1. Reservoirs, Spillways, & Energy
Dissipators
CE154 – Hydraulic Design
Lecture 3
Fall 2009 1CE154

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. Fall 2009 3
Hoover Dam – downstream face
CE154

4. Fall 2009 4
Hoover Dam – Lake Mead
CE154

5. Fall 2009 5
Hoover Dam – Spillway Crest
CE154

6. Fall 2009 6
Hoover dam – Outflow Channel
CE154

7. Fall 2009 7
Hoover Dam – Outlet Tunnel
CE154

8. Fall 2009 8
Hoover Dam – Spillway
CE154

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. 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. 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. 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. 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. Reservoir Capacity Curve
Fall 2009 14CE154

15. Spillway Discharge Rating
Fall 2009 15CE154

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. Spillway Capacity vs. Surcharge
Fall 2009 17CE154

18. Spillway Cost Analysis
Fall 2009 18CE154

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. 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. 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. Sabo Dam, Japan – Drop Chute
Fall 2009 22CE154

23. New Cronton Dam NY – Stepped Chute
Spillway
Fall 2009 23CE154

24. Sippel Weir, Australia – Drop Spillway
Fall 2009 24CE154

25. Four Mile Dam, Australia – Ogee
Spillway
Fall 2009 25CE154

26. Upper South Dam, Australia – Ogee
Spillway
Fall 2009 26CE154

27. Winnipeg Floodway - Ogee
Fall 2009 27CE154

28. Hoover Dam – Gated Side Channel
Spillway
Fall 2009 28CE154

29. Valentine Mill Dam - Labyrinth
Fall 2009 29CE154

30. Ute Dam – Labyrinth Spillway
Fall 2009 30CE154

31. Matthews Canyon Dam - Chute
Fall 2009 31CE154

32. Itaipu Dam, Uruguay – Chute Spillway
Fall 2009 32CE154

33. Itaipu Dam – flip bucket
Fall 2009 33CE154

34. Pleasant Hill Lake – Drop Inlet (Morning
Glory) Spillway
Fall 2009 34CE154

35. Monticello Dam – Morning Glory
Fall 2009 35CE154

36. Monticello Dam – Outlet - bikers heaven
Fall 2009 36CE154

37. Grand Coulee Dam, Washington – Outlet
pipe gate valve chamber
Fall 2009 37CE154

38. Control structure – Radial Gate
Fall 2009 38CE154

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. 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. Overflow Spillway
Fall 2009 41CE154

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. Overflow Spillway
overcrestheadenergydesign
crestoverheadenergytotal
spillwayofwidtheffectiveL
esubmergencdownstreamPfC
CLQ
H
H
H
H
H
o
e
o
e
e
=
=
=
=
=
),,,(
2/3
θ
Fall 2009 43CE154

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. 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. 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. C vs. P/Ho
Fall 2009 47CE154

48. C vs. He/Ho
Fall 2009 48CE154

49. C. vs. θ
Fall 2009 49CE154

50. Downstream Apron Effect on C
Fall 2009 50CE154

51. Tailwater Effect on C
Fall 2009 51CE154

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. 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. 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. 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. Hydraulic Jump
Fall 2009 56CE154

57. Hydraulic Jump
yy11 VV11
VV22 yy22
LLjj
Fall 2009 57CE154

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. 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. Type IV Stilling Basin – 2.5<Fr<4.5
Fall 2009 60CE154

61. Stilling Basin – 2.5<Fr<4.5
Fall 2009 61CE154

62. Stilling Basin – 2.5<Fr<4.5
Fall 2009 62CE154

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. 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. Type III Stilling Basin – Fr>4.5
Fall 2009 65CE154

66. Type III Stilling Basin – Fr>4.5
Fall 2009 66CE154

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. 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. Type II Stilling Basin – Fr>4.5
Fall 2009 69CE154

70. Type II Stilling Basin – Fr>4.5
Fall 2009 70CE154

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. 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. 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. 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. 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