The document discusses the design of grassed waterways. It defines grassed waterways as natural or manmade open channels protected with grasses or shrubs that are constructed along slopes to drain runoff from terraces or graded bunds. The key design parameters discussed include peak runoff rate, permissible flow velocity, waterway cross-sectional area, length, width, depth, grade, and shape. Equations for calculating time of concentration, rainfall intensity, and peak runoff rate using nomographs are also provided. A step-wise design procedure is outlined that involves sizing the waterway based on runoff calculations and checking the velocity using Manning's formula.

1. Grassed Waterways
Course Instructor: Prof. Dr. Muhammad Shoaib
Course: Advanced Soil and Water Conservation Engineering

2. Principles of Soil Conservation
• Generally, Soil Erosion strategies be based on following points:
• Development of Soil Cover to protect the soil from raindrop impact
• Increasing Infiltration capacity of soil to reduce the runoff
• Improving aggregate stability of soil and
• Increasing surface roughness to reduce runoff velocity
• The Soil Conservation objectives can be achieved by various
measures
1. Agronomic measures
2. Soil management
3. Mechanical/Engineering measures

3. Principles of Soil Conservation
• The Soil Conservation objectives can be achieved by various
measures
1. Agronomic measures
• Utilizes vegetation to minimize erosion
2. Soil management
• Refers to the practices for preparing the soil to generate a vigorous vegetative growth
and improved soil structures-results in good resistant of soil particles detachment
3. Mechanical measures
• Manipulate the land topography for controlling soil loss
• Bunds and terraces are main structures under mechanical measures for the control of
flow of water and reducing the soil loss

4. Mechanical/Engineering Measures
• Engineering methods deal with the physical structures that stops or
try to prevent the happening of soil erosion.
• There are different methods of engineering for soil erosion control.
They
• are as follows:
• a) Check dams
• b) Retaining walls
• c)Water ways
• d) Terracing
• e) Embankment
• f) Spurs and soil ways

5. Grassed Waterways
• Means to drain/divert the runoff from the catchment
• Natural or manmade open channels protected with
suitable grasses or shrubs.
• It is simply a drainage channel.
• It is constructed along slope and act as outlet for
terraces or graded bunds.
• Its size depends on expected run off.
• Vegetated waterways are built to protect soil against
the erosive forces of concentrated runoff from
sloping lands.

6. Grassed/Vegetative
Waterways
Grassed Waterway Purpose
• Used as outlets to prevent rill and gully formation.
• The vegetative cover slows the water flow, minimizing channel surface
erosion
• The vegetation improves the soil aeration and water quality (impacting
the aquatic habitat) due to its nutrient removal (nitrogen, phosphorus,
herbicides and pesticides) through plant uptake and sorption by soil.
• The waterways can also provide a wildlife habitat.

7. Grassed Waterways
Advantages
• Carries large flows and
make it suitable for
large watersheds
• Farm machineries can
easily cross them
• Its maintenance is less
once the grasses are
established
Disadvantages
• Use of farm machinery
around them is difficult
• Restricts the installation
of tile drainage outlet
• Establishment of
vegetation is quite
difficult at some places

8. Grassed Waterways
Site Selection Criteria
• Land slope less than 20%
• Water flow velocity not more than 1.8 m/s
Objectives
• Remove excess runoff from terraced field, bunded
field
• Carry rainwater down the slope to a safe slope
• Reduce soil erosion caused by high velocity of
runoff

9. Major types
of Waterways
1. Grassed waterway
2. Grassed waterway with drop
structures
3. Ballasted (gravel or coarse material
used) waterway
4. Prefabricated concrete waterway
(parabolic and v-notch chute)
5. Stepped waterway
6. Waterway and road ditch (complex)
7. Footpath and chute (complex)
Stepped
Ballasted
Road ditch
Footpath chute
Grassed waterways with drop structures

10. Different types of waterways and their uses
Sr.
No.
Type Shape Protection
measure
Velocity
(m/s)
Slope
(%)
Uses
1 Grassed waterway Parabolic By grass 1.8 < 20 • For new waterway
• Uniform slope
depression
2 Grassed waterway
with drop
structures
Parabolic By grass or concrete or
masonry structures
1.8 3% (between
two structures)
80% (overall)
• For discontinuous
type channel
3 Ballasted
waterway
Parabolic By stones or stones+
wire mesh
3 <20% • Where stones are
available
4 Prefabricated
concrete
waterway
• A stilling basin is
needed at the end
4 (a) Parabolic
waterways
Parabolic By concrete and grass 6 <30% • Where rainfall is
frequent, and flows
are constant

11. Different types of waterways 1. Grassed waterway
2. Grassed waterway with
drop structures
3. Ballasted waterway
4. Prefabricated concrete
waterway
4a. Parabolic waterway
4b. V-notch chute

12. Different types of waterways and their uses
Sr.
No.
Type Shape Protection
measure
Velocity
(m/s)
Slope
(%)
Uses
4(b) V-notch chute 90o V-notch By concrete and grass 6 <35% • Where rainfall is
frequent, and flows
are constant
5 Stepped
waterway
Parabolic
and
rectangular
By grass, concrete or
masonry drops
1.8 (on
grass
part)
<35% • For 4-wheel
mechanization and
in the middle of
bench terraces
6 Waterway and
road ditch
Parabolic By grass and stone
ballasting
3 <14% • For tractor crossing
and 4-wheel
mechanization
7 Chute Trapezoid or
rectangular
By concrete or masonry
structure
6 <30% • For paths on small
farms or steep
slope

13. Different types of waterways 5. Stepped waterway
6. Waterway with road ditch
7. Chute

14. Design of Grassed
waterways
Similar to the design of the irrigation channels grassed waterways are
designed based on their functional requirements.
Generally, these waterways are designed for carrying the maximum runoff for
a 10- year recurrence interval period.
The rational formula is invariably used to determine the peak runoff rate.
Waterways can be shorter in length or sometimes, can be even very long.
For shorter lengths, the estimated flow at the waterways outlets forms the
design criterion
For longer lengths, a variable capacity waterway is designed to account for the
changing drainage areas.

15. Design of Grassed waterway
1. Data required
2. Design considerations
3. Design parameters
1. Size of waterway
2. Shape of waterway
• Factors affecting shape of the waterway
3. Flow velocity
4. Length of waterway
5. Width and depth of waterway
6. Grades/slope of waterway
7. X-sectional area of waterway

16. Design of Grassed waterways
1. Data required
• Watershed area
• Soil characteristics
• Crop cover
• Topography
• Above data is mainly required for peak discharge computation
• Grade of proposed waterway (fixed by considering outlet of the outlet)
• Information on existing vegetal cover for roughness coefficient estimation
• Erodibility of proposed grassed waterway. Used to predict soil erosion till the vegetation/grassed fully established
• Allowable flow velocity
• Allowance to be provided for accommodating the space occupied by vegetation
• Freeboard allowance for removing chances of overtopping of flow

17. Design of Grassed waterways
2. Design Considerations
• Design of waterway requires proper determination of channel dimensions which ensure that
• The velocity of flowing water will not cause erosion. It depends on
• Type, condition and density of vegetation
• Erosive characteristics of soil
• Capacity of the waterway is sufficient to carry water without overtopping. Capacity of waterway depends on:
• Waterway width, depth and grade
• Soil erodibility
• Vegetative cover
• Waterway dimensions vary with the following factors
• Waterway size: Width and depth as per volume of surface runoff
• Grade: Steep grade channel require shallow and wide channel
• Soil erodibility: A high erodible soil require wide and shallow waterway

18. Design of Grassed waterways
Design Parameters
Size of waterway
Shape of waterway
Flow velocity
Length of waterway
Width and depth of waterway
Grade of waterway
X-Sectional area of waterway

19. Design of Grassed waterways
Design Parameters
1. Size of waterway
• Depends upon the expected runoff.
• A 10-year recurrence interval is used to calculate the maximum expected runoff to the waterway.
• As the catchment area of the waterway increases towards the outlet, the expected runoff is
calculated for different reaches of the waterway and used for design purposes.
• The waterway is to be given greater cross-sectional area towards the outlet as the amount of
water gradually increases towards the outlet. The cross-sectional area is calculated using the
following formula:
𝑄 = 𝐴𝑉
𝐴 =
𝑄
𝑉

20. Design of Grassed waterways
Design Parameters
2. Shape of waterway
• Depends upon the field conditions and type of the
construction equipment used.
• The three common shapes adopted are
• trapezoidal, triangular, and parabolic shapes.
• In course of time due to flow of water and sediment
depositions, the waterways assume an irregular shape
nearing the parabolic shape.
• If the farm machinery has to cross the waterways, parabolic
shape or trapezoidal shape with very flat side slopes are
preferred.

21. Design of Grassed waterways
Design Parameters
2. Shape of waterway

22. Design of Grassed waterways
Design Parameters
2. Shape of waterway
Factors Affecting Shape
• Available construction equipment; Trapezoidal channel needs blade type machines
provided that bottom width of the channel should be more than minimum cut width of
machines
• Triangular and parabolic shapes with side slope 4:1 constructed using suitable machines
• Trapezoidal channel can easily be constructed suitable to act as terrace outlet
• Triangular and Trapezoidal channels become parabolic due to siltation and scouring

23. Design of Grassed waterways
Design Parameters
3. Flow Velocity
• Velocity is dependent on the condition of the vegetation and the soil erodibility.
• It is recommended to have a uniform cover of vegetation over the channel
surface to ensure channel stability and smooth flow.
• The velocity of flow through the grassed waterway depends upon the ability of
the vegetation in the channel to resist erosion.
• Even though different types of grasses have different capabilities to resist erosion;
an average of 1.0 m/sec to 2.5 m/sec are the average velocities used for design
purposes.

24. Design of Grassed
waterways
Design Parameters
3. Flow Velocity
• Average velocity of flow is higher than the actual velocity in
contact with the bed of the channel.
• Velocity distribution in a grassed lined channel
• Permissible values of Velocity
Cover Condition Permissible Velocity
(m/s)
Sparse grass cover 0.9 to 1.2
Good grass cover 1.5 to 1.8
Sod of excellent
grass cover
2.0 to 2.5

25. Design of Grassed waterways
Design Parameters
3. Flow Velocity
• When waterways are not lined with
vegetation, critical velocity is used instead of
permissible velocity
• The velocity at which neither siltation nor soil
scouring occurred
• Recommended values of Critical velocity
Name of the
soil
Critical flow Velocity
(m/s)
Earth 0.3 to 0.6
Ordinary
murram
0.6 to 0.9
Hard murram 1.2
Boulders 1.5 to 1.8
Soft rock 1.8 to 2.4
Hard rock >3.0

26. Design of Grassed waterways
Design Parameters
4. Length of waterway
• Length of waterway considered suitable if the velocity of runoff flow does not cause
erosion
• Drop structures are provided on moderate slope if the velocity is more than the safe limit
• Low check dams and drop structures generally placed at 30 to 40 m interval to slow the
velocity

27. Design of Grassed waterways
Design Parameters
5. Width and depth of waterway
• Width depends on
• Flow rate
• Slope
• Available land
• Width of waterways may be 20 to 60 m or more on large cultivated fields
• Width of waterways may be 1 to 2 m on small fields
• Waterway depth affects field operations as
• Deeper waterways causes inconvenience in field operations
• Deeper waterways also causes problems for crossing of farm machines
• Width and depth must ensure safe drainage of runoff from the area
• A freeboard of 10 to 20 cm is also provided for safety point

28. Design of Grassed waterways
Design Parameters
6. Grades of waterway
• Depends on land slope
• Grades within 5 percent is generally preferred
• Should not exceed 10 percent

29. Design of Grassed waterways
Design Parameters
7. X-sectional area of waterway
𝐴 =
𝑄
𝑉
• Used for X-sectional area computation
• Next flow velocity is computed using
𝑉 =
𝑄
𝐴
• Velocity is further checked by the Manning’s formula using trial method
𝑉 =
𝑅
2
3𝑆
1
2
𝑛
• Manning’s roughness coefficient is to be selected depending on the existing and proposed vegetation to be established in the bed of the channel.
• Velocity is not an independent parameter. It will depend on n which is already fixed according to vegetation,
• R which is a function of the channel geometry and slope S for uniform flow.
• Slope S has to be adjusted. If the existing land slope gives high velocity, alignment of the channel has to be changed to get the desired velocity.

30. Design of Grassed waterways
Step-Wise Procedure
1. Determine Peak runoff rate using rational formula 𝑄𝑝 =
𝐶𝐼𝐴
360
2. Select permissible flow velocity
3. Compute X-sectional area of waterway to suit value of peak runoff rate, Q and
permissible flow velocity v
• Size of waterway should be increase towards outlet to account for increases
in catchment area and runoff
4. Compute depth, width, side slope to suit
5. Calculate hydraulic radius (R) 𝑅 =
𝐴
𝑃

31. Design of Grassed waterway-
Stepwise Procedure
6. Compute grade (S) of waterway using Manning’s formula
𝑉 =
1
𝑛
𝑅
2
3𝑆
1
2
𝑆 =
𝑉. 𝑛
𝑅
2
3
7. Roundoff the values of S for convenience of layout. Find out value of
outlet elevation based on grades
8. Again, compute value so value of velocity, if it is same to the one
obtained in (2). Design is correct
9. Design should be such as to pass the computed peak runoff rate with
safe velocity

32. • Peak runoff rate can be computed
using rational formula 𝑄𝑝 =
𝐶𝐼𝐴
360
• Rational method requires the
followings for accurate estimation
• Rainfall intensity for given frequency and
• Time of concentration
• Depends on length of waterway and
• Slope of waterway
• Nomographs are developed for time of
concentration estimation
Design of grassed waterway-
Time of concentration computation

33. Design of grassed waterway-
Time of concentration computation
Example
Given :
L= 500 m
H = 44 m
Find = Time of concentration
Solution:
Draw a straight 'Line from L axis
(500) to H axis (44).
The line intercepts one point at T
axis. This is the answer: 0.1 hr or 6
min

34. • After time of concentration
estimation, rainfall intensity can be
estimated using nomograph
Example
Given: R24 (24-hr rainfall depth) =
250 mm (9-8 in)
t (time of concentration)= 12 min
or 0.2 hr
Find rainfall intensity figure for 12
min
Design of grassed waterway-
Rainfall intensity computation

35. Example
Given: R24 (24-hr rainfall depth) =
250 mm (9-8 in)
t (time of concentration)= 12 min
or 0.2 hr
Find rainfall intensity figure for 12
min
Solution
Drawing a straight line as it
intercept at 183 mm/hr .This is the
answer
Design of grassed waterway-
Rainfall intensity computation

36. • Finally, peak runoff
rate computed
using the values of
by nomograph
• Rainfall intensity
• Time of
concentration and
• Contributing area
Design of grassed waterway-
Peak runoff rate computation

37. Example
C = 0.8, i = 210 mm/h, A = 0.5 ha
Solution:
1. Connect C (0.8) to i (210
mm/h)
2. Connect C (0.8) with A (0.5 ha)
and obtain intercept point on
Q (0.24 cumecs)
Design of grassed waterway-
Peak runoff rate computation

38. • Once peak runoff rate to
be handled is estimated
• The flow velocity can be
calculated using
Manning’s formula or
nomograph
• Connect R with n by
straight line up
• to pivot line
• Connect S by straight line
to point on pivot line
obtain intercept point on
Q obtain intercept point
on V
Design of grassed waterway-
Flow velocity computation

39. Example
For a cultivated area of 0.5 ha, determine the type of waterway and safe flow
velocity. Take field slope of 18% and peak flow as 0.237 m3/s
Solution: A = 0.5 ha, S = 18% , Qp = 0.237 m3/s
1. Type of waterway: Since slope is 18%, for this slope, the shape should be
PARABOLIC
2. Size and Shape: Trial 1
Step-1 For small farm, a narrow waterway is normally preferred. Assuming
top width =t= 1.0 m and depth = d =15 cm = 0.15 m
1. The X-sectional area is calculated using formula of parabolic
𝐴 =
2
3
𝑡𝑑 =
2
3
× 1.0 × 0.15 = 0.1 m2
2. Wetted parameter is computed as: 𝑃 = 𝑡 +
8𝑑2
3𝑡
= 1 +
8×0.152
3×1
= 1.05 𝑚
3. The Hydraulic radius is calculated as: 𝑅 =
𝐴
𝑃
=
0.1
1.05
= 0.094 m

40. Example
For a cultivated area of 0.5 ha, determine the type of waterway and
safe flow velocity. Take field slope of 18% and peak flow as 0.237 m3/s
Solution: A = 0.5 ha, S = 18% , Qp = 0.237 m3/s
1. Size and Shape:
Step-2: (iv) Flow velocity using Manning’s formula 𝑉 =
1
𝑛
𝑅
2
3𝑆
1
2
𝑉 =
1
0.05
0.094
2
30.18
1
2 = 1.75 m/s (n = 0.05 for dense grass)
(v) Discharge capacity is computed Q = AV = 0.1X1.75 = 0.175 m3/s
As calculated discharge is less than the peak discharge of 0.237 m3/s,
therefore the waterway needs to be either widened or deepened

41. Example
For a cultivated area of 0.5 ha, determine the type of waterway and safe flow
velocity. Take field slope of 18% and peak flow as 0.237 m3/s
Solution: A = 0.5 ha, S = 18% , Qp = 0.237 m3/s
TRIAL 2
1. Size and Shape:
Step-1 Assuming width = 1.2 m and depth = 15 cm = 0.15 m
1. The X-sectional area is calculated using formula of parabolic
𝐴 =
2
3
𝑡𝑑 =
2
3
× 1.2 × 0.15 = 0.12 m2
2. Wetted parameter is computed as: 𝑃 = 𝑡 +
8𝑑2
3𝑡
= 1 +
8×0.152
3×1.2
= 1.00625 𝑚
3. The Hydraulic radius is calculated as: 𝑅 =
𝐴
𝑃
=
0.12
1.00625
= 0.119 m

42. Example
For a cultivated area of 0.5 ha, determine the type of waterway and safe flow
velocity. Take field slope of 18% and peak flow as 0.237 m3/s
Solution: A = 0.5 ha, S = 18% , Qp = 0.237 m3/s
1. Size and Shape:
Step-2: (iv) Flow velocity using Manning’s formula 𝑉 =
1
𝑛
𝑅
2
3𝑆
1
2
𝑉 =
1
0.05
0.119
2
30.18
1
2 = 1.8 m/s (n = 0.05 for dense grass)
(v) Discharge capacity is computed Q = AV = 0.12X1.80=0.216 m3/s
As calculated discharge is less than the peak discharge of 0.237 m3/s,
therefore the waterway needs to be either widened or deepened again
Trial 3: Assume b = 1. 3 m d = 0.15 m and solve

43. Example
Design a grassed waterway with trapezoidal cross-section. The relevant data is
given as under:
Peak runoff = Qp = 4 m3/s , Grade to be used = S = 0.3 %
Manning’s roughness coefficient = n = 0.04, Side slope = 2:1
Solution
Step 1: Assume a trial value of bottom width = 2.0 m
Step 2: Compute x-sectional area = 𝐴 = 𝑏𝑑 + 𝑧𝑑2
= 2 × 1 + 2 × 12 = 4 𝑚2
For a given slope of 2:1, d = 1, 𝑧 = 2
Step 3: Compute wetted perimeter, 𝑃 = 𝑏 + 2𝑑 1 + 𝑧2 = 6.47 𝑚 𝑧 = 2
Step 4: Calculate R, R = A/P = 0.62 m
Step 5: Calculate V using Manning’s Formula, 𝑉 =
1
0.04
× 0.62
2
3 × 0.0030.5
= 1.0
𝑚
𝑠
Step 6: Calculate Q, Q = AV = 4x1 = 4 m3/s

44. Design Procedure for Parabolic grassed
Waterways
Step (1): Determine the peak runoff rate using most appropriate method
Step (2): Fix grade of waterway as per defined criteria. Preference is 5%. However,
it may be up to 10%
Step (3): Maximum permissible flow velocity (Table 15.3, page 571)
Step (4): Select a suitable value of Manning’s roughness coefficient according to the
vegetation retardness class (CI) using the equation
𝑪𝑰 = 𝟐. 𝟓 𝒉 𝑴
𝟏
𝟑
, h is length or height of plant (m), M is stem density
(Table 15.4 for h and M values, Table 15.5 for n values)

45. Material of
waterway
Max. velocity on the cover expected after two seasons
(m/s)
Bare Medium grass cover Very good grass cover
Very light silty sand 0.30 0.75 1.50
Light loose sand 0.50 0.90 1.50
Coarse sand 0.75 1.25 1.70
Sandy soil 0.75 1.50 2.0
Firm clay loam 1.0 1.70 2.30
Firm clay or stiff
gravel
1.50 1.80 2.50
Coarse gravel 1.50 1.80
Unlikely to form very
good grass
Shale, hard pan, soft
rock etc.
1.80 2.10
Hard cemented 2.50 Unlikely to form very
medium grass
Table 15. 3: Maximum permissible flow
velocity based on the condition of
vegetative cover

46. Table 15. 4: Values of stem density (M) for different grasses
Cover group Estimated cover
factor (cf)
Cover tested Reference stem
density (M)
(stem/m2)
Creeping grasses 0.90 Bermuda grass 5380
Centipede grass 5380
Sod forming grasses 0.87 Buffalo grass 4300
Kentucky blue grass 3770
Blue grama 3770
Bunch grasses 0,50 Weeping love grass 3770
Yellow blue stem 2690
Legumes 0.50 Alfalfa 5380
Lespedeza sericea 3230
Annuals 0.50 Common Lespedeza 1610
Sudan grass 538
• The values of M given
in the table apply to
good uniform strand of
cover
• To get value of M for
Poor, Fair, Good and
Excellent cover, the
values of the Table to
be multiplied by 1/3,
2/3, 4/3 and 5/3,
respectively.

47. CI Description n
10.0 Very long dense grass (more than 600 mm) 0.06-0.02
7.6 Long grass (250-600 mm) 0.04-0.15
5.6 Medium grass (150-200 mm) 0.03-0.08
4.4 Short grass (50-150 mm) 0.03-0.06
2.9 Very short grass (less than 50 mm) 0.02-0.04
Table 15.5: Manning’s n for vegetated waterways

48. Design Procedure for Parabolic grassed
Waterways
Step (5): Calculate hydraulic radius R of the waterway using Manning’s
equation 𝑅 =
𝑉.𝑛
𝑆0.5
1.5
Step (6): Compute x-sectional area of waterway, A =Q/V
Step (7): Determine the design depth of waterway using d = 1.5 R
Step (8): Calculate grassed width of waterway using 𝑡 =
𝐴
0.67𝑑
Step (9): Check, whether the capacity obtained under design is
adequate or not Q = A.V. = 0.67.t.d.V
Step (10): Add 20% depth as the free board

49. Example
For a cultivated area of 0.5 ha determine type of waterway and safe flow velocity. Take
field slope 18% and peak flow as 0.237 m3/s
Solution
A = 0.5 ha, S = 18%, Q = 0.237 m3/s
1. Shape :For steep slope of 18%, parabolic waterway is selected
2. Size and velocity: For a small field of 0.5 ha, a narrow waterway is selected. Assume
width 1 m, depth = 15 cm
i. Compute X-sectional area of a parabolic channel
𝐴 =
2
3
𝑡𝑑 =
2
3
× 1 × 0.15 = 0.1 𝑚2
ii. Compute wetted parameter 𝑃 = 𝑡 +
8𝑑2
3𝑡
= 1 +
8𝑋0.152
3𝑋1
= 1.05 m
iii. Hydraulic radius = R = A/P = 0.1/1.05 = 0.094 m
iv. Compute flow velocity using Manning’s Equation 𝑽 =
𝟏
𝒏
𝑹
𝟐
𝟑𝑺
𝟏
𝟐=
𝟏
𝟎.𝟎𝟓
𝟎. 𝟎𝟗𝟒
𝟐
𝟑𝟎. 𝟏𝟖
𝟏
𝟐 = 𝟏. 𝟕𝟓
𝒎
𝒔
(Safe velocity)
v. Calculate Discharge capacity of channel , Q = AV = 0.1x 1.75 = 0.175 m3/s < 0.237

50. Example
For a cultivated area of 0.5 ha determine type of waterway and safe flow velocity.
Take field slope 18% and peak flow as 0.237 m3/s
Solution
Waterway needs to be either widened or deepened, Increasing depth will change
the velocity which is already safe. Increasing width is required
1st Trial, Assume width 1.2 m, A = 0.12 m2, Q = 0.12x1.75 = 0.216 m3/s< 0.237 m3/s
2nd Trial Assume width = 1.3 m, A = 0.13 m2, Q = 0.13x1.75 = 0.2367 m3/s= 0.237 m3/s
𝑃 = 𝑡 +
8𝑑2
3𝑡
= 1.3 +
8..0.152
3.1.3
=1.346 m, R = 0.097 m
𝑽 =
𝟏
𝒏
𝑹
𝟐
𝟑𝑺
𝟏
𝟐=
𝟏
𝟎.𝟎𝟓
𝟎. 𝟎𝟗𝟕
𝟐
𝟑𝟎. 𝟏𝟖
𝟏
𝟐 = 𝟏. 𝟕𝟗
𝒎
𝒔
3. Add free board of 10 cm to the waterway depth and calculate the top width (T)
using formula 𝑇 = 𝑡
0.25
0.15
1
2
= 1.678 𝑚

51. Example 15.3
Compute the Manning’s roughness coefficient (n) for the buffalo grass which has to be
established in the section of waterway for lining purpose. The density of stem per m2 area
is to be used as 4300 and the stem height as 6 cm. Also calculate the n for excellent grass
cover:
Solution:
M = 4300 stem/m2, h= 6 cm,
𝐶𝐼 = 2.5 (0.06 4300)
1
3= 3.93 𝑚 Using Table 15.5, n = 0.027 (Interpolation for the lowest
values of n)
(Solved Examples 15. 4 to 15.8)

52. Example 15.4
Design a parabolic shape grassed waterway to convey the peak runoff
rate of 6 m3/s on 1% land slope. The waterway soil is sandy and to be
covered with Bermuda grass. The grass height would be kept 0.06 m
from the soil surface.
Solution: Q = 6 m3/s, S = 1%, Soil = sandy, Grass = Bermuda, h = 0.06 m
1. Select the maximum flow velocity, V = 1.5 m/s
2. Determine n using the formula 𝑉𝐼 = 2.5 ℎ 𝑀
1
3
=
2.5 5380
1
3=6.75 and n = 0.034 (by interpolation)
3. Determine Hydraulic Radius using 𝑅 =
𝑉×𝑛
𝑆0.5
1.5
=
1.5×0.034
0.010.5
1.5
= 0.364 𝑚
4. Calculate X-sectional area = 𝐴 =
𝑄
𝑉
=
6
1.5
= 4𝑚2

53. Example 15.4
Design a parabolic shape grassed waterway to convey the peak runoff rate of
6 m3/s on 1% land slope. The waterway soil is sandy and to be covered with
Bermuda grass. The grass height would be kept 0.06 m from the soil surface.
Solution: Q = 6 m3/s, S = 1%, Soil = sandy, Grass = Bermuda, h = 0.06 m
5. Compute the depth of parabolic shape waterway
𝑑 = 1.5𝑅 = 1.5 × 0.364 = 0.55 𝑚
6. Calculate top width of the waterway 𝑡 =
𝐴
0.67𝑑
=
4
0.67×0.55
= 𝟏𝟎. 𝟖𝟔 𝒎
7. Check capacity of waterway
𝑄 = 𝐴𝑉 =
2
3
𝑡𝑑 𝑉 =
2
3
× 10.86 × 0.55 × 1.5 = 6.0 m3/s
6. Add 20% free board to the waterway depth
𝑑 = 0.55 + 0.55 × 0.2 = 𝟎. 𝟔𝟔 𝒎

54. Example 15.5
Design a grassed waterway of parabolic shape to carry the peak runoff rate
of 2.6 m3/s on a slope of 3%. The waterway has to be a good grass cover. The
allowable velocity may be taken as 1.75 m/s. Assume n=0.03
Solution: Q = 2.6 m3/s, S = 3%, Good grass cover, V = 1.75 m/s, n = 0.03
1. Calculate R, 𝑅 =
𝑉×𝑛
𝑆0.5
1.5
=
1.75×0.03
0.030.5
1.5
= 0.17 𝑚
2. Calculate X-sectional area = 𝐴 =
𝑄
𝑉
=
2.6
1.75
= 1.48 𝑚2
3. Compute the depth of parabolic shape waterway
𝑑 = 1.5𝑅 = 1.5 × 0.17 = 0.26 𝑚
4. Calculate top width of the waterway 𝑡 =
𝐴
0.67𝑑
=
1.48
0.67×0.26
= 𝟖. 𝟓𝟎𝒎
6. 5. Add 20% free board to the waterway depth
𝑑 = 0.26 + 0.26 × 0.2 = 𝟎. 𝟑𝟏 𝒎

55. Example 15.7
Determine the dimensions of trapezoidal shape grassed waterway to carry
the peak runoff rate of 4. m3/s The grade of the waterway to be used as
0.3%. Assume permissible flow velocity of 0.9 m/s.
Solution: Q = 4 m3/s, S = 00.3 %, V = 0.9 m/s, n = 0.034
1. Let side slope of the channel is 2:1 (d = 1.0 m & b = 2.0 m)
𝐴 = 𝑏𝑑 + 𝑧𝑑2 = 2 × 1 + 2 × 12 = 4 𝑚2
2. 𝑃 = 𝑏 + 2𝑑 1 + 𝑧2 = 2 + 2 × 1 1 + 4 = 6.47 𝑚
3. 𝑅 =
𝐴
𝑃
=
4
6.47
= 0.62 𝑚
4. 𝑉 =
1
𝑛
𝑅
2
3𝑆
1
2 =
1
0.045
× 0.62
2
3× 0.003
1
2 = 0.885 m/s
5. 𝑄 = 𝐴𝑉 = 4 × 0.9 = 3.54 m3/s As calculated discharge < 4 m3/s
6. Increase the either bottom width or depth, So b = 2.5 m (instead of 2.0 m)

56. Example 15.7
Determine the dimensions of trapezoidal shape grassed waterway to
carry the peak runoff rate of 4. m3/s The grade of the waterway to be
used as 0.3%. Assume permissible flow velocity of 0.9 m/s.
Solution: Q = 4 m3/s, S = 00.3 %, V = 0.9 m/s, n = 0.034
𝐴 = 𝑏𝑑 + 𝑧𝑑2 = 2.5 × 1 + 2 × 12 = 4.5𝑚2
𝑃 = 𝑏 + 2𝑑 1 + 𝑧2 = 2.5 + 2 × 1 1 + 4 = 6.97 𝑚
𝑅 =
𝐴
𝑃
=
4.5
6.97
= 0.645 𝑚
𝑉 =
1
𝑛
𝑅
2
3𝑆
1
2 =
1
0.045
× 0.645
2
3× 0.003
1
2 = 0.91 m/s
𝑄 = 𝐴𝑉 = 4.5 × 0.9 = 4.09 m3/s

57. Example 15.8
(A) Calculate the peak runoff rate expected to occur once in 10-years
for design of grassed waterway for draining 35 ha area of watershed.
The other details are as: Time of concentration of watershed = 30 min.,
Runoff coefficient of the watershed is 0.40, The maximum rainfall
during 10-years duration is 6.0 cm for the storm of duration of 30-
minutes.
(B) Using the peak runoff, determine the mean velocity of runoff and
flow depth passing through a trapezoidal grassed waterway. The other
details are as: side slope = 4:1, Bottom width = 5 m, Slope of the
waterway = 2% and n = 0.04

58. Example 15.8
(A) Calculate the peak runoff rate expected to occur once in 10-years for design of grassed
waterway for draining 35 ha area of watershed. The other details are as: Time of concentration
of watershed = 30 min., Runoff coefficient of the watershed is 0.40, The maximum rainfall
during 10-years duration is 6.0 cm for the storm of duration of 30-minutes.
Solution: Q = ? A = 35 ha, TC = 30 min, C = 0.40, Rainfall Depth= 6 cm, Rainfall duration = 30 min =
30/60 = 0.5 hr
1. I = 6/0.5 = 12 cm/hr = 120 mm/hr
2. 𝑄 =
𝐶𝐼𝐴
360
=
0.4×120×35
360
= 4.67 m3/s

59. Example 15.8
(B) Using the peak runoff, determine the mean velocity of runoff and flow depth passing
through a trapezoidal grassed waterway. The other details are as: side slope = 4:1, Bottom
width = 5 m, Slope of the waterway = 2% and n = 0.04
Solution: Q = 4.67 m3/s, z = 4, b = 4 m, S = 2%, n = 0.04
𝐴 = 𝑏𝑑 + 𝑧𝑑2
= 5 × 𝑑 + 4 × 𝑑2
= 5𝑑 + 4𝑑2
(1)
𝑃 = 𝑏 + 2𝑑 1 + 𝑧2 = 5 + 2 × 𝑑 1 + 4 = 5 + 4.47𝑑 (2)
𝐴 =
𝑄
𝑉
=
4.67
𝑉
, 𝑉 =
1
𝑛
𝑅
2
3𝑆
1
2 =
1
0.04
× 𝑅
2
3× 0.002
1
2 = 3.54 × 𝑅
2
3
Trial 1 Assume Velocity = V = 1.5 m/s , Put in Eq. (1)
𝐴 =
𝑄
𝑉
=
4.67
1.5
= 5𝑑 + 4𝑑2, 3.11 = 5𝑑 + 4𝑑2, 4𝑑2 + 5𝑑 − 3.11 = 0, d = 0.45 m
𝐴 = 𝑏𝑑 + 𝑧𝑑2
= 5 × 0.45 + 4 × 0.452
= 3.06 m2
𝑄 = 𝐴𝑉 = 3.06 × 1.5 = 4.59 m3/s < 4.67 m3/s
Trial 2 Assume V = 1.55 m/s
𝑄 = 𝐴𝑉 = 3.06 × 1.55 = 4.74 m3/s > 4.67 m3/s

60. Waterway location
• Located at a point where water can drain from all sides of
watershed
• locations which require minimum earthwork are preferred
• Soils are deep, fertile, moisture favorable for grass growth
• Soil surface is in depression
• Waterway constructed along the field boundary, if depression is not
available
• Waterway can also be located in the middle of the field if there is
good outlet is available.

61. Waterway location
• Need of the waterway can be accessed based on climatic conditions of the
area as per followings:
• In humid regions, where rainfall intensity is high, and yield of excess
runoff is very frequent
• In semi-arid or regions, where infiltration rate is very low, and yield of
concentrated runoff is there
• The area provided with terraces and drainage of rainwater is essential
• In uncultivated lands to provide a safe connection of runoff water to
the structures like diversions, cutoff ditches etc.

62. Selection of suitable grasses
• Selection of grasses/vegetation of the waterway depends on
• soil and
• climatic conditions
• Other factors include
• Duration of growth of vegetation/grasses
• Quantity and velocity of runoff
• Ease of growth of vegetation/grasses
• Time required to develop a good cover
• Suitability to the farmers for seed and spreading into adjoining areas
• Cost and availability of vegetation/grasses
• Generally, Rhizomatous grasses are preferred as spread quickly
• Deep rooted legumes are avoided as they tend to lose the soil

63. Layout of the waterways
• Layout is prepared under following considerations:
• The elevation of the waterway should be lower than the drainage or
terraced outlet
• All the outlet connected to the waterway (as straight and uniform as
possible)
• Small structures or basins should be provided in the waterway at
regular interval where there is change in direction or slope
• A well protected area at the downstream side of waterway is provided
for safe disposal of runoff

64. Construction of grassed waterway-Steps
Step 1. Shaping (soil digging)
• Ensure even shaping/digging
• Sudden fall or sharp turn must be eliminated
• Channel grade should be planned as per design plan
• Stones or stumps must be removed as they interfere the discharge rate

65. Construction of grassed waterway-Steps
Step 2. Grass planting
• Grass planting is important after shaping
• Local grass species must be preferred
• Rhizome grasses are preferred
• Seeding is found cheaper than sodding in large waterways
• Seeded area should be mulched
• Runoff should not be allowed immediately after planting grass

66. Construction of grassed waterway-Steps
Step 3. Ballasting (Stability)
• Waterway gradient is very steep
• In locations where rocks are readily available adjected to the construction site
• Generally, recommended for waterways on small farms
• Stones for ballasting should be 15 to 20 cm diameter
• Placed firmly on the ground
• Wire mesh should also be used for protection on steep sloped area
• Partial ballasting in center leaving sides is preferred on parabolic waterways
Step 4. Placing of Structure

67. Construction of grassed waterway-Steps
Step 4. Placing of Structure
• Drop structures are needed where there is sudden fall because of soil
scouring due to falling of water from high to low evelvation
• Water must flow from top of structure not below or around the structure
• The structure should be constructed on firm soil and have strong and deep
foundation
• Stilling basin of the structure should be sufficiently strong for dissipating
energy

68. Maintenance of grassed waterway
• Grass in the waterway must be kept short and flexible
• Grasses must be mowed three or four times in a year
• Mowed grasses must be removed
• Cattles grazing may also be for keep the grasses short
• Waterway should not be used as road for livestock
• Proper care should be taken for implements to cross the waterway
• Any damage to the waterway should be repaired immediately

69. Maintenance of grassed waterway
• Outlets should be safe and open to ensure free flow
• Waterways should not be used as animal tracks, footpaths or as pastures
• Newly constructed waterways should be watched properly
• Frequent crossing of vehicles should not be allowed
• Waterways must be frequently inspected for checking any break in channel
or structure
• Bushes or large plants growth should be immediately removed

70. Cost of Construction
• Cost depends on many factors
• Equipment
• Labor used for construction
• Costs of grading
• Seed and fertilizer cost
• Cost of establishing grass in the waterway

71. Diversions
• Individual channels constructed across the slope to intercept the
surface runoff and disposed it off safely
• Diversions are used for the following purposes
• To reduce length of the slope
• To divert runoff from the protected area
• To protect bottom land from overflow
• To intercept the water coming from the top terraced area

72. Diversions-Design Steps
1. Determine the alignment of diversion drain on topographic map
2. Determine catchment area of the diversion drain
3. Compute the runoff generated from the watershed area using rational
formula:𝑄𝑃 =
𝐶𝐼𝐴
360
4. Calculate x-sectional area of drain using 𝑄 = 𝐴𝑉, 𝐴 =
𝑄
𝑉
5. Determine non-erosive grade of the channel using Manning’s
formula:𝑉 =
1
𝑛
𝑅
2
3𝑆
1
2, 𝑆 =
𝑉.𝑛
𝑅
2
3
6. Determine depth of diversion drain, 𝑑 =
3
2
𝑅
7. Calculate the top width. For shallow parabolic channel 𝑇 = 1.5
𝐴
𝑑

73. Example
• Design a parabolic shape diversion drain to divert the water of upper
catchment area of 20 ha. The channel slope to be used is 8%. The
other details are as follows:
• Diversion has good vegetation
• Runoff coefficient (C) = 0.70
• Rainfall intensity (I) = 20 mm/h
• Manning’s roughness coefficient n = 0.04

74. Safe Flow velocity in diversion drain
Sr.
No.
Soil Condition Safe Velocity
(m/s)
1 Bare channel sand 0.5
2 Other 0.6
3 Poor vegetation 1.0
4 Fair vegetation 1.25
5 Good vegetation 1.50

75. Example-Solution
1. 𝑄𝑃 =
𝐶𝐼𝐴
360
=
0.75𝑋20𝑋20
360
= 0.83 m3/s
2. Select safe velocity = v = 1. 50 m/s
3. A = Q/V = 0.83/1.50 = 0.55 m2
4. 𝑉 =
1
𝑛
𝑅
2
3𝑆
1
2,
𝑅 =
𝑉.𝑛
𝑆0.5
1.5
= 0.097
5. d = 1.5 R = 1.5x0.097 = 0.15 m
6. 𝑇 = 1.5
𝐴
𝑑
= 1.5(0.55/0.15)= 5.51 m
•