Barrage 215

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

z Conjugate Depth Cofficients Conjugate Depths

z` d1 = z x E d2 = z` x E

Jump Submergency = Dpool - d2 Remarks

10. SCOUR PROTECTION Assume 20 % Concentration, q = 1.2x qweir R = 0.9(q2/f)1/3=

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 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

Vo

ho=Vo2/2g

Ho=USWLCL

(ft) 3.5 2.5 1.5 3.5

(ft) 604 602.5 599 594.5

(ft) 22 20.5 17 12.5

(ft/sec) 8.59 7.69 4.63 3.15

(ft) 1.15 0.92 0.33 0.15

(ft) 14.25 12.75 9.25 4.75

For Retrogressed State 7 6 6.5

603 601.6 597

21 19.6 15

9.00 8.04 5.25

1.26 1.00 0.43

13.25 11.85 7.25

1.05 0.80 0.24 8

15.25 14.25 12.25 9

For Accreted State 2 1.5 0.5 4

605 604 602 5

23 22 20 6

NDERSLUICES 89602 cfs DSWL 600.5 596 603

AFFLUX 3.5 6.5 2 360 586.75 ft

Retrogressed 20.5 12.14 2.29 9.25 15.75 18.04 76.64 0.51

Accreted 23 10.82 1.82 16.25 18.25 20.07 89.93 0.81

0.94

0.78

3.57 273.76 98554.82

2.96 266.55 95958.10

USWL 604 602.5 605

8.22 7.16 3.94 7

OF D/S GLACIS AND D/S FLOOR.

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 604.26 602.60 597.43

DSEL 597.26 596.60 590.93

USEL 606.05 604.80

DSEL 604.05 603.30

hL

E2

7.00 6.00 6.50

21 18.8 12.3

hL

E2

2.00 1.50

18.5 16.5

DSFL 576.26 577.80 578.63

DSFL 585.55 586.80

602.24

601.74

50% Discharge at

0.50

10

FALSE

591.74

State.

576.00

LOOR LEVELS FOR UNDERSLUICES qclear

USEL 605.99 604.79 606.82

275.83 273.76 266.55

DSEL 602.49 598.29 604.82

hL

E2

3.50 6.50 2.00

23.9 25.5 22.4

DSFL 578.59 572.79 582.42

572.00 BARRAGE SECTION USING CRUMP`S METHOD AND DETERMINATION OF FLOOR LENGTH. 453750.00

cfs 602.5 604 604.80 582 589.75 576.00 26.50 ft 5.27 ft/sec 0.43 602.93 15.05 1.87 157.55

ft ft ft ft cfs

9.169 ft 0.20

1.94 2.738 Jump with Glacis= Crest Level -F =

587.008

ntersection of Jump with Glacis =

15.92

= 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

WATER CURVE

TABLE FOR LENGTH OF ft ft

ft/sec2

D 1 16 16 16 16 16 16 16 16 16 16 16 16

d1 2 22 21.5 21 20.5 20 19.5 19 18.5 18 17.5 17 16.5

d2 3 21.5 21 20.5 20 19.5 19 18.5 18 17.5 17 16.5 16.1

d1-d2/S 4 2500 2500 2500 2500 2500 2500 2500 2500 2500 2500 2500 2000

1/S-C2/g 5 4843.4 4843.4 4843.4 4843.4 4843.4 4843.4 4843.4 4843.4 4843.4 4843.4 4843.4 4843.4

miles

Note. For Determination of f (d 1 D) and f (d 2 /D) see Sh

om 3.1(b)Curve om 3.1(c) Curve From Blench Curve From Crump`s Curve rom Table for Cojugate Depths (In 3rd Sheet)

st Crest height such that %difference in cell # E52 is 2%.

Page 85 Page 85 Page 76 Page75 Page 73-74

Eo=Ho+ho h=DSWL-CL

h/Eo

C/C'

C'

qclear=C'Eo3/2

Qclear

% Difference

(ft) 15.40 13.67 9.59 4.91

(ft) 10.75 10.25 7.75 1.25

0.70 0.75 0.81 0.26

0.86 0.83 0.78 0.98

3.268 3.154 2.964 3.724

197.5 159.4 88.0 40.5

568840 459161 253412 116624

4.5 1.2 11.7 2.8

14.51 12.86 7.68

6.25 5.85 0.75

0.43 0.46 0.10

0.95 0.94 0.99

3.61 3.572 3.762

199.6 164.7 80.1

574803 474278 230701

5.6 4.5 1.7

16.30 15.05 12.49 10

13.25 12.75 11.75 11

0.81 0.85 0.94 12

0.78 0.73 0.5 13

2.964 2.774 1.9 14

195.1 162.0 83.9 15

561931 466468 241684 16

3.2 2.8 6.5

. 586.64

586.75 ft

ft

ft 586.75 ft

ft

Sheet CIVIL 03F

% % if not adjust d 2

FOR LENGTH OF BACKWATER CURVE T1 = d1/D 6 1.375 1.344 1.313 1.281 1.250 1.219 1.188 1.156 1.125 1.094 1.063 1.031

T2 = d2/D 7 1.344 1.313 1.281 1.250 1.219 1.188 1.156 1.125 1.094 1.063 1.031 1.006

f1(d1/D)

f2(d2/D)

f1-f2

8 0.3163 0.3399 0.3614 0.3906 0.4198 0.4573 0.5001 0.5507 0.6207 0.7052 0.8284 1.0480

9 0.3399 0.3614 0.3906 0.4198 0.4573 0.5001 0.5507 0.6207 0.7052 0.8284 1.0480 1.5881

10 0.0237 0.0215 0.0292 0.0292 0.0375 0.0427 0.0506 0.0700 0.0845 0.1231 0.2197 0.5401

(d 1 D) and f (d 2 /D) see Sheet CIVIL 03F

L 5x10x1 1836 1665 2262 2262 2906 3313 3923 5427 6550 9542 17023 41854

L 11+4 4336 4165 4762 4762 5406 5813 6423 7927 9050 12042 19523 43854

Total =

128062

ft

24.26 miles

USEL= USWL DESL= + ho DSWL +ho

605.15 603.42 599.33 594.65

601.65 60.4388 600.92 50.54883 597.83 29.68631 591.15 10.87392

604.26 602.60 597.43

597.26 55.28657 596.60 46.103 590.93 21.29307 0 0 0 0 604.05 65.82822 603.30 58.3879 601.74 44.16738

606.05 604.80 602.24

% 120% 100% 50%

Q 544500 453750 226875

Normal 582.00 584.00 586.00

DSFL Retrogressed Accreted 576.30 585.00 577.00 586.00 578.00 591.00

6.2 578.59 572.79 582.42

OK NOT OK

DESIGN OF BARRAGE PROFILE FOR SUB-SURFACE FLOW COND 15. FIXING THE DEPTH OF SHEET PILES Scour Depth = LLC= W a= 67N+35= N=

19.15 ft 1.6 30.6 ft 30.6 -1

-1 Bays @60ft= -4Piers @7ft= Fish 1 ladder = Divide 2 Walls =

-60 -28 26 30

ft ft ft ft

W clear

Total W a=

-32 ft Discharge b/w Abutments=qabt=Qmax/Total W#REF! a

ft2/sec

Dischargeover the crest =qweir=Qmax/W clear

ft2/sec

#REF!

16. CALCULATION OF EXIT GRADIENT f=

#REF!

17. CALCULATION OF UPLIFT PRESURES AFTER APPLYING CORRECTIONS 17.1 U/S PILE LINE Ho=R-P=

#REF!

ft

Vo=qabt/R

#REF!

ft/sec

ho=Vo /2g=

#REF!

ft

Eo=Ho+ho

#REF!

ft

#REF! #REF!

ft ft

2

E1=Do+ho +AFFLUX Crest Level =

Level of E1-Eo

Maximum D/s Water Level = h=D/s WL - CL= h/Eo=

C'/C= C'= Q=C'xW clearxEo3/2

600 ft #REF! #REF!

ft

0.84 3.192 #REF! cusecs

17.2 INTERMEDIATE SHEET PILE AT TOE OF D/S GLACIS

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 LLC =Total W a/Pw

-1.67

17.3 D/S SHEET PILE AT END OF IMPERVIOUS FLOOR

18. CLACULATION FOR FLOOR THICKNESS Fix Crest level

3

feet below the main weir.

Crest level of undersluices= bays to act as Undersluices on both sides b1 = Assuming qus=120% of main weir

#REF!

ft

#REF!

ft2/sec

R=0.9(qus2/f)1/3

#REF!

ft

Vo=qus/R

#REF!

ft/sec

ho=Vo /2g

#REF!

ft

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=

#REF! #REF! #REF! #REF! 0.79 3.002 #REF!

ft ft ft

Qmain weir = C'x(W clear-2b1)x(Eo)3/2= Total Discharge = Check =Total Q>Qmax %age water passing through undersluices=

#REF! #REF! #REF! #REF!

cusecs cusecs

0 ft

2

cusecs

%

5. DETERMINATION OF WATER LEVELS AND ENERGY LEVELS 5.1 CHECH FOR MAIN WEIR Wclear=

-60

Q

DSWL

Afflux

(cusecs) #REF! #REF! #REF! #REF!

(ft) 601 600 597.5 591

(ft) #REF! #REF! 1 #REF!

CL=

#REF!

USWL=D R=USWLSWL+Affl RBL ux (ft) (ft) #REF! #REF! #REF! #REF! 598.5 #REF! #REF! #REF!

Vo (ft/sec) #REF! #REF! #REF! #REF!

#REF! #REF! #REF!

595.5 595.4 590

6 6 6.1

601.5 601.4 596.1

#REF! #REF! #REF!

#REF! #REF! #REF!

#REF! #REF! #REF! #REF!

603.5 603 600.5 594.5

2 1 0.5 0.5

605.5 604 601 595

#REF! #REF! #REF! #REF!

#REF! #REF! #REF! #REF!

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

f(d1/D) d1/D 1.06 0.8382 1.1 0.6806 1.07317 0.7863

f(d2/D) d2/D 1.06 0.8382 1.1 0.6806 1.06341 0.82475

d/D

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

d1/D 1.3 1.35 1.375 1.3 1.35 1.344 1.3 1.25 1.313 1.3 1.25 1.281 1.25 1.2 1.250 1.2 1.25 1.219 1.15 1.2 1.188 1.15 1.2 1.156 1.1 1.15 1.125 1.06 1.1 1.094 1.06 1.1

f(d1/D) 0.3731 0.3352 0.3163 0.3731 0.3352 0.3399 0.3731 0.4198 0.3614 0.3731 0.4198 0.3906 0.4198 0.4798 0.4198 0.4798 0.4198 0.4573 0.5608 0.4798 0.5001 0.5608 0.4798 0.5507 0.6806 0.5608 0.6207 0.8382 0.6806 0.7052 0.8382 0.6806

f(d2/D) d2/D 1.005 1.6486 1.006 1.5881 1.006 1.5881

d/D

1.063 0.8284 1.03 1.0596 1.04 0.9669 1.031 1.0480

the desired value of

f(z)and also put

Page 73-74

the desired value of Page 105

d/D and f(d/D).

the desired value of

d/D and f(d/D).

f(z)

z 0.8

0.85

z` 0.9

0.95

1

0.008 0.0161 0.0241 0.0321 0.04

0.001 0.002 0.003 0.004 0.005

0.050 0.071 0.086 0.099 0.110

0.053 0.075 0.092 0.106 0.117

0.056 0.080 0.097 0.112 0.124

0.059 0.084 0.103 0.118 0.131

0.063 0.089 0.108 0.125 0.139

0.048 0.056 0.0639 0.0719 0.0799

0.006 0.007 0.008 0.009 0.01

0.121 0.130 0.138 0.147 0.154

0.128 0.138 0.147 0.156 0.164

0.136 0.147 0.156 0.166 0.174

0.144 0.155 0.165 0.175 0.184

0.151 0.163 0.174 0.184 0.194

0.0996 0.1195 0.1391 0.1588 0.1784

0.0125 0.015 0.0175 0.02 0.0225

0.171 0.187 0.201 0.214 0.226

0.183 0.199 0.214 0.228 0.241

0.194 0.211 0.227 0.242 0.256

0.205 0.224 0.240 0.256 0.271

0.216 0.236 0.253 0.270 0.285

0.1981 0.2175 0.2371 0.2759 0.3145

0.025 0.0275 0.03 0.0325 0.035

0.238 0.248 0.258 0.289 0.318

0.253 0.264 0.275 0.308 0.339

0.269 0.281 0.292 0.327 0.360

0.284 0.297 0.309 0.346 0.381

0.300 0.313 0.326 0.366 0.402

0.3528 0.3911 0.4291 0.4668 0.5055

0.045 0.05 0.055 0.06 0.065

0.310 0.325 0.338 0.351 0.364

0.331 0.346 0.361 0.375 0.389

0.351 0.368 0.384 0.399 0.413

0.372 0.390 0.407 0.422 0.438

0.393 0.412 0.429 0.446 0.463

0.5417 0.5789 0.6157 0.6524 0.689

0.07 0.075 0.08 0.085 0.09

0.375 0.386 0.396 0.406 0.415

0.400 0.412 0.423 0.433 0.444

0.426 0.438 0.450 0.461 0.472

0.451 0.464 0.477 0.489 0.501

0.476 0.491 0.504 0.517 0.529

0.7253 0.7614 0.8327 0.9036 0.9727

0.095 0.1 0.11 0.12 0.13

0.424 0.433 0.449 0.464 0.477

0.453 0.463 0.480 0.496 0.510

0.482 0.492 0.511 0.528 0.544

0.512 0.522 0.542 0.560 0.577

0.541 0.552 0.573 0.593 0.610

1.0418 1.1092 1.1604 1.2428 1.3077

0.14 0.15 0.16 0.17 0.18

0.490 0.501 0.504 0.522 0.531

0.524 0.536 0.540 0.559 0.569

0.558 0.572 0.576 0.596 0.607

0.593 0.607 0.611 0.634 0.645

0.627 0.643 0.647 0.671 0.683

1.3722 1.4352 1.4976 1.5592 1.6192

0.19 0.2 0.21 0.22 0.23

0.540 0.548 0.555 0.562 0.568

0.579 0.587 0.595 0.603 0.609

0.617 0.627 0.636 0.644 0.651

0.656 0.666 0.676 0.685 0.693

0.695 0.706 0.716 0.726 0.734

1.6789 1.7372 1.7947 1.851 1.9064

0.24 0.25 0.26 0.27 0.28

0.574 0.579 0.584 0.588 0.592

0.616 0.622 0.627 0.632 0.636

0.658 0.664 0.670 0.675 0.680

0.700 0.707 0.713 0.719 0.724

0.742 0.750 0.757 0.763 0.769

Table 5.1

FOR NORMAL STATE

Q

(cusecs)

DSWL

(ft)

Afflux

(ft)

USWL=

R=

DSWL+

USWL-

Afflux

RBL

(ft)

(ft)

Vo

(ft/sec)

ho=

H o =US

Eo=

V o 2 /2g

WL-CL

H o +h o

(ft)

(ft)

(ft)

h= DSWLCL

(ft)

FOR RETROGRESSED STATE

FOR ACCRETED STATE

ATE

STATE

ATE

h/E o

C/C'

C'

q clear = C'E o 3/2

Q clear

(ft2/sec) (ft3/sec)

% Diff