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 CE154

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 CE154

C vs. P/Ho Fall 2009 CE154

C vs. He/Ho Fall 2009 CE154

C. vs.  Fall 2009 CE154

Downstream Apron Effect on C Fall 2009 CE154

Tailwater Effect on C Fall 2009 CE154

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 CE154

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 apron elevation! Fall 2009 CE154

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 CE154

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 jump Fall 2009 CE154

Hydraulic Jump Fall 2009 CE154

Hydraulic Jump V2 y2 y1 V1 Lj Fall 2009 CE154

Hydraulic Jump Jump in horizontal rectangular channel y2/y1 = ½ ((1+8Fr12)1/2 -1) - see figure y1/y2 = ½ ((1+8Fr22)1/2 -1) Loss of energy E = E1 – E2 = (y2 – y1)3 / (4y1y2) Length of jump Lj  6y2 Fall 2009 CE154

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 CE154

Type IV Stilling Basin – 2.5<Fr<4.5 Fall 2009 CE154

Stilling Basin – 2.5<Fr<4.5 Fall 2009 CE154

Stilling Basin – 2.5<Fr<4.5 Fall 2009 CE154

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 CE154

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 CE154

Type III Stilling Basin – Fr>4.5 Fall 2009 CE154

Type III Stilling Basin – Fr>4.5 Fall 2009 CE154

Type III Stilling Basin – Fr>4.5 Calculate impact force on baffle blocks: F = 2  A (d1 + hv1) where F = force in lbs  = 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 CE154

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 CE154

Type II Stilling Basin – Fr>4.5 Fall 2009 CE154

Type II Stilling Basin – Fr>4.5 Fall 2009 CE154

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

Example Use Manning’s equation to determine the normal flow condition in the upstream channel. V = 1.486R2/3S1/2/n Q = 1.486 R2/3S1/2A/n A = 20y R = A/P = 20y/(2y+20) = 10y/(y+10) Q = 400 = 1.486(10y/(y+10))2/3S1/220y/n Fall 2009 CE154

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

Example For a simple hydraulic jump basin, y2/y1 = ½ ((1+8Fr12)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 A simple basin with an end sill may work well. Fall 2009 CE154

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 out Fall 2009 CE154