Question:
Answer:
Solution-1
As given in question and graph below
We have to put these two point in the equation,
As per calculation the bottom most point of channel = coordinate point is (0, -3.375)
Putting the value of a in equation
a = 0.844
Equation for the channel profile = y = 0.844x2 – 3.375
Now I have to calculate embankment point i.e. B & C. The tangents are [X = -2 and 2]
Solution-2
Differentiating the equation, we get,
Similarly,
Solution 4
Suppose in the given figure the depth of the channel is = h
The cross section of channel is parabola, and top width of channel is 4m and depth is = h,
But minimum area of channel is = 9 m2
Therefore h = 3.375
As given in question the excavated earth from channel will be used for double embankment of same profile, In this condition the cross-section area of embankment = ½ of cross section area of channel.
Cross section area of channel = 9 m2
∴ Area of each embankment = 4.5 m2
∴ Base of the embankment, B = 0.25 + 2
B = 0.25 + 0.593h
Solution 5
∴ Area =
0.593h2 + 0.05h – 18 = 0
∴ h= 5.1 m
Solution 6
The full height of channel = 5.1 +3.375 = 8.475 m
The maximum safe flow height = 0.7 x 8.475 = 5.9325
The width of channel at ground level = 4 m
After ground level the shape of flow cross section is trapezoidal (Shaded portion).
∴ Water top width=
∴ Area of safe maximum flow = parabolic + trapezoidal area
= 9 +
= 21.125 m2 Ans
Solution 7
If the channel is full of water at any point of time, then wetted perimeter of the channel will be
= Full length of parabolic portion+ 2x(side of embankment) as a symmetrical cross section determine only single half of the section.
Parabolic portion
∴ Total length = (Parabolic portion)
Now side wall of trapeziums
For two embankments = = 2 x 5.32 = 10.64
Total wetted embankment = 10.64+7.84 = 18.48 m
∴ per 1 meter length of channel side wall or lining area = 18.48×1 =18.48 m2
∴ Total cost of lining per unit length = 18.48 * 40 = $ 739.12
Solution 8.
Parabolic portion
∴ Total length = (Parabolic portion)
Now side wall of trapeziums
For two embankments = = 2 x 5.32 = 10.64
Total wetted embankment = 10.64+7.84 = 18.48 m
∴ per 1 meter length of channel side wall or lining area = 18.48×1 =18.48 m2
∴ Total cost of lining per unit length = 18.48 * 40 = $ 739.12
References
Christopher Gorse, D, J, M, P , 2012. Dictionary of Construction, Surveying, and Civil Engineering. 1st ed. Oxford: Oxford University Press.
Hill, H. W, 2015, An Historical Review of Waterways and Canal Construction. 2nd ed. New York: BiblioLife.
Wood, D. M., 2012. Civil Engineering: A Very Short Introduction. 1st ed. Oxford: Oxford University Press.
