AEG-101-Class-14 : Grassed Waterways

Introductory Soil and Water Conservation Engineering

II Semester 3^rd Feb to 30^th June 2020, 2019-20

Teacher Information

Professor	Email	Phone
Dr. K. C. Shashidhar	shashidhar.kumbar@gmail.com	9448103268

Class-14 – Reference Material

Grassed Waterways

Classroom Video

 

Grassed waterways and outlets are natural or constructed waterways shaped to required dimensions and vegetated for safe disposal of runoff from a field, diversion terrace or other structures.

The grass lined waterway is one of the basic conservation practices. Waterways subject to constant or prolonged flows require special supplemental treatment, such as grade control structures, stone center or subsurface drainage capable of carrying such flows. After establishment, protective vegetative cover must be maintained. Vegetated outlets and waterways are used for the following purposes:

• as outlets for diversion and terraces;
• as outlets for surface and subsurface drainage systems on sloping land;
• to dispose of water collected by road ditches or discharged through culverts; and
• To rehabilitate natural drains carrying concentration of runoff.

The waterway or outlet may be protected by using a combination of the following steps:

• Construct the waterway in advance of any other channel that will discharge into it, and divert the

Flow during the period of stabilization.

• Establish and maintain the vegetative cover.

Shape

Grassed waterways may be built to three general shapes-parabolic, trapezoidal or V-shaped. Parabolic waterways are the most common. The successful vegetated waterway is dependent on good conservation treatment on its watershed, which reduce the peak rate of runoff and the volume of runoff to be carried by the waterway.

Profile and cross-section 

The original ground surface should be surveyed for longitudinal and cross-section in detail, to permit dividing the waterway into reaches of approximately uniform slope and shape.

Design

In designing a grassed waterway, care should be taken to see that its size is sufficient to carry all the runoff water from the contributing catchment and the gradient is such that the runoff will flow non-erosive velocities.

Design data

The following information is required for designing a waterway:

• Water shed area (ha) together with soil characteristics, cover and topography.

This information is needed to estimate peak rate of runoff.

• Grade of the proposed waterway (in per cent slope). This is fixed considering the outlet elevation.
• Vegetal covers adapting to site condition (for selecting roughness coefficient).
• Erodibility of the soil in the waterway (the information is necessary for protecting the waterway till the vegetation or grass gets established).
• Permissible velocity for the condition encountered.
• Allowance for space that will be occupied by vegetative lining.
• Free-board.

Non-erosive velocity of flow

The non-erosive velocity of flow depends usually on site conditions. However, following velocities of flow can be considered safely for design purposes.

• a velocity of 0.9 m/sec should be maximum where only a sparse cover can be established or maintained;
• a velocity of 0.9 to 1.2 m/sec should be used where vegetation is to be established by seeding;
• velocity of 1.2 to 1.5 m/sec should be used only in areas where dense, vigorous sod is established quickly;
• a velocity of 1.5 to 1.8 m/sec may be used on well- established sod of excellent quality; and
• Velocity of 1.8 to 2.1 m/sec may be used on well- established quality and conditions under which the flow cannot be handled at lower velocity. Also special maintenance measures are needed.

For situations where waterways are without vegetation, following values of critical velocity may be used (critical velocity is the velocity of water flowing in the channel, such that no silting or scouring takes place).

Nature of soil	Critical velocity (m/sec)	Nature of soil	Critical velocity (m/sec)
Earth	0.3-0.6	Boulder	1.5-1.8
Ordinary murrum	0.6-0.9	Soft rock	1.8-2.4
Hard murrum	1.2	Hard rock	More than 3.0

Cross-section

The area of cross-section (A), wetted perimeter (P) and top width (T) of trapezoidal section can be obtained by using the formula:

Trapezoidal
A = bd + Zd² 

P = b + 2d √(1 + Z²)

T = b + 2dZ

where, b = bottom width; d = depth of channel; z = side slope.

In area of cross-section (A), wetted perimeter(P) and top (T) of parabolic and triangular sections can be obtained, by using the following formulae:

Parabolic

A = (2/3) Td

And P = T + (8/3) x (d²/ T)

T= Top width

Triangular (Fig.3.36)

A = Zd² 

P = 2d √(1 + Z²)

T = 2dZ

The wetted perimeter (P) in the above equations is the length of line of inter-section of the plane of the cross-section with the wetted surface of the channel.

The hydraulic radius (R) is the ratio which is obtained by dividing area of cross-section (A) by the wetted perimeter (P). This is necessary to determine (R) to compute the velocity of flow of water in the channel using Manning’s formula.

STEPS FOR DESIGN

Step 1. The area drained, A (ha) may be obtained from the contour maps.

Step 2. Estimate the peak rate of runoff(cumecs) for the area to be drained using the rational formula.

Q=CIA/360

Step 3. Permissible velocity of flow, V (m/sec) in the vegetated and non-vegetated waterways can be obtained as discussed earlier.

Step 4. Compute approximate area of cross-section of the channel using the formula, Q =A.V., where V = the permissible velocity as obtained under Step 3 and q = peak rate of runoff.

Step 5. Knowing the cross-section from Step 4, determine the channel dimension in such a way that the area of cross-section equals the area of cross-section computed as in Step 4. (Use of different formulae given to compute area of cross-section for different shaped channels).

Step 6. Compute hydraulic radius (R) from the cross-section obtained in Step 5.

Step 7. Compute the grade of the channel using the Manning’s formula,

V = (1 / n) R^2/3S^1/2

where, V = permissible velocity (m/sec) (Step 3),

R = hydraulic radius (m)

n = roughness coefficient (refer Tables 1 and 2)

Step 8. The channel gradient obtained under Step 7 may be rounded for convenience of layout. The outlet elevation obtained by computing with this channel gradient should coincide with field outlet elevation.

Step 9. From the rounded off value of grade (S), we calculate the velocity of flow for the section under consideration. If needed, cross-section is to be adjusted (by adopting the channel dimensions) and it is to be verified whether the computed velocity is approximately equal or less than the velocity as assumed under Step 3.

Example 20: Determine the dimensions of a grassed waterway for stability and capacity, with a trapezoidal cross-section using the following data:

Peak rate of runoff = 3.5 cumec

Grade = 0.3 per cent

Vegetative cover = Blue grass

n = 0.045

Classroom Video

Table 1 Values of Manning’s roughness co-efficient, n. To be used with Manning’s formula

Surface	Best	Good	Fair	Bad
Uncoated cost iron pipe	0.012	0.013	0.014	0.015
Coated cast iron pipe	0.011	0.012	0.013*
Commercial wrought iron pipe, black	0.012	0.013	0.014*	0.015
Commercial wrought iron pipe, galvanized	0.013	0.014	0.0.15	0.017
Smooth brass and glass pipe	0.009	0.010	0.011	0.013
Rivoted and spiral steel pipe	0.013	0.015*	0.017*
Vitrified sewer pipe	0.010	0.013*	0.015	0.017
	0.011
Common clay drainage till	0.011	0.012*	0.014*	0.017
Glazed brick work	0.011	0.012	0.013*	0.015
Brick in cement mortar, brick sewers	0.012	0.013	0.015*	0.017
Neat cement surfaces	0.010	0.011	0.012	0.013
Cement mortar surfaces	0.011	0.012	0.013*	0.015
Concrete pipe	0.012	0.013	0.015*	0.016
Concrete lined channels	0.012	0.014	0.016*	0.018
Cement rubble surface	0.025	0.030	0.033	0.035
Dry rubble surface	0.025	0.030	0.033	0.035
Semi-circle metal flumes, smooth	0.011	0.012	0.013	0.015
Semi-circle metal flumes, corrugated	0.0225	0.025	0.0275	0.030
Canals and ditches
Earth, straight and uniform	0.017	0.020	0.0225*	0.025
Rock cuts, smooth and uniform	0.025	0.030	0.033*	0.035
Rock cuts, jagged and irregular	0.035	0.040	0.045
Winding sluggish canals	0.0225	0.025*	0.0275	0.030
Dredged earth channels	0.025	0.0275*	0.030	0.033
Canals with rough stony beds, Weeds on earth banks	0.025	0.030	0.035*	0.040
Earth bottom, rubble sides	0.028	0.030*	0.033*	0.035
Natural Stream Channels
Clean, straight, bank, full stage, No rifts or deep pools	0.025	0.0275	0.030	0.033
Same as above but some weeds and stones	0.030	0.033	0.035	0.040

*Values commonly used in designing.

Soil moderately erodible
Side slope, Z=2.
Solution
Assume B = 2m
Area of cross-section,

A = BD + ZD²

Wetted Parameter

P = 2D√(1+Z²)

A + (2x1) + (2x1²) = 4m²

P=2 + 2√(1+2²) = 2 + 4.47 = 6.47m

Hydraulic radius

R = A / P = 4 / 6.47 = 0.62m

Using the Nomograph below

Velocity of flow in the channel section is approximately equal to 0.9m/sec

Q = A x V = 4.0 x 0.9 = 3.6cumec 

Hence, the design can be accepted.

Table 2 Values of Manning’s roughness co-efficient, n. To be used with Manning’s formula

Kind of pipe	Variation		Use in designs
	From	To	From	To
Clean uncoated cast iron pipe	0.011	0.015	0.013	0.015
Clean coated cast iron pipe	0.010	0.014	0.012	0.014
Dirty or tuberculated cast iron pipe	0.0.15	0.035
Riveted steel pipe	0.013	0.017	0.015	0.017
Lockbar and welded pipe	0.010	0.013	0.012	0.013
Galvanized iron pipe	0.012	0.017	0.015	0.017
Brass and glass pipe	0.009	0.013
Wood stave pipe	0.010	0.014
Wood stave pipe, small diameter			0.011	0.012
Wood stave pipe, large diameter			0.012	0.013
Concrete pipe	0.010	0.017
Concrete pipe with rough joints			0.016	0.017
Concrete pipe, ‘dry mix’ rough forms			0.015	0.016
Concrete pipe, ‘wet mix’ steel forms			0.012	0.014
Concrete pipe, very smooth			0.011	0.012
Vitrified sewer pipe	0.010	0.017	0.013	0.015
Common clay drainage tile	0.011	0.017	0.012	0.014

Source: Handbook of Hydraulics, King, McGraw-Hill Book Company, INC(Fourth edition) (1954) pages 6.12)

Example 21: Design a grassed waterway of parabolic shape to carry a flow of 2.6 m3/s, down a slope of 3 per cent. The waterway has a good stand of grass and a velocity of 1.75 m/s can be allowed. Assume the value of n in Manning’s formula as 0.04.

Solution
Using Q = AV, for a velocity of 1.75 m/s, a cross-section of 2.6/1.75 = 1.5 m2 is needed.
Assuming

t = 4m, d = 60cm

A = (2/3) t d = (2/3) x 4 x 60 = 1.6m² 

P = t + (8d² / 3t) = 4 + (8.062 / (3 x 4)) = 4.24m

R = A / P = 1.6 / 4.24 = 0.377

V = R^2/3S^1/2 / n = ((03.77)^2/3 x (0.03)^1/2) / 0.04 = (0.522 x 0.173) / 0.04 
   = 2.6m/s

The velocity exceeds the permissible limit. Assuming a revised value of t=6 m and d=0.4m

A = (2/3) x 6 x 0.4 = 1.6m² 

P = 6 + (80.4² / (3 x 6)) = 6.45m

R = 1.6 / 6.45 = 0.248

V = (0.248^2/3 x 0.03^1/2 ) / 0.04 =1.70m/s

The velocity is within permissible limits

Q = 1.6 x 1.7 = 2.72m³ /s


Hence Satisfactory. A suitable freeboard to the depth is to be given in the final dimensions.

Classroom Video

Example 22: Design a parabolic shaped grassed waterway to carry a peak flow of 3.0 m3/sec down a slope of 4.0%. An excellent stand of dub grass is maintained in the waterway.

Manning’s Coefficient n = 0.04
initially assume a top width T = 4.0 m (Fig. 3.38)
Depth of flow d = 0.60

The section is much higher in size. Therefore, assuming that T=4.15 m and depth of flow as 0.475 m and adopting the above method, Q = 3.007 cumec. This is satisfactory.

DESIGN OF A DIVERSION CHANNEL
A case is met where in an area to be improved by contour bunding receives runoff from an outside catchment. It is necessary to devise a diversion channel before undertaking contour bunding project.

Classroom Video

Design Example 23:
Outside catchment = 15 ha
Slope = 8%
Coefficient of runoff may be assumed as 0.75
Intensity of rainfall = 25 mm/hr.
Assume a coefficient of roughness = 0.05
Peak rate of runoff,

Q = CIA / 360 = (0.75 x 25 x 15) / 360 = 0.78cumec

Solution design of diversion drain

First trail-Assume b=2m, d=0.3m,Z=2
Area of cross section= bd+Zd2 = (2 x 0.3 ) + (2 x 0.3² ) = 0.78m²

Wetted perimeter = b + 2d√(1+Z²) = 2 + ( 2 x 0.3√(1+2² )) = 3.338m



Hydraulic radius, R = 0.78 / 3.338 = 0.233m

V = (1/n) x (R^2/3 x S^1/2 ) = (1/0.05) x (0.233^2/3 x 0.08^1/2 ) = 2.12m/s

Q = A x V = 0.78 x 2.12 = 1.65cumec





The requirement being 0.78cumec the size of diversion drain is very big. Hence the dimension can be lowered

2nd trail-b=0.6m: d=0.3m, Z=2

Area of cross section of flow,
A= bd+Zd² =  (0.6 x 0.3) + (2 x 0.3 x 0.3) 
  = 0.36m²

Wetted perimeter P = b +  2d√(1+Z²) = 0.6 + ( 2 x 0.3√(1+4) ) = 1.94m

Hydralic radius  R = A / P = 0.36 / 1.94 = 0.185m

V = (1/n) (R^2/3 x S^1/2 ) = (1 / 0.05) x 0.185^2/3 x 0.08^1/2  = 1.05m/s


Manning’s formula

Q = A V = 0.78 x 1.05 = 0.81cumec

This is satisfactory. Here channel dimensions of bottom width of 0.6m and depth 0.3m with side slopes of 2:1 can be adopted.