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ENGINEERING DESIGNS-CANAL STRUCTURES GENERAL DESIGN PRINCIPLES

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Presentation on theme: "ENGINEERING DESIGNS-CANAL STRUCTURES GENERAL DESIGN PRINCIPLES"— Presentation transcript:

1 ENGINEERING DESIGNS-CANAL STRUCTURES GENERAL DESIGN PRINCIPLES
TRAINING PROGRAMME ON ENGINEERING DESIGNS-CANAL STRUCTURES GENERAL DESIGN PRINCIPLES Off Take Sluices, Canal outlets and Modules BY ROUTHU SATYANARAYANA FORMER CHIEF ENGINEER FORMER ADVISOR, GOVERNMENT OF A.P

2 Modular Outlets DESIGN - EXAMPLES HYDRAULIC DESIGN OFM – APM - PSM

3 Cross Masonry Works- Modular Outlets
Semi-Modular outlets or Flexible modules Open Flume Outlet (OFMs) Crump’s Adjustable proportional Module (Crump’s APM) Adjustable Orifice Semi module (AOSM) Pipe Semi Module – Pipe Out let (PSM)

4 Modular Outlets The suitability of the type of the semi module outlet is determined based on the ratio of parent canal discharge (Q) to the discharge of the out let (q) and the throat width (Bt) as detailed below. i) for (Q/q ) < or = 20 and B t ≥ 6 cm Open Flume Module( OFM) ii) for (Q/q ) < or = 20 and B t < 6 cm Adjustable Proportional module ( APM ) for (Q/q ) > 20 If the above requirements do not suit the site condition, provide pipe semi module (where possible) with diaphragm of required diameter inserted at the first joint. The minimum diameter of pipe used will be 150 mm.

5 Modular Outlets The above conditions are further explained as below
Arrive at the ratio of parent channel / out let channel. If it is < or = 20, select OFM. Calculate the Bt ( throat width ), using weir formula. If Bt is > 6 cm it is ok. Otherwise select A.P.M. Work out the Bt using the sluice formula setting the crest of outlet at less than 0.80 D from FSL of Parent Channel and adjusting the height of outlet opening. If Bt = or > 6 it is ok Otherwise go for pipe semi module (PSM), if it is possible to do so. Check for proportionally

6 Canal Outlets & Modular Outlets
Open Flume Module(OFM): A smooth weir with a throat constricted sufficiently to ensure velocity above critical, and long enough to ensure that the controlling section remain with in the parallel throat at all discharges up to the maximum. A gradually expanded flume is provided at the outfall to recover the head. The entire work can be built in brick, but the controlling section is provided with cast iron or steel bed and check plates. Throat width Bt never kept less than 6cm hence it is necessary to raise the crest level of the out let much above bed level. Working head required with a 1 in 5 glacis and side walls splaying at 1 in 5 is 20% of depth of water above the crest of out let. Formation of hydraulic jump makes the out let independent of water level in the out let channel

7 Canal Out lets & Modular Outlets
Open Flume Module (OFM): Q in Cumecs = K Bt H3/2 Where K = a constant depending upon the width of the flume Bt = throat width of the weir (minimum 6 cm) For Bt , 6 cm to 9 cm K=1.6 9 cm to 12 cm K= 1.64 > 12cm K= 1.66 H = head over crest in m. It is most suited to tail cluster and proportionate distribution. Length of throat (crest) = 2H Setting H = 0.9 D, where D is FS dept of Parent canal Minimum working head (modular ) = 0.2 H Crest level = U/S FSL – 0.9 D U/S approach to throat one curved and diverging and another straight D/S expansion splayed to 1 in 10 to meet the bed width of the out let channel

8 Canal Out lets & Modular Outlets
HYHDRAULIC DESIGN OF OFM

9 Canal Out lets & Modular Outlets
Open Flume Module (OFM): HYDRAULIC PARTICULARS S. No. Details Unit Parent Canal Off Take Kopparam Major 13 R Minor 1 Discharge (Required) Cumecs 0.939 0.080 2 Bed Width m 1.200 0.300 3 F.S. Depth 0.730 0.310 4 Free Board 0.610 0.460 5 Bed Fall 1 in 2640 2000 6 Side Slopes 1.5 : 1 7 Velocity m/sec 0.561 0.339 8 n 0.020 9 CBL 81.589 81.662 10 FSL 82.319 81.972 11 TBL 82.929 82.432 12 GL

10 Design of OFM

11 Canal Out lets & Modular Outlets
Open Flume Module (OFM): Hydraulic Design: Ratio of Q/q = / 0.08 = < 20 Hence O.F.M. is Proposed. FSD in the parent channel, D = m Depth of flow in the parent channel at 2/3rd discharge: Dn = m (assumed) For 2/3rd Discharge in Parent Canal Qn = 2/3 x Q = cumecs Let Dn = m , Area = m2 Perimeter = m, R = 0.372, and R2/3 = 0.517 Velocity = m/sec Discharge = cumecs Hence Dn i.e., depth of flow of m in the parent channel assumed for 2/3rd discharge is correct. Difference in depths of flow for full discharge and 2/3rd discharge x = D - Dn = = m

12 Canal Out lets & Modular Outlets
Open Flume Module (OFM): The discharge of the outlet q (in cumecs) is given by the formula: q = K x Bt x H3/2 Where K = Constant depending upon the width of flume Bt = Throat width of weir H = Head over crest in meters Hn = Head over crest at 2/3 Q = ( H – x ) For full discharge q = K x Bt x H3/ ( 1 ) For 2/3 discharge q = K x Bt x (Hn)3/2 = K x Bt x (H-x)3/2 - ( 2 ) By dividing equation (1) by equation (2) q / 2/3 q = K x Bt x H3/2 / 2/3 (q) K x Bt x (H - x)3/2 3/2 = (H3/2 /(H - x)3/2 ) 2/3 (3/2)2/3 = (H3/2 /(H - x)3/2 ) 2/3 1.310 = H / (H - x) H = x H = x x = m Hence, H = x = m Crest Level = F.S.L. of Parent Channel - H = = m

13 Canal Out lets & Modular Outlets
Open Flume Module (OFM): Minimum Head required (modular) = 0.2 x H = 0.2 x = m Available Working Head = Difference of FSL's in Parent Canal and Off - Take Canal = = m OK q = K x Bt x H3/2 Value of "K" as per IS: = 1.64 Bt = q / (K . H3/2) = / 1.64 x (0.575)3/2 = m = 1.64 x x /2 = cumecs OK For 2/3rd q: qn = (2/3 x 0.080) = cumecs Hn = H - x = = m qn = K x Bt x Hn3/2 = 1.64 x x /2 = cumecs OK In order to maintain the modularity of the outlet, the following parameters are adopted. Bt = m H = m Throat Length = m Crest Level = m

14 Design of OFM

15 Open Flume Module

16 OPEN FLUME ModuleT

17 Canal Out lets & Modular Outlets
HYHDRAULIC DESIGN OF APM

18 Canal outlets & Modules
Adjustable Orifice Semi Module (A.O.S.M) or Adjustable Proportional Module (APM)  Discharge through outlet in cumecs Q = Bt Y Hs1/2 Y =Height of opening in meters. Bt =Throat width (minimum m ) H =Depth of water in parent canal over the crest in meters Hs = Depth to under side of the roof block below FSL of parent canal. Hs = H – Y , Hs ≤ 0.80 D y > (2/3 ) H Setting of crest, H = x D , where D = Full supply depth in the parent canal Setting of crest shall not be below D/S B.L.

19 Canal outlets & Modules
Setting of crest shall not be below D/S B.L. Minimum modular head Hm = Hs for modularity between full supply and any fraction of full supply. Crest level ≈ U/S FSL D Length of throat = width of roof block + H U/S slope of glacis = curve with radius 2H. U/s approach wings =one curved and the other straight, top at FSL m D/S expansion =1 in 10 to meet bed width of outlet channel

20 Design of APM Hydraulic Particulars:
S.No. Details Kuppam Major DP 17-R Kuppam major DP 17-R At FS Discharge Condition At 2/3 FS Discharge Condition Discharge in Cumecs Bed width in m FSD in m Bed fall 1 in Value of ‘ n ‘ Velocity in m/s Side slopes 1.5 : : : : 1 Free board in m CBL m FSL m TBL m GL m Propose APM if Q / q < 20, or Q / q > 20, and Bt < 60 mm Ratio of Q / q = / = > 20 Hence APM is proposed.

21 Design of APM

22 Design of APM Hydraulic Design of APM
Propose APM if Q/q < 20 and Bt < 6 or Q/q >20 Ratio of Q/q = / = > 20 Hence A P M is proposed q = 4.03 x Bt x y x Hs1/2 Where Y = Height of opening in meters Bt = Throat width H =Depth of water in parent canal over the crest in meters. Hs = Depth of under side of the roof block below FSL of parent canal For 2/3 full supply Parent canal Off take canal Qn =2/3Q = cum qn =2/3*q = cum             Let Dn = m Let dn =               Area = sq.m   sq. m             perimeter = m   m             R = m    m  R3/2 =                   velocity = m/sec   m/sec             discharge =                  

23 Design of APM Discharge through the outlet in cumecs.
q = x Bt x y x ( Hs ) ½ (Page 11, Clause of IS : ) Throat width Bt = q / (4.03 x (Hs)½ x Y) Let Bt = 0.06 m q = 4.03 x Bt x y x Hs1/ equation (1) 2/3 (q) = = 4.03 xBt x y x (Hsn)1/ equation (2) Where Hsn = Hs - (D - Dn) = 4.03 x Bt x y x (Hs - ( ))1/2 = 4.03 x Bt x y x (Hs )1/2 Dividing equation (1) by (2) q/ = 2/3 (q) = x Bt x y x Hs1/2 /  4.03 x Bt x y x (Hs ) ½ 3/2 = Hs1/2 / (Hs )1/2 = 2 x Hs1/2 = 3 x (Hs )1/2 4 x Hs = 9x (Hs ) 5 Hs = Hs = m and Hs1/2 = 0.455 Hsn = Hs – ( D – Dn ) = – ( – ) = 0.092m

24 Design of APM Y = q / 4.03xBtx(Hs)1/2 = / 4.03 x 0.06 x Y = m say m Q = 4.03 Bt Y ( Hs ) ½ = 4.03 x 0.06 x x = cumecs qn = 4.03 Bt Y ( Hsn ) ½ = 4.03 x 0.06 x x = cumecs The following Parameters may be adopted Bt = m Crest Level = U/S FSL - H = H = m Y = m Hs = m W= Setting forward of the D/S wing wall of the approach =0.15 The setting of the Wing wall W = K q/Q (Bu + D/2)

25 Design of APM

26 CRUMP’S ADJUSTABLE PROPORTIONAL MODULE

27 OPEN FLUME OUTLET WITH ROOF BLOCK

28 Canal Outlets & Modules
Pipe semi module (PSM) Design criteria The discharge through pipe semi module is given by Q = Cd . A (2g hc )1/2   Where Cd = 0.62 for free pipe out let hc = head on U/S above the centre of pipe  hc should be more than 1.5 times the dia of the pipe proposed. The above formulae can be reduced to Q = 0.62 x √ (2x 9.81 ) A √ (hcnt)   =2.746 A hc 1/2   For free fall condition set the F.S.L of OT Channel below the pipe sill level keeping in view the command under the pipe sluice .It is a simplest type and the users will appreciate.

29 Modular Outlets DESIGN OF PIPE SEMI MODULE (for DIRECT OUTLET )
HYDRAULIC PARTICULARS DESCRIPTION UNIT ` PARTICULARS PARENT CHANNEL OUTLET CHANNEL   1 Ayacut Acres   2 Discharge (Required/Designed) Cumecs / /   3 Bed width m   4 Full Supply Depth m   5 Free board m   6 Side slopes (Inner / Outer) / 2 : / 2 : 1   7 Top width of banks L/R m / /   8 C.B.L. m   9 F.S.L. m   10 T.B.L. m   11 G.L. M  

30 Modular Outlets   Discharge through the Pipe Outlet by using Orifice formula: Q = CA (2g hcnt)1/2 where C = Q = Discharge of Out Let A = Cross section Area of Pipe h(cnt) = Head of U/s water surface over the center of pipe Where C = 2.476 Provide Dia of Pipe = 16 cm Area A = m2 Q = cumec hcnt = Q CA 2 = x0.02 = m > 1.5 times dia of pipe Provide did. of pipe = 200mm

31 Hydraulic Design of PSM
The Center line of the Pipe = – = m Pipe sill level (Invert Level) = (pipe radius) = m Height of pipe above CBL = – = m Pipe invert level = m FSL of 17-R = m Hence the pipe is at free fall condition

32 Design of PSM

33 PIPE OUTLET

34 THANK YOU


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