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.