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 watersurface profiles?
1 > F ^{2} < (S_{o} / S_{c})
What is the rule for the Type III family of watersurface profiles?
1 < F ^{2} > (S_{o} / S_{c})
Which watersurface profiles are completely horizontal?
C_{1} and C_{3}.
What are the five limits to the water surface profiles?
S_{o}, S_{c}, 0, + ∞ and  ∞.
What is the typical application of the M_{1} watersurface profile?
Flow in a mild channel, upstream of a reservoir.
What is the typical application of the S_{1} watersurface profile?
Flow in a steep channel, upstream of a reservoir.
What is the typical application of the S_{3} watersurface profile?
Flow in a steep channel, downstream of a steeper channel carrying supercritical flow.
What is the main difference between the directstep and standardstep methods
of watersurface profile computations?
The directstep method advances without iteration; the standardstep method advances through iteration.
PROBLEMS
A perennial stream has the following properties: discharge
Q = 30 m^{3}/s, bottom width b = 55 m, side slope z = 2,
bottom slope S_{o} = 0.0004, and Manning's n = 0.035.
A 2m 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 M_{1} watersurface profile. Use n = 400 and m = 400.
 Calculate the partial length of the M_{1} 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. 721 A diversion dam.


 Use ONLINE CHANNEL 01 to calculate the normal depth.
y_{n} = 0.968 m ANSWER.
Use ONLINE WSPROFILES 21 to calculate the M_{1} 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 M_{1} profile
is L = 7510.7 m. ANSWER.
The u/s depth to calculate the partial length of the M_{1} 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 ft^{3}/s, bottom width b = 150 m, side slope z = 2,
bottom slope S_{o} = 0.00038, and Manning's n = 0.035.
A 6ft 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 M_{1} watersurface profile. Use n = 400 and m = 400.
 Calculate the partial length of the M_{1} profile,
from the dam to a location upstream where the normal depth is exceeded by 1%.
Use ONLINE CHANNEL 01
and ONLINE WSPROFILES 21.
 Use ONLINE CHANNEL 01 to calculate the normal depth.
y_{n} = 3.476 ft ANSWER.
Use ONLINE WSPROFILES 21 to calculate the M_{1} 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 M_{1} profile
is L = 25,854.7 ft. ANSWER.
The u/s depth to calculate the partial length of the M_{1} 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 M_{2}
upstream of the brink. The hydraulic conditions in the channel are:
Q = 28 m^{3}/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.
 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 M_{2} 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.
A mild stream flows into a steep channel, producing an M_{2}
upstream of the brink. The hydraulic conditions in the channel are:
Q = 500 ft^{3}/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.
 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 M_{2} 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.
An overflow
spillway flows into a mild channel, producing a hydraulic jump.
The channel is rectangular, with Q = 3 m^{3}/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 y_{2} (the normal depth downstream of the hydraulic jump),
 the length L_{j} of the jump, assuming L_{j} = 6.2 y_{2}
 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.
An overflow
spillway flows into a mild channel, producing a hydraulic jump.
The channel is rectangular, with Q = 100 ft^{3}/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 y_{2} (the normal depth downstream of the hydraulic jump),
 the length L_{j} of the jump, assuming L_{j} = 6.2 y_{2}
 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.
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 m^{3}/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 y_{2}
using ONLINE CHANNEL 11.
The type of watersurface profile is S_{1}.
Therefore,
run ONLINE WSPROFILES 24 for the given data. ANSWER.
The length of the S_{1} watersurface profile is L = 29.79 m. ANSWER.
The sequent depths are y_{1} = 0.397 m, and y_{2} = 0.668 m. ANSWER.
The Froude number of the upstream flow [normal depth] is F_{1} = 1.5. ANSWER.
Using ONLINE CHANNEL 11
with y_{1} = 0.397 m and v_{1} = 2.962 m/s,
it is verified that y_{2} = 0.667 m. ANSWER.
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 m^{3}/s, bottom width b = 5 m, side slope z = 0,
and Manning's n = 0.015. What is the normaldepth Froude number in the steep channel?
Use m = 100 and n = 100.
The type of watersurface profile is S_{2}. 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 S_{2} watersurface 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 normaldepth Froude number in the steep channel is F_{n} = 2.634. ANSWER.
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 m^{3}/s, bottom width b = 4 m, side slope z = 2,
and Manning's n = 0.015. What are the normaldepth 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 watersurface profile is S_{3}. 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 S_{3} watersurface profile to a depth of 0.260 m is approximately L = 74 m.
The normaldepth Froude number in the downstream [milder steep]
channel is F_{n} = 1.8. ANSWER.
The normaldepth 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 S_{3} watersurface 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 m^{3}/s, bottom width b = 8 m, side slope z = 2,
bottom slope S_{o} = 0.0025, and Manning's n = 0.035.
A 2.0m 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 M_{1} watersurface 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 M_{1} 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. 721 A diversion dam.


 Use ONLINE CHANNEL 01 to calculate the normal depth.
y_{n} = 1.117 m ANSWER.
 Use ONLINE WSPROFILES 21 to calculate the M_{1} 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 M_{1} 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 M_{1} 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 M_{1} profile is L = 1079.8 m. ANSWER.
At higher resolution, the total length of the M_{1} profile increases. ANSWER.
 The u/s depth to calculate the partial length of the
M_{1} 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.
An overflow spillway flows into a mild channel, producing a hydraulic jump.
The channel is rectangular, with Q = 3.6 m^{3}/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 y_{2} (the normal depth downstream of the hydraulic jump),
 the length L_{j} of the jump, assuming L_{j} = 6.2 y_{2}
 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.
For a bottom width b = 7 m:
 Length from toe of spillway to critical depth L_{tc} = 19.91 m.
 Sequent depth y_{2} = 0.808 m.
 Length of the jump L_{j} = 6.2 × 0.808 = 5.01 m
 Minimum length of the stilling basin L_{sb} =
L_{tc} + L_{j} = 19.91 + 5.01 = 24.92 m. ANSWER.
A 5m high diversion dam is planned on a channel operating at critical flow.
The channel is rectangular, with
Q = 100 m^{3}/s, bottom width b = 4.7 m,
bottom slope S_{o} = 0.01, and Manning's n = 0.023.
Calculate the length of the C_{1} watersurface profile.
Assume n = 100 and m = 100.
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 watersurface profile is: L = 139.71 m. ANSWER.
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 m^{3}/s, bottom width b = 4.7 m,
bottom slope S_{o} = 0.01, and Manning's n = 0.024.
Calculate the length of the C_{3} watersurface profile.
Assume n = 100 and m = 100.
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 watersurface profile is: L = 99.59 m. ANSWER.
A horizontal channel of length L = 500 m
is designed to convey Q = 5 m^{3}/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 watersurface profile is H_{2}. 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 H_{2} 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.
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 m^{3}/s, the
bottom width b = 5 m, and Manning's n = 0.015.
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.
The type of watersurface profile is H_{3}. 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 H_{3} 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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