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1.design of dam [mfsa]
- 4. 27.Eβ...............................................................................................................14 28...........................................................................................15 29.ΫΆβ...................................................................................................................15 31.Ϋ²β...........................................................................................................................15 31.ΫΆβ.....................................................................................................................15 32.Ϋ²β................................................................................................................................15 33...............................................................................................................................16 34..............................................................................................18 35.β..............................................................................19 36.β.........................................................19 37.β............................................................................................19 38.β............................................................................................................21 39.β.........................................................................................................21 41.β............................................................................21 41.β....................................21 42.β..........................................22 43.β..............................................................................................22 44...............................................................................................23 45................................................................................................................26 46....................................................................................26 47................................................................................27 48.( Minimum surface ( Min. S)..............................................................................27 49.(free Board (HfB)...........................................................................................................27 51........................................................................................................................................27 51.( Reservoir General Capacity ( R.G.C ).......................................................28 52.Dead capacity (D.C).........................................................................................................28 53.Live capacity (Liv. C)............................................................................................28 54.Surplus Capacity (S.C)......................................................................................28
- 5. 55.Inactive Capacity........................................................................................28 56.Active Capacity..................................................................................................................28 57..................................................................................................................................31 58................................................................................................................................31 59............................................................................................31 61..........................................................................................................................31 61........................................................................31 62............................................................................................................................32 63......................................................................................................32 64...........................................................................................33 65.( Corrosion ).......................................................................33 66................................................................................................34 67.................................................................................................35 68.................................................35 69.....................................................................................36 71.............................................................................................37 71................................................................................................................................................37 72.ο§...............................................................................................................................37 73.ο§...............................................................................................................37 74.ο§.....................................................................................................................37 75.ο§...............................................................................................................37 76.ο§..................................................................................................38 77.ο§......................................................................................................................38 78........................................................................................................................................41 79.......................................................................41 81....................................................................................................................................41 81..............................................................................................................41 82.........................................................................................................43
- 6. 83................................................................................................43 84. .......................................................................43 85..........................................................................44 86.............................................................44 87..........................................................................................................46 88............................................................................................46 89..........................................................................................................47 91.................................................................................................................................47 91...........................................................................47 92........................................................49 93............................................................................49 94............................................................................................51 95..........................................................................................................................52 96.........................................................................................................................................54 97...................................................................................54 98......................................................................................................................54 99..........................................................................................................................................54 111...............................................................................................................................55 111..............................................................................................................56 112...........................................................................................................................................56 113.58 114...............................................................................................................................59 115......................................................................................................................59 116............................................................................................................................63 117...........................................................................................................................................63 118.Hydraulic Calculation for spillways..................................................................................................74 119.Hydraulic calculation for top and bottom of spillway.....................................................................76 111.Shape of ogee β crest......................................................................................................................78
- 7. 111.Design of Barge................................................................................................................................89 112.Slab Design.......................................................................................................................................89 113.Force acting on a Dam:..................................................................................................................112 114.Water pressure:.............................................................................................................................112 115.Uplift pressure:..............................................................................................................................113 116.Wave pressure:..............................................................................................................................114 117.Ice pressure:..................................................................................................................................115 118.Earth quake pressure:....................................................................................................................115 119.Checking........................................................................................................................................111 121.Case I. Reservoir full case:.............................................................................................................111 121.(a) Normal Load Combinations......................................................................................................111 122.Case II. Reservoir empty case :......................................................................................................112 123.(2} By crushing..............................................................................................................................112 124.Case 1. Reservoir Empty:...............................................................................................................114 125.Case 2 when Reservoir is full:........................................................................................................116 126.Case 2 (b) Reservoir full, without up lift........................................................................................118 127........................................................................................................................121 128.........................................................................................................................................121 129.................................................................................................................................122 131................................................................................122 131.............................................................................................................................123 132.............................................................................................................127 133...................................................................................................................................128 134....................................................................................................................................131 135...........................................................................................................131 136.................................................132 137............................................................................................133 138..................................................................................................134 139..........................................................................................136 141................................................................................................................137 141......................................................................................................................138 142...............................................................................................................................138
- 8. 143.Design of Kaplan turbine...............................................................................................................139 144.Section head:.................................................................................................................................141 145.Generator Design:.........................................................................................................................143 146.Design at civil structure:................................................................................................................144
- 10. ΫΆ Ϋ² ΫΉ
- 14. 3
- 15. 4 (Water Economy) (Water Resource Engineering)
- 16. 5 (Water Supply)
- 18. 7 (Hydraulic structures Engineering) (Hydraulic structures) ΫΉΫ΄ΫΉMWΫ±Ϋ± ΫΆΫΉΫΉMWΫΉΫΉ ΫΆΫΉΫΉMWΫΆΫΉΫΉ Ϋ΄ΫΉMWΫ²Ϋ²
- 19. 8 ΫΆΫ΅MWΫΆΫΆ Ϋ΄ MW ΫΆΫΉ Ϋ΅ MWΫ² Ϋ΅ MWΫ² Ϋ΅ MWΫ² ΫΉ MWΫ² Ϋ΄ MWΫΉ Ϋ± (Micro Hydropower)
- 20. 9 β (Locks)
- 21. 10 βSingle purpose (Multipurpose Hydraulic structures)
- 23. 12 Small DamsΫΆΫ΄ Medium DamsΫΆΫ΄Ϋ΄ΫΉ High DamsΫ΄ΫΉ A β Earth fill dams Rock fill dams Earth Rock fill dams
- 24. 13 B β ( Concrete dams) Gravity dams Ϋ²Ϋ΄ΫΉΫΉ (Buttress Dams) Arch dams Arch dams
- 25. 14 C β ( steel dams)ΫΆΫ³Ϋ΄ΫΉ Direct strutted type Cantilever type Dams D β Timber E β
- 26. 15 Ϋ±β ( Non-overflow Dams) Ϋ²β ( Overflow Dams) Ϋ±β ( Rigid ) Ϋ²β
- 27. 16 ο§ ο§ ο§
- 28. 17 ΫΉΫΉ
- 29. 18 β
- 30. 19 β β β
- 32. 21 β β β
- 33. 22 β β
- 38. 27 Maximum water surface (Max. W.S or flood surface ( FI.S) ( Minimum surface ( Min. S) ( Minimum surface) (free Board (HfB) parapet ΫΆΫΆΫΉΫΆΫ΄ΫΉ
- 39. 28 ( Dead Surface D.S) ( Reservoir General Capacity ( R.G.C ) Dead capacity (D.C) Live capacity (Liv. C) Surplus Capacity (S.C) ( N.W.S.el) ( Max. W.S.el) Inactive Capacity Active Capacity
- 40. 29
- 41. 30
- 43. 32 (20x20x20) cm Ϋ²Ϋ΅ΫΆΫ΅ΫΉ ΫΆΫ΄ ΫΉΫΉΫ²Ϋ΅ ο§M500, M400, M350, M300, M250, M200, M150, M100 ο§P35, P31, P27, P23, P20, P18, P15, P11 ( Water Cement ratio = 0.5-0.55 )
- 44. 33 B8,B6,B4,B2 ( Samples) 5- Ax109- Ax10 300,200,150,100,50400( Samples ) ( 25% ) ( Corrosion ) 2Ca (OH)
- 45. 34 10 m/sec ( Abrasion) Cavitation 12 m/sec 400,350500 40 ( Water Cement ratio )0.400.45
- 46. 35 Ϋ΅ΫΉ Ϋ²Ϋ΅ ΫΆβ ( 210 k Jul/Kg)(252 k. Jul/Kg) Ϋ²β3 1m ΫΉβ
- 47. 36 Ϋ΅β 0.05 mm/m 0.07 mm/m Ϋ²Ϋ΅0.3 mm/ mΫΆΫ΅ΫΉ0.7 mm/ m Ϋ²Ϋ΅0.1 mm/mΫΆΫ΅ΫΉ0.3 mm/m Ϋ² Ϋ²Ϋ΅ Ϋ΄Ϋ΅ ΫΆ Ϋ΅ΫΆΫ²
- 48. 37 ο§ ο§400,300500 (20-15)% ΫΆΫ² ο§( Hydro phobia cement )400 β 300 ο§( Pozzolana Portland cement )
- 49. 38 ( 70 β 60 ) % ( 40 β 25 ) % ο§( Sulphate resisting cement ) 400 β 300 ο§Ϋ²ΫΉΫΉΫΉΫΉΫΉ( Slag Portland cement) (60 β 30 )% -0.150.6 -0.60.48
- 53. 42 )f( T )f(T)1(P 1 2 = 0.5 * Β₯ * H1P P1β = G*ffTβ Fβ )1>Pf( T (P1) ΫΆΫ³Ϋ±ΫΉΫΆΫ³Ϋ±Ϋ΅
- 54. 43 ο·By head ( height) ο·By type of foundation materials ο·By forms of cross profile ο· 1ΫΆΫ΄ 2ΫΆΫ΄Ϋ΄ΫΉ 3Ϋ΄ΫΉ A
- 55. 44 B Ϋ±ΫΉ
- 57. 46 ( Penitence )ΫΆΫ±Ϋ³ΫΆ ( Grobo )ΫΆΫ΅ΫΉΫ΅
- 58. 47 A B A( Dead load) B( Live load )
- 59. 48 ο· ο·
- 60. 49
- 61. 50 )nNβ( Normative force )n( N)n( N(n) )c( N n*nN=cN ( n = 1.0 β 1.2 ) (Nc)Ϋ²ΫΉ ( Nn)( n = 0.9 β 0.8 ) RR Nc R β€R m/SFc* Ncn nc=c( n 1.0 )=0.9)c(n mβ m1.0
- 62. 51 SFβ Grid SF=1.25SF=1.20 SF=1.15SF=1.10 SF allβ€ Οcal; Οallβ€ Ξ/cal; Ξ/allβ€ Β£calΒ£ Ξ/ Β£Ο
- 65. Dam
- 67. 55 Ϋ±Ϋ²Ϋ³Ϋ· ΫΉΫ·Ϋ· Ϋ²Ϋ± ΫΉΫΆ π π πππ = 28 β 103 ππ ππ2β (10β15m)
- 68. 56 (6-7)
- 70. 58
- 71. 59 1 2 3 4 5
- 73. 61
- 74. 62
- 75. 63 IIIIII IVV VI π΄ = ππππ₯ β ππππ π΄ = 331 β 10.8 = 320.2 π3 π ππ C=10-20 VII βQ = π΄ πΆ = 320.2 15 = 21.35 π3 π ππ βΉ βπ = 21.35 π3 π ππ βQ331 β 21.35 = 309.65IXV XIX V X 25 β 100% 1 β π π₯ = 100% 25 = 4%
- 80. 68 P Qave πππ£π = β ππ π=1 π = 4708.2 25 = 188.33 π3 π KiQi/Qave Ki-1 ,(Ki-1)2,m-0.3,n+0.4 mn P π = π β 0.3 π + 0.4 β 100 Cs Timing P=(m-0.3)/(n+0.4)*100 1 1959 0 1972 331 188.33 1.76 0.76 0.57 0.7 25.4 2.76 2 1960 168 1973 306 188.33 1.62 0.62 0.39 1.7 25.4 6.69 3 1961 179 1969 290 188.33 1.54 0.54 0.29 2.7 25.4 10.63 4 1962 129 1965 272 188.33 1.44 0.44 0.20 3.7 25.4 14.57 5 1963 216 1968 258 188.33 1.37 0.37 0.14 4.7 25.4 18.50 6 1964 175 1967 246 188.33 1.31 0.31 0.09 5.7 25.4 22.44 7 1965 272 1966 223 188.33 1.18 0.18 0.03 6.7 25.4 26.38 8 1966 223 1963 216 188.33 1.15 0.15 0.02 7.7 25.4 30.31 9 1967 246 1975 213 188.33 1.13 0.13 0.02 8.7 25.4 34.25 10 1968 258 1978 190 188.33 1.01 0.01 0.00 9.7 25.4 38.19 11 1969 290 2007 189 188.33 1.00 0.00 0.00 10.7 25.4 42.13 12 1970 137 1961 179 188.33 0.95 -0.05 0.00 11.7 25.4 46.06 13 1971 128 1964 175 188.33 0.93 -0.07 0.01 12.7 25.4 50.00 14 1972 331 1974 174 188.33 0.92 -0.08 0.01 13.7 25.4 53.94 15 1973 306 2009 174 188.33 0.92 -0.08 0.01 14.7 25.4 57.87 16 1974 174 1977 170 188.33 0.90 -0.10 0.01 15.7 25.4 61.81 17 1975 213 1960 168 188.33 0.89 -0.11 0.01 16.7 25.4 65.75 18 1976 164 1976 164 188.33 0.87 -0.13 0.02 17.7 25.4 69.69 19 1977 170 2010 139 188.33 0.74 -0.26 0.07 18.7 25.4 73.62 20 1978 190 1970 137 188.33 0.73 -0.27 0.07 19.7 25.4 77.56 21 1979 10.8 2008 136 188.33 0.72 -0.28 0.08 20.7 25.4 81.50 22 1980 0 1962 129 188.33 0.68 -0.32 0.10 21.7 25.4 85.43 23 2007 189 1971 128 188.33 0.68 -0.32 0.10 22.7 25.4 89.37 24 2008 136 2011 90.4 188.33 0.48 -0.52 0.27 23.7 25.4 93.31 25 2009 174 1979 10.8 188.33 0.06 -0.94 0.89 24.7 25.4 97.24 26 2010 139 1959 0 188.33 0.00 -1.00 1.00 25.7 25.4 101.18 27 2011 90.4 1980 0 188.33 0.00 -1.00 1.00 26.7 25.4 105.12 28 2012 0 2012 0 188.33 0.00 -1.00 1.00 27.7 25.4 109.06 YearsNO n+0.4m-0.3(KI-1)2 Ki-1 Ki=Qi/Qa v Q av.Flow sortYearsFolw
- 81. 69 πΆπ£ = β β (πΎπβ1)2π=π π=1 πβ1 = β 6.39 25β1 = 0.516 Cs Cs Cs Cs=2Cv Cs=(2-2.5)Cv Cs=(3-4)Cv Cs=(3-4)Cv Cs=4Cv Cs=4Cv Cs upCs downCs Cs Cs upΓp upCs downΓP downΓP KpQp%
- 82. 70 Qp% Real Number Cs up Cs Cs Down Γp up Γp Γp Down 0.01 2 2.064 2.1 8.21 8.21 0.516 5.23631 188.33 986.155 0.1 2 2.064 2.1 5.91 5.9932 6.04 0.516 4.09244 188.33 770.729 0.5 2 2.064 2.1 4.3 4.3384 4.36 0.516 3.23858 188.33 609.922 1 2 2.064 2.1 3.6 3.632 3.65 0.516 2.87408 188.33 541.276 3 2 2.064 2.1 2.51 2.5228 2.53 0.516 2.30175 188.33 433.488 5 2 2.064 2.1 2 2.0064 2.01 0.516 2.03529 188.33 383.306 10 2 2.064 2.1 1.3 1.2936 1.29 0.516 1.66749 188.33 314.039 20 2 2.064 2.1 0.61 0.5972 0.59 0.516 1.30815 188.33 246.365 25 2 2.064 2.1 0.39 0.3772 0.37 0.516 1.19464 188.33 224.986 30 2 2.064 2.1 0.2 0.1872 0.18 0.516 1.09660 188.33 206.522 40 2 2.064 2.1 -0.08 -0.093 -0.1 0.516 0.95212 188.33 179.312 50 2 2.064 2.1 -0.31 -0.316 -0.32 0.516 0.83674 188.33 157.583 60 2 2.064 2.1 -0.49 -0.496 -0.5 0.516 0.74386 188.33 140.091 70 2 2.064 2.1 -0.64 -0.64 -0.64 0.516 0.66976 188.33 126.137 75 2 2.064 2.1 -0.7 -0.706 -0.71 0.516 0.63550 188.33 119.684 80 2 2.064 2.1 -0.78 -0.767 -0.76 0.516 0.60413 188.33 113.775 90 2 2.064 2.1 0.9 -0.23 -0.867 0.516 0.88108 188.33 165.934 95 2 2.064 2.1 -0.95 -0.927 -0.914 0.516 0.52169 188.33 98.250 97 2 2.064 2.1 -0.97 -0.944 -0.93 0.516 0.51269 188.33 96.555 99 2 2.064 2.1 -0.99 -0.961 -0.945 0.516 0.50402 188.33 94.922 99.9 2 2.064 2.1 -1 -0.969 -0.952 0.516 0.49985 188.33 94.137 Qp%=Qav*Kp Timing Real Numbers Cs coifecents ΓP Foster Coifecent Cv Kp=Γp*Cv+1 Qav
- 83. 71 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 0.01 0.1 0.5 1 3 5 10 20 25 30 40 50 60 70 75 80 90 95 97 99 99.9
- 84. 72 10.000 30.000 50.000 70.000 90.000 110.000 130.000 150.000 170.000 190.000 210.000 230.000 250.000 270.000 290.000 310.000 330.000 350.000 0.01 0.1 0.5 1 3 5 10 20 25 30 40 50 60 70 75 80 90 95 97 99 99.9 Cs up Cs Cs Down Γp up Γp Γp Down 0.01 2.1 2.120 2.2 8.21 0.530 5.35078 57.66 308.526 0.1 2.1 2.120 2.2 6.04 6.0597 6.14 0.530 4.21129 57.66 242.823 0.5 2.1 2.120 2.2 4.36 4.3718 4.42 0.530 3.31681 57.66 191.247 1 2.1 2.120 2.2 3.65 3.6559 3.68 0.530 2.93741 57.66 169.371 3 2.1 2.120 2.2 2.53 2.532 2.54 0.530 2.34179 57.66 135.027 5 2.1 2.120 2.2 2.01 2.012 2.02 0.530 2.06622 57.66 119.138 10 2.1 2.120 2.2 1.29 1.2861 1.27 0.530 1.68153 57.66 96.957 20 2.1 2.120 2.2 0.59 0.5861 0.57 0.530 1.31057 57.66 75.567 25 2.1 2.120 2.2 0.37 0.3661 0.35 0.530 1.19398 57.66 68.845 30 2.1 2.120 2.2 0.18 0.1761 0.16 0.530 1.09330 57.66 63.039 40 2.1 2.120 2.2 -0.1 -0.104 -0.12 0.530 0.94491 57.66 54.484 50 2.1 2.120 2.2 -0.32 -0.322 -0.33 0.530 0.82937 57.66 47.822 60 2.1 2.120 2.2 -0.5 -0.5 -0.5 0.530 0.73503 57.66 42.382 70 2.1 2.120 2.2 -0.64 -0.64 -0.64 0.530 0.66084 57.66 38.104 75 2.1 2.120 2.2 -0.71 -0.706 -0.69 0.530 0.62584 57.66 36.086 80 2.1 2.120 2.2 -0.76 -0.758 -0.75 0.530 0.59829 57.66 34.498 90 2.1 2.120 2.2 -0.867 -0.862 -0.842 0.530 0.54316 57.66 31.319 95 2.1 2.120 2.2 -0.914 -0.908 -0.882 0.530 0.51899 57.66 29.925 97 2.1 2.120 2.2 -0.93 -0.923 -0.895 0.530 0.51082 57.66 29.454 99 2.1 2.120 2.2 -0.945 -0.937 -0.905 0.530 0.50340 57.66 29.026 99.9 2.1 2.120 2.2 -0.952 -0.944 -0.909 0.530 0.50000 57.66 28.830 Qp%=Qav*Kp TimingReal Numbers Cs coifecents ΓP Foster Coifecent Cv Kp=Γ*Cv+1 Qav
- 85. 73
- 86. Dam
- 87. 74 Hydraulic Calculation for spillways Width of spillways: B = ππππ₯ πππ₯π»0π₯β2π π Cd = coefficient of discharge (C=0.6, 0.67) K = pier Construction coefficient (k=1) , g= gravity acceleration g=9.81m/sec2 H0 = Total head on crest (H0 = βπ» + ππ£02 2π ) βπ»= design head m a = Karialos coefficient (a = 1.05) V0 = water velocity over spillway crest (1-2 m/s) v0= 1.5m/sec βπ»= 2m H0 = 2+ 1.05π₯1.52 2π₯9.81 = 2.12π, Qmax= 360m3/sec B = 360 0.65π₯1π₯2.123/2 π₯β2π₯9.81 = 40.5 π ππ¦ 41π Now we can find width of every span by equation: B = n*I + (n-1) *bp (m) n= number of span (n=5) I= width of span bp= width of pier (2-3m) accept bp=2.5m 41= 5xI+(5-1) x2.5, I = 6.2 say 6.5m B = 5x6.5+ (5-1) x2.5 = 42.5
- 88. 75
- 89. 76 Hydraulic calculation for top and bottom of spillway Discharge per meter width of spillway crest: q = cv x h1 x β2π₯π( π»0 + π) β β1 m3/sec x m cv= coefficient of velocity (0.85-0.95) accept cv=0.9 P = height of spillway (p=29m) q = 9.81 h1 = first hydraulic jump, also q= Qmax/B => 360/42.5 = 8.47 m3/sec2 Now we can find h1 from equation: q = cv x h1 β2π₯π (π»0+ π) β β1 if h1=0.5m=>q=0.9x0.45β2π₯9.81π₯(2.12 + 29) β 0.5 =11.11m2/sec if h1=0.5m =>q = 0.9x0.45 β2π₯9.81(2.12 + 29) β 0.45 =10 m2/sec if h1 = 0.4 => q = 0.9x0.4β2π₯9.81(2.12 + 29) β 0.4 = 8.89 m2/sec if h1 = 0.39 => q = 0.9x0.39 β2π₯9.81(2.12 + 29) β 0.39 = 8.67 m2/sec if h1 = 0.3818 => q = 8.488 m2/sec accept h1 = 0.381 m Height of second hydraulic jumps: h2 = β1 2 β 1+8ππ2 πβ1 3 π = 0.381 2 β1 + 8 = 6.35π
- 90. 77 Length of stilling basin is finding from equation: L = 5(h2-h1) => 5(6.35-0.381) = 29.8 m
- 91. 78 Shape of ogee β crest 1. Based on latest research of US corps up streams curve is drawing by following formula: ο§ Y = 0.724(π₯+0.27π»0)1.8 π»0 0.5 + 0.126 π»0 β 0.4315π»0 0.375 (π₯ + 0.27π»0)0.652 ο§ BY empirical formulas r1=0.5H0, r2=0.2H0, a= 0.175H0, b= 0.282H0 2. For drawing down stream curve we use the following formulas: ο§ X1.85 = 2H0 0.85y For upstream curve: r1 =0.5x2.12= 1.06m, r2 = 0.2x2.12= 0.424, a = 0.175x2.12 = 0.371 b= 0.282x2.12 = 0.6 For downstream curve X1.85 = 2H0 0.85y => y = π1.85 2π»00.85 X = 1m, y = 11.85 2π₯120.85 = 0.264m X = 3m, y = 31.85 2π₯120.85 = 2.621m X = 5m, y = 51.85 2π₯120.85 = 5.184m X = 7m, y = 71.85 2π₯120.85 = 9.661m X = 9m, y = 91.85 2π₯120.85 = 15.38m X = 12m, y = 121.85 2π₯120.85 = 26.186m X = 12.7m, y = 12.71.85 2π₯120.85 = 29.08m
- 92. 79
- 93. 80 new formula empircal X2 =2HY H0 X Y1 Y2 y 3 2.12 0 -0.312566732 0 0 2.12 1 -1.624164718 -0.2639882 -0.23585 2.12 2 -3.515713249 -0.951678 -0.9434 2.12 3 -5.963986392 -2.0149248 -2.12264 2.12 4 -8.944382643 -3.4308008 -3.77358 2.12 5 -12.43668006 -5.1841676 -5.89623 2.12 6 -16.42422483 -7.2638077 -8.49057 2.12 7 -20.89302808 -9.6608624 -11.5566 2.12 8 -25.83111178 -12.368043 -15.0943 2.12 9 -31.22805079 -15.379179 -19.1038 2.12 10 -18.688934 -23.5849 2.12 11 -22.292614 -28.5377 2.12 12 -26.18604 -35 -30 -25 -20 -15 -10 -5 0 0 2 4 6 8 10 12 14 16 Chart Title Y1 Y2 y 3
- 94. 81 H0
- 95. 82 R ho 200 X Y X Y -52.8 -25.4 1 -16.4 -49.4 47 -29.4 -4.2 2 0 -106 106 0 0 3 0 -165 165 43.4 -5.8 4 -30.8 -277.8 282 116.6 -37.4 5 -176 -515 560 246 -146.8 6 -733.6 -1001.4 1300 368 -311.12 7 -1665.8 -1585.4 2400 551.6 -667.6 Center R
- 96. 83 XY
- 97. 84 Stability Calculation for Spillway Cross Section Gravity center of Spillway Section is calculation in Auto CAD easily. Tools, inquiry, Region/mass, Properties Area: 439.22m2 Perimeter: 108.21m Centroid: X: 9.23m Y: 10.85m Center of Gravity
- 98. 85 Forces acting on a Spillway: 1. Water pressure: a) = External water pressure per length of spillway p1 = π€π₯(βπ»2) 2 T/m w = Unit weight of water (w= 1000 kg/m3) = 1ton/m3 ΞH = Water head on spillway crest (ΞH = 2m) P2= π€π₯(βπ»2) 2 T/m = 1π₯(22) 2 = 2 ton/m b) Water pressure β sensitive on spillway vertical wall. P3= wxΞHxH = 1x2x24 = 48 T/m P4= π€π₯π»π₯π» 2 = 1π₯24π₯24 2 = 288 T/m 2. Total pressure on vertical wall:
- 99. 86 P = p2 +p3 = 48 + 288 = 336 T/m Position of resultant pressure: Depth of the point of the resultant pressure from base is calculating by taking moment of both the pressure about point A and equation the source> Pxh = (p2 x H/2) + (p3 x H/3) => 336xh = (48 x 24/2) + (288 x 24/3) 336h = 2880, h = 8.57m 3. Water pressure bellow the base of spillways or uplift pressure: P = π€π₯π»π₯π 2 T/m => Puplift = 1π₯24π₯27.48 2 = 329.76 ton/m
- 100. 87 4. Psoil = 1 2 8h2 1βπ πππ 1+π πππ t/m h = the height of the silt is depositor (h = 2) πΎ = the submerged unit of silt (1.2 t/m) Q = the angle of internal fraction (π = 15 π ) Psoil = 1 2 8h2 1βπ ππ15 1+π ππ15 = 28.62 t/m 5. Self-weight of spillway: The weight of the dam is major resisting force for analysis purpose, generally unit length of the dam is considered. G = A x r x 1m t/m, G = weight of spillways A = spillways cross section Area r = unit weight of spillway marital A = 439.22 m2 G = 439.22 x 2.7 x 1 = 1185.89 t/m G = G β Puplift = 1185.89 β 329.76 = 856.13 t/m Ptotal = P(p2+p3) + Psoil = 336+28.62 = 364.62 t/m Safety calculation for spillways: R = β(πΊ12) + (ππ‘ππ‘ππ)2 = β856.132 + 364.622 = 930.54 t/m TanΞ± = π π‘ππ‘ππ πΊ1 = 364.62 856.13 = 1.42589, Ξ± = 23.17o B β€ π΅ 6 b = ππ₯πβ² πΊ1 = 364.62 π₯ 10.85 856.13 = 4.62 say b = 4.6209
- 101. 88 π΅ 6 = 27.76 6 = 4.6267 b β€ π΅ 6 , b = 4.6209 < 4.6267 0k Resultant line calculation for Rotation taking moment at point C : Mc = p(p2+p3) x π» 2 + psoil x π» π πππ 3 + psoil x π΅ 3 Mc = 336 x 33 2 + 28.62 x 9 3 + 329.76 x 27.76 3 Mc = 8681.24 T*m Mcβ = G1 x Xβ = 856.13 x 9.23 = 7912.1 T*m ππ ππβ² = 8681 7902 = 1.1 β₯ 1.1 safe against overturning. Calculation for slipper or sliding: πΊ1 π π‘ππ‘ππ x n = 856.13 364.62 x 0.7 = 1.644 > 1.1 ok Safe against sliding n = righty coefficient of material
- 102. 89 Design of Barge Slab Design LL = 60 T/m2 L = 1.4 h = 25 cm self-weight of slab = 0.25 x 24 = 0.6 T/m2 self-weight of pcc = 0.2 x 2.9 = 0.44 T/m2 Total = 1.04 T/m2 Qu = 1.2 (1.04) + 1.6 (60) = 97.24 T/m2 Wu = 97.248 T/m per 1m strip Wu = 97.248 x 1.4 = 136.2 T/m, b = 1.4 -veMo = π€πΏ2 β24 = 136.2 π₯ 1.42 24 = 11.123 T x m = 109.1 KN.m -veMo = π€πΏ2 10 = 136.2 π₯ 1.42 10 = 261.7 Txm = 261.84 KN.m +veMo = π€πΏ2 14 = 136.2 π₯ 1.42 14 = 19.1 T x m = 187.31 KN xm Note: negative moment at face of all support for a slabs with spans not exceeding 10 ft moment is π€π2 12 From Table 14.1 ACI coefficients Design for 11.123 T x m = 109.1 KN x m Rn = ππ’ β ππ2 = 109.1 π₯ 106 0.9 π₯ 1000 π₯ 2202 = 2.505 mpa π = 0.85 π₯ πΉβ²π ππ¦ ( 1 - β1 β 2π π 0.85 π₯ πΉβ²π ) π = 0.85 π₯ 28 420 ( 1 - β1 β 2 π₯ 2.2505 0.85 π₯ 28 ) = 0.00632 As = πbd = 0.00632 x 1000 x 220 = 1390.4 m2 /sec Trial and error Method: β Mn = β Asfy (d-a/2)
- 103. 90 Mn = Asfy (d β a/2) Req As = ππ’ β ππ¦ (πβ π 2 ) Assume z = 0.9d First trial Z = 0.9 x (22) = 19.8 cm As = 109.1 π₯ 106 0.9 π₯ 420 π₯ = 1457.7 mm2 a = 1457.7 π₯ 420 0.85 π₯ 28 1000 = 25.724mm z = d - π 2 = 220 - 25.724 2 = 207.138 mm 2nd trial: Z = 207.138 mm As = 109.1 π₯ 106 0.9 π₯ 420 π₯ 207.138 = 1393.39 mm2 a = 1393.39 π₯ 420 0.85 π₯ 28 1000 = 24.589 mm z = 220 - 24.589 2 = 207.71 3rd trail: Z = 207.71 As = 109.1 π₯ 106 0.9 π₯ 420 π₯ 207.71 = 1389.55 mm2 closed Req As = 1390 mm2 Try 14β mm Ab = ππ2 4 = π π₯ 142 4 = 153.86 mm2 N = 1390 153.86 = 9.03 say 10 S = π΄π π ππ π΄π π₯ 1000
- 104. 91 As = 9 x 153.86 = 1538.6 S = 153.86 1538.6 π₯ 1000 = 1000mm β Mn = β Asfy ( d β a/2 ) a = π΄π ππ¦ 0.85 πΉβ² π π = 1538.6 π₯ 420 0.85 π₯ 28 1000 = 27.15 mm βt = πβπ π x (0.003) > 0.005 Tension control C = π π½1 = 27.15 0.85 = 31.94 βt = 220β31. 31.94 x (0.003) =0.0176 tension control β Mn = 0.9 x 1588.6 x 420 (220 - 27.15 2 ) β Mn = 120054880.9 = 120.1 KN .m > 109.1 KN*M ok Design for 26.7 T x m = 261.84 KN x m Rn = ππ’ β ππ2 = 261.84 π₯ 106 0.9 π₯ 1000 π₯ 2202 = 6.01 π = 0.85 π₯ πΉβ²π ππ¦ ( 1 - β1 β 2π π 0.85 π₯ πΉβ²π ) = 0.85 π₯ 28 ππ¦420 ( 1 - β1 β 2 π₯ 6.011 0.85 π₯ 28 ) = 0.0168 As = πbd = 0.0168 x 1000 x 220 = 3696 mm2 Trial and Error method: β Mn = β Asfy (d-a/2) Mn = β Asfy (d-a/2) Req As = ππ’ β ππ¦ (πβ π 2 ) Assume Z = 0.9 d a = 0.9 x 220 = 198 mm first trial:
- 105. 92 As = 201.84 π₯ 106 0.9 π₯ 420 π₯ 198 = 3498.48 mm2 a = 3498.48π₯ 420 0.85 π₯ 28 1000 = 61.74mm z = d - π 2 = 220 - 61.74 2 = 189.13 mm 2nd trail: As = 261.84 π₯ 106 0.9 π₯ 420 π₯189.13 = 3662.55 mm2 a = 3662.55π₯ 420 0.85 π₯ 28 1000 = 64.62 mm z = d - π 2 = 220 - 64.62 2 = 187.69 mm 3rd trail: As = 261.84 π₯ 106 0.9 π₯ 420 π₯ 187.69 = 3690.65 mm2 a = 3690.65 π₯ 420 0.85 π₯ 28 1000 = 65.13 mm z = d - π 2 = 220 - 65.13 2 = 187.435 mm 4TH trail: As = 261.84 π₯ 106 0.9 π₯ 420 π₯ 187.435 = 3695.67 mm2 a = 3695.67 π₯ 420 0.85 π₯ 28 1000 = 65.22mm z = d - π 2 = 220 - 65.22 2 = 187.39 mm 5th trial: As = 261.54 π₯ 106 0.9 π₯ 420 π₯ 187.39 = 3696.56 mm2 Req As = 3700mm2 Try β 32 mm Ab = ππ2 4 = π322 4 = 804.25 mm2
- 106. 93 S = π΄π π ππ π΄π π₯ 1000 = 804.25 3700 = 217.36 mm say 200 mm β Mn = β Asfy ( d β a/2 ) 200 1 1000 X X=5 As = 5 x 804.25 = 4021.25 mm2 a = 4021π₯ 420 0.85 π₯ 28 1000 = 70.96 mm βt = πβπ π x (0.003) > 0.005 Tension control C = π π½1 = 70.96 0.85 = 83.48 βt = 220β83.48. 83.48 x (0.003) = 0.005 tension control β Mn = 0.9 x 4021 x 420 (220 - 70.96 2 ) = 280458959.8 N.mm = 280.46 KN.m > 261.84 0k Design for positive moment: Mo = 19.7 Txm = 187.3 kn.m Rn = ππ’ β ππ2 = 187.3 π₯ 106 0.9 π₯ 1000 π₯ 2202 = 4.3 π = 0.85 π₯ πΉβ²π ππ¦ ( 1 - β1 β 2π π 0.85 π₯ πΉβ²π ) = 0.85π₯ 28 420 ( 1 - β1 β 2 π₯ 4.3 0.85 π₯ 28 ) = 0.0114 Trial and error method: β Mn = β Asfy (d-a/2) Mn = β Asfy (d-a/2) Req As = ππ’ β ππ¦ (πβ π 2 ) Assume
- 107. 94 Z = 0.9 d a = 0.9 x 220 = 198 mm first trial: As = 187.3 π₯ 106 0.9 π₯ 420 π₯ 198 = 2502.54 mm2 a = 2502.54π₯ 420 0.85 π₯ 28 1000 = 44.162mm z = d - π 2 = 220 - 44.162 2 = 197.919 mm 2nd trail: As = 187.3 π₯ 106 0.9 π₯ 420 π₯197.919 = 2503.56 mm2 Req As = 2504 mm2 Try 250 mm Ab = ππ2 4 = π252 4 = 490.87 mm2 S = π΄π π ππ π΄π π₯ 1000 = 490.87 2504 π₯ 1000 = 196.03 say 200mm N = 5 from table B4 As = 2550 mm2 a = 2550 π₯ 420 0.85 π₯ 28 1000 = 45 mm C = 45/0.85 = 52.94 βt = 220β52.94. 52.94 x (0.003) = 0.0095 > 0.005 Tension control β Mn = 0.9 x 2550 x 420 (220 - 45 2 ) = 190.37 KN . m > 187.84 OK Shrinkage Reinforcement: Try 100mm, Ab = 78.54 Req As = πbd = 0.00181 x 1000 x 220 = 398.2 m S = π΄π π ππ π΄π π₯ 1000 = 78.54 398.2 π₯ 1000 = 197.2 say 190
- 108. 95 S = 165mm, As = 430 mm2 ok Design of Beam Mu = π€πΏ2 8 = 136.2 π₯ 8.52 8 = 1230 T*m = 12062.2 KN*m
- 109. 96 Fβc = 35, fy = 421 πmax is single reinforcement = 0.0216 from table B.9 As1 = 0.0216 x 700 x 1430 = 20020 mm2 ππ’1 β ππ2 = ? π = 0.85 π₯ πβ²π ππ¦ (1 - β1 β 2ππ 085πβ²π We want to find Rn 0.0216 = 0.85 π₯ 35 420 x (1 - β1 β 2π π/0.85) 0.0216 = 0.070833 x (1 - β1 β 2π π 29.75 ) 0.0216 = 0.070833 β0.070833 x (1 - β1 β 2π π 29.75 ) - 0.049233 = - 0.070833 x (1 - β1 β 2π π 29.75 ) 0.002424 = 0.005017 (1 - β1 β 2π π 29.75 ) 0.002424 = 0.005017 - 0.01π π 29.75 0.077742 = 0.01 Rn Rn = 7.7142 π = 0.85 π₯ 35 420 x (1 - β1 β 2π₯7.7142 0.85 π₯ 35 ) = 0.0217 R = 7.714 ππ’1 β ππ2 = 7.714 Mu1 = β ππ2 x 7.714 = 0.9 π₯ 700 π₯ 14302 π₯ 7.714 106 = 9937.84 KN.m Mn1 = 9937.84 0.9 = 11042.1 Kn.m
- 110. 97 Mn2 = Mn - Mn1 = 13402.44 β 11042.1 = 2360.34 Kn.m Mn = 12062.2 0.9 = 13402.49 Checking to see compression steel yields: a = 20020 π₯ 420 0.85 π₯ 35 π₯ 700 = 403.76 C = π π½1 = 403.76 0.814 = 496.025 from table B7 B= 0.81 , fβc = 35 Es = 496.025β70 496.025 x (0.003) = 0.00256 > 420 2000 πππ = 0.002 Compression steel yields Asβ Req = ππ2 ππ¦ (πβπβ²) = 2360.34 π₯ 106 420 (1430β70) = 4132.25 mm2 Use 7 # 29 mm bars (4515) spicin 100mm As = As1 + As2 = 20020 + 4132.25 = 24152.25 Use 10# 57 bar (25810) spicing 70 mm
- 111. 98 Checked As1 = As β As2 As = 25810 mm2 As2 = 4515 mm2 As1 = 25810 β 4515 = 21295 mm2 Fs = fy a = 21295 π₯ 420 0.85 π₯ 35 π₯ 700 = 429.479 mm c = 429.479 0.814 = 527.615 mm Es = = 527.615β70 527.615 x (0.003) = 0.0026 > 0.0021 Fy =fs, As2 = As Mn1 = As1fy (d β a/2) = 21295 x 420 (1430 - 429.48 2 ) = 10869.16 Kn.m Mn2 = As2fy (d-dβ) = 4515 x 421nx (1431 β 70) = 2578.92 Kn.m Mn = Mn1 + Mn2 = 10869.16 + 2578.97 = 13448.128 KN.m βt = πβπ π x 0.003 = 1430β527.615 527.615 x 0.003 = 0.00513 > 0.005 tension control β Mn = 0.9 x 13448.128 = 12103.31 > Mu = 12062.2 OK
- 112. 99 Design of Beam for shear VA = VB = ππ 2 = 136.2 π₯ 8.5 2 = 518.85 T = 5676.58 KN Fβc = 35 Fy = 420 d = 1430 mm = 1.43 m 5676.58 4.25 = ππ’ (4.25β1.43) Vu = 3766.6 Kn β vc = β βπΉβ² π π ππ€ π₯ π 6 = 0.75 β35 π₯ 700 π₯ 1430 6 = 740.249 KN Vs = ππ’β β ππ β = 3766.6β740.25 0.75 = 4035.133 KN Use 19β mm Av = 283.53 mm2 S = π΄π£ π₯ πΉπ¦π ππ = 283.53 π₯ 420 π₯ 1430 4035.133 = 42.2 mm
- 113. 100 Smax = 3π΄π£ π₯ πΉπ¦ ππ€ = 3 π₯ 283.53 π₯ 420 700 = 510 mm 1. If VS < 1/3 βπΉβ²πππ€π Smax = d/2 2. VS > 1/3 βπΉβ²πππ€π Smax = d/4 3. VS > 2/3 βπΉβ²πππ€π Not good 1/3 β35 π₯ 700 π₯ 1430 =1974 KN < Vs 2/3β35 x 700 x 1430 = 3948 KN < Vs If we change d to up 1460 then its ok but now we accepted this ok The d/4 = 1430/4 = 357.5 S = 42mm Av 283.53 Fy 420 Fc 35 Bw 700 d 1430 Vu 5676.58 L 4.25 oVc 740.249 Distance 1.43 1.5 1.6 1.7 1.8 1.9 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 Vu 3767 3673 3540 3406 3272 3139 3005 2738 2471 2204 1937 1670 1402 1135 868 Vs 4035 3910 3732 3554 3376 3198 3020 2664 2308 1951 1595 1239 883 527 171 S 42 44 46 48 50 53 56 64 74 87 107 137 193 323 998 Smax 510 510 510 510 510 510 510 510 510 510 510 510 510 510 510 1 con 1974 1974 1974 1974 1974 1974 1974 1974 1974 1974 1974 1974 1974 1974 1974 2 con 1974 1974 1974 1974 1974 1974 1974 1974 1974 1974 1974 1974 1974 1974 1974 3 con 3948 3948 3948 3948 3948 3948 3948 3948 3948 3948 3948 3948 3948 3948 3948 Spacing 42 44 46 48 50 53 56 64 74 87 107 137 193 323 715 Design for shear with deferent distance
- 114. 101 B = (1.25 Γ· 1.75) H B = (1.75 Γ· 2) H B = (2 Γ· 2.25) H B = (2 Γ· 2.5) H π΅ π»1 = m π΅ π»1 = 1 β πΎπ πΎπ€ βπ H = 32m πΎπ= (2.3 β 2.4) T/m3 πΎπ€ = 1.0 T/m3 π΅ π»1 = m => B = m x H1 = 0.7 x 32 = 22.4 π΅ π»1 = 1 β πΎπ πΎπ€ βπ = 32 β 2.4 1 β0.7 , B = 24.5 say 25m b = 7m, ΞH = 3m X = π» π₯ π π΅ = 32 π₯ 7 25 = 9m H5 = H4 + ΞH = 9 + 3 = 12m
- 115. 102 Force acting on a Dam: 1. Water pressure: 2. Uplift pressure 3. Pressure date earthquake force 4. Silt pressure 5. Wave pressure 6. Ice pressure 7. The stabilising force is the weight of the dam itself> Water pressure: P1 = Β½ πΎπ€ H2 P1 = Β½ x 1 x 322 = 512 T/m P4 = Β½ x 1 x 2.52 = 3.125 T/m
- 116. 103 26 β 18 2.5 β X X = 2.5 π₯ 18 26 = 1.73 m Uplift pressure: P2 =2.5* 25*1 = 62.5 T/m P3= Β½ *25*29.5 = 368.5 T/m
- 117. 104 Wave pressure: Waves are generated on the surface of the reservoir by the blowing winds, which causes a pressure towards the downstream side. Wave pressure depends upon the wave height. Wave height may be given by the equation, Pw = 1 2 (2.9 πΎπ€ hw) 5 3 hw Pw = 2 πΎπ€ hw2 hw = 0.032 β π. πΉ + 0.763 β 0.271 (F)3/4 , F<32km hw = 0.032 β π. πΉ for F > 32 km hw = height of water from top of crest to bottom of trough in meters V = wind velocity in Km/hr F = fetch or straight length of water expanse in km V = 124 km/hr F = 7km hw = ? hw = 0.032 β124 + 0.763 β 0.271(73/4) hw = 1.75 β 1.167 = 0.543m pw = 2 x πΎ x hw2 = 2x 1 x 0.5432 = 0.6 T/m 3/8 x 0.543 = 0.204m Aram = 32 + 0.204 = 32.204 Moment = 19.32 T*m
- 118. 105 Ice pressure: The ice which may be formed on the water surface of the reservoir in cold countries, may sometimes melt and expand. The dam face has then to resist the thrust exerted by the expanding ice. This force acts linearly along the length of the dam and at the reservoir level. The magnitude of this force varies from 250 to 1500 kN/m 2 depending upon temperature variations. On an average, a value of 500 kN/m 2 may be allowed under ordinary conditions. Earth quake pressure: If the dam to be designed, is to be located in a region which is susceptible to earthquakes, allowance must be made for the stresses generated by the earthquakes. An earthquake produces waves which are capable of shaking the Earth upo~ which the dam is resting, in every possible direction. The effect of an earthquake is, therefore, equivalent to imparting an acceleration to. the foundations of the dam iri the direction in which the wave is travelling at the moment. Earthquake wave may move in any direction, and for design purposes, it has to be resolved in vertical and horizontal components. Hence, two accelerations, i.e. one horizontal acceleration (ah) and one vertical acceleration (av) are induced by an earthquake. The values of these accelerations are generally expressed as percentage of the acceleration due Β·tO gravity (g} πΌ = 0.1π ππ 0.2π ππ‘π ln India, the entire country has been divided into five seismic zones depending upon the severity of the earthquakes. Zone Vis the most serious zone and includes Himalayan regions of NortJ::i India. A map and description of theseΒ·zones is available in "Physical and Engineering Geology" (1999 edition) by the same author, and can be referred to, in order to obtain an idea of the value of the a which should be chosen for designs. On an average, a value of a equal to 0.1 to 0. 15 g is generally sufficient for high daqs in seismic zones. A vaiue equal to 0.15 g has been used in ~hakra dam design, and 0.2 g in Ramganga -dam design. However, for areas not subjected to extreme earthquakes,
- 119. 106 πΌβ = 0.1 g and πΌ π£ = 0.05 g may by used. In areas of no earthquakes or very less earthquakes, these forces may be neglected. In extremely seismic regions and in con-servative designs, even a value upto 0.3 _g may sometimes be adopted. - Effect of vertical acceleration (CXy). A vertical acceleration may either act downward or upward. When it is acting in the upward direction, then the foundation of the dam will be lifted upward and becomes closer to the body of the dam, and thus the effective weight of the dam will increase and hence, the ~tress developed will increase. When the vertical acceleration is acting downward, the foundation shall try to move downward away from the dam body ; thus reducing the effective weight and the stability of the dam, and hence is the worst case for designs . Such acceleration will, therefore, exert an inertia force given by π π πΌ π£ (i.e. force= Mass x Acceleration) where W is the total weight of the darn. :. The net effective weight of the dam= π β π π πΌ π£ If where kv is the fraction of gravity adopted for ver-tical acceler-at:ion, such as 0.1 or 0.2, etc.]. Then, the net effective weight of the darn
- 120. 107 In other words, vertical acceleration reduces the unit weight of the darn material and that of water to (I - ky) times their original unit weights. Effect at vertical acceleration: av = Kv*g Kv = is the fraction of gravity adopted for vertical acceleration such as 0.1 or 0.2 ok w/g*av The net effect weight of dam = w π€ π av => w π€ π kv = w [1-kv] Av = 0.2 x 1 = 0.2 t/sec Effect of horizontal acceleration: ah = horizontal acceleration may cause the following two force: I. Hydrodynamic pressure II. Horizontal inert force A. Hydrodynamic pressure: Pe = 0.555 KN rw H2 If act at the height of 4H/3π above the base where kh is the fraction at gravity adopted for horizontal , such 0.1 , 1.2 etcβ¦ πΎπ€ = unit weight of water ππππππ‘ ππ π‘βππ ππππππ :
- 121. 108 Mc = pe 4π» 3π = 0.424 pe *H Zanger formula: Zanger.has given certain-big.formulas for evaluating theamoi.mLoLthis force and_ itsΒ· position, etc. on the vertical as well as on an inclined faces. The results of these big formulas are quite comparable to those given by Von- Karman equation and hence, for average ordinary purposes, the Von-Karman equation (19.1) is sufficient. Zanger's formula for hydrodynamic force. According to Zanger Pe= 0.726pe Β· H where Pe =-Cm KN Β· Yw Β· H Pe= 0.726 Cm. kh. Yw. H2 Pe = 0.555 x 0.1 x 1 x 322 = 56.83 T/m Me= 0.299 PeΒ· H2 or Me=0.412Pe Β·H Me = 56.83 ( 4π₯32 3π ) = 771.1 txm Silt pressure: It has been explained under 'Reservoir Sedimentation' in chapter 18 that silt gets deposited against the upstream face of the dam. If his the height of silt Β· deposited, then the force exerted by this silt in addition to external water pressure, can be represented by Rankine's formula as : Β· -
- 122. 109 Psilt = 1 2 πΎπ π’π π₯ β2 ππ And its acts at h/3 from base: Ka = is the coefficient at active earth pressure of silt = 1βπ πππ 1+π πππ where π is the angles at internal fraction of soil, and cohesion is neglected. πΎπ ππ = submerged unit weight of salt material h = height of silt deposited unit wt equal to 3.6 KN/m3 for horizontal direction and vertical 9.2 KN/m3 the total horizontal force will be 1.8h2 KN/m 1.8 KN/m = 0.184 T/m, 4.6 KN/m = 0.469 T/m
- 123. 110 No Name of the force Symbol Deminsion (m) Area (m2 ) Ξ΄w (T) F (T) Downword=+ve, Upword =-ve Arm (m) Moment (T*m) 1 W1 7*35 245 2.7 661.5 21.5 14222.25 2 W2 (18*26)/2 234 2.7 631.8 12 7581.6 βV1,βM1 1293.3 21803.85 3 weight of Water W3 (1.73*2.5)/2 2.1625 1 2.1625 0.5767 1.247041667 βV2,βM2 2.1625 1.247041667 4 U1 (-)25*2.5 -62.5 1 -62.5 12.5 -781.25 5 U2 (-)(25*29.5)/2 -245.8 1 -245.8333333 16.667 -4097.222222 βV3,βM3 -308.3333333 -4878.472222 6 P1 (32*32)/2 -341.3 1 -341.3333333 10.667 -3640.888889 7 P2 (1.73*2.5)/3 1.0813 1 1.08125 1.6667 1.802083333 βH1,βM4 -340.2520833 -3639.086806 8 Wave pressure P(wave) 1 -0.6 32.204 -19.3224 βH2,βM5 -0.6 -19.3224 9 Upword vertical earthquak force (-)1.15*βV1 -64.665 -109.01925 βV4,βM6 -64.665 -109.01925 10 Horizontal hydrodynamic pressure pe -56.83 -771.1 βH3,βM7 -56.83 -771.1 11 (-)0.1*W1 -66.15 17.5 -1157.625 12 (-)0.1*W2 -63.18 8.6667 -547.56 13 βH4,βM8 -129.33 -1705.185 Uplift force Horizontal hydrostatic pressure weight of dam Horizantal inertia force due to earthquak total moments 10682.91136 total of vertical force 922.4641667 Total of positive moment 21805.09704 total of nagitive moment -11122.18568 Total of horzoantal forces -527.0120833
- 124. 111 Checking Combination of forces for Designs. The design of a gravity dam should be checked for two cases, i.e. (i) when Reservoir is full; and (ii) when Reservoir is empty. Case I. Reservoir full case: When reservoir is full, the m~jor forces acting are: weight of the dam, external water pressure, uplift pressure, and earthquake forces in serious seismic zones. The minor forces are: silt pressure, ice pressure and wave pressure. For the most conserva-tive designs, and from purely theoretical point of view, one can say that a situation may arise when all the forces may act together. But such a situation will never arise and hence, all the forces are not generally taken together. U.S.B.R. has classified the 'normal load combinations' and 'extreme load combination, as given below: (a) Normal Load Combinations (i) Water pressure upto normal pool level, normal uplift, silt pressure and ice pressure. This class of loading is taken when ice force is serious. (ii)Water pressure upto normal pool level, normal uplift, earthquake force. s, and silt pressure. (iii)Water pressure upto maximum reservoir level (maximum pool level), normal uplift, and silt pressure. (b)Extreme Load Combinations (i)Water pressure due to maximum pool levcl, extreme uplift pressure witho_ut any reduction due to drainage and silt pressure. -
- 125. 112 Case II. Reservoir empty case : (i)- Empty rese-rvo1r - w-ithout eartlicfuake- fotce-s-ro-be- computed-for-determining-bending diagrams, etc. for reinforcement design, for grouting studies or other puiyoses. (ii) Empty reservoir with a horizontal earthquake force produced towards the upstream has to be checked for non- development of tension at toe. Modes of Failure and Criteria for Structural Stability of Gravity Dams A gravity dam may fail in the following ways: (1) By overturning (or rotation) about the toe (2} By crushing. (3) By development of tension, causing ultimate failure by crushing. ( 4) By shear failure called sliding. The failure may occur at the foundation plane (i.e. at the base of the dam) or at any other plane at higher level. (lj Over-turning. If the resultant of all the- forces acting on a dam at any of its sections, passes outside the toe, the dam shall rotate and overturn about the toe. Practi-cally, such a condition shall not arise, as the dam will fail much earlier by compression. The ratio of the righting moments about toe (anti clockwise) to the over turning moments about toe (clock-wise) is called the factor of safety against overturning. Its value, generally varies between 2 to 3.
- 126. 113 (2)Compression or crushing. A dam may fail by the failure of its materials, i.e. the compressive stresses produced may exceed the allowable stresses, and the dam-material may get crushed. The vertical direct stress distribution at the baseis given by the equation: Pmax/min = β π£ π΅ (1 Β± 6π π΅ ) Note. Resultant is nearer theΒ· toe and hence, maximum compressive stress is produced at the toe
- 127. 114 (Reservoir full case) Not.,, The resultant is nearer the heel and hence, maximum compressive stress (+ve stress) is produced at the heel (Reservoir empty with horizontal earthquake wave moving away from reservoir-case). If Pmin comes out to be negative, it means that tension shall be produced at the appropriate end. If Pmin exceeds the allowable compressive stress of dam material [generally taken as 3000kN/m2 (30 kg/cm2 ) for concrete]; the dam may crush and fail by crushing. Case 1. Reservoir Empty: β π = Moment due to weight or dam M1+ due to horizontal earthquake force moment due to vertical earth quack force. β π = M1 + M2 +M3 β π = 21803.85 + 1705.185 +109 = 23618 Txm β π£ = β π£ 1 + β π£ 4 β π£ = 1293.3 + 64.665 = 1357.965 T/m Case 1. Reservoir empty and vertical earthquake force acting down word.
- 128. 115 When the reservoir is empty, the only single force acting on it is the self- weight CW) of the dam and it acts at a distance B/3 from the heel. This is the maximum possible innermost position of the resultant for no tension to develop. Hence, such a line of action of Wis the most ideal, as it gives the maximum possible stabilising moment about the toe without causing tension at toe, when the reservoir is empty. The vertical stress distribution at the base, when the reservoir is empty, is given as: Xβ = β π β π£ = 23618 1357.965 = 17.39m e = π΅ 2 β Xβ = 25 2 β 17.39 = -4.89m > π΅ 2 < 4.1667 m Resultant acts near the heel and slight tension will be develop at toe Pmax/min = β π£ π΅ (1 Β± 6π π΅ ) = 1357.965 25 (1 Β± 6 π₯ 489 25 ) = 54.32 ( 1Β± 1.174 ) Pmax = 118.1 Pmin =-9.45 Pv at heel = 118.1 T/m2 = 1158.165 KN/m β€ 3000 (safe) The allowable compressive stress of dam material: (generally taken as 3000 KN/m2 = 306 T/m2 for concrete) R at toe = -9.45 T/m2 = -92.67 KN/m2 < 420 (safe) 420 KN = 42.82 T Average vertical stress: β π£ π΅ = 1357.965 25 = 54.32 T/m2 < 306 t/m2 (safe) Principle stress at toe π = pv(heel) sec2 π π = 48.1 = 118 T/m2 < 306 (safe) Shear stress at toe To (toe) = p toe tanΞ± To(toe) = -9.45 x 0.7 = -6.615 < 42.82 (safe) Shear stress at heel β 0 (heel) = Pv (heel) tanπ To (heel) = 1357.965 x 0 = 0 < 306 (safe) Case 1. (b) Reservoir empty and vertical earthquake force are acting up word. Then β π£ = β π£ 1 - β π£ 4
- 129. 116 β π£ =1293.3 β 64.665 = 1228.635 T/m β π = β π 1 + β π 8 - β π 6 β π = 21803.85 + 1705.185 β 109 = 23400.035 Txm Xβ = β π β π£ = 23400.035 1228.635 = 19.05 m e = π΅ 2 β xβ => 25 2 β 19.05 = -6.55m e = -6.55 < π΅ 6 = 4.167 Negative sign shows that resultant lies near heel and therefore, tension will develop at toe Average vertical stress: β π£ π΅ = 1228.635 25 = 49.14 < 306 (safe) Pmax/min = β π£ π΅ (1Β± 6π π΅ ) Pmax/min = 1228.635 25 (1Β±1.572) Pmax = Pv at heel = 49.14 x 2.572 = 126.39 < 306 (safe) Pmin = Pv at toe = 49.14 x (0.572) = -28.11 < 42.82 (safe) Principal stress act on toe π = Pv (toe) tanΞ± = -28.11 x 0.7 = -19.68 < 42.82 (safe) Shear stress act on toe: = pv (toe) tanΞ± = -28.11 x 0.7 = -19.68 < 42.82 (safe) Stresses at heel remain critical in this 1st case. Case 2 when Reservoir is full: Horizontal earthquake moving towards the reservoir causing upstream acceleration, and thus producing horizontal forces towards downstream is considered, as it is the worst case for this condition. Similarly, a vertical earthquake moving downward and thus,
- 130. 117 producing forces upward, i.e. subtractive to the weight of the dam is considered. The uplift coefficient C is taken as equal to 0.6, as given in the equation, and thus uplift pressure diagram as shown in Fig. 19.20 (c), is developed. Case 2 (a) Reservoir full with all forces including uplift. β π= β π 1 + β π 2 - β π 3 - β π 4 - β π 5 - β π 6 - β π7 - β π 8 β π = 21803.85 + 1.25 β 4878.5 β 19.32 β 109 β 771.1 β 1705.185 = 10683 Txm β π£ = β π£ 1 + β π£ 2 - β π£ 3 - β π£ 4 β π£ = 1293.3 + 2.16 β 308.33 β 64.665 = 922.465 T/m Xβ = β π β π£ 10683 922.5 = 11.58 m e = π΅ 2 β xβ = 25 2 β 11.58 = 0.92 < π΅ 6 = 4.1667 Average vertical stress: β π£ π΅ = 922.5 25 = 36.9 T/m2 Pmax/min = β π£ π΅ (1Β± 6π π΅ ) = 922.5 25 (1Β± 6 π₯ 0.92 25 ) =36.9 x (1 Β± 0.2208) Pmax = Pv (toe) = 36.9 x 1.2208 = 45.05 T/m < 306 (safe) Pmin = Pv (heel) = 36.9 x 0.7792 = 28.75 T/m < 42.82 (safe) P = πΎπ€ x H = 1 x32 = 32 T/m, Pβ= πΎπ€ x Hβ = 1x2.5 = 2.5 T/m Principle stress at toe: π = Pv(toe) x Sec2 Ξ± β Pβ tan2 Ξ± π = 45.05 x (1+0.49) β 2.5 x 0.49 = 65.9 T/m2 < 306 T/m2 (safe) Principle stress of heel is: π1 = Pv (heel) sec2 π β (P + Pe) tan2 π, tan π = 0, Pe = 56.83 T/m π1 = 28.75 x ( 1+ tan2 0) β (32 + 56.83) x tan (0)2
- 131. 118 π = 28.75 T/m2 < 42.82 T/m2 (safe) Shear stress at toe β 0 (toe) = (Pv(toe) β Pβ) tanΞ± = (45.05 β 2.5) x 0.7 = 29.785 T/m < 42.82 (safe) Shear stress at heel: β 0 (heel) = - (Pv(heel) β (P+Pe)tan π β 0 (heel) = - (28.75) β (32 + 56.83) x tan0 = -28.75 < 306 T/m (safe) Factor of safety against overturning: = β π (+) β π (β) = 21805.1 11122.18 = 1.96 > 1.5 (safe) Factor of safety against sliding: = π β π£ β π» = 0.7 π₯ 922.47 527.013 = 1.22 > 1 (safe) Shear fraction factor: S.F.F = π β π£+π΅π β π» q = Average shear strength of the joint which varies from about 1400 KN /m2 for poor rocks to about 4000 KN/m2 for good rocks. Β΅ = 0.15 β 0.75 S.F.F = 0.75 π₯ 922.46+25 π₯ 1400 527.013 = 3.93 >3 (safe) Case 2 (b) Reservoir full, without up lift Sometimes, values of stresses at toe and heel are worked out when there is no uplift, as the vertical downward forces. is maximum in this case? For this case, we shall calculate 2:M and I: V by ignoring' the corresponding values of 2: V3 and 2:M3 caused by uplift. β π = β π 1 + β π 2 - β π 4 - β π 5 - β π 6 - β π 7 - β π 8 β π = 21803.85 + 1.25 β 3639.086 β 19.32 β 109 β 771.1 β 1705.185 β π = 15561.41 T*m
- 132. 119 β π£ = β π£ 1 + β π£ 2 - β π£ 4 = 1293.3 + 2.16 β 64.665 = 1230.8 T/m = 1230.8 T/m Xβ = β π β π£ = 15561.41 1230.8 = 12.64m e = π΅ 2 β xβ = 25 2 - 12.64 = -0.14 < π΅ 6 Pmax/min = β π£ π΅ (1Β± 6π π΅ ) = 1230.8 25 (1Β± 6 π₯ 0.4 25 ) = 49.23 (1 Β± 0.096) Pmax = Pv at toe = 53.96 T/m < 306 T/m2 (safe) Pmin = Pv at heel = 44.50 T/m2 < 306 T/m2 (safe) Principal stress at toe: π = Pv x Sec2 Ξ± β Pβ tan2 Ξ± Pβ = 2.5 T/m tan Ξ± = 0.7 π = 53.96 x (1 + tan2 Ξ±) β 2.5 tan2 Ξ± = 79.175 T/m2 < 306 (safe) Principal stress at heel: π1 = Pv (heel) sec2 π β (P + Pe) tan2 π π1 = 49.23 T/m > 42.86 Shear stress at toe: β 0 (toe) = (Pv(toe) β Pβ) tanΞ± = 53.5 -2.5 x 0.7 = 35.7 T/m2 < 1400 KN/m2 (safe)
- 133. 120 Dam
- 136. 123 ( h = p/πΎ + π2 /2π) π»π = π π₯ βπ 75 Q hβ Ξ³β)3 (8 = 1000kg/m eβ ( Eddy losses) Ϋ΅Ϋ΄Ϋ³ΫΉ
- 137. 124 Ϋ±β Ϋ²β Ϋ³β Ϋ΄β (Francis)(Kaplan)(Propeller) (Tangential flow)(Radial flow)(Axial flow) (Mixed flow) (e)
- 138. 125 (en)(em)(ev) v*em*ehe = e he meβ veβ Ϋ² Ϋ°ΫΉ (out put) = β β2πβπ1 β = π1 β2πβ = ππΎ1 β2πβ = ( 2ππ 60 )( π· 2 ) β2πβ = π·π 84.6ββ Dβ Nβ
- 141. 128 ππ(Ns) Ϋ²ΫΉΫΉΫ±ΫΆΫΉΫ±Ϋ²ΫΉΫ°ΫΉΫ΄ΫΉNs Ϋ±Ϋ΅ΫΉΫΉ Ϋ°ΫΉΫΉ Ϋ΅Ϋ΅ΫΉ Ϋ΄ΫΉΫΉ Ϋ±ΫΉππ nsh h o 27 Ϋ±β Ϋ²β Ϋ³β Ϋ³ΫΉ Ϋ³ΫΉΫ±ΫΆΫΉ
- 143. 130 Ϋ±Ϋ±ΫΆΫ± kw/hf)9 ( 3800 x 10Ϋ±Ϋ±
- 145. 132 Ϋ΅
- 146. 133 Ϋ°ΫΆΫ° Ϋ°ΫΉΫ³ΫΉ Ϋ΄ΫΉ (Cross Head) (Hydraulic efficiency) ΫΆΫΉΫ±ΫΉ Ϋ± Ϋ³Ϋ΄ (firm power) (Surplus)
- 148. 135 Ϋ±ΫΉΫ±ΫΉΫΉ
- 152. 139 Design of Kaplan turbine Q = 15 m3 /sec, H = 29.5 m, nh = 0.9, π = 998 kg/m3 , g = 9.81 m/sec2 Power and specific speed: P= Q x H x nh x π x g P = 15 x 29.5 x 0.9 x 998 x 9.81 = 3899018.835w => 3899kw => 3.9Mw Specific speed: NQE = 2.294 π»π0.486 H = Gross head Hn = Net head Hn = H x nh => 0.9 x 29.5 = 26.55m NQE = 2.294 26.550.486 = 0.467 Rotational speed:
- 153. 140 NQE = π π₯ βπ πΈ3/4 => n = π ππΈ π₯ πΈ3/4 βπ E = specific hydraulic anergy : E = Hn x g = 26.55 x 9.81 = 260.45 J/kg There for: n = 0.467 π₯ 260.453/4 β15 = 7.82 5-1 = 469 Rpm Run away speed: The runaway speed is the max speed which the turbine can theoretically itβs achieve during load reaction The following guideline can be use to determine the runaway speed. Turbine type Runaway speed max/n Single regulated Kaplan turbine 2.0 β 2.6 Doled regulated Kaplan turbine 2.8 β 3.2 Choosing double regulated turbine nmax = 3.2 x 7.82 = 25 5-1 Runner Diameter: De = 84.5 x (0.79 + 1.602 x nQE) x β π» 60 π₯ π De = 84.5 x (0.79 + 1.602 x 0.467) x β26.55 60 π₯ 7.82 = 1.43m Hub Diameter: Di = ( 0.25 + 0.0951 π ππΈ ) x De Di = ( 0.25 + 0.0951 0.467 ) x 1.43 = 0.65m
- 154. 141 Section head: The section head, itβs the head where the turbine is installed it the section head is positive the section head turbine is located above the trail water if the section head is negative , the turbine is located under the trail water, to avoid cavitation the range of the section head limited. The maximum allowed section head can be calculated using the following formula or equation: Hs = πππ‘ππβππ£ π π₯ π + πΆπ’2 2 π₯ π β 6 x Hn Where Patom = atmospheric pressure (Pa) Pv = water vaper pressure (Pv) P = water density (kg/m3 ) g = acceleration of gravity
- 155. 142 Cu = outlet average velocity (m/sec) π = Conation coefficient (-) Hn = net head (m) Patom = 101300 pu, Cu = 2m/sec Pv = 2985.7 pa, Hn = 26.55m π = 1.5241 x nQE 1.46 + 2 2 π₯ π π₯ π»π π = 1.5241 x 0.4671.46 + 2 2 π₯ 9.81 π₯ 26.55 = 0.51 Hs = 101300β2985.7 998 π₯ 9.81 + 22 2 π₯ 9.81 β 0.51 x 26.55 = -3.29m
- 156. 143 Generator Design: Number of poles Np = 120 x F/N Where Np = Number of poles F = frequency of supply ie 50 Hz in Afghanistan N = Rotational speed (RPM) Np = 120 x 50 469.2 = 12.79 β 13 poles The table of standard rotational speed of Generator with explanation: The ideal speed = 1500 RPM1- N = 25 5 Hence Np = 120 x f/N => 120 x 50 1500 = 4 poles Exciter of Generator Explanation on page : 27 Type of exciter 1. Brush type 2. Brushless type Explanation on page 27 β 28 Generator out put: The output of generator is shown in VA and calculation following Pg (KVA) = 9.81 x H x Q x q x P/Pf Pg = required power output, KVA = kilo volt ampere. H = net head Q = design discharge
- 157. 144 Mo = over all efficiency is turbine efficiency transmission efficiency, nm x generator efficiency. π = density of water PπΎ = power factor Pg (KVA) = 9.81 π₯ 26.55 π₯ 15 π₯ 1000 π₯ 0.89/0.8 1000 = 4346.351 KVA Speed Governor: Explanation on page 28 β 29 Pd = pg x pt x s.f Where Pd = capacity of the dummy load Pg = rated output of the generator S.F = safety factor according the cooling method being employed (1.2 β 1.4) Pd (KW) = 4.4MVA x 0.8 x 1.4 = 4.93 MW Design at civil structure: Penstock flow velocity = 4.5 m/sec This is form exβ¦ Penstock rang (2m/s to 5m/s) Find internal diameter A = a/v = 15 4.5 = 3.33 Area of circle A = πr2 => A = πD2 /4 D2 = 4A/π => D = 2 x β π΄ π = 2 x β 3.33 π = 2.1 m Determination of the desntock thickness, tp tp = p x πΎ/6 p = pn + ps where p = total pressure
- 158. 145 ph = pressure due to water hammer ps = static water pressure π = stress Ph = πw x cp x v Cp = for water under ordinary condition Cp = 1120 Ph = 1000 x 1120 x 4.5 = 5.04 Mpa Static pressure ps Ps = πw x g x H Ps = 1000 x 9,81 x 29.5 = 289395 = 0.2894 Mpa Factor of safety n = 4 πgp = 957 Mpa But P = hn + ps = 5.04 + 0.2894 = 5.3294 Mpa π Allowable = p/n = 957 x 106 /4 π Allowable = 239.25 Mpa Hence tp = P x πΎ/ π Allowable = = 5.3294 π₯ 106 π₯ ( 21 2 ) 239.25 π₯ 106 = 0.023389m tp = 23.4 mm β 24 mm Head losing penstock: hv = K x V2 /2xg But k = 0.2 hv = 0.2 π₯ 4.52 2 π₯ 9.81 = 0.2064 m but two values lie at the entry and exct hut = 2 x 0.2064 = 0.42 m
- 159. 146 head loss date bend, hb hb = C x V2 /2g for the deflection angle of 45 Co = 0.09 hb = 0.09 π₯ 4.52 2 π₯ 9.81 = 0.093 m head loss due to fraction hf hf = f x (lp/Dp) x (V2 /2g) f = π4/Re for Re β€ 2111 (laminar flow) Re = VD/πΎ f = 1.325 βln( π 3.7π· + 5.74 π π0.9)β2 for 5111 β€ Re β€ 118 and 10-6 β€ e/D β€ 11-2 (Turbulent flow) hf = 0.009 x 150/2.1 x 4.52 /2x9.81 = 0.663 total head loss in the ht ht = 0.42 + 0.093 + 0.663 = 1.987 β 2m ht = 1.176m Net head is 29.5 β 1.176 = 28.3m For the design rule: hl β€ 1.15 x 29.5 m 1.176 β€ 1.15 x 29.5 1.176 β€ 1.475 From the above rule head loss comply with it hence the design is safe.
- 160. Dam
- 161. 148 Estimation of penstock pipes Internal Diameter Di = 21m Lengh of penstock lp = 150m Thickness tp = 24mm Density of steel 8000 kg /m3 Price of steel = 50 Afg per kg Volume of steel = π x lp ( π·π+2 π₯ π‘π 2 )2 β ( π·π 2 )2 V = π x 150 x ( 2.1+2 π₯ 0.029 2 )2 β ( 2.1 2 )2 = 24m3 Mass = density x voulume = 8000 x 24 = 192175 kg Cost = (price/kg) (m) Cost = 50 x 192975 = 1921750 Afg 1T = 60000 Af
- 162. 149 18000 12335.8 308.50.71.5267.75 6122.53540 37331.15 58.570.2574.375 21091890 30092700Ϋ΅Ϋ· 2208.51870 43751505 43751505 3187.53.8412240 87032610 1200 124 807 75551 25425 96360816 987216 34 37405 52922110 8222110 8218194049 31363880 99 22662 0722536 7918554866 29371097 33
- 163. 150 βΫ±M310335 86807028344 Ϋ²T on124 94682539506 174936170 51 βΫ²M3371097 33320 9877119117677 Ϋ³M322536 79370 37048346959 3 βΫ³M322536 79370 37048346959 3 Ϋ΄K g24000501200000 βΫ΄3000000 Ϋ΅30000000 βΫ΅80000000 ΫΆ20000000 383487509 8281976110
- 164. 151 1Cum405030tunalexcuvation Cum30cofferdam Cum30MobilzationofSiteCamp 2Sqm90060SitePreparation 3Cum3304105ExcavationofFoundation 4Sqm330430Compactionoffoundation 5Cum244180LeanConcretePCC 6Cum262.7415Steelworkandframework 7Cum436.3515RCCCloumn 8Cum23815Slabframeworkandsteel 9Sqm2119.8315slabconcrete 10Sqm2119.8315Plastering 11Sqm4239.3630Painting 12Sqm19890powerhouse 13LumpSumLumpSum60penstock 14LumpSumLumpSum30turbinework 15LumpSumLumpSum15generatorwork 16LumpSumLumpSum45electranics 810 15th Month KhurasanUniversity,EngineeringFaculty,CivilDepartment,2017EngineeringFacultyGraduationCermony ConstructionWorkPlanforadamonpeachrevirKonarpravence 1styear Noofdays 11th Month 12th Month 13th Month 16th Month 17th Month 18th Month 19th Month 20th Month 2ndyear 21st Month 22nd Month 23rd Month 24th Month 14th Month 6th Month 7th Month 8th Month 9th Month 10th Month 1st Month 2nd Month 3rd Month 4th Month 5th Month ActiviityDiscriptionS/NUnitQuanity totalday
- 165. 152 ΫΆ
- 166. 153 REFERENCES: 1. βDesing of gravity damsβ Denver Colorado, 1976 2. CelsoPenche, (1998) European Small Hydropower Association (ESHA), Guide on How to Develop a Small Hydropower Plant, ESHA 2004 3. Kenya Bureau of Standard, manual guide 4. H.C. Huang and C.E. Hita, βHydraulic Engineering Systemsβ, Prentice Hall Inc., Englewood Cliffs, New Jersey 1987. 5. Allen R. Inversin, βMicro-Hydropower Sourcebookβ, NRECA International Foundation, Washington, D.C. 6. USBR, βDesign of Small Canal Structureβ, Denver Colorado, 1978a. 7. USBR, βHydraulic Design of Spillways and Energy Dissipatersβ, Washington DC, 19ΞΈ4. 8. T. Moore, βTLC for small hydro: good design means fewer headachesβ, HydroReview, April 1988. 9. ASCE, Committee on Intakes, βGuidelines for the Design of Intakes for Hydroelectric Plantsβ, 199Ξ·. 10. G. Munet y J.M. Compas, βPCH de recuperation dβenergie au barrage de βLe Pouzinββ, Actas de HIDROENERGIA 93, Munich. 11. G. Schmausser& G. Hartl, βRubber seals for steel hydraulic gatesβ, Water Power & Dam Construction September 1998. 12. Electrobras (CentraisElΓ©ctricasBrasileiras S.A.) βManual de MinicentraisHidrelΓ©tricas.β