Hydraulics for Causeway
Here is an example for quick calculations of Hydraulics of Causeway, which can be done immediately for estimate purpose.
HYDRAULIC DATA
1 | Catchment Area | 11.25 | sq.mile |
---|---|---|---|
2 | Bed width of define cross section | 25 | metres |
3 | Bank width of define cross section | 42 | metres |
4 | H.F.L. | 102.42 | metres |
5 | Lowest bed level | 98.905 | metres |
6 | Discharge by Inglis formula | 508 | Cumecs |
7 | Protected bed level | 100.9 | metres |
8 | R.T.L. | 99.96 | metres |
9 | Annual rainfall | 1250 | mm |
DESIGN OF VENT
Calculation of area on define C/S at RTL = 100.765Chainage | G.L. | R.T.L. | Diff. between R.T.L. and G.L. | Mean Diff. | Length | Weighted area |
---|---|---|---|---|---|---|
-13.8 | 100.765 | 100.765 | 0 | - | - | - |
-10 | 100.125 | 100.765 | 0.64 | 0.32 | 3.8 | 1.21 |
-5 | 100.045 | 100.765 | 0.71 | 0.67 | 5 | 3.35 |
0 | 99.955 | 100.765 | 0.81 | 0.76 | 5 | 3.8 |
5 | 99.695 | 100.765 | 1.07 | 0.94 | 5 | 4.7 |
10 | 99.365 | 100.765 | 1.4 | 1.23 | 5 | 6.15 |
15 | 99.995 | 100.765 | 0.76 | 1.08 | 5 | 5.4 |
19.8 | 100.76 | 100.765 | 0 | 0.38 | 4.8 | 1.82 |
Total = | 26.43 |
1. Provide vent area 30% of area on defined C/S of R.T.L. i.e. 26.43 sq.m
= 26.43 x 30% = 7.977 Sq.m.
2. No.of 1000 mm. dia. pipe.
= 7.977/ (ã„«/4 x 1.00 2) = 10.15 Nos
Say 11 Nos.
Calculation of area on define C/S at RTL = 101.705
Chainage | G.L. | R.T.L. | Diff. between R.T.L. and G.L. | Mean Diff. | Length | Weighted area |
---|---|---|---|---|---|---|
-22.2 | 101.705 | 101.705 | 0 | - | - | - |
-20 | 101.445 | 101.705 | 0.26 | 0.13 | 2.2 | 0.28 |
-15 | 100.695 | 101.705 | 1.01 | 0.63 | 5 | 3.15 |
-10 | 100.125 | 101.705 | 1.58 | 1.29 | 5 | 6.45 |
-5 | 100.045 | 101.705 | 1.66 | 1.62 | 5 | 8.1 |
0 | 99.955 | 101.705 | 1.75 | 1.7 | 5 | 8.5 |
5 | 99.695 | 101.705 | 2.01 | 1.88 | 5 | 9.4 |
10 | 99.365 | 101.705 | 2.34 | 2.17 | 5 | 10.85 |
15 | 99.995 | 101.705 | 1.7 | 2.02 | 5 | 10.1 |
20 | 100.795 | 101.705 | 0.9 | 1.3 | 5 | 6.5 |
25 | 100.605 | 101.705 | 1.09 | 0.99 | 5 | 4.95 |
30 | 101.165 | 101.705 | 0.53 | 0.81 | 5 | 4.05 |
35 | 101.615 | 101.705 | 0.09 | 0.31 | 5 | 1.55 |
45 | 101.705 | 101.705 | 0 | 0.04 | 10 | 0.4 |
Total = | 74.28 |
Calculation of area between R.T.L. & P.B.L. on existing C/S
Chainage | G.L. | R.T.L. | Diff. between R.T.L. and G.L. | Mean Diff. | Length | Weighted area |
---|---|---|---|---|---|---|
-34 | 100.9 | 100.9 | 0 | - | - | - |
-16 | 99.96 | 100.9 | 0.94 | 0.47 | 18 | 8.46 |
0 | 99.96 | 100.9 | 0.94 | 0.94 | 16 | 15.04 |
16 | 99.96 | 100.9 | 0.94 | 0.94 | 16 | 15.04 |
34 | 100.9 | 100.9 | 0 | 0.47 | 18 | 8.46 |
47 |
3. Provide 16 No. of 1000 mm. dia. pipe as pipes can be accommodated in bed width i.e. 25 metres at define cross section.
PERCENTAGE OBSTRUCTION
A1 = Area provided through vent
16 x 0.7854 x 1.00 2 = 12.56 Sq.m.
A2 = Area of protected bed
i.e. ares between Road top and top of Protected bed.
B = X + Y = Area of define cross section corresponding to protected bed top.
B = 74.24 Sq.m.
%bstruction = (1 - (A1 + A2)/B) x 100
= (1-(12.56 + 47.00)/74.24) x 100
= 19.77% < 30% hence Safe
0 Comments
If you have any doubts, suggestions , corrections etc. let me know