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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