QUESTIONS
What determines the surface roughness
in an artificial canal?
The surface roughness in an artificial
canal is determined by the type and finish of the material lining the canal boundary.
Generally, coarser surfaces have higher friction than smoother surfaces.
What is the freeboard in the design of a canal?
The freeboard is the vertical depth measured
above the design channel depth, up to the total channel depth. It is intended
to account for a safety factor and to minimize wave overtopping.
What is the abscissa in the Shields diagram?
The boundary Reynolds number, i.e,, a Reynolds number
based on sediment particle size.
What is the ordinate in the Shields diagram?
The dimensionless shear stress.
What Froude number will generally assure initiation of motion?
F ≥ 0.087 (Eq. 68).
How are the minimum and maximum permissible velocities
reconciled in a channel design?
The minimum permissible velocity is necessary to avoid
clogging of the channel by sediment deposition on the channel bed.
The maximum permissible velocity is necessary to avoid
erosion of the channel boundary. They are totally different concepts.
How are the maximum permissible velocities and shear stresses
related?
The maximum permissible velocities and shear stresses
are related though the quadratic shear stressmean velocity equation, Eq. 611.
What is the maximum value of the coeeficient C_{s} for shear stress
on the channel sides?
C_{s} = 0.78 (Fig. 621).
What is the range of angle of repose of noncohesive materials?
From Fig. 623, the angle of repose varies between 19° and 41°.
What is the tractive force ratio?
The tractive force ratio is the ratio of tractive stress on the channel side
to tractive force on the channel bottom. It is used in design of channels by the permissible tractive force method.
How does the content of fine sediment in the water affect the value of
permissible unit tractive force?
The cleaner the water (the less amount of fine sediment),
the more likely it is to pick up sediment from the boundary and, therefore, the lower the value of the permissible unit tractive force.
When is a grade control structure justified?
In situations where it is desirable to fix the channel bottom, usually to control degradation.
PROBLEMS
What is the minimum permissible velocity for a sediment particle diameter d_{s} = 0.6 mm
and a dimensionless Chezy friction factor f = 0.004?
Using Eq. 69:
V_{min} = [(0.066) (9.806) (0.0006) / (0.004) ] ^{1/2} = 0.31 m/s.
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?
Using ONLINE SHIELDS VELOCITY: F_{min} = 0.09, and
V_{min} = 0.28 m/s.
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?
Using ONLINE SHIELDS VELOCITY: F_{min} = 0.084, and
V_{min} = 0.83 ft/s.
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.
Given: Q = 330 cfs; z = 2; S= 0.0018; n = 0.025;
Sides and bottom: noncohesive material,
d_{25} = 0.9 in.
(a) moderately angular, and
(b) moderately rounded.
A. MODERATELY ANGULAR.
Assume b /y = 6.
Assume tractive force on sides is critical (as opposed to tractive force on level ground).
With b /y and z, enter Fig. 77 left to determine C_{s} = 0.78.
With d_{25}
= 0.9 in, and grain shape moderately angular, find angle of repose θ from Fig. 79: θ = 37^{o}.
Calculate φ from:
tanφ = 1/z φ = tan^{1} (1/z) = 26.656^{o}.
Calculate K from: K = [1  (sin^{2}φ /sin^{2}θ)]^{1/2} = 0.667
Determine permissible unit tractive force on level ground τ_{L} from Fig. 710.
τ_{L} = 0.4 × d_{25} (in) = 0.4 × 0.9 = 0.36 psf.
Calculate the permissible unit tractive force on the sides:
τ_{s} = K τ_{L} = 0.667 × 0.36 = 0.24 psf.
Set permissible and acting unit tractive forces equal:
τ_{s} = T_{s}
τ_{s} = 0.24 = C_{s} γ y S = (0.78) (62.4) y (0.0018)
Solve for flow depth y:
y = τ_{s} /(C_{s} γ S) = 0.24 / (0.78 × 62.4 × 0.0018) = 2.74 ft.
With y and b /y, calculate b = y (b /y) = 2.74 × 6 = 16.4 ft. Assume b = 17 ft.
With Q = 330, b = 17, z = 2, S = 0.0018, and n
= 0.025 known, use
ONLINECHANNEL01 to find y_{n} = 3.16 ft.
Test to confirm that: y_{n} = 3.16 > y = 2.74. Normal depth too high!
If not satisfied, assumed b /y is too small. Assume a greater value and return to step 11.

Assume b /y = 8.5. Then b = 23.2 ≅ 23 ft.
With Q = 330, b = 23, z = 2, S = 0.0018, and n =
0.025 known, use
ONLINECHANNEL01 to find y_{n} = 2.72 ≅ y = 2.74. Normal depth OK now.
With b /y = 8.5, enter Fig. 77 to
determine C_{b} = 1.0
Calculate T_{L} = C_{b} γ y_{n} S = 1.0 × 62.4 × 2.72 × 0.0018 = 0.306 psf.
Compare acting unit tractive force T_{L}
with permissible unit tractive force on level ground τ_{L} calculated in step 7.
T_{L} = 0.306 < τ_{L} = 0.36. Therefore, the sides control the design.
The design is OK.
B. MODERATELY ROUNDED.
Assume b /y = 6.
Assume tractive force on sides is critical (as opposed to tractive force on level ground).
With b /y and z, enter Fig. 77 left to determine C_{s} = 0.78.
With d_{25} = 0.9 in, and grain shape moderately rounded,
find angle of repose θ from Fig. 79: θ = 32.5^{o}.
Calculate φ from: tan φ = 1/z φ = tan^{1} (1/z) = 26.656^{o}.
Calculate K from: K = [1  (sin^{2}φ / sin^{2}θ)]^{1/2} = 0.55
Determine permissible unit tractive force on level ground τ_{L} from Fig. 710.
τ_{L} = 0.4 × d_{25} (in) = 0.4 × 0.9 = 0.36 psf.
Calculate the permissible unit tractive force on the sides:
τ_{s} = K τ_{L} = 0.55 × 0.36 = 0.198 psf.
Set permissible and acting unit tractive forces equal:
τ_{s} = T_{s}
τ_{s} = 0.198 = C_{s} γ y S = (0.78) (62.4) y (0.0018)
Solve for flow depth y:
y = τ_{s} / (C_{s} γ S ) = 0.198 / (0.78 × 62.4 × 0.0018) = 2.26 ft.
With y and b /y, calculate b = y (b /y)
= 2.26 × 6 = 13 ft. Assume b = 13 ft. Deemed too low.
Assume b /y = 14. Then b = 31.6 ≅ 32 ft.
With Q = 330, b = 32, z = 2, S = 0.0018, and n = 0.025 known,
use
ONLINECHANNEL01 to find y_{n} = 2.28 ≅ y = 2.26. Normal depth OK now.
With b /y = 14, enter Fig. 77 to
determine C_{b} = 1.0
Calculate T_{L} = C_{b} γ y_{n} S =
1.0 × 62.4 × 2.28 × 0.0018 = 0.256 psf.
Compare acting unit tractive force T_{L}
with permissible unit tractive force on level ground τ_{L} calculated in step 7.
T_{L} = 0.256 < τ_{L} = 0.36. Therefore, the sides control the design.
The design is OK.
When the particle shape changes from moderately angular to moderately rounded,
the angle of repose decreases, resulting in an increase
in channel width and an associated decrease in flow depth.
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.
Given: Q = 220 m^{3}/s; z = 2; S = 0.0006; n = 0.03;
Case A:
Sides: noncohesive material, slightly angular, d_{25} = 35 mm;
Bottom: noncohesive material, with d_{50} = 5 mm,
with low content of fine sediment in the water, no channel sinuosity.
Case B:
Sides: noncohesive material, slightly angular, d_{25} = 35 mm;
Bottom: noncohesive material, with d_{50} = 4 mm,
with low content of fine sediment in the water, no channel sinuosity.
CASE A: Bottom: 5 mm
 Assume b/ y = 6.
Assume tractive force on sides is critical (as opposed to tractive force on level ground).
With b/y and z, enter Fig. 77 left to determine C_{s} = 0.78.
With d_{25} = 35 mm = 1.38 in, and grain shape slightly angular, find angle of repose θ from Fig. 79: θ = 37.6^{o}
Calculate φ from: tan φ = 1/z φ = tan^{1} (1/z) = 26.656^{o}.
Calculate K from: K = [1  (sin^{2}φ/sin^{2}θ)]^{1/2} = 0.678
Determine permissible unit tractive force on level ground τ_{L} from Fig. 710.
τ_{Ls} = 0.4 × d_{25} (in) = 0.4 × 1.38
= 0.552 psf.
τ_{Lb} = 0.17 psf.
Calculate the permissible unit tractive force on the sides:
τ_{s} = K τ_{Ls} = 0.678 × 0.552 = 0.374 psf.
Set permissible and acting unit tractive forces equal: τ_{s} = T_{s}
τ_{s} = 0.374 = C_{s} γ y S = (0.78) (62.4) y (0.0006)
Solve for flow depth y:
y = τ_{s} /(C_{s} γ S) = 0.374 / (0.78 × 62.4 × 0.0006) = 12.8 ft.
With y and b/ y, calculate b = y (b/ y) = 12.8 × 6 = 76.8 ft. Assume b = 77 ft.
With Q = 220, b = 77 = 23.5 m, z = 2, S = 0.0006, and n = 0.03
known, use
ONLINECHANNEL01 to find y_{n} = 4.04 m = 13.25 ft.
Test to confirm that:
y_{n} = 13.25 > y = 12.8. Normal depth too high!
If not satisfied, assumed b/y is too low. Assume a larger value and return to step 11.
Assume b/y = 7. Then b = 7 × 12.8 = 89.6 ft. = 27.3 m.
With Q = 220, b = 27.3 m, z = 2, S = 0.0006, and n = 0.03 known,
use
ONLINECHANNEL01 to find y_{n} = 3.75 m = 12.3 ft
Test to confirm that: y_{n} = 12.3 ≤ y = 12.8 Normal depth now OK!
With b/y = 7, enter Fig. 77 to
determine C_{b} = 0.99
Calculate T_{L} =
C_{b} γ y_{n} S = 0.99 × 62.4 × 12.3 × 0.0006 = 0.46 psf.
Compare acting unit tractive force T_{L} with permissible unit tractive force on level ground τ_{L} calculated in step 7.
T_{L} = 0.46 > τ_{Lb} = 0.17. Therefore, the bottom controls the design.
The design is not OK.
Force T_{L} = 0.17.
Then: 0.17 = C_{b} γ y_{n} S =
(0.99) (62.4) y_{n} (0.0006)
Solve for new y_{n}: y_{n} = 0.17/(0.99 × 62.4 × 0.0006) = 4.59 ft.
Solve for new b by trial and error:
Assume b/y = 110; new C_{b} = 1.0
y_{n} = 0.17/(1.00 × 62.4 × 0.0006) = 4.54 ft.
b = 499.4 ft. = 152 m.
With Q = 220, b = 152 m, z = 2, S = 0.0006, and n = 0.03 known, use
ONLINECHANNEL01 to find y_{n} = 1.41 m = 4.62 ft OK!
CASE B: Bottom: 4 mm
 Assume b/y = 6.
Assume tractive force on sides is critical (as opposed to tractive force on level ground).
With b/ y and z, enter Fig. 77 left to determine C_{s} = 0.78.
With d_{25} = 35 mm = 1.38 in, and grain shape slightly angular, find angle of repose θ from Fig. 79: θ = 37.6^{o}
Calculate φ from: tan φ = 1/z φ = tan^{1} (1/z) = 26.656^{o}.
Calculate K from: K = [1  (sin^{2}φ/sin^{2}θ)]^{1/2} = 0.678
Determine permissible unit tractive force on level ground τ_{L} from Fig. 710.
τ_{Ls} = 0.4 × d_{25} (in) = 0.4 × 1.38
= 0.552 psf.
τ_{Lb} = 0.13 psf.
Calculate the permissible unit tractive force on the sides:
τ_{s} = K τ_{Ls} = 0.678 × 0.552 = 0.374 psf.
Set permissible and acting unit tractive forces equal: τ_{s} = T_{s}
τ_{s} = 0.374 = C_{s} γ y S = 0.78 × 62.4 × y × 0.0006
Solve for flow depth y :
y = τ_{s} /(C_{s} γ S) = 0.374 / (0.78 × 62.4 × 0.0006) = 12.8 ft.
With y and b/y, calculate b = y (b/ y) = 12.8 × 6 = 76.8 ft. Assume b = 77 ft.
With Q = 220, b = 77 = 23.5 m, z = 2, S = 0.0006, and n = 0.03
known, use
ONLINECHANNEL01 to find y_{n} = 4.04 m = 13.25 ft.
Test to confirm that: y_{n} = 13.25 > y = 12.8. Normal depth too high!
If not satisfied, assumed b/ y is too low. Assume a larger value and return to step 11.

Assume b/ y = 7 Then b = 12.8 × 7 = ≅ 89.6 ft.
With Q = 220, b = 89.6 ft = 27.3 m, z = 2, S = 0.0006, and
n = 0.03 known, use
ONLINECHANNEL01 to find y_{n} = 3.75 m = 12.3 ft
Test to confirm that: y_{n} = 12.3 < y = 12.8 Normal depth now OK!
With b/ y = 7.0, enter Fig. 623 to
determine C_{b} = 0.99
Calculate T_{L} = C_{b}
γ y_{n} S = 0.99 × 62.4 × 12.3 × 0.0006 = 0.46 psf.
Compare acting unit tractive force T_{L} with permissible unit tractive force on level ground τ_{L} calculated in step 7.
T_{L} = 0.46 > τ_{Lb} = 0.13. Therefore, the bottom controls the design.
The design is not OK.
Force T_{L} = 0.13.
Then: 0.13 = C_{b} γ y_{n} S = 0.99
× 62.4 × y_{n} × 0.0006
Solve for new y_{n} : y_{n} = 0.13/(0.80 × 62.4 × 0.0006) = 3.5 ft.
Solve for new b :
Assume b/ y = 230
New C_{b} = 1.0
y_{n} = 0.13/(1.00 × 62.4 × 0.0006) = 3.5 ft.
b = 230 × 3.5 = 805
With Q = 220, b = 805 ft = 245 m, z = 2, S = 0.0006, and n = 0.03 known, use
ONLINECHANNEL01 to find y_{n} = 1.08 m = 3.54 ft OK!
When the bottom mean particle size decreases from 5 mm to 4 mm, the required channel width increases, with an associated decrease in flow depth.
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?
Use the quadratic law relating bottom shear stress and mean velocity:
τ_{o} = ρ f v^{ 2}
At the beginning of erosion:
τ_{c} = ρ f v_{c}^{2}
v_{c}^{2} = τ_{c} / (ρ f )
v_{c}^{2} = (g τ_{c}) / (γ f )
v_{c}^{2} = [ (9.81 m/s^{2}) (30 N/m^{2}) ] / [ (9810 N/m^{3}) (0.0075) ] = 4 m^{2}/s^{2}
v_{c} = 2 m/s
http://openchannelhydraulics.sdsu.edu 

140829 14:30 
