1.1 Introduction In Amhara region farming has been simply a way of life and was not a business enterprise till recent years. Farmers are supposed to produce everything that they need, if possible, for their family. This recurrent phenomenon though time be come an inherent responsibility of the farmers. On the other hand frequent drought, land degradation and an increase of extra mouth per year pushes the farmers to start irrigation agriculture. Farmers divert rivers and irrigate their plots of land. Even though this traditional activity contributes a lot in Regional total food production, it faced numerous problems. The major problem is yearly erosion of traditional diversion structures. Hence to alleviate this problem the Amhara National Regional State Bureau of Water Resource (ANRS- BoWRD) carried out the study design and construction of irrigation projects in the Region. This technical study document presents the engineering study of Chacha-2 diversion irrigation project. 1.1 1.2 Objective The objectives of this paper are: Hydrology study of the project area Weir design Canal layout and design and Engineering estimation of volume of work and cost Preparing working drawings 1.2 1.3 Methodology 1.2.1 1.3.1 Office woke Available data collection about the project area. The data include: Meteorological data Study document revision Locate the project area on 1:50,000 top sheet 1.2.2 1.3.2 Field work Identification of the proposed area Quantifying physical resources such as water and land resources measuring(estimating )base flow weir site selection collecting and geotechnical information of the area(with shallow test pit) surveying data collection(Total station) study existing farming practice 1.2.3 1.2.3 Design stage- office work data compiling, analysis and organization design with the help of pre-designed TOR preparing working drawing preparing technical document 2 Chapter-2 Project description 2.1 The project area Chacha-2 diversion irrigation project is found in North Shewa Zone, Baso Worana Woreda, Angolela- Asagret Kebel. It is south of chacha Village along Debre-Berihan to Mendida all weather road. Geographically is can be described as 390 N and 9032’ E. The average altitude is 2740 +MASL. The river Chacha-2 divides Oromiya and Amhara Region. Hence the command area found in the two Regions ( left side and right side of the river). It is estimated that 50 ha in Amhara side and 40 ha to Oromiya side. Downsteream of the proposed weir site, a traditional rock type weir were constructed by Mendida woreda office of agriculture. However this weir is non-functional due to undermining and excess seepage. 2.2 Physical Resources 2.1.1 2.2.1 Climate The agro-climatic zonation of the project area is under Dega (wet) zone. The minimum and maximum temperature at Deber-Birhan meteorological station is 50C and 19.80C.The area receives uni-modal rainfall. The rainy season is between June and October with a maximum precipitation of 365mm on Augest. 2.1.2 2.2.2 Topography and soil The irrigable area is narrow at the outlet and wider to the tail. The average slope across the contour is about 2.5%.The land is highly and repeatedly cultivated. More over the traditional practice of the farmers of burning the soil called “gaye” severely affect the soil structure. For this reason the soil is clay and has some water logging effects. For irrigation practice to reduce water lodging the field is tried to drain with intensive drain canals. 2.1.3 2.2.3 Land and water resources The proposed irrigable land is 90 ha which is irrigated by a perinial base flow of the 150 lit/ sec as measured on May 2001 EC. 2.1.4 2.3 BASE FLOW WATER BALANCE U/S & D/S UTILIZATION ALLOWANCE CONSIDERATION The base flow at the proposed diversion site (Chacha-2) has been estimated and measured to be 0.150m3/sec when the upstream irrigation structures of Chacha-1 existing modern irrigation project were functioning on the existing situation. It has been observed that the flow at Chacha-2 gets better recharges from springs and additional tributaries. Observation and Consideration of the recharging capacity of the downstream river, there are different additional tributaries for downstream users just after just the proposed irrigation project (chacha- 2) which are released and flow towards to down stream with the discharge of more than 0.035m3/sec of water. As to the downstream consideration, the river flows to Abay river basin and no existing irrigation practice has been seen / not suitable / however for sustaining the existing down stream ecology more than 35lit/sec of the flow has been released downstream. Chacha-2 was planned after detail analysis of above issues Measured base flow of all above this project from previous studies can be shown in the table below. Project Catchment Discharg Command Irrigation Down Stream Down Stream area (km2) e ed area Requirem Available Available (l/s) (ha) ent Water Water (Lit/Sec) (Lit/Sec) (M3/Sec) Chacha-1 385.60 140 88 Diversion 150 40 0.04 Chacha-2 539.64 150 90 Diversion 90 30 0.03 Chacha 35 Down Stream The net water balance in the catchments is shown on the above table to describe the amount of available water in the down stream of the project. Chapter-3 Hydrology 3.1 Available water resource The base flow (150 l/s) available in Chacha-2 River is the main source of irrigation water. There is small spring at the start of the main canal. It is added to the main canal. The spring potential is 10 to 15 l/s. 3.2 Available Hydrological data The river is not a gauged one. The Debr-birham metrological station used as the source of data. Month Min Temp Max Temp Humidity Wind Sunshine Rain °C °C % m/s hours mm January 2.8 19.4 71 2.7 9.2 0.00 February 3.3 21.5 72 2.8 8 0.00 March 5.6 20.5 73 2.9 7.5 26.50 April 5.2 21.8 75 2.6 6.7 2.80 May 5.8 23.6 71 2.7 8.4 11.80 June 6.3 22.4 68 2.2 7.7 48.90 July 8.6 17.1 78 1.8 5.3 362.40 August 8.6 18.1 83 1.6 5.6 365.10 September 6.2 18.5 80 1.8 6.1 52.40 October 4.4 17.9 77 2 8.3 59.60 November 0.6 17.7 77 2.2 9.2 1.40 December 2.4 18.5 75 2.4 9 0.00 Average 5 19.8 75 2.3 7.6 77.58 3.3 Dependable and effective Rainfall Only seven year data is available for this design. From this data 75% dependable rainfall is considered. Effective rainfall recommended by FAO/AGLW formula on monthly step is computed according to: Peff = 0.6 * P - 10 for Pmonth ? 70 mm Peff = 0.8 * P - 24 for Pmonth > 70 mm D.Rain Eff. rain Month mm mm January 0 0 February 0 0 March 26.5 5.9 April 2.8 0 May 11.8 0 June 48.9 19.3 July 362.4 265.9 August 365.1 268.1 September 52.4 21.4 October 59.6 25.8 November 1.4 0 December 0 0 Total 930.9 606.4 2.2 3.4 Design rainfall Daily highest rainfall year Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Height RF 1994 0 0 20.6 0 17.5 25 60.2 30 22.2 31.1 0 60.2 1995 0 13.3 3.5 16.4 8.2 6.8 40 15.1 2.6 0 1 40 1996 6.7 1.7 19.8 5.4 38.4 41.5 26.8 36 16 0 2.8 0 41.5 1997 14.6 4 18 23.3 19 19.5 31.6 29.8 7.6 23.1 31.6 1998 9 10.1 9.9 17.7 14.2 7.8 40 26.3 13 3.2 0 0 40 1999 6.9 0 10.8 2.7 6.5 12.9 42.4 45.7 11.8 13.4 1.4 0 45.7 2000 0 0 21.9 15.9 15.7 13.4 41.1 43.8 23 13.1 16.9 4.8 43.8 SUM 302.8 MEAN,R 43.25714 Design Rainfall Year Height RF,Rx (Rx)2 Mean,ý (x-ý)^2 Unbiased 1994.00 60.20 3624.04 43.26 286.96 sn-1 1995.00 40.00 1600.00 43.26 10.63 8.69 1996.00 41.50 1722.25 43.26 3.10 1997.00 31.60 998.56 43.26 135.96 1998.00 40.00 1600.00 43.26 10.63 1999.00 45.70 2088.49 43.26 5.95 2000.00 43.80 1918.44 43.26 0.29 SUM 302.80 13551.78 453.52 MEAN,R 43.26 From Gamble extreme statically estimation YT = -ln[-ln(1 -1/T) , T= 50 years Y50 = -ln[-ln(1 -1/50) = 3.9 RT = R + (sn-1)-1[0.78YT - 0.45] = 65.8mm 3.5 Design flood Catchment characteristics Elivation Max. Length El. Difference 3418-2880 12250 538 2880-2800 7000 80 2800-2739 31000 61 Time of Concentration,Tc 13.44 Hrs Curve No-3 90.274 Runoff Cofficient, Cf 0.3577 Average stream slope 135.00% % Stream length 50250 M Catchment area 539.64 km2 1. Increamental Hour,D D= Tc/5 = 2.688hrs take 2hrs. 2. Direct runoff,S S= (25400/CN)-254 S= 27.36 3. Unit hydrograph Tc = 13.44hrs D = 2hrs 4. Peak time, Tp Tp = (D/2)+0.6*Tc Tp = 9.064hr 5. Base time, Tb Tb = 2.67*Tp Tb = 24.2hrs 6. Peak descharge , q q = (0.21*Area*1)/Tp q = 12.5 m3/s 7. Design Rainfall Adjustment 8. Direct Runoff,Qmm Qmm = [I-0.2S]2/[I+ .8S] S =25400/CN -254; S= 27.36 Areal RF profile Re- Design RF % RF mm Incremental arranged Incremental Cumulative D RF RF (mm) order RF (mm) (CRF) or I 00 – 02 65.79 45 29.61 29.61 6 1.97 1.97 02—04 65.79 67 44.08 14.47 4 2.63 4.6 04—06 65.79 77 50.66 6.58 3 6.58 11.18 06—08 65.79 81 53.29 2.63 1 29.61 40.79 08—10 65.79 84 55.27 1.97 2 14.47 55.26 10—12 65.79 87 57.24 1.97 5 1.97 57.23 12—24 65.79 100 65.79 Cumulativ Qincrementa q(m3/s Q Hydrograph D e (CRF) or Qmm l ) (m3/s Begin Peak End I ) 00 – 02 1.97 0.51 0.51 12.5 6.39 0 9.06 24.2 02—04 4.6 0.03 0.00 12.5 0.00 2 11.06 26.2 04—06 11.18 0.99 0.96 12.5 11.96 4 13.06 28.2 06—08 40.79 19.90 236.4 6 15.06 30.2 18.92 12.5 5 08—10 55.26 32.13 152.8 8 17.06 32.2 12.23 12.5 7 10—12 57.23 33.86 1.73 12.5 21.60 10 19.06 34.2 12—24 9. Complex Hydrograph Time U/t ordinate H1 H2 H3 H4 H5 H6 Base flow Cumulative m3/sec 0 0 0.00 0.150 0.10 1 0.71 0.150 0.81 2 1.41 0.00 0.150 1.51 3 2.12 0.00 0.150 2.22 4 2.82 0.00 0.00 0.150 2.92 5 3.53 0.00 4.58 0.150 8.21 6 4.23 0.00 5.49 0.00 0.150 9.83 7 4.94 0.00 6.41 109.90 0.150 121.35 8 5.64 0.00 7.33 125.60 0.00 0.150 138.67 9 6.35 0.00 8.24 141.30 80.65 0.150 236.64 9.06 12.5 6.39 0.00 8.30 142.25 81.18 0.150 238.21 10 5.99 0.00 9.16 157.01 89.61 0.00 0.150 261.86 11.06 5.54 0.00 10.13 173.65 99.11 12.53 0.150 301.06 13.06 4.70 11.96 205.05 117.03 14.80 0.150 353.64 15.06 3.86 0.00 10.38 236.45 134.95 17.07 0.150 402.80 17.06 3.01 0.00 8.80 205.21 152.87 19.33 0.150 389.33 19.06 2.17 0.00 7.22 173.98 132.68 21.60 0.150 337.74 24.2 0 0.00 0.00 3.16 93.71 80.78 14.27 0.150 192.01 26.2 0.00 1.58 62.47 60.58 11.41 0.150 136.15 28.2 0.00 31.24 40.39 8.56 0.150 80.28 30.2 0.00 20.19 5.71 0.150 26.00 32.2 0.00 2.85 0.150 2.95 34.2 0.00 0.00 900.00 800.00 700.00 600.00 Cumulative m3/sec Q, in m3/s 500.00 H6 H5 400.00 H4 300.00 H3 200.00 H2 100.00 H1 0.00 9.06 24.2 26.2 28.2 30.2 32.2 34.2 0 1 2 3 4 5 6 7 8 9 10 11.06 13.06 15.06 17.06 19.06 Time in hrs Project Type And Location Project Name – Chacha-2 Project Type – Diversion Location o Administrative Zone – Shewa Zone o Administrator- Woreda Baso Worana Woreda o Kebele - Angolela-Asagret Kebel o Grid Location Altitude-2740 +Masl Location -9032’ E -390 N 2 Watershed Outputs Watershed Size ,Km2 –539.6352 Average Stream Slope,% – 3.697814 3 Weir Body Weir type – Ogee WES standard weir To irrigate targeted command the off take level is made at = 2743.0m. River bed level = 2740.3m The water depth in the main canal =0.49m Therefore the total crest level is 2743.6m Total weir height = 2743.6-2740.3 = 3.3m The total crest length available at 3.3m height is 46m. Out of this total length 1.5m is provided for operation slab of offtake arrangement. Hence the effective weir length is 44.5m 5. Irrigable land (ha): 90 ha CHAPTER THREE 3. HEAD WORK 3.1 General A division head work weir is constructed across river to rise water level to that the required amount of water on a irrigation field can be directed also the off taking canals sufficiently.Diversion head work is some is sometimes is known as canal head work .the purpose of diversion head work mainly for head maintenance with less storage capacity. There are two type of diversion head work . A diversion head work serves the following function 1. It raises the water level on its upstream side. 2. It regulates the supply of water in to the canal. 3. It controls the entry of silts in to the canal. 4. It creates temporary storage up stream of the weir(NN Basak) 3.2 Location of diversion weir Diversion head work is generally located in boulder stage or through stage of a river at the site near to the common area. In diversion head work selection cost is governing for appropriate decision. The following point should be remembered while selecting the location of the head work. At the site the river should be straight and narrow The river bank should be well defined The valuable land should not be submerged when the weir is constructed. The elevation of the site should be higher than the area to be irrigated The site should be easily accessible by roads. 3.3 Selection of type of weir There are various types of weir in use. The classification of weirs may be done is number of ways. The classification of weir according to the function: Storage weir Intake weir Waste Weirs Pick up weirs Pick-up weirs: In a storage project the reservoir water is discharge in the river through supply sluices. This released water is picked up by constructing a weir a cross the river where canals take off a series of such pick-up weirs may be constructed to utilize the available water Under Kesem irrigation project pick up (vertical drop weir) weir is selected because of the following reasons. The simplicity for construction. The relation to the function of the weir. The skill and the experience of the workers availability. The suitability problems of the structure which arises due to earth quake. In the present situation also vertical drop pick up weir is selected in KIP 3. 4 Hydraulic design of a weir Hydraulic data of weir means fixing the dimension of vertical drop weir. The following is the procedure of the design. 1. Determination of the crest level Determination of the crest level of the weir is closely related to afflux, water way or intensity of discharge and the pond level. It is determined by the permissible maximum value of afflux, attained during the maximum flood. A crest level can be worked out as below. Average level of the highest field =780m. Water depth required=1.43m Head loss at the turn out =0.1m Head loss at the head regulator = 0.45m (given) Slope from the highest field level to the weir =0.002 Distance from the highest field to the weir =400m bed level of the weir =779.5m The slope of the canal at 400m distance from the weir =0.002*400 = 0.8m Crest level = 780 + 0.8 +0.45 +0.1 +1.43 = 782.70m 2) Full supply level The water level of the canal corresponding to full supply discharge is full supply level (FSL). FSL of the canal depends on the following factor. Design discharge of the canal. Head loss in the head regulator and turn out. Topography of the area. FSL = Maximum elevation in the field + All head loss OR FSL = pond level – modular head Modular head means a sum of head of at the regulator and the head loss at the Modular head = 0.1 +0.45 = 0.55m Thus FSL =782.70-0.55 = 782.15m Weir height = crest level of the weir – bed level of the weir = 782.70 – 779.5 = 3.2m 1) Water way (L) It is the width provided at the site for the river water to flow. In other word, it is the length of the weir. Approximate water way to provide between the abutments may be calculated from lacey’s regime perimeter formula. Pw = 4.75√𝑄 Where Pw – is the wetted perimeter at the site at the river, but in this case it denotes the length of the weir between the abutments in meter and denoted by L. Pw = 4.75√𝑄 Q – The design discharge in m3/s Pw = 4.75√632.29 = 119.49𝑚 ≈ 120𝑚 2) Discharge intensity (q) 𝑑𝑒𝑠𝑖𝑔𝑛 𝑑𝑖𝑐ℎ𝑎𝑟𝑔𝑒 (𝑄) 632.29m3/s Discharge intensity (q) = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑤𝑎𝑦(𝐿) = 120𝑚 = 5.269m2/s ≈ 5.27 m2/s 3) Head over the crest (He) Can be determined from broad crusted weir formulas q = 1.71(He) 3/2 𝑞 Where He = ( 1.71) 2/3 =2.12m He-Total energy level above the crest of the weir He =Hsd + ha, ha – velocity head. Hd – water head above the crest of the 4) Normal scour depth (R) 𝑞 𝑅 = 1.35( ) 1/3 where, q – discharge intensity. 𝑓 5.27 1/3 𝑅 = 1.35[ ] f – silt factor for standard silt 1.5 1 =4.09m R – Scour depth 5) Regime velocity Under the flood condition when the weir expected to attain regime condition 𝑞 Regime velocity (v) = 𝑅 where R – scour depth 5.27 𝑉= = 1.288 ≈ 1.29 m/s 4.09 𝑉2 , (1.29)2 Approach velocity head = ℎ𝑎 = 2𝑔 2∗9.81 Thus water head above the crest of the weir. Hd = He – ha = 2.12 - 0.085 = 2.04m 6) Energy level Upstream TEL = crest level + He = 782.7 +2.12 = 784.82m Upstream HFL = u/s TEL – ha = 784.82 – 0.085 = 784.74m D/S TEL before retrogression = D/S TEL before retrogression – head loss at the regulator. = 783.35 – 0.085 = 783.26m 2.2.1 2.2.2 3.5 Design of weir wall The weir wall is trapezoidal cross section and the u/s face of the weir is vertical. A) Top Width The largest of the following is adopted (Arora, 2003). 𝑑 1) 𝑎 = Where, d = maximum depth over the water. √𝐺 2.04 𝑎= = 1.36m, G = specific gravity of the wall material (2.24for masonry) √2.24 3𝑑 3∗2.04 𝑎 = 2𝐺 = 2∗2.24 =1.37m, therefore Top width (a) =1.37m Adopt a = 1.5m 2) a = S + 1, where shutter height (No shutter provided). B) Bottom width (B) This should be fixed so that there is development of tension on the base of the weir. 𝐻+𝑑 3.2+2.04 i.For primary design 𝐵 = = = 4.88 ≈ 5𝑚 √𝐺−1 √2.24−1 However the actual width of the base should be determined as the maximum of the following situation. ii. No flow condition. This occurs when the u/s water level at the pond level and there is no tail water on the downstream. The over turning moment (Mo) about toe the weir 𝛾𝐻𝑠3 𝑀𝑜 = , where Hs = H + s 6 H – Height of shutter (in our case shutter is not provided) 𝛾 − Weight of the water (9.81KN/m2) The resting moment is due to the weight of the weir for a vertical up stream face of the weir. 𝛾𝐻𝐺 Mr= (B2 + aB – a2) 6 9.81∗3.2∗2.24 = (B2 + 1.5B – 1.52) 6 =12.45 (B2 + 1.5B – 1.52) Equating the over turning and the resisting moment, we get 64.24 = 12.45 (B2 + 1.5B – 1.52) ⟹ 𝐵 = 2.07𝑚 iii.High flood condition (submerged weir ) During high flood the over turning moment is from difference between upstream and downstream water pressure diagrams. 𝛾𝐻 3 𝛾𝑑𝐻 2 𝛾𝐻 3 𝛾(𝑑−ℎ)𝐻 2 Mo = ( + )–( + ) 6 2 6 2 𝛾ℎ𝐻 2 Mo = ( ) 2 Where d is the depth of water over the crest on the upstream and h is the difference of the water level on the upstream and downstream. The maximum over turning occurs when the value of h is max. I.e. when the weir is submerged i.e. h=d head over the crest and given by 2 d = 3 ∗Cd√2𝑔d2/3 where, Cd = 0.58 3∗q d=[ ] 2/3 MO = 9.81*2.12*3.22 2Cd√2g 3∗5.269 = [20.58 ] 2/3 = 2.12 MO = 120.21KN-m ………….. (3) √2∗9.81 The resisting moment (Mr) at outer middle third point 𝛾𝐻(𝐺−1) Mr = 6 (B2 +aB +a2) 9.81∗3.2(2.24−1) = (B2 +aB +a2) 6 = 6.89(B2 +aB +a2) …………………………………………. (4) Equating equation (3) and (4) 120.21 = 6.89(B2 +aB +a2) B2 + 1.5B – 2.25 - 17.45 = 0 B = 3.75m B = max [5, 2.07, 3.75] ⟹ take B = 5m 2.3 2.4 2.5 2.6 3.5 Structural Analysis of the Weir Structural analysis deals with checking the stability of the weir against. i. - Over turning ii. - Sliding Force acting on the weir. Static pressure Friction force at the base (up lift pressure) to counter balance the sliding Horizontal force Weight of the weir Silt-pressure Figure 3.1 Force acting on the weir Analysis of the weir for three conditions should be taken: i. There is no flow downstream ii. Submerged flow iii. When the weir is empty But it is observed that from the above three conditions the static conditions ( when there is no flow downstream is considered the critical one , therefore analysis of the weir only for static condition is carried out . Table 3.1 Force and moment acting on the weir Item Vertical force (KN) Horizontal force (KN) Lever arm Moment about from the the toe toe Pw ½*9.81*3.22=50.23 h/3=3.2/3= -53.75 1.07 W1 3.2*1.5*22.4=107.52 3.5 +1.5/2 = +450.96 4.25 W2 ½*3.2*3.5*22.4=125.44 2/3*3.5 +296.27 =2.33 PS ½*8.19*0.33*0.82 0.8/2 =0.4 -0.35 =0.873 PW ½*9.81*3.2*5=78.48 2/3*5 =3.33 -261.34 Σ 311.44 51.103 ΣM+=749.23 ΣM-=315.44 Note:- PS = ½* 𝛾𝑠𝑢𝑏 * ka * hs 2 where hs silt height = 0.8m (assumed) 1− sinϕ Ka = active earth pressure = 1+ sinϕ 𝜙 = 400 where 𝜙 − angle of internal response 1−sin 400 Ka = 1+sin 400 = 0.33 𝛾𝑠 𝑡𝑜𝑡 = 18 KN / m3 𝛾𝑠𝑢𝑏 = 𝛾𝑠 𝑡𝑜𝑡 − 𝛾𝑤𝑎𝑡𝑒𝑟 = 18KN/m3 – 9.81KN/m3 = 8.19KN/m3 I) Check for over turning ∑ 𝑀+ 749.23 ∑ 𝑀− = = 3.37 > 1.5 … OK! 315.44 II) Check for sliding ∑𝐻 51.103 Friction factors = ∑𝑉 = = 0.164 < 0.75 OK! 311.44 III) Check for stress + − ∑𝑀 − ∑𝑀 749.23− 315.44 𝑋̅ = ∑𝑉 = = 2.39 311.44 𝐵 5 e = ( 2 –X̅ ) = (2 - 2.39) = 0.11 < B/6 *(0.83) …… hence no tensions develop ∑𝑉 6𝑒 311.44 1∓6∗0.11 p= (1 + )= − ( ) = 62.28 (1∓ 0.132) 𝐵 𝐵 5 5 P1 = 70.50 KN/m2 P2 = 54.06 KN/m2 < 440 KN/m hence ok! Therefore the weir is safe against over stress. 2.6.1 3.6 Provision of sheet pile At the end of the impervious floor sheet piles are provided in order to increase the seepage length. U/S sheet pile level = U/S HFL -1.5R D/S sheet pile level = D/S HFL -2.0R (Lecture note) U/S sheet pile level = 784.74- 1.5*4.09 =778.605m D/S sheet pile level = 783.26m + 2.0*4.09 = 775.08m Depth of U/S sheet pile = Bed level – level of the pile = 779.5 – 778.605 d1 = 0.89, say d1 =1m Depth of D/S sheet pile = 779.5 – 775.08 = 4.42m – 775.08m = 4.42m ≈ say d2 = 4.5m 2.7 3.7 Impervious floor 2.7.1 3.7.1 Impervious floor length a) The length of the following equation Length of D/s impervious floor from the hydraulic jump point of view Ld = 5 (y2 –y1). Where, y1 & y2 initial & equate depth result. 𝑦1 Y2 = 2 (-1+ √1 + 8𝐹𝑟 2 ) 0.65 = 2 [−1 + √1 + 8(3.21)] = 1.35 Ld = 5(1.35m – 0.65) = 3.52m From Bligh’s formula 𝐻 Ld = 2.21C√ 𝐵𝑆 - - - - - for weir shutter provided above the crest Where 𝐻𝑆 = H +S (but not shutter provided in the weir B = bottom width (m) C = 12, Bligh’s coefficient for medium soil (Sahasrabudhe) 𝐻 Ld = 2.218*C*√10 - - - for weir without shutter. Where H – height of the weir above bed level = 2.21*12*√3.2/10 = 15.46 ≈ 16m Thus the length of d/s impervious floor (Ld=16m) Total creep length (L) = CHs from Bligh’s formula Hs=seepage head =pond level – bed level =782.70-779.5 =3.2m L = 12*3.2 = 38.4 adopt L=39 b) Length of u/s impervious floor (Lu) Lu = L – (Ld+B+2d1+2d2) =39 – (16+5+2*1+2*4.5) =39−32 =7m 2.7.2 3.7.2 Thickness of floor The floor thickness and weight should be sufficient to withstand the up lift pressure. Considering a horizontal floor of length L subjected to seepage head H, the residual head, h at any point p is given by h = H – l*(H/L) OR Residual head at any point ‘x’ is given by h= Hs – (𝐻𝑠/𝐿)*(Lx) Where, Lx – seepage length from u/s end towards point x Residual head at toe (point A) of weir (hA) hA= Hs*(1−LA/L) =3.2*(1-15/39) = 1.97m LA = 2*d1 + LU + B = 2*1 + 7+5 =15m The thickness of the floor at any point should be sufficient to with stand the up lift pressure. ℎ tA = 4/3*(𝐺−1) ℎ𝐴 1.27 tA = 4/3*(𝐺−1) = 4/3(2.24−1) = 2.12 take, tA = 2.4m Residual head floor at point ‘B’= 5m from toe. Lc hc = hs (1- 𝐿 ) 24 = 3.2 (1- 39) =1.23 1.23 tc= 4/3*( )= 1.32 = 1.5m 2.24−1 Provide the upstream with a nominal thickness of 1m 2.7.3 3.6.3 Checking thickness of floor by khosla’s theory The most hydraulic structure consists of all the elementary forms the simple form have to be solved independently and the pressure can be supper imposed to determine the pressure at a key points. In the khosla’s theory of independent variables the composite variable of the structure in to the simple form. The key points are the junctions of the floor and the pile line on either side and the bottom point of the pile line, and the bottom corner of the case of the depressed floor. The percent pressure at these key points for the simple form in to in to which the complex profile itself. If corrected for: a) Correction for the mutual interference of profile. b) correction for thickness of the floor c) Correction for the slope of the floor. I. Uplift pressure at the pile a) Upstream pile Total length of the floor, b= 39m Depth of U/s pipe line = d1 = 779.5m – 778.605m ≈ 1m ∝ = b/d = 28m/1m =28 1+√(1+∝2 ) 1+√1+282 𝜆= = = 14.5 2 2 100 𝜆−2 ∅∈ = 𝜋 𝑐𝑜𝑠𝑐 −1 ( 𝜆 ) 100 14.5−2 = 𝑐𝑜𝑠 −1 ( )= 16.92% 𝜋 14.5 𝜙𝑐1 = 100 - ∅∈ ⇒ 100% -16.92%=83.08 % 100 𝜆−2 100 14.5−2 ∅𝐷 = 𝑐𝑜𝑠 −1 ( ) = 𝑐𝑜𝑠 −1 ( ) = 11.89% 𝜋 𝜆 𝜋 14.5 𝜙𝐷 = 100 - ∅𝐷 = 100% -11.89%= 88.12% Correction 1. Due to the thickness of: the correction is positive downstream of the flow and negative upstream of the flow of the relation to sheet pile. (𝜙 D1−𝜙C1)t For 𝜙C1 = where t - thickness of upstream floor. d1 (88.12−83.08)1 = = 5.04 % 1 2. Due to mutual interference of the line: this correction is positive for the point in the rear or back water; and subtractive for the point forward in the direction of flow. 𝐷 𝑑+𝐷 For 𝜙𝑐1 =19√𝑏1 ( ) Where D – depth off pile No 2 𝑏 4.5 1+4.5 For 𝜙𝑐1 = 19√ ( ) D = 779.5 – 775.08 = 4.5m 28 28 For 𝜙C1 = 1.49% d- depth of pile No 1 = 779.5 – 778.605 = 0.89 ≈ 1𝑚 b′ = stance b/n two pile =28m b = total floor length = 28m Corrected 𝜙𝑐1 =83.08 %+5.04 %+1.49% 𝜙𝑐1 =89.61% b) Downstream pile Total length of the floor=b=28m Depth of downstream pile line=d2=779.5-775.08=4.5m 𝑏 28 𝛼= = =6.22 𝑑2 4.5 1+√1+𝛼2 1+√1+6.222 𝜆 = 2 = 2 = 3.65 100 𝜆−2 100 3.65−2 𝜙𝐸 = COS −1 ( )= COS −1 ( ) = 35.06% 𝜋 𝜆 𝜋 3.65 100 𝜆−1 100 3.65−1 𝜙𝐷 = COS−1 ( )= COS −1 ( ) = 24.14% 𝜋 𝜆 𝜋 3.65 1) Correction due to thickness 𝜙𝐸−𝜙𝐷 For 𝜙𝐸 = -( )*t 𝑑2 35.06−24.14 = -( 4.5 ) *1.5 = -3.64 % 2) Due to mutual interference 𝑑+𝐷 For 𝜙 𝐸 = - 19( ) * √𝐷⁄ ′ 𝑏 𝑏 Where D = depth of pile no 1 =1m d= depth of no 2 = 4.5 m 4.5+1 = - 19( 28 ) * √1⁄28 b= total floor length =39m = -0.71% b′ = length b/n two piles = 39m Corrected 𝜙 𝐸 = 29.97 % - 3.64% - 0.71% =25.62% II. Exit gradient The hydraulic gradient at the downstream end where water emerges out is called exit gradient. When the pressure due to particular exit gradient is such that is equals submerged weight of soil particles the exit gradient is called a critical exit gradient. 1 For safety exit gradient should always be less than this critical value. The critical gradient varies from 4 1 to 7 for different formations. [Sahasrabudhe, 1994) Total length of the impervious floor b=28m Depth of d/s stream pile (d2) = 4.5m 𝑏 28 𝛼 = 𝑑 = 4.5 = 6.22 2 1+√1+∝2 1+√1+6.222 λ= 2 = 2 = 3.65 𝐻 1 3.2 1 GE = 𝑑𝑆 × = 4.5 × 𝜋√3.65 2 𝜋√𝜆 1 GE = 0.12 < --------------- 0K! 7 Percentage pressure at any point, with in the impervious soil 𝜙𝑐1 −𝜙𝐸 = 𝜙𝐸 + ∗ 𝑙𝑎 𝑏 Where 𝑙𝑎-length from d/s & towards the point 𝜙𝑐1 −𝜙𝐸 Percentage pressure at ‘A’ = 𝜙𝐸 + 𝑏 ∗ 𝑙𝑎 (89.61−25.62) =25.62+ 28 ∗16 =62.18% 62.18∗3.2 Residual head (h) = 100 =1.99m ℎ 1.99 Thickness of floor at ‘A’ = 𝐺−1 = 2.24−1 = 1.61 < 2.21--------ok! 𝜙𝑐1 −𝜙𝐸 Percentage pressure at ‘B’ = 𝜙𝐸 + 𝑏 ∗ 𝑙𝑏 (89.61−25.62) = 25.62+ 28 ∗11 =50.76% 50.76∗3.2 Residual head (h) = 100 =1.63m < 1.76m ----------------ok! 𝜙𝑐1 −𝜙𝐸 Percentage pressure at ‘C’ = 𝜙𝐸 + 𝑏 ∗ 𝑙𝐶 (89.61−25.62) = 25.62+ 28 *6 = 39.33% 39.33∗3.2 Residual head (h) = 100 = 1.26 ℎ 1.26 Thickness of the floor at ‘C’ = 𝐺−1 = 1.24 =1.02m < 1.32m 2.8 3.7 Design Of Under Sluice Protection They are opening provided in the body of a weir or any cut at low levels. They are located in the smaller compartment in front of still pond. These sluices are perfectly control by means of gates which are operated from top. Naturally the sluices should be located quite below the crest of the head regulator (Generally 1.8m below). Thus the scouring sluices maintenance the channel clear and defined in front of the head regulator. The sluices can also be used during the peak flood period to the lower discharge over the crest of the weir. Under sluices is used for quick lower of u/s high flood level because the discharge intensity over the sluice portion is greater than that is the weir portion [Sahasrabudhe, 1994] 2.8.1 a) Design consideration Sill of the under-sluice pocket is kept at or slightly above the deepest river bed and about 0.9 to 1.8m below the sill of the canal head regulator. The length of the under-sluice pocket between the divide wall and the head regulator may be taken as 1.5times the upstream length of divide wall. However, this length is governed by discharging capacity of the under-sluices, which should be sufficient to enable them to serve their main functions, described above. The discharging capacity of under-sluices may be selected as follows: i. It should be able to ensure sufficient scouring capacity for which the discharge Capacity should at least twice the full supply discharge of the main canal at its head. ii. It should be able to pass the dry weather – flow and low floods during the Months excluding the rainy season .without the necessity of dropping the weir Shutters. iii. It should be able to dispose of 10 to 15% the high flood discharge during several Floods. [Garge 2003] - Head over under sluice (He) = U/S Tel - Crest level under sluice = 784.82 - 779.5 = 5.32m - Discharge intensity qu = 1.705 𝐻𝑒 3/2 = 1.705*(5.32)3/2 = 20.92 𝑚3 / 𝑠𝑒𝑐⁄𝑚 - Qu = under sluice discharge = qu*Lu where Lu under sluice - Water way length(assume 4m) [NN Basak] Qu = 20.92*4m = 83.68 𝑚3 /𝑠𝑒𝑐 Weir head = u/s TEL – crest level = 784.82- 782.7 = 2.21m Q = 1.075 𝐻𝑒 3/2 = 1.770 *(2.12)3/2 = 5.27m^3/sec/m Q = weir discharge = q *Lw where Lw weir water way Lw = 120 – 4 = 116m 𝑚3 Q = 5.27𝑠𝑒𝑐⁄ ∗ 116𝑚 = 611.32𝑚3 /𝑠𝑒𝑐 𝑚 Qu + Q = 611.32 + 83.68 = 695𝑚3 /𝑠𝑒𝑐 > Qd = 632.29𝑚3 /𝑠𝑒𝑐 -------------- ok! The minimum length of d/s floor (Ld) 𝐿𝑑 = 3.87√ℎ/10 = 3.87√3.2/10 = 2.2𝑚 [For under sluice without shutter] Total length of d/s impervious floor and d/s protection given by 𝐻∗𝑞 3.2∗20.92 L = 27√10∗75 = 27√ = 8.1m 10∗75 Depth of sheet pile from scour consideration: 𝑞2 Depth of scour R =1.35( 𝑓 )1/3 20.922 1 R = 1.35( )3 = 10m 1 Bottom level of D/s cut off = D/S HFL - 1.5R = 783.26-1.5*10 = 768.6m Bottom level of U/S cut off = U/S HFL-1.25R = 784.74 -1.25*10 =772.24m 2.9 b) Head Regulator of canal design It is a structure constructed at the entrance (the head) of the canal where it takes off from the river. The regulator serves the following purpose. It regulates the flow of irrigation water entering in to the canal. It can be used as a meter for measuring the discharge. It regulates and prevents excessive silt entry in to the canal. [Sahasrabudhe 1994] It is a aligned at 90° to the weir but slightly larger angles up to about 110° are now considered preferable as they provided a smoother entry. The height flood in the river may rise much higher than the pond level and the prevent water from spilling over the gates during flood RC breast wall is provided from the pond level to above the high flood level. [Arora, 2003] Design procedure - Full supply of off taking canal = 5.86 𝑚3 /𝑠𝑒𝑐 - Full supply of canal = Pond level-Modular head = 782.7m - 0.55m = 782.15m - Water depth in the canal at head = 1.43m - Safe exit gradient for canal bed material = 1/6 - Crest level of head regulator = crest level of under sluice +1.8 = 779.5 +1.8 = 781.3m [Sahasrabudhe 1994] The water way for regulator is for the full supply discharge of 7.03m^3/sec can pass through it. Discharge Q through the regulator is given by [Garge 2003] 3 2 Q = 3 ∗ 𝑐𝑑1 ∗ √2𝑔 ∗ 𝐵(ℎ + ℎ𝑟)2 + 𝑐𝑑2 ∗ 𝐵ℎ1 √2𝑔(ℎ + ℎ𝑣) Where Cd1 = 0.577 & Cd2 = 0.8 h1- depth of D/S water level in the channel above the crest Hv - head due to velocity H - difference of water level U/S and D/S of the crest. Neglecting Velocity head hv we get 2 Q=3 ∗ 0.577 ∗ √2𝑔 ∗ 𝐵ℎ3/2 + 0.8ℎ1𝐵√2𝑔ℎ1 H=0.55 H1=0.85 3 2 5.86 = 3 ∗ 0.577 ∗ √2 ∗ 9.81 ∗ 𝐵(0.55)2 + 0.8 ∗ 0.85√2 ∗ 9.81 ∗ 0.85 ∗ 𝐵 5.86 = 0.695B+2.78B B= 1.68m adopt B= 1.7m Provide 2 piers of 1.5m thickness each on the side (Garge 2003) Overall waterway of the regulator = 1.5*2+2=4.5m Figure 3.2 section of head regulator for the pond level flow condition 2.10 C) Divide Wall It is constructed at right angle to the weir axis. It divides the river channel in to compartment. Thus in the smaller compartment, which is nearer to the head regulator, a still pond is created. Divide wall have the following functions: It separates the floor of the scouring sluice from that of the weir proper. It provides comparatively still pocket in front of the canal head regulator so that silt gets deposited in it and relatively clear water enters the canal. It helps in the minimizing the effect of the main river current on the flow condition in the head regulator. Dimensions of divide wall - Width of the head regulator, B= 1.7.0m - The length of divide wall on the upstream = 0.58*B=0.58*1.7 = 0.99m - D/S extends up to the end weir glacis(toe) = 6m - Dived wall height = 4m - The top width of the wall =1.5m (slope 0.5H:1V) [Arora 2002] 2.11 d) Silt Excluder It is a structure which excludes the silt from irrigation water as the name implies. It separates the lower silt laden portion of the water from the upper silt free portion. It consists of a series of parallel tunnels of low height. The tunnels are constructed in the pocket parallel to the flow of water in the river. The height of the tunnels depends upon the silt distribution in the flow of the water. 2.12 Design of silt Excluder It is design that the bottom layer of water which is highly charged with silt and sediment will pass down the tunnels and escape over the floor of the under-sluice way(s),since the gates of the under sluice way(s) shall be kept open up to the top of the tunnels. The clearer water over the top of the roof of the excluder tunnels will thus enter the canal through the head regulator. Design procedure - Full supply discharge of canal = 5.86𝑚3 /𝑠𝑒𝑐 - Crest level of the under sluice = 779.5m - Crest level of the head regulator = 781.3m - Usually, two or three bays of under sluice o f the weir are covered by the excluder. However excluder covering only one bay has been designed [Garge 2003] - Design discharge =15% to 20% of canal discharge [Garge 2003] 20 Q = 100 ∗ 5.86 = 1.17𝑚3 /𝑠𝑒𝑐 - A minimum velocity of 2 to 4.5m/sec must be maintained through the tunnels in Order to keep them free from sediment. 2m/sec- adopted for the design 1.17𝑚3 /𝑠𝑒𝑐 Area of cross section A = = 0.58𝑚2 2𝑚/𝑠𝑒𝑐 Height of tunnels generally varies from 0.5 to 0.6m for Sandy River and 0.8 to 1.2m for boulder Stage River. [Garge 2003] Assuming thickness of roof slab =0.2m Height of tunnel (h) =781.3-0.2-779.5=1.6m 𝐴 0.58 Total clear width =ℎ = 1.6 = 0.36𝑚 For clear span of 0.4m (assume) 0.36 Number of tunnel = 0.4 = 1.1 ≈ 1𝑡𝑢𝑛𝑛𝑒𝑙 Assume thickness of divide wall =0.3m Overall width =0.4*1+0.3=0.7m Only one bay sluice will be used for silt excluder. 3.00-3.50 DCe 3.510 0.006732 3.52-4.00 DCf 4.000 0.007671 4.134-4.50 DCg 4.500 0.007863 6.00 DCh 6.000 0.008630 Sample calculation Data available: Drainage coefficient(Dc)=16.57mm/day Drainage area(A)=0.742ha Drainage discharge(Qdr)=Dc*A=0.00135m3/sec 1 1 Bed slope of surface drain, S= to (B.Dass) 165 330 1 In this case, S= is adopted. 200 Side slope, m=0.5 Manning’s roughness coefficient, n For open drain, n>=0.03 Then take n=0.25 From Manning’s equation 2 1 1 Q= * AR 3 S 2 n 2 Qn 0.00135 * 0.25 AR 3 1 =0.004773……………………..(1) 2 0.005 s Let; B=0.20m A = BD + mD2 = 0.2D + 0.5D2 P = B+2D (m 2 1) =B+ 5D , R= A P 2 0.2 D 0.5D 2 AR 3 (0.2 D 0.5D 2 ) ………………………..(2) 0.2 5D Combining (1) and (2) 0.2D 0.5D 5 2 3 0.004773= 0.2D 2 5*D 3 By trial and error, D=0.066m FB D Fig 6.1 B Drain Qdr slope N M B A P D Adapted FB type (m3/sec) (m) m2 (m) (m) m (m) DCa 0.001350 0.005 0.25 0.5 0.2 0.0135 0.332 0.066 0.07 0.5 DCb 0.002723 0.005 0.25 0.5 0.2 0.0245 0.431 0.115 0.12 0.56 DCc 0.004219 0.005 0.25 0.5 0.2 0.0341 0.507 0.152 0.16 0.5 DCd 0.004795 0.005 0.25 0.5 0.2 0.0435 0.578 0.187 0.19 0.5 DCe 0.006732 0.005 0.25 0.5 0.2 0.0441 0.579 0.186 0.20 0.5 DCf 0.007671 0.005 0.25 0.5 0.2 0.0506 0.678 0.208 0.21 0.5 DCg 0.007863 0.005 0.25 0.5 0.2 0.0599 0.612 0.234 0.24 0.5 DCh 0.008630 0.005 0.25 0.5 0.2 0.0902 0.841 0.306 0.31 0.5 3 6 HEAD WORK DESIGN 6.1 General Weir is an obstruction or a barrier constructed across a river. The obstruction is of smaller in comparison with the dam. It raises the water level locally and supports the water against its face. Thus, the diversion of the water from the river into the canal takes place. 6.2 Selection of Weir type The weir may be broadly divided in to three: 1. vertical drop weir This type of weir was used in most case, particularly suitable for land clay and consolidated gravel foundation. 2. Rock fill weir Is suitable for fine sandy foundation. Such weir requires huge quantity of stone and is economical only when the stone is easily available. 3. Concrete glacis or sloping weir. This type of weir is used on permeable foundation and is generally provided with low crest, with counter balance get. In deciding the type of the weir, the following conditions should be considered. Economy of construction Foundation condition Size of the project Head across the weir and practically during implementation Taking all the above factors & future expansion of the project into account, the case for construction & suitability of foundation masonry weir of vertical drop is selected for this particular irrigation project. 6.3 Weir Design Available Data: m3 Q peak =250 (refer data from hydrology, Chapter two) s River bed level=1846.87m (data from top map) Assumed data: Afflux=1.0m Retrogression=0.5m (Source: Irrigation Structure hand out; by HABTAMU ITAFA) Hydraulic Design of Weir Determination of the crest level Average Project Type And Location Project Name – Chacha-2 Project Type – Diversion Location o Administrative Zone – Shewa Zone o Administrator- Woreda Baso Worana Woreda o Kebele - Angolela-Asagret Kebel o Grid Location Altitude-2740 +Masl Location -9032’ E -390 N 2 Watershed Outputs Watershed Size ,Km2 –539.6352 Average Stream Slope,% – 3.697814 3 Weir Body Weir type – Ogee WES standard weir To irrigate targeted command the off take level is made at = 2743.0m. River bed level = 2740.3m The water depth in the main canal =0.49m Therefore the total crest level is 2743.6m Total weir height = 2743.6-2740.3 = 3.3m The total crest length available at 3.3m height is 46m. Out of this total length 1.5m is provided for operation slab of offtake arrangement. Hence the effective weir length is 44.5m 5. Irrigable land (ha): 90 ha CHAPTER THREE a) 3. HEAD WO level of highest field = 1844 b) Head loss across the field = 0.1 m c) Head loss at the turn out = 0.15 m d) Head loss at the head regulator = 0.32 m e) Water depth required = 0.65 m(data from canal design part) f) Slope of the canal * distance of the highest point from the weir = 0.008 * 600 =4.8m Therefore, the crest level of the weir=1844+0.1+0.15+0.32++0.65+4.8=1850.02m Weir height = Crest Level of the Weir – River Bed Level =1850.02-1846.87 =3.15m Water Way It should be adequate to pass the design flood safely. From the topographic map, a measurement of the river cross section at weir axis (length of weir) is found to be 44.5m.Upstream(U/S) TEL=Crest level+He 2 3 Q He= where He=Head over the crest Cle Q=Design flood discharge Le=44.5m C=coefficient of discharge=1.50 23 805.6 He= Q=805.6m3/s 1.5 * 44.5 =5.26m U/S TEL=2743.6+5.26 =2748.86m Regimes scour depth(R) 13 q2 R=1.35 where q=Q/L=805.6/44.5=18.1 m3 s m f and f=1.5 13 q2 R=1.35 f =8.13m Regime velocity (Va) =q/R Va=18.1/8.13=1.23 m/s Velocity head (ha) =Va 2 /2g=1.23 2 /2*9.81=0.077m Wáter head abovecrest of weir(Hd)=He-ha =5.26-0.077 =5.18m Energy level U/s TEL=crest level+He =2783.6+5.26 =2748.86m U/s HFL=U/S TEL-ha =2748.86-0.077 =2748.783m Down stream (D/S) HFL =U/S HFL – Afflux Assume afflux=1m(Assumption) Retrogression=0.5m(source .KRArora) D/S HFL=2748.783-1 =2747.783m D/S HFL before construction =D/S HFL – Retrogression =2747.783 – 0.5 =2747.283m Design of Weir Wall The weir wall is proposed to be trapezoidal cross-section with u/s face vertical and d/s face with slope 1:1. The top width of weir wall (B ' ) is given as H B'= where, B ' = Top width of weir wall and is G 1 Generally 1.5 to 1.8m H=Head of water over the weir wall at the time of max. flood G=Specific gravity of floor material H=He-ha =5.18-0.077 =5.1m 5.1 B'= 2.24 1 =4.58m Note: Since B ' =4.58m is out of range (1.5-1.8m), take B ' =1.5m Top width (B ' ) =1.5m The bottom width (B) of the weir wall should not be less than H Heightofweir B= G 1 5.1 3.3 B= 2.24 1 =7.5m 6.6 Design of weir wall Depth of water over crest (Hd )=U/S HFL-crest level =1194.1-1192.02=2.1m Top width(a) Can be fixed as the largest of the following three equations Bottom width(B) 1. No flow condition This condition is when the water level on the U/S is at the pond level. The over turning movment is due to Hs=H+s=2.22+0=2.22m Where Hs is seapage analysis head Water resoursces and irrigation engineering | HEAD WORK DESIGN 64 √√ 2. a=s+1=0+1=1m Where s=height of the shutter G=specfic gravity Hd=maximum depth of water So maximum top width=1.9m Bottom width(B) 1. No flow condition This condition is when the water level on the U/S is at the pond level. The over turning movment is due to Hs=H+s=2.22+0=2.22m Where Hs is seapage analysis head ) 2 High flow condition The weir is submerged when the flood is passing over the crest and the weir is submerged. The overturning moment at the base is obtained from the water pressure diagram on two sides. When the d/s water level,(He=d=2.2m) thus, MO=Mr Head over the crest Cd=0.98 Depth of Sheet Piles R.L of bottom of upstream (U/S) pile= U/s HFL-1.5R(R=8.13) =2748.783-1.5*8.13 =2736.6m Therefore, depth of U/S pile (d 1 )=River bed level-level of pile at u/s d 1 = 2740.3-2736.6 =3.7m R.L of bottom of downstream (D/S) pile =D/S HFL after retrogression – 2R =2748.783-2*8.13 =2732.523m Depth of D/S pile (d 2 ) =2740.3-2732.523 =7.78m Impervious Floor Seepage head, Hs= Pond level(crest weir level) – Bed Level =2743.6-2740.3 =3.3m ' By Bligh s theory, the total creep length (L) is given by: L=CHs where, C=Bligh ' s Creep coefficient taken as (5-9) for grave l foundation Let us take C=9 L=9*3.3m =29.7m Length of downstream impervious floor, l 2 Hs L 2 =2.21*C 10 3.3 =2.21*9 10 =11.4m Length of upstream impervious floor, L 1 L 1 =L- (L 2 +B+2d 1 +2d 2 ) =29.7-(11.4 +1.5+2*3.7+2*7.78) =-ve For the u/s impervious floor let us take nominal value of L 1 = 3m Therefore, total length of impervious floor b, will be b=L 1 +B+L 2 =3+7.5+11.4 =22m Total creep length changed into=b+2d 1 +2d 2 =22+2*3.7+2*7.78 =45m Protection Work a) D/S protection work The total length of d/s floor and d/s protection work is given by =L 2 +L 3 Hsq =18C 10q s 3.3 *18.1 =18*9 10 * 75 =45.7m Length down stream protection=L1+L2-L3 =45.7-11.4=34.3m Hence provide 1m thickness d/s loose talus of 35m in length . Note: This length of 35m can be partly provided as blocks over inverted filter and partly as launching apron. Up stream Protection Work The length of upstream talus (L4) may be kept equal to half the length of talus. L3 L4= =34.3/2=17.15m 2 Thickness of the impervious floor by Bligh’s theory. Seepage head=3.3m Creep length=45m The maximum ordinate of the H.G.Line above the bottom of the floor for the downstream portion at the function of weir wall: 3.3 H= * 29.7 2.2m 45 The thickness of D/S floor at this point is then obtained by; H 2.2 t= 1.33 ( ) =1.33*( ) 2.35m G 1 2.24 1 Thickness of D/s Floor after 5m from the function of the weir wall. HS H=HS- (16.1 5) L =3.3-3.3/45(16.1+5) =1.75m 1.75 t=1.33*( ) =1.87m 1.24 NOTE: The up stream impervious floor thickness is assumed to be nominal i.e. t=.8m and for floor length beneath the weir 1m thickness is provided. 3.1 7.4 Stilling Basin Design During the flood season, when high flood occurs over the weir crest water falls from the maximum reservoir level of u/s to the d/s tail water and the difference b/n the u/s and d/s energy grade line becomes very high. Therefore ,the energy must be dissipated before it reaches the natural river source: other wise it causes damage to d/s of the apron. The energy tends to dissipate through a hydraulic jump d/s of the weir .To control the location of the jump stilling basin is designed. Conventional Method It is experimental formula to determine the length and depth of stilling basin. L=3* hf 1 D= hf 2 Where :f/2 >=d>=f/3 L=Length of basin D=depth of basin h=over flow depth f=u/s water level +velocity head –d/s water level. f=2748.783+0.077-2747.783=1.077m h=5.18m Length of basin, L=3* 5.18 *1.077 =7.08m 1 Length of basin, d= 5.18 * 1.077 2 =1.18m f/2>=d>=f/3, 0.54<d<0.36 Correction for mutual interference D d D Correction=19* * b' b Where, D=Depth of pile whose effect is required on the another pile (D=8-1.87=6.13) ' b =Distance b/n two piles =22m b=Total floor length, b=22m d=the depth of the pile on which the influence occur (d=2.9-0.8=2.1m) 6.26 2.1 6.26 Correction=19* * 21.5 21.5 =3.99% (+ve) Therefore, corrected C 1 (67.72 2.75 3.99) =74.46% b) Downstream pile: b=21.5m, d 2 8.0m , 2.69m, 1.93 100 2 E cos 1 =51.15% 100 1 D cos 1 =34% Thickness correction for E D E E *1.74 d2 =3.73%(-ve) Correction for mutual interference D d D Correction=-19* * b' b 6.26 2.1 6.26 =-19* * 21.5 21.5 =-3.99%(-ve) Corrected E 51.15 3.73 3.99 =43.43% Percentage pressure at A C1 ECorrected = ECorrected * lengthofd / sfloor b =43.43+ 74.46 44.43 *11.3 21.5 =59.21% Residual head, h=0.59*3.15 =1.87m h 1.87 Thickness of the floor = 1.5 2.2m………….Ok G 1 1.24 Percentage pressure at B =43.43+ 74.46 44.43 * 5 21.5 Residual head, h=0.5*3.15 =1.59m 1.59 Thickness of the floor = 1.28 1.74m…………….Ok 1.24 Hence the floor is safe by Khosla ' s theory Checking of the thickness of the floor by Kohsla’s theory 1. Exit gradient Total length of the impervious floor, b=21.5m Depth of down stream pile, d2=8m 21.5 1 1 2 = 2.69 , 1.93 8 2 H S *1 =0.09 < 0.11 ………..ok 1 GE= d2 * * 9 2 Up lift pressure a. Up stream pile 29.5 b=21.5m, d1=2.9m, 7.41 , 4.24 2 .9 t=0.8(assumed) E1 C1 d1=2.9m d2=8m D1 Fig 7.1 D2 100 2 E COS 1 , C1 100 E =67.72% =32.28% 100 1 D cos 1 D1 100 D =22.32% = 77.68% Thickness correction for c1 d C1 c1 1 * t =2.75% d1 3.2 7.5 Stability analysis of weir Dynamic case Uplift pressure is considered for the weir wall. Water wedge weight is considered for weir crest only Unit weight of water and masonry is taken to be 9.81 and 22.4 KN/m 3 respectively. Moment is taken about the toe per meter widt HFL wHd Hd=He-Ha 4.65m PH1 PH1 H=3.3m PH2 W1 W2 PH2 y1 PH3 PH3 w(H+Hd)= wHt Pu 9m P2 P1 Fig.7.2 He=5.26m Ha=0.077m Hd=He-Ha=5.18m and y1=1.54m,y2=5.8m(by trial and error) Table 7.1 No Item Forces(KN) Lever Moments(KN- arm(m m) ) Vertical Horizo Overturning Restor ntal ing 1 PH1 5.18 * w * 3.3 167.69 1.65 276.7 2 PH 2 0.5 * w * 3.32 53.4 1.1 58.74 3 PU 0.5 * w 5.18 3.3 * 9 -374.35 6.00 2246.10 4 PH 3 0.5 * w *1.542 -11.63 1.03 11.94 5 WW w * Aw 504.53 9.77 4929.2 58 6 W1 m * A1 343.73 6.675 2294.4 7 W2 m * A2 160.8 2.9 466.32 8 Foundation reaction 3.65P1 P2 8.882P1 2P2 V 634.71 H 209.46 M R 7701.91 M O 2581.54 Safety factors MR 7701.91 Overturning stability, S o 2.98 1.5 Safe M O 2581.54 Sliding safety factor, Ss H 209.46 0.33 0.75 Safe V 634.71 Check for tension, M 5120.37 8.06 And for no tension e B / 6 V 634.71 B 9 e X 8.06 3.56 2 2 B e=0.24 1.22 No tension, ok! 2 Foundation reaction, 3.65P1 P2 V 421.06 P1 115.36 P2 8.882P1 2P2 M 755.4905 P1 2P2 85.06 115.36-P 2 +2P 2 =85.06 P2 31.36KN Down ward p1 146.66KN Upward Static case W1 PH1 PH W2 H H wH Pu Fig. 7.3 Table 7.2 Forces and moments acting on weir at static case N Item Forces(KN) Lever Moments(K o arm(m) N-m) Vertical Horizo Overturning Restorin ntal g 1 P 0.5 * * 3.15 H w 2 48.67 1.05 51.1 2 Pu 0.5 * w * 3.15 * 7.3 -112.79 4.87 548.91 3 W1 m * 3.15 *1.5 105.84 6.55 693.25 4 W2 0.5 * w * 5.8 * 3.15 204.62 3.87 791.2 V 197.67 H 48.67 M 600O M 1484.45 R Safety factors Overturning stability, S o M R 1484.45 2.47 1.5 Safe M O 600 Sliding stability, S s H 48.67 0.25 0.75 OK! V 197.67 Check for tension, x M 884.62 4.48 V 197.67 B 7.3 B 7.3 e x 4.48 0.83, 1.22 2 2 6 6 B e=0.83 1.22 OK! No tension. 6 3.3 7.6 Design of under sluice This structure has crest at low level to develop a deep channel pocket, which will help to bring low dry weather discharge to wards this pocket, there by ensuring easy division of water in to the canal through the head regulator. This opening will also help in scouring and removing the deposited silt from the under sluice pocket. Designed with the discharge of; 1) Twice the discharge of the off taking canal capacity Q=2*0.542=1.08m*3/sec 2) 20% of the max. flood, Q=0.2*250=50m*3/sec Therefore, Qsluice will be max. of the above. Qsluice=50m*3/sec Providing one under sluice with 2m width (divide wall is provided between the proper weir and the under sluice). Q 50 q 25m 3 / sec L 2 Scoured depth for the sluice section (R) 1 q2 3 R 1.35 , for f=2.74 f 1 25 2 3 R 1.35 8.25m 2.74 RL of bottom of scour depth on u/s side=U/S HFL-1.5R=1854.93-1.5*8.25=1842.56m. Therefore, the depth of the u/s pile, d 1 1846.87-1842.56=4.31m. RL of bottom of scour pile on d/s side= D/S HFL-2R=1853.93-2*8.25=1837.43m. Therefore, the depth of the d/s pile, d 2 =1846.87-1837.43=9.44m. Impervious floor Hs Min. length of d/s impervious floor, L2 3.87 10 Where H=Hs=3.15mC=9 (for boulder foundation Dr.K.A.Arora, 2002) 3.15 L2 3.87 * 9 =19.55m 20m. 10 Min. length of u/s impervious floor, L1 L L2 B 2d1 2d 2 28.35 20 7.3 2 * 4.31 2 * 9.44 26.47m Therefore, take nominal value of 2m for u/s length. Protection work Total length of d/s impervious floor and protection work H q 3.15 25 L2 L3 27C s * 27 * 9 * * 78.74m 10 75 10 75 Length of the d/s protection work, L3 L2 L3 L2 78.74 20 58.74m. this length is both inverted filter and launching apron. L3 58.74 Length of the u/s protection work, L4 29.37m. 2 2 3 Note; using broad crested weir formula, QS C d LH 2 Where H=weir height + He=3.15+5.32=8.47m. L=2m and Cd=1.7 Qs 1.7 * 2 * 8.47 2 84.06m 3 / sec Discharge through the under sluice. 3 3 3 And Qw C d LH 2 1.7 *10 * 8.47 2 209.21m / sec Discharge through the proper weir with 3 length, L=10m. Therefore, the total discharges, Qs Qw Qs 209.21 84.06 293.27m 3 / sec Q 250m 3 / sec ok 7.7 Design of head regulator It is provided at the head of the off taking canal and has the following objectives; To regulate the supply of water in to the canal To completely shutout the high flood from entering to the canal. To control the entry of silt to the canal. The regulation is provided by the gate which is fixed in such a way that, the discharge or desired capacity of water can easily flow in to the intake canal. The intake canal is placed so as the top level should be less than or equal to the crest level of the proper weir. Crest levels 1) Under sluice=the crest level of under sluice is equal to the river bed level=1846.87m. 2) Head regulator=is kept 1.2 to 1.5m higher than the crest level of the under sluice (say 1.5m) =1846.87+1.5=1848.37m. Bed level of canal=crest level of head regulator-canal flow depth =1848.37-0.65=1847.72m. 7.8 Design of retaining wall (Guide wall) To avoid out flanking of the river due to the control structure across the river a masonry guide wall is provided. Considerations; Analysis per meter span and moment heel Earth pressure at rest was considered m 22.4 KN / m , w 10 KN / m , soil 18KN / m 3 3 3 Drained angle of interval friction was considered The u/s wing walls are kept segmental with radius of 5 to 6 times head of water over the crest and subtending angle of 45 0 to 60 0 Therefore, u/s wing; R=5(He-Ha) =5*(5.32-0.41) =25m. Say 45 0 r * 25 * 45 Arclength 20m. 180 180 The d/s wing walls are kept straight for a length of 5 to 8 times the square root of the product of head of water over the crest and difference between U/S HFL and D/S HFL. Therefore, d/swing; L 5 H e H a * U / SHFL D / SHFL 5 5.53 0.41 * 1854.93 1853.93 11m. Upstream retaining wall Data available River bed level=1846.87m U/S HFL=1854.93m w 10 KN / m , m 22.4 KN / m , soil 18KN / m 3 3 3 Angle of repose 30 0 Top width=1.0m (source soil mechanics Arora) Free board(FB)=0.5m(assumed) Anchored depth below river bed =0.5m (source soil mechanics Arora) Therefore, height of wing wall H=Anchored depth+ (U/S HFL- river bed level) +FB H=0.5+ (1854.93-1846.87) +0.5=9m. Bottom width, B=50% TO70% Of H, Say 70% B=0.7*9=6.3m Ws1 Ww1 8.5m Fig. 7.4 Ws2 H=9m WW3 Ps1 5.3m 1m Table 7.3 Forces and moments acting on u/s retaining wall No Item Forces Lever Moments(KN- (KN) arm(m) m) Vertical horizontal Overturning restoring 1 Ws1 siol * A1 328.95 3.86 1271.94 2 Ws 2 siol * A2 153 5.8 887.4 3 Wm1 m * Am1 190.4 0.5 95.2 4 Wm 2 m * Am 2 409.36 2.43 996.1093 5 Wm3 m * Am3 70.56 3.15 222.264 6 Ps 0.5 * K o * soil * H 2 243 3 729 V 1152.27 H 243 M R 729 M o 3472.913 Safety factors Overturning stability, S o M o 3472.913 4.764 1.5, OK ! M R 729 H 243 0.211 0.65, OK ! Sliding stability, S S V 1152.27 Check for tension. x M 2743.913 2.38m; e B x 6.3 2.38 0.77 V 1152.27 2 2 B/6=6.3/6=1.05m B Since e=0.77m 1.05m no tensión. 6 Downstream retaining wall D/S HFL=1853.93m Free board (FB) =0.4m (assumed) Top width=1.0m (source soil mechanics Arora) Anchored depth below river bed=0.6m (source soil mechanics Arora) Therefore, H=0.6+ (1853.93-1846.87) +0.4=8m. Bottom width, B=0.7*8=5.6 1m WW1 WW2 Wm1 7.4m W M2 PS W M3 0.6m 4.6m Fig 7.5 forces acting on d/s retaining wall Table 7.4 Forces and moments acting on d/s retaining wall No Item Forces(KN) Lever Moments arm(m) (KN-m) Vertical Horizontal Overturning restoring 1 Ws1 soil * As1 239.36 3.4 815.184 2 Ws 2 soil * AS 2 133.2 5.1 679.32 3 Wm1 m * Am1 165.76 0.5 82.88 4 Wm 2 m * AM 2 298.368 2.2 656.41 5 Wm3 m * Am3 75.264 2.8 210.7392 6 Ps 0.5 * K o * soil * H 2 192 2.67 512 V 912.352 H 192 M O 2444.5332 M 512R Safety factors Overturning stability, S O M O 2444.5332 4.77 1.5, OK ! M R 512 H 192 0.21 0.65, OK! Sliding stability, S S V 912.352 Check for tension, x M 1932.5332 2.12m, e B x 5.6 2.12 0.682m V 912.352 2 2 B/6=5.6/6=0.93m