design of barrage(has)

8/12/2019 Design of Barrage(HAS)

1/22

DESIGN OF BARRAGE

Maximum Discharge Qmax = 300000 CusecsMinimum Discharge Qmin = 12000 Cusecs

River Bed Level RBL = 582 Ft

Highest Flood Level HFL = 600 Ft

Lowest Water Level LWL = 587 Ft

Minimum Pond Level = 598

No of Canals on Right

Bank= 2

No of Canal on Left Bank = 1

Maximum Discharge For

1 Canal= 3500 Cusecs

Slope of River = 1 Ft/mile

DESIGN OF BARRAGE PROFILE FOR OVERFLOW CONDITION

1.Minimum stable Wetted PerimeterWetted Parameter Pw = 2.666(sqrt(300000))

= 1460 Ft

Using Lacy Looseness

CoeffientLCC = 1.8

Width b/w Abutment Wa = 1.8 x 1460

2628 Ft

Try:,

40bays @ 60 = 2400 Ft

25 Piers @ 7 = 175 Ft

1 Fish Ladder = 20 Ft


2/22

2 Divide Walls = 35 Ft

Total Wa = 2630 Ft

Discharge b/w

Abutments

qabt = (300000/2630)

= 114.07 Cfs

Discharge b/w Weir qweir = (300000/2400)

= 125 Cfs

2. Calculation of Lacys Silt Factor(f).

f = [(1844/5000)(300000) ]

= 1.94

3. Fixation of Crest Level.

Assume

Afflux = 3 Ft

Height of Crest P = 6 Ft

Maximum Scour Depth R = 0.9 x (114.07 /1.94)

= 16.97 Ft

H0 = 16.97-6= 10.97 Ft

Velocity Head V0 = 114.07/16.97

= 6.72 Ft/sec

H0 = 6.72 /(2 x 32.2)

0.7 Ft

E0 = 10.97 + 0.7

= 11.67 Ft

E1 = 18 + 0.7 +3

21.7 Ft

Level of E1 = 582+21.7

= 603.7 Ft


3/22

Crest Level = 603.711.67

= 592.03 Ft

Maximum D/S Water

Level= 597.5 Ft

(From Discharge Curve)

h = 597.5592.03

= 5.47 Ft

h/E0 = 5.47/11.67

= 0.47

C/C = 0.95

Assume C = 3.5

C = 3.5 x 0.95= 3.325

Q = 3.325 x 2400 x 11.67

= 318133.33 Cfs

Q > Qmax OK

Lacys Looseness

CoefficientLCC = 2630/1660

= 1.58

4. Design of Undersluices.

Fix Crest Level of undersluices 3 ft below the main weir.

Crest Level of

Undersluices= 592.03-3

= 589.03 ft

Assume 5 Bays acts as Undersluices on both sides

b1 = 5 x 60= 300 Ft

Assuming,

Discharge ofUndersluices

qus = 120% of main weir.

= 1.2 x 125


4/22

= 150 Cfs

Maximum Scour Depth R = 0.9 x (150 /1.94)

= 20.37 Ft

D0 = 20.37 Ft

Velocity Head V0 = 150/20.37

= 7.36 Ft/sec

H0 = 7.36 /(2 x 32.2)

0.84 Ft

Maximum USEL = 603.00 + 0.84

603.84 Ft

E0 = 603.84589.03

= 14.81 Ft

h = 14.81-3

= 11.81 Ft

h/E0 = 11.81/14.81

= 0.79

C/C = 0.80

Assume C = 3.5

C = 3.5 x 0.80

= 2.8

Q1 & Q3 = 2.8 x (300 x 2) x 14.81

= 95750.70 Cfs

Qmain weir = 3.325 x (2400-600) x 11.67

238600.00 Cfs

Total Discharge QTotal = 95750.70 + 238600.00= 334350.7 Cfs

QTotal > Qmax(300,000) OK

Now %age Water Passing through

Undersluices= [95750.70/300000] x 100

= 32 %


5/22

The Undersluices are fixed at Crest = 589.03 ft

Number of Bays on each side = 5

5. Determination of Water Level and Energy Level.

We shall determine the water level and energy levels of the barrage for three states of the river

i.e

a) Retrogressed Stateb) Normal Statec) Accreted State

The Curve of the three states of the river are obtained by the actual stage-discharge relationship

at the barrage site.

5.1 Check for Main Weir

Check for Normal StateCheck for Retrogressed State

Check for Accreted State


6/22

FOR UNDERSLUICES:-

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 Crumps method and determination

of floor length.

a)

Discharge Q = 300000 Cfs

Maximum DSWL = 603 Ft

USWL = 604 FtUSEL = 604.5 Ft

RBL = 582 Ft

Crest Level = 592.03 Ft

DSFL = 576 Ft

Dpool = 603-576

= 27 Ft

d/s velocity V = 300000/(27 x 2630)

= 4.23 Ft/sec

d/s velocity head h0 = 4.23 /(2 x 32.2)

= 0.28 Ft

DSEL = 603+0.28= 603.28 Ft

K = 604.5-592.03

= 12.47 Ft

L = 604.5-603.28

= 1.22 Ft

q = 300000/2400

= 125 Cfs/ft


7/22

Critical Depth C = [125/32.2]

= 7.86 Ft

L/C = 1.22/7.86

= 0.15

(K+F)/C = 1.92(From Crumps Curve)

F = 1.92 x 7.8612.47

= 2.62

Level of Interaction

of jump with glacis= Crest levelF

= 592.03-2.62

= 589.41 Ft

E2 = 603.28589.41= 13.87 Ft

Submergency ofJump

= 589.41576

= 13.41 Ft

Length of glacis d/s

of jump= 3 x 13.41

= 40.23 Ft

Length of silting pool = 4.5 x 13.87

= 62.42 Ft

Length of d/s floor = 62.42 - 40.23

= 23

say = 25 Ft

b)

Discharge Q = 300000 Cfs

Minimum DSWL = 593 Ft

USWL = 598 FtUSEL = 598.95 Ft

RBL = 582 Ft


DSFL = 576 Ft

Dpool = 593-576

= 17 Ft

d/s velocity V = 300000/(17 x 2630)


8/22

= 6.71 Ft/sec


= 0.70 Ft

DSEL = 593+0.7

= 593.7 Ft

K = 598.95-592.03= 6.92 Ft

L = 598.95-593.7

= 5.25 Ft

q = 300000/2400

= 125 Cfs/ft

Critical Depth C = [125/32.2]

= 7.86 Ft

L/C = 5.25/7.86

= 0.67

(K+F)/C = 2.52(From Crumps Curve)

F = 2.52 x 7.866.92

= 12.89

Level of Interaction of jump with

glacis= Crest levelF

= 592.03-12.89

= 579.14 Ft

E2 = 593.7579.14

= 14.56 Ft

Submergence of

Jump= 579.14576

= 3.14 Ft

Length of glacis d/s of jump = 3 x 3.14

= 9.42 Ft

Length of silting pool = 4.5 x 14.56= 65.52 Ft

Length of d/s floor = 65.52- 9.42

= 56

say = 60 Ft

Hence we shall provide d/s floor 60 ft long.


9/22

6.1 Fixation of d/s floor levels for undersluices section using Crumps method and determination

of floor length

95750.70

a)

Discharge Q = 114900 CfsMaximum DSWL = 603.5 Ft

USWL = 604.5 Ft

USEL = 605.19 Ft

RBL = 582 Ft


DSFL = 573 Ft

Dpool = 603.5-573

= 30.5 Ft

d/s velocity V = 114900/(30.5 x 600)

= 6.28 Ft/sec

d/s velocity head h0 = 6.28 /(2 x 32.2)= 0.61 Ft

DSEL = 603.5+0.61

= 604.11 Ft

K = 605.19-589.03

= 16.16 Ft

L = 605.19-604.11

= 1.08 Ft

q = 114900/600

= 191.5 Cfs/ft

Critical Depth C = [191.5 /32.2]

= 10.44 Ft

L/C = 1.08/10.44

= 0.10

(K+F)/C = 1.97

(From Crumps Curve)

F = 1.97x 10.4416.16

= 4.40

Level of Interactionof jump with glacis

= Crest levelF

= 589.03-4.4

= 584.63 Ft

E2 = 604.11584.63

= 19.48 Ft


10/22

Submergence of

Jump= 584.63573

= 11.63 Ft


of jump

= 3 x 11.63

= 34.89 Ft

Length of silting pool = 4.5 x19.48

= 87.66 Ft

Length of d/s floor = 87.6634.89

= 53

say = 55 Ft

b)

Discharge Q = 114900 CfsMinimum DSWL = 598 Ft

USWL = 603 Ft

USEL = 603.79 Ft

RBL = 582 Ft


DSFL = 573 Ft

Dpool = 598-573

= 25 Ft

d/s velocity V = 114900/(25 x 600)

= 7.66 Ft/sec


= 0.91 Ft

DSEL = 598+0.91

= 598.91 Ft

K = 603.79-589.03

= 14.76 Ft

L = 603.79-598.91

= 13.88 Ft

q = 114900/600

= 191.5 Cfs/ft

Critical Depth C = [191.5 /32.2]

= 10.44 Ft

L/C = 13.88/10.44

= 1.33

(K+F)/C = 2.9

(From Crumps Curve)


11/22

F = 2.9 x 10.4414.76

= 15.5

Level of Interactionof jump with glacis

= Crest levelF

= 589.03-15.5= 573.53 Ft

E2 = 598.91573.53

= 25.38 Ft

Submergence of

Jump= 573.53573

= 0.53 Ft


of jump = 3 x 0.53= 1.59 Ft

Length of silting pool = 4.5 x 25.38

= 114.21 Ft

Length of d/s floor = 114.21-1.59

= 113

say = 115 Ft

Hence we shall provide d/s floor 115 ft long.

7. Scour Protection.

Assume 20% concentration,

q = 125 x 1.2

= 150 Cfs

R = 0.9 x [(150 )/1.94]

= 20.37 ft

7.1 D/S Scour Protection.

Safety Factor =1.75 for d/s floor critical

condition.

= 1.75 x 20.37

= 35.65 Ft

Minimum D/S Water = 593 Ft


12/22

Level for 0.3 million

discharge.

D/s Apron Level = 576 Ft

Depth of Water onApron

= 593576

= 17 Ft

Add 0.5 ft increase in depth for concentration.

D =Depth of Water with

concentration.

= 17.5 Ft

R D = 35.6517.5

= 18.15 Ft

Length of Apron to cover a surface

of scour at 1:3 slope= Sq.rt (3

2+ 1

2) x (18.15)

57.40 Ft

Therefore the length of D/S stoneapron in horizontal position

= 57.40 x {1.25t/1.75t]

41 Ft

7.2 U/S Scour Protection.

Safety Factor = 1.25 for U/S Floor

R = 1.25 x 20.37

= 25.46 Ft

Minimum U/S Water Level forDischarge of 0.3 million cfs.

= 598 Ft

U/S Apron Level = 582 Ft

Depth of Water onApron = 598582

= 16 Ft



concentration.

= 16.5 Ft


13/22

R D = 25.4616.5

= 8.96 Ft

Length of Apron to

cover a surface ofscour at 1:3 slope

= Sq.rt (32

+ 12

) x (8.96)

= 28.32 ft

= 30 (Say)

Therefore the length

of U/S stone apron in

horizontal position

= 30 x {1.25t/1.75t]

= 21.45 ft

7.3 Thickness of Aprons.

The Basic Idea in determining the thickness is that there should be sufficient volume of stoneswhen launching that it fully covers the surface of scour at maximum possible slope (1:3).Now t

for medium sand and a slope of 12/mile is 34.

Thickness of Stone Apron inHorizontal position

= 1.75 x (34/12)

= 5 Ft

While the size of the concrete blocks over the filter will be 4 x 4 x 4 , the thickness of the

stone apron shall be 5 ft.

Summary,

Total Length of D/S Stone Apron = 41 Ft

4 thick block Apron = 13 (1/3 of Total Length) Ft

5 thick block Apron = 28 Ft

Total Length of U/S Stone Apron = 21.45 Ft

4 thick block Apron = 7 (Block Size 4 x 4x 4) Ft

5 thick block Apron = 14.45 Ft

7.4 Scour Protection For Undersluices.

Proceeding on similar lines above, the length and thickness for the U/S & D/s loose stone aprons


14/22

may be calculated for the undersluices portion. In this case since the discharge intensity will be

higher, the maximum scour will be more than that in previous case & consequently the length

will be greater as compared to those for the normal weir section.

Assume 20% concentration,

q = 159.58 x 1.2= 191.5 Cfs

R = 0.9 x [(191.5 )/1.94]

= 23.97 ft

7.4.1 D/S Scour Protection.

Safety Factor =1.75 for d/s floor criticalcondition.

= 1.75 x 23.97

= 41.96 Ft

Minimum D/S Water

Level for 0.3 million

discharge.

= 593 Ft

D/s Apron Level = 576 Ft

Depth of Water onApron

= 593576

= 17 Ft



concentration.

= 17.5 Ft

R D = 41.9617.5

= 24.46 Ft

Length of Apron to cover a surface

of scour at 1:3 slope= Sq.rt (3

2+ 1

2) x (24.46)

= 77.35 Ft

Therefore the length of D/S stone

apron in horizontal position= 77.35 x {1.25t/1.75t]

= 55.25 Ft

7.4.2 U/S Scour Protection.


15/22

Safety Factor = 1.25 for U/S Floor

R = 1.25 x 23.97

= 29.96 Ft

Minimum U/S Water Level for

Discharge of 0.3 million cfs.= 598 Ft

U/S Apron Level = 582 Ft

Depth of Water on

Apron= 598582

= 16 Ft


D = Depth of Water withconcentration.

= 16.5 Ft

R D = 29.9616.5

= 13.46 Ft

Length of Apron tocover a surface of

scour at 1:3 slope

= Sq.rt (32+ 1

2) x (13.46)

= 42.57 ft

= 45 (Say)

Therefore the length

of U/S stone apron inhorizontal position

= 45 x {1.25t/1.75t]

= 32.15 ft

7.4.3 Thickness of Aprons.

The Basic Idea in determining the thickness is that there should be sufficient volume of stoneswhen launching that it fully covers the surface of scour at maximum possible slope (1:3).Now t

for medium sand and a slope of 12/mile is 34.

Thickness of Stone Apron in

Horizontal position= 1.75 x (34/12)

= 5 Ft

While the size of the concrete blocks over the filter will be 4 x 4 x 4 , the thickness of the


16/22

stone apron shall be 5 ft.

Summary,

Total Length of D/S Stone Apron = 55.25 Ft

4 thick block Apron = 18 (1/3 of Total Length) Ft


Total Length of U/S Stone Apron = 32.15 Ft

4 thick block Apron = 10 (Block Size 4 x 4x 4) Ft


8. Inverted Filter Design.

Extensive experiments have shown that it is not necessary for a filter to restrain all the particles

of the soil. Instead it may restrain 15% coarser or the D85of the soil. These voids will createsmaller openings to trap finer soil particles. Therefore the diameter of the openings of the filter

must be less then D85 of the soil. Since Effective pore diameter is about 1/5D15.

Therefore: D85(Filter) > 5D85(Soil)

If Filter provide free drainage than it much more pervious than the soil and in such case thecriterion is.

D15(Filter) > 5 D15 (Soil)

The Concrete Block of 4 x 4 x 4 will be placed over 2ft thick inverted filter which consists of a9 layer of Single (3-6) over 9 layer of coarse shingle (3/4 -3) over 6 layer of fine shingle

(3/16 ).there are 2 inches wide spacings on the sides of the blocks which are filled withfine shingle to provide free seepage flow.

9. Design of Guide Bank.

1)

Length of each guide bank measured in a straight line along the barrage U/S is

LU/S = 1.5 x 2630

= 3945 ft

2)


17/22

Length of Guide bank D/s of Barrage

LD/S = 0.2 x 2630

= 526 Ft

3)For the nose of the S/S guide

bank & the full length of D/S guide

bank use Laceys depth.

= 1.75 x 20.37

= 15.28 Ft

For remaining U/s guide bank

Laceys depth= 1.25 x 20.37

= 25.46 Ft

4)

Possible Slope ofScour

= 1 : 3

5)

Free Board U/S = 7 (Above HFL Level) Ft

Free Board D/S = 6 (Above HFL Level) ft

These Free board also contains allowance for Accretion.

6)

Top of Guide Bank Width = 40 Ft

7)

Side Slope of Guide Bank = 1 : 2

8)

Minimum Apron Thickness = 4 Ft

Length of Barrage = 2630 Ft

Length of U/S Guide Bank = 3945 ft

Length of D/S Guide Bank = 526 ft

Radius of U/S Curved Part = 600 Ft

Radius of D/s Curved Path = 400 Ft


18/22


19/22

Assume d2 At the nose of the guide Bank & Calculate l. If the assumed value is correct, the L

will come out to be equal to the Length of the Guide Bank.

Assume,

d2 = 18.8 Ft

d1/D = 19 / 18

= 1.056 ft

( d1/D) = 0.8568

=From Bresse Back WaterFunction Table.

d2/D = 18.8/18

= 1.05 Ft

( d2/D) = 0.8968

Substituting the Values in Above Formula

L =( (19-18.8) / (1/5000) ) + 18((50000(71

2/32.2))[0.8968

0.8568]

L = 3987 ft

>= 3945 Ft

(OK)

L = 3987 Ft

Rise I River Bed Level = 3945 / 5000

0.789

From The Level At Barrage

water level along h/w axis at3945; U/s of Barrage.

= 582 + 0.789 + 18.8

= 601.589 Ft

Level At the Nose of U/S Guide

Bank

= 601.589 + free board

= 601.589 + 7

= 608.598 Ft

Level at the Barrage = HFL + Free Board

= 600 + 7

= 607 Ft


20/22


21/22

Transition From Nose

to Straight= 1.5 x 20.04

= 30 Ft

Straight Reach ofGuide Bank

= 1.25 x 20.04

= 25 Ft

D = 18.9 ft

Length of unlaunched (horizontal)

Apron= 2.5 (R D)

= 2.5 x (20.0418.9)

= 2.85 Ft

Length of Launched Apron at 1:3Slope

= Sq.rt(10) (R D)

= Sq.rt (10) x (20.0418.9)

= 3.6 Ft

Volume of Stone in Apron = 9.5 (R D)

= 9.5 x ( 20.0418.9)

= 10.83 Ft

11. Design of Marginal Bank.

Top width (With NoDowel)

=


22/22

design of barrage(has)

Documents