Discharge of Parent Channel 140 CumecsDischarge of Distributary 15 CumecsFSL of parent channel, u/s 210 mFSL of parent channel, d/s 209.8 mBed width of parent channel, u/s 52 mBed width of parent channel, d/s 46 m

2.5 mFSL of Distributary 209.1 mSilt factor 0.8 mAssume safe exit-gradient 1 : 5

Depth of water in the parent channel, d/s and u/s

Discharge of Parent Channel 140 CumecsDischarge of Distributary 15 CumecsFSL of parent channel, u/s 210 mFSL of parent channel, d/s 209.8 mBed width of parent channel, u/s 52 mBed width of parent channel, d/s 46 m

2.5 mFSL of Distributary 209.1 mSilt factor 0.8 mAssume safe exit-gradient 1/5Glacis slope 2 :1

Design of Cross Regulator

1 Crest Level = 207.5 m

Q=h= u/s FSL - d/s FSL = 0.2 m

d/s FSL - Crest Level = 2.3 m

Clear waterway required B= = 36.92 m

Provide 5 Bays of 7 each with a clear waterway

2 Downstream Floor Level or Cistern LevelDischarge intensity q= 3.7837838 cumecs/m

0.2 m

From Plate 10.1 1.85 m

d/s floor level = 207.95 md/s Bed level = d/s FSL - W d 207.3 m

Provide The Cistern or d/s floor at R.L 207.3

3 Length of d/s floor

From Plate 10.2 1.85 m 1.8 m

2.05 0.8 m

Length Required = 5 m

4 Vertical cut-offs

U/s Cut-offs = 1.43 m

206.07 m

D/s Cut-offs = 1.85 m

205.45 m

Depth of water in the parent channel, d/s and u/s

B.√h(1.69h + 3.54h1)

h1=

Q / h1/2*(1.69h + 3.54h1)

HL =

Ef2 =

d/s FSL - Ef2

For Ef2 = y2 =

For Ef1 = Ef1 + HL m y1 =

5 (y2- y1)

(yu/3)+0.6

Bottom level of u/s cutoff=

(yd/2)+0.6

Bottom level of d/s cutoff=

5 Total Floor Length from Exit Gradient Considerations

H =u/s FSL -d/s bed level 2.7 md= Depth of d/s cutoff 1.85 m

(1/π*√λ) 0.137From Plate 10.2

For (1/π*√λ) 0.137 α = 9Floor Length, b = α*d 16.65 m say 17 m

Min d/s floor length required = (2/3)*b 11.3 m

Glacis length = 0.4 mU/s floor length = 5.3 m

6 Calculations for Uplift Pressuresa Upstream Cut-off

d= 1.43 mb= 17 m

0.084

From khosla Curve l = 6.45

= 18 % f E = 26 %

100%82 % 74 %

Assuming minimum floor thickness (Ft) at U/s end = 0.5 m

2.72 %

76.96 %

b Downstream Cut-offd= 1.85 mb= 17 m

0.109

From khosla Curve l = 5.12

= 29 % = 20 %

= 0 %

Assuming minimum floor thickness (Ft) at D/s end = 0.7 m

GE = (H/d)*(1/π*√λ)

(1/α)=(d/b)

f D

f E1 =f D1 = 100 - f D

f C1 = 100 - f E

Correction for floor thickness at u/s =((D1-C1)xFt)/d

f C1 (corrected)=

(1/α)=(d/b)

f E2 = f Ef D2 = f D

f C2

Correction for floor thickness at u/s =((E2-D2)xFt)/d 3.37 %

25.77 %

7 Floor Thicknesses : D/s FloorAt Toe of Glacis% pressure at toe of the Glacis 58.08 %

Max unbalanced head at toe of glacis due to max static head 1.57 m

Head Due to Dynamic action can be taken as == 0.62 m

Thickness required at toe of glacis = 1.3 m

Provide thickness for a distance 3 from toe of Glacies% Pressure = 50.9 %Max unbalanced head = 1.37 mThickness required = 1.1 m

Provide thickness for a distance 9 from toe of Glacies% Pressure = 32.8 %Max unbalanced head = 0.89 mThickness required = 0.7 m

8 Upstream protectionScour depth D = 1.43 mLaunching apron thickness t = 1.2 mLaunching apron Required L = (2.25*D)/t

2.7 mDownstream protectionScour depth D = 1.85 mLaunching apron Required L = 3.5 m

f E2 (corrected)=

50% (y2-y1)+f*HL

InputsOutputs

= 37 m

Discharge of Parent Channel 140 CumecsDischarge of Distributary 15 CumecsFSL of parent channel, u/s 210 mFSL of parent channel, d/s 209.8 mBed width of parent channel, u/s 52 mBed width of parent channel, d/s 46 m

2.5 mFSL of Distributary 209.1 mSilt factor 0.8 mAssume safe exit-gradient 1/5Glacis slope 2 :1

Design of Distrbutary Head Regulator

From Lacey's Regime diagrams

Bed width of such a channel = 15 mDepth of water in this channel = 1.5 mBed level of distributary = 207.6 m

1 Crest Level Crest Level is kept 0.6 m higher then bed level parent channel

Crest Level = 208.1 m

2 Water Way Q =h = u/s FSL - d/s FSL

= d/s FSL - Crest Level

Clear waterway required B =

Provide 2 Bays of 3 each with a clear waterway

3 Downstream Floor Level or Cistern Levelq= 2.5 cumecs/m

0.9 m

From Plate 10.1 1.8 m

2.7 mFrom Plate 10.2

For 2.7 m y1 = 0.4

For 1.8 m 1.7

R.L of Cistern = 207.3 m

Length Required = 6 m

Depth of water in the parent channel, d/s and u/s

B.√h(1.69h + 3.54h1)

h1

Q / h1/2*(1.69h + 3.54h1)

HL =

Ef2 =

Ef1 = Ef2 + HL

Ef1 =

Ef2 = y2 =

d/s FSL - Ef2

5 (y2- y1)

4 Vertical cut-offs

U/s Cut-offs = 1.43 m

206.07 m

D/s Cut-offs = 1.4 m

205.95 m

5 Total Floor Length from Exit Gradient Considerations

H = u/s FSL -d/s bed level 2.7 md = Depth of d/s cutoff 1.4 m

(1/π*√λ) 0.100From Plate 10.2

For (1/π*√λ) 0.100 α = 13Floor Length, b = α*d 17.55 m say 18 m

Min d/s floor length required = (2/3)*b 12.0 m

Length of d/s Glacis = 1.6 mLength of crest = 1.0 mLength of u/s Glacis = 0.6 m

6 Calculations for Uplift Pressuresa Upstream Cut-off

d= 1.43 mb= 18 m

0.080

From khosla Curve l = 6.80

= 17 % f E = 25 %

100%83 % 75 %

Assuming minimum floor thickness (Ft) at U/s end = 0.5 m

2.64 %

77.58 %

Downstream Cut-offd= 1.4 mb= 18 m

0.075

From khosla Curve

(yu/3)+0.6

Bottom level of u/s cutoff=

(yd/2)+0.6

Bottom level of d/s cutoff=

GE = (H/d)*(1/π*√λ)

(1/α)=(d/b)

f D

f E1 =f D1 = 100 - f D

f C1 = 100 - f E

Correction for floor thickness at u/s =((D1-C1)xFt)/d

f C1 (corrected)=

(1/α)=(d/b)

l = 7.19= 24 % = 17 %

= 0 %

Assuming minimum floor thickness (Ft) at D/s end = 0.7 m

Correction for floor thickness at d/s =((E2-D2)xFt)/d 3.81 %

20.53 %

7 Floor Thicknesses : D/s FloorAt Toe of Glacis% pressure at toe of the Glacis 58.57 %

Max unbalanced head at toe of glacis due to max static head 1.58 m

Head Due to Dynamic action can be taken as == 1.18 m

Thickness required at toe of glacis = 1.3 m

Thickness at beyond 3.00 from toe of Glacies% Pressure = 49.1 %Max unbalanced head = 1.32 mThickness required = 1.1 m

Thickness at beyond 6.00 from toe of Glacies% Pressure = 39.5 %Max unbalanced head = 1.07 mThickness required = 1 m

Thickness at beyond 9.00 from toe of Glacies% Pressure = 26.9 %Max unbalanced head = 0.73 mThickness required = 0.6 m

8 Upstream protectionScour depth D = 1.10 mLaunching apron thickness t = 1.2 mLaunching apron Required L = (2.25*D)/t

2.1 mDownstream protectionScour depth D = 0.6 mLaunching apron Required L = 1.1 m

f E2 = f Ef D2 = f D

f C2

f E2 (corrected)=

50% (y2-y1)+f*HL

InputsOutputs

= 0.9 m

= 1 m

= 3.12 m

each with a clear waterway 6 m

m

m

U/S F.S.L. 210.00

D/S F.S.L. 209.10

CREST 208.1

2 :1 d/s glacis

207.50 U/S Bed Level :1 u/s Cistern Level 207.30

0.50 0.6

1.3 1.1 0.9

206.07

2.800.60 1.00

1.60 3.00 3.00 3.00

12.00 m

18.00 m

HEAD REGULATOR

U/S F.S.L. 210.00

D/S F.S.L. 209.80

CREST 207.5

Cistern Level 207.30

0.50

1.3 1.1 0.7

206.07

205.45

2.44 0.00 0.40 3 5.67 2.83

11.3 m

17 m

CROSS REGULATOR

205.95

3.00