PROBLEMS

1. A perennial stream has the following properties:   discharge Q = 30 m³/s, bottom width b = 55 m, side slope z = 2, bottom slope S_o = 0.0004, and Manning's n = 0.035. A 2-m high diversion dam is planned on the stream to raise the head for an irrigation canal.

   • Calculate the normal depth.

   • Calculate the total length of the M₁ water-surface profile. Use n = 400 and m = 400.

   • Calculate the partial length of the M₁ profile, from the dam to a location upstream where the normal depth is exceeded by 1%.

   Use ONLINE CHANNEL 01 and ONLINE WSPROFILES 21.

   Fig. 7-21  A diversion dam.

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   • Use ONLINE CHANNEL 01 to calculate the normal depth.

     y_n = 0.968 m  ANSWER.

   • Use ONLINE WSPROFILES 21 to calculate the M₁ water surface profile. The normal depth is confirmed to be y_n = 0.968 m.

     From the output table, for y_n = 0.968 m, the total length of the M₁ profile is L = 7510.7 m.  ANSWER.

   • The u/s depth to calculate the partial length of the M₁ profile is y(1.01 yn) = 1.01 × 0.968 = 0.978 m.

     From the output table, for y = 0.979 m, the partial length is L' = 5144.1 m. For y = 0.976, the partial length is L' = 5348.3 m. The difference is 204.3 m. By linear interpolation, for y = 0.978, the partial length is L' = 5144.1 + (1/3) 204.3 = 5212.2 m.  ANSWER.

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2. A perennial stream has the following properties:   discharge Q = 1000 ft³/s, bottom width b = 150 m, side slope z = 2, bottom slope S_o = 0.00038, and Manning's n = 0.035. A 6-ft high diversion dam is planned on the stream to raise the head for an irrigation canal.

   • Calculate the normal depth.

   • Calculate the total length of the M₁ water-surface profile. Use n = 400 and m = 400.

   • Calculate the partial length of the M₁ profile, from the dam to a location upstream where the normal depth is exceeded by 1%.

   Use ONLINE CHANNEL 01 and ONLINE WSPROFILES 21.

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   • Use ONLINE CHANNEL 01 to calculate the normal depth.

     y_n = 3.476 ft  ANSWER.

   • Use ONLINE WSPROFILES 21 to calculate the M₁ water surface profile. The normal depth is confirmed to be y_n = 3.476 ft.

     From the output table, for y_n = 3.476 ft, the total length of the M₁ profile is L = 25,854.7 ft.  ANSWER.

   • The u/s depth to calculate the partial length of the M₁ profile is y(1.01 yn) = 1.01 × 3.476 = 3.511 ft.

     From the output table, for y = 3.508, the partial length is L' = 16,348.5 ft. For y = 3.514, the partial length is L' = 15,857.4 ft. By linear interpolation, for y = 3.511, the partial length is L' = = 16,102.9 ft.  ANSWER.

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3. A mild stream flows into a steep channel, producing an M₂ upstream of the brink. The hydraulic conditions in the channel are:   Q = 28 m³/s, bottom width b = 12 m, side slope z = 2.5, bottom slope S_o = 0.0007, and Manning's n = 0.04. Assume critical depth near the change in slope.

   • Calculate critical and normal depth.

   • Calculate the length of the M₂ water-surface profile using n = 100 and m = 100.

   • Calculate the length of the M₂ water-surface profile using n = 200 and m = 200.

   • Calculate the length of the M₂ water-surface profile using n = 400 and m = 400.

   • Comment on the results of b, c, and d.

   • Determine the design channel length at 99% of the normal depth. Use n = 400.

     Use ONLINE CHANNEL 05 and ONLINE WSPROFILES 22.

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   • Use ONLINE CHANNEL 05 to calculate critical depth y_c = 0.777 m and normal depth y_n = 1.948 m.  ANSWER.

   • Use ONLINE WSPROFILES 22 to calculate the M₂ water surface profile, with given data. Set n = 100 and m = 100. The critical and normal depths calculated are confirmed to be the same as in a.

     From the output table, for y_n = 1.948 m, the length is L = 3815.6 m.  ANSWER.

   • For n = 200, and m = 200, for y_n = 1.948 m, the length is L = 4321.1 m.  ANSWER.

   • For n = 400, and m = 400, for y_n = 1.948 m, the length is L = 4823.6 m.  ANSWER.

   • A higher resolution (2X and 4X) caused the asymptotic profile to extend from 3815.6 to 4321.1 to 4823.6 m.  ANSWER.

   • The channel depth at 99% of normal is y_(0.99yn) = 1.928 m.

     For n = 400, and m = 400, the partial length for y = 1.928 is 2027.6 m

     Therefore, the design channel length at 99% of normal depth is L = 2028 m.  ANSWER.

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4. A mild stream flows into a steep channel, producing an M₂ upstream of the brink. The hydraulic conditions in the channel are:   Q = 500 ft³/s, bottom width b = 30 ft, side slope z = 1.5, bottom slope S_o = 0.00075, and Manning's n = 0.04. Assume critical depth near the change in slope.

   • Calculate critical and normal depth.

   • Calculate the length of the M₂ water-surface profile using n = 100 and m = 100.

   • Calculate the length of the M₂ water-surface profile using n = 200 and m = 200.

   • Calculate the length of the M₂ water-surface profile using n = 400 and m = 400.

   • Comment on the results of b, c, and d.

   • Determine the design channel length at 99% of the normal depth. Use n = 400.

     Use ONLINE CHANNEL 05 and ONLINE WSPROFILES 22.

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   • Use ONLINE CHANNEL 05 to calculate critical depth y_c = 1.983 ft and normal depth y_n = 5.161 ft.  ANSWER.

   • Use ONLINE WSPROFILES 22 to calculate the M₂ water surface profile, with given data. Set n = 100 and m = 100. The critical and normal depths calculated are confirmed to be the same as in a.

     From the output table, for y_n = 5.161 ft, the length is L = 9,903.4 ft.  ANSWER.

   • For n = 200, and m = 200, for y_n = 5.161 ft, the length is L = 11,214.1 ft.  ANSWER.

   • For n = 400, and m = 400, for y_n = 5.161 ft, the length is L = 12,516.8 m.  ANSWER.

   • A higher resolution (2X and 4X) caused the asymptotic profile to extend from 9,903.4 to 11,214.1 to 12,516.8 ft.  ANSWER.

   • The channel depth at 99% of normal is y_(0.99yn) = 5.109 ft.

     For n = 400, and m = 400, the partial length for y = 5.106 is 5,266.4 ft. The partial length for y = 5.114 is 5,547.0 ft. The difference is: 280.6. Interpolating, the partial length for y = 5.109 ft is: 5,266.4 + (3/8) 280.6 = 5,371.6 ft.

     Therefore, the design channel length at 99% of normal depth is L = 5,372 ft.  ANSWER.

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5. An overflow spillway flows into a mild channel, producing a hydraulic jump. The channel is rectangular, with Q = 3 m³/s, bottom width b = 8 m, and Manning's n = 0.015. The flow depth at the toe of the spillway is 0.1 m and the approximate slope at the toe of the spillway is 0.1. The slope of the [mild] downstream channel, which functions as a stilling basin, is 0.0001. Calculate:

   • the critical depth,

   • the length L_tc from the toe of the spillway to critical depth downstream,

   • the Froude number at the toe of the spillway,

   • the Froude number downstream of the hydraulic jump,

   • the sequent depth y₂ (the normal depth downstream of the hydraulic jump),

   • the length L_j of the jump, assuming L_j = 6.2 y₂

   • the minimum length of the stilling basin L_sb = L_tc + L_j

   Use ONLINE CHANNEL 02 and ONLINE WSPROFILES 23. In the latter, use n = 100 and m = 100.

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   • Use ONLINE CHANNEL 02 to calculate the critical depth.

     y_c = 0.243 m  ANSWER.

     Use ONLINE WSPROFILES 23 to calculate the M₃ water surface profile, with given data.

   • Length from toe of spillway to critical depth L_tc = 19.36 m.  ANSWER.

   • Froude number at the toe of the spillway (Froude number of upstream flow) F₁ = 4.439  ANSWER.

   • Froude number downstream of the jump F₂ = 0.181  ANSWER.

   • Sequent depth y₂ = 0.759 m.  ANSWER.

   • Length of the jump L_j = 6.2 × 0.759 = 4.706 m  ANSWER.

   • Minimum length of the stilling basin L_sb = L_tc + L_j = 19.36 + 4.706 = 24.066 m.  ANSWER.

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6. An overflow spillway flows into a mild channel, producing a hydraulic jump. The channel is rectangular, with Q = 100 ft³/s, bottom width b = 20 ft, and Manning's n = 0.015. The flow depth at the toe of the spillway is 0.4 ft and the approximate slope at the toe of the spillway is 0.1. The slope of the [mild] downstream channel, which functions as a stilling basin, is 0.0001. Calculate:

   • the critical depth,

   • the length L_tc from the toe of the spillway to critical depth downstream,

   • the Froude number at the toe of the spillway,

   • the Froude number downstream of the hydraulic jump,

   • the sequent depth y₂ (the normal depth downstream of the hydraulic jump),

   • the length L_j of the jump, assuming L_j = 6.2 y₂

   • the minimum length of the stilling basin L_sb = L_tc + L_j

   Use ONLINE CHANNEL 02 and ONLINE WSPROFILES 23. In the latter, use n = 100 and m = 100.

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   • Use ONLINE CHANNEL 02 to calculate the critical depth.

     y_c = 0.919 ft  ANSWER.

     Use ONLINE WSPROFILES 23 to calculate the M₃ water surface profile, with given data.

   • Length from toe of spillway to critical depth L_tc = 70.76 ft.  ANSWER.

   • Froude number at the toe of the spillway (Froude number of upstream flow) F₁ = 4.507  ANSWER.

   • Froude number downstream of the jump F₂ = 0.176  ANSWER.

   • Sequent depth y₂ = 2.927 ft.  ANSWER.

   • Length of the jump L_j = 6.2 × 2.927 = 18.147 ft  ANSWER.

   • Minimum length of the stilling basin L_sb = L_tc + L_j = 70.76 + 18.147 = 88.907 ft.  ANSWER.

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7. A diversion dam of height H = 1.8 m is planned on a steep stream with bottom slope S_o = 0.035. A hydraulic jump is expected upstream of the dam. Identify the type of water surface profile. Using ONLINE CALC, calculate the length of the water surface profile, from the location of the diversion dam, in the upstream direction, to the [downstream end of the] hydraulic jump. The channel has Q = 4 m³/s, bottom width b = 3 m, side slope z = 1, and Manning's n = 0.03. What are the sequent depths? What is the Froude number of the upstream flow? Use m = 100 and n = 100. Verify the sequent depth y₂ using ONLINE CHANNEL 11.

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   The type of water-surface profile is S₁. Therefore, run ONLINE WSPROFILES 24 for the given data.  ANSWER.

   The length of the S₁ water-surface profile is L = 29.79 m.  ANSWER.

   The sequent depths are y₁ = 0.397 m, and y₂ = 0.668 m.  ANSWER.

   The Froude number of the upstream flow [normal depth] is F₁ = 1.5.  ANSWER.

   Using ONLINE CHANNEL 11 with y₁ = 0.397 m and v₁ = 2.962 m/s, it is verified that y₂ = 0.667 m.  ANSWER.

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8. A mild channel enters into a steep channel of slope S_o = 0.03. Identify the type of water surface profile in the steep channel. Using ONLINE CALC, calculate the normal depth in the steep channel, and the length of the water surface profile to within 2% of normal depth. The channel has Q = 3 m³/s, bottom width b = 5 m, side slope z = 0, and Manning's n = 0.015. What is the normal-depth Froude number in the steep channel? Use m = 100 and n = 100.

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   The type of water-surface profile is S₂. Therefore, run ONLINE WSPROFILES 25 for the given data.  ANSWER.

   The normal depth in the steep channel, using ONLINECHANNEL01, is 0.174 m.  ANSWER.

   1.02 of the normal depth is: 0.174 × 1.02 = 0.1775 m.

   The length of the S₂ water-surface profile to within 2% of the normal depth is:

   L_(0.1775) = 25.78 + 0.75 (29.91 - 25.78) = 25.78 + 3.10 = 28.88 m.  ANSWER.

   The normal-depth Froude number in the steep channel is F_n = 2.634.  ANSWER.

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9. A steep channel of slope S_o = 0.035 enters into a milder steep channel of slope S_o = 0.012. Identify the type of water surface profile in the milder steep channel. Using ONLINE CALC, calculate the normal depth in the downstream channel, and the length [to normal depth] of the water surface profile. The channel has Q = 3.2 m³/s, bottom width b = 4 m, side slope z = 2, and Manning's n = 0.015. What are the normal-depth Froude numbers? What is the normal depth in the upstream channel? What would be the length of the water surface profile if Manning's n was instead estimated at 0.013? Use m = 100 and n = 100.

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   The type of water-surface profile is S₃. Therefore, run ONLINE WSPROFILES 26 for the given data.  ANSWER.

   The normal depth in the downstream [milder steep] channel is 0.260 m.  ANSWER.

   Based on the output, the length of the S₃ water-surface profile to a depth of 0.260 m is approximately L = 74 m.

   The normal-depth Froude number in the downstream [milder steep] channel is F_n = 1.8.  ANSWER.

   The normal-depth Froude number in the upstream [steep] channel is F_n,u/s = 2.94.  ANSWER.

   The normal depth in the upstream [steeper] channel is y_n,u/s = 0.19 m.  ANSWER.

   Run ONLINE WSPROFILES 26 for the same data, but change Manning's n to 0.013.

   Then, based on the output, the length of the S₃ water-surface profile to a [new] normal depth in the downstream channel of 0.239 m is approximately L = 96 m.  ANSWER.

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10. A perennial stream has the following properties:   discharge Q = 15 m³/s, bottom width b = 8 m, side slope z = 2, bottom slope S_o = 0.0025, and Manning's n = 0.035. A 2.0-m high diversion dam is planned on the stream to raise the head for an irrigation canal.

    • Calculate the normal depth.

    • Calculate the total length of the M₁ water-surface profile. Use three resolutions: (a) n = 100 and m = 100, (b) n = 200 and m = 200, and (c) n = 400 and m = 400. Comment on the results.

    • Using the higher resolution results, calculate the partial length of the M₁ profile, from the dam location to a point upstream where the normal depth is exceeded by 1%.

    Use ONLINE CHANNEL 01 and ONLINE WSPROFILES 21.

    Fig. 7-21  A diversion dam.

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    • Use ONLINE CHANNEL 01 to calculate the normal depth.

      y_n = 1.117 m  ANSWER.

    • Use ONLINE WSPROFILES 21 to calculate the M₁ water surface profile. The normal depth is confirmed to be y_n = 1.117 m.

      From the output table, for y_n = 1.117 m, n = m = 100, the total length of the M₁ profile is L = 940.2 m.  ANSWER.

      From the output table, for y_n = 1.117 m, n = m = 200, the total length of the M₁ profile is L = 1009.7 m.  ANSWER.

      From the output table, for y_n = 1.117 m, n = <>m = 400, the total length of the M₁ profile is L = 1079.8 m.  ANSWER.

      At higher resolution, the total length of the M₁ profile increases.  ANSWER.

    • The u/s depth to calculate the partial length of the M₁ profile is y_(1.01yn) = 1.01 × 1.117 = 1.128 m.

      From the output table, for n = m = 400, the partial length which is 1% in excess of normal L_(1.128) = 712.1 m.  ANSWER.

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11. An overflow spillway flows into a mild channel, producing a hydraulic jump. The channel is rectangular, with Q = 3.6 m³/s, bottom width b = 5 m, and Manning's n = 0.015. The flow depth at the toe of the spillway is 0.15 m and the approximate slope at the toe of the spillway is 0.1. The slope of the downstream mild channel, which functions as a stilling basin, is 0.00016. Use n = 100 and m = 100. Calculate:

    • the critical depth,

    • the length L_tc from the toe of the spillway to critical depth downstream,

    • the Froude number at the toe of the spillway,

    • the Froude number downstream of the hydraulic jump,

    • the sequent depth y₂ (the normal depth downstream of the hydraulic jump),

    • the length L_j of the jump, assuming L_j = 6.2 y₂

    • the minimum length of the stilling basin L_sb = L_tc + L_j

    What would be the length of the stilling basin if the bottom width were to be increased to 7 m?

    Use ONLINE CHANNEL 02 and ONLINE WSPROFILES 23. In the latter, use n = 100 and m = 100.

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    • Use ONLINE CHANNEL 02 to calculate critical depth y_c = 0.375 m.  ANSWER.

      Use ONLINE WSPROFILES 23 to calculate the M₃ water surface profile, with given data.

    • Length from toe of spillway to critical depth L_tc = 33.30 m.  ANSWER.

    • Froude number at the toe of the spillway (Froude number of upstream flow) F₁ = 4.653  ANSWER.

    • Froude number downstream of the jump F₂ = 0.215  ANSWER.

    • Sequent depth y₂ = 1.046 m.  ANSWER.

    • Length of the jump L_j = 6.2 × 1.046 = 6.48 m  ANSWER.

    • Minimum length of the stilling basin L_sb = L_tc + L_j = 33.30 + 6.48 = 39.78 m.  ANSWER.

    For a bottom width b = 7 m:

    1. Length from toe of spillway to critical depth L_tc = 19.91 m.

    2. Sequent depth y₂ = 0.808 m.

    3. Length of the jump L_j = 6.2 × 0.808 = 5.01 m

    4. Minimum length of the stilling basin L_sb = L_tc + L_j = 19.91 + 5.01 = 24.92 m.  ANSWER.

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12. A 5-m high diversion dam is planned on a channel operating at critical flow. The channel is rectangular, with Q = 100 m³/s, bottom width b = 4.7 m, bottom slope S_o = 0.01, and Manning's n = 0.023. Calculate the length of the C₁ water-surface profile. Assume n = 100 and m = 100.

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    Run ONLINE CHANNEL 02 to calculate the critical depth: y_c = 3.587 m.  ANSWER.

    Run ONLINE WSPROFILES 31 for the given data. The critical depth is confirmed to be y_c = 3.587 m. The length of the water-surface profile is: L = 139.71 m.  ANSWER.

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13. A steep channel, with bottom slope S_o = 0.03, flows into a channel operating at critical flow. The channel is rectangular, with Q = 100 m³/s, bottom width b = 4.7 m, bottom slope S_o = 0.01, and Manning's n = 0.024. Calculate the length of the C₃ water-surface profile. Assume n = 100 and m = 100.

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    Run ONLINE CHANNEL 05 to calculate the normal depth in the upstream channel y_c = 2.571 m, and the critical depth in the donwstream channel: y_c = 3.587 m.  ANSWER.

    Run ONLINE WSPROFILES 34 for the given data. The critical depth in the downstream channel is confirmed to be y_c = 3.587 m. The normal depth in the upstream channel is confirmed to be y_c = 2.571 m. The length of the water-surface profile is: L = 99.59 m.  ANSWER.

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14. A horizontal channel of length L = 500 m is designed to convey Q = 5 m³/s from a reservoir to a free overfall. The bottom width is b = 2 m, and side slope z = 1.5. The channel is lined with gabions and the Manning's n recommended by the manufacturer is n = 0.028.

    • What is the headwater depth (accuracy to 1 cm) required to pass the design discharge?

    • What is the tailwater depth, that is, the critical flow depth at the downstream boundary?

    Use ONLINE_WSPROFILES_32; assume n = 100 and m = 100.

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    The type of water-surface profile is H₂. Therefore, run ONLINE WSPROFILES 32 for the given data.  ANSWER.

    The computation proceeds by trial and error.

    • Assume a headwater depth such that the length of the H₂ backwater profile is equal to the length of the channel L = 500 m. By trial and error, the headwater depth is calculated to be y_HW = 1.5505 ≅ 1.55 m.  ANSWER.

    • The tailwater depth is y_TW = 0.714 m.  ANSWER.

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15. A sluice gate is designed to release supercritical flow into a stilling basin, where a hydraulic jump will occur. The basin channel bottom is horizontal, of rectangular cross section. The design discharge is Q = 5 m³/s, the bottom width b = 5 m, and Manning's n = 0.015.

    • What should be the gate opening (accuracy to 1 mm) to ensure that the length of the downstream water surface profile is not greater than 10 m?

    • What is the critical depth?

    Use ONLINE_WSPROFILES_35; assume n = 100 and m = 100. Assume S_o,u/s = 0.05 so that the flow through the sluice gate remains supercritical.

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    The type of water-surface profile is H₃. Therefore, run ONLINE WSPROFILES 35 for the given data.  ANSWER.

    The computation proceeds by trial and error.

    • Assume a flow depth y_u such that the calculated length of the H₃ backwater profile is slightly less than 10 m. By trial and error, this depth is calculated to be y_u = 0.351 m.  ANSWER.

    • The critical depth, at the end of the profile, immediately upstream of the jump, is y_c = 0.467 m.  ANSWER.

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