13. CONCEPT OF STORAGE ZONING Crest RL MDDL Surcharge Storage MWL Dead Storage Conservation Storage Flood Control Storage Top of dam Spillway Undersluices Dam River Free board FRL Rule Level
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16. Rule Curves A reservoir is operated according to prevailing water level, demands for various purposes, and the elevation of rule levels for different purposes To increase the flexibility of operation, different rule curves may be derived for different purposes depending on their priority.
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18. Min. required flow Natural flow D I S C H A R G E Non-damaging flow Water deficit Flood damage Demand Time Min. flow deficit
24. CONSERVATION OPERATION Operation of a reservoir using Rule Curves: If reservoir level falls below LRC, meet only partial highest priority demands Basic principle – It is always better to meet reduced demands throughout the year than to meet full demands in the beginning with no water at the end.
61. Computer Programming For Gate Operation ( PGCP.EXE ) Data File : PGDAT Line 1 : Name of Scheme Line 2 : Hd CRL FRL Avg Bed RL No.of Gate Wg Hg 0 Line 3 : X-ordinate of Gate Seat Radius of Gate Xt Yt 1 0 Example : PGDAT SURVO WATER RESOURCES PROJECT 20 307 575 327 588 303 17 16 30 20 0 5.712 24 28.661 6.0367 1 0 Output File : File 1 PGCPO File 2 PGOUT Note : Conversion Factor from meter to ft. is only 3.28 Correct. Because 3.28 is used for programming.
64. Plate - 1 Note : 1.Spillway is equipped with 7 No.of Vertical Lift Gate of Size 9.14 m X 3.05 m.( 30’ X 10’ ) 2. All Gates are assumed to be opened uniformly. 3. Gate opening Represents vertical distance of lower edge of gate above gate seat in cms. 4. As the FRL & MWL of Kaniyad W.R.Project are same ( 02.25m.) Free & Partial discharge curves up to RL 102.25m.have been drawn. 5. Partial Curve indicates discharge through all Gates. 6. Transition Zone in corporate here is subject to modification on model Run.
65. (A)Operation guidelines for reservoir condition below FRL When flood starts impounding reservoir, the following guide lines may be observed. 1) Note the rule level fixed for the current period. 2) Note the PWL and its corresponding discharging capacity at current elevation from free discharge curve in Plate-1. 3) If the PWL is below rule level , fill the reservoir up to scheduled rule level. Example: Current rule level = 46.00m Current reservoir level = 43.00m Operation: Fill the reservoir up to RL 46.00m
66. 4) Thereafter, release outflow corresponding to inflow rate by gate operation either partially or under free discharge from the graph as shown in Plate-1 or Table no.1 Example: Radial gates 22 No. Size 14.93m x 10.67m FRL = 53.10m Current rule level = 46.00m Current reservoir level = 46.0m Q free at RL 46.00m = 3965 m 3 /s Operation: Generally there will be two cases (a) Inflow rate is less than free discharging capacity. ( Q = 2790 m 3 /s ) (b) Inflow rate is more than free discharging capacity.( Q = 6000 m 3 /s ) So for each case, it would be necessary to maintain current rule level by controlling the discharge in the following way. a)Example: For inflow rate = 2790 m 3 /s (Less than free discharging capacity 3965 m 3 /s), reduce outflow equal to inflow by opening all gates partially with gate opening =180cm b) Example: For inflow rate = 6000 m 3 /s (more than free discharging capacity 3965 m3/s), allow free discharge of 6000 m 3 /s by opening all gates above water profile( above RL 46.00m). So that the level will build up resulting in to increased free outflow.
67. 5) Repeat this procedure for making outflow equal to inflow approximately until reservoir elevation stops raising and information is received from u/s wireless system that the recession of flood has been started. 6) Operate gates at 1 hr. to 2 hr. intervals , following Plate-1 and decrease outflow gradually when the flood starts receding and rate of rise in water level decreased or fall in water level has been observed. During the process of fall in water level of reservoir, take care that the water level does not fall below scheduled rule level, by suitable gate operation , keeping the outflow restricted within maximum safe carrying capacity of river channel d/s of dam. 7) Close all the gates gradually when water level in the reservoir is again back to the scheduled rule curve level. 8) Know Inflow from forecast station in u/s of dam or u/s projects. 9) To know inflow from rise/fall of water level in reservoir when there is no existence of forecast station in u/s of dam for information of inflow rate incoming when inflow rate can be decided earlier observed rate of rise or fall in the reservoir water levels by using Plate-2
68. How Plate-2 can be used is shown in the below illustrated example .
69. Example : Survo.dat SURVO WATER RESOURCES PROJECT 16 93.75 99.85 93.75 94.0 94.5 95.0 95.5 96.0 96.5 97.0 97.5 98.0 98.5 99.0 99.5 100.0 100.5 101.0 0.87 1.01 1.48 1.95 2.56 3.16 4.11 5.05 6.20 7.35 8.93 10.50 12.50 14.61 17.18 19.75 For Rate of Change of Storage per Hour in Equivalent Discharge Computer Programme : RMK.EXE Line 1 : Name of Scheme Line 2 : No.of RL-Cap Ordinates CRL FRL Line 3 : RL Ordinates. Line 4 : Capacity Ordinates.
72. USE OF PLATE ; 2 ( Table Showing Rate of Change of Storage per Hour in Equivalent Discharge) CASE-I a)Observations:---- Reservoir level at 10:00hr 46.00m Reservoir level at 10:30hr 46.15m So, rise in water level = (46.15 – 46.00)/ 0.50 hr = 0.30m / hr (30cm/hr.) Outflow released between 10:00 to 10:30hr –Nil Earlier observed rate of rise in the reservoir 2)Actions:- i) Read change in storage. Flow rate from Plate-2 or Table no.3 corresponding to current elevation of 46.00m and rise/fall rate of 0.30m/hr @ 10:00hr is 240 m3/s. ii) Now inflow rate on average can be found out by using the following storage equation I – O = ∆ S ∆ S +ve for Rise Where, I = Average Inflow rate I = ∆ S + O --ve for Fall O= Average Outflow rate = 240 + Nil = 240 m 3 /s Therefore, average Inflow rate during 10:00hr to 10:30 hr =240 m 3 /s. Therefore from Plate-1 or Table no.2 release 240m 3 /s with 15cm by partial opening of all gates if conservation of storage is not required.
73. CASE-II Observations: Reservoir level @ 12:00hr 47.00m Reservoir level @ 12:30hr 47.60m Therefore, rate of rise is (47.60 – 47.00) / 0.5hr = 1.20m/hr (120cm/hr.) Outflow released between 12:00 to 12:30 hr =500 m 3 /s From Plate-2 or Table no.3, flow rate corresponding to RL 47.00m and rise/fall rate of 1.20m/hr is read as 1150 m 3 /s. So, by equation I = ∆S + O (Where O is the average outflow rate) = 1150+500 = 1650m 3 /s Therefore average inflow rate between 12:00 to 12:30 hr = 1650m 3 /s 8.1.0 Release outflow as shown in Plate-1 corresponding to inflow rate worked out. To release Q = 1650m 3/ s , open all gates, if conservation of storage is not required. 8.2.0 When water level starts to decrease after peak inflow, then flood is on recession and therefore decrease outflow gradual say at an interval of 1hr to 2hr and allow water level to decrease up to rule level prescribed, but not below the Rule curve level. 8.3.0 Close all the gates where inflow rate under observation become equal to base flow rate and ensure that no successive flood is incoming in to reservoir.