Skip to contentUnited States Department of Transportation - Federal Highway AdministrationFHWA HomeFeedback

Hydraulics Engineering

 
<< Previous Contents Next >>

Introduction to Highway Hydraulics

Chapter 4 - Open-Channel Flow

4.1 Introduction

Open-channel flow is more complex than closed-conduit flow flowing full because the water surface is determined by the mechanics of motion. In addition, if the bottom boundary is movable (alluvial boundary) another complexity is introduced. When the channel is mobile, the resistance to flow is a function of the flow.

In this chapter the concepts and equations for the simplest flow condition (steady, uniform flow) will be described, as well as the bedform conditions that occur in an alluvial channel. Flow conditions and equations for solving problems of increasing flow complexity will be given. The one-dimensional method will be used in the descriptions of the equations.

4.2 Alluvial Channel Flow

4.2.1 Alluvial Channels

Alluvial channels are channels formed in material that has been and can be transported by the flow. They are commonly made up of bed material composed of sand-, gravel-, and cobble-sized material. These materials are important in drainage design because they affect resistance to flow and erosion. Concrete channels and culverts may have an alluvial boundary because of deposition of bed material in the invert.

4.2.2 Bedforms in Sand Channels

The predominant material in sand-bed streams ranges from coarse silt to sand. There may be finer or coarser material in the bed, but the dominant size will be sand (50 percent or more). In sand-bed streams, the bed material is easily eroded and continually being moved and shaped by the flow. Interaction between the flow of water sediment mixture and the sand bed creates different bed configurations which change the resistance to flow, velocity, water surface elevation, and sediment transport. Consequently, it is necessary to understand what bedforms will be present so that the resistance to flow can be estimated and flood stages, depth of flow, and water surface profiles can be computed in order to design drainage channels.

4.2.3 Flow Regime

Flow in alluvial channels is divided into two regimes separated by a transition zone. Forms of bed roughness in sand channels are shown in Figure 4.1. The flow regimes are:

  • Lower flow regime, where resistance to flow is large and sediment transport is small. The bedform is either ripples or dunes or some combination of the two. Water surface undulations are out of phase with the bed surface, and there is a relatively large separation zone downstream from the crest of each ripple or dune. The velocity of the downstream movement of the ripples or dunes depends on their height and the velocity of the grains moving up their backs.

  • The transition zone, where the bed configuration may range from that typical of the lower flow regime to that typical of the upper flow regime, depending mainly on antecedent conditions. If the antecedent bed configuration is dunes, the depth or slope can be increased to values more consistent with those of the upper flow regime without changing the bedform; or, conversely, if the antecedent bed is plane, depth and slope can be decreased to values more consistent with those of the lower flow regime without changing the bedform.

    Resistance to flow and sediment transport also have the same variability as the bed configuration in the transition. This phenomenon can be explained by the changes in resistance to flow and, consequently, the changes in depth and slope as the bedform changes.

  • Upper flow regime, in which resistance to flow is small and sediment transport is large. Usual bedforms are plane bed or antidunes. Water surface is in phase with the bed surface except when an antidune breaks, and normally the fluid does not separate from the boundary.

Eight sketches of bed forms in sand channels: one, typical ripple pattern; two, dunes with ripples superposed and weak boil; three, dunes with boil; four, washed out dunes; five, plane bed; six, antidune standing wave; seven, antidune breaking wave; eight, chutes and pools
Figure 4.1. Forms of bed roughness in sand channels.

Resistance to flow for the different bedforms and coarser bed material will be given later in this section. For information on sediment transport and additional information on bedforms the reader is referred to HDS 6 (Richardson et al. 2001).

At high flows, most sand-bed stream channels shift from a dune bed to a transition or a plane bed configuration. If the slope is steep antidune flow may occur. Resistance to flow is then decreased to one half to one third of that preceding the shift in bedform. The increase in velocity and corresponding decrease in depth may increase erosion and scour around bridge piers and abutments and increase the required size of riprap. If flow transitions to antidune flow significant wave action may occur.

4.2.4 Coarse-Bed Material

At low flow, coarse alluvial bed material may not move, but at moderate or large flows, the material may become mobile. With the movement of coarse-bed material, large bars may form which will be residual at low flow. These bars can re-direct flow and cause bank erosion, scour holes, and clog drainage channels. Resistance to flow for coarse-bed material is caused by the grain roughness of the material and the form loss caused by the bars. However, coarse-bed material in drainage channels can have a beneficial effect by decreasing erosion by armoring of the bed. Information on armoring is given in HEC-20 (Lagasse et al. 2001) and HDS 6 (Richardson et al. 2001). The determination of Manning's n for coarse-bed material is given later.

4.3 Steady Uniform Flow

In steady, uniform open-channel flow, there are no accelerations, streamlines are straight and parallel, and the pressure distribution is hydrostatic. The slope of the water surface Sw, the bed surface So, and the energy gradient Sf are equal (Figure 3.2). It is the simplest flow condition to analyze. Steady uniform flow is an idealized concept for open-channel flow and is difficult to obtain even in laboratory flumes. For many applications, the flow is essentially steady and changes in width, depth, or direction (resulting in nonuniform flow) are so small that the flow can be considered uniform. In other cases, the changes occur over such a long distance the flow is a gradually varied flow.

Depth in steady uniform flow is called the normal depth and the symbol for it is given the subscript o as in Yo. Velocity (V) is often given the same subscripts, i.e., Vo. Other variables of interest for steady uniform flow are (1) the discharge (Q), (2) the velocity distribution vy in the vertical, (3) the headloss hL through the reach, and (4) the shear stress, both local and at the bed τo. All these variables are interrelated. In the following section, engineering equations will be given along with example problems for obtaining values for these variables.

4.3.1 Manning's Equation for Mean Velocity and Discharge

Water flows in a sloping drainage channel because of the force of gravity. Flow is resisted by the friction between the water and wetted surface of the channel. The quantity of water flowing (Q), the depth of flow (y), and the velocity of flow (V) depend upon the channel shape, roughness (n), and slope (S0). Various equations have been devised to determine the velocity and discharge in open channels. A useful equation is the one that is named for Robert Manning, an Irish engineer. The Manning's equation for the velocity of flow in open channels is:

(4.1)

Equation 4.1: Velocity V equals (Units conversion factor K sub u divided by Manning's n) times Hydraulic radius R to the two thirds power times Energy Slope S to the power one-half   

where:

V =  Mean velocity, m/s (ft/s)
N =  Manning's coefficient of channel roughness
R =  Hydraulic radius, m (ft)
S =  Energy slope, m/m (ft/ft)
For steady uniform flow S = S0
Ku =  Units conversion factor equal to 1 (1.49 in English units)

Over many decades, typical Manning's n values have been compiled allowing an engineer to estimate the appropriate value by knowing the general nature of the channel boundaries. Most hydraulics textbooks and drainage design manuals provide tables of typical Manning's n values. An abbreviated list of such Manning's roughness coefficients is given in Appendix B, Table B.2. Several pictorial guides are also available showing the Manning's n value for different types of channels and floodplains (Barnes 1967 and Acrement and Schneider 1984). Special considerations exist for very steep channels (Jarrett 1985).

A numerical approach for n value estimates consists of the selection of a base roughness value for a straight, uniform, and smooth channel in the materials involved, and then adding values for the channel under consideration:

(4.2)

n = (n0 + n1 + n2 + n3 +n4) m5   

where:

no =  Base value for straight uniform channels
n1 =  Additive value due to cross-section irregularity
n2 =  Additive value due to variations of the channel
n2 =  Additive value due to obstructions
n4 =  Additive value due to vegetation
m5 =  Multiplication factor due to sinuosity

A discussion of this method and coefficients can be found in Cowan (1956) and Chow (1959). This method may be useful for natural channels, but has limited application for most roadway drainage design work.

For rock riprap channels the Manning's n is often described as some function of the rock size. Several equations are provided in HEC-15, including:

(4.3)

n = (Ku)(y1/6) / (2.25 + 5.23 log (y/D50))

where:

y =  Flow depth (average) in the channel, m (ft)
D50 =  Median riprap/gravel size, m (ft)
Ku =  Unit conversion constant, 0.319 (SI) and 2.262 (English)

This equation is valid for y /D50 ranging from 1.5 to 185 which should be typical of most conditions encountered in roadside and other small channels. For conditions outside this range see HEC-15.

Roughness characteristics on the floodplain are complicated by the presence of vegetation, natural and artificial irregularities, buildings, undefined direction of flow, varying slopes, and other complexities. Resistance factors reflecting these effects must be selected largely on the basis of past experience with similar conditions. In general, resistance to flow is large on the floodplains. In some instances, conditions are further complicated by deposition of sediment and development of dunes and bars which affect resistance to flow and direction of flow.

The presence of ice affects channel roughness and resistance to flow in various ways. When an ice cover occurs, the open channel is more nearly comparable to a closed conduit. There is an added shear stress developed between the flowing water and ice cover. This surface shear is much larger than the normal shear stresses developed at the air water interface. The ice water interface is not always smooth. In many instances, the underside of the ice is deformed so that it resembles ripples or dunes observed on the bed of sand-bed channels. This may cause overall resistance to flow in the channel to be further increased. With total or partial ice cover, the drag of ice retards flow, decreasing the average velocity and increasing the depth.

The hydraulic radius, R, is a shape factor that depends only upon the channel dimensions and the depth of flow. It is computed by the equation:

(4.4)

R = A / P    

where:

A =  Cross-sectional area of the flowing water perpendicular to the direction of flow
P =  Wetted perimeter or the length, of wetted contact between a stream of water and its containing channel, perpendicular to the direction of flow

The discharge (Q) is determined from the equation of continuity (see Chapter 3). The equation is:

Q = V A

where:

Q =  Discharge, m3/s (ft3/s)
A =  Cross-sectional area, m2 (ft2)
V =  Mean velocity, m/s (ft/s)

By combining Equations 4.1 and 4.4, Manning's equation can be used to compute discharge directly:

(4.5)

Equation 4.5: Discharge Q equals (Units conversion factor K sub u divided by Manning's n) times cross-sectional Area A times hydraulic radius R to the two thirds power times Energy Slope S to the power one-half    

In some computations, it is convenient to group the cross-sectional properties into a term called conveyance, K,

(4.6)

Equation 4.6: Conveyance K equals (Units conversion factor K sub u divided by Manning's n) times Area A times hydraulic radius R to the two thirds power   

then

Q = K S1/2

When a channel cross section is irregular in shape such as one with a relatively narrow deep main channel and wide shallow overbank area, the cross section must be subdivided and the flow computed separately for the main channel and overbank area. The same procedure is used when different parts of the cross section have different roughness coefficients. In computing the hydraulic radius of the subsections, the water depth common to the two adjacent subsections is not counted as wetted perimeter (see Example Problem 4.3).

Conveyance can be computed and a curve drawn for any channel cross section. The area and hydraulic radius are computed for various assumed depths and the corresponding value of K is computed from the equation. Values of conveyance are plotted against the depths of flow and a smooth curve connecting the plotted points is the conveyance curve. If the section was subdivided, the conveyance of each subsection (Ka, Kb,...Kn) is computed and the total conveyance of the channel is the sum of the conveyances of the subsections. Discharge can then be computed using Equation 4.7 Example Problem 4.3 illustrates a conveyance curve for a compound cross section. The concept of channel conveyance is useful when computing the distribution of overbank flood flows in the stream cross section and the distribution through the openings in a proposed stream crossing. The discharge through each opening can be assumed to have the same ratio to the total discharge as the ratio of conveyance of the opening bears to the total conveyance of the channel.

4.3.2 Aids in the Solution of Manning's Equation

Equations for the computation of Area, A, wetted perimeter, P, and hydraulic radius, R, in rectangular and trapezoidal channels (Figure 4.2) are:

(4.8)

A = By + Zy2    

(4.9)

Equation 4.9: Perimeter P equals Base width B plus 2 times depth y times the square root of (1 plus side slope Z squared)    

(4.10)

Equation 4.10: Hydraulic radius R equals ( B times y plus Z times y squared) divided by ( B plus 2 times y times the square root of (1 + Z squared))    

Variables are defined in Figure 4.2.

Typical trapezoidal open channel, water depth y, bottom width B, side slope one vertical to two horizontal
Figure 4.2. Trapezoidal channel.

Example Problem 4.1 (SI Units)

Given: Trapezoidal earth channel B = 2 m, sideslope 1V:2H, S = 0.003 m/m, normal depth y = 0.5 m, n = 0.02.

Sketch of trapezoidal open channel, water depth y equals zero point 5 meters, bottom width two meters, side slope one vertical to two horizontal

Find: Velocity (V) and discharge (Q)

Solution:

Velocity V equals (Units conversion factor K sub u divided by Manning's n) times Hydraulic radius R to the two thirds power times Energy Slope S to the power one-half      SI units; Ku = 1

R equals (B times y plus Z times y squared) divided by ( B plus 2 times y times the square root of (1 + Z squared))      R equals ( 2 times 0.5 plus 2 times 0.5 squared) divided by ( 2 plus 2 times 0.5 times the square root of (1 + 2 squared)) equals 1.5 divided by 4.24 equals 0.35 meters

V = (1/0.02) (0.352/3) (0.0031/2) = 1.36 m/s

Q = V A =1.36 (2(0.5) + 2(0.5)2) = 2.0 m3/s

Example Problem 4.1 (English Units)

Given: Trapezoidal earth channel B = 6.5 ft, sideslope 1V:2H, S = 0.003 ft/ft, normal depth y = 1.6 ft, n = 0.02.

Sketch of trapezoidal open channel, water depth y equals one 1.6 feet, bottom width 6.5 feet, side slope one vertical to two horizontal

Find: Velocity (V) and discharge (Q)

Solution:

Velocity V equals (Units conversion factor K sub u divided by Manning's n) times Hydraulic radius R to the two thirds power times Energy Slope S to the power one-half      English units; Ku = 1.49

R equals (B times y plus Z times y squared) divided by ( B plus 2 times y times the square root of (1 + Z squared))      R equals ( 6.5 times 1.6 plus 2 times 1.6 squared) divided by ( 6.5 plus 2 times 1.6 times the square root of (1 + 2 squared)) equals 15.52 divided by 13.66 equals 1.14 feet

V = (1.49/0.02) (1.142/3) (0.0031/2) = 4.45 ft/s

Q = A V = (6.5(1.6) + 2 (1.6)2) 4.45 = 69.06 ft3/s

Example Problem 4.2 (SI Units)

Given: A concrete trapezoidal channel B = 1.5 m, sideslopes = 1V:2H, n = 0.013, slope = 0.002, Q = 3 m3/s

Sketch of trapezoidal open channel carrying a discharge of 3 cubic meters per second in a bottom width of 1.5 meters, and a side slope one vertical to two horizontal

Find: Depth (y) and velocity (v)

Solution:

1. Use Manning's equation

Discharge Q equals (Units conversion factor K sub u divided by Manning's n) times cross-sectional Area A times hydraulic radius R to the two thirds power times Energy Slope S to the power one-half

where Ku = 1 and relationships for A and R are

A = By + Zy2 = 1.5 y + 2y2

R equals ( B times y plus Z times y squared) divided by ( B plus 2 times y times the square root of (1 + Z squared))      R equals ( 1.5 times y plus 2 times y squared) divided by ( 1.5 plus 4.47 times y

substitute A and R into Manning's equation

Manning's equation wiht A and R substituted into i.

2. Trial and error solution for y to find a depth where Q = 3 m3/s

Try y = 0.70;     Q equals (1 divided by 0.013 ) times (1.05 plus 0.98) times the quotient of (1.05 plus 0.98) and ( 4.629 ) to the two thirds power times 0.002 to the power one-half equals 4.03 = 4.03 m3/s

Since 4.03 > 3.0, the assumed value for y is too large. Try a smaller value such as 0.60.

Try y = 0.6;    Q equals (1 divided by 0.013 ) times (0.9 plus 0.72) times the quotient of (0.9 plus 0.72) and ( 4.18 ) to the two thirds power times 0.002 to the power one-half equals 2.96 = 2.96 m3/s

since 2.96 = 3.0, the assumed value for y is okay.

Therefore, use y = 0.60 m and use continuity to find the velocity (V = Q/A)

V = 3.0 / (0.9 + 0.72) = 1.85 m/s

Example Problem 4.2 (English Units)

Given: A concrete trapezoidal channel B = 5.0 ft, sideslopes = 1V:2H, n = 0.013, slope = 0.002, Q = 105 ft3/s

Sketch of trapezoidal open channel carrying a discharge of 105 cubic feet per second in a bottom width of 5 feet, and a side slope one vertical to two horizontal

Find: Depth (y) and velocity (v)

Solution:

1. Use Manning's equation

Q equals (K sub u divided by n) times a times R to the two thirds power times S to the power one-half

where Ku = 1.49 and relationships for A and R are

A = B y + Z y2 = 5y + 2y2

R equals the quotient of ( B times y plus Z times y squared) and ( B plus 2 times y times the square root of (1 + Z squared)) Equals (5 times y plus 2 times y squared) divided by ( 5 plus 4.47 times y)

substitute A and R into Manning's equation

Q equals (1.49 divided by 0.013) times (5 times y plus 2 times y squared) times the quotient of (5 times y plus 2 times y squared) and (5 plus 4.47 times y) to the power two thirds times 0.002 to the power one-half

2. Trial and error solution for y to find a depth where Q = 105 ft3/s

Try y = 2.5

Q equals (1.49 divided by 0.013) times (5 times 2.5 plus 2 times 2.5 squared) times the quotient of (5 times 2.5 plus 2 times 2.5 squared) and ( 5 plus 4.47 times 2.5) to the power two thirds times 0.002 to the power one-half equals 171.30 cubic feet per second

Try y = 2.0

Q equals (1.49 divided by 0.013) times (5 times 2 plus 2 times 2 squared) times the quotient of (5 times 2 plus 2 times 2 squared) and ( 5 plus 4.47 times 2) to the power two thirds times 0.002 to the power one-half equals 109.40 cubic feet per second

109 is close to 105, try 1.96 for y-

Q equals (1.49 divided by 0.013) times (5 times 1.96 plus 2 times 1.96 squared) times the quotient of (5 times 1.96 plus 2 times 1.96 squared) divided by ( 5 plus 4.47 times 1.96) to the power two thirds times 0.002 to the power one-half equals 105.12 cubic feet per second

Since 1.96 ft gave a Q of 105.11, the y of 1.96 is good, therefore use y = 1.96 and use continuity to find the velocity (V = Q/A).

V equals 105 divided by (5 times 1.96 plus 2 times 1.96 squared) equals 6.01 feet per second

Example Problem 4.3 (SI Units)

Given: A compound channel as illustrated, with an n value of 0.03, a longitudinal slope of 0.002 m/m and sideslopes of 1V:1H.

Sketch of a compound trapezoidal channel with side slopes of 1 vertical to 1 horizontal and total bottom width of 50 meters - section B the first 30 meters 1 m deep and section A next 20 meters 2 meters deep

Find: Discharge (Q)

Solution:

  1. Subsection A
    • A = (2)(20) + ½ (2)(2) = 42 m2

    • WP = 20 + (2)(2)1/2 + 1 = 23.83 m

    • R = 42 / 23.83 = 1.76 m

    • V = (1/0.03) (1.76)2/3 (0.002)1/2 = 2.17 m/s

    • Q = 42(2.17) = 91.14 m3/s

  2. Subsection B
    • A = 30(1) + (½)(1)(1) = 30.5 m2

    • WP = 30 + 1(2)1/2 = 32.41 m

    • R = (30.5) / (31.41) = 0.97 m

    • V = (1/0.03) (0.97)2/3(0.002)1/2 =1.46 m/s

    • Q = 30.5 (1.46) = 44.53 m3/s

  3. For the entire channel
    • A = 42 + 30.5 = 72.5 m2

    • Q = 91.14 + 44.53 = 135.67 m3/s, say 136 m3/s

  4. If the channel had been considered as a whole without subdividing, the following results would have been obtained.
    • A = 42 + 30.5 = 72.5 m2

    • WP = 23.82 + 31.41 = 55.23 m

    • R = 72.5 / 55.23 = 1.31 m

    • V = (1/0.03) (1.31)2/3(0.002)1/2 = 1.78 m/s

    • Q = 72.5 (1.78) = 129.05 m3/s

    The discharge using the whole channel is considerably less than the discharge obtained by subdividing the channel.

  5. Plot the conveyance curve for this cross section and calculate the total discharge using Equation 4.7
  • With a subdivided cross section, the conveyance at a given flow depth should be calculated for each subsection, and then added together to get the total conveyance at that depth. This calculation and a plot of the conveyance are illustrated below:

    Nonlinear graph of y axis stage in meters verses x axis conveyance K in cubic meters per second, positive slope decreasing, followed by a table of results with headings of stage and conveyance through section A through section B and total conveyance

  • At a main channel flow depth of 2 m, the total conveyance is 3037.1. The total discharge is then (Equation 4.7)

Q = K S1/2 = 3037.1 (0.002)1/2 = 135.8 m3/s, say 136 m3/s

This result matches the correct discharge value for a 2 m flow depth as calculated above in item 3.

Example Problem 4.3 (English Units)

Given: A compound channel as illustrated, with an n value of 0.03, a longitudinal slope of 0.002 ft/ft and sideslopes of 1V:1H.

Sketch of a compound trapezoidal channel with side slopes of 1 vertical to 1 horizontal and total bottom width of 164.05 feet - the first 98.43 feet 3.28 feet deep and the next 65.62 feet 6.56 feet deep.

Find: Discharge (Q)

Solution:

  1. Subsection A
    • A = (65.62)(6.56) + (½)(6.56)(6.56) = 451.98 ft2

    • WP = 65.62 + (6.56)(2)1/2 + 3.28 = 78.18 ft

    • R = 451.98 / 78.18 = 5.78 ft

    • V = (1.49/0.03) (5.78)2/3(0.002)1/2 = 7.16 ft/s

    • Q = 451.98 (7.16) = 3235 ft3/s

  2. Subsection B
    • A = (98.43)(3.28) + (½)(3.28)(3.28) = 328.23 ft2

    • WP = 98.43 + 3.28 (2)1/2 = 103.07 ft

    • R = 328.23 / 103.07 = 3.18 ft

    • V = (1.49/0.03) (3.18)2/3(0.002)1/2 = 4.81 ft/s

    • Q = 328.23 (4.81) = 1577 ft3/s

  3. For the entire channel
    • A = 451.98 + 328.23 = 780.21 ft2

    • Q = 3235 + 1577 = 4812 ft3/s

  4. If the channel had been considered as a whole without subdividing, the following results would have been obtained.
    • A = 451.98 + 328.23 = 780.21 ft2

    • WP = 78.18 + 103.07 = 181.25 ft

    • R = 780.21 / 181.25 = 4.30 ft

    • V = (1.49/0.03) (4.30)2/3(0.002)1/2 = 5.88 ft/s

    • Q = 780.21 (5.88) = 4588 ft3/s

    The discharge using the whole channel is considerably less than the discharge obtained by subdividing the channel.

  5. Plot the conveyance curve for this cross section and calculate the total discharge using Equation 4.7.
  • With a subdivided cross section, the conveyance at a given flow depth should be calculated for each subsection, and then added together to get the total conveyance at that depth. This calculation and a plot of the conveyance are illustrated below:
  • At a main channel flow depth of 6.56 ft, the total conveyance is 107,655. The total discharge is then (Equation 4.7)

Q = K S1/2 = 107655 (0.002)1/2 = 4814 ft3/s

This result matches closely to the 4,812 ft3/s calculated in item 3. The small difference is the result of rounding in the conveyance calculations.

4.3.3 Velocity Distribution

There are times in the design of highway drainage facilities that knowledge of the velocity distribution in the vertical is needed (e.g., the design of riprap for scour and erosion control). As a result of boundary roughness, the velocity varies vertically from some minimum value along the bed to a maximum value near the water surface (Figure 4.3). In this section, the Einstein form of the Karman-Prandtl velocity distribution in the vertical and mean velocity equations will be given for steady uniform flow (Einstein 1950). For their derivation the reader is referred to any standard fluid mechanics text or HDS 6 (Richardson et al. 2001).

Nonlinear graph of y axis stage in meters verses x axis conveyance K in cubic feet per second, positive slope decreasing, followed by a table of results with headings of stage and conveyance through section A through section B nd total conveyance

Sketch of vertical velocity profile showing velocity approximately zero at zero depth increasing parabolically with maximum at greatest distance from the stream bed of roughness k sub s
Figure 4.3. Schematic of vertical velocity profile.

The equations for velocity distribution (v) and mean velocity (V) can be written in the following dimensionless forms:

(4.11)

Equation 4.11: Local mean velocity, lower case v, divided by shear velocity, upper case v sub * equals 5.75 log ( 30.2 times X times y divided by k sub s) with variables described in table below     

(4.12)

Equation 4.12: Depth averaged velocity upper case V divided by shear velocity, upper case V sub * equals 5.75 log ( 12.27 times X times y sub naught divided by k sub s) with variables described in table below    

X =  Coefficient given in Figure 4.4
ks =  Measure of the roughness height, ks varies from the D84 size for pure sand bed channels, to 3.5 times D84 for graded coarse-bed streams; for practical application use 3.5 times D84, m (ft)
y =  Depth to specified location, m (ft)
v =  Local mean velocity at depth y, m/s (ft/s)
yo =  Depth of flow, m (ft)
V =  Depth-averaged velocity, m/s (ft/s)
V* =  Shear velocity, (τo/ ρ)1/2, m/s (ft/s)
τo =  Shear stress at the boundary, N/m2 (lb/ft2)
δ' =  Thickness of the viscous sublayer, 11.6 μ/V*, m (ft)
μ =  Dynamic viscosity of water, Table A.6, N-s/m2 (lb-s/ft2)
ρ =  Density of water, kg/m3 (lb-s2/ft4)

4.3.4 Shear Stress

Shear stress is the force water exerts on the bed and bank of a channel as it flows over them. The following equations can be used to determine the shear stress on the boundary of the channel that results from the force of flowing water. For the derivations of these equations refer to fluid mechanics texts or HDS 6 (Richardson et al. 2001). The first equation (Equation 4.13) is an exact equation, giving the average shear stress over the wetted perimeter. The next equations are semi-empirical and result from solving the Karman-Prandtl velocity equation.

A graph of Einstein's factor X on the y axis verses the ratio of measure of the roughness height k sub s to the thickness of the viscous sublayer on the x axis. Factor X ranges from approximately 0.6 at 0.2 x axis to a maximum of approximately 1.6 at 1.0 x axis then drops to 1.0 for values greater than about 9 on the x axis
Figure 4.4. Einstein's multiplication factor X in the logarithmic velocity equations (Einstein 1950).

(4.13)

τo = γ R So    

where:

τo =  Average shear stress on the wetted perimeter, N/m2 (lb/ft2)
γ =  Unit weight of water, N/m3 (lb/ft3)
R =  Hydraulic radius, m (ft)
So =  Slope of the channel, m/m (ft/ft).  In gradually varied flow the slope is of the energy grade line,  So = Sf
(4.14)

Equation 4.14: Shear stress tau sub naught equals density rho times the square of ( v sub 1 minus v sub 2) divided by the square of [5.75 times log ( y sub 1 divided by y sub 2)] with variables described in the text     

(4.15)

Equation 4.15: Shear stress tau sub naught equals density rho times the square of ( v sub 1 minus v sub 2) divided by the square of [5.75 times log ( y sub 1 divided by y sub 2)] with variables described in the text     

where τo is the shear stress at a point in the flow, N/m2 (lb/ft2), v1 and v2 are point velocities in the vertical at y1 and y2, respectively; V is the mean velocity in the vertical with a depth of yo; and the other terms have been defined previously.

Example Problem 4.4 (SI Units)

Determine the shear stress along the wetted perimeter of a trapezoidal channel. Also determine the shear stress on a particle along the bottom of the same channel.

Given: Trapezoidal channel as illustrated with So = 0.005, γ = 9800 N/m3, V = 1.8 m/s, D84 = 0.15m

Sketch of a rough boundary trapezoidal channel with bottom width of 5 meters, depth of 1.25 meters, and side slope ratio of 1 vertical to 3 horizontal.

Find:

(1) τo along wetted perimeter
(2) τo along bed

Solution:

(1) the shear stress along the wetted perimeter is given by

τo = γ R S0

where:

  • R = A / P

  • A = 5 (1.25) + 3 (1.25)2 = 10.94 m2

  • P = 5 + (2)(1.25)(10)1/2 = 12.91 m

  • R = 10.94 / 12.91 = 0.85 m

  • τo = 9800 (0.85) (0.005) = 41.7 N/m2

(2) shear stress along the bottom at a point is

Shear stress tau sub naught equals density rho times the square of ( mean velocity V) divided by the square of {5.75 times log [12.27 times( y sub naught divided by k sub s ) ] } with variables described in the text

tau sub naught equals 1000 times 1.8 squared divided by the square of {5.75 log [12.27 times ( 1.25 divided by (3.5 times 0.15)) ] } equals 45.6 Newton per square meter

Example Problem 4.4 (English Units)

Determine the shear stress along the wetted perimeter of a trapezoidal channel. Also determine the shear stress on a particle along the bottom of the same channel.

Given:Trapezoidal channel as illustrated with So = 0.005, γ = 62.4 lb/ft3, V = 5.9 ft/s, D84 = 0.49 ft

Sketch of a rough boundary trapezoidal channel with bottom width of 16.4 feet, depth of 4.10 feet, and side slope ratio of 1 vertical to 3 horizontal.

Find:

(1) τo along wetted perimeter
(2) τo along bed

Solution:

(1) the shear stress along the wetted perimeter is given by

τo = γ R S0

where:

  • R = A / P

  • A = 16.4 (4.10) + 3 (4.10)2 = 117.67 ft2

  • P = 16.4 +(2)(4.10)(10)1/2 = 42.33 ft

  • R = 117.67/42.33 = 2.78 ft

  • τo = 62.4 (2.78) (0.005) = 0.87 lb/ft2

(2) the shear stress along the bottom at a point is

Shear stress tau sub naught equals density rho times the square of ( mean velocity V) divided by the square of {5.75 times log [12.27 times( y sub naught divided by k sub s ) ] } with variables described in the text

tau sub naught equals 1.94 times 5.9 squared divided by the square of {5.75 log [12.27 times ( 4.10 divided by ( 3.5 times 0.49)) ] } equals 0.95 pounds per square foot

4.3.5 Froude Number and Relationship to Subcritical, Critical, and Supercritical Flow

An extremely important dimensionless parameter in open-channel flow is the Froude Number, defined as the ratio of the inertia forces to the gravity forces. It is normally expressed as:

(4.16)

Equation 4.16: Froude number Fr equals V divided by the square root of ( g times y) with variables explained in the text    

where:

Fr =  Froude Number
V =  Velocity of flow, m/s (ft/s)
g =  Acceleration of gravity, m/s2 (ft/s2)
y =  Depth of flow, m (ft)

V and y can be the mean velocity and depth in a channel or the velocity and depth in the vertical. If the former are used, then the Froude Number is for the average flow conditions in the channel. If the latter are used, then it is the Froude Number for that vertical at a specific location in the cross section. The Froude Number uniquely describes the flow pattern in open-channel flow. For example, in alluvial channel flow with sand-bed material, ripples and dunes only form when the Froude Number is less than 1.0 (subcritical flow); whereas, antidunes only form when the Froude Number is greater than 1.0. Plane bed formation is independent of the Froude Number. The Froude Number is the scaling parameter that is used in modeling open-channel flow structures in the laboratory.

When the Froude Number is 1.0, the flow is critical; values of the Froude Number greater than 1.0 indicate supercritical or rapid flow and smaller than 1.0 indicate subcritical or tranquil flow. Velocity and depth at critical flow are called the critical velocity and critical depth. Channel slope which produces critical depth and critical velocity is the critical slope. The change from supercritical to subcritical flow is often abrupt (particularly if the Froude Number is larger than 2.0) resulting in a phenomenon known as the hydraulic jump.

Critical depth and velocity for a particular discharge are only dependent on channel size and shape and are independent of channel slope and roughness. Critical slope depends upon the channel roughness, channel geometry, and discharge. For a given critical depth and velocity, the critical slope for a particular roughness can be computed by Manning's equation.

Supercritical flow is difficult to control because abrupt changes in alignment or in cross section produce waves which travel downstream, alternating from side to side, sometimes causing the water to overtop the channel sides. Changes in channel shape, slope, alignment, or roughness cannot be reflected upstream. In supercritical flow, the control of the flow is located upstream. Supercritical flow is common in steep flumes, channels, and mountain streams.

Subcritical flow is relatively easy to control for flows with Froude Numbers less than 0.8. Changes in channel shape, slope, alignment, and roughness affect the flow for small distances upstream. The control in subcritical flow is located downstream. Subcritical flow is common in channels, flumes and streams located in the plains regions and valleys where slopes are relatively flat.

Critical depth is important in hydraulic analysis because it is always a hydraulic control. The flow must pass through critical depth in going from subcritical flow to supercritical or going from supercritical flow to subcritical. Although, in the latter case a hydraulic jump usually occurs. Typical locations of critical depth are:

  1. At abrupt changes in slope when a flat (subcritical) slope is sharply increased to a steep (supercritical) slope.
  2. At channel constrictions such as a culvert entrance, flume transitions, etc., under some conditions.
  3. At the unsubmerged outlet of a culvert or flume on a subcritical slope, discharging into a wide channel, steep slope channel (supercritical), or with a free fall at the outlet.
  4. At the crest of an overflow dam, weir, or embankment.
  5. At bridge constrictions where the bridge chokes the flow.

Location and magnitude of critical depth and the determination of critical slope for a cross section of a given shape, size, and roughness are important in channel design and analysis. The equations for determining the critical depth are provided in the discussion of specific discharge and specific energy in steady rapidly varied flow (Section 4.6).

4.4 Unsteady Flow

Unsteady flows of interest to the highway drainage engineer or designer are:

  1. Waves resulting from disturbances of the water surface by wind and boats.
  2. Waves resulting from the surface instability that exists for flows with Froude Numbers close to 1.0.
  3. Waves resulting from flow disturbance due to change in direction of flow with Froude Numbers greater than about 2.0.
  4. Surges or bores resulting from sudden increase or decrease in the flow by opening or closing of gates or the movement of tides on coastal streams.
  5. Standing waves and antidunes that occur in alluvial channel flow.
  6. Flood waves resulting from the progressive movement downstream of stream runoff or gradual release from reservoirs.

Waves are an important consideration in bridge hydraulics when designing slope protection of embankments and dikes, and channel improvements. In the following paragraphs, only the basic one dimensional analysis of waves and surges is presented. Other aspects of waves are presented in other sections.

4.4.1 Gravity Waves

For shallow water waves (long waves - Figure 4.5) where the normal depth (yo) is small in comparison to the wave length, the basic equation for the celerity (velocity of the wave relative to the velocity of the flow) is given by:

(4.17)

Equation 4.17: Celerity c equals the square root of the product g times y sub naught    

Note that the celerity of a shallow water wave of small amplitude is the same as the denominator of the Froude Number.

(4.18)

Equation 4.18: Froude number Fr equals V divided by the square root of (g times y sub naught)    

As explained in the discussion of the Froude Number (Section 4.3.5), when Fr < 1 (subcritical or tranquil), a small amplitude wave moves upstream. When Fr > 1 (supercritical or rapid flow), a small amplitude wave moves downstream and when Fr = 1 (critical flow), a small amplitude wave is stationary. The fact that waves or surges cannot move upstream when the Froude Number is equal to or greater than 1.0 is important to remember when determining when the stage discharge relation at a cross section can be affected by downstream conditions.

Sketch defining wave amplitude equally above and below the normal depth, wavelength as from wave peak to wave peak and indicating the direction of wave Celerity c right to left.
Figure 4.5. Definition sketch for small amplitude waves.

4.4.2 Surges

A surge is a rapid increase in the depth of flow (Figure 4.6). A surge may result from the sudden release of water from a dam or an incoming tide. The lifting of a gate in a channel not only causes a positive surge to move downstream, it also causes a negative surge to move upstream (Figure 4.6). As it moves upstream, a negative surge quickly flattens out. See HDS 6 for more detail and the basic surge equation (Richardson et al. 2001).

4.5 Steady Nonuniform Flow

Steady nonuniform flow occurs when the quantity of water (discharge) remains constant, but the depth of flow, velocity, or cross section changes from section to section. From the continuity equation, the relation of all cross sections will be:

(4.19)

Q = A1 V1 = A2 V2 = An Vn    

Velocity in steady nonuniform flow can be computed using Manning's equation if the change in velocity from section to section is small so that the effect of acceleration is small.

The hydraulic design engineer needs a knowledge of nonuniform flow in order to determine the behavior of the flowing water when changes in channel resistance, size, cross section, shape, or slope occur. Typical examples might include determining water surface elevation changes in a channel of constant slope that goes through a short transition from a concrete trapezoidal cross section (with a low Manning's n) to a larger grass-lined trapezoidal cross section (a high Manning's n), or a stream with constant slope and Manning's n that is a long distance upstream of a culvert that constricts the flow.

One sketch of control volume from a shallow upstream water depth y sub 1 at the toe of the wave front through the rising wave to the higher downstream water surface y sub 2, stream velocity equals zero. A second sketch of a submerged gate opening with negative surge in the upstream direction on the upstream side of the gate, and a positive surge downstream of the gate opening in the downstream direction.
Figure 4.6. Sketch of positive and negative surges.

These two situations define two basic cases of steady nonuniform flow. The first case is for relatively short distances (a few meters (feet) to several hundred meters (feet)) where accelerations are more important than friction. This case is called STEADY RAPIDLY VARIED FLOW. The effect of friction, if it is important, is taken into account by subdividing the distance into shorter segments and using Manning's equation along these shorter segments. The second case is for long distances (hundreds to thousands of meters (feet)), where friction losses are more important than accelerations. This case is called GRADUALLY VARIED FLOW. Method of analysis and equations for these two cases will be given in the next two sections.

4.6 Steady Rapidly Varied Flow

4.6.1 Introduction

Steady flow through relatively short transitions where the flow is uniform before and after the transition can be analyzed using the energy equation. Energy loss due to friction may be neglected, at least as a first approximation. Refinement of the analysis can be made in a second step by including friction loss. For example, the water surface elevation through a transition is determined using the energy equation and then modified by determining the friction loss effects on velocity and depth in short subsections through the transition. However, energy losses resulting from flow separation cannot be neglected, and transitions where separation may occur need special treatment which may include model studies. Contracting flows (converging streamlines) are less susceptible to separation than expanding flows. Also, any time a transition changes velocity and depth such that the Froude Number approaches unity, problems such as waves, blockage, or choking of the flow may occur. If the approaching flow is supercritical, a hydraulic jump may result. Transitions for supercritical flow are discussed in the next section.

Transitions are used to contract or expand a channel width (Figure 4.7a), to increase or decrease bottom elevation (Figure 4.7b), or to change both the width and bottom elevation. The analysis or design of transitions is aided by the use of the depth of flow and velocity head terms in the energy equation (see Chapter 3). The sum of the two terms is called the specific energy or specific head, H, and defined as:

(4.20)

Equation 4.20: Specific head, H equals V squared divided by ( 2 times g ) plus y equals unit discharge q squared divided by ( 2 times g times y squared) plus y  Equation 20: Specific head, H equals V squared divided by ( 2 times g ) plus y equals unit discharge q squared divided by ( 2 times g times y squared) plus y    

where:

H =  Specific energy, m (ft)
V =  Velocity, m/s (ft/s)
q =  Unit discharge, defined as the discharge per unit width m3/s/m (ft3/s/ft) in a rectangular channel
g =  Acceleration of gravity, 9.81 m/s2 (32.2 ft/s2)
y =  Depth of flow, m (ft)

The specific energy, H, is the height of the total energy above the channel bed.

The relationship between the three terms in the specific energy equation, q, y, and H, are evaluated by considering q constant and determining the relationship between H and y (specific energy diagram) or considering H constant and determining the relationship between q and y (specific discharge diagram). These diagrams for a given discharge or energy are then used in the design or analysis of transitions or flow through bridges. They are explained in the next two sections.

Sketch A showing a plan view stream width contraction, width W sub 1 upstream and reduced width W sub 2 downstream. Profile shows upstream depth y sub 1 greater than contracted section depth y sub 2. Velocity head upstream less than velocity head in contracted section with total energy line horizontal. Sketch B shows a bed rise in plan view, width w sub 1 equals width W sub 2. Profile shows depth y sub 1 greater than bed rise delta Z plus depth y sub 2. Velocity head upstream less than velocity head in bed rise section with total energy line horizontal.
Figure 4.7. Transitions in open-channel flow (subcritical flow).

4.6.2 Specific Energy Diagram and Evaluation of Critical Depth

For a given q, Equation 4.20 can be solved for various values of H and y. When y is plotted as a function of H, Figure 4.8 is obtained. There are two possible depths called alternate depths for any H larger than a specific minimum. Thus, for specific energy larger than the minimum, the flow may have a large depth with small velocity or small depth with large velocity. Flow for a given unit discharge (q) cannot occur with specific energy less than the minimum. Single depth of flow at the minimum specific energy is called the critical depth, yc, and the corresponding velocity, the critical velocity, Vc = q/yc. The relation for yc and Vc for a given q (for a rectangular channel) is:

(4.21)

Equation 4.21: Critical depth y sub c equals (unit discharge q squared divided by g) to the power one-third equals 2 times the critical velocity V sub c squared divided by ( two times g )     

Note that for critical flow:

(4.22)

Equation 4.22: V sub c divided by the square root of the product g times y sub c equals 1 equals the Froude number    

(4.23)

Equation 4.23: H sub min equals V squared divided by ( 2 times g) plus y sub c equals (3 divided by two) times y sub c    

Graph of flow depth y on the y axis verses specific energy H on the x axis. Specific energy curves for various unit discharges are shown with minimum specific head occurring at critical depth and flows with specific head above the minimum showing two possible depth values
Figure 4.8. Specific head diagram.

Thus, flow at minimum specific energy has a Froude Number equal to 1. Flows with velocities larger than critical (Fr > 1) are called rapid or supercritical and flow with velocities smaller than critical (Fr < 1) are called tranquil or subcritical. These flow conditions are illustrated in Figure 4.9 where a rise in the bed causes a decrease in depth when the flow is tranquil and an increase in depth when the flow is rapid. Furthermore there is a maximum rise in the bed for a given H1 where the given rate of flow is physically possible. If the rise in the bed is increased beyond Δzmax for Hmin then the approaching flow depth y1 would have to increase (increasing H) or the flow would have to be decreased. Thus, for a given flow in a channel, a rise in the bed level can occur up to a Δzmax without causing backwater.

Two Sketches of changes in water surface resulting from an increase in bed elevation. Small increase in bed elevation changes sub critical approach flows with depth greater than critical depth to depths less than approach flows but still greater than critical depth; super critical approach flows with depths less than critical depth are still at depths less than critical depth. The two resulting depths are shown in a specific head diagram as having the same energy. Large increase in bed elevation changes: sub critical flows with depth greater than critical depth to a depth equal to critical depth; super critical flows with depth less than critical depth to a depth equal to critical depth. The two resulting depths are shown in a specific head diagram as having the same depth - critical depth and hence the same energy.
Figure 4.9. Changes in water surface resulting from an increase in bed elevation.

Distinguishing between the types of flow and how the water surface reacts with changes in cross section is important in channel design; thus, the location of critical depth and the determination of critical slope for a cross section of given shape, size, and roughness becomes necessary. Equations for direct solution of the critical depth are available for several prismatic shapes (Brater and King 1976); however, some of these equations were not derived for use in the metric system.

For any channel section, regular or irregular, critical depth may be found by a trial-and-error solution of the following equation:

(4.24)

Equation 4.24: A sub c cubed divided by T sub c equals Q squared divided by g    

where Ac and Tc are the area and topwidth at critical flow. Nomographs are available to solve this equation, and are particularly useful for circular sections (Figures 4.10a and b). An expression for the critical velocity (Vc) of any cross section at critical flow conditions is:

(4.25)

Equation 4.25: V sub c equals square root of (g times y sub c)    

(4.26)

where: yc = Ac/ Tc    

Uniform flow within about 10 percent of the critical depth is unstable and should be avoided in design. The reason for unstable flow can be seen by referring to the specific head diagram (Figure 4.8). As the flow approaches the critical depth from either limb of the curve, a very small change in energy is required for the depth to abruptly change to the alternate depth on the opposite limb of the specific head curve. If the unstable flow region cannot be avoided in design, the least favorable type of flow should be assumed for the design.

Chart 4A Three graphs of critical depth in meters on y axis verses discharge in cubic meters per second on x axis. Range of round pipe sizes covered is from 300 mm to 4500 mm and flows to approximately 110 cubic meters per second.
Figure 4.10a. Critical depth in a circular pipe, SI units (from HDS-5).

Chart 4B: Three graphs of critical depth in feet on y axis verses discharge in cubic feet per second on x axis. Range of round pipe sizes covered is from 1 foot to 15 foot and flows to approximately 4000 cubic feet per second.
Figure 4.10b. Critical depth in a circular pipe, English units (from HDS-5).

4.6.3 Specific Discharge Diagram

Equation 4.20 can be rearranged to determine q, the unit discharge, as a function of H, the specific energy, and y, the depth of flow.

(4.27)

Equation 4.27: Unit discharge q equals y times the square root of (2 times g times (H minus y))    

For a constant H, q can be solved as a function of y and the specific discharge diagram will result (Figure 4.11).

Sketch shows for constant specific energy H the maximum unit discharge q occurs at Froude number equals 1
Figure 4.11. Specific discharge diagram.

For any discharge smaller than a specific maximum q for the given H, two depths of flow are possible. The depth at maximum q for a given specific energy (H) is the critical depth (yc) and the velocity is the critical velocity (Vc).

(4.28)

Equation 4.28: y sub c equals two thirds H equals 2 times V sub c squared divided by (2 times g)    

and

(4.29)

Equation 4.29: V sub c divided by the square root of the product g times y sub c equals 1 equals the Froude number    

Example Problem 4.5 (SI Units)

Given: Determine the critical depth in a trapezoidal shaped swale with z = 1, given a discharge of 9.2 m3/s and a bottom width, B = 6 m. Also, determine the critical velocity.

Cross section sketch of a trapezoidal channel with a 6 meter bottom width carrying 9.2 cubic meters per second.

Find: Critical depth

Velocity at critical depth

Solution:

For a Q of 9.2 m3/s

Ac3 / Tc = Q2 / g

Ac3 / Tc = (9.2)2 / 9.81 = 8.63

For a trapezoidal channel area substituting A = y (B + Z y) and T = B + 2Z y gives

8.63 = [y (6 + y)]3 / (6 + 2y)

A trial and error solution yields y = 0.6 m.

Vc = (g yc)1/2 yc = A / T = [0.6 (6 + 0.6)] / (6 + 1.2) = 0.55 m

Vc = [9.81 (0.55)]1/2 = 2.3 m/s

Example Problem 4.5 (English Units)

Given: Determine the critical depth in a trapezoidal shaped swale with z = 1, given a discharge of 325 ft3/s and a bottom width, B = 20 ft. Also, determine the critical velocity.

Cross section sketch of a trapezoidal channel with a 20 foot bottom width carrying 325 cubic feet per second.

Find: Critical depth

Velocity at critical depth

Solution:

For a Q of 325 ft3/s

Ac3 / Tc = Q2 / g

Ac3 / Tc = 3252 / 32.2 = 3280.28

For a trapezoidal channel area substituting A = y (B + Z y) and T = B + 2Z y gives

3280 = [y (20 + y)]3 /(20 + 2y)

A = 1.95 (20 + (1) 1.95) = 42.80 ft2

T = 20 + 2 (1) (1.95) = 23.90 ft

A trial and error solution yields y = 1.95 ft.

Vc = (g yc)1/2 yc = A / T = 42.80 / 23.9 = 1.79 ft

Vc = [32.2 (1.79)]1/2 = 7.59 ft/s

(end example problem)

For maximum discharge at constant H, the Froude Number is 1.0, and the flow is critical. The relation between yc, Vc, H, and qmax for a constant H is:

(4.30)

Equation 4.30: y sub c equals two thirds H equals 3 times the square root of ( unit discharge q max squared divided by( 2 times g)) equals 2 times V sub c squared divided by (2 times g)    

Flow conditions for constant specific energy for a width contraction are illustrated in Figure 4.12 assuming no geometrical effects such as eccentricity, skew, piers, scour, and expansion. Contraction causes a decrease in flow depth when the flow is tranquil and an increase when the flow is rapid. The maximum possible contraction without causing backwater effects occurs when the Froude Number is 1.0, the discharge per unit of width q is a maximum, and yc is 2/3 H. A further decrease in width will cause backwater. That is, an increase in depth upstream will occur to produce a larger specific energy and increase yc in order to get the flow through the decreased width.

The flow in Figure 4.12 can go from point A to C and then either back to D or down to E depending on the downstream boundary conditions. An increase in slope of the bed downstream from C and no separation would allow the flow to follow the line A to C to E. Similarly, the flow can go from B to C and back to E or up to D depending on boundary conditions. Figure 4.12 is drawn with the side boundary forming a smooth streamline. If the contraction was due to bridge abutments, the upstream flow would follow a natural streamline to the point of maximum constriction, but then downstream, the flow would probably separate. Tranquil approach flow could follow line A to C but the downstream flow probably would not follow either line C to D or C to E, but would have an undulating hydraulic jump. There would be interaction of the flow in the separation zone and considerable energy would be lost. If the slope downstream of the abutments was the same as upstream, then the flow could not be sustained with this amount of energy loss. Backwater would occur, increasing the depth in the constriction and upstream, until the flow could go through the constriction and establish uniform flow downstream.

Sketch illustrating the use of the specific energy concept to evaluate the change in water surface elevation resulting from a change in width. Conditions are illustrated for both subcritical and supercritical flow which are described in the text.
Figure 4.12. Change in water surface elevation resulting from a change in width.

4.6.4. Hydraulic Jump

A hydraulic jump will occur when the flow velocity (V1) is rapid or supercritical and the slope is decreased to a slope for subcritical flow, or an obstruction such as an energy dissipator is placed in the flow. The supercritical depth is changed to a subcritical depth, called the sequent depth. Depending on the magnitude of the Froude Number, a considerable amount of energy is changed to heat. The larger the Froude Number, the more energy that is lost. The existence of a jump assumes adequate tailwater conditions exist. Many engineers/ designers assume that a jump will always occur when a change from a steep grade to a flat grade is encountered, such as near the outlet end of a culvert (i.e., a broken-back culvert). The jump will only occur with adequate downstream tailwater to maintain the sequent depth just below the culvert grade break. Without adequate tailwater, the jump will be swept downstream out of the culvert, causing a potentially large scour hole at the culvert outlet.

The relation between the supercritical depth and the sequent depth for a rectangular flat channel is:

(4.31)

Equation 4.31: y sub 2 divided by y sub 1 equals one-half times [ the square root of (1 plus 8 times Froude number Fr sub 1 squared) minus 1]    

The corresponding energy loss in a hydraulic jump is the difference between the two specific energies. It can be shown that this headloss is:

(4.32)

Equation 4.32: delta H equals headloss h sub L equals ( y sub 2 minus y sub 1) cubed divided by ( 4 times y sub 1 times y sub 2)    

Equation 4.32 has been experimentally verified along with the dependence of the jump length Lj and energy dissipation (headloss hL) on the Froude Number of the approaching flow (Fr1). The results of these experiments are given in Figure 4.13.

When the Froude Number for rapid flow is less than 2.0, an undulating jump with large surface waves is produced. The waves are propagated for a considerable distance downstream. In addition, when the Froude Number of the approaching flow is less than 3.0, the energy dissipation of the jump is not large and jets of high velocity flow can exist for some distance downstream of the jump. These waves and jets can cause erosion a considerable distance downstream of the jump. For larger values of the Froude Number, the rate of energy dissipation in the jump is very large and Figure 4.13 is recommended.

Example Problem 4.6 (SI Units)

Given: A hydraulic jump occurs in a 5-m wide rectangular channel at a flow depth of 0.5m. Determine the downstream water surface elevation needed to cause the jump. Also calculate the headloss due to the jump. Given Q = 20 m3/s.

Graph showing the ratio of downstream depth y sub 2 to upstream depth y sub 1 and ratio of length of hydraulic jump L sub j to downstream depth y sub 2 on primary y axis zero to 25. Secondary y axis on the right side with ratio of headloss to upstream depth y sub 1 zero to 50. Froude number on x axis ranges from zero to 20. The ratio of downstream depth to upstream depth is an increasing linear function from zero to a Froude number of approximately 17. The ratio of headloss to upstream depth shows a nonlinear increase with Froude number from zero to approximately 11.5. The ratio of hydraulic jump length to downstream depth is a relatively flat nonlinear curve with peak at a Froude number of approximately 8.
Figure 4.13. Hydraulic jump characteristics as a function of the upstream Froude Number.

Profile sketch of open channel flow carrying 20 cubic meters per second going from supercritical flow 0.5 meters deep to an unknown downstream subcritical flow depth.

Find:

(1) Determine the required downstream WSEL to initiate a jump
(2) Determine the headloss across the jump

Solution:

(1) Using the formula for a hydraulic jump (Equation 4.31), find y2.

From continuity V1 = Q / A1 = 20 / [5 (0.5)] = 8 m/s

Equation 4.31 solved for y sub 2: y sub 2 equals (y sub 1 divided by 2) times [ the square root of (1 plus 8 times Froude number Fr sub 1 squared) minus 1]    Fr = V1 / (g y1)1/2 = 8 / [9.81 (0.5)]1/2 = 3.6

y sub 2 equals 0.5 divided by 2 times [ the square root of (1 plus 8 times 3.6 squared) minus 1] equals 2.3 meters

(2) Find the headloss, hL, across the jump (Equation 4.32)

headloss h sub L equals ( y sub 2 minus y sub 1) cubed divided by ( 4 times y sub 1 times y sub 2)

hL = (2.3 - 0.5)3 / [4 (2.3) (0.5)] = 1.27 m

Example Problem 4.6 (English Units)

Given: A hydraulic jump occurs in a 16.4 ft wide rectangular channel at a flow depth of 1.64 ft. Determine the downstream water surface elevation needed to cause the jump. Also calculate the headloss due to the jump. Given Q = 700 ft3/s.

Profile sketch of open channel flow carrying 700 cubic feet per second going from supercritical flow 1.64 feet deep to an unknown downstream subcritical flow depth.

Find:

(1) Determine the required downstream WSEL to initiate a jump
(2) Determine the headloss across the jump

Solution:

(1) Using the formula for a hydraulic jump (Equation 4.31), find y2.

From continuity V1 = Q / A1 = 700 / [1.64 (16.4)] = 26.03 ft/s

Equation 4.31 solved for y sub 2: y sub 2 equals (y sub 1 divided by 2) times [ the square root of (1 plus 8 times Froude number fr sub 1 squared) minus 1]    Fr = V1 / (g y1)1/2 = 26.03 / [32.2 (1.64)]1/2 = 3.58

y sub 2 equals 1.64 divided by 2 times [ the square root of (1 plus 8 times 3.58 squared) minus 1] equals 7.52 feet

(2) Find the headloss, hL, across the jump (Equation 4.32)

headloss h sub L equals ( y sub 2 minus y sub 1) cubed divided by ( 4 times y sub 1 times y sub 2)

hL = (7.52 - 1.64)3 / [4 (7.52) (1.64)] = 4.12 ft

4.6.5 Subcritical Flow in Bends

When subcritical flow goes around a bend, the water surface is elevated on the outside of the bend and lowered on the inside of the bend (Figure 4.14). The approximate difference in elevation (ΔZ) between the water surface along the sides of the curved channel can be found by the following equations.

(4.33)

Equation 4.33: Change in water surface elevation delta Z equals Z sub o minus Z sub i equals V squared divided by (g times r sub c ) times ( r sub o minus r sub i)    

where:

Z =  Elevation of the water surface, m (ft)
V =  Average velocity in the channel, m/s (ft/s)
g =  Acceleration of gravity, 9.81 m/s2 (32.2 ft/s2)
rc =  Radius of curvature to the centerline of the channel, m (ft)
r0 =  Radius of curvature to the outside flow line around the bend, m (ft)
ri =  Radius of curvature to the inside flow line around the bend, m (ft)

Cross section sketch showing superelevation with a water surface lower on the inside of the bend and higher on the outside of the bend resulting in an overall elevation difference delta Z across the channel
Figure 4.14. Superelevation of water surface in a bend.

In channel design, superelevation is accounted for by adding ΔZ/2 to the normal depth to define the maximum water surface depth at the outside of the bend.

This equation gives values of ΔZ somewhat lower than will occur in natural channels because of the assumption of uniform velocity and uniform curvature, but the computed value will be generally less than 10 percent in error. Equation 4.34 (HDS 6, Richardson et al. 2001) is more accurate, but the difference in superelevation obtained by using the two equations is small, and in alluvial channels the resulting erosion of the concave bank and deposition on the convex bank leads to further error in computing superelevation.

(4.34)

Equation 4.34: Change in water surface elevation delta Z equals V sub max squared divided by ( 2 times g) times [ 2 minus ( r sub i divided by r sub c ) squared minus ( r sub c divided by r sub o ) squared]    

It is recommended that Equation 4.33 be used to compute superelevation in alluvial channels. For lined canals with strong curvature and large velocities, superelevation should be computed using Equation 4.34.

Other problems introduced by curved alignment of channel in subcritical flow include spiral flow, changes in velocity distribution, and increased friction losses within the curved channel as contrasted with the straight channel. For more information on flow-around bends see Rouse (1950), Chow (1959), or Richardson et al. (2001).

4.6.6 Supercritical Flow in Bends

Changes in alignment of supercritical flow are difficult to make. Water traveling at supercritical velocities around bends builds up waves which may "climb out" of the channel. Waves that are set up may continue downstream for a long distance. Also, sharp changes in alignment may set up a hydraulic jump with the flow overtopping the banks. Changes in alignment, whenever possible, should be made near the upper end of the section before the supercritical velocity has developed. If a change in alignment is necessary in a channel carrying supercritical flow, the channel should be rectangular in cross section, and preferably enclosed. On small chutes, experiments have shown that an angular variation (α) of rectangular flow boundaries (expansion) should not exceed that produced by the equation:

(4.35)

Equation 4.35: tan alpha equals 1 divided by ( 3 times Froude number)    

Changes in alignment of open channels can and should be designed to reduce the wave action, resulting from the change in direction in flow (see Richardson et al. 2001). Often designs involving supercritical flow should be model tested to develop the best design, or even a design that will work.

Example Problem 4.7 (SI Units)

Given: During high runoff, a 2.0 m deep mountain stream flows near bank full with a normal depth and velocity of 1.8m and 3.4 m/s, respectively. At a sharp bend ro = 12 m, rc = 10 m, ri = 8 m. Will flow overtop the bend?

Plan view schematic of situation in example problem 4.7 SI units depicting the given radii of curvative and velocity.

Find: ΔZ

Solution: Use the superelevation formula

Change in water surface elevation delta Z equals V squared divided by (g times r sub c ) times ( r sub o minus r sub i)    Change in water surface elevation delta Z equals 3.4 squared divided by (9.81 times 10 ) times ( 12 minus 8) equals 0.47 meters

The water surface raises approximately (0.47/2) m on the outside of the bend and lowers by that same amount on the inside of the bend. The maximum flow depth in the bend will be:

youtside = 1.8 + (0.47 / 2) = 2.04 m

which is greater than the channel depth (2.0 m) and overtopping will occur.

Example Problem 4.7 (English Units)

Given: During high runoff, a 6.56 ft deep mountain stream flows near bank full with a normal depth and velocity of 5.91 ft and 11.15 ft/s, respectively. At a sharp bend ro = 39.37 ft, rc = 32.81 ft, ri = 26.25 ft. Will flow overtop the bend?

Schematic of situation in example problem 4.7 English units depicting the given radii of curvature and velocity.

Find: ΔZ

Solution: Use the superelevation formula

Change in water surface elevation delta Z equals V squared divided by (g times r sub c ) times ( r sub zero minus r sub 1)equals 11.15 squared divided by (32.2 times 32.81 ) times ( 39.97 minus 26.25) equals 1.54 feet

The water surface raises approximately (1.54/2) ft on the outside of the bend and lowers by that same amount on the inside of the bend. The maximum flow depth in the bend will be

youtside = 5.9 + (1.54 / 2) = 6.67 ft

which is greater than the channel depth (6.56 ft) and overtopping will occur.

4.7 Gradually Varied Flow

4.7.1 Introduction

In this section, a second type of steady nonuniform flow is considered, gradually varied flow. In gradually varied flow, changes in depth and velocity take place slowly over large distances, resistance to flow dominates, and acceleration forces are neglected. Analysis of gradually varied flow involves: (1) the determination of the general characteristics of the water surface; and (2) the elevation of the water surface or depth of flow.

In gradually varied flow, the actual flow depth, y, is either larger or smaller than the normal depth, yo, and either larger or smaller than the critical depth, yc. The water surface profiles, which are often called backwater curves, depend on the magnitude of the actual depth of flow, y, in relation to the normal depth, yo, and the critical depth, yc. Normal depth, yo, is the depth of flow that would exist for steady uniform flow as determined using Manning's velocity equation, and the critical depth is the depth of flow when the Froude Number equals 1.0. Reasons for the depth being different than the normal depth are changes in slope of the bed, changes in cross section, obstruction to flow, and imbalances between gravitational forces accelerating the flow and shear forces retarding the flow. In working with gradually varied flow, the first step is to determine what type of water surface profile would exist. The second step is to perform the numerical computations. An excellent gradually flow reference is Chow (1959).

4.7.2 Types of Water Surface Profiles

The types of water surface profiles are obtained by analyzing the change of the various terms in the gradually varied flow equation:

(4.36)

Equation 4.36: slope of the water surface equals S sub naught times [ 1 minus (n divided by n sub naught ) squared times ( y sub naught divided by y ) to the power ten thirds divided by [1 minus ( y sub c divided by y) cubed]    

The slope of the water surface dy/dx depends on the slope of the bed So, the ratio of the normal depth yo to the actual depth y and the ratio of the critical depth yc to the actual depth y. The difference between flow resistance for steady uniform flow no to flow resistance for steady nonuniform flow n is small and the ratio is taken as 1.0. With n = no, there are 12 types of water surface profiles. The 12 types are subdivided into 5 classes which depend on the bed slope. These are illustrated in Figure 4.15 and summarized in Table 4.1.

Five classes based on bed slope with each class further classified by water surface profile at that slope: Mild slope with M1, Horizontal, M2, M3; Steep slope with S1, S2, S3 and horizontal; Critical slope with C1, C2 and horizontal; horizontal slope with H2 and H3; Adverse slope with A2 and A3
Figure 4.15. Classification of water surface profiles.

When y —> yc, the assumption that acceleration forces can be neglected no longer holds. Equation 4.36 indicates that dy/dx is perpendicular to the bed slope when y —> yc. For locations close to the cross section where flow is critical, a distance from 3 to 20 m, (10 to 65 ft) curvilinear flow analysis and experimentation must be used to determine the actual values of y. When analyzing long distances, 30 to 100 m or longer, (100 to 300 ft or longer) one can assume qualitatively that y reaches yc. In general, when the flow is rapid (Fr > 1), the flow cannot become tranquil or subcritical without a hydraulic jump occurring. In contrast, subcritical flow can become rapid, or supercritical, (cross the critical depth line). This is illustrated in Figure 4.16.

When there is a change in cross section or slope or an obstruction to the flow, the qualitative analysis of the flow profile depends on locating the control points, determining the type of water surface profile upstream and downstream of the control points, and then sketching these profiles. It must be remembered that when flow is supercritical (Fr > 1), the control depth is upstream and the water surface profile analysis proceeds in the downstream direction. When flow is subcritical (Fr < 1), the control depth is downstream and the computations must proceed upstream. Water surface profiles that result from a change in slope of the bed are illustrated in Figure 4.16.

Examples of water surface profiles and bed slope changes they can occur on, M1 on mild slope before transition to milder slope, M2 on mild slope before transition to steeper mild slope, M2 on mild slope before transition to critical slope, M2 on mild slope before transition to steep slope then S2 profile, C1 on critical slope before transition to mild slope change, S2 after transition from critical slop to steep slope, steep slope with hydraulic jump then S1 profile on steep slope before mild slope, steep slope then M3 profile and hydraulic jump on transition to mild slope, S3 on transition from steep slope to milder steep slope, S2 on steep slope transitioning to steeper slope, A2 on adverse slope before transitioning to mild slope, A2 on adverse slope before transition to steep slope with S2 profile
Figure 4.16. Examples of water surface profiles.

Table 4.1. Characteristics of Water Surface Profiles.
Class Bed Slope Depth Type Classification
Mild So > O Y > yo > yc 1 M1
Mild So > O yo > y > yc 2 M2
Mild So > O yo > yc > y 3 M3
Critical So > O Y > yo = yc 1 C1
Critical So > O Y < yo = yc 3 C3
Steep So > O Y > yc > yo 1 S1
Steep So > O yc > y > yo 2 S2
Steep So > O yc  > yo > y 3 S3
Horizontal So = O y > yc 2 H2
Horizontal So = O yc > y 3 H3
Adverse So < O y > yc 2 A2
Adverse So < O yc > y 3 A3

Note:

  1. With a type 1 curve (M1, S1, C1), the actual depth of flow y is greater than both the normal depth yo and the critical depth yc.  Because flow is tranquil, control of the flow is downstream.
  2. With a type 2 curve (M2, S2, A2, H2), the actual depth y is between the normal depth yo and the critical depth yc.  The flow is tranquil for M2, A2, and H2 and thus the control is downstream.  Flow is rapid for S2 and the control is upstream.
  3. With a type 3 curve (M3, S3, C3, A3, H3), the actual depth y is smaller than both the normal depth yo and the critical depth yc.  Because the flow is rapid control is upstream.
  4. For a mild slope, So is smaller than Sc and yo > yc.
  5. For a steep slope, So is larger than Sc and yo < yc.
  6. For a critical slope, So equals Sc and yo = yc.
  7. For an adverse slope, So is negative.
  8. For a horizontal slope, So equals zero.
  9. The case where y —> yc is of special interest because the denominator in Equation 4.36 approaches zero.

To determine if the hydraulic jump occurs on the steep or mild slope, calculate the sequent depth (y2) for the y1 depth using the hydraulic jump equation. If y2 from the hydraulic jump is larger than the normal depth y0 from Manning's equation on the mild slope, then there will be an M3 curve on the mild slope until the y2 equals the depth that corresponds to the initial depth needed for the jump to occur. If y2 is smaller than the depth that would balance with the downstream depth, the jump will occur on the steep slope and an S1 curve will occur to connect with the normal depth at the control section.

Example Problem 4.8 (SI Units)

Given: A 5 m wide, rectangular channel goes from a very steep grade to a mild slope. The design discharge is 24.8 m3/s and the normal depth and velocity on the steep slope were calculated to be 0.33 m and 15 m/s, respectively. On the mild slope, the normal depth and velocity were calculated to be 2.96 m and 1.68 m/s, respectively. Determine the type of flow occurring in both channels. If a hydraulic jump occurs, evaluate the depth downstream of the hydraulic jump, the location of the jump, and the water surface profile classification.

Sketch of Example Problem 4.8 in SI Units as described in the text with potential water surface profiles S 1 and M 3 shown

Find:

1. Find the critical depth, yc, on the steep slope.

Critical depth y sub c equals ( unit discharge q squared divided by g) to the power one-third    q = Q/B where B is the channel width

Critical depth y sub c equals [ ( total discharge Q divided by channel width B) squared divided by g ] to the power one-third   Critical depth y sub c equals [ ( 24.8 divided by 5) squared divided by 9.81 ] to the power one-third equals 1.36 meters

On the steep slope, the normal depth is 0.33 m. Since y < yc, supercritical flow occurs on the steep slope. Note that the unit discharge (q) is the same for the mild slope and hence, yc, is the same for the steep and mild slope sections. On the mild slope, the normal depth is 2.96 m. Since y > yc, subcritical flow occurs on the mild slope. Therefore, a hydraulic jump should occur.

2. Next, determine if the jump will occur on the steep slope or on the mild slope.

To determine if the hydraulic jump occurs on the steep or mild slope, calculate the sequent depth (y2) for the steep slope y1 depth using the hydraulic jump equation. If y2 from the hydraulic jump is larger than the normal depth y0 from Manning's equation on the mild slope, then there will be an M3 curve on the mild slope until the y2 equals the critical depth. If y2 is smaller than y0 on the mild slope, then the jump may occur on the steep slope and an S1 curve will occur to connect with the normal depth at the control section.

y sub 2 equals y sub 1 divided by 2 times [ the square root of (1 plus 8 times Froude number Fr sub 1 squared) minus 1]

Froude number Fr equals V divided by the square root of (g times y)   Froude number Fr equals 15 divided by the square root of (9.81 times 0.33) equals 8.34

y sub 2 equals 0.33 divided by 2 times [ the square root of (1 plus 8 times 8.34 squared) minus 1] equals 3.73 m

Compare parameters and determine the type of water surface classification.

Since the sequent depth (y2) is greater than the mild slope normal depth (i.e., 3.73 > 2.96), the mild slope channel will have an M3 curve until the hydraulic jump occurs.

Example Problem 4.8 (English Units)

Given: A 16 ft wide, rectangular channel goes from a very steep grade to a mild slope. The design discharge is 875 ft3/s and the normal depth and velocity on the steep slope were calculated to be 1.0 ft and 54.7 ft/s, respectively. On the mild slope, the normal depth and velocity were calculated to be 9.71 ft and 5.6 ft/s, respectively. Determine the type of flow occurring in both channels. If a hydraulic jump occurs, evaluate the depth downstream of the hydraulic jump, the location of the jump, and the water surface profile classification.

Profile sketch of Example Problem 4.8 in English Units as described in the text with potential water surface profiles S 1 and M 3 shown

Find:

1. Find the critical depth, yc, on the steep slope.

Critical depth y sub c equals ( unit discharge q squared divided by g) to the power one-third    q = Q/B where B is the channel width

Critical depth y sub c equals [ ( total discharge Q divided by channel width B) squared divided by g ] to the power one-third   Critical depth y sub c equals [ ( 875 divided by 16) squared divided by 32.2 ] to the power one-third equals 4.52 feet

On the steep slope, the normal depth is 1.0 ft. Since y < yc, supercritical flow occurs on the steep slope. Note that the unit discharge (q) is the same for the mild slope and hence, yc, is the same for the steep and mild slope sections. On the mild slope, the normal depth is 9.71 ft. Since y > yc, subcritical flow occurs on the mild slope. Therefore, a hydraulic jump should occur.

2. Next, determine if the jump will occur on the steep slope or on the mild slope.

To determine if the hydraulic jump occurs on the steep or mild slope, calculate the sequent depth (y2) for the steep slope y1 depth using the hydraulic jump equation. If y2 from the hydraulic jump is larger than the normal depth y0 from Manning's equation on the mild slope, then there will be an M3 curve on the mild slope until the y2 equals the critical depth. If y2 is smaller than y0 on the mild slope, then the jump may occur on the steep slope and an S1 curve will occur to connect with the normal depth at the control section.

y sub 2 equals y sub 1 divided by 2 times [ the square root of (1 plus 8 times Froude number Fr sub 1 squared) minus 1]

Froude number Fr equals V divided by the square root of (g times y)   Froude number Fr equals 49 divided by the square root of the product 32.2 times 1 equals 8.64

y sub 2 equals 1 divided by 2 times [ the square root of (1 plus 8 times 8.64 squared) minus 1] equals 11.72 feet

Compare parameters and determine the type of water surface classification.

Since the sequent depth (y2) is greater than the mild slope normal depth (i.e., 13.14 > 9.71), the mild slope channel will have an M3 curve until the hydraulic jump occurs.

4.7.3 Standard Step Procedure

The standard step method is a simple computational procedure to determine the water surface profile in gradually varied flow. Prior knowledge of the type of water surface profile as determined in the preceding section would be useful to determine whether the analysis should proceed up- or downstream.

The standard step method is derived from the energy equation. The equation is:

Delta L equals ( H sub 1 minus H sub 2) divided by ( S sub naught minus S sub f)

where:

ΔL =  Distance between Sections 1 and 2, m (ft)
H =  Specific energy at Sections 1 and 2, m (ft)
So =  Slope of the bed
Sf =  Friction slope

The above equation is used in the standard step method. An example of the use of the standard step method is given Chow (1959) and Richardson et al. (2001).

Although computer programs (such as HEC-RAS) are commonly used to compute water surface profiles, it is recommended that a qualitative sketch of the water surface profiles be made using the information given in the preceding section. This is particularly useful in complicated profiles where the channel slopes change from steep to mild or mild to steep.



<< Previous Contents Next >>

Contact:

Veronica Ghelardi
Resource Center (Lakewood)
720-963-3240
veronica.ghelardi@dot.gov

Updated: 08/29/2012

FHWA
United States Department of Transportation - Federal Highway Administration