QUESTIONS
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.
What is retarded flow?
Retarded flow has a positive value of the
depth gradient.
What is accelerated flow?
Accelerated flow has a negative value of the
depth gradient.
What is subnormal flow?
Subnormal flow is a flow where the depth is greater than normal.
What is supernormal flow?
Supernormal flow is a flow where the depth is smaller than normal.
How many profiles are possible in steady gradually varied flow?
12 profiles.
What is the rule for the Type I family of water-surface profiles?
1 > F 2 < (So / Sc)
What is the rule for the Type III family of water-surface profiles?
1 < F 2 > (So / Sc)
Which water-surface profiles are completely horizontal?
C1 and C3.
What are the five limits to the water surface profiles?
So, Sc, 0, + ∞ and - ∞.
What is the typical application of the M1 water-surface profile?
Flow in a mild channel, upstream of a reservoir.
What is the typical application of the S1 water-surface profile?
Flow in a steep channel, upstream of a reservoir.
What is the typical application of the S3 water-surface profile?
Flow in a steep channel, downstream of a steeper channel carrying supercritical flow.
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
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.
Use ONLINE CHANNEL 01
and ONLINE WSPROFILES 21.
Fig. 7-21 A diversion dam. 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.
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.
Use ONLINE CHANNEL 01
and ONLINE WSPROFILES 21.
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.
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.
Use ONLINE CHANNEL 05
and ONLINE WSPROFILES 22.
From the output table, for yn = 1.948 m, the length is L = 3815.6 m. ANSWER.
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.
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.
Use ONLINE CHANNEL 05
and ONLINE WSPROFILES 22.
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.
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:
Use ONLINE CHANNEL 02
and ONLINE WSPROFILES 23. In the latter, use n = 100 and m = 100.
yc = 0.243 m ANSWER.
Use ONLINE WSPROFILES 23 to calculate the
M3 water surface profile, with given
data.
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:
Use ONLINE CHANNEL 02
and ONLINE WSPROFILES 23. In the latter, use n = 100 and m = 100.
yc = 0.919 ft ANSWER.
Use ONLINE WSPROFILES 23 to calculate the
M3 water surface profile, with given
data.
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.
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.
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.
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.
Use ONLINE CHANNEL 01
and ONLINE WSPROFILES 21.
Fig. 7-21 A diversion dam. yn = 1.117 m ANSWER.
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.
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.
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:
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 WSPROFILES 23 to calculate the M3 water surface profile, with given
data.
For a bottom width b = 7 m:
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.
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 30 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 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.
Use ONLINE_WSPROFILES_28; assume n = 100 and m = 100.
The type of water-surface profile is H2. Therefore,
run ONLINE WSPROFILES 28 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.
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_31; 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 31 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.
A weir is located at the downstream end of a wide rectangular
channel of bottom slope
The Froude number in the upstream channel is: F 2 =
So / f = 1. Therefore: F = 1, and the type of water-surface
profile immediately upstream of the weir is C1. The corresponding flow depth for the upstream flow is: yu = (q 2 /g )1/3 = 0.254 m.
From geometry: So = Δy /Δx, where Δx is the length of the profile. Therefore, the length of the C1 water-surface profile is: Δx = Δy /So = (1.5 - 0.254) / 0.0035 = 356 m.
A hydraulically wide channel with So = 0.025 flows
into a critical flow channel of So = 0.004. The dimensionless
Chezy friction factor f
is the same is both channels.
The unit-width discharge is 2.2 m3/s/m.
What is the type and length of the water-surface profile in the downstream channel?
The Froude number in the upstream channel is: F 2 =
So / Sc = 0.025 / 0.004 = 6.25. Therefore: F = 2.5, and the type of water-surface
profile in the downstream channel is C3. The corresponding flow depth
for the upstream flow is: yu = [q 2 /(6.25 g )]1/3 = 0.429 m.
The flow depth
for the downstream (critical) flow is: yu = (q 2 /g )1/3 = 0.790 m.
From geometry: So =
Run
ONLINE TRANSITION DESIGN
using the data of the example given in Section 7.6. Confirm the results shown in Tables 7-10 and 7-11.
The results are shown in the following table (click on top of the table to expand).
This table confirms the results of the example.
Design a warped inlet transition from canal to flume
for the following data: Design discharge = 300 cfs;
canal bottom width b = 12 ft;
canal side slope = 2 H : 1 V;
canal bottom slope = 0.0012;
canal Manning's n = 0.020;
flume bottom width b = 8 ft;
flume side slope = 0 H : 1 V;
flume bottom slope = 0.01;
flume Manning's n = 0.012;
water surface elevation immediately upstream of the transition Z = 100 ft.
Use the inverse parabolas option of ONLINE TRANSITION DESIGN.
The results are shown in the following table (click on top of the table to expand).
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