QUESTIONS

1. What is steady gradually varied flow?

The flow is steady gradually varied when the discharge Q is constant (in time and space) but the other hydraulic variables vary gradually in space, but not in time.

2. What is retarded flow?

Retarded flow has a positive value of the depth gradient.

3. What is accelerated flow?

Accelerated flow has a negative value of the depth gradient.

4. What is subnormal flow?

Subnormal flow is a flow where the depth is greater than normal.

5. What is supernormal flow?

Supernormal flow is a flow where the depth is smaller than normal.

6. How many profiles are possible in steady gradually varied flow?

12 profiles.

7. What is the rule for the Type I family of water-surface profiles?

1 > F 2 < (So / Sc)

8. What is the rule for the Type III family of water-surface profiles?

1 < F 2 > (So / Sc)

9. Which water-surface profiles are completely horizontal?

C1 and C3.

10. What are the five limits to the water surface profiles?

So, Sc, 0, + ∞ and - ∞.

11. What is the typical application of the M1 water-surface profile?

Flow in a mild channel, upstream of a reservoir.

12. What is the typical application of the S1 water-surface profile?

Flow in a steep channel, upstream of a reservoir.

13. What is the typical application of the S3 water-surface profile?

Flow in a steep channel, downstream of a steeper channel carrying supercritical flow.

14. What is the main difference between the direct-step and standard-step methods of water-surface profile computations?

The direct-step method advances without iteration; the standard-step method advances through iteration.

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 So = 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%.

Fig. 7-21  A diversion dam.

• Use ONLINE CHANNEL 01 to calculate the normal depth.

yn = 0.968 m  ANSWER.

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

From the output table, for yn = 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.

2. A perennial stream has the following properties:   discharge Q = 1000 ft3/s, bottom width b = 150 m, side slope z = 2, bottom slope So = 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 to calculate the normal depth.

yn = 3.476 ft  ANSWER.

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

From the output table, for yn = 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.

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 So = 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 to calculate critical depth yc = 0.777 m and normal depth yn = 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 yn = 1.948 m, the length is L = 3815.6 m.  ANSWER.

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

• For n = 400, and m = 400, for yn = 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.

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 So = 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 to calculate critical depth yc = 1.983 ft and normal depth yn = 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 yn = 5.161 ft, the length is L = 9,903.4 ft.  ANSWER.

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

• For n = 400, and m = 400, for yn = 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.

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 Ltc 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 Lj of the jump, assuming Lj = 6.2 y2

• the minimum length of the stilling basin Lsb = Ltc + Lj

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

• Use ONLINE CHANNEL 02 to calculate the critical depth.

yc = 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 Ltc = 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 Lj = 6.2 × 0.759 = 4.706 m  ANSWER.

• Minimum length of the stilling basin Lsb = Ltc + Lj = 19.36 + 4.706 = 24.066 m.  ANSWER.

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 Ltc 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 Lj of the jump, assuming Lj = 6.2 y2

• the minimum length of the stilling basin Lsb = Ltc + Lj

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

• Use ONLINE CHANNEL 02 to calculate the critical depth.

yc = 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 Ltc = 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 Lj = 6.2 × 2.927 = 18.147 ft  ANSWER.

• Minimum length of the stilling basin Lsb = Ltc + Lj = 70.76 + 18.147 = 88.907 ft.  ANSWER.

7. A diversion dam of height H = 1.8 m is planned on a steep stream with bottom slope So = 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.

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.

8. A mild channel enters into a steep channel of slope So = 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.

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 Fn = 2.634.  ANSWER.

9. A steep channel of slope So = 0.035 enters into a milder steep channel of slope So = 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.

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 Fn = 1.8.  ANSWER.

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

The normal depth in the upstream [steeper] channel is yn,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.

10. A perennial stream has the following properties:   discharge Q = 15 m3/s, bottom width b = 8 m, side slope z = 2, bottom slope So = 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%.

Fig. 7-21  A diversion dam.

• Use ONLINE CHANNEL 01 to calculate the normal depth.

yn = 1.117 m  ANSWER.

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

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

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

From the output table, for yn = 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.

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 Ltc 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 Lj of the jump, assuming Lj = 6.2 y2

• the minimum length of the stilling basin Lsb = Ltc + Lj

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.

• Use ONLINE CHANNEL 02 to calculate critical depth yc = 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 Ltc = 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 Lj = 6.2 × 1.046 = 6.48 m  ANSWER.

• Minimum length of the stilling basin Lsb = Ltc + Lj = 33.30 + 6.48 = 39.78 m.  ANSWER.

For a bottom width b = 7 m:

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

2. Sequent depth y2 = 0.808 m.

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

4. Minimum length of the stilling basin Lsb = Ltc + Lj = 19.91 + 5.01 = 24.92 m.  ANSWER.

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 So = 0.01, and Manning's n = 0.023. Calculate the length of the C1 water-surface profile. Assume n = 100 and m = 100.

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

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

13. A steep channel, with bottom slope So = 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 So = 0.01, and Manning's n = 0.024. Calculate the length of the C3 water-surface profile. Assume n = 100 and m = 100.

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

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

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.

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 yHW = 1.5505 ≅ 1.55 m.  ANSWER.

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

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 So,u/s = 0.05 so that the flow through the sluice gate remains supercritical.

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 yu 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 yu = 0.351 m.  ANSWER.

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

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