HYDRAULIC DESIGN OF DELIVERY CISTERN (Designing cistern as a vertical drop)

Design Data :

Total Discharge = 79.2 Cumecs

No. of Pipes = 2 No's

Dia. of the Pipe = 4 m

Spacing b/w Pipes = 17 m

i. Details of D/S Canal :

F.S.L. = + 588.525 m

C.B.L. = + 585.025 m

Bed width = 17.40 m

Average G.L. = + 588.500 m

FSD = 3.50 m

1. Hydraulic Calculations :

Width of Cistern = 25 m

M.W.L. in Cistern = + 590.275 m

Delivery level for syphon action = + 592.275 m

C.L. of Delivery Pipe in cistern (D.L.) = M.W.L. - Dia. = + 586.275 m

Cistern Floor Level = D.L. - 1.5 * Dia. = 586.275 - 1.5 * 4

= + 580.275 m

Spillway Crest Level = + 588.525 m

Discharge through each pipe = 39.600 Cumecs

Area of the Pipe = 12.566

Velocity in the pipe = 3.15 m/sec

= 0.51 m

= + 588.781 m 586.78

Depth of water (DC) = 10.00 m

= 2.00 m -1.744

Head of water ( DC) = 6.00 m= 17.321 m

Say 17.500 m

2. Check for Straight length :

Jet Length (X) = V ( 2*H / g ) = 3.49 m

m2

Velocity head ( Vh = V2/2g )

T.E.L ( D.L.+ Dia. + Vh )

Head Loss ( HL )

Length of Cistern = 5 * sqrt ( HL * DC )

25

17.40

+ 580.275 m

17.50 m

Plan of Cistern :

MWL + 590.275

Crest + 588.525 FSL + 588.525

+ 586.275

CBL + 585.025

+ 584.025 m

+ 580.275

Longitudianal Section : -2.814

DESIGN OF OGEE SPILLWAY

Discharge Computations :

Discharge to be considered (Q) = 79.20 Cumecs= 2797 Cusecs

MWL = + 590.275 mCrestLevel = + 588.525 mFloor Level = + 580.275 mTWL = + 588.525 m

The discharge over the spillway is calculated as per the following formula

Q =

= (2/3)sqrt(2g)C

Height of the spillway crest measured from the Floor level ( P ) P = Crest Level - Floor Level

= 588.525 - 580.275= 8.25 m

= 1.75MWL = +588.525 + 1.75

= +590.275

Proposing U/S face of the spillway is vertical :

As per fig.3 of IS 6934:1998, for = 8.25 / 1.75= 4.71

C = 0.741 ( As per IS: 6934 fig.3)

= 2/3 x sqrt(2 x 9.81) x0.741

= 2.2

Length of the spillway required to discharge the maximum flood == 15.55 m

However provide length of spillway = 17.4 m

The discharge over the spillway Q == 2.2 * 17.4 * ((1.75) ^1.5)= 88.62 cumecs Hence OK

Effect of tail water on discharge coefficient (Effect of submergence) .As per fig.5A of IS 6934:1998:

= Discharge coefficient as affected by submergence of crest

= Head of overflow= 1.75

h = MWL - tail water level= +590.275 - +588.525= 1.75

h = 1.75

1.75

As per fig.5A of IS 6934:1998, for h = 1.75 / 1.75

= 1.00

= 0.90

Cd L H 3/2

where Cd

Assume design head (Hd)

P/Hd

Cd

Cd

Q/(CdH3/2)

Cd L H 3/2

Cs

Hd

Hd

Hd

Cs

= 0.90 x 2.2= 2.0

By Using the corrected coefficient, the discharge formula

79.2 = 2.2 x L x 1.75

L = 79.2 / (1.98 * (1.75 ^1.5))= 17m

Shape of the Profile :

Upstream quadrant:

The upstream quadrant of the crest may conform to the ellipse:

+ = 1

As per fig.2of IS 6934:1998, for = 4.714

= 0.29

hence = 0.5075

As per fig.2of IS 6934:1998, for = 4.71

= 0.164

hence = 0.287

Upstream co-ordinates:

0.000 0.2870.050 0.286 X1 = 0.5080.100 0.281 Y1 = 0.2870.150 0.2740.200 0.2640.250 0.2500.300 0.2310.350 0.2080.400 0.1770.405 0.1730.508 0.000

RL of upstream tangent point (TP) = Crest level - depth of U/s quadrant= + 588.525 - 0.287= + 588.238

Downstream Profile:

The downstream quadrant of the crest may conform to the equation:

= ….(1)

= 4.714

Cd

Cs

X12 Y1

2

A12 B1

2

P/Hd

A1/Hd

A1

P/Hd

B1/Hd

B1

X1 Y1

X21.85 K2 Hd

0.85 Y2

P/Hd

As per fig.2of IS 6934:1998, for = 4.714

Hence = 2.0

Substituting the above values in the equation (1)

= 2.0 x 1.609 x

= 3.218

= 0.311 ….(2)

To get tangent point, differentiating and equating to glacis slope

= 0.311 x 1.85 x

= 0.575

Proposing glacis slope of 0.8 : 1(H) (V)

0.575 = 1 = 1.250.8

= 1.2500 = 2.1740.575

= 2.174 1.176

= 2.494 m

= 0.311 x 2.494 1.85

= 1.686 m

Downstream co-ordinates:

0.000 0.0000.500 0.0861.000 0.3111.500 0.6582.000 1.1202.494 1.685

== 588.5 - 1.686

= + 586.839 m

+ 588.525 OCrest level

0.287+ 588.238 0.508

RL of TP 1.686 m

+ 586.839

2.494 RL of TP1

Base width of spillway at TP1 + 586.839 = 0.508 + 2.494= 3.002 m

P/Hd

K2

X21.85 Y2

X21.85 Y2

Y2 X2 1.85

dy2 / dx2 X2 0.85

dy2 / dx2 X20.85

x20.85

x20.85

X2

X2

Substituting this x2 value in Equation (2) above

Y2

Y2

X2 Y2

R.L. of TP1 Crest level - depth D/s profile (Y2)

R.L. of TP1

Y1

X1

Y2

X2

STABILITY ANALYSIS OF OGEE SPILLWAY :

MWL +590.28

1.81.8 FTL +588.53

0.287

1.686

0.508

8.25 2.494 TWL 588.525

0.8

7.96 1 6.564

8.25

+580.28 θ8.25

0.508 2.494 5.251

8.253

8.253

10.00 8.25

Uplift Pressure diagram

Case 1 : When water is up to crest level @ U/S & No water @ D/STaking moments about "A"

Sl Load Description Force L A MomentH V M

1 0.508 x 7.963 x 2.40 9.699 0.254 2.461

2 2.494 x 6.56 x 2.40 39.289 1.755 68.933

3 0.5 x 5.25 x 6.56 x 2.40 41.363 4.752 196.551

4 0.67 x 2.494 x 1.686 x 2.40 6.728 1.443 9.707

5 0.67 x 0.508 x 0.287 x 2.40 0.233 0.317 0.074

6 0.33 x 0.508 x 0.287 x 1.00 0.049 0.127 0.006

7 0.50 x 8.253 x 8.250 x 1.00 -34.042 2.751 -93.647

8 0.5 x 8.25 x 8.25 x 1.00 34.031 2.750 93.58634.031 97.360 277.671

Lever arm = 277.671 / 97.36= 2.852 m

Eccentricity = 2.852 - 8.253 / 2= 1.274 m

Allowable 'e' = 8.253 / 6 = 1.375 No tensionMaximum stress (at Toe ) = 277.671 / 8.253 x ( 1+ 6 x 1.274 / 8.253 )

64.82 Minimum stress (at heel) = 277.671 / 8.253 x ( 1- 6 x 1.274 / 8.253 )

= 2.47

W7

W5

W6 W4

WH1

WH

WH W1 W2

WH1

W3

WU

W1

W2

W3

W4

W5

W6

WU

WH

å H å V å M

t/m2

t/m2

A

Case 2 : When water is up to MWL @ U/STaking moments about "A"S.No Load Description Force L A Moment

H V H V

1 0.508 x 7.963 x 2.40 9.699 0.254 2.461

2 2.494 x 6.56 x 2.40 39.289 1.755 68.933

3 0.5 x 5.25 x 6.56 x 2.40 41.363 4.752 196.551

4 0.67 x 2.494 x 1.686 x 2.40 6.728 1.443 9.707

5 0.67 x 0.508 x 0.287 x 2.40 0.233 0.317 0.074

6 0.333 x 0.508 x 0.287 x 1.00 0.049 0.127 0.006

7 0.508 1.750 x 2.400 x 1.00 2.132 0.254 0.541

8 0.50 x 8.253 x 10.000 x 1.00 -41.264 2.751 -113.512

9 0.5 x 8.25 x 8.25 x 1.00 34.031 2.750 93.586

10 1.75 x 8.25 x 1.00 14.438 4.125 59.55548.469 58.228 317.902

Lever arm = 317.902 / 58.228= 5.460 m

Eccentricity = 5.46 - 8.253 / 2= 1.333 m

Allowable 'e' = 8.253 / 6 1.375 No tensionMaximum stress (at toe) = 317.902 / 8.253 x ( 1+ 6 x 1.333 / 8.253 )

= 75.86 Minimum stress (at heel) = 317.902 / 8.253 x ( 1- 6 x 1.333 / 8.253 )

= 1.18

Stresses @ Case1 Case2

Maximum stress (at heel ) 64.82 75.86Minimum stress (at Toe) 2.47 1.18

Factor of Safety against sliding

F = [((w – u) tan f ) / Ff) + (C.A./Fc)] / P

F = Factor of safety against slidingw = Total mass of the damu = Total uplift force

tan f = Co-efficient of internal frictionC = Cohesion of the material at the plane consideredA = Area under consideration for cohesionFf = Partial factor of safety inrespect of frictionFc = Partial factor of safety inrespect of cohesion andP = Total horizontal force

C = 100 t/m2

f = 30 Tan f = 0.577A = 8.2527 Sqm

Allowable factor of Safety = 1Table-1 of IS 6512-1984

= 1.5

= 3.6

Sl åW P Ff (åW - åU)Tanf/Ff

1 A 97.36 - 1.5 37.474 3.60 - -

2 B 97.36 34.031 1.5 64.907 3.60 229.24 8.643 OK

W1

W2

W3

W4

W5

W6

W7

WU

WH

WH1

å H å V å M

t/m2

t/m2

t/m2 t/m2

O

Ff

Fc

Load Combin

ationFC CA / FC

Factor of

safety

CHECK

3 C 58.23 48.469 1.5 22.412 3.60 229.24 5.192 OK

ENERGY DISSIPATION ARRANGEMENTS ON D/S OF SPILLWAY

Stilling Basin Type energy dissipation arrangement is proposed as per IS:4997: 1985

Spillway Crest level = + 588.525 mCistern Top level = + 591.275 m

Cistern Floor Level = + 580.275 mTWL = + 588.525 mMWL = + 590.275 m

Let

= Depth of overflow over spillway (m)

q =

= Depth of flow entering to Stilling basin =

= Critical water Depth of flow =

= Sequent Depth

= Theoretical velocity of flow in m/s =

= Actual velocity of flow in m/s

= D/s Retrogressed water Level = 588.025 m

Assuming Apron Level = 584.025 m

= 1.8

= + 589.40 - +584.025= 5.375

Q

q = QL

q = 1.98 x 1.8 1.5= 4.584

== sqrt 2 x 9.81 x 5.375= 10.27 m/s

From fig 7 of IS 7365 - 1985

1.750 m and = 5.375 m

= 0.98

= 0.98 x Vt= 0.98 x 10.270= 10.065 m/s

= q

(As per IS : 7365 - 1985 Clause 4.3.2.1)= 4.584

10.065

= 0.46 m

Hd

Discharge intensity per metre width in m3/s/m

d1 q / Va

dc (q2 / g) 1/3

d2

Vt sqrt ( 2 g H2 )

Va

H2 Depth from (FTL+Hd/2) to Apron Level (m)

Hd

H2

= Cd x L x Hd3/2

= Cd x Hd3/2

m3/s/m

Vt sqrt ( 2 g H2 )

for Hd = H2

Va

Vt

Va

d1

Va

d1

( As per IS: 4997-1968 Cla.4.3.2 )

g

= 21.01 = 1.289 m9.81

= 1.29 m

sqrt(g* d1)

= 10.065 = 4.7382.124

From IS : 4997 - 1985 Clause 4.3.2

= sqrt 1 + - 12

= 0.46 13.44 - 12

= 3.157 m=

Level at which jump will form = D/s Retrogressed water level -= + 588.025 - 3.16= + 584.868

say + 584.868 m

Provide Cistern Level at = + 584.025 m

From IS : 4997 - 1985 Clause 4.3.2

= -

4

= 3.157 - 0.464 x 0.46 x 3.16

= 19.625.81

= 3.38 m

Length of Cistern: Length of Cistern = 5 (d2 - d1)

= 5 x 3.16 - 0.460= 14.00 m

As per Cl 4.3.4.2 of IS : 4997 - 1968 and fig 8A (Recommended Length for Basin 1)As per Cl 4.3.5.1 of IS : 4997 - 1968 and fig 9A (Recommended Length for Basin 2)

= 4.738 (Basin 2 stilling basin)

= 4.8

= 4.8 x 3.16= 15.15 m

Critical Depth (dc) = q2 1/3

dc 1/3

dc

Froud Number of jet entering bucket ( Fr1) = Va

Fr1

d2 d1 8 Fr 2

d2

d2

Loss of Head (HL)

HL d2 d13

x d1 x d2

3

HL

For Froude Number (Fr)

Lb

d2

Lb

However provide 16.00 m length of stilling Basin at the level of + 584.025 m

As per Cl 4.3.5.2 of IS : 4997 - 1968 and fig 9B (Recommended Height of Basin blocks)

= 1.5

Height of Basin Blocks = 1.5 x 0.46= 0.69 m

SCHEMATIC DIAGRAM OF BASIN-I

0.47355 DentatedSill

0.6314

2.0 : 1

16.00 m

hb

d1

hb

Calculation of Uplift Pressure and Floor Thickness :

MWL +590.281.75

Crest +588.53

+588.5258.25

0.8 4.500

1 0.460 0.10 m thick wearing coat

+580.28 + 584.0250.51 2.494 5.2512 0.80

3.00 1.90

3.00+577.28 1

0.5581.03

0.5 8.253 8.000 8.000 0.5Solution :

Discharge over the Spillway = 79.2 CumecsLength of Spillway = 17.4 mWidth of Canal = 17.4 m

d2 =

d1=

Scour Depth Calculations :1. At Upstream side :

Discharge per m width of canal, q = Q/b = 79.2 / 17.4= 4.551724 cumecs/m

Silt factor,f = 2

Mean Scour depth , R = 1.34 x ( 4.552 ^2 / 2 ) ^ (1/3) ) = 2.921 m

1.5 x 2.921 = 4.382 m

= 590.275 - 4.382 585.893Depth of cut-off = 580.275 - 585.893 = -5.62 m

Depth of U/S Curtain wall = U/S FSD /3 10 / 3 = 3.333 m

However provide 3.000 m below the U/S bed level

2. At Downstream side :Discharge per m width of canal, q = Q/b = 79.2 / 17.4

= 3.958 cumecs/m

Silt factor , f = 2

Mean Scour depth,R = 1.34 x ( 3.958 ^2 / 2 ) ^ (1/3) ) = 2.661 m

2.0 x 2.661 = 5.322 m

= 588.525 - 5.322 =583.20Depth of cut-off = 584.025 - 583.2 = 0.822 m

Depth of D/S Curtain wall = D/S FSD / 2 = 4.5 / 2 = 2.250 mHowever provide 3.000 m below the D/S bed level

Mean Scour depth R = 1.34 x (q2 / f )1/3

Max Scour depth , Rmax = 1.5 R =

Scour Level = MWL - Rmax

Mean Scour depth R = 1.34 x (q2 / f )1/3

Max Scour depth , Rmax = 2.0 R =

Scour Level = MWL - Rmax

d = Depth of D/s cut-off = 3.000 mH = 8.250 m

b = floor length 0.5 +8.253+ 8 + 8 + 0.5 = 25.253 mα = b/d = = 8.418

= 4.738

= 0.402 Re

i). At U/S cutt off wall :α = b/d = 25.2527 / 3 = 8.418

= 4.738

= 30 .4%

= 70 %

ii). At D/S cut off wall :α = b/d = = 8.418

= 4.738

= 30%

25.25 30 %

70% D/s cut-offafter Prejump length

D/S of Body wall 8.500U/s cut-off 16.500

Calculation of Floor Thickness :1. At D/S Body wall :i. Overflow condition :Uplift presure at the D/S of body wall = 30 + ( 70 - 30 ) x 16.5 / 25.253

= 56 %The Cistern is resting on Impermeable foundations, hece we consider 50% of uplift pressure.

50 % Uplift Pressure = 28 % Pressure head = ( H + d - d2 ) = 8.25 + 1.75 - 4.5

= 5.50 mEffective uplift = Pressure Head x Uplift = 5.5 x 28

100= 1.540 m

Thickness of floor required = 1.54 / 1.4= 1.10 m

ii. Non-Overflow condition : Pressure head = H = 8.25Effective uplift = Pressure Head x Uplift = 8.25 x 28

100= 2.311 m

Thickness of floor required = 2.311 / 2.4= 0.96 m

Floor Thickness is provided for Maximum of above 2 conditions .Thickness of floor provided = 2.00 m including wearing coat

EXIT GRADIENT (GE)

λ ={1+sqrt(1+α2)}/2

GE = 1x H/(sqrt(λ)xΠxd

Uplift Pressures :

λ ={1+sqrt(1+α2)}/2

ØE =(1/Π) x cos-1 x ((λ-2)/λ)

Øc1 = 100-ØE =

λ ={1+sqrt(1+α2)}/2=

ØE =(1/Π) x cos-1 x ((λ-2)/λ) =

2. After Prejump length :Uplift presure at after prejump length = 30 + ( 70 - 30 ) x 8.5 / 25.253

= 43.59 %The Cistern is resting on Impermeable foundations, hece we consider 50% of uplift pressure.

50 % Uplift Pressure = 22 % Pressure head = ( H + d - d2 ) x U = 8.25 + 1.75 - 4.5

= 5.500 mEffective uplift = 5.5 x 21.8

= 1.199 mThickness of floor required = 1.199 / 1.4

= 0.856 mThickness of floor provided = 0.90 m including wearing coat