Barrage 215

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)=


qclear =


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)=


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 =


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 =


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=


E2 = DSEL - Level of Intersection of hydraulic jump =


USEL =USWL +ho =

Dpool = Max DSWL - DSFL =

D/S Velocity = Q/(Dpoolxb1)=


q =Q/b1x2=


E2 = DSEL - Level of Intersection of hydraulic jump =


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` =


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



Minimum U/S Water Level for the 0.48 million cfs Discharge =

Add 0.5 ft increase in depth for concentration


q = 1.2x qweir

R = 0.9(q2/f)1/3=


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


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


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


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


= 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


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


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


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! %


5.1 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!


#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


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


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


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



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)

Documents

Barrage 215