Fundamentals of Open-channel Hydraulics, Chapter 1, Introduction, Victor Miguel Ponce, San Diego State University, 2014

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. 6-8).

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 stress-mean velocity equation, Eq. 6-11.

What is the maximum value of the coeeficient C_s for shear stress on the channel sides?

C_s = 0.78 (Fig. 6-21).

What is the range of angle of repose of noncohesive materials?

From Fig. 6-23, 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

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 d₂₅ = 0.9 in.
- Same as in (a), but the particles are moderately rounded.
Discuss how the particle shape affects the design.

Given:
Q = 330 cfs;
z = 2;
S= 0.0018;
n = 0.025;
Sides and bottom: noncohesive material, d₂₅ = 0.9 in.
(a) moderately angular, and
(b) moderately rounded.
A. MODERATELY ANGULAR.
1. Assume b/y = 6.
2. Assume tractive force on sides is critical (as opposed to tractive force on level ground).
3. With b/y and z, enter Fig. 7-7 left to determine C_s = 0.78.
4. With d₂₅ = 0.9 in, and grain shape moderately angular, find angle of repose θ from Fig. 7-9:
  θ = 37^o.
5. Calculate φ from:
  tanφ = 1/z
  φ = tan^-1 (1/z) = 26.656^o.
6. Calculate K from:
  K = [1 - (sin²φ/sin²θ)]^1/2 = 0.667
7. Determine permissible unit tractive force on level ground τ_L from Fig. 7-10.
  τ_L = 0.4 × d₂₅ (in) = 0.4 × 0.9 = 0.36 psf.
8. Calculate the permissible unit tractive force on the sides:
  τ_s = K τ_L = 0.667 × 0.36 = 0.24 psf.
9. Set permissible and acting unit tractive forces equal: τ_s = T_s
  τ_s = 0.24 = C_sγyS = 0.78 × 62.4 × y × 0.0018
10. Solve for flow depth y:
  y = τ_s/(C_sγS) = 0.24 / (0.78 × 62.4 × 0.0018) = 2.74 ft.
11. With y and b/y, calculate b = y (b/y) = 2.74 × 6 = 16.4 ft. Assume b = 17 ft.
12. With Q = 330, b = 17, z = 2, S = 0.0018, and n = 0.025 known, use ONLINECHANNEL01 to find y_n = 3.16 ft.
13. 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.
14. With b/y = 8.5, enter Fig. 7-7 to determine C_b = 1.0
15. Calculate T_L = C_bγy_nS = 1.0 × 62.4 × 2.72 × 0.0018 = 0.305 psf.
16. Compare acting unit tractive force T_L with permissible unit tractive force on level ground τ_L calculated in step 7.
17. T_L = 0.306 < τ_L = 0.36. Therefore, the sides control the design. The design is OK.
B. MODERATELY ROUNDED.
1. 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.
2. With b/y = 14, enter Fig. 7-7 to determine C_b = 1.0
3. Calculate T_L = C_bγy_nS = 1.0 × 62.4 × 2.28 × 0.0018 = 0.256 psf.
4. Compare acting unit tractive force T_L with permissible unit tractive force on level ground τ_L calculated in step 7.
5. T_L = 0.256 < τ_L = 0.36. Therefore, the sides control the design. The design is OK.
A channel has the following data: Q = 60 m³/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 d₂₅ = 35 mm; the particles on the bottom are of size d₅₀ = 5 mm.
- The particles on the sides are the same as in (a), but the particles on the bottom are smaller, of size d₅₀ = 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.

Given:
Q = 60 m³/s;
z = 2;
S= 0.0006;
n = 0.03;
Case A:
Sides: noncohesive material, slightly angular, d₂₅ = 35 mm;
Bottom: noncohesive material, with d₅₀ = 5 mm, with low content of fine sediment in the water, no channel sinuosity.
Case B:
Sides: noncohesive material, slightly angular, d₂₅ = 35 mm;
Bottom: noncohesive material, with d₅₀ = 4 mm, with low content of fine sediment in the water, no channel sinuosity.
CASE A: Bottom: 5 mm
1. With Q = 60, b = 77 = 23.4 m, z = 2, S = 0.0006, and n = 0.03 known, use ONLINECHANNEL01 to find y_n = 1.934 m = 6.3 ft.
2. Test to confirm that: y_n = 6.3 < y = 12.8. Normal depth too low!
  If not satisfied, assumed b/y is too large. Assume a smaller value and return to step 11.
  - Assume b/y = 0.9 Then b = 11.5 ≅ 12 ft.
    With Q = 60, b = 12 ft = 3.6 m, z = 2, S = 0.0006, and n = 0.03 known, use ONLINECHANNEL01 to find
    y_n = 3.9 m = 12.8 ft
    Test to confirm that: y_n = 12.8 ≤ y = 12.8 Normal depth now OK!
3. With b/y = 0.9, enter Fig. 7-7 to determine C_b = 0.79
4. Calculate T_L = C_bγy_nS = 0.79 × 62.4 × 12.8 × 0.0006 = 0.38 psf.
5. Compare acting unit tractive force T_L with permissible unit tractive force on level ground τ_L calculated in step 7.
6. T_L = 0.38 > τ_Lb = 0.17. Therefore, the bottom controls the design. The design is not OK.
7. Force T_L = 0.17. Then: 0.17 = C_bγy_nS = 0.79 × 62.4 × y_n × 0.0006
  Solve for new y_n:
  y_n = 0.17/(0.79 × 62.4 × 0.0006) = 5.7 ft.
8. Solve for new b by trial and error:
  - y_n = 0.17/(1.00 × 62.4 × 0.0006) = 4.5 ft.
    b = 135 ft.
    With Q = 60, b = 135 ft = 41 m, z = 2, S = 0.0006, and n = 0.03 known, use ONLINECHANNEL01 to find y_n = 1.4 m = 4.6 ft OK!
CASE B: Bottom: 4 mm
1. With Q = 60, b = 77 = 23.4 m, z = 2, S = 0.0006, and n = 0.03 known, use ONLINECHANNEL01 to find y_n = 1.934 m = 6.3 ft.
2. Test to confirm that: y_n = 6.3 < y = 12.8. Normal depth too low!
  If not satisfied, assumed b/y is too large. Assume a smaller value and return to step 11.
  - Assume b/y = 1 Then b = 12.8 ≅ 13 ft.
    With Q = 60, b = 13 ft = 4 m, z = 2, S = 0.0006, and n = 0.03 known, use ONLINECHANNEL01 to find
    y_n = 3.83 m = 12.6 ft
    Test to confirm that: y_n = 12.6 < y = 12.8 Normal depth now OK!
3. With b/y = 0.9, enter Fig. 7-7 to determine C_b = 0.79
4. Calculate T_L = C_bγy_nS = 0.79 × 62.4 × 13.2 × 0.0006 = 0.39 psf.
5. Compare acting unit tractive force T_L with permissible unit tractive force on level ground τ_L calculated in step 7.
6. T_L = 0.39 > τ_Lb = 0.13. Therefore, the bottom controls the design. The design is not OK.
7. Force T_L = 0.13. Then: 0.13 = C_bγy_nS = 0.79 × 62.4 × y_n × 0.0006
  Solve for new y_n:
  y_n = 0.13/(0.79 × 62.4 × 0.0006) = 4.4 ft.
8. Solve for new b:
  - b = 60 X 3.5 = 210
    With Q = 60, b = 210 ft = 64 m, z = 2, S = 0.0006, and n = 0.03 known, use ONLINECHANNEL01 to find y_n = 1.08 m = 3.5 ft OK!
A certain type of lawn has a critical shear stress of 30 N/m². The dimensionless Chezy friction factor f is 0.0075. What is a good estimate of the critical velocity?

Use the quadratic law relating shear stress and velocity:
τ_o = ρ f v²
At the beginning of motion:
τ_c = ρ f v_c²
v_c² = τ_c / (ρ f)
v_c² = (g τ_c) / (γ f)
v_c² = [ (9.81 m/s²) (30 N/m²) ] / [ (9810 N/m³) (0.0075) ]
v_c = 2 m/s
x

The

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140825 06:40

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