barrage 215
TRANSCRIPT
Roll No. 104 M.A.B.F.
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=
1. MINIMUM STABLE WETTED PERIMETER
1800
LLC= 1.8
3240.067N+35= 3260.0N= 48
48 Bays @60ft=
45 Piers @7ft=
1 Fish ladder =
2 Divide Walls =
3249
2. CALCULATION OF LACEY`S SILT FACTOR
f = 2.03
3. FIXATION OF CREST LEVEL
USEL RBL+E1
Maximum Discharge = Q max=
Do=HFL-RBL=
Pw=
Wa=
Total Wa=
Discharge b/w Abutments=qabt=Qmax/Total Wa
Dischargeover the crest =qweir=Qmax/Wclear
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
Level of E1=
Crest Level = Level of E1-Eo
Maximum D/s Water Level =
h=D/s WL - CL=h/Eo=
C'/C=C'=
% 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 2% then decrease P to make it near to 2%
4. DESIGN OF UNDERSLUICES
Fix Crest level 3
Crest level of undersluices=
3 bays to act as Undersluices on both sides b1 = Assuming
Maximum USEL= HFL+Afflux+hoEo=Maximum USEL-CL of Undersluices=h=DSEL-CL of undersluices=h/Eo=C''/C=C''=
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
Q=C'xWclearxEo3/2
LLC =Total Wa/Pw
qus=120% of main weir
R=0.9(qus2/f)1/3
Vo=qus/R
ho=Vo2/2g
Q1 and Q2= C''x(b1x2)x(Eo)3/2=
Qmain weir = C'x(Wclear-2b1)x(Eo)3/2=
Wclear=
For Normal State
% Q DSWL
(cusecs) (ft)120 544500 600.5100 453750 60050 226875 597.525 113437.5 591
For Retrogressed State
120 544500 596100 453750 595.650 226875 590.5
For Accreted State
120 544500 603100 453750 602.550 226875 601.51 2 3
5.2 CHECK FOR UNDERSLUICES
Normal State
Retrogressed State
Accreted State
Total width of Undersluices = Crest Level of Undersluices =
Items NormalD 22
11.31
1.99h 13.75
17.25
19.24
84.40
0.71
C`/C 0.86
with 20%Concentration, Q = 1.2 x Q1 and Q2 =
Vo
ho
Ho
Eo
Eo3/2
h/Eo
C` 3.268
275.83Q 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 SECTION USING BLENCH CURVES.
I) Normal state of rivera) For
b) For
c) For
II) Retrodressed state of river
Q
544500 199.58
453750 164.68
226875 80.10
qclear
qclear =
USEL = USWL + ho =
DSEL = DEWL + ho =
hL =USEL-DSEL =
E2 =
DSFL = DSEL - E2 =
qclear =
USEL = USWL + ho =
DSEL = DEWL + ho =
hL =USEL-DSEL =
E2 =
DSFL = DSEL - E2 =
qclear =
USEL = USWL + ho =
DSEL = DEWL + ho =
hL =USEL-DSEL =
E2 =
DSFL = DSEL - E2 =
qclear
III) Accreted state of river
Q
544500 195.11
453750 161.97226875 83.92
Worst Condition occurs at
D/S Floor Level =
6.2 FIXATION OF FLOOR LEVELS FOR UNDERSLUICES
Q
Normal 99298.78
Retrogressed 98554.82
Accreted 95958.10
Therefore undersluices floor level will be fixed at
Undersluice floor level =
7. FIXATION OF D/S FLOOR LEVEL FOR NORMAL BARRAGE SECTION USING CRUMP`S METHOD AND DETERMINATION OF FLOOR LENGTH.
a) Q =Maximum DSWL=USWL =USEL =RBL =Crest Level =DSFL =
DSEL= Max DSWL+D/S Velocity HeK= USEL-CL =L = USEL - DSEL =
L/C =
(K+F)/C =F=Level of Intersection of Jump with Glacis= Crest Level -F =
qclear
Dpool =
D/S Velocity = Q/(DpoolxTotal Wa)=
D/S Velocity Head = V2/2g =
q clear = Q/Wclear =
C=Critical Depth =(q2/g)1/3
E2 = DSEL - Level of Intersection of Jump with Glacis =
Submergency of Jump = Level of Intersection of Jump with Glacis-D/S Floor level =
Length of Glacis D/S Jump = 3xSubmergency of jump =
Length of D/S Floor= Length of Stilling pool - Length of Glacis D/S of Jump =
b) Q =Minimum DSWL =USWL=USEL =
DSEL = Min DSWL + D/S Velocity Head =K =USEL - CL =L=USEL -DSEL =
L/C =
(K+F)/C =F =Level of Intersection of hydraulic jump with Galcis = CL -F =
Submergency of Jump = Level of Intersection of Jump with Glacis-D/S Floor level =Length of Glacis D/S of Intersection = 3xSubmergency of Jump =
Length of D/S Floor = Length of Stilling Pool - Length of Glacis D/S of Intersection =
Hence we shall provide D/S Floor
8. FIXATION OF D/S FLOOR LENGTH FOR UNDERSLUICES.
a) Maximum DSWL =Q =USWL =
DSFL =
DSEL =Max DSWL + D/S Velocity Head =
Length of Stilling Pool =4.5E2 =
Dpool = Min DSWL - D/S Floor Level =
D/S Velocity = Q/(DpoolxTotal Wa)=
D/S Velocity Head = V2/2g =
qclear =
C=Critical Depth =(q2/g)1/3
E2 = DSEL - Level of Intersection of hydraulic jump with Galcis=
Length of Stilling Pool = 4.5xE2 =
USEL =USWL +ho =
Dpool = Max DSWL - DSFL =
D/S Velocity = Q/(Dpoolxb1)=
D/S Velocity Head = V2/2g =
K =USEL - CL of undersluices =L=USEL -DSEL =
L/C =
(K+F)/C =F =Level of Intersection of hydraulic jump = CL -F =
Submergency of Jump = Level of Intersection of Jump-D/S Floor level =Length of Glacis D/S of Intersection = 3xSubmergency of Jump =
Length of D/S Floor = Length of Stilling Pool - Length of Glacis D/S of Intersection =
b) Minimum DSWL =Q =USWL =
DSFL =
DSEL =Max DSWL + D/S Velocity Head =K =USEL - CL of undersluices =L=USEL -DSEL =
L/C =
(K+F)/C =F =Level of Intersection of hydraulic jump = CL -F =
Submergency of Jump = Level of Intersection of Jump-D/S Floor level =Length of Glacis D/S of Intersection = 3xSubmergency of Jump =
Length of D/S Floor = Length of Stilling Pool - Length of Glacis D/S of Intersection =
Hence we shall provide D/S Floor
9. CHECK FOR ADEQUACY FOR D/S FLOOR LEVELS USING CONJUGATE DEPTH METHOD
9.1 FOR NORMAL WEIR SECTION
q =Q/b1x2=
C=Critical Depth =(q2/g)1/3
E2 = DSEL - Level of Intersection of hydraulic jump =
Length of Stilling Pool = 4.5xE2 =
USEL =USWL +ho =
Dpool = Max DSWL - DSFL =
D/S Velocity = Q/(Dpoolxb1)=
D/S Velocity Head = V2/2g =
q =Q/b1x2=
C=Critical Depth =(q2/g)1/3
E2 = DSEL - Level of Intersection of hydraulic jump =
Length of Stilling Pool = 4.5xE2 =
Note. For Determination of z and z` see Sheet CIVIL 03F
Q = Discharge in river (cfs)
USELE = USEL - DSFL
Conjugate Depth Cofficientsz
z`
Conjugate Depths
Remarks
9.2 FOR UNDERSLUICES SECTION
D/S Floor level of undersluices = `
Q = Discharge in river (cfs)
USELE = USEL - DSFL (for undersluices)
Conjugate Depth Cofficientsz
z`
Conjugate Depths
Q1 = Discharge through the main weir = 80% of Q
q = Intensity od Discharge on D/S Floor = Q1/2400
Dpool = Depth in Stilling Pool =DSWL - DSFL
E3/2
f(z) = q/E3/2
d1 = z x E
d2 = z` x E
Jump Submergency = Dpool - d2
Q1 = Discharge through undersluices with 20% concentration
q = Intensity od Discharge on D/S Floor = Q1/(2xb1)
Dpool = Depth in Stilling Pool =DSWL - DSFL
E3/2
f(z) = q/E3/2
d1 = z x E
d2 = z` x E
Remarks
10. SCOUR PROTECTION
Assume 20 % Concentration,
10.1 D/S SCOUR PROTECTION
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 concentrationD` = 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 concentrationD` = 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 3Sand Classification Thickness of Flexible protection at launched position
Very Coarse 16Coarse 22
Medium 28
Jump Submergency = Dpool - d2
q = 1.2x qweir
R = 0.9(q2/f)1/3=
Fine 34VeryFine 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,
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 =
q = 1.2x qweir
R = 0.9(q2/f)1/3=
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
II) Length of guide bank D/S of barrage
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 = 7 Free Board D/S = 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 =
L u/s = 1.5x Total Wa =
L d/s = 0.2x Total Wa =
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 =
Slope of river bed= 1/5000 =
Assume
Substituting the values in the formula,L =
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 bankWater 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 = 34
d1 = U/S HFL -BL =
d2 =
d1/D =
d2/D =
f(d1/D) =
f(d2/D) =
Once d2 is fixed the levels of guide bank can be determined
say 3
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 Rage of R'
Nose of Guide bank 2.0 R to 2.5 RTransition from nose to straight 1.25R to 1.75RStraight reaches of guide bank 1.0R to 1.5R
14. DESIGN OF MARGINAL BUNDS
I) Top Width = 20II) 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 1:6V) 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
22
D = Minimum Pond Level - RBL = 16Slope of Canal = 0.0002C= Chezy`s Coefficient = 71
g = 32.2
Length of Backwater Curve = 24.26
Volume of Stone in Apron = 3x(32+12)1/2(R'-D) =9.5(R'-D)
d1 = U/S HL with Accretion - RBL =
M.A.B.F. Blue Value Take From 3.1(b)Curve Green Value take from 3.1(c) Curve
453750 cusecs Brown Value Take From Blench Curve 12000 cusecs Orange Value Take From Crump`s Curve
582 ft Violet Value Take From Table for Cojugate Depths (In 3rd Sheet)600 Red Value Put Yourself
211 ft/canal mile 0.0002
3 ft
6 ft NOTE. First Adjust Crest height such that %difference in cell # E52 is 2%.18 ft
ft
ft 2880
2880 ft
315 ft
24 ft
30 ft
ft
139.7
157.55
19.15 ft
13.15 ft
7.29 ft/sec
0.83 ft
14.08 ft
21.83 ft
603.83 ft
Wclear
ft2/sec
ft2/sec
589.75 ft
600 ft
10.25 ft0.73
0.843.192
485690 cusecs
7.04 %
If this %age difference is 2 then its OK and if greater than 2% 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
189.06
23.43 ft
8.07 ft/sec
1.01 ft
604.01 ft17.27 ft14.26 ft
0.830.762.89
74669 cusecs
424713 cusecs 499381 cusecs 9.1%
OK16.5 %
586.753
5. DETERMINATION OF WATER LEVELS AND ENERGY LEVELS
2880 CL= 589.75
ft2/sec
For Normal State
Afflux R=USWL-RBL
(ft) (ft) (ft) (ft/sec) (ft) (ft)3.5 604 22 8.59 1.15 14.252.5 602.5 20.5 7.69 0.92 12.751.5 599 17 4.63 0.33 9.253.5 594.5 12.5 3.15 0.15 4.75
For Retrogressed State
7 603 21 9.00 1.26 13.256 601.6 19.6 8.04 1.00 11.85
6.5 597 15 5.25 0.43 7.25
For Accreted State
2 605 23 8.22 1.05 15.251.5 604 22 7.16 0.80 14.250.5 602 20 3.94 0.24 12.254 5 6 7 8 9
5.2 CHECK FOR UNDERSLUICES
89602 cfs
DSWL AFFLUX USWL
600.5 3.5 604
596 6.5 602.5
603 2 605
360586.75 ft
Retrogressed Accreted 20.5 23
12.14 10.82
2.29 1.829.25 16.25
15.75 18.25
18.04 20.07
76.64 89.93
0.51 0.81
0.94 0.78
USWL=DSWL+Afflux
Vo ho=Vo2/2g
Ho=USWL-CL
3.57 2.96
273.76 266.5598554.82 95958.10
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 SECTION USING BLENCH CURVES.
0.54450 million cfs
197.51
605.15
601.65
3.50 ft
19.4
582.25
0.454 million cfs
159.43
603.42
600.92
2.50 ft
16.5
584.42
0.2269 million cfs
87.99
599.33
597.83
1.50 ft
11.2
586.63
USEL DSEL DSFL
604.26 597.26 7.00 21 576.26
602.60 596.60 6.00 18.8 577.80
597.43 590.93 6.50 12.3 578.63
hL E2
USEL DSEL DSFL
606.05 604.05 2.00 18.5 585.55
604.80 603.30 1.50 16.5 586.80602.24 601.74 0.50 10 591.74
50% Discharge at 0% State.
576.00
6.2 FIXATION OF FLOOR LEVELS FOR UNDERSLUICES
USEL DSEL DSFL
275.83 605.99 602.49 3.50 23.9 578.59
273.76 604.79 598.29 6.50 25.5 572.79
266.55 606.82 604.82 2.00 22.4 582.42
572.00
7. FIXATION OF D/S FLOOR LEVEL FOR NORMAL BARRAGE SECTION USING CRUMP`S METHOD AND DETERMINATION OF FLOOR LENGTH.
453750.00 cfs602.5
604604.80
582589.75576.00
26.50 ft
5.27 ft/sec
0.43 ft602.93 ft
15.05 ft1.87 ft
157.55 cfs
9.169 ft0.20
1.942.738
Level of Intersection of Jump with Glacis= Crest Level -F = 587.008
15.92
hL E2
qclear hL E2
= DSEL - Level of Intersection of Jump with Glacis =
Submergency of Jump = Level of Intersection of Jump with Glacis-D/S Floor level = 11.01 ft
Length of Glacis D/S Jump = 3xSubmergency of jump = 33.02 ft
71.65 ftLength of D/S Floor= Length of Stilling pool - Length of Glacis D/S of Jump = 38.63 ft
Say 39.00 ft
453750 cfs595.6601.6
602.60
19.60
7.13 ft/sec
0.79 ftDSEL = Min DSWL + D/S Velocity Head = 596.39 ft
12.86 ft6.21 ft
157.55 cfs
9.17 ft0.68
2.812.82 ft
Level of Intersection of hydraulic jump with Galcis = CL -F = 576.93
19.46 ftSubmergency of Jump = Level of Intersection of Jump with Glacis-D/S Floor level = 0.93Length of Glacis D/S of Intersection = 3xSubmergency of Jump = 2.79 ft
87.56 ftLength of D/S Floor = Length of Stilling Pool - Length of Glacis D/S of Intersection = 84.78 ft
Say 85.00 ft
Hence we shall provide D/S Floor 85.00 ft, long.
603 ft CL of Undersluices=89602 cfs
605
606.82 ft572.00 ft
31.00 ft
8.03 ft/sec
1.00 ftDSEL =Max DSWL + D/S Velocity Head = 604.00 ft
= Min DSWL - D/S Floor Level =
= DSEL - Level of Intersection of hydraulic jump with Galcis=
20.07 ft2.82 ft
248.90 cfs/ft
12.44 ft0.23
1.954.18 ft
Level of Intersection of hydraulic jump = CL -F = 582.57
21.44 ftSubmergency of Jump = Level of Intersection of Jump-D/S Floor level = 10.57 ftLength of Glacis D/S of Intersection = 3xSubmergency of Jump = 31.71 ft
96.46 ftLength of D/S Floor = Length of Stilling Pool - Length of Glacis D/S of Intersection = 64.75
Say 65.00
596 ft CL of Undersluices=89602 cfs602.5
604.79 ft572.00 ft
24.00 ft
10.37 ft/sec
1.67 ftDSEL =Max DSWL + D/S Velocity Head = 597.67 ft
18.04 ft7.12 ft
248.90 cfs/ft
12.41 ft0.57
2.614.21 ft
Level of Intersection of hydraulic jump = CL -F = 572.53
25.14 ftSubmergency of Jump = Level of Intersection of Jump-D/S Floor level = 0.54 ftLength of Glacis D/S of Intersection = 3xSubmergency of Jump = 1.62 ft
113.11 ftLength of D/S Floor = Length of Stilling Pool - Length of Glacis D/S of Intersection = 111.49 ft
Say 112.00
Hence we shall provide D/S Floor 113.00 ft long.
9. CHECK FOR ADEQUACY FOR D/S FLOOR LEVELS USING CONJUGATE DEPTH METHOD
9.1 FOR NORMAL WEIR SECTION
= DSEL - Level of Intersection of hydraulic jump =
= DSEL - Level of Intersection of hydraulic jump =
1Floor level of Stilling Pool = 576.00
453750 226875
363000 181500
Maximum Minimum Maximum Minimum
604.80 602.60 602.24 597.4328.80 26.60 26.24 21.43
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 3.80 1.83 2.06
17.34 16.81 12.49 11.59
9.16 2.79 13.01 2.91
Jump is submurged in all cases
9.2 FOR UNDERSLUICES SECTION
572.00
45375089602
Maximum Minimum
606.82 604.7934.82 32.79
248.90 248.90
31.00 24.00
205.5 187.81.211 1.3260.166 0.183
0.6650 0.6860
5.78 6.00
23.15 22.49
F =
7.85 1.51
189.06 cfs/ft
23.43 ft
10.1 D/S SCOUR PROTECTION
Safety Factor =1.75 for D/S Floor Critical Condition.41.01 ft
595.6576.00
19.60 ft
D` = Depth of Water with Concentration = 20.10 ft20.91 ft
66.12 ft
47.23 ft
10.2 U/S SCOUR PROTECTION.
Safety Factor =1.25 for U/S Floor Critical Condition.29.3 ft
601.6582.00
19.60 ft
D` = Depth of Water with Concentration = 20.10 ft9.2 ft
29.09 ft
20.79 ft
9 12 18 24Thickness of Flexible protection at launched position
19 22 25 2825 28 31 34
31 34 37 40
(32+12)1/2x(R`-D`)
(32+12)1/2x(R`-D`)
37 40 43 4643 45 49 52
Thickness of Stone Apron in Hz. Position =1.75 x 34/12 = 5 ft
47.23 ft15.74 ft
31.5 ft20.79 ft
6.9 ft13.9 ft
10.4 SCOUR PROTECTION 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
Minimum D/S Water Level for the 0.48 million cfs Discharge = 596D/S Apron Level = 572.00Depth of Water on Apron = 24.00 ft
Add 0.5 ft increase in depth for concentrationD` = Depth of Water with Concentration = 24.50 ftR` - D` = 31.13 ft
Length of Apron to Cover a surface of scour at 1:3 Slope =98.4 ft
Therefore the length of 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
Minimum U/S Water Level for the 0.48 million cfs Discharge = 602.5U/S Apron Level = 582.00Depth of Water on Apron = 20.50 ft
Add 0.5 ft increase in depth for concentrationD` = Depth of Water with Concentration = 21.00 ftR` - D` = 18.7 ft
Length of Apron to Cover a surface of scour at 1:3 Slope =59.26 ft
(32+12)1/2x(R`-D`)
(32+12)1/2x(R`-D`)
Therefore the length of D/S Stone Apron in Horizontal Position = 42 ft
10.4.3 THICKNESS OF APRONS
Total length of D/S Stone Apron = 70 ft23.44 ft
46.9 ft42 ft
14.1 ft28.2 ft
I) Length of guide bank measured in a straight line along th barrage U/S is
4873.5 ft
649.8 ft
III) For the nose of U/S Glacis bank and the full length of D/S guide bank use Lacey`s Depth =1.75xR = 41.01 ft
29.3 ft
ft (Above HFL)ft (Above HFL)
These freeboards also include allowance for Accretion.
40 ft
4 ft
3249 ft4873.5 ft
649.8 ft600 ft400 ft
12.1 DETERMINATION OF LEVELS OF GUIDE BANK
582 ft602.5 ft
20.5 ft71
604
22 ft0.0002
21.8 ft
1.0732
1.0634
0.7863 Note.
0.8247
4816.63639751553 ft OK
Percentage Difference 1.18Note.
0.97470 from the level at the barrage
ft U/S of baarage = 604.77
611.77 ft
607 ft
602.56
608.5 ft
inches
140o
57o- 80o
For Determination of f(d1D) and f(d2/D) see Sheet CIVIL 03F
Check that this difference should be witin 10 % if not adjust d2
ft
3.2 ft3.8 ft4.4 ft4.5 ft
3 ft
Mean R'2.25R1.5R
1.25R
ftII) Top Level to be 5 ft above estimated HFL after allowing for 1.5 ftof Accretion.
IV) Back Slope to be such as to provide a minimum of 2ft cover, over a hydraulic gradient of 1:6ft
5 ft616.77 ft
CALCULATION OF LENGTH OF BACKWATER CURVE
TABLE FOR LENGTH OF BACKWATER CURVE ft
ft D1 2 3 4 5
16 22 21.5 2500 4843.4
16 21.5 21 2500 4843.416 21 20.5 2500 4843.416 20.5 20 2500 4843.416 20 19.5 2500 4843.416 19.5 19 2500 4843.416 19 18.5 2500 4843.416 18.5 18 2500 4843.416 18 17.5 2500 4843.4
16 17.5 17 2500 4843.416 17 16.5 2500 4843.416 16.5 16.1 2000 4843.4
miles
Note.
d1 d2 d1-d2/S 1/S-C2/g
ft/sec2
For Determination of f(d1D) and f(d2/D) see Sheet CIVIL 03F
Blue Value Take From 3.1(b)Curve Page 85
Green Value take from 3.1(c) Curve Page 85
Brown Value Take From Blench Curve Page 76
Orange Value Take From Crump`s Curve Page75
Violet Value Take From Table for Cojugate Depths (In 3rd Sheet) Page 73-74
First Adjust Crest height such that %difference in cell # E52 is 2%.
For Normal State
h=DSWL-CL C/C' C' % Difference
(ft) (ft)15.40 10.75 0.70 0.86 3.268 197.5 568840 4.513.67 10.25 0.75 0.83 3.154 159.4 459161 1.29.59 7.75 0.81 0.78 2.964 88.0 253412 11.74.91 1.25 0.26 0.98 3.724 40.5 116624 2.8
For Retrogressed State
14.51 6.25 0.43 0.95 3.61 199.6 574803 5.612.86 5.85 0.46 0.94 3.572 164.7 474278 4.57.68 0.75 0.10 0.99 3.762 80.1 230701 1.7
For Accreted State
16.30 13.25 0.81 0.78 2.964 195.1 561931 3.215.05 12.75 0.85 0.73 2.774 162.0 466468 2.812.49 11.75 0.94 0.5 1.9 83.9 241684 6.5
10 11 12 13 14 15 16
.
586.64
Eo=Ho+ho h/Eo qclear=C'Eo3/2 Qclear
586.75 ft
ft
ft
586.75 ft
ft
%
/D) see Sheet CIVIL 03F
Check that this difference should be witin 10 % if not adjust d2
TABLE FOR LENGTH OF BACKWATER CURVE
L L6 7 8 9 10 5x10x1 11+4
1.375 1.344 0.3163 0.3399 0.0237 1836 4336
1.344 1.313 0.3399 0.3614 0.0215 1665 41651.313 1.281 0.3614 0.3906 0.0292 2262 47621.281 1.250 0.3906 0.4198 0.0292 2262 47621.250 1.219 0.4198 0.4573 0.0375 2906 54061.219 1.188 0.4573 0.5000 0.0427 3313 58131.188 1.156 0.5000 0.5507 0.0506 3923 64231.156 1.125 0.5507 0.6207 0.0700 5427 79271.125 1.094 0.6207 0.7052 0.0845 6550 90501.094 1.063 0.7052 0.8284 0.1231 9542 120421.063 1.031 0.8284 1.0480 0.2197 17023 195231.031 1.006 1.0480 1.5881 0.5401 41854 43854
Total = 128062 ft24.26 miles
T1 = d1/D T2 = d2/D f1(d1/D) f2(d2/D) f1-f2
(d1D) and f(d2/D) see Sheet CIVIL 03F
605.15 601.65 60.438804603.42 600.92 50.548829599.33 597.83 29.686314594.65 591.15 10.873923
For Retrogressed State
604.26 597.26 55.286566602.60 596.60 46.102997597.43 590.93 21.293068
00
For Accreted State00
606.05 604.05 65.828216604.80 603.30 58.387896
602.24 601.74 44.167378
USEL= USWL + ho
DESL= DSWL +ho
DSFL
% Q Normal Retrogressed Accreted 6.2
120% 544500 582.00 576.30 585.00 578.59100% 453750 584.00 577.00 586.00 572.7950% 226875 586.00 578.00 591.00 582.42
OKNOT OK
DESIGN OF BARRAGE PROFILE FOR SUB-SURFACE FLOW CONDITION
15. FIXING THE DEPTH OF SHEET PILES
Scour Depth = 19.15 ftLLC= 1.6
30.6 ft67N+35= 30.6N= -1
-1s @60ft= -60 ft
-4ers @7ft= -28 ft
1 ladder = 26 ft
2e Walls = 30 ft
-32 ft
#REF!
#REF!
16. CALCULATION OF EXIT GRADIENT
f = #REF!
17. CALCULATION OF UPLIFT PRESURES AFTER APPLYING CORRECTIONS
17.1 U/S PILE LINE
#REF! ft
#REF! ft/sec
#REF! ft
#REF! ft
#REF! ftCrest Level = Level of E1-Eo #REF! ft
Maximum D/s Water Level = 600 ft
h=D/s WL - CL= #REF! fth/Eo= #REF!
C'/C= 0.84C'= 3.192
#REF! cusecs
17.2 INTERMEDIATE SHEET PILE AT TOE OF D/S GLACISNOTE.If this %age difference is 2 then its OK and if greater than 2% then increase P(Crest Level) and if less than 2% then decrease P to make it near to 2%
-1.67
17.3 D/S SHEET PILE AT END OF IMPERVIOUS FLOOR
Wa=
Wclear
Total Wa=
Discharge b/w Abutments=qabt=Qmax/Total Wa ft2/sec
Dischargeover the crest =qweir=Qmax/Wclear ft2/sec
Ho=R-P=
Vo=qabt/R
ho=Vo2/2g=
Eo=Ho+ho
E1=Do+ho
+AFFLUX
Q=C'xWclearxEo3/2
LLC =Total Wa/Pw
18. CLACULATION FOR FLOOR THICKNESS
Fix Crest level 3 feet below the main weir.
Crest level of undersluices= #REF! ftbays to act as Undersluices on both sides b1 = 0 ftAssuming
#REF!
#REF! ft
#REF! ft/sec
#REF! ft
Maximum USEL= HFL+Afflux+ho #REF! ftEo=Maximum USEL-CL of Undersluices= #REF! fth=DSEL-CL of undersluices= #REF! fth/Eo= #REF!C''/C= 0.79C''= 3.002
#REF! cusecs
#REF! cusecs Total Discharge = #REF! cusecsCheck =Total Q>Qmax #REF!%age water passing through undersluice #REF! %
5. DETERMINATION OF WATER LEVELS AND ENERGY LEVELS
5.1 CHECH FOR MAIN WEIR
-60 CL= #REF!
Q DSWL Afflux
(cusecs) (ft) (ft) (ft) (ft) (ft/sec)#REF! 601 #REF! #REF! #REF! #REF!#REF! 600 #REF! #REF! #REF! #REF!#REF! 597.5 1 598.5 #REF! #REF!#REF! 591 #REF! #REF! #REF! #REF!
#REF! 595.5 6 601.5 #REF! #REF!#REF! 595.4 6 601.4 #REF! #REF!#REF! 590 6.1 596.1 #REF! #REF!
qus=120% of main weir ft2/sec
R=0.9(qus2/f)1/3
Vo=qus/R
ho=Vo2/2g
Q1 and Q2= C''x(b1x2)x(Eo)3/2=
Qmain weir = C'x(Wclear-2b1)x(Eo)3/2=
Wclear=
USWL=DSWL+Affl
ux
R=USWL-RBL
Vo
#REF! 603.5 2 605.5 #REF! #REF!#REF! 603 1 604 #REF! #REF!#REF! 600.5 0.5 601 #REF! #REF!#REF! 594.5 0.5 595 #REF! #REF!
5.2 CHECK FOR UNDERSLUICES
#REF! cfs
DSWL AFFLUX USWL
Normal State 601 2 603
with 20%Concentration, Q = 1.2 x Q1 and Q2 =
DESIGN OF BARRAGE PROFILE FOR SUB-SURFACE FLOW CONDITION
17. CALCULATION OF UPLIFT PRESURES AFTER APPLYING CORRECTIONS
If this %age difference is 2 then its OK and if greater than 2% then increase P(Crest Level) and if less than 2% then decrease P to make it near to 2%
Note.
respective values z and z` values for interpolation.
Violet Value Take From Table for Cojugate Depths Page 73-74
9.1 FOR NORMAL WEIR SECTION
For f(z)= 0.979 For f(z)= 1.102
f(z) z f(z) z
1.0418 0.14 0.6275 1.1604 0.16 0.65760.9727 0.13 0.6107 1.1092 0.15 0.6432
0.979 0.131 0.6122 1.102 0.149 0.6413
For f(z)= 0.563 For f(z)= 0.762
f(z) z f(z) z
0.6157 0.08 0.5041 0.8327 0.11 0.57320.5789 0.075 0.4906 0.7614 0.1 0.5521
0.563 0.073 0.4846 0.762 0.100 0.5524
9.2 FOR UNDERSLUICES SECTION
For f(z)= 1.211 For f(z)= 1.326
f(z) z f(z) z
1.3077 0.18 0.683 1.4352 0.2 0.70621.2428 0.17 0.671 1.3722 0.19 0.6953
1.211 0.165 0.6652 1.326 0.183 0.6872
Note.
Grey Value Take From Table for Bresse`s Backwater Function Page 105
12.1 DETERMINATION OF LEVELS OF GUIDE BANKS
d1/D d2/D1.06 0.8382 1.06 0.83821.1 0.6806 1.1 0.6806
Write one Value greater and one value smaller than the desired value of f(z)and also put
z`(f=1.0) z`(f=1.0)
z`(f=1.0) z`(f=1.0)
z`(f=1.0) z`(f=1.0)
Write one Value greater and one value smaller than the desired value of d/D and f(d/D).
f(d1/D) f(d2/D)
1.073171 0.7863 1.063415 0.824746
Note.
Grey Value Take From Table for Bresse`s Backwater Function
14. DESIGN OF MARGINAL BUNDS
d1/D d2/D1.3 0.3731 1.005 1.64861.35 0.3352 1.006 1.5881
1.375 0.3163 1.006 1.58811.3 0.37311.35 0.3352
1.344 0.33991.3 0.37311.25 0.4198
1.313 0.36141.3 0.37311.25 0.4198
1.281 0.39061.25 0.41981.2 0.47981.250 0.4198
1.2 0.47981.25 0.4198
1.219 0.45731.15 0.56081.2 0.47981.188 0.5000
1.15 0.56081.2 0.47981.156 0.5507
1.1 0.68061.15 0.5608
1.125 0.62071.06 0.83821.1 0.6806
Write one Value greater and one value smaller than the desired value of d/D and f(d/D).
f(d1/D) f(d2/D)
1.094 0.70521.06 0.83821.1 0.68061.063 0.8284
1.03 1.05961.04 0.9669
1.031 1.0480
Write one Value greater and one value smaller than the desired value of f(z)and also put
Write one Value greater and one value smaller than the desired value of d/D and f(d/D).
Write one Value greater and one value smaller than the desired value of d/D and f(d/D).
f(z) z z`0.8 0.85 0.9 0.95 1
0.008 0.001 0.050 0.053 0.056 0.059 0.0630.0161 0.002 0.071 0.075 0.080 0.084 0.0890.0241 0.003 0.086 0.092 0.097 0.103 0.1080.0321 0.004 0.099 0.106 0.112 0.118 0.125
0.04 0.005 0.110 0.117 0.124 0.131 0.139
0.048 0.006 0.121 0.128 0.136 0.144 0.1510.056 0.007 0.130 0.138 0.147 0.155 0.163
0.0639 0.008 0.138 0.147 0.156 0.165 0.1740.0719 0.009 0.147 0.156 0.166 0.175 0.1840.0799 0.01 0.154 0.164 0.174 0.184 0.194
0.0996 0.0125 0.171 0.183 0.194 0.205 0.2160.1195 0.015 0.187 0.199 0.211 0.224 0.2360.1391 0.0175 0.201 0.214 0.227 0.240 0.2530.1588 0.02 0.214 0.228 0.242 0.256 0.2700.1784 0.0225 0.226 0.241 0.256 0.271 0.285
0.1981 0.025 0.238 0.253 0.269 0.284 0.3000.2175 0.0275 0.248 0.264 0.281 0.297 0.3130.2371 0.03 0.258 0.275 0.292 0.309 0.3260.2759 0.0325 0.289 0.308 0.327 0.346 0.3660.3145 0.035 0.318 0.339 0.360 0.381 0.402
0.3528 0.045 0.310 0.331 0.351 0.372 0.3930.3911 0.05 0.325 0.346 0.368 0.390 0.4120.4291 0.055 0.338 0.361 0.384 0.407 0.4290.4668 0.06 0.351 0.375 0.399 0.422 0.4460.5055 0.065 0.364 0.389 0.413 0.438 0.463
0.5417 0.07 0.375 0.400 0.426 0.451 0.4760.5789 0.075 0.386 0.412 0.438 0.464 0.4910.6157 0.08 0.396 0.423 0.450 0.477 0.5040.6524 0.085 0.406 0.433 0.461 0.489 0.5170.689 0.09 0.415 0.444 0.472 0.501 0.529
0.7253 0.095 0.424 0.453 0.482 0.512 0.5410.7614 0.1 0.433 0.463 0.492 0.522 0.5520.8327 0.11 0.449 0.480 0.511 0.542 0.5730.9036 0.12 0.464 0.496 0.528 0.560 0.5930.9727 0.13 0.477 0.510 0.544 0.577 0.610
1.0418 0.14 0.490 0.524 0.558 0.593 0.6271.1092 0.15 0.501 0.536 0.572 0.607 0.6431.1604 0.16 0.504 0.540 0.576 0.611 0.6471.2428 0.17 0.522 0.559 0.596 0.634 0.6711.3077 0.18 0.531 0.569 0.607 0.645 0.683
1.3722 0.19 0.540 0.579 0.617 0.656 0.6951.4352 0.2 0.548 0.587 0.627 0.666 0.706
1.4976 0.21 0.555 0.595 0.636 0.676 0.7161.5592 0.22 0.562 0.603 0.644 0.685 0.7261.6192 0.23 0.568 0.609 0.651 0.693 0.734
1.6789 0.24 0.574 0.616 0.658 0.700 0.7421.7372 0.25 0.579 0.622 0.664 0.707 0.7501.7947 0.26 0.584 0.627 0.670 0.713 0.7571.851 0.27 0.588 0.632 0.675 0.719 0.763
1.9064 0.28 0.592 0.636 0.680 0.724 0.769
Table 5.1
FOR NORMAL STATE
Q DSWL Afflux
(cusecs) (ft) (ft) (ft) (ft) (ft/sec) (ft) (ft) (ft) (ft)
FOR RETROGRESSED STATE
FOR ACCRETED STATE
USWL=DSWL+Afflux
R= USWL-
RBLVo
ho= Vo
2/2gHo=USWL-CL
Eo= Ho+ho
h= DSWL-
CL
Table 5.1
FOR NORMAL STATE
C/C' C' % Diff
FOR RETROGRESSED STATE
FOR ACCRETED STATE
h/Eoqclear= C'Eo
3/2 Qclear
(ft2/sec) (ft3/sec)