QUESTIONS

  1. 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.

  2. 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.

  3. What is the abscissa in the Shields diagram?

    The boundary Reynolds number, i.e,, a Reynolds number based on sediment particle size.

  4. What is the ordinate in the Shields diagram?

    The dimensionless shear stress.

  5. What Froude number will generally assure initiation of motion?

    F ≥ 0.087 (Eq. 6-8).

  6. 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.

  7. 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.

  8. What is the maximum value of the coeeficient Cs for shear stress on the channel sides?

    Cs = 0.78 (Fig. 6-21).

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

    From Fig. 6-23, the angle of repose varies between 19° and 41°.

  10. 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.

  11. 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.

  12. When is a grade control structure justified?

    In situations where it is desirable to fix the channel bottom, usually to control degradation.

PROBLEMS

  1. What is the minimum permissible velocity for a sediment particle diameter ds = 0.6 mm and a dimensionless Chezy friction factor f = 0.004?

    Using Eq. 6-9:

    Vmin = [(0.066) (9.806) (0.0006) / (0.004) ] 1/2 = 0.31 m/s.

  2. What is the minimum Froude number and minimum permissible velocity for a sediment particle diameter ds = 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: Fmin = 0.09, and Vmin = 0.28 m/s.

  3. What is the minimum Froude number and minimum permissible velocity for a sediment particle diameter ds = 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: Fmin = 0.084, and Vmin = 0.83 ft/s.

  4. 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. Verify with ONLINE TRACTIVE FORCE.

    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 Cs = 0.78.

    4. With d25 = 0.9 in, and grain shape moderately angular, find angle of repose θ from Fig. 7-9: θ = 37o.

    5. Calculate φ from: tanφ = 1/z

      φ = tan-1 (1/z) = 26.656o.

    6. Calculate K from:

      K = [1 - (sin2φ /sin2θ)]1/2 = 0.667

    7. Determine permissible unit tractive force on level ground τL from Fig. 7-10.

      τL = 0.4 × d25 (in) = 0.4 × 0.9 = 0.36 psf.

    8. Calculate the permissible unit tractive force on the sides:

      τs = K τL = 0.667 × 0.36 = 0.24 psf.

    9. Set permissible and acting unit tractive forces equal: τs = Ts

      τs = 0.24 = Cs γ y S = (0.78) (62.4) y (0.0018)

    10. Solve for flow depth y:

      y = τs /(Cs γ S) = 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 yn = 3.16 ft.

    13. Test to confirm that: yn = 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 yn = 2.72 ≅ y = 2.74. Normal depth OK now.

    14. With b /y = 8.5, enter Fig. 7-7 to determine Cb = 1.0

    15. Calculate TL = Cb γ yn S = 1.0 × 62.4 × 2.72 × 0.0018 = 0.306 psf.

    16. Compare acting unit tractive force TL with permissible unit tractive force on level ground τL calculated in step 7.

    17. TL = 0.306 < τL = 0.36. Therefore, the sides control the design. The design is OK.

    B. MODERATELY ROUNDED.

    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 Cs = 0.78.

    4. With d25 = 0.9 in, and grain shape moderately rounded, find angle of repose θ from Fig. 7-9: θ = 32.5o.

    5. Calculate φ from: tan φ = 1/z

      φ = tan-1 (1/z) = 26.656o.

    6. Calculate K from:

      K = [1 - (sin2φ / sin2θ)]1/2 = 0.55

    7. Determine permissible unit tractive force on level ground τL from Fig. 7-10.

      τL = 0.4 × d25 (in) = 0.4 × 0.9 = 0.36 psf.

    8. Calculate the permissible unit tractive force on the sides:

      τs = K τL = 0.55 × 0.36 = 0.198 psf.

    9. Set permissible and acting unit tractive forces equal: τs = Ts

      τs = 0.198 = Cs γ y S = (0.78) (62.4) y (0.0018)

    10. Solve for flow depth y:

      y = τs / (Cs γ S ) = 0.198 / (0.78 × 62.4 × 0.0018) = 2.26 ft.

    11. 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 yn = 2.28 ≅ y = 2.26. Normal depth OK now.

    12. With b /y = 14, enter Fig. 7-7 to determine Cb = 1.0

    13. Calculate TL = Cb γ yn S = 1.0 × 62.4 × 2.28 × 0.0018 = 0.256 psf.

    14. Compare acting unit tractive force TL with permissible unit tractive force on level ground τL calculated in step 7.

    15. TL = 0.256 < τL = 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.

  5. A channel has the following data:   Q = 220 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. Verify with ONLINE TRACTIVE FORCE.

    Given:

    Q = 220 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. 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 Cs = 0.78.

    4. With d25 = 35 mm = 1.38 in, and grain shape slightly angular, find angle of repose θ from Fig. 7-9:

      θ = 37.6o

    5. Calculate φ from: tan φ = 1/z

      φ = tan-1 (1/z) = 26.656o.

    6. Calculate K from: K = [1 - (sin2φ/sin2θ)]1/2 = 0.678

    7. Determine permissible unit tractive force on level ground τL from Fig. 7-10.

      τLs = 0.4 × d25 (in) = 0.4 × 1.38 = 0.552 psf.

      τLb = 0.17 psf.

    8. Calculate the permissible unit tractive force on the sides:

      τs = K τLs = 0.678 × 0.552 = 0.374 psf.

    9. Set permissible and acting unit tractive forces equal: τs = Ts

      τs = 0.374 = Cs γ y S = (0.78) (62.4) y (0.0006)

    10. Solve for flow depth y:

      y = τs /(Cs γ S) = 0.374 / (0.78 × 62.4 × 0.0006) = 12.8 ft.

    11. With y and b/ y, calculate b = y (b/ y) = 12.8 × 6 = 76.8 ft. Assume b = 77 ft.

    12. With Q = 220, b = 77 = 23.5 m, z = 2, S = 0.0006, and n = 0.03 known, use ONLINECHANNEL01 to find yn = 4.04 m = 13.25 ft.

    13. Test to confirm that: yn = 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 yn = 3.75 m = 12.3 ft

        Test to confirm that: yn = 12.3 ≤ y = 12.8 Normal depth now OK!

    14. With b/y = 7, enter Fig. 7-7 to determine Cb = 0.99

    15. Calculate TL = Cb γ yn S = 0.99 × 62.4 × 12.3 × 0.0006 = 0.46 psf.

    16. Compare acting unit tractive force TL with permissible unit tractive force on level ground τL calculated in step 7.

    17. TL = 0.46 > τLb = 0.17. Therefore, the bottom controls the design. The design is not OK.

    18. Force TL = 0.17. Then: 0.17 = Cb γ yn S = (0.99) (62.4) yn (0.0006)

      Solve for new yn:

      yn = 0.17/(0.99 × 62.4 × 0.0006) = 4.59 ft.

    19. Solve for new b by trial and error:

      • Assume b/y = 110; new Cb = 1.0

      • yn = 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 yn = 1.41 m = 4.62 ft OK!

    CASE B: Bottom: 4 mm

    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 Cs = 0.78.

    4. With d25 = 35 mm = 1.38 in, and grain shape slightly angular, find angle of repose θ from Fig. 7-9: θ = 37.6o

    5. Calculate φ from: tan φ = 1/z

      φ = tan-1 (1/z) = 26.656o.

    6. Calculate K from: K = [1 - (sin2φ/sin2θ)]1/2 = 0.678

    7. Determine permissible unit tractive force on level ground τL from Fig. 7-10.

      τLs = 0.4 × d25 (in) = 0.4 × 1.38 = 0.552 psf.

      τLb = 0.13 psf.

    8. Calculate the permissible unit tractive force on the sides:

      τs = K τLs = 0.678 × 0.552 = 0.374 psf.

    9. Set permissible and acting unit tractive forces equal: τs = Ts

      τs = 0.374 = Cs γ y S = 0.78 × 62.4 × y × 0.0006

    10. Solve for flow depth y :

      y = τs /(Cs γ S) = 0.374 / (0.78 × 62.4 × 0.0006) = 12.8 ft.

    11. With y and b/y, calculate b = y (b/ y) = 12.8 × 6 = 76.8 ft. Assume b = 77 ft.

    12. With Q = 220, b = 77 = 23.5 m, z = 2, S = 0.0006, and n = 0.03 known, use ONLINECHANNEL01 to find yn = 4.04 m = 13.25 ft.

    13. Test to confirm that: yn = 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 yn = 3.75 m = 12.3 ft

        Test to confirm that: yn = 12.3 < y = 12.8 Normal depth now OK!

    14. With b/ y = 7.0, enter Fig. 6-23 to determine Cb = 0.99

    15. Calculate TL = Cb γ yn S = 0.99 × 62.4 × 12.3 × 0.0006 = 0.46 psf.

    16. Compare acting unit tractive force TL with permissible unit tractive force on level ground τL calculated in step 7.

    17. TL = 0.46 > τLb = 0.13. Therefore, the bottom controls the design. The design is not OK.

    18. Force TL = 0.13. Then: 0.13 = Cb γ yn S = 0.99 × 62.4 × yn × 0.0006

      Solve for new yn :

      yn = 0.13/(0.80 × 62.4 × 0.0006) = 3.5 ft.

    19. Solve for new b :

      • Assume b/ y = 230

      • New Cb = 1.0

      • yn = 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 yn = 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.

  6. A certain type of lawn has a critical shear stress τc = 30 N/m2. 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:

    τo = ρ f v 2

    At the beginning of erosion:

    τc = ρ f vc2

    vc2 = τc / (ρ f )

    vc2 = (g τc) / (γ f )

    vc2 = [ (9.81 m/s2) (30 N/m2) ] / [ (9810 N/m3) (0.0075) ] = 4 m2/s2

    vc = 2 m/s

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140829 14:30

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