design of barrage requires lot of considerations and practices this is a guide for itFull description
Descripción: sim 7 ael 215
Descripción completa
jurnal
Descripción: z
Les barrages sont des ouvrages en terre ou en béton établis au travers d’un lit de cours d’eau entre l’amont et l’aval. Ils servent à l’alimentation en eau potable, à la production d’énergie…Description complète
104
Roll No.
M.A.B.F.
Maximum Discharge = Q max= Minimum Discharge = River Bed Level (RBL)=
Highest Flood Level (HFL)= No. of canals on right bank= No. of canals on left bank= Slope of Canal=
AFFLUX= Crest Height P= D o =HFL-RBL=
1. MINIMUM STABLE WETTED PERIMETER Pw=
1800
LLC=
1.8
W a= 67N+35= N=
3240.0 3260.0 48 48 45 1 2
Bays @60ft= Piers @7ft= Fish ladder = Divide Walls =
Total W a=
3249
Discharge b/w Abutments=qabt=Qmax/Total W a Dischargeover the crest =qweir=Qmax/W clear
2. CALCULATION OF LACEY`S SILT FACTOR f=
2.03
3. FIXATION OF CREST LEVEL Maximum Scour Depth R=0.9(qabt2/f)1/3 Ho=R-P= Vo=qabt/R ho=Vo2/2g= Eo=Ho+ho E1=Do+ho +AFFLUX USEL
Level of E1= Crest Level =
Maximum D/s Water Level =
RBL+E1 Level of E1-Eo
h=D/s WL - CL= h/Eo=
C'/C= C'= Q=C'xW clearxEo3/2
% Differene =
NOTE. If this %age difference is 2 then its OK and if greater than 2% then increase P(Crest Level) and if less than LLC =Total W a/Pw
4. DESIGN OF UNDERSLUICES 3
Fix Crest level Crest level of undersluices=
3 bays to act as Undersluices on both sides b1 = Assuming qus=120% of main weir R=0.9(qus2/f)1/3 Vo=qus/R ho=Vo2/2g Maximum USEL= HFL+Afflux+ho Eo=Maximum USEL-CL of Undersluices= h=DSEL-CL of undersluices= h/Eo= C''/C= C''= Q1 and Q2= C''x(b1x2)x(Eo)3/2= Qmain weir = C'x(W clear-2b1)x(Eo)3/2= Total Discharge = Check =Total Q>Qmax %age water passing through undersluices= Hence The Undersluices are fixed at crest = No. of Bays on Each Side =
5. DETERMINATION OF WATER LEVELS AND ENERGY LEVELS 5.1 CHECH FOR MAIN WEIR Wclear= %
Q
DSWL
120 100 50 25
(cusecs) 544500 453750 226875 113437.5
(ft) 600.5 600 597.5 591
120 100 50
544500 453750 226875
596 595.6 590.5
120 100 50 1
544500 453750 226875 2
603 602.5 601.5 3
5.2 CHECK FOR UNDERSLUICES with 20%Concentration, Q = 1.2 x Q1 and Q2 =
Normal State Retrogressed State Accreted State Total width of Undersluices = Crest Level of Undersluices =
Items D Vo ho h Ho Eo Eo3/2 h/Eo
Normal 22 11.31 1.99 13.75 17.25 19.24 84.40 0.71
C`/C
0.86
C` qclear Q
3.268 275.83 99298.78
6. FIXATION OF D/S FLOOR LEVEL AND LENGTH OF D/S GLACIS AND D/S FLOOR.
6.1 FIXATION OF D/S FLOOR LEVELS FOR NORMAL WEIR I) Normal state of river a) For qclear = USEL = USWL + ho = DSEL = DEWL + ho = hL =USEL-DSEL =
E2 = DSFL = DSEL - E2 = b) For qclear = USEL = USWL + ho = DSEL = DEWL + ho = hL =USEL-DSEL =
E2 = DSFL = DSEL - E2 = c) For qclear = USEL = USWL + ho = DSEL = DEWL + ho = hL =USEL-DSEL =
E2 = DSFL = DSEL - E2 = II) Retrodressed state of river Q 544500 453750 226875
qclear 199.58 164.68 80.10
III) Accreted state of river Q 544500 453750
qclear 195.11 161.97
226875
83.92
Worst Condition occurs at
D/S Floor Level = 6.2 FIXATION OF FLOOR LEVELS FOR UNDERSLUICES Q 99298.78 98554.82 95958.10
Normal Retrogressed Accreted Therefore undersluices floor level will be fixed at
Undersluice floor level =
7. FIXATION OF D/S FLOOR LEVEL FOR NORMAL BARRAGE SECTION USING CRUMP a) Q = Maximum DSWL= USWL = USEL = RBL = Crest Level = DSFL = Dpool = D/S Velocity = Q/(DpoolxTotal W a)= D/S Velocity Head = V2/2g = DSEL= Max DSWL+D/S Velocity Head = K= USEL-CL = L = USEL - DSEL = q clear = Q/W clear = C=Critical Depth =(q2/g)1/3 L/C =
(K+F)/C = F= Level of Intersection of Jump with Glacis= Crest Level -F = E2 = DSEL - Level of Intersection of Jump with Glacis =
Submergency of Jump = Level of Intersection of Jump with Glacis-D/S Length of Glacis D/S Jump = 3xSubmergency of jump =
Length of Stilling Pool =4.5E2 = Length of D/S Floor= Length of Stilling pool - Length of Glacis D/S of J
b) Q = Minimum DSWL = USWL= USEL = Dpool = Min DSWL - D/S Floor Level = D/S Velocity = Q/(DpoolxTotal W a)= D/S Velocity Head = V2/2g = DSEL = Min DSWL + D/S Velocity Head = K =USEL - CL = L=USEL -DSEL = qclear = C=Critical Depth =(q2/g)1/3 L/C =
(K+F)/C =
F= Level of Intersection of hydraulic jump with Galcis = CL -F = E2 = DSEL - Level of Intersection of hydraulic jump with Galcis= Submergency of Jump = Level of Intersection of Jump with Glacis-D/S Length of Glacis D/S of Intersection = 3xSubmergency of Jump = Length of Stilling Pool = 4.5xE2 = Length of D/S Floor = Length of Stilling Pool - Length of Glacis D/S of
Hence we shall provide D/S Floor 8. FIXATION OF D/S FLOOR LENGTH FOR UNDERSLUICES. a) Maximum DSWL = Q= USWL = USEL =USWL +ho = DSFL = Dpool = Max DSWL - DSFL = D/S Velocity = Q/(Dpoolxb1)= D/S Velocity Head = V2/2g = DSEL =Max DSWL + D/S Velocity Head = K =USEL - CL of undersluices = L=USEL -DSEL = q =Q/b1x2= C=Critical Depth =(q2/g)1/3 L/C =
(K+F)/C = F= Level of Intersection of hydraulic jump = CL -F =
E2 = DSEL - Level of Intersection of hydraulic jump = Submergency of Jump = Level of Intersection of Jump-D/S Floor level Length of Glacis D/S of Intersection = 3xSubmergency of Jump = Length of Stilling Pool = 4.5xE2 = Length of D/S Floor = Length of Stilling Pool - Length of Glacis D/S of
b) Minimum DSWL = Q= USWL = USEL =USWL +ho = DSFL = Dpool = Max DSWL - DSFL = D/S Velocity = Q/(Dpoolxb1)= D/S Velocity Head = V2/2g = DSEL =Max DSWL + D/S Velocity Head = K =USEL - CL of undersluices = L=USEL -DSEL = q =Q/b1x2= C=Critical Depth =(q2/g)1/3 L/C =
(K+F)/C =
F= Level of Intersection of hydraulic jump = CL -F = E2 = DSEL - Level of Intersection of hydraulic jump = Submergency of Jump = Level of Intersection of Jump-D/S Floor level Length of Glacis D/S of Intersection = 3xSubmergency of Jump = Length of Stilling Pool = 4.5xE2 = Length of D/S Floor = Length of Stilling Pool - Length of Glacis D/S of
Hence we shall provide D/S Floor
9. CHECK FOR ADEQUACY FOR D/S FLOOR LEVELS USING CONJUGATE DEPTH MET 9.1 FOR NORMAL WEIR SECTION Note. For Determination of z and z` see Sheet CIVIL 03F Q = Discharge in river (cfs) Q1 = Discharge through the main weir = 80% of Q USEL E = USEL - DSFL
q = Intensity od Discharge on D/S Floor = Q1/2400 Dpool = Depth in Stilling Pool =DSWL - DSFL E3/2 f(z) = q/E3/2
z Conjugate Depth Cofficients
z` d1 = z x E d2 = z` x E
Conjugate Depths
Jump Submergency = Dpool - d2 Remarks
9.2 FOR UNDERSLUICES SECTION D/S Floor level of undersluices = ` Q = Discharge in river (cfs) Q1 = Discharge through undersluices with 20% concentration USEL E = USEL - DSFL (for undersluices) q = Intensity od Discharge on D/S Floor = Q1/(2xb1) Dpool = Depth in Stilling Pool =DSWL - DSFL E3/2 f(z) = q/E3/2
Safety Factor =1.75 for D/S Floor Critical Condition. Dept R` =1.75xR = Minimum D/S Water Level for the 0.457 million cfs Discharge = D/S Apron Level = Depth of Water on Apron = Add 0.5 ft increase in depth for concentration D` = Depth of Water with Concentration = R` - D` = Length of Apron to Cover a surface of scour at 1:3 Slope =
Therefore the length of D/S Stone Apron in Horizontal Position =
10.2 U/S SCOUR PROTECTION Safety Factor =1.25 for U/S Floor Critical Condition. Dept R` =1.25xR = Minimum U/S Water Level for the 0.457 million cfs Discharge = U/S Apron Level = Depth of Water on Apron = Add 0.5 ft increase in depth for concentration D` = Depth of Water with Concentration = R` - D` = Length of Apron to Cover a surface of scour at 1:3 Slope =
Therefore the length of U/S Stone Apron in Horizontal Position =
10.3 THICKNESS OF APRONS Fall in inch /mile Sand Classification Very Coarse Coarse Medium Fine VeryFine
3 Thickness of Flexible protection at 16 22 28 34 40 Thickness of Stone Apron in Hz. Position =1.75 x 34/12 =
SUMMARY Total length of D/S Stone Apron = 4' Thick Block Apron = 5' Thick Stone Apron = Total length of U/S Apron = 4' Thick Block Apron = 5' Thick Stone Apron =
10.4 SCOUR PROTECTION FOR UNDERSLUICES Assume 20 % Concentration, q = 1.2x qweir R = 0.9(q2/f)1/3=
Minimum D/S Water Level for the 0.48 million cfs Discharge =
Add 0.5 ft increase in depth for concentration
Length of Apron to Cover a surface of scour at 1:3 Slope =
Therefore the length of D/S Stone Apron in Horizontal Position =
Minimum U/S Water Level for the 0.48 million cfs Discharge =
Add 0.5 ft increase in depth for concentration
Length of Apron to Cover a surface of scour at 1:3 Slope =
Therefore the length of D/S Stone Apron in Horizontal Position =
SUMMARY Total length of D/S Stone Apron = 4' Thick Block Apron = 5' Thick Stone Apron = Total length of U/S Apron = 4' Thick Block Apron = 5' Thick Stone Apron =
11. INVERTED FILTER DESIGN Note.
Write from Page 100 of Book by Dr. Iqbal Ali. 12. DESIGN OF GUIDE BANKS I) Length of guide bank measured in a straight line along th barrage U/S is L u/s = 1.5x Total W a = II) Length of guide bank D/S of barrage L d/s = 0.2x Total W a = III) For the nose of U/S Glacis bank and the full length of D/S guide bank use Lacey`s Depth =1.75xR =
For the remaining U/S Guide bank Lacey`s Depth = 1.25 xR = IV) Possible Slope of Scour = 1: 3 V) Free Board U/S = Free Board D/S =
7 6 These freeboards also include allowance for Accretion.
VI) Top of Guide Bank width = VII) Side Slope of Guide Bank =1:2 VIII) Minimum Apron Thickness = Length of Barrage = Length of U/S Guide Bank = Length of D/S Guide Bank = Radius of U/S Cueved Part = Radius of D/S Cueved Part = Maximum U/S Angle Protected = Maximum U/S Angle Protected =
12.1 DETERMINATION OF LEVELS OF GUIDE BANK Bed Level = D/S HFL with Accretion = D = D/S HFL - BL = C= Chezy`s Coefficient = U/S HFL with Accretion = d1 = U/S HFL -BL =
Slope of river bed= 1/5000 = Assume d2 = d1/D = d2/D = f(d1/D) = f(d2/D) = Substituting the values in the formula, L= Once d2 is fixed the levels of guide bank can be determined Rise in RBL = Length of U/S guide bank /5000 = Water level along h/w axis at
4873.500
I) Level at the nose of U/S guide bank = II) Level at the barrage = HFL + Freeboard = III) At D/S guide bank Water level D/S of Barrage = Freeboard = Level of uide Bank D/S =
13. DESIGN OF GUIDE BANK APRON Working on the same lines as in section 10. Length of Unlauched (horizontal) Apron = 2.5 (R' - D) Length of lauched Apron at 1:3 Slope = 31.6 (R' - D) As Calculated Previously t= say
34 3
Volume of Stone in Apron = 3x(32+12)1/2(R'-D) =9.5(R'-D) Minimum thickness of unlaunched apron = 1.07 x3 = Mean Thickness of unlaunched Apron =9.5(R'-D)/2.5(R'-D) = Maximum thickness of unlaunched apron = 2 x 3.8 -3.2 = say t= Area Nose of Guide bank Transition from nose to straight Straight reaches of guide bank
Rage of R'
2.0 R to 2.5 R 1.25R to 1.75R 1.0R to 1.5R
14. DESIGN OF MARGINAL BUNDS I) Top Width =
20
II) Top Level to be 5 ft above estimated HFL after allowing for 1.5 ftof Accretion. III) Front Slope of Marginal Bunds is 1:3 (notpitched) IV) Back Slope to be such as to provide a minimum of 2ft cover, over a hydraulic gradient of 1:6 V) U/S water level at nose of guide bank = 611.77 Free board of Marginal Bund = Level of Marginal Bund =
CALCULATION OF LENGTH OF BACKWATER CURVE d1 = U/S HL with Accretion - RBL =
22
D = Minimum Pond Level - RBL = Slope of Canal = C= Chezy`s Coefficient = g=
16 0.0002 71 32.2
Length of Backwater Curve =
24.26
Blue Value Take From 3.1(b)Curve Green Value take from 3.1(c) Curve Brown Value Take From Blench Curve Orange Value Take From Crump`s Curve Violet Value Take From Table for Cojuga Red Value Put Yourself
M.A.B.F. 453750 cusecs 12000 cusecs 582 ft
600 2 1 1 ft/canal mile
0.0002
3 ft 6 ft
NOTE.
First Adjust Crest height such th
18 ft
ft ft
2880
2880 315 24 30
ft ft ft ft
W clear
ft 2 139.7 ft /sec 2 157.55 ft /sec
19.15 ft 13.15 ft 7.29 ft/sec 0.83 ft 14.08 ft 21.83 ft 603.83 ft 589.75 ft
600 ft
10.25 ft 0.73
0.84 3.192 485690 cusecs
7.04
%
then increase P(Crest Level) and if less than 2% then decrease P to make it near to 2% 1.81
feet below the main weir. 586.75 ft 180 ft 2 189.06 ft /sec
23.43 ft 8.07 ft/sec 1.01 ft 604.01 17.27 14.26 0.83 0.76 2.89 74669
ft ft ft
cusecs
424713 cusecs 499381 cusecs OK 16.5 %
9.1%
586.75 3
ERGY LEVELS
2880 Afflux
CL=
589.75 For Normal State USWL=DSWL+ R=USWL-RBL Afflux
= Level of Intersection of Jump with Glacis-D/S Floor level =
ump = 3xSubmergency of jump =
ength of Stilling pool - Length of Glacis D/S of Jump =
11.01 ft 33.02 ft
71.65 ft 38.63 ft
Say
39.00 ft 453750 cfs 595.6 601.6 602.60
S Floor Level =
19.60 7.13 ft/sec 0.79 ft 596.39 ft
D/S Velocity Head = 12.86 ft 6.21 ft 157.55 cfs 9.17 ft 0.68
2.8
12.82 ft hydraulic jump with Galcis = CL -F = ntersection of hydraulic jump with Galcis= = Level of Intersection of Jump with Glacis-D/S Floor level = f Intersection = 3xSubmergency of Jump =
576.93 19.46 ft 0.93 2.79 ft 87.56 ft
Length of Stilling Pool - Length of Glacis D/S of Intersection =
Say
84.78 ft
85.00 ft
l provide D/S Floor
85.00 ft, long.
SLUICES. 603 89602 605 606.82 572.00 31.00
ft cfs
CL of Undersluices=
ft ft ft
8.03 ft/sec
D/S Velocity Head =
1.00 604.00 20.07 2.82 248.90
ft ft ft ft cfs/ft
12.44 ft 0.23
1.95 4.18 ft hydraulic jump = CL -F =
582.57
ntersection of hydraulic jump = = Level of Intersection of Jump-D/S Floor level = f Intersection = 3xSubmergency of Jump =
21.44 ft 10.57 ft 31.71 ft
96.46 ft Length of Stilling Pool - Length of Glacis D/S of Intersection =
64.75
Say 596 89602 602.5 604.79 572.00 24.00
ft cfs
65.00
CL of Undersluices=
ft ft ft
10.37 ft/sec 1.67 597.67 18.04 7.12 248.90
D/S Velocity Head =
ft ft ft ft cfs/ft
12.41 ft 0.57
2.6 14.21 ft
hydraulic jump = CL -F = ntersection of hydraulic jump = = Level of Intersection of Jump-D/S Floor level = f Intersection = 3xSubmergency of Jump =
572.53 25.14 ft 0.54 ft 1.62 ft
113.11 ft Length of Stilling Pool - Length of Glacis D/S of Intersection =
111.49 ft
Say
provide D/S Floor
113.00 ft long.
LS USING CONJUGATE DEPTH METHOD
WEIR SECTION F =
1
Floor level of Stilling Pool =
576.00
453750
226875
363000
181500
Maximum
Minimum
Maximum
Minimum
604.80 28.80
602.60 26.60
602.24 26.24
597.43 21.43
112.00
151.25
151.25
75.63
75.63
26.50
19.60
25.50
14.50
154.5
137.2
134.4
99.2
0.979
1.102
0.563
0.762
0.125
0.143
0.070
0.096
0.6022
0.6317
0.4760
0.5410
3.60 17.34
3.80 16.81
1.83 12.49
2.06 11.59
9.16
2.79
13.01
2.91
Jump is submurged in all cases
UICES SECTION
ROTECTION
572.00 453750 89602 Maximum
Minimum
606.82 34.82
604.79 32.79
248.90
248.90
31.00
24.00
205.5 1.211 0.166
187.8 1.326 0.183
0.6650
0.6860
5.78 23.15 7.85
6.00 22.49 1.51
189.06 cfs/ft 23.43 ft
r D/S Floor Critical Condition. 41.01 ft 595.6 576.00 19.60 ft
ith Concentration =
20.10 ft 20.91 ft (32+12)1/2x(R`-D`) 66.12 ft 47.23 ft
ROTECTION .
r U/S Floor Critical Condition. 29.3 ft 601.6 582.00 19.60 ft
ith Concentration =
20.10 ft 9.2 ft 2 2 1/2 (3 +1 ) x(R`-D`) 29.09 ft 20.79 ft
F APRONS 9 12 Thickness of Flexible protection at launched position 19 22 25 28 31 34 37 40 43 45
Position =1.75 x 34/12 =
5 ft
47.23 15.74 31.5 20.79 6.9 13.9
ft ft ft ft ft ft
18
24
25 31 37 43 49
28 34 40 46 52
ECTION FOR UNDERSLUICES
298.67 cfs/ft 31.79 ft
10.4.1 D/S SCOUR PROTECTION Safety Factor =1.75 for D/S Floor Critical Condition. Dept R` =1.75xR = 55.63 ft
evel for the 0.48 million cfs Discharge = D/S Apron Level = Depth of Water on Apron = depth for concentration D` = Depth of Water with Concentration = R` - D` = ver a surface of scour at 1:3 Slope =
596 572.00 24.00 ft 24.50 ft 31.13 ft (32+12)1/2x(R`-D`) 98.4 ft
D/S Stone Apron in Horizontal Position =
70 ft
10.4.2 U/S SCOUR PROTECTION . Safety Factor =1.25 for U/S Floor Critical Condition. Dept R` =1.25xR = 39.7 ft
evel for the 0.48 million cfs Discharge = U/S Apron Level = Depth of Water on Apron = depth for concentration D` = Depth of Water with Concentration = R` - D` = ver a surface of scour at 1:3 Slope =
602.5 582.00 20.50 ft 21.00 ft 18.7 ft (32+12)1/2x(R`-D`) 59.26 ft
D/S Stone Apron in Horizontal Position =
42 ft
10.4.3 THICKNESS OF APRONS
one Apron =
70 23.44 46.9 42 14.1 28.2
ft ft ft ft ft ft
arrage U/S is 4873.5 ft
649.8 ft
guide bank use Lacey`s Depth =1.75xR =
41.01 ft
29.3 ft
ft ft
(Above HFL) (Above HFL)
include allowance for Accretion. 40 ft
4 3249 4873.5 649.8 600 400 140o 57o- 80o
ft ft ft ft ft ft
ION OF LEVELS OF GUIDE BANK 582 602.5 20.5 71 604 22
ft ft ft
ft
0.0002
21.8 ft 1.0732 1.0634 0.7863 0.8247
Note. For Determination of f(d1D) and f(d2/D) see Sheet CIVIL 03F
4816.636398 ft
OK Percentage Difference=
Note. Check that this difference should be witin 10 % if not adjust d 2 from the level at the barrage
0.97470 ft U/S of baarage =
604.77 611.77 ft 607 ft
602.5 6 608.5 ft
inches ft
3.2 3.8 4.4 4.5 3 Mean R'
2.25R 1.5R 1.25R
ft
1.18
ft ft ft ft ft
1.5 ftof Accretion.
ver, over a hydraulic gradient of 1:6 ft 5 ft 616.77 ft
5.2 CHECK FOR UNDERSLUICES with 20%Concentration, Q = 1.2 x Q1 and Q2 =#REF! cfs
Normal State
DSWL 601
AFFLUX 2
USWL 603
ACE FLOW CONDITION
RECTIONS
est Level) and if less than 2% then decrease P to make it near to 2%
Note. Write one Value greater and one value smaller than the desired value of respective values z and z` values for interpolation. Violet Value Take From Table for Cojugate Depths
f(z)
Page 73-74
9.1 FOR NORMAL WEIR SECTION For f(z)= f(z)
0.979 z`(f=1.0)
z
1.0418 0.9727
0.14 0.13 0.131
0.979
For f(z)= f(z)
For f(z)=
0.6275 0.6107 0.6122
0.563
0.08 0.075 0.073
0.563
0.5041 0.4906 0.4846
z`(f=1.0)
z
1.1604 1.1092
0.16 0.15 0.149
1.102
For f(z)= z`(f=1.0)
z
0.6157 0.5789
f(z)
1.102
f(z)
0.6576 0.6432 0.6413
0.762 z`(f=1.0)
z
0.8327 0.7614
0.11 0.1 0.100
0.762
0.5732 0.5521 0.5524
9.2 FOR UNDERSLUICES SECTION
For f(z)= f(z)
1.211 z`(f=1.0)
z
1.3077 1.2428 1.211
For f(z)=
0.18 0.17 0.165
0.683 0.671 0.6652
f(z)
1.326 z`(f=1.0)
z
1.4352 1.3722 1.326
0.2 0.19 0.183
0.7062 0.6953 0.6872
Note. Write one Value greater and one value smaller than the desired value of Grey Value Take From Table for Bresse`s Backwater Function 12.1 DETERMINATION OF LEVELS OF GUIDE BANKS
Note. Write one Value greater and one value smaller than the desired value of Grey Value Take From Table for Bresse`s Backwater Function 14. DESIGN OF MARGINAL BUNDS
