Design of Aqueduct
Design of Aqueduct
Design of Aqueduct
1). CANAL
C.B.L. = 300.641 m
Q = 141.20 cusec
T.B.L = 307.141 m
Bed Width = 12.0 m
F.S.D. = 5.50 m
F.B. = 1.0 m
2). DRAINAGE
Bed Slope 1 in = 765
Q = 2358.00 cusec
H.F.L. (Calculated) = 287.450 m
Bed Level = 281.325 m
HYDRAULIC DESIGN
1). DRAINAGE WATERWAY
In the Transition, the side slopes of the section will be warped from 1.5 : 1 to vertical.
At Section 4 - 4
2
Area of Section = (B + nD) x D = 111.375 m
Velocity = Q/A = 1.268 m/s
Velocity Head = V2/2g = 0.082 m
R.L. of Bed (Given) = 300.641 m
R.L. of Water Surface = 306.141 m
R.L. of T.E.L. = 306.223 m
At Section 3 -3
2
Area of Section = 49.500 m
Velocity = Q/A = 2.853 m/s
2
Velocity Head = V /2g = 0.415 m
Loss of Head in Expansion from 3 - 3 to 4 - 4 = 0.3 x (V32- V42)/2g
= 0.100 m
R.L. of T.E.L. = 306.323 m
R.L. of Water Surface = 305.908 m
R.L. of Bed (to maintain constant water depth) = 300.408 m
At Section 2 -2
R.L. of T.E.L = 306.481 m
R.L. of Water Surface = 306.066 m
R.L. of Bed = 300.566 m
At Section 1 - 1
2 2
Loss of head in Contraction from section 1 - 1 to 2 - 2 = 0.2 x ( V2 - V1 ) / 2g
= 0.067 m
R.L. of T.E.L. = 306.547 m
R.L. of Water Surface = 306.465 m
R.L. of Bed ( to maintain constant depth) = 300.965 m
(B). TRANSITIONS
a) Expansion Transition
The transitions shall be worked out on the basis of Mitra's formula as given below:
Bx = BC x BF x L / ( L x BC - (Bc - Bf) X )
where,
Bf = 9.0 m BC = 12.0 m L = 4.5 m
Substituting we get,
Bf = 486/(54 - 3 X)
X = 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 m
BX = 0.0 9.3 9.5 9.8 10.1 10.5 10.8 11.2 11.6 12.0 m
b) Contraction Transition
( C ). TROUGH
The Trough shall be divided in two equal compartments of 4.5 m wide by providing an intermediate wall of 0.5
m thick.
The road shall be carried out on the top of the left cpmpartments. A free board of 1.0 m above normal water depth would
be suffecient.
The bottom level of the slab over left compartment shoul therefore kept at = 6.5 m above the bed level
of trough