CHAPTER 7: STEADY GRADUALLY VARIED FLOW 
7.1 EQUATION OF GRADUALLY VARIED FLOW
The flow is gradually varied when the discharge Q is constant but the other
hydraulic variables (A, V, D, R, P, and so on) vary gradually in space.
The basic assumptions of gradually varied flow are:
The flow is steady, i.e., none of the hydraulic variables vary in time.
The streamlines are essentially parallel; thus,
the pressure distribution in the vertical direction is hydrostatic, i.e., proportional to the flow depth.
The head loss is the same as that corresponding to uniform flow; therefore, the uniform
flow formula may be used to evaluate the energy slope.
The value of Manning's n is the same as that of uniform flow.
Other assumptions of gradually varied flow are:
The slope of the channel is small.
The pressure correction factor cosθ ≅ 1.
There is negligible air entrainment.
The conveyance is an exponential function of the flow depth (except for the case of circular culverts).
The roughness (Manning's n) is independent of the flow depth (only an approximation) and is constant
throughout the reach under consideration.
Fig. 71 Definition sketch for energy in openchannel flow.


Under gradually varied flow, the gradient of hydraulic head is (Fig. 71):
dH d V^{ 2}
^{_____} = ^{___} ( z + y + ^{_____} ) =  S_{f}
dx dx 2g
 (71) 
The negative sign in front of the friction slope S_{f} is required because the flow direction is from left to right,
while the derivative is taken from right to left, by convention. By definition, the friction slope is:
h_{f}
S_{f} = ^{_____}
ΔL
 (72) 
in which ΔL = length of the channel reach under consideration. The gradient of specific energy is:
dE d V^{ 2} dz
^{_____} = ^{___} ( y + ^{_____} ) =  ^{____}  S_{f}
dx dx 2g dx
 (73) 
The gradient of the channel bed, or channel slope (bottom slope), is:
dz z_{2}  z_{1}
^{_____} = ^{________}
dx ΔL
 (74) 
dz z_{1}  z_{2}
 ^{_____} = ^{________} = S_{o}
dx ΔL
 (75) 
Therefore, the gradient of specific energy is:
dE d V^{ 2}
^{_____} = ^{___} ( y + ^{_____} ) = S_{o}  S_{f}
dx dx 2g
 (76) 
Under steady flow: Q = V A = constant. Therefore:
d Q^{ 2}
^{____} ( y + ^{_______} ) = S_{o}  S_{f}
dx 2g A^{2}
 (77) 
dy d Q^{ 2}
^{_____} + ^{_____} ( ^{_______} ) = S_{o}  S_{f}
dx dx 2g A^{2}
 (78) 
dy Q^{ 2} dA
^{_____}  ( ^{______} ) ^{_____} = S_{o}  S_{f}
dx g A^{3} dx
 (79) 
dy Q^{ 2} dA dy
^{_____}  ( ^{______} ) ^{_____} ^{_____} = S_{o}  S_{f}
dx g A^{3} dy dx
 (710) 
Using Eq. 311:
dy Q^{ 2} T dy
^{_____}  ( ^{________} ) ^{_____} = S_{o}  S_{f}
dx g A^{3} dx
 (711) 
Therefore, the flowdepth gradient is:
dy S_{o}  S_{f}
^{_____} = ^{_______________________}
dx 1  [(Q^{ 2} T ) / (g A^{3})]
 (712) 
The friction slope based on the Chezy equation (Eqs. 510 and 24) is:
Q^{ 2}
S_{f} = ^{____________}
C^{ 2} A^{2} R
 (713) 
Since R = A / P :
Q^{ 2} P
S_{f} = ^{_________}
C^{ 2} A^{3}
 (714) 
Substituting Eq. 714 into Eq. 712, the flowdepth gradient is:
dy S_{o}  [(Q^{ 2} P ) / (C^{ 2} A^{3})]
^{_____} = ^{___________________________}
dx 1  [(Q^{ 2} T ) / (g A^{3})]
 (715) 
dy S_{o}  (g/C^{ 2}) (P / T ) [(Q^{ 2} T ) / (g A^{3})]
^{_____} = ^{_______________________________________}
dx 1  [(Q^{ 2} T ) / (g A^{3})]
 (716) 
The square of the Froude number is (Eq. 312):
Q^{ 2} T
F^{ 2} = ^{_________}
g A^{3}
 (717) 
Substituting Eq. 717 into Eq. 716:
dy S_{o}  (g/C^{ 2}) (P / T ) F^{ 2}
^{_____} = ^{_________________________}
dx 1  F^{ 2}
 (718) 
Substituting Eq. 512 into Eq. 718:
dy S_{o}  f (P / T ) F^{ 2}
^{_____} = ^{_____________________}
dx 1  F^{ 2}
 (719) 
Therefore, the depth gradient (dy/dx) is a function of:
Channel slope S_{o},
Friction coefficient f,
Ratio of wetted perimeter to top width P / T, and
Froude number.
For dy/dx = 0, Eq. 719 reduces to a statement of uniform flow:
S_{o} = f (P / T ) F^{ 2}
 (720) 
For F = 1, Eq. 720 reduces to a statement of critical uniform flow:
S_{o} = f (P_{c} / T_{c} ) = S_{c}
 (721) 
in which S_{c} = critical slope, i.e., the channel slope for which the flow is critical.
In terms of critical slope (Eq. 721), the flowdepth gradient is:
dy S_{o}  (P / T ) (T_{c} / P_{c} )
S_{c} F^{ 2}
^{_____} = ^{________________________________}
dx 1  F^{ 2}
 (722) 
For (P / T ) ≅ (P_{c} / T_{c} ), i.e., for a constant
ratio (P / T) with flow depth, Eq. 722 reduces to:
dy S_{o} 
S_{c} F^{ 2}
^{_____} = ^{_______________}
dx 1  F^{ 2}
 (723) 
For conciseness, the flowdepth gradient can be written as:
dy
S_{y} = ^{_____}
dx
 (724) 
Substituting Eq. 724 into Eq. 723, the flowdepth gradient is:
S_{y} (S_{o} / S_{c})  F^{ 2}
^{____} = ^{__________________}
S_{c} 1  F^{ 2}
 (725) 
Equation 725 (or 723) is the steady gradually varied flow equation (Fig. 72). The depth gradient S_{y} is a function
only of: (1) channel slope S_{o}, (2) critical slope S_{c}, and (3) Froude number F.
Fig. 72 Steady gradually varied flow.


Note on the applicability of Eq. 725
Strictly speaking,
Eq. 725 applies only for
the case (P / T ) (T_{c} / P_{c} ) = 1, which is the same as
(P / T ) = (P_{c} / T_{c} ); that is,
for a constant ratio (P / T ), regardless
of flow depth. This condition is less restrictive that the (asymptotic) hydraulically wide
channel condition, for which (P / T ) = 1.
Therefore, for a hydraulically wide channel, for which P ≅ T,
it follows that: (P / T ) (T_{c} / P_{c} ) ≅ 1. Thus, it is concluded that Eq. 725 applies for a hydraulically wide channel.

7.2 CHARACTERISTICS OF FLOW PROFILES
In Eq. 725, the sign of the lefthand side (LHS) is that of S_{y}
(numerator), since S_{c} (denominator)
is always positive (friction is always positive). The sign of S_{y} (i.e, the sign of
the LHS) may be one of three possibilities:
 A positive value, leading to RETARDED FLOW (BACKWATER),
 A zero value, leading to UNIFORM FLOW (NORMAL), or
 A negative value, leading to ACCELERATED FLOW (DRAWDOWN).
In the righthand side (RHS) of Eq. 725, there are three possibilities for the numerator (USDA Soil Conservation Service, 1971):
 S_{o} / S_{c} > F^{ 2}, leading to SUBNORMAL FLOW,
 S_{o} / S_{c} = F^{ 2}, leading to NORMAL FLOW, or
 S_{o} / S_{c} < F^{ 2}, leading to SUPERNORMAL FLOW.
There are three possibilities for the denominator:
 1 > F^{ 2}, leading to SUBCRITICAL FLOW,
 1 = F^{ 2}, leading to CRITICAL FLOW, or
 1 < F^{ 2}, leading to SUPERCRITICAL FLOW.
Given the above inequalities,
there arise three types (or families) of watersurface profiles, shown in Table 71.
The total number of profiles is 12.
A summary is shown in Table 72.

Table 71 Types of watersurface profiles.
 Type  Description
 Numerator and denominator of RHS of Eq. 725 
Sign of LHS  Flow profile

I
 Subnormal/subcritical flow 
Both numerator and denominator are positive

+ 
Retarded 
II  A
 Subnormal/supercritical flow  Numerator is positive and denominator is negative 
 
Accelerated  B  Supernormal/subcritical flow  Numerator is negative and denominator is positive 
 
Accelerated 
III  Supernormal/supercritical flow
 Both numerator and denominator are negative 
+ 
Retarded 
Table 72 Summary of watersurface profiles.

Family
 Character
 Rule
 S_{o} > S_{c}
 S_{o} = S_{c}
 S_{o} < S_{c}
 S_{o} = 0
 S_{o} < 0

I
 Retarded (Backwater)
 1 > F^{ 2} < (S_{o} / S_{c})
 S_{1}
 C_{1}
 M_{1}
 
 

II  A
 Accelerated (Drawdown)
 1 < F^{ 2} < (S_{o} / S_{c})
 S_{2}
 
 
 
 

B
 Accelerated (Drawdown)
 1 > F^{ 2} > (S_{o} / S_{c})
 
 
 M_{2}
 H_{2}
 A_{2}

III
 Retarded (Backwater)
 1 < F^{ 2} > (S_{o} / S_{c})
 S_{3}
 C_{3}
 M_{3}
 H_{3}
 A_{3}

Type I
In the Type I family, the flow is subnormal/subcritical. Therefore, the rule is:
1 > F^{ 2} <
(S_{o} / S_{c} )
 (726) 
which is the same as:
(S_{o} / S_{c} ) > < 1
 (727) 
Equation 727 states that S_{o} may be lesser than, equal to, or greater than S_{c} .
This gives rise to three types of profiles:
Since
(S_{o} / S_{c} ) > F^{ 2}
 (728) 
and
it follows that
(S_{o} / S_{c} ) > 0
 (730) 
Thus:
Therefore, no horizontal (H) or adverse (A) profiles are possible in the Type I family of watersurface profiles.

Type II A
In the Type II A family, the flow is subnormal/supercritical. Therefore, the rule is:
1 < F^{ 2} <
(S_{o} / S_{c} )
 (732) 
which is the same as:
(S_{o} / S_{c} ) > 1
 (733) 
Equation 733 states that S_{o} may only be greater than S_{c} . This gives rise
to only one profile:
Since
(S_{o} / S_{c} ) > F^{ 2}
 (734) 
and
it follows that
(S_{o} / S_{c} ) > 0
 (736) 
Thus:
Therefore, no horizontal (H) or adverse (A) profiles are possible in the Type II A family of watersurface profiles.

Type II B
In the Type II B family, the flow is supernormal/subcritical. Therefore, the rule is:
1 > F^{ 2} >
(S_{o} / S_{c} )
 (738) 
which is the same as:
(S_{o} / S_{c} ) < 1
 (739) 
Equation 739 states that S_{o} may be lesser than S_{c} ,
equal to 0, or lesser than 0.
This gives rise to three types of profiles:

Type III
In the Type III family, the flow is supernormal/supercritical. Therefore, the rule is:
1 < F^{ 2} >
(S_{o} / S_{c} )
 (744) 
which is the same as:
(S_{o} / S_{c} ) > < 1
 (745) 
Equation 745 states that S_{o} may be lesser than, equal to, or greater than S_{c} .
This gives rise to five types of profiles:

Figure 715 shows a graphical representation of flowdepth gradients
in watersurface profile computations. The
arrows indicate the direction of computation.
Fig. 715 Graphical representation of flowdepth gradient
ranges in watersurface profile computations.


7.3 LIMITS TO WATER SURFACE PROFILES
The flowdepth gradients vary between five (5) limits (Fig. 715):
The channel slope S_{o}
The critical slope S_{c}
Zero.
+ ∞
 ∞
The theoretical limits to the watersurface profiles may be analyzed using Eq. 725, repeated
here in a slightly different form:
S_{o}  S_{c} F^{ 2}
S_{y} = ^{______________}
1  F^{ 2}
 (746) 
Operating in Eq. 746:
S_{y} (1  F^{ 2}) = S_{o}  S_{c} F^{ 2}
 (747) 
S_{o}  S_{y}
F^{ 2} = ^{____________}
S_{c}  S_{y}
 (748) 
For uniform (normal) flow: S_{y} = 0, and Eq. 748 reduces to:
S_{o} = S_{c} F^{ 2}
 (749) 
For gradually varied flow: S_{y} ≠ 0, and Eq. 748 is subject to three (3) cases:

F^{ 2} > 0
S_{o} > S_{y} and S_{c} > S_{y}
The following inequality is satisfied: S_{o} > S_{y} < S_{c}
S_{o} < S_{y} and S_{c} < S_{y}
The following inequality is satisfied: S_{o} < S_{y} > S_{c}
It is concluded that S_{y} has to be either less than
both S_{o} and S_{c}, or greater than both.

F^{ 2} = 0
This leads to:
S_{o} = S_{y}
z_{1} 
z_{2}
S_{o} = ^{_________}
L
 (751) 
y_{2} 
y_{1}
S_{y} = ^{_________}
L
 (752) 
Combining Eqs. 751 and 752:
z_{1} +
y_{1} = z_{2} +
y_{2}
 (753) 
Equation 753 depicts a true reservoir (Fig. 715).
Fig. 716 True reservoir condition.



F^{ 2} < 0: Since F ≥ 0, this condition is impossible.
The following inequality is NOT satisfied: S_{o} > S_{y} > S_{c}
The following inequality is NOT satisfied: S_{o} < S_{y} < S_{c}
It is concluded that S_{y} cannot be less than S_{o} and greater than
S_{c} , or
less than S_{c} and greater than
S_{o}
. Thus, S_{y} has to be less than both S_{o } AND S_{c},
or greater than both (See Case 1).

Uses of watersurface profiles
Table 73 and Fig. 717 show the typical occurrence of mild watersurface profiles.
Table 73 Occurrence of mild watersurface profiles. 
M_{1}
 Flow in a mild channel, upstream of a reservoir.

M_{2}
 Flow in a mild channel, upstream of an abrupt change in grade or a steep channel carrying supercritical flow.

M_{3}

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

Fig. 717 Typical occurrence of mild profiles.


Table 74 and Fig. 718 show the typical occurrence of steep watersurface profiles.
Table 74 Occurrence of steep watersurface profiles. 
S_{1}
 Flow in a steep channel, upstream of a reservoir.

S_{2}

Flow in a steep channel, downstream of a mild channel carrying subcritical flow.

S_{3}
 Flow in a steep channel,
downstream of a steeper channel carrying supercritical flow.

Fig. 718 Typical occurrence of steep profiles.


7.4 METHODOLOGIES
There are two ways to calculate watersurface profiles:
The direct step method.
The standard step method.
The direct step method is applicable to prismatic channels, while the standard step method
is applicable to any channel, prismatic and nonprismatic (Table 75). The direct step method is direct,
readily amenable to use with a spreadsheet, and relatively straight forward in its solution.
The standard step method is iterative and complex
in its solution. In practice, the standard step method is represented by the Hydrologic Engineering Center's
River Analysis System, referred to as HECRAS (U.S. Army Corps of Engineers, 2014).
The direct step method applies particularly where
data is scarce and resources are limited.
The standard step method applies for comprehensive projects. The use of a widely accepted government program
such as HECRAS enhances credibility.
The required number of cross sections in the standard step method increases with the channel slope.
Steeper channels may require more cross sections. Lesser cross sectional variability
results in more reliable and accurate results. Note that extensive two and threedimensional flow features
may not be accurately represented in the onedimensional watersurface profile model.
Table 75 Comparison between direct step and standard step methods. 
No.  Characteristic  Direct step method 
Standard step method 
1  Crosssectional shape  Prismatic 
Any (prismatic or nonprismatic) 
2  Ease of computation  Easy (hours) 
Difficult (months) 
3  Calculation advances ⇒  Directly 
By iteration (trial and error) 
4  Type of crosssection input  One typical
cross section (prismatic) 
Several cross sections (nonprismatic) 
5  Data needs  Minimal 
Extensive  6  Accuracy increases with ↠  A smaller flow depth increment 
More cross sections and/or lesser crosssectional variability 
7  Independent variable  Flow depth 
Length of channel 
8  Dependent variable  Length of channel 
Flow depth 
9  Tools  Spreadsheet or programming 
HECRAS 
10  Reliability  Answer is always possible 
Answer is sometimes not possible, depending
on the type of crosssectional input data 
11  Cost  Comparatively small 
Comparatively large 
12 
Public acceptance  Average 
High 
7.5 DIRECT STEP METHOD EXAMPLE
Calculation of M_{2} and S_{2} profiles, upstream and downstream
of a change in grade, from mild to steep
Input data:
 Discharge Q = 2000 m^{3}
 Bottom width b = 100 m.
 Side slope 2 H : 1 V
 Upstream channel slope = 0.0001
 Upstream channel Manning's n = 0.025
 Downstream channel slope = 0.03
 Downstream channel Manning's n = 0.045
Solution
Calculate the normal depth and velocity,
and critical depth and velocity, in the upstream and downstream channels.
Use ONLINE CHANNEL 05.
For the upstream channel:
Normal depth = 10.098 m
Normal velocity = 1.648 m/s
Normal Froude number = 0.179
Critical depth = 3.364 m
Critical velocity = 5.571 m/s
For the downstream channel:
Normal depth = 2.669 m
Normal velocity = 7.113 m/s
Normal Froude number = 1.425
Critical depth = 3.364 m
Critical velocity = 5.571 m/s
Calculation of critical slope for the upstream channel:
y_{c} = 3.364 m
V_{c} = 5.571 m/s
A_{c} = (b + zy_{c} ) y_{c} = 359.033 m^{2}
P = b + 2 y_{c} (1 + z^{ 2})^{1/2} = 115.044 m
R_{c} = A_{c} / P_{c} = 3.121 m
S_{c} = n^{ 2} V_{c}^{2} / R_{c}^{4/3} = (0.025)^{2}
(5.571)^{2} (3.121)^{4/3} = 0.00425
Verify the critical slope with ONLINE CHANNEL 04: S_{c} = 0.004254.
Calculation of critical slope for the downstream channel:
S_{c} = n^{ 2} V_{c}^{2} / R_{c}^{4/3} = (0.045)^{2}
(5.571)^{2} (3.121)^{4/3} = 0.0138
Verify the critical slope with ONLINE CHANNEL 04: S_{c} = 0.01378.
The calculation of the M_{2} watersurface profile is shown in Table 76.
The following instructions are indicated:
The calculation moves in the upstream direction, starting at critical depth at the downstream end.
Column [1]: The second depth is set at 4 m; subsequently, the depth interval is set at 1 m.
Column [2]: The flow area is: A = (b + zy ) y . . . (1)
Column [3]: The mean velocity is: V = Q / A . . . (2)
Column [4]: The velocity head is: V^{ 2} / (2g) . . . (3)
Column [5]: The specific head is: H = y + V^{ 2} / (2g) . . . (4)
Column [6]: The wetted perimeter is: P = b + 2 y (1 + z^{ 2})^{1/2} . . . (5)
Column [7]: The hydraulic radius is: R = A / P . . . (6)
Column [8]: The friction slope is: S_{f} = n^{ 2} V^{ 2} / R^{ 4/3} . . . (7)
Column [9]: The average friction slope is: S_{f} _{ave} = 0.5 (S_{f} _{1}
+ S_{f} _{2}) . . . (8)
Column [10]: The specific head difference is: ΔH = H_{2}  H_{1} . . . (9)
Column [11]: The channel length increment ΔL,
explained in the box below.
Column [12]: The cumulative channel length, i.e., the cumulative sum of ΔL increments.
Derivation of the formula for the channel length increment ΔL
With reference to Fig. 719, the average friction slope is:
h_{f}
S_{f ave} = ^{______}
ΔL
 (754) 
Fig. 719 Definition sketch for the calculation of channel length increment ΔL.


H_{1} +
z_{1} = H_{2} +
z_{2} + S_{f ave} ΔL
 (755) 
z_{1} 
z_{2} = H_{2} 
H_{1} + S_{f ave} ΔL
 (756) 
z_{1} 
z_{2}  S_{f ave} ΔL = H_{2} 
H_{1}
 (757) 
z_{1} 
z_{2}
S_{o} = ^{_________}
ΔL
 (758) 
S_{o} ΔL  S_{f ave} ΔL = H_{2} 
H_{1}
 (759) 
(S_{o}  S_{f ave}) ΔL = H_{2} 
H_{1}
 (760) 
H_{2} 
H_{1}
ΔL = ^{______________}
S_{o}  S_{f ave}
 (761) 
ΔH
ΔL = ^{______________}
S_{o}  S_{f ave}
 (762) 
Equation 762 enables the calculation of the channel length increment ΔL, i.e., Col. 11 of Table 76.
When ΔL is negative, the calculation moves upstream, as in the M_{2} profile (Table 76).
Conversely, when ΔL is positive, the
calculation moves downstream, as in the S_{2} profile (Table 77).

Table 76 Calculation of M_{2} watersurface profile (S_{o} = 0.0001). 
[1]
 [2]
 [3]
 [4]
 [5]
 [6]
 [7]
 [8]
 [9]
 [10]
 [11]
 [12]

y
 A
 V
 V^{ 2}/(2g )
 H
 P
 R
 S_{f}
 S_{f} _{ave}
 ΔH
 ΔL
 ∑ ΔL

 (1)
 (2)
 (3)
 (4)
 (5)
 (6)
 (7)
 (8)
 (9)
 Box


3.364
 359.033
 5.570
 1.581
 4.945
 115.044
 3.1208
 0.00425
 
 
 
 0

4.000
 432.000
 4.630
 1.092
 5.092
 117.888
 3.664
 0.00237
 0.00331
 0.147
 45.794
 45.794

5.000
 550.000
 3.636
 0.674
 5.674
 122.360
 4.495
 0.00111
 0.00174
 0.528
 354.878
 400.672

6.000
 672.000
 2.976
 0.451
 6.451
 126.833
 5.298
 0.00060
 0.000855
 0.777
 1029.139
 1429.811

7.000












8.000












9.000












10.000












10.097






 0.0001





In the direct step method, the accuracy of the computation depends on the size of depth interval (Col. [1] of Table 76).
The smaller the interval, the more accurate the computation will be. This is because the average friction slope for a subreach (Col. [9])
is the arithmetic average of the friction slopes at the grid points (linear assumption). In practice, a computation
using a computer would have a much finer interval than that shown in Table 76.
The calculation of the S_{2} watersurface profile is shown in Table 77.
Table 77 Calculation of S_{2} watersurface profile (S_{o} = 0.03). 
[1]
 [2]
 [3]
 [4]
 [5]
 [6]
 [7]
 [8]
 [9]
 [10]
 [11]
 [12]

y
 A
 V
 V^{ 2}/(2g )
 H
 P
 R
 S_{f}
 S_{f} _{ave}
 ΔH
 ΔL
 ∑ ΔL

 (1)
 (2)
 (3)
 (4)
 (5)
 (6)
 (7)
 (8)
 (9)
 Box


3.364
 359.033
 5.570
 1.581
 4.945
 115.044
 3.1208
 0.00425
 
 
 
 0

3.300
 351.780
 5.685
 1.647
 4.9475
 114.758
 3.0654
 0.0147
 0.01425
 0.002
 0.127
 0.127

3.200
 340.48
 5.874
 1.759
 4.9586
 114.311
 2.979
 0.0163
 0.0155
 0.0111
 0.765
 0.892

3.100
 329.22
 6.075
 1.881
 4.981
 113.864
 2.891
 0.0181
 0.0172
 0.0224
 1.750
 2.642

3.000












2.900












2.800












2.700












2.669






 0.03





Online calculations
The M_{2} watersurface profiles may be calculated online using ONLINE_WSPROFILES_22 (Fig. 720).
Using the number of computational intervals n = 100, and number of tabular output intervals m = 100,
the length of the M_{2} watersurface profile is calculated to be: ∑ ΔL = 147,691.5 m.
Fig. 720 Definition sketch for M_{2} watersurface profile.


The S_{2} watersurface profiles may be calculated online using ONLINE_WSPROFILES_25 (Fig. 721).
Using the number of computational intervals n = 100, and number of tabular output intervals m = 100,
the length of the S_{2} watersurface profile is calculated to be: ∑ ΔL = 152.02 m.
Fig. 721 Definition sketch for S_{2} watersurface profile.


QUESTIONS
What is steady gradually varied flow?
What is retarded flow?
What is accelerated flow?
What is subnormal flow?
What is supernormal flow?
How many profiles are possible in steady gradually varied flow?
What is the rule for the Type I family of watersurface profiles?
What is the rule for the Type III family of watersurface profiles?
Which watersurface profiles are completely horizontal?
What are the five limits to the water surface profiles?
What is the typical application of the M_{1} watersurface profile?
What is the typical application of the S_{1} watersurface profile?
What is the typical application of the S_{3} watersurface profile?
What is the main difference between the directstep and standardstep methods
of watersurface profile computations?
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 (Fig. 722).
 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. 722 A diversion dam.


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.
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.
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.
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.
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.
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.
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 (Fig. 723).
 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. 723 A diversion dam.


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.
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.
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.
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.
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.
REFERENCES
Chow, V. T. 1959. Openchannel Hydraulics. McGraw Hill, New York.
U.S. Army Corps of Engineers. (2014). HECRAS: Hydrologic Engineering Center River Analysis System.
USDA Soil Conservation Service. (1971). Classification system for varied flow in prismatic channels. Technical Release No. 47 (TR47), Washington, D.C.
http://openchannelhydraulics.sdsu.edu 

140830 08:00 
