DESIGN OF BARRAGE

Maximum Discharge Qmax = 300000 Cusecs

Minimum 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 Perimeter
Wetted Parameter Pw = 2.666(sqrt(300000))
= 1460 Ft

Using Lacy Looseness
Coeffient
LCC = 1.8
Width b/w Abutment Wa = 1.8 x 1460
2628 Ft

Try:,

40 bays @ 60 = 2400 Ft
25 Piers @ 7 = 175 Ft
1 Fish Ladder = 20 Ft
2 Divide Walls = 35 Ft

Total Wa = 2630 Ft

Discharge b/w
Abutments
q
abt
= (300000/2630)
= 114.07 Cfs

Discharge b/w Weir q
weir
= (300000/2400)
= 125 Cfs

2. Calculation of Lacys Silt Factor (f).


f = [(1844/5000)(300000)
1/6
]
3/5

= 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
2
/1.94)
1/3

= 16.97 Ft

H
0
= 16.97-6
= 10.97 Ft

Velocity Head V
0
= 114.07/16.97
= 6.72 Ft/sec

H
0
= 6.72
2
/(2 x 32.2)
0.7 Ft

E
0
= 10.97 + 0.7
= 11.67 Ft

E
1
= 18 + 0.7 +3
21.7 Ft

Level of E
1
= 582+21.7
= 603.7 Ft

Crest Level = 603.7 11.67
= 592.03 Ft
Maximum D/S Water
Level
= 597.5 Ft
(From Discharge Curve)

h = 597.5 592.03
= 5.47 Ft

h/E
0
= 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
3/2

= 318133.33 Cfs

Q > Qmax OK

Lacys Looseness
Coefficient
LCC = 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
b
1
= 5 x 60
= 300 Ft

Assuming,

Discharge of
Undersluices
q
us
= 120% of main weir.
= 1.2 x 125
= 150 Cfs
Maximum Scour Depth R = 0.9 x (150
2
/1.94)
1/3

= 20.37 Ft

D
0
= 20.37 Ft

Velocity Head V
0
= 150/20.37
= 7.36 Ft/sec

H
0
= 7.36
2
/(2 x 32.2)
0.84 Ft

Maximum USEL = 603.00 + 0.84
603.84 Ft

E
0
= 603.84 589.03
= 14.81 Ft

h

= 14.81-3
= 11.81 Ft

h/E
0
= 11.81/14.81
= 0.79

C/C = 0.80

Assume C = 3.5

C = 3.5 x 0.80
= 2.8

Q
1
& Q
3
= 2.8 x (300 x 2) x 14.81
3/2

= 95750.70 Cfs

Q
main weir
= 3.325 x (2400-600) x 11.67
3/2

238600.00 Cfs

Total Discharge Q
Total
= 95750.70 + 238600.00
= 334350.7 Cfs

Q
Total
> Q
max
(300,000) OK

Now %age Water Passing through
Undersluices
= [95750.70/300000] x 100
= 32 %

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 State
b) Normal State
c) 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 State
Check for Retrogressed State
Check for Accreted State











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 Ft
USEL = 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 h
0
= 4.23
2
/(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
Critical Depth C = [125/32.2]
1/3

= 7.86 Ft
L/C = 1.22/7.86
= 0.15

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

F = 1.92 x 7.86 12.47
= 2.62

Level of Interaction
of jump with glacis
= Crest level F
= 592.03-2.62
= 589.41 Ft

E2 = 603.28 589.41
= 13.87 Ft

Submergency of
Jump
= 589.41 576
= 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 Ft
USEL = 598.95 Ft
RBL = 582 Ft
Crest Level = 592.03 Ft
DSFL = 576 Ft
Dpool = 593-576
= 17 Ft
d/s velocity V = 300000/(17 x 2630)
= 6.71 Ft/sec
d/s velocity head h
0
= 6.71
2
/(2 x 32.2)
= 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]
1/3

= 7.86 Ft
L/C = 5.25/7.86
= 0.67

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

F = 2.52 x 7.86 6.92
= 12.89

Level of Interaction of jump with
glacis
= Crest level F
= 592.03-12.89
= 579.14 Ft

E2 = 593.7 579.14
= 14.56 Ft

Submergence of
Jump
= 579.14 576
= 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.

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 Cfs
Maximum DSWL = 603.5 Ft
USWL = 604.5 Ft
USEL = 605.19 Ft
RBL = 582 Ft
Crest Level = 589.03 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 h
0
= 6.28
2
/(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
2
/32.2]
1/3

= 10.44 Ft
L/C = 1.08/10.44
= 0.10

(K+F)/C = 1.97
(From Crumps Curve)

F = 1.97x 10.44 16.16
= 4.40

Level of Interaction
of jump with glacis
= Crest level F
= 589.03-4.4
= 584.63 Ft

E2 = 604.11 584.63
= 19.48 Ft

Submergence of
Jump
= 584.63 573
= 11.63 Ft

Length of glacis d/s
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 = 87.66 34.89
= 53
say = 55 Ft

b)
Discharge Q = 114900 Cfs
Minimum DSWL = 598 Ft
USWL = 603 Ft
USEL = 603.79 Ft
RBL = 582 Ft
Crest Level = 589.03 Ft
DSFL = 573 Ft
Dpool = 598-573
= 25 Ft
d/s velocity V = 114900/(25 x 600)
= 7.66 Ft/sec
d/s velocity head h
0
= 7.66
2
/(2 x 32.2)
= 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
2
/32.2]
1/3

= 10.44 Ft
L/C = 13.88/10.44
= 1.33

(K+F)/C = 2.9
(From Crumps Curve)

F = 2.9 x 10.44 14.76
= 15.5

Level of Interaction
of jump with glacis
= Crest level F
= 589.03-15.5
= 573.53 Ft

E2 = 598.91 573.53
= 25.38 Ft

Submergence of
Jump
= 573.53 573
= 0.53 Ft

Length of glacis d/s
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
2
)/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
Level for 0.3 million
discharge.

D/s Apron Level = 576 Ft

Depth of Water on
Apron
= 593 576
= 17 Ft

Add 0.5 ft increase in depth for concentration.
D =
Depth of Water with
concentration.

= 17.5 Ft

R D = 35.65 17.5
= 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 stone
apron 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 for
Discharge of 0.3 million cfs.
= 598 Ft

U/S Apron Level = 582 Ft

Depth of Water on
Apron
= 598 582
= 16 Ft

Add 0.5 ft increase in depth for concentration.
D =
Depth of Water with
concentration.

= 16.5 Ft

R D = 25.46 16.5
= 8.96 Ft

Length of Apron to
cover a surface of
scour at 1:3 slope
= Sq.rt (3
2
+ 1
2
) 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 stones
when 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
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
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
2
)/1.94]
1/3

= 23.97 ft

7.4.1 D/S Scour Protection.

Safety Factor =
1.75 for d/s floor critical
condition.

= 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 on
Apron
= 593 576
= 17 Ft

Add 0.5 ft increase in depth for concentration.
D =
Depth of Water with
concentration.

= 17.5 Ft

R D = 41.96 17.5
= 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.

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
= 598 582
= 16 Ft

Add 0.5 ft increase in depth for concentration.
D =
Depth of Water with
concentration.

= 16.5 Ft

R D = 29.96 16.5
= 13.46 Ft

Length of Apron to
cover a surface of
scour at 1:3 slope
= Sq.rt (3
2
+ 1
2
) x (13.46)
= 42.57 ft
= 45 (Say)

Therefore the length
of U/S stone apron in
horizontal 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 stones
when 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
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

5 thick block Apron = 37.25 Ft

Total Length of U/S Stone Apron = 32.15 Ft

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

5 thick block Apron = 22.15 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 D
85
of the soil. These voids will create
smaller 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 1/5D
15
.

Therefore: D
85
(Filter) > 5D
85
(Soil)

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

D
15
(Filter) > 5 D
15
(Soil)

The Concrete Block of 4 x 4 x 4 will be placed over 2ft thick inverted filter which consists of a
9 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 with
fine 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

L
U/S
= 1.5 x 2630
= 3945 ft

2)
Length of Guide bank D/s of Barrage

L
D/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 of
Scour
= 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

Maximum U/S Angle Protected = 140
0


Maximum D/S Angle Protected = 57
0
-80
0

9.1 Determination Levels of Guide Bank.


L =
[d
1
-d
2
}/S +D(1/S C
2
/g) x
[(d
1/
D) (d
2
/D)]


Normal Depth of Flow Without
Weir (Assuming a wide rectangular
channel)
= D

Depth Flow U/S Side of Weir = d
1


Depth of Water At Distance L U/S = d
2


Chezys Coefficient C =
71 (Maximum For Earthen
Channel)


g = 32.2 Ft/sec
2


Bed Slope S = 1/5000

Function Which has been
Calculated By Bresse For Various
Values of d
1
/D
= (d
1
/D)

Using the Above
Formulas


Bed Level = 582.00 Ft

D/S High Flood Level With
Accretion
= 603.00 Ft

D = 603.00 582.00
= 18 Ft

U/S High Flood
Level
= 604.00 Ft

d
1
= 604 - 582
= 19 Ft

S = 1/5000

Assume d
2
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,
d
2
= 18.8 Ft

d
1
/D = 19 / 18
= 1.056 ft

( d
1
/D) = 0.8568
=
From Bresse Back Water
Function Table.


d
2
/D = 18.8/18
= 1.05 Ft

( d
2
/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 at
3945; 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

At D/S Guide Bank:-

Water Level D/S of Barrage = 603 Ft

Free Board = 6 Ft

Level Of Guide bank At D/S = 609 ft

10. Design of Guide Bank Apron.

Length of unlaunched (horizontal)
Apron
= 2.5 (R D)

Length of Launched Apron at 1:3
Slope
= Sq.rt(10) (R D)

As Calculated Previously
t = 34 In
= 3 (Say) Ft

Volume of Stone in
Apron
= 3 x 3.16 (R D)
= 9.5 (R D)

Minimum Thickness
of Unlaunched Apron
= 1.07 x 3
= 3.2 Ft

Mean Thickness Of
Launched Apron
=
[ 9.5 (R D) ] /
[ 2.5 (R D) ]

= 3.80 Ft

Maximum Thickness
of Unlaunched Apron
= 3 x 3.8 -3.2
= 4.2
= 4 (Say) Ft

t = 3 Ft

R = [ q
2
/f ]
1/3

[125
2
/1.94]
1/3

= 20.04 Ft

Nose of Guide Bank:- = 2.25 x 20.04
= 45 Ft

Transition From Nose
to Straight
= 1.5 x 20.04
= 30 Ft

Straight Reach of
Guide Bank
= 1.25 x 20.04
= 25 Ft

D = 18.9 ft

Length of unlaunched (horizontal)
Apron
= 2.5 (R D)
= 2.5 x (20.04 18.9)
= 2.85 Ft

Length of Launched Apron at 1:3
Slope
= Sq.rt(10) (R D)
= Sq.rt (10) x (20.04 18.9)
= 3.6 Ft

Volume of Stone in Apron = 9.5 (R D)
= 9.5 x ( 20.04 18.9)
= 10.83 Ft

11. Design of Marginal Bank.

Top width (With No
Dowel)
=