CHAPTER 6: CHANNEL DESIGN 
6.1 NONERODIBLE CHANNELS
There are two types of channels: (1) nonerodible, and (2) erodible. Nonerodible channels are typically lined
with a hard construction material, such as concrete or stone masonry. Erodible channels are dug on the ground surface and, therefore, directly in contact
with the underlying soil. The design of erodible channels is much more complex
than that of nonerodible channels.
The design of nonerodible channels begins with the choice of a uniform flow formula (Chapter 5).
The following factors are considered in the design of nonerodible channels:
Surface roughness
The surface roughness (boundary friction) is determined by the type and finish of the material lining the channel boundary.
Generally, coarser surfaces have higher friction than smoother surfaces (Fig. 61).
Fig. 61 A masonrylined channel.


Minimum permissible velocity
All channels carry a certain amount of sediments (sand, silt, clay). Therefore,
flow in channels of very mild bottom slope may lead to excessive sediment deposition.
Thus, a minimum permissible velocity, i.e., a minimum bottom slope, is needed to avoid clogging.
Maximum bottom slope
All nonerodible channels will have a tendency to develop roll waves if the Vedernikov V > 1 (Chapter 1).
To avoid roll waves, the bottom slope must be kept below a maximum value which is a function of the hydraulics
of the flow.
For instance, Fig. 62 shows a series of drops implemented in a channel to reduce the bottom slope.
Fig. 62 Channel drops implemented in a channel to decrease the channel slope, La Joya Canal, Arequipa, Peru.


Shape of the cross section
The channel section of maximum conveyance and, therefore, maximum discharge,
is that which has the minimum wetted perimeter for a given flow area.
This shape of cross section is referred to as the best hydraulic section.
The best hydraulic circular section is a semicircle.
The best hydraulic trapezoidal section is a half regular hexagon inscribed in a circle (Chow, 1959)
(Fig. 63). In practice, however, other design considerations are usually more important
than that of maximum conveyance.
Fig. 63 Best hydraulic section of trapezoidal shape.


Side slopes
The side slope is z H : 1 V, wherein z = 0 for a rectangular section and z > 0 for a trapezoidal section, is a design decision that varies with local conditions (Fig. 64). Typical side slopes vary from z = 0 to z = 3.
Fig. 64 Channel transition from rectangular to trapezoidal shape.


Freeboard
The freeboard is the vertical depth measured above the design channel depth, up to the total channel depth (Fig. 65).
It is intended to account for a safety factor and to minimize
wave overtopping.
Fig. 65 Freeboard in the San Luis Canal, California.


Slide risk reduction
In hillslope alignments,
care should be taken to reduce
the risk of
local slides, which, depending on their extent,
may render the channel inoperable.
Some hillslopes are prone to slides, while others are not.
A thorough geological/geotechnical site investigation is required to reduce the risk of slides.
Figure 66 shows a slidegenerated breach in the La Cano Canal,
Arequipa, Peru, which occurred on November 4, 2010 (see
related video).
Fig. 66 Slidegenerated breach, La Cano Canal,
Arequipa, Peru, which occurred on November 4, 2010.


Channel lining
Channels or canals may be lined with one of several materials, including concrete, stone masonry, steel,
cast iron, timber, glass, and plastic.
Channel lining reduces boundary friction and minimizes channel maintenance costs.
Figure 67 shows the Santa Ana river, at Huntington Beach,
California, paved with concrete to reduce flood stage and
control floods.
Fig. 67 Paved Santa Ana river, Huntington Beach,
California.


Figure 68 shows Alamar Creek, in Tijuana, Mexico, recently lined up with concrete (2013).
It is noted that current environmental design practices strongly discourage
the lining of natural channels
with concrete.
Fig. 68 Alamar Creek, Tijuana, Baja California, Mexico.


Minimum permissible velocity
All water carries of certain amount of suspended solids, in the form of sediments
that have entrained somewhere upstream and are being transported by the flow.
Typical values of suspended load are 200300 parts per million (or ppm,
equivalent, at this range, to mg/L) (Ponce, 1989).
The noslip condition at the channel boundary
(i.e., the local longitudinal velocity is zero at the channel boundary) produces shear stresses and
results in sediment entrainment.
Once entrained, this sediment
needs to be transported through the channel; otherwise, it will settle and eventually clog
the channel. The minimum velocity to avoid
settling/clogging is the minimum permissible velocity.
The evaluation of minimum permissible velocity may be accomplished with the Shields criterion for initiation of motion.
This criterion is embodied in the Shields diagram, shown in Fig. 69, which relates the dimensionless shear stress
τ_{*}
with the boundary Reynolds number R_{*}.
The solid curve in Fig. 69 separates
no motion (below the curve) from motion (above the curve).
Fig. 69 Shields diagram for initiation of motion (American Society of Civil Engineers, 1975).


Minimum Froude number and minimum permissible velocity
The quadratic friction law, Eq. 53, is recast here with τ_{o} as the bottom shear stress:
The Shields criterion for initiation of motion is:
τ_{o}
τ_{*} = ^{_____________} ≥
τ_{*}_{c}
(γ_{s}  γ ) d_{s}
 (62)

in which τ_{*}
= dimensionless shear stress, γ_{s} = specific weight of
sediment particles, γ = specific weight of water, d_{s} = sediment
particle diameter, and
τ_{*}_{c}
= dimensionless critical shear stress.
Figure 69 shows the Shields curve, i.e., the variation of
dimensionless critical shear stress τ_{*}_{c} with the boundary Reynolds number
R_{*}:
U_{*} d_{s}
R_{*} = ^{_________}
ν
 (63)

in which U_{*} = shear velocity = (τ_{o} /ρ)^{1/2} =
( f )^{1/2} V,
and ν = kinematic viscosity.
The Froude number F is:
V
F = ^{___________}
(g D )^{1/2}
 (64)

in which V = mean velocity, D = hydraulic depth, and g = gravitational acceleration.
Replacing Eqs. 61 and 64 into Eq. 62:
f D F^{ 2}
^{_____________________} ≥ τ_{*}_{c}
[ (γ_{s} / γ )  1 ] d_{s}
 (65)

In most cases of practical interest, the ratio of specific weights of sediment and water
γ_{s} /γ is equal to 2.65. Therefore, in terms of Froude number,
the Shields criterion for initiation of motion reduces to:
1.65
τ_{*}_{c} (d_{s} / D )
F ≥ [ ^{__________________} ]^{ 1/2}
f
 (66)

For practical applications, a constant value of dimensionless critical shear stress
τ_{*}_{c} = 0.04
may be considered. From Fig. 69, it is seen that this value
encompasses the range: 4 ≤ R_{*} ≤ 60. Therefore:
0.066
(d_{s} / D )
F ≥ [ ^{________________} ]^{ 1/2}
f
 (67)

Equation 67 is the practical Shieldsbased Froude criterion for initiation of motion, applicable through a wide range of boundary Reynolds numbers.
As a practical example of the use of Eq. 67,
assume d_{s} = 0.4 mm, D = 1 m, and a midrange dimensionless Chezy friction factor
f = 0.0035 (equivalent to a DarcyWeisbach friction factor f = 0.028).
The application of Eq. 67 leads to:
Combining Eqs. 64 and 67:
0.066
(g d_{s})
V ≥ [ ^{______________} ]^{ 1/2}
f
 (69)

From Eq. 69, the minimum permissible velocity for this flow condition is: V_{min} = 0.27 m/s.
A somewhat more precise value of minimum Froude number and minimum permissible velocity
may be obtained by using the actual value of dimensionless critical shear stress in Eq. 67,
in lieu of the convenient approximation τ_{*}_{c} = 0.04.
This procedure, however, requires an iteration.
For this purpose, the following algorithm is suggested:
Assume R_{*}
Using Fig. 69, find τ_{*}_{c}
Using Eq. 62, calculate τ_{o}
 Using Eq. 61, calculate the shear velocity U_{*} = (τ_{o} /ρ)^{1/2}
Using Eq. 63, calculate the new value of R_{*}
Stop if the new value of R_{*} is the same as that assumed in Step 1 (within a certain small tolerance), and use the last
value of τ_{*}_{c} (calculated in Step 2) in Eq. 66;
Otherwise, return to Step 1, using the new value of R_{*} as the assumed value, and repeat the iteration.

Example 61.
Assume the following channel data: particle diameter d_{s} = 0.4 mm, hydraulic depth D = 1 m, dimensionless DarcyWeisbach friction factor
f = 0.0035, and water temperature T = 20°C.
Using the Shields criterion for initiation of motion, calculate the minimum Froude number and minimum permissible velocity.
 
ONLINE CALCULATION. Using
ONLINE SHIELDS VELOCITY,
the minimum Froude number is F_{min} = 0.081, and the minimum permissible velocity is V_{min} = 0.25 m/s.
Note that these results are slightly more conservative than those obtained using Eqs. 67 and 69.



Channel slope
The design channel slope is usually governed by the chosen alignment and prevailing topography.
The actual design channel slope may depend on the purpose of the channel. For example, channels used for
irrigation and hydropower require small slopes, in order not to lose too much head during conveyance.
Side slopes depend on local soil and construction conditions, usually as steep as practicable.
For lined canals, a side slope of 1.5 H : 1 V is recommended as a standard by the U.S. Bureau of Reclamation, for use in most canals.
For lined canals and steep slopes, when the Vedernikov number exceeds 1, the possibility
arises for the development of roll waves (Chapter 1).
Freeboard
Freeboard is the vertical distance from the top of the channel to the water surface at the design condition. This distance
should be sufficient to prevent waves or fluctuations in the water surface from overflowing the channel sides.
This feature becomes important, particularly in the design of elevated flumes, for the flume substructure may be endangered
by an overflow (Fig. 610).
Fig. 610 The Dulzura conduit, in San Diego County, California,
overflowing after heavy rain, on March 5, 2005.


There is no universally accepted rule for the determination of freeboard.
Windgenerated waves or tidal action may induce high waves, which would need to be kept within channel bounds.
Freeboards ranging from less than 5% to more than 30% of the design flow depth are in common practice.
For smooth, interior, semicircular flumes on tangents, carrying water at velocities no greater than the critical
velocity, with a maximum of 8 ft/s (2.4 m/s),
experience has indicated that a freeboard of 6% of the flume diameter is appropriate.
For flumes on curves, freeboard must be increased to prevent water from sloping over (Chow, 1959).
Note that hydraulic design criteria commonly allow for the use of all or part of the freeboard to accomodate the
Probable Maximum Flood (Ponce, 1989).
According to the U.S. Bureau of Reclamation, the approximate range
of freeboard extends from 1 ft (0.3 m)
for small laterals with shallow depths, to 4 ft (1.2 m) in canals up to 3,000 cfs (85 m^{3}/s) or more.
The following formula is applicable:
F_{b} = ( C y ) ^{1/2}
 (610) 
in which F_{b} = freeboard, in ft; y = depth of water in the canal, in ft; and C = coefficient,
varying from C = 1.5
for a canal capacity of 20 cfs (0.57 m^{3}/s) to C = 2.5 for a canal capacity of 3,000 cfs (85 m^{3}/s) or more.
Figure 611 shows U.S. Bureau of Reclamation recommended
height of lining and height of bank, applicable for the design of freeboard.
Fig. 611 U.S. Bureau of Reclamation recommended
height of lining and height of bank (Chow, 1959).


Section dimensions
The following steps are taken to choose the dimensions of a channel or canal cross section:
Select the design discharge Q
Select the bottom width b
Select the side slope z [z H : 1 V] (Fig. 612)
Fig. 612 Definition sketch for bottom width b and side slope z.


Select the bottom slope S
Estimate the value of Manning's n (Chapter 5: Manning's n)
Using Q,
b, z, S, and n [Steps 1 to 5],
compute the normal depth y_{n}, normal velocity
v_{n}, and normal Froude number F_{n}
(Chapter 5: Computation)
Check to see if the normal velocity v_{n} and normal Froude number F_{n}
are large enough to avoid clogging of the canal with sediment (see Minimum permissible velocity in this Chapter)
Select an appropriate freeboard F_{b} .
Figure 613 shows U.S. Bureau of Reclamation recommended
bottom width and water depth of lined channels.
Fig. 613 U.S. Bureau of Reclamation recommended
bottom width and water depth of lined channels (Chow, 1959).


Example 62.
Design a channel for the following flow conditions: Q = 10 m^{3}/s; b = 5 m; z = 2; S = 0.0016; n = 0.025.
Using Eq. 555, by iteration, the normal depth is y_{n} = 1.049 m and the normal velocity is v_{n} = 1.342 m/s.
The Froude number based on normal depth is F_{n} = 0.418.
The calculated Froude number is shown to be much larger than 0.087 (Eq. 68), the minimum value required to avoid sediment deposition.
A freeboard F_{b} = 0.6 m is assumed.
 
ONLINE CALCULATION. Using the
ONLINECHANNEL01 calculator,
the normal depth is y_{n} = 1.049 m; the normal velocity is v_{n} = 1.342 m/s;
the Froude number (based on hydraulic depth D = 0.81 m) is: F_{n} = 0.476.


Note that the true Froude number is that based on hydraulic depth (Eq. 46). For a hydraulically wide channel, for which
D ≅ R, both Froude numbers are approximately the same.

Figure 614 shows a major canal built in 1993 in Ceara, Brazil, for the purpose of combating a regional drought.
The hydraulic characteristics are: Q = 8.32 m^{3}/s, b = 5 m, z = 1.5, S = 0.00005, and n = 0.015.
Using the
ONLINECHANNEL01 calculator,
the normal depth is y_{n} = 1.915 m, the normal velocity is v_{n} = 0.552 m/s, and
the Froude number is F_{n} = 0.149.
Fig. 614 The Workers' Channel (Canal do Trabalhador), Ceara, Brazil (1993).

6.2 ERODIBLE CHANNELS
The design of an erodible channel is much more complex than that of a nonerodible channel.
The longterm stability of the channel boundary depends on the properties of the material lining the boundary (sand, silt, clay, and combinations thereof),
and of the properties of the sediment being transported by the flow. Changes in water and sediment discharge
result in crosssectional changes, either channel depth, channel width, or both.
The stability of the channel boundary is normally assessed in terms of two criteria:
Maximum permissible velocity, and
Maximum permissible tractive stress.
Given Eq. 61, these two methods are related as follows:
τ_{o max} = ρ f (V_{max })^{2}
 (611) 
Example 63.
Calculate the maximum permissible velocity
corresponding to a maximum permissible tractive
stress of 20 N/m^{2}.
Assume f = 0.0035, i.e., DarcyWeisbach f = 0.028.
Using Eq. 611:
V_{max } = [τ_{o max} / (ρ f ) ] ^{1/2} =
[ 20 N/m^{2} / (1000 N s^{2}/m^{4} × 0.0035 ) ] ^{1/2} = 2.39 m/s.

Example 64.
Calculate the maximum permissible velocity
corresponding to a maximum permissible tractive stress of 0.1 lb/ft^{2}.
Assume f = 0.005, i.e., DarcyWeisbach f = 0.04.
Using Eq. 611:
V_{max } = [τ_{o max} / (ρ f ) ] ^{1/2} =
[ 0.1 lb/ft^{2} / (1.94 lbss^{2}/ft^{4} × 0.005 ) ] ^{1/2} = 3.21 ft/s.

Vegetative channels
Table 61 shows typical values of maximum permissible tractive stress for various types of vegetative channel linings.
Table 61 Typical values of maximum permissible tractive stress.

Channel lining 
Maximum permissible tractive stress  (N/m^{2}) 
(lbs/ft^{2}) 
Lawn (shorttime loaded) 
2030 
0.420.63 
Lawn (longtime loaded) 
1518 
0.310.38 
Fascine sausage 
6070 
1.251.46 
Fascine roll 
100150 
2.093.13 
Weighted fascine 
60100 
1.252.09 
Brush mattress 
150300 
3.136.26 
Live staking in rip rap 
> 140 
> 2.92 
Willows/alder 
80140 
1.672.92 
Gabions 
80140 
1.672.92 
Figure615 shows details of the design of a channel lining using a brush mattress.
Fig. 615 (a) Design of brush mattress: Top view.


Fig. 615 (b) Design of brush mattress: Side view.


Fig. 615 (c) Design of brush mattress: Side view after a few months.


Fig. 615 (d) Design of brush mattress: View after project completion.


Figure 616 shows details of a channel lining using live staking in riprap.
Fig. 616 (a) Live staking: Schematic.


Fig. 616 (b) Live staking: Soon after installation.


Fig. 616 (c) Live staking: Sometime after installation.


Fig. 616 (d) Live staking: 25 years after installation.


Fig. 616 (e) Live staking: Established project.


Gabionlined channels
A gabion system is wireenclosed riprap consisting of mats or baskets fabricated with wire mesh, filled with small
riprap, and anchored to a slope (Fig. 617). Wrapping the
riprap enables the use of smaller stone sizes for the same resistance to displacement by water energy. This is a particular advantage
when constructing rock lining in areas of difficult access. The wire basket also allows steeper (up to vertical) channel linings
to be constructed from commercially available wire units or from wirefencing material.
Fig. 617 Placement of gabion mattress.


Gabion systems provide an effective way to control erosion in streams, rivers and canals.
They are normally designed to sustain large channel velocities [5 m/s (15 ft/s) or higher]. Gabions are
constructed by individual units that vary in length from 2 m (6 ft)
to nearly 30 m (100 ft); therefore, applications can range
anywhere from small ditches to large canals (Fig. 618).
Fig. 618 Dimensions of gabions boxes and mattresses.


Gabion channels are a compromise between riprap and concrete channels.
When the samesize rocks are used in gabions and riprap, the acceptable velocity for gabions is three to four times that of riprap.
Unlike concrete, gabions can be
vegetated to blend into the natural landscape (Fig. 619).
Gabion channels with vegetation have the following advantages:
Allow infiltration and exfiltration
Filter out contaminants
More flexible than paved channels
Provide greater energy dissipation than concrete channels
Improve habitat for flora and fauna
More aesthetically pleasing
Lower cost to install, although some maintenance is required.
Values of Manning's n
for gabionlined channels are normally in the range 0.025 ≤ n ≤ 0.030.
Fig. 619 Established channel lining with gabions.


6.3 TRACTIVE FORCE
The tractive force is the summation of the (tractive) shear stresses over (an area of) the channel boundary.
Under equilibrium conditions, to develop the tractive stress equation,
the acting gravitational force is made equal to the resisting frictional force. The gravitational force is (Fig. 620):
W sin θ ≅ W tan θ = W (ΔH /Δx )
 (612) 
in which W = weight of the control volume of area A and length Δx,
ΔH = drop within the control volume,
and θ = channel bottom slope.
Fig. 620 Definition sketch for derivation of the tractive force equation.


Since S = ΔH /Δx, it follows that the acting gravitational force is:
The resisting frictional force is:
F_{f} = τ_{o} P Δx
 (614) 
in which τ_{o} = bottom shear stress, and P = wetted perimeter.
Equating gravitational and frictional forces:
in which R = A /P = hydraulic radius.
For hydraulically wide channels, for which R ≅ D ≅ y,
The tractive stress varies along the wetted perimeter, reaching a peak value along the
middle of the channel. Figure 621 shows a typical variation for a trapezoidal cross section, in which b/y = 4 and z = 1.5.
Fig. 621 Variation of tractive stress in a trapezoidal channel, with b/y = 4 and z = 1.5 (Chow, 1959).


At the bottom center,
the tractive stress approaches asymptotically
its full (maximum) value: τ_{b} = γyS , as shown
in Fig. 622.
Along the channel sides, the tractive stress
approaches asymptotically a fraction of the full value: τ_{s} = 0.78 γyS , as shown in
Fig. 623.
Fig. 622 Variation
of maximum tractive stress on channel bottom with aspect ratio b/y (Chow, 1959).


Fig. 623 Variation
of maximum tractive stress on channel sides
with aspect ratio b/y (Chow, 1959).


Tractive force ratio
Figure 624 shows three particles on the channel boundary, one on the left side, another on the right side, and the third one
on level ground (bottom). The tractive force ratio K is defined as:
a τ_{s} τ_{s}
K = ^{_______} = ^{______}
a τ_{L} τ_{L}
 (617)

in which a = effective area of the particle, τ_{s} = shear stress
on the side, and τ_{L} = shear stress on level ground.
Fig. 624 Definition sketch
for the forces acting on a particle resting on the surface of a channel bed (Chow, 1959).


The ratio K is a function of the channel side slope φ and of the angle of repose θ of the
material forming the channel bed.
To derive K, consider the two forces that are acting on a particle of submerged weight W_{s}
resting on the side of the channel, with tractive stress τ_{s} (Fig. 620):
The tractive force: aτ_{s}
The gravitational force component along the side: W_{s} sin φ
The resultant force acting along the side plane is:
F_{a} = (W_{s}^{2}
sin^{2}φ + a^{2}τ_{s}^{2})^{1/2}
 (618) 
At equilibrium, the resisting force is equal to the acting force. From solid mechanics, the resisting force is equal to the
normal force (W_{s} cosφ )
times the coefficient of friction (tanθ ):
F_{r} =
W_{s} cosφ tanθ
 (619) 
Therefore:
W_{s} cosφ tanθ = (W_{s}^{2}
sin^{2}φ + a^{2}τ_{s}^{2})^{1/2}
 (620) 
Squaring both sides:
W_{s}^{2} cos^{2}φ tan^{2}θ = W_{s}^{2}
sin^{2}φ + a^{2}τ_{s}^{2}
 (621) 
a^{2}τ_{s}^{2} = W_{s}^{2} cos^{2}φ tan^{2}θ  W_{s}^{2}
sin^{2}φ
 (622) 
τ_{s}^{2} = (W_{s} /a)^{2} cos^{2}φ tan^{2}θ  (W_{s} /a)^{2}
sin^{2}φ
 (623) 
tan^{2}φ
τ_{s}^{2} = (W_{s} /a)^{2} cos^{2}φ tan^{2}θ [ 1 
^{_________} ] tan^{2}θ
 (624) 
tan^{2}φ
τ_{s} = (W_{s} /a) cosφ tanθ [ 1 
^{_________} ] ^{1/2} tan^{2}θ
 (625) 
Equation 625 is the shear stress on the side of a channel with side slope angle φ and repose angle θ.
For a level surface, with φ = 0:
sinφ = 0 and cosφ = 1. Thus, the force balance of Eq. 620 reduces to:
W_{s} tanθ =
aτ_{L}
 (626) 
Therefore, the shear stress that causes impending motion on a level surface is:
τ_{L} = (W_{s} /a) tanθ
 (627) 
Combining Eqs. 617, 625, and 627, the tractive force ratio is:
tan^{2}φ
K = cosφ [ 1 
^{_________} ] ^{1/2} tan^{2}θ
 (628) 
It is seen that K is only a function of φ and θ. Equation 628 is equivalent to (See box below):
sin^{2}φ
K = [ 1 
^{__________} ] ^{1/2} sin^{2}θ
 (629) 
Derivation of Eq. 629
Equation 628 is simplified by resorting to established trigonometric identities:
tan^{2}φ
K = cosφ [ 1 
^{_________} ] ^{1/2} tan^{2}θ
 (628a) 
cos^{2}φ tan^{2}φ
K = [ cos^{2}φ 
^{________________} ] ^{1/2} tan^{2}θ
 (628b) 
sin^{2}φ
K = [ cos^{2}φ 
^{_________} ] ^{1/2} tan^{2}θ
 (628c) 
sin^{2}φ cos^{2}θ
K = [ cos^{2}φ 
^{________________} ] ^{1/2} sin^{2}θ
 (628d) 
sin^{2}φ (1  sin^{2}θ)
K = [ 1  sin^{2}φ 
^{_____________________} ] ^{1/2} sin^{2}θ
 (628e) 
sin^{2}φ
K = [ 1 
^{__________} ] ^{1/2} sin^{2}θ
 (629) 

Angle of repose
Figure 625 shows the angle of repose of noncohesive materials, i.e., the angle θ in Eq. 629.
The particle size (in inches) is that for which 25% of the material, by weight, is larger.
Fig. 625 Angle of repose
of noncohesive materials (Chow, 1959).


6.4 PERMISSIBLE TRACTIVE FORCE
The permissible tractive force (unit force, or stress) method is used in the design of
erodible channels and canals.
The permissible unit tractive force is the maximum shear stress
that will not cause erosion of the material forming the channel bed on a level surface. The value, obtained
from laboratory experiments, is referred to as the critical unit tractive force, or critical shear stress.
Figure 626 shows permissible unit tractive force (unit tractive stress) for canals in noncohesive materials.
The figure applies through a range of (average) particle diameter from 0.1 to 100 mm.
The left side of the graph applies for fine noncohesive material (sand), with diameter ranging from 0.1 to 5 mm. Three curves are shown:
The upper curve, applicable for canals with high content of fine sediment
(silt) in the water;
The middle curve, applicable for canals with low content of fine sediment
(silt) in the water;
The lower curve, applicable for canals with clear water.
Note that the cleaner the water, the more likely it is to pick up sediment from the boundary and, therefore,
the lower the value of the permissible unit tractive force.
The right side of Fig. 626 applies for canals in coarse noncohesive material, with diameter ranging from
5 to 100 mm (fine to very coarse gravel, to small cobbles). The line shown in Fig. 626
is equal to 0.4 times the particle diameter (in inches) for which 25% of the material is larger,
by weight, i.e., the permissible unit tractive force, or permissible tractive stress, is equal
to
0.4 × d_{25} (in).
Fig. 626 Permissible unit tractive force for canals in
noncohesive materials (Chow, 1959).


For cohesive materials, the critical shear stress is a function of the type of soil (sandy clays to lean clays)
and the voids ratio, as shown in Fig. 627.
Fig. 627 Permissible unit tractive force for canals in
cohesive materials (Chow, 1959).


The values of Figs. 626 and 627 are applicable to straight channels.
For sinuous channels, the values shown in the figures should be reduced to account for bank scour.
Table 62 shows recommended percentage reduction in critical shear stress
to account for channel sinuosity.
Table 62 Reduction in critical shear stress to account for channel sinuosity.

Sinuosity 
Reduction (%) 
Low 
10 
Moderate 
25 
High 
40 
Permissible tractive force method
The permissible tractive force method is a method to design erodible channels
based on the estimated critical shear stress for the material(s) forming the channel boundary. The value of critical shear stress is determined
experimentally or from established experience. The computational procedure is described below,
along with a worked example, for two possible cases:
Use of the
same material
on sides and bottom, and
Use of one material on the sides
and another on the bottom.
A. The same
material on sides and bottom
Input data: Discharge Q, side slope z, channel slope S, and Manning's n; particle d_{25} and grain shape; material on sides and bottom are
the same.
Assume b/y = 6, i.e., a reasonable value to start.
Assume that
the tractive force on the sides is controlling the design. This is the usually case when
the material on the sides and bottom are the same.
With b/y and z, enter Fig. 623 to determine the value of
C_{s} (C_{s} is the ordinate of Fig. 623) on the expression for the acting unit tractive force on the sides:
T_{s} = C_{s} γ y S.
With d_{25} and grain shape, use Fig. 625 to find the angle of repose θ.
Calculate side slope: φ = tan^{1} (1/z)
Use Eq. 629 to calculate K.
Use Fig. 626 to determine the permissible unit tractive force on level ground τ_{L}.
Calculate the permissible unit tractive force on the sides: τ_{s} = K τ_{L}.
Set the permissible unit tractive force
greater than or equal to the acting unit tractive force: τ_{s} ≥ T_{s}; i.e.,
τ_{s} ≥ C_{s} γ y S
Solve for flow depth: y = τ_{s} / (C_{s} γ S)
Calculate b: b = (b/y) y
With Q, b, z, n, and S known, solve for normal depth y_{n}.
Test to confirm that y_{n} ≤ y. If not, assumed value of b/y is too small.
Assume a larger value and return to Step 11 and iterate. Once satisfied, set y = y_{n} and go to next step.
With last b/y and z, enter Fig. 622 to determine the value of
C_{b} (the ordinate of Fig. 622) on the expression for the acting unit tractive force on level ground T_{L} = C_{b} γ y S.
Calculate T_{L} = C_{b} γ y_{n} S
Compare T_{L} calculated in Step 15 with τ_{L} calculated in Step 7.
If T_{L} ≤ τ_{L}, the sides control the design, as initially assumed. Stop here.


A. Worked example
Input data: Q = 600 cfs; z = 2, channel slope S = 0.001, and Manning's n = 0.022; particle d_{25} = 0.7 in, and slightly angular grain shape;
material on sides and bottom are the same.
Assume b/y = 6.
Assume that the tractive force on the sides is controlling the design.
With b/y = 6
and z = 2, enter Fig. 623 to determine the value of
C_{s} = 0.78.
With d_{25} = 0.7 in
and slightly angular grain shape, use Fig. 625 to find θ = 34°.
φ = tan^{1} (1/z) = 26.565°
Use Eq. 629 to calculate K = 0.6.
Use d_{25} = 0.7 inches in Fig. 626 to determine τ_{L} = 0.28 lbs/ft^{2}.
τ_{s} = K τ_{L} = 0.6 × 0.28 = 0.168 lbs/ft^{2}.
Set the permissible unit tractive force greater than or equal to the
acting unit tractive force:
τ_{s} = 0.168 ≥ C_{s} γ y S = (0.78) (62.4) y (0.001)
Solve for flow depth: y ≤ 3.45 ft.
Set y = 3.45 as the target flow depth.
b = (6) (3.45) = 20.7. Round to b = 21 ft.
With Q = 600, b = 21,
z = 2, n = 0.022, and
S = 0.001, solve for normal depth: y_{n} = 4.536 ft.
y_{n} = 4.536 > y = 3.45.
The width b is too short.
Assume (by trial and error) b/y = 10. b = (10) (3.45) = 34.5. Round to b = 35 ft.
With new b, solve for normal depth: y_{n} = 3.376 ft.
Now y_{n} < y. Set y = 3.376 ft and go to next step.
With b/y = 10
and z = 2, enter Fig. 622 to determine the value of
C_{b} = 1.0.
Calculate T_{L} = (1.0) (62.4) (3.376) (0.001) = 0.211 lbs/ft^{2}.
Compare T_{L} = 0.211 calculated in Step 15 with τ_{L} = 0.28 calculated in Step 7.
T_{L} = 0.211 < τ_{L} = 0.28; therefore, the sides control the design, as initially assumed. Stop here.


B. One material on the sides
and another on the bottom
Input data: Discharge Q, side slope z,
channel slope S, and Manning's n; material on sides and bottom are
different, specify type of material, particle size and grain shape, and fine sediment content if required.
Assume b/y = 6, i.e., a reasonable value to start.
Assume that
the tractive force on the sides is controlling the design. This is the usually case when
the material on the sides and bottom are the same.
With b/y and z, enter Fig. 623 to determine the value of
C_{s} (C_{s} is the ordinate of Fig. 623) on the expression for the acting unit tractive force on the sides:
T_{s} = C_{s} γ y S.
With d_{25} and grain shape, use Fig. 625 to find the angle of repose θ.
Calculate side slope: φ = tan^{1} (1/z)
Use Eq. 629 to calculate K.
Use Fig. 626 to determine the permissible unit tractive force on level ground based on the material from the sides τ_{Ls} (right of Fig. 626), and the permissible unit tractive force on level ground based
on the material from the bottom τ_{Lb} (left of Fig. 626)
Calculate the permissible unit tractive force on the sides: τ_{s} = K τ_{Ls}.
Set the permissible unit tractive force
greater than or equal to the acting unit tractive force: τ_{s} ≥ T_{s}; i.e.,
τ_{s} ≥ C_{s} γ y S
Solve for flow depth: y = τ_{s} / (C_{s} γ S)
Calculate b: b = (b/y) y
With Q, b, z, n, and S known, solve for normal depth y_{n}.
Test to confirm that y_{n} ≤ y. If not, assumed value of b/y is too small.
Assume a larger value and return to Step 11 and iterate. Once satisfied, set y = y_{n} and go to next step.
With last b/y and z, enter Fig. 622 to determine the value of
C_{b} (the ordinate of Fig. 622) on the expression for the acting unit tractive force on level ground T_{L} = C_{b} γ y S.
Calculate T_{L} = C_{b} γ y_{n} S
Compare T_{L} calculated in Step 15 with τ_{Lb} calculated in Step 7.
If T_{L} ≤ τ_{L}, the sides control the design, as initially assumed. Instead, if T_{L} > τ_{L}, the bottom controls the design.
If T_{L} > τ_{L}, force T_{L} = τ_{L}.
Solve for new y, confirming bottom control.


B. Worked example
Input data: Q = 600 cfs; z = 2, channel slope S = 0.001, and
Manning's n = 0.022; material on the sides: noncohesive, d_{25} = 0.7 in,
slightly angular grain shape; material on the bottom:
noncohesive, d_{60} = 0.8 mm, with
high content of fine sediment in the water.
Assume b/y = 6.
Assume that the tractive force on the sides is controlling the design.
With b/y = 6
and z = 2, enter Fig. 623 to determine the value of
C_{s} = 0.78.
With d_{25} = 0.7 in
and slightly angular grain shape, use Fig. 625 to find θ = 34°.
φ = tan^{1} (1/z) = 26.565°
Use Eq. 629 to calculate K = 0.6.
Use Fig. 626 (right)
to determine τ_{Ls} = 0.28 lbs/ft^{2};
use Fig. 626 (left)
to determine τ_{Lb} = 0.09 lbs/ft^{2}.
τ_{s} = K τ_{Ls} = 0.6 × 0.28 = 0.168 lbs/ft^{2}.
Set the permissible unit tractive force greater than or equal to the
acting unit tractive force:
τ_{s} = 0.168 ≥ C_{s} γ y S = (0.78) (62.4) y (0.001)
Solve for flow depth: y ≤ 3.45 ft.
b = (6) (3.45) = 20.7. Round to b = 21 ft.
With Q = 600, b = 21,
z = 2, n = 0.022, and
S = 0.001, solve for normal depth: y_{n} = 4.536 ft.
y_{n} = 4.536 > y = 3.45.
The width b is too short.
Assume (by trial and error) b/y = 10. b = (10) (3.45) = 34.5. Round to b = 35 ft.
With new b, solve for normal depth: y_{n} = 3.376 ft.
Now y_{n} < y. Set y = 3.376 ft and go to next step.
With b/y = 10
and z = 2, enter Fig. 622 to determine the value of
C_{b} = 1.0.
Calculate T_{L} = (1.0) (62.4) (3.376) (0.001) = 0.211 lbs/ft^{2}.
Compare T_{L} = 0.211 calculated in Step 15 with τ_{Lb} = 0.09 calculated in Step 7.
T_{L} = 0.211 > τ_{L} = 0.09;
therefore, the bottom controls the design.
Force
T_{L} = C_{b} γ y_{n} S = (1.0) (62.4) y (0.001) = 0.09.
Therefore: y = 1.44 ft. Assume (by trial and error)
b/y = 106. Therefore: b = 154 ft.
With Q = 600, b = 152, z = 2, S = 0.001, and n = 0.022, calculate y = 1.44 ft. Design now OK!


6.5 OTHER FEATURES
Other features in hydraulic channel design
include canal drops and creek crossings, dissipation
structures, and grade control structures. Figures 628 to 634 show
some illustrative examples.
Fig. 628 A series of canal drops
for the purpose of controlling flow instability, CabanaMaņazo
irrigation project, Puno, Peru.


Fig. 629 Crossing of
Tinajones feeder canal through
Chiriquipe Creek, Lambayeque, Peru.


Fig. 630 A small creek bypass,
WelltonMohawk Canal, Wellton, Arizona.


Fig. 631 A canal crossing a stream by means
of a siphon, CabanaMaņazo
irrigation project, Puno, Peru.


Fig. 632 A large
siphon crossing the New River, Imperial valley, California.


Fig. 633 Dissipation structure, Mashcon river,
Cajamarca, Peru.


Fig. 634 Grade control structure, Caqueza river,
Cundinamarca, Colombia.


QUESTIONS
What determines the surface roughness
in an artificial canal?
What is the freeboard in the design of a canal?
What is the abscissa in the Shields diagram?
What is the ordinate in the Shields diagram?
What Froude number will generally assure initiation of motion?
How are the minimum and maximum permissible velocities
reconciled in a channel design?
How are the maximum permissible velocities and shear stresses
related?
What is the maximum value of the coeeficient C_{s} for shear stress
on the channel sides?
What is the range of angle of repose of noncohesive materials?
What is the tractive force ratio?
How does the content of fine sediment in the water affect the value of
permissible unit tractive force?
When is a grade control structure justified?
PROBLEMS
What is the minimum permissible velocity for a sediment particle diameter d_{s} = 0.6 mm
and a dimensionless Chezy (modified DarcyWeisbach) friction factor f = 0.004?
What is the minimum Froude number and minimum permissible velocity for a sediment particle diameter d_{s} = 0.6 mm,
dimensionless Chezy friction factor f = 0.004, hydraulic depth D = 1 m, and water temperature T = 20°C?
What is the minimum Froude number and minimum permissible velocity for a sediment particle diameter d_{s} = 0.3 mm,
dimensionless Chezy friction factor f = 0.003, hydraulic depth D = 3 ft, and water temperature T = 68°F?
A channel has the following data: Q = 330 cfs, z = 2, n = 0.025, and S = 0.0018.
Use the tractive force method to calculate the bottom width and depth under the following conditions:
 The particles are the same in sides and bottom; they
are moderately angular and of size d_{25} = 0.9 in.
 Same as in (a), but the particles are moderately rounded.
Discuss how the particle shape affects the design. Verify with ONLINE TRACTIVE FORCE.
A channel has the following data: Q = 220 m^{3}/s, z = 2, n = 0.03, and S = 0.0006.
Use the tractive force method to calculate the bottom width and depth under the following conditions:
 The particles on the sides
are slightly angular and of size d_{25} = 35 mm; the particles on the bottom are of size d_{50} = 5 mm.
 The particles on the sides
are the same as in (a), but the particles on the bottom are smaller, of size d_{50} = 4 mm.
Both cases (a) and (b) have low content of fine sediment
in the water and the channel sinuosity is negligible.
Discuss how the bottom particle size affects the design. Verify with ONLINE TRACTIVE FORCE.
A certain type of lawn has a critical shear stress τ_{c} = 30
N/m^{2}.
The dimensionless Chezy friction factor f = 0.0075.
What is a good estimate of the critical velocity?
REFERENCES
American Society of Civil Engineers, 1975. Sedimentation Engineering. Manual of Practice No. 54.
Chow, V. T. 1959. Openchannel Hydraulics. McGraw Hill, New York.
Ponce, V. M. 1989. Engineering Hydrology, Principles and Practices. Prentice Hall,
Englewood Cliffs, New Jersey.
http://openchannelhydraulics.sdsu.edu 

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