Chapter 5.pdf

Chapter

5 Hydraulic Processes: Open-Channel Flow Open-channel flow refers to that flow whose top surface is exposed to atmospheric pressure. The topic of open-channel flow is covered in detail in textbooks such as Chow (1959), Henderson (1966), French (1985), Townson (1991), Chaudhry (1993), Jain (2001), and Sturm (2001).

5.1 STEADY UNIFORM FLOW Thissection describes thecontinuity, energy, and momentum equations for steady uniform flow in open channels. Consider the control volume shown in Figure 5.1.1 in which the channel crosssection slope and boundary roughness are constant along the length of the control volume. For uniformfiow thevelocity is uniform throughout thecontrol volume, so that VI = Vl for thecontrol volume in Figure 5.1.1. Hencefor a uniform flow, QI = Ql,A I = Al, VI = Vl, and YI = Yl. The depth of flow in uniform open-channel flow is also referred to as the normal depth. Figure 5.1.2 shows an open-channel flow, in an aqueduct of the Central Arizona Project.

5.1.1 Energy Theenergy equation foropen-channel flow canbederived in a similar manner astheenergy equation forpipeflow (equation 4.2.13) using thecontrol volume approach. In section 3.4,thegeneral energy equation forsteady fluid flow wasderived asequation (3.4.20). Considering open-channel flow inthe control volume in Figure 5.1.1, the energy equation can be expressed as

(5.1.1)

Assume the energy correction factor (section 3.7)is a = 1.0. Referto equation (3.6.4) for the definition of a. The shaft work term is dWs/dt = 0 because no pump or turbine exists. Because hydrostatic conditions prevail, the terms (p/ p + eu +gz) can be taken outside the integral in equation (5.1.1):

(5.1.2)

113

114 Chapter 5 Hydraulic Processes: Open-Channel Flow

CD

-- -- ................ .....................

®

.....................

....................

Uniform flow

CD

-- -- -- ................

-- -- ................--

-- ---

IFigure 5.1.1 Open-channel flow: uniform and nonuniform flow.

J

The terms PVd4

J(pV

3/2)d4

= m are the mass rate of flow at sections 1 and 2 and the terms

= (pV 3/2)A = m ~2, so that equation (5.1.2) becomes

A

dH (P2p+eUz + gz2)m+mT. . Vi (P1 . . Vf di= p+eU1+ gz1)m-mT

(5.1.3)

Dividing through by mg and rearranging yields P1 Vf P2 Vi -+Zl+-=-+Z2+-+ Y 2g Y 2g

[euz - e --.--1 dH] Uj

g

mg dt

(5.1.4)

Similar to equation (4.2.10), the terms in square brackets represent the headloss hL due to viscous stress (friction). This energy loss due to friction effects per unit weight of fluid is denoted as hL . The energy equation for one-dimensional flow in an open-channel is

-Y +Zl +a1 -Vf = P2 - +Z2 + a2-Vi +hL 2g Y 2g

P1

(5.1.5)

5.1 Steady Uniform Flow

115

Figure 5.1.2 Hayden-Rhodes Aqueduct, Central Arizona Project. (Courtesy of the U.S. Bureau of Reclamation (1985), photograph by Joe Madrigal Jr.)

where we have put back in the energy correction factor (see section 3.7). Pressure is hydrostatically distributed, and thus p /y + z is constant at each section in the control volume, so that pdY = Yl and pl/y = Y2· The energy equation for nonuniform open-channel flow is expressed as

(5.1.6)

For uniform flow, VI

= V2 and Yl = Y2, so (5.1.7)

By dividing both sides by L, the length ofthe control volume (channel), the following headloss per unit length of channel, Sf' is obtained as

(5.1.8) so that the friction slope equals the channel bottom slope. The channel bottom slope So = tan e, where e is the angle of inclination. If e is small « 100), then tan e ~ sin e =

(ZI - Z2)/L.

116 Chapter 5 Hydraulic Processes: Open-Channel Flow

5.1.2 Momentum Theforces acting uponthe fluid control volume in Figure5.1.1 are friction, gravity, and hydrostatic pressure. Thefriction force, Ff , is the product of the wallshearstress 'to and the area overwhich it acts, PL, where P is the wetted perimeter of the cross-section, thus

(5.1.9)

Ff = -'toPL

where the negative signindicates that the friction force acts opposite to the direction of flow. The gravity force Fg relates to the weight of the fluid yAL, where y is the specific weight of the fluid (weight perunitvolume). Thegravity force on the fluid is thecomponent ofthe weightacting in the direction of flow, that is,

r, =yALsine

(5.1.10)

The hydrostatic forces are denoted as F, and F2 , and are identical for uniform flow so that F j -F2

= o.

For a steady uniform flow, the general form of the integral momentum equation (3.5.6) in the x direction is

(5.1.11) cs

or

(5.1.12) where I>x(pV. A) = O. Because F, = F2, then by equation (5.1.12) Fg +Ff = 0, or cs

yALsine - 'toPL

For e small, So

~

=0

(5.1.13)

sin e so

yALSo

= 'toPL

(5.1.14)

which statesthat for steady uniform flow the friction andgravity forces are in balance and So Solving equation (5.1.14) for the wall shear stress (for steady uniform flow) yields

= Sf.

yALSo 'to=-PL

(5.1.15)

'to = yRSo = yRSj

(5.1.16)

or

where R = AlP is the hydraulic radius. Equation (5.1.16) expresses the effects offriction through the wallshearstress 'toas represented from a momentum viewpoint and through the rate of energy dissipation Sf represented from an energy viewpoint. Consequently, equation (5.1.16) expresses a linkage between the momentum and energy principles. Theshearstress 'to forfullyturbulent flow canbeexpressed as a function of density, velocity, and resistance coefficient Cf as

(5.1.17)

5.1 Steady Uniform Flow 117 Equating (5.1.16) and (5.1.17) yields

(5.1.18) and solving for the velocity gives

v= yc; (2g~

(5.1.19)

Defining C = V2g/Cf, then equation (5.1.19) can be simplified to the well-known Chezy equation

V=C~

(5.1.20)

where C is referred to as the Chezy coefficient. Robert Manning (1891, 1895) derived the following empirical relation for C based upon experiments:

(5.1.21) where n is the Manning roughness coefficient. Values of n are listed in Table5.1.1. Values of n for natural channels have been also published by the U.S. Geological Survey (Barnes, 1962). Substituting C from equation (5.1.21) into equation (5.1.20) results in the Manning equation V = ~ R2 / 3 S~/2 n

(5.1.22)

which is valid for SI units and So = Sf. Table 5.1.1 Values of the Roughness Coefficient n (Boldface figures are values generally recommended in design) Type of channel and description

Minimum Normal Maximum

A. Closed conduits flowing partly full

A-I. Metal a. Brass, smooth b. Steel 1. Lockbar and welded 2. Riveted and spiral c. Castiron 1. Coated 2. Uncoated d. Wrought iron 1. Black 2. Galvanized e. Corrugated metal 1. Subdrain 2. Storm drain A -2. Nonmetal a. Lucite b. Glass c. Cement 1. Neat, surface 2. Mortar

0.009

0.010

0.013

0.010 0.013

0.012 0.016

0.014 0.017

0.010 0.011

0.013 0.014

0.014 0.016

0.012 0.013

0.014 0.016

0.015 0.017

0.017 0.021

0.019 0.024

0.021 0.030

0.008 0.009

0.009 0.010

0.010 0.013

0.010 0.011

0.011 0.013

0.013 0.015 (Continued)

Table 5.1.1 (Continued) Type of channel and description

Minimum Normal Maximum

d. Concrete 1. Culvert, straight and free of debris 2. Culvert with bends, connections, and some debris 3. Finished 4. Sewerwith manholes, inlet, etc., straight 5. Unfinished, steel form 6. Unfinished, smooth wood form 7. Unfinished, rough woodform e. Wood 1. Stave 2. Laminated, treated f Clay 1. Common drainage title 2. Vitrified sewer 3. Vitrified sewer with manholes, inlet, etc. 4. Vitrified subdrainwith openjoint g. Brickwork 1. Glazed 2. Lined with cement mortar h. Sanitary sewers coated with sewage slimes, with bends and connections i. Pavedinvert, sewer, smooth bottom j. Rubblemasonry, cemented B. Lined or built-up channels B-1. Metal a. Smooth steel surface 1. Unpainted 2. Painted b. Corrugated B-2. Nonmetal a. Cement 1. Neat, surface 2. Mortar b. Wood 1. Planed, untreated 2. Planed, creosoted 3. Unplaned 4. Plank with battens 5. Lined with roofing paper c. Concrete 1. Trowel finish 2. Float finish 3. Finished, with gravelon bottom 4. Unfinished 5. Gunite, good section 6. Gunite, wavy section 7. On good excavated rock 8. On irregular excavated rock d. Concrete bottom float finished with sides of 1. Dressed stone in mortar 2. Random stone in mortar

118

0.010 0.011 0.011 0.013 0.012 0.012 0.015

0.017

0.013 0.014 0.014 0.017 0.014 0.016 0.020

0.010 0.015

0.012 0.017

0.014 0.020

0.011 0.011 0.013 0.014

0.013 0.014 0.Ql5 0.016

0.017 0.017 0.017 0.018

0.011 0.012 0.012

0.013 0.015 0.013

0.Ql5 0.017 0.016

0.016 0.018

0.019 0.025

0.020 0.030

0.011 0.012 0.021

0.012 0.013 0.025

0.014 0.017 0.030

0.010 0.011

0.011 0.013

0.013 0.Ql5

0.010 0.011 0.011 0.012 0.010

0.012 0.012 0.013 0.015 0.014

0.014 0.Ql5 0.015 0.018 0.017

0.011 0.013 0.015 0.014 0.016 0.018 0.017 0.022

0.013 0.015 0.017 0.017 0.019 0.022 0.020 0.027

0.015 0.016 0.020 0.020 0.023 0.025

0.015 0.Q17

0.017 0.020

0.011

0.013 0.012 0.015 0.013

0.014

0.020 0.024

Table 5.1.1 (Continued) Type of channel and description

Minimum

3. Cement rubble masonry, plastered 4. Cement rubble masonry 5. Dry rubble or riprap e. Gravel bottom with sides of I. Formed concrete 2. Random stone in mortar 3. Dry rubble or riprap f Brick I. Glazed 2. In cement mortar g. Masonry I. Cemented rubble 2. Dry rubble h. Dressed ashlar i. Asphalt I. Smooth 2. Rough j. Vegetal lining C. Excavated or dredged a. Earth, straight and uniform I. Clean, recently completed 2. Clean, after weathering 3. Gravel, uniform section, clean 4. Withshortgrass, few weeds b. Earth, winding and sluggish I. No vegetation 2. Grass, someweeds 3. Dense weeds or aquatic plants in deep channels 4. Earthbottom and rubble sides 5. Stony bottom and weedy banks 6. Cobble bottom and clean sides c. Dragline-excavated or dredged I. No vegetation 2. Lightbrush on banks d. Rockcuts I. Smooth and uniform 2. Jagged and irregular c. Channels not maintained, weeds and brush uncut I. Dense weeds, high as flow depth 2. Clean bottom, brush on sides 3. Same, highest stage of flow 4. Dense brush, high stage D. Natural streams D-l. Minor streams (top width at flood stage 100ft). The n value is less than that for minor streams of similar description, because banks offerless effective resistance. a. Regular section with no boulders or brush b. Irregular and rough section

0.045 0.050 0.075

0.050 0.070 0.100

0.060 0.080 0.150

0.030 0.040

0.040 0.050

0.050 0.070

0.025 0.030

0.030 0.035

0.035 0.050

0.020 0.025 0.030

0.030 0.035 0.040

0.040 0.045 0.050

0.035 0.035 0.040 0.045 0.070

0.050 0.050 0.060 0.070 0.100

0.070 0.060 0.080 0.110 0.160

0.110 0.030 0.050 0.080

0.150 0.040 0.060 0.100

0.200 0.050 0.080 0.120

0.100

0.120

0.100

0.Q25 0.035

0.060 0.100

Source: Chow (1959).

Manning's equation in SI units can also be expressedas

Q = ! AR2/ 3 S6/ 2 n

(5.1.23)

For V in ft/sec and R in feet (U.S. customary units), equation (5.1.22) can be rewritten as

V = 1.49R2/3S6/2 n

(5.1.24)

and equation (5.1.23) can be written as

_ 1.49 2/3 1/2 Q--AR SO n

(5.1.25)

where A is in ft2 and So = Sf. Table 5.1.2 lists the geometric function for channel elements.

5.1 Steady Uniform Flow 121 Table 5.1.2 Geometric Functions for Channel Elements Section:

Rectangle

Trapezoid

B-;1

r-B--j

r-

I ~ II

U z

Wetted perimeter P Hydraulic radius R

1

L.l

f-Bw-l

f-Bw-J Area A

Triangle

BwY

(Bw+zy)y

zl

1 . g(8-sm 8)d~

Bw+2y

Bw+2y~

2Y\h +z2

18do

BwY Bw+2y

(Bw+zy)y Bw+2YVl +z2

2V1 +z2

1

zy

~ (1- sin 8)d

4

8

0

[sin(~) ]d

o

Top width B

Bw

Bw+2zy

2zy

or

2Jy(do-Y) 4(2 sin 8 + 38 - secas 6) 3do8(8 - sin e) sin (8/2)

(Bw+ 2zy)(SBw+ 6yVl + z2) 2dR 1M 3Rdy + Ady

SBw+6y 3y(Bw+2y)

\-4z/V1 +Z2

8

3y(B w+ zy)(Bw+ 2yVl + z2)

3y

where 8 = 2 cas - 1 ( 1 -

~)

Source: Chow (1959) (with additions).

Todetermine the normaldepth(forunifonn flow), equation(5.1.23)or (5.1.25)canbe solvedwith a specified discharge. Becausethe original shear stress "to in equation (5.1.17)is for fully turbulent flow, Manning's equation is valid only for fully turbulent flow. Henderson (1966) presented the following criterion for fully turbulent flow in an open channel:

n6

JRSf ? 1.9 x 10 -13

n6 JRSf ? 1.1 x 10 -13

EXAMPLE 5.1.1

(S.1.26a)

(R in meters)

(S.1.26b)

An 8-ft widerectangular channel witha bed slopeof 0.0004 ftlft has a depthof flow of 2 ft. Assuming steady uniform flow, determine the discharge in the channel. The Manning roughness coefficient is n=0.015.

SOLUTION

(R in feet)

Fromequation (5.1.25), the discharge is Q

= 1.49 AR2/3S~/2 n

=

1.49 (8)(2)[ (8)(2) ]2/3(00004)1/2 0.015 8 + 2(2) .

=

38.5 ft3/s

122 Chapter 5 Hydraulic Processes: Open-Channel Flow EXAMPLE 5.1.2 SOLUTION

Solve example 5.1.1 using SI units. Thechannel width is 2.438 m,with a depthof flow of 0.610 m.Using equation (5.1.23), thedischarge is

?

/3

_ 1 [ (2.438)(0.610) ]- (0.0004)1/2 - 0.015 (2.438)(0.610) 2.438 + 2(0.610)

EXAMPLE 5.1.3

SOLUTION

Determine thenormal depth (foruniform flow) if thechannel described inexample 5.1.1 hasaflow rateof 100 cfs. Thisproblem is solved using Newton's method wish Q; defined by equation (5.1.25): 5/3

1.49 1/2 (Bwy; ) (Bw + 2y;) 2/3

Q--S J n 0

(8

5/3

)5/3

Q. = 1.49 (0.0004)1/2 (8y;) = 1.987 Y; 3 J 0.Q15 (8 + 2y;) 2/3 (8 +2yi/ Using a numerical method such as Newton's method (see Appendix A), the normal depth is 3.98 ft.

5.1.3 Best Hydraulic Sections for Uniform Flow in Nonerodible Channels The conveyance of a channel section increases with an increase in the hydraulic radius or with a decrease in the wetted perimeter. Consequently, the channel section with the smallest wetted perimeterfor a given channel section area will have maximumconveyance, referredto as the best hydraulic section or the cross-section of greatest hydraulic efficiency. Table 5.1.3 presents the geometricelementsof the best hydraulic sectionsfor six cross-section shapes. These sectionsmay not always be practical because of difficulties in construction and use of material. The concept of Table 5.1.3 Best Hydraulic Sections Wetted perimeter

Hydraulic radius

p

R

Top width T

v'3i

2v'3y

%Y

1V3y

%Y

Rectangle, half of a square

2i

4y

Y2Y

2y

Y

Triangle, half of a square

i

2V2y

!V2y

2y

Y2Y

Semicircle

Ii

1ty

%Y

2y

~y

13/21

'l3/2y

Y2Y

2V2y

%Y

1.39586i

2.9836y

0.46784y

1.917532y

0.72795y

Cross-section Trapezoid, half of a hexagon

Parabola, T

= 2V2y

Hydrostatic catenary Source: Chow (1959).

Area A

Hydraulic depth D

5.1 Steady Uniform Flow 123 besthydraulic section is onlyfornonerodible channels. Eventhough thebesthydraulic section gives the minimum area for a given discharge, it may not necessarily have the minimum excavation.

EXAMPLE 5.1.4

SOLUTION

Determine the cross-section of greatest hydraulic efficiency for a trapezoidal channel if the design discharge is 10.0 m3/sec, the channel slope is 0.00052, and Manning's n = 0.025. From Table 5.1.3, the hydraulic radius should be R = y /2, so that thewidth B and area A are 2V3y B = -3- = 1.155y (becauseB

1

= Jpfor halfof a hexagon)

A = v'3l = 1.732l

Manning's equation (5.1.23) is used to determine the depth: 1

1

( )

Q = -;:,AR2/3S~/2 = 0.025 (1.732 y2) ~

2/3

(0.00052)1/2

= 10

so 10 x 0.025 X 22/3 _ 8/3 1.732(0.00052)1/2 - y Thus, y

= 2.38 m, so that B = 2.75 m andA = 9.81 m2.

5.1.4 Slope-Area Method The slope-area method can be used to estimate the flood discharge through a channel or river reach of length L1x with known cross-sectional areas of flow at the upstream, A w and downstream, Ad, endsof the reach.The use of high-water marksfrom a flood and a survey of the cross sections allow computation of the cross-sectional areas of flow. Manning's equation (5.1.25) can be expressed as

(5.1.27)

Q=KVSo

where K is the conveyance factor expressed as K

1.49 2/3 . Conveyance is a measure of the = --AR n

carrying capacity of a channel since it is directly proportional to the discharge Q. The average conveyance factor is the geometric mean of the conveyance factors at the upstream, K w and the downstream, Kd , ends of the channel reach, i.e.,

K = VKuKd

(5.1.28)

The discharge is then expressed as

(5.1.29) where S is the waterslope given as S = (zu - Zd), Zu and Zd are the watersurface elevations at the ~x

upstream and downstream ends of the reach, respectively. Alternatively the friction slope, Sf could be used in equation (5.1.29), Q = K /Sf, where (Chow, 1959)

Sf=

[(ZU-Zd)-k(au~; -ad~D]/~x

(5.1.30)

124 Chapter 5 Hydraulic Processes: Open-Channel Flow Thedifference in water surface elevations is referred to as thefall. The k is a factor to account for a contraction and expansion of a reach. For a contracting reach Vu < Vd so k = 1.0 and for an expanding reach (Vu> Vd ) so k = 0.5. Thefirst approximation would compute thedischarge using Q = K/Sf with the friction slope computed ignoring the velocity heads. Using the first approximation of Q, theupstream anddownstream velocity heads arecomputed forthenextapproximation of the friction slope, which is used to compute the second approximation of the discharge. The procedure continues computing the new friction slope using the last discharge approximation to compute the new discharge. This process continues until the discharges approximations do not change significantly.

5.2 SPECIFIC ENERGY, MOMENTUM, AND SPECIFIC FORCE 5.2.1 Specific Energy The total head or energy head, H, at any location in an open-channel flow can be expressed as V2 H=y+z+ 2g

(5.1.6)

which assumes thatthevelocity distribution is uniform (i.e., a = 1) andthe pressure distribution is hydrostatic (i.e., p = yy). Using thechannel bottom as thedatum (i.e., z = 0) then define the total head above the channel bottom as the specific energy

V2

E=y+2g

(5.2.1)

Using continuity (V = Q/A), the specific energy can be expressed in terms of the discharge as Q2

E=Y+2gA2

(5.2.2)

Specific energy curves, such as are shown in Figure 5.2.1 and5.2.2, can be derived using equation (5.2.2).

y

ConstantQ

Q2 E=y+ --2 2gA

Figure5.2.1 Specific energy.

5.2 Specific Energy, Momentum, and Specific Force 125 y

Ql

E=y+2gA2

Figure 5.2.2 Specific energy showing subcritical and supercritical flow ranges.

Criticalflow occurswhenthe specific energy is minimum for a givendischarge (i.e., dEIdy = 0), so that

dE=l_~dA=O dy

gA3 dy

(5.2.3)

Referring toFigure5.2.1,thetop-width is defined as T = dAIdy soequation (5.2.3)canbe expressed as

(5.2.4) or

(5.2.5) To denote critical conditions use Te, An Ve, and Ye' so

(5.2.6) or V~

Ae

g

t:

(5.2.7)

Equation (5.2.6) or (5.2.7) can be used to determine the critical depth and/or the critical velocity. Rearranging equation (5.2.7) yields

(5.2.8) The hydraulic depth is defined as D

= AIT so equation (5.2.7) becomes

V~ = 1

gDc

(5.2.9)

126 Chapter 5 Hydraulic Processes: Open-Channel Flow or

(5.2.10)

This is basically the Froude number, Fr , which is I at critical flow:

< 1 subcritical flow F, = ~ = 1 critical flow { gD > 1 supercritical flow

(5.2.11)

Figure 5.2.2 illustrates the range of subcritical flow and the range of supercritical flow along with the location of the critical states. Note the relationship of the specificenergy curves and the fact that Q3 > Q2 > Q1. Figure 5.2.1 illustrates the alternate depths YI and Y2 for which £1

= £2 or Y1

V

2

V2

1 2 + -=Y2+2g 2g

(5.2.12)

For a rectangular channel De = Ac/Te = Ye, so equation (5.2.10) for critical flow becomes

(5.2.13) If welet q be the flow rate per unit widthof channelfor a rectangularchannel, i.e., q = Q/B where T = B, the width of the channel (or q = Q/n then equation (5.2.6) can be rearranged, TeQ2 / (gT~yn = q2 / (gYe) = 1, and solved for Yc to yield _ (q2) 1/3 Ye g

EXAMPLE 5.2.1 SOLUTION

(5.2.14)

Compute the critical depth for the channel in example 5.1.1 using a discharge of 100 cfs. Using equation (5.2.13), Ve = ,;gy; = Q/A = 100/8Ye, so 100

(100)2/3

y~/2 = 8/8 or Yc = 8/8

= 1.69 ft

Alternatively, using equation (5.2.14) yields

Ye

EXAMPLE 5.2.2

=(

(100/ 8)2) 1/3 g

=

1.69 ft

Fora rectangular channel of 20 ft width, construct a family of specific energy curves for Q = 0,50, 100, and300cfs. Draw thelocus of thecritical depth points on these curves. Foreach flow rate, what is the minimum specific energy found from these curves?

5.2 Specific Energy, Momentum, and SpecificForce 127

SOLUTION

The specific energy is computed using equation (5.2.1):

Computing critical depths for the flow rates using equation (5.2.14) with q = Q/B yields Q=O:

Q = SOds:

Q = 100ds:

Q = 300ds: Computed specific energies are listed in Table 5.2.1. The specific energy curves are shown in Figure 5.2.3. The minimum specific energies are: Q=SOds:

Emin

Q = 100ds:

Emin =

Q=300ds:

Emin

= 0.868 1.379

= 2.868

Table 5.2.1 Computed Specific Energy Values for Example 5.2.2 Specific energy, E (ft-1b/lb) Depth, y (ft)

Q=O

Q = 50

Q = 100

Q = 300

0.5 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.5 4.0 4.5 5.0

0.50 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00 2.20 2.40 2.60 2.80 3.00 3.50 4.00 4.50 5.00

0.89 0.87 0.95 1.10 1.27 1.45 1.64 1.83 2.02 2.22 2.42 2.61 2.81 3.01 3.51 4.01 4.50 5.00

2.05 1.68 1.41 1.39 1.47 1.60 1.75 1.92 2.10 2.28 2.47 2.66 2.85 3.04 3.53 4.02 4.52 5.02

14.86 10.57 6.41 4.59 3.69 3.23 3.00 2.91 2.90 2.94 3.02 3.13 3.26 3.40 3.79 4.22 4.68 5.14

128 Chapter 5 Hydraulic Processes: Open-Channel Flow y(ft) 7

7 E(ft)

Figure 5.2.3 Specific energy curves for example 5.2.2.

EXAMPLE 5.2.3

SOLUTION

Arectangular channel 2mwide hasaflow of2.4m3/s atadepth of 1.0m.Determine whether critical depth occurs at (a)a section where a hump ofilz = 20 em high is installedacrossthe channelbed, (b) a side wall constriction (with no humps)reducingthe channel width to 1.7 m, and (c) both the hump and side wall constrictions combined. Neglect headlosses of the hump and constriction caused by friction, expansion, and contraction. (a) The computation is focused on determining the critical elevation change in the channel bottom (hump) L1zcrit that causes a critical depth at the hump. The energy equation is E = Emin +L1zct it or L1zct it = E - Emin, where E is the specific energy of the channel flow and E rnin is the minimum specific energy, which isatcritical depth by definition. If 1 Mild: Yn>Yc or Yn Yc

(S.3.12a)

Yn 1 Steep: Yn 0 Adverse So < 0

Designation Zone I

Zone 2

Relation of Y to Yn and Yc Zone 3

Zone I

H3

Y > Yn Yn Yn

> Yc >Y >Yc > Yc > Y

None Drawdown Backwater

None Subcritical Supercritical

M3

Y > Yn Yn Yn

> Yc >Y >Yc > Yc > Y

Backwater Drawdown Backwater

Subcritical Subcritical Supercritical

C3

Y > Yc Yn Yc

Backwater Parallel to channel bottom Backwater

Subcritical Uniform-critical Supercritical

S3

Y > Yc Yc Yc

> Yn >Y > Yn > Yn > Y

Backwater Drawdown Backwater

Subcritical Supercritical Supercritical

A3

Y> (Yn)* (Yn)* (Yn)*

> Yc >Y >Yc > Yc > Y

None Drawdown Backwater

None Subcritical Supercritical

None

H2 MI M2 CI C2 SI S2 None

A2

*Yn in parentheses is assumed a positive value.

Source: Chow (1959).

Zone 2

Zone3

Yn ==y

== Yc Yn > Y

General type of curve

Type of flow

{In)

dyldx "'+

--_ ...'---~ -~-

¥

j

"':'

\

--

\

I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I

I 1

1

9 FIll"'" 5.3.3 1'\OW "",Jiles (fiOm ChOW (\ 59)).

5.3 Steady, Gradually Varied Flow 139 The three zones for mild slopes are defined as Zone 1:

Y>Yn>Ye

Zone 2:

Yn>Y>Yc

Zone 3:

Yn>Ye>Y

The energygrade line, water surface, and channel bottomare all parallelfor uniformflow, i.e., Sf = So = slopeofwatersurfacewheny = Yn.FromManning'sequationforagivendischarge,SfYn' Now consider the qualitative characteristics using the three zones. Zone 1 (Ml profile):

Y > Yn; then Sf Ye, so 1 - F; = ·

+ +

So -Sf = -1-= -+ = + 2 x -F, +

dy

by equation (5.3.11), -d

then Y increases with x so that Y --+ Yn Zone 2 (M2profile):

Y < Yn; then Sf > So or So - Sf

=-

F, < 1 since Y > Ye, so 1 - F; = + · dy So -Sf by equation (5.3.11), -d = -1-= - = x -F,2 + then Y decreases with x so that Y --+ Yc Zone 3 (M3 profile):

Y < Yn; then Sf > So or So - Sf = -

F, > 1 since Y < Ye so 1 - F; = ·

dy

by equation (5.3.11), -d x

So -Sf -F,

-

= -1--2 = - = +

then Y increases with x so that Y --+ Yc This analysis can be made of the other profiles. The results are summarized in Table 5.3.1.

.EXAMPLE 5.3.2

For the rectangular channel described in examples 5.1.1, 5.1.3, and 5.2.1, classify the type of slope. Determine thetypesof profiles thatexistfordepths of5.0 ft, 2.0 ft, and1.0ft witha discharge of 100fels.

SOLUTION

In example 5.1.3, thenormal depth is computed as Yn = 3.97ft, andin example 5.2.1, thecriticaldepthis computed as Ye = 1.69ft. Because Yn > Ye' this is a mild channel bed slope. For a flow depth of 5.0 ft, 5.0 > Yn > Ye, so that an Ml profile with a backwater curve exists (refer to Table 5.3.1 and Figure 5.3.3). The flow is subcritical, For a flow depth of 2.0 ft, Yn > 2.0 > Ye' so that an M2 profile with a drawdown curveexists. The flow is subcritical. For a flow depthof 1.0 ft, Yn > Ye > 1.0, so that an M3 profile witha backwater curveexists. The flow is supercritical.

Prismatic Channels with Changes in Slope Consider a channel that changes slope from a mild slope to a steep slope. The critical depth is the same for each slope; however, the normal depth changes, for the upstream mild slope, Ynl > Ye, and for the downstreamsteep slope, Yn2 < Ye' The only control is the critical depth at the break in slopes where flow transitions from a subcritical flow to a supercritical flow. The flow profiles are an M2

140 Chapter 5 Hydraulic Processes: Open-Channel Flow

Figure 5.3.4 Water surface profile for prismatic channel with slope change from mild to steep.

profile for the upstream reach andan S2profile for thedownstream reachas shown in Figure 5.3.4. Thespecific energy forthereaches, En l and En2 , andthespecific energy forcriticalconditions canbe computed. Achannel thatchanges slope from a steep slope toa mildslope is morecomplicated inthat a hydraulic jumpforms. Thisjumpcould form on the upper steep slope or on the lowermildslope.

5.3.3 Direct Step Method In example 5.3.1 we computed the location where a specified depth occurred, using the energy equation. Thisprocedure canbe extended to compute reachlengths for specified depths at eachend of a reach. Computations are performed stepby stepfrom oneend of the channel reachby reach to the otherend. This procedure is calledthe direct step method and is applicable only to prismatic channels. The gradually varied flow equation can be expressed in termsof the specific energy. Equation (5.3.1) can be rearranged to

(5.2.12) where Sf is the average friction slope for the channel reach. Solving for Llx,

(5.3.13) Thedirectstepmethod is basedonthisequation. Manning's equation is usedto compute thefriction slope at the upstream and downstream ends of each reach for the specified depths using

n2 y2 Sf = -2.-22-R-2-j 3

(5.3.14)

=!

The average friction slope, Sf (Sfl +Sf2 ) , is used in equation (5.3.13). The computation procedure must be performed fro~ downstream to upstream if the flow is subcritical and from upstream to downstream if the flow is supercritical.

EXAMPLE

A trapezoidal channel has theflowing characteristics: slope = 0.0016; bottom width = 20ft; and side slopes = 1vertical to2 horizontal (z = 2).Water atadownstream location is anembankment where the water depth is5.0ft justupstream oftheembankment and the discharge is400cfs.Determine thedistance upstream to where the flow depth is 4.60 ft. (Adapted from Chow, 1959.) (Assume a = 1.10.)

5.4 Gradually Varied Flow for Natural Channels

SOLUTION

141

Thecritical depth isYc = 2.22 ftandthenormal depth is Yn = 3.36 ft, sotheflow issubcritical. Because the depth of5.0ft isgreater than thenormal depth of3.36 ft,thisisanMl profile. Then thedepth will decrease proceeding in the upstream direction because dy/ dx = + in thedownstream direction. In other words, going upstream thedepth will approach normal depth. Sowewill consider thefirst reach, going upstream to a depth of 5.0 to 4.8 ft andthe second reach upstream is from a depth of 4.8 to 4.6 ft depth. First reach; AtY = 5.0ft then A = 150 fr', R = 3.54 ft, V = 2.67 ftIsec, Sf = 0.000037, andE = 5 + (I.I) (2.67)z/[2(32.2)] = 5.123 ft,andat Y = 4.80ft then A = 142.1 ftz, R = 3.43 ft, V = 2.82 ft/sec, Sf = 0.00043, and E = 4.936 ft. Using equation 5.3.14, 1 where Sf = 2: (0.00037 +0.00043) = 0.00040, Ez -E1 then ~x = So _ Sf

(5.123 -4.938) 0.0016 _ 0.00040 = 156 ft.

Second reach: Aty = 4.60ft,A = 134.3 ftZ,R = 3.3lft, V = 2.98 ftIsec,Sf = 0.00051,andE 1 Sf = 2: (0.00043 + 0.00051) = 0.00047, and ~x = 163 ft. So thedistance upstream to a depth of 4.60 ft is 156 upstream to where the normal depth occurs.

= 4.752ft.

+ 163 = 319ft. Theprocedure canbe continued

5.4 GRADUALLY VARIED FLOW FOR NATURAL CHANNELS 5.4.1 Development of Equations As an alternate to the procedure presented above, the gradually varied flow equation can be expressed in termsof the watersurfaceelevation for application to naturalchannelsby considering w = z + y where w is the water surfaceelevation above a datum such as mean sea level. The total energy H at a section is

V2 2g

V2 2g

H = z+y+a- = w+a-

(5.4.1)

including the energycorrection factor a. The changein total energyhead with respectto location along a channel is

(5.4.2) The total energy loss is due to friction losses (Sf) and contraction-expansion losses (Se):

dH

- = -Sf-Se

dx

(5.4.3)

S, is the slope term for the contraction-expansion loss. Substituting (5.4.3) into (5.4.2) results in

(5.4.4) The friction slope Sf can be expressed using Manning's equation (5.1.23) or (5.1.25):

(5.4.5) where K is defined as the conveyance in SI units K = !AR 2/ 3

n

(5.4.6a)

142 Chapter 5 Hydraulic Processes: Open-Channel Flow or in U.S. customary units as

(5.4.6b) for equations (5.1.23) or (5.1.25), respectively. The friction slope (from equation 5.4.5) is then

Q2

Sf = K2

Q2[1

=T

Kr + Ki1]

(5.4.7)

withtheconveyance effect ~ = ~ [~ + ~] , where K, andK2 aretheconveyances, respectively, K 2 K1 K2 at the upstream and the downstream ends of the reach. Alternatively, the friction slope can be determined using an average conveyance, i.e., X = (Kl + K2) /2 and Q = XS}/2; then

Q2

(5.4.8)

Sf=-2

X

The contraction-expansion loss term S; can be expressed for a contraction loss as for d ( aV2) 2g

2

2)

V -a1.-l V = (a2~ 2g

2g

>0

(5.4.9)

and for an expansion loss as

(5.4.10) The gradually variedflow equation for a natural channel is defined bysubstituting equation (5.4.7) into equation (5.4.4):

_ Q2

[~ + ~] _ s, = dw + !!..-

2 Kr Ki

dx

dx

(a V2) 2g

(5.4.11a)

or

(5.4.11b)

Rearranging yields

(5.4.12)

EXAMPLE 5.4.1

SOLUTION

Derive an expression forthechange in water depth as a function of distance along a prismatic channel (i.e., constant alignment and slope) for a gradually varied flow. We start with equation (5.3.11):

5.4 Gradually Varied Flow for Natural Channels 143 with

Then

dy

So -Sf

dx

(1- Q2T) gA3

To determine a gradually varied flow profile, this equation is integrated.

EXAMPLE 5.4.2

SOLUTION

Fora riversection witha subcritical discharge of6500ft3/S, thewatersurface elevation atthedownstream section is 5710.5 ft witha velocity headon.72 ft.Thenextsection is 500ft upstream witha velocity head of 1.95 ft. The conveyances for the downstream and upstream sections are 76,140 and 104,300, respectively. Using expansion and contraction coefficients of 0.3 and 0.1, respectively, determine the water surface elevation at the upstream section. Using equation (5.4.12), the objective is to solve for the upstream water surface elevation

WI:

The friction slope term is

Q2[~ ~]Lix=65002[ 1 2 + 1 2 ]500=2.79ft 2 Ki + KI 2 76,140 104,300

Because ~[a(v2 /2g)] > 0, a contraction exists for flow from cross-section 1 to cross-section 2. Then

The water surface elevation at the upstream cross-section is then WI

= 5710.5 + 1.77 +2.79 +0.177 = 5715.2 ft

5.4.2 Energy Correction Factor In section 3.7, the formula for the kinetic energy correction factor is derived as

(3.7.4)

144 Chapter 5 Hydraulic Processes: Open-Channel Flow

C (channel)

Figure 5.4.1 Compound channel section.

which can be approximated as

(5.4.13) where V is the mean velocity. Consider a compound channel section as shown in Figure 5.4.1 thathasthree flow sections. The objective is to derive an expression for the energy coefficient in terms of the conveyance for a compound channel, sothatthevelocity head fortheentire channel is u(V 2 /2g) where Vis themean velocity in the compound channel. Equation (5.4.13) can be expressed as N

EV/A; ;=1

(5.4.14)

U;:::j--N-

V3

E A; ;=1

where N is the number of sections (subareas) of thechannel (e.g., in Figure 5.4.1,N = 3), Vis the mean velocity in each section (subarea), andAi is the cross-sectional area of flow in each section (subarea). The mean velocity can be expressed as N

EViAi V=~ N

(5.4.15)

EA i i=l

Substituting Vi

= Q;/Ai and equation (5.4.15) for V into (5.4.14) and simplifying yields

EN (Q-.!. )3Ai i=l

Ai

U=----...,,----3 '"

Z:: ;~1 N

[

Q; iii

t.',

A;

N 1 EA-

C,)

(5.4.16)

5.4 Gradually Varied Flow for Natural Channels

145

Now using equation (5.4.5) for each section, we get

Qi = «s;1/2

(5.4.17)

and solving for S)./2 yields

(5.4.18) Assuming thatthefriction slopeis the samefor all sections, Sf; to equation (5.4.18)

= SfU = 1, ... , N), thenaccording (5.4.19)

This leads to

(5.4.20) and the total discharge is

(5.4.21) Substituting the above expression for

L Qi and (5.4.22)

into equation (5.4.16) and simplifying, we get

N(K3)(N )2

~ a=

At

~Ai

(tK;)3

(5.4.23)

1=1

or

(5.4.24) whereAt and K, are the totals. The friction slope for the reach is S 'f,

= (2:Qi)2

2:Ki

(5.4.25)

by eliminating QN/ KNfrom equations (5.4.19) through (5.4.21).

EXAMPLE 5.4.3

Forthecompound cross-section at river mile 1.0shown in Figure 5.4.2, determine theenergy correction factor a. The discharge is Q = 11,000 cfs andthe water surface elevation is 125 ft.

146 Chapter 5 Hydraulic Processes: Open-Channel Flow 50'

115' n = .025

I

n =.0451

.1

60'

In =.045

150'

---

aV2

2C

--

u

.

-

!"""-------~-------

Vhf

70'

70'

n=.04

n=.04

-f-a 2

-------~-------

r----1115'

I

......-

125' ... 110'

105'

100' L

Elev.---l Cross-section atriver mile 1.5 Q 10,500 cfs WS=? ft

L= 2640ft

=

Elev.---l

AM 1.0

AM 1.5

C,=.3

Cross-section at river mile 1.0 Q =11,000cfs WS=125ft

C,=.3

=

Cc = .1

Cc .1

Figure 5.4.2 Cross-section and reach length data for example 5.4.3 (from Hoggan (1997».

SOLUTION

Step 1 Compute thecross-sectional areas offlow fortheleftoverbank (L), channel (C), and rightoverbank (R):

AL = 1050ft2,Ac = 3000fr,A R = 1050fr Step2 Compute the hydraulic radius for L, C, and R: RL = 1050/85 = 12.35 ft; Rc = 3000/140 = 21.40 ft; RR = 1050/85 = 12.35 ft Step3 Compute the conveyance factor for L, C, and R: )( )2/3 K = 1.486ALRL2/3 = 1.486 ( 1050 12.35 = 208 700 L n 0.04 '

= 1.486(3000)(21.4)2/3 = 1 716300

K c

0.02

"

KR = 208,700(KR = Kd Step 4 Compute totals At and K; K,

= KL + Kc + KR = 2,133,700

At = AL +A c +A R = 5100 ft 2 Step5 Compute K3/A2 and LK3/A2: KiiAI

= 8.25 x 109

K~/A~

= 561.8 x 109

KUA~ = 8.25 x 109

LK3 /A2 = 578.3

X

109

Step 6 Use equation (5.4.24) to compute a: a

ExAMPLE5.4.4

=

E(%)

(At? = (578.3 x 109)(5100)2 = 1.55 (Kt)3 (2,133,700)3

ForthedatainFigure 5.4.2, startwith theknown water surface elevation atrivermile 1.0anddetermine the water surface at rivermile 1.5 (adapted from Hoggan, 1997).

5.4 Gradually Varied Flow for Natural Channels

SOLUTION

147

Computations are presented in Table 5.4.1. Table 5.4.1 Standard Step Backwater Computation

(10) (11) (12) (13) (14) (15) (16) (4) (5) (6) (7) (8) (9) (2) (3) Energy Average Water Hydraulic Manning Average friction Friction correction River surface v2 factor Velocity a"2g loss mile elevation Area radius roughness Conveyance conveyance slope 3/A2 V hL a R n K Wk A K h0 ** Wk *** K Sf (ftlsec) (ftlsec) (ftlsec) (ft) (ft) (£1) (109 ) (ft) (ft2) (1)

Ll(*)

1.0

125.0

1.5

126.1

1.5

125.0

1050 3000 1050 5100 666.0 2426.5 666.0 3758.5 600 2300 600 3500

12.35 21.43 12.35

0.040 0.020 0.040

9.37 0.045 17.97 0.025 9.37 0.045 8.57 0.045 17.04 0.025 8.57 0.045

208,700 1,716,300 208.700 2,133,700 97,650 989,400 97,650 1,184,700 1,659,200 0.000042 83,000 905,300 83,000 1,071,300 1,602,350 0.000043

8.25 561.80 8.25 578.30 2.10 164.50 2.10 0.111 168.70 1.59 140.22 1.59 0.113 143.30

1.55

2.16

0.11

1.43

2.79

0.17 -0.060.02 125.07

1.43

3.00

0.20 -0.09 0.03 125.05

3/A2 *a= (At )z"w KJ I

(Kt )3

"h o = Cel~(aVZ /2g) I for Ll(aV2j2g) < 0 (loss due to channel expansion); ho = Ccl~(av2 /2g) I for ~(aVZ /2g) > 0 (loss due to channel contraction).

'''Wz

=

WI

+~(aVz/2g)

+hdho = 125.0+( -0.06)+0.111+0.02 = 125.066 ~ 125.07.

Source: Hoggan (1997).

5.4.3 Application for Water Surface Profile The change in head withrespect to distance x along the channel has been expressed in equation (5.4.2) as 2

dH dw d ( V ) dx = dx + dx a 2g

(5.4.2)

Thetotalenergy losstermis dHj dx = - Sf - Se, where Sfis thefriction slope defined by equation (5.4.7) or equation (5.4.8) and S, is the slope of thecontraction or expansion loss. Thedifferentials 2 dw and d[a(V 2 j2g)] aredefined overthechannel reachas dw = Wk - Wk+ 1 and d[a(V j2g)] = V2 V2 ak + 1 ~; 1 - ak 2;' where wedefine k + 1 at thedownstream and k at theupstream. River crosssections are normally defined from downstream to upstream for gradually varied flow. Equation (5.4.2) is now expressed as

Vf Vf+l Wk +ak 2g = Wk+l +ak+l ~

+Sfdx +Se dx

(5.4.26)

The standard step procedure for water surface computations is described in the following steps: a. Startata pointinthechannel where thewatersurface is known orcanbe approximated. Thisis

the downstream boundary condition forsubcritical flow andthe upstream boundary condition

148 Chapter 5 Hydraulic Processes: Open-Channel Flow

b.

c. d. e.

forsupercritical flow. Computation proceeds upstream forsubcritical flow anddownstream for supercritical flow. Why? Choose a water surface elevation Wk at the upstream end of the reach for subcritical flow or Wk + 1 at the downstream endof the reach for supercritical flow. Thiswatersurface elevation will be slightly lower or higher depending upon the type of profile (see Chow (1959); Henderson (1966); French (1985); or Chaudhry (1993)). Next compute the conveyance, corresponding friction slope, and expansion and contraction loss terms in equation (5.4.26) using the assumed water surface elevation. Solve equation (5.4.26) for wk+ 1 (supercritical flow) or Wk (subcritical flow). Compare thecalculated water surface elevation W withtheassumed water surface elevation w'. If thecalculated andassumed elevations donotagree within anacceptable tolerance (e.g.,0.01 ft), thensetwk+l = Wk+l (forsupercritical flow) andWk = Wk (forsubcriticalflow)andretum to step (c).

Computer models for determining water surface profiles using the standard step procedure include the HEC-2 model and the newer HEC-RAS model. HEC-RAS River Analysis System (developed by the U.S. Army Corps of Engineers (USACE) Hydrologic Engineering Center) computes water surface profiles for one-dimensional steady, gradually varied flow in rivers of any cross-section (HEC, 1997a-c). HEC-RAS can simulate flow through a single channel, a dendritic system of channels, or a full network of openchannels (sometimes calleda fullylooped system).

100 0 100 200 I"II!

I

I

Scale (ft)

Figure5.4.3 Map oftheRed Fox River indicating cross-sections forwater surface profile analysis (from U.S. Bureau of Reclamation (1957)).

5.4 Gradually Varied Flow for Natural Channels 149 HEC-RAS can model sub- or supercritical flow, or a mixture of each within the same analysis. A graphical user interface providesinput data entry,data modifications, and plots of stream crosssections, profiles, and other data. Program options include inserting trapezoidal excavations on cross-sections, and analyzing the potential for bridge scour. The water surface profile through structuressuch as bridges,culverts,weirs,andgatescan be computed.TheWorldWideWebaddress to obtain the HEC-RAS model is www.hec.usace.army.mil.

,EXAMPLE 5.4.5

A planview of the RedFoxRiver in California is shown in Figure 5.4.3, along with the location of four cross-sections. Perform the standard stepcalculations to determine the water surface elevation at crosssection 3 for a discharge of 6500ft3/s. Figures 5.4.4a, b, and c are plotsof cross-sections at 1, 2, and 3, respectively. Figures 5.4.5a, b, andcaretheareaandhydraulic radius curves forcross-sections 1,2,and3, respectively. Useexpansion andcontraction coefficients of 0.3 and0.1, respectively. Manning's roughness factors are presented in Figure 5.4.4. The downstream starting water surface elevation at crosssection 1is 5710.5 ft above mean sealevel. Thisexample wasoriginally adapted bytheU.S. Army Corps ofEngineers frommaterial developed bytheU.S. Bureau ofReclamation (1957). Distance between crosssections 1and2 is 500ft, between cross-sections 2 and3 is 400 ft, andbetween cross-sections 3 and4 is 400 ft.

SOLUTION

The computations for this example are illustrated in Table 5.4.2.

(20,25)

5725 5720

6

~ m

1:\

'"

A

A3

f

n 0.10

I

I

(1635,25)

,

AI

A,

n

n 0.05

n 0.10

0.05

1--1= 0.03

-f+

(110,18)

5715

I

f A.

(415,17) (650,14) (1020,14)

(710,13)

5710

(1590,14)

5705 (675,1)

5700

o

200

(71°1 1)

\

600(690, 0) 800

400

1000

1500

1700

Station (It) Cross-section No. 1 (a)

5730 I

I

5725

g

~,25)

5720

c

.2 5715

~

ijj

(110,20)

A,

n 0.05

A3

=

(200~ 20)

I

~

I-

~7)'\

n

n

=0.10

J =0.03

(640,18)

r\. I (58,,4)

5710

v,L"J,(1195,18)

(575,9.5)

5705

o

A

A,

n =0.10

200

400

(61

1,4)

600

800

1000

1200

1400

Station (It) Cross-section No.2 (b)

Figure 5.4.4 Cross-sections of the Red Fox River (fromU.S. Bureau of Reclamation (1957)).

ISO Chapter 5 Hydraulic Processes: Open-Channel Flow

1

A,

1 ~=0.11

5725

~ ~

5720 -

I

~

""" (260,22) .....

(875,25)WI (850,22 _I...-

1--,,=0.10

5710

I

f<

(600.20) (560,17:3)

l370,18.7)\ I 1\ (420.15)1 \

w 5715

A2

,,=o.OS

" =0.03

(40.25)

~

~

A3

I

5730

J \ I) (s+.7.5)

(500,7.1)

5705

0

200

400

600

1000

800

Station (tt) Cross-section No.3 (c)

I

~

5725

,g

5720

~

AJ

A3

10

o os .

I,,-..0. ..,.., "i

5730

1

30,26) (130,24) 330,23) I - -

~ W 5715

""r

5710 5705

0

200

"i

TJ " .. 0.10

A2

A,

,,= 0.036

o os .

I

(700.26)

1~0,22)

7

(460.22

i 1/

(400'1 10)

400

600

800

1000

Station (tt) Cross-section No.4 (Ii)

Figure 5.4.4 (Continued)

572118

o

16

5720

~

5716

.9 1ii ~

m

5710

~~

~

400

~

~ 1~ 1~ Area(tt2) (a)

1400

1~ 1~

Figure 5.4.5 Areaelevation andhydraulic radius-elevation curves forcross-sections I to 4. (a) Crosssection 1; (b) Cross-section 2; (c) Cross-section 3; (d) Cross-section 4 (from U.S. Bureau of Reclamation (1957)).

16 57211r-8---T----r--=~..-_Tjr--r-v~:::;::;;;;p,_{_;~ 5720

o

5710 5708 0

800 1000 Area (tt2) (b)

400

600

16

14

12

10

8

4

0

200

400

600

800 1000 1200 Area (fI2) (c)

18 5721

16

14

~

400

572118

A2

1200

1400

200

1600

1800

1400

1600

1800

4

2

0

A1

5720 5718

g c

5716

0

il>

CD

iii

5714 5712 5710 5708

5720 5718

g

.~ c

5716

0

iii

5714

~~

~

~

1~ 1~

1400

1~ 1~

Area (tt2) (d)

Figure 5.4.5 (Continued)

151

152 Chapter 5 Hydraulic Processes: Open-Channel Flow Table 5.4.2 Standard Step Backwater Computation for Red Fox River (1)

(4) (7) (2) (3) (5) (6) Water Average Cross- surface Hydraulic Manning roughness Conveyance conveyance section elevation Area radius K R n K A Wk + !

1 2

5710.5* 5714.7

(ftl)

(ft)

420 470 260 730

7.0 7.6 2.5

0.03 0.03 0.05

(12) (13) (14) (15) (11) (10) (9) (8) Energy Average correction friction Friction J factor Velocity slope loss ~ K V u 2g U 2g A2 ho a S/1O- 3) hL 6 (ft) (ft) (ft/sec) (ft) (ft) (10 ) (ft/ft)

E ( E)

1.0

76,100 90,100 14,200 104,300

90,200

5.19 2.60

3311.1 42.0 3353.5

15.5

(16)

Wk (ft)

3.72

1.58

8.90 1.95 + 1.77 0.18 5715.07**

1.59 1.0 1.0

8.13 1.63 +2.09 0.21 5715.1 5.68 0.50 + 1.13 0.115717.1 6.70 0.70 +.93 0.095717.1

5715.0

3

500 300 800 5718.0 1145 5717.1 970

7.85 2.7

0.03 0.05

5.85 5.6

0.03 0.03

97,800 17,300 115,100 184,100 151,500

95,600 149,600 133,300

4.62 2.31 1.89 0.76 2.38 0.95

*Known starting watersurface elevation. **Wk+1 = 5710.7+ 1.77+2.60 + 0.18 = 5715.07 = 5715.1; a. = (A,/ ,£Kl!Af /(K,)3;

channel expansion); ho

= Ccl~(aV2 /2g) I for ~(aV2 /2g) >0

3741.8 57.5 3799.3

h, =

eel (a.V2/2g)I for ~(aV22g) < 0 (loss due to

(lossdue to channel contraction); Wk+ 1 = Wk + ~(av2 /2g) + hL + h.,

Source: Hoggan (1997).

5.5 RAPIDLY VARIED FLOW Rapidly variedfiow occurs when a water flow depth changes abruptly overa veryshortdistance. The following are characteristic features of rapidly varied flow (Chow, 1959): • Curvature of the flow is pronounced, so that pressure distribution cannot be assumed to be hydrostatic. • The rapid variation occurs over a relatively short distance so that boundary friction is comparatively small and usually insignificant. • Rapid changes of water area occur in rapidly varied flow, causing the velocity distribution coefficients a and pto be much greater than 1.0. Examples of rapidly varied flow are hydraulic jumps, transitions in channels, flow over spillways, flow in channels of nonlinear alignment, and flow through nonprismatic channel sections such as flow in channel junctions, flow through trash racks, and flow between bridge piers. The discussion presented in this chapter is limited to the hydraulic jump. The hydraulic jump occurs when a rapid change in flow depth occurs from a small depth to a large depth such that there is an abrupt rise in water surface. A hydraulic jump occurs wherever supercritical flow changes to subcritical flow. Hydraulic jumps can occur in canals downstream of regulating sluices, at the foot of spillways, or where a steepchannel slope suddenly becomes flat. Figure 5.5.1 illustrates a hydraulic jumpalong with thespecific energy andspecific force curves. The depths of flow upstream anddownstream of thejump are called sequent depths or conjugate depths. Because hydraulic jumps are typically short in length, the losses due to shear along the wetted perimeter aresmall compared to thepressure forces. Neglecting these forces andassuming a horizontal channel (Fg = 0), the momentum principle can be applied as in section 5.2.3 to derive

5.5 Rapidly Varied Flow 153 y

y

Hydraulic jump

Specific force curve

Figure 5.5.1 Hydraulic jump.

equation (5.2.24):

Q2 _ Q2 _ - +AlYl = - +A2Y2 gA2 gAl Consider a rectangular channel of width B > 0, so Q = Al VI = A2V2,A l )11 = yI!2, and )12 = y2/2:

(5.2.24)

= BYl,A2 = BY2,

(5.5.1) Simplifying yields

Q2 g

(~_~) = !B2(y~ -yi) Yl

2

Y2

Q2 ( ) 1 2 (2 2) g Y2 -Yl ="i B Y1Y2 Y2 -Yl

B2y2V2 1 ~ (Y2 - Yl) = "iB2YIY2(Y2 +Yl)(Y2 - vr) YlVf 1 -g- = "iY2(Y2 +Yl) Dividing by

Yi, we get 2Vf gYl

= Y2 (Y2 + Yl Yl

1)

The Froude number for a rectangular channel is Fr 1 equation (5.5.2) reduces to

(5.5.2)

= VI! J gDl = VI! Viii;

therefore

(5.5.3) or

(5.5.4) using the quadratic formula and discarding the negative roots.

154 Chapter 5 Hydraulic Processes: Open-Channel Flow Alternatively, Q = BY2 V2 and Frz = V2//iYi. could have been used to derive

~~ =~( -1+V1+ 8F;z)

(5.5.5)

Equations (5.5.4)and (5.5.5) can be used to findthe sequent depths of a hydraulicjump. The use of hydraulic jumps as energy dissipaters is further discussed in Chapters 15 and 17.

EXAMPLE 5.5.1

SOLUTION

EXAMPLE 5.5.2 SOLUTION

Consider the8-ft widerectangular channel usedinexamples 5.1.1,5.1.3,and5.3.1 witha discharge of 100 cfs. If a weirwereplacedin the channel and the depth upstream of the weirwere5 ft, would a hydraulic jump form upstream of the weir? Forthedischarge of 100cfs,thenormal depthis Yn = 3.97ft from example 5.1.3,andthecriticaldepthis Ye = 1.69. Because Ye < Yn < 5 ft, a hydraulic jump would not form. As a resultof Yn > Ye, a mild slope exists. For a jump to form, Yn < Ye, whichis a steep slope.

For example 5.3.1, determine whether a hydraulic jump will occur. The normal depthandcriticaldepthmustbe computed andcompared. Usingequation (5.2.14) with q = 4.646 m3/s per meter of width, we get _ (q2) 1/3 _ (4.646 Ye g 9.81

2

)

1/3 _

-1.30m.

The depths of flow at Ya = 0.457 m and at Yb = 0.5 m, so this flow is supercritical flow. Next, the normal depth is computed using Manning's equation (5.1.23): Q=

!n AR2f3 56 / 2

or

4.646 = 0.~20y~f3(0.0003)1/2 Thus Yn = (5.365)3 / 5 = 2.74 m. Underthese conditions Yn > Ye > 0.5 m, an M3 water surface profile existsand a hydraulic jump occurs. If normal depth occurs downstream of the jump, what is the depth before the jump? Usingequation (5.5.5) with Y2 = 2.74 m, we find that Yl is 0.5 m.

EXAMPLE 5.5.3

SOLUTION

A rectangular channel is 10.0 ft wide and carries a flow of 400 cfs at a normal depth of 3.00. Manning's n = 0.017. An obstruction causesthe depthjust upstream of the obstruction to be 8.00 ft deep. Will a jump form upstream from the obstruction? If so, how far upstream? What type of curve will be present? Firsta determination mustbe madewhether ajumpwillformbycomparing the normal depthandcritical depth. Using equation (5.2.14), we find q2) 1/3

Ye

= (g

=

( 402 ) 1/3 32.2

= 3.68 ft

5.5 Rapidly Varied Flow 155 andYc>Yn = 3.0 ft, therefore thechannel issteep. Because Yn Ynz. thejumpoccurs downstream of the break in slope with control at B. If (En! - tilli !) > Enz• then a hydraulic jump occurson the downstream mild slope. If Enz > (En! - tilliI ) . then the hydraulic jump occurson the upstream steepslope. tilliI is the energy lossin thejump in reach 1. Another wayto lookat thisis if Ey' > Enz controlis at B because the energy loss tilliI in the hydraulic jump in reach 1 is not large enough to decrease the energy from En! to Enz.

Yn, Mildslope

Figure 5.5.2 Rapidly varied flow caused by slope change from steep to mild.

158 Chapter 5 Hydraulic Processes: Open-Channel Flow

5.6 DISCHARGE MEASUREMENT

5.6.1 Weir A weir is a device (or overflow structure) that is placed normal to the direction of flow. The weir essentially backs up water so that in flowing overthe weir, the water goes through critical depth. Weirs have been usedforthemeasurement ofwater flow inopenchannels for many years. Weirs can generally beclassified as sharp-crested weirs andbroad-crested weirs. Weirs arediscussed indetail in Bas et al. (1984), Brater et al. (1996), and Replogle et al. (1999). A sharp-crested weir isbasically athinplatemounted perpendicular totheflow withthetopofthe plate having a beveled, sharp edge, which makes the nappe spring clear from the plate (see Figure 5.6.1). Therateof flow is determined by measuring thehead, typically in a stilling well(see Figure 5.6.2) at a distance upstream from the crest. The head H is measured using a gauge.

Suppressed Rectangular Weir These sharp-crested weirs areaswide asthechannel, andthewidth ofthenappe is thesamelengthas thecrest. Referring toFigure 5.6.1, consider anelemental areadA = Bdhandassume thevelocity is then the elemental flow is

J2ih;

(5.6.1) Thedischarge isexpressed byintegrating equation (5.6.1) overtheareaabove thetopof the weir crest: H

H

J

Q = dQ =

Vii B Jh1/ 2dh = ~ Vii BH3/ 2

o

0

va2

Drawdown

2g

Figure 5.6.1 Flow over sharp-crested weir. u

u

~

-

Stilling well

C

•tdh ~

Crest

tr

1

I·- - - - - - - - B - - - - - - - + I·1

1

Figure5.6.2 Rectangular sharp-crested weir without endcontraction.

(5.6.2)

5.6 Discharge Measurement

159

Friction effects havebeenneglected inderivation ofequation (5.6.2). Thedrawdown effectshown in Figure 5.6.1 and the crest contraction indicate thatthe streamlines are not parallel or normal to the area in the plane. To account for these effects a coefficient of discharge Cd is used, so that

~gBH3/2 Q = Cd~3 v-«

(563) ..

where Cdis approximately 0.62. Thisis thebasicequation for a suppressed rectangular weir, which can be expressed more generally as Q = C BH3/2 (5.6.4) w

2

w~ere c, is theweircoefficient, c, = c, "3 /li. ForU.S. customary units, c, : : ; 3.33, andfor SI units Cw ::::; 1.84. If thevelocity of approach Va where H is measured is appreciable, thenthe integration limitsare v2

Q = /liB

H

+J*h1/ 2dh = CwB [(H + ~;2) 3/2 - (2)3/2] ~;

(5.6.5a)

v2

...!L

2g

V2) 3/2 When ( 2a ::::; 0, equation (5.6.5a) can be simplified to g V2 3/2 Q = CwB ( H +

2;)

(5.6.5b)

Contracted Rectangular Weirs Acontracted rectangular weiris another sharp-crested weirwithacrestthatis shorter thanthewidth of the channel and one or two beveled end sections so that water contracts both horizontally and vertically. This forces the nappe width to be less than B. The effective crest length is

B' = B - 0.1 nH

(5.6.6)

where n = 1 if the weiris placed against one sidewallof the channel so that the contraction on one side is suppressed and n = 2 if the weir is positioned so that it is not placed against a side wall. Triangular Weir Triangular or V-notch weirs aresharp-crested weirs thatareusedforrelatively smallflows, buthave theadvantage thattheycanalsofunction forreasonably largeflows aswell. Referring toFigure5.6.3,

Figure 5.6.3 Triangular sharp-crested weir.

160 Chapter 5 Hydraulic Processes: Open-Channel Flow the rate of discharge through an elemental area, dA, is

(5.6.7)

dQ = Cd/fih dA where dA

= 2xdh, and, x = (H-

h)tan~ so dA = 2(H- h)tan(~)dh. Then

dQ = Cd/fih[2(H -

h)tan(~)dh]

(5.6.8)

and H

Q=

Cd2y1Lgtan(~)J(H-h)hl/2dh o

(5.6.9)

= CdC85) fig tan(~)HS/2

= C",H S/ 2 The value of Cw for a valueof e = 90° (themostcommon) is Cw = 2.50for U.S. customary units and Cw = 1.38 for SI units. Broad-Crested Weir Broad-crested weirs (refer to Figure5.6.4) are essentially critical-depth weirs in that if the weirs are high enough, critical depth occurs on the crest of the weir. For critical flow conditions 2 1/3 3 Ye = (q jg) and E = 2Ye for rectangular channels:

2 )3

~

Q=B.q=Bj;i=Byg\3 E) =B

(2)3/2 3 y'gE

3 2 /

or, assuming the approach velocity is negligible:

Q= B

2)3/2 (3 y'gH

32 /

Q = CwBH3/ 2 Figure5.6.5 illustrates a broad-crested weir installation in a concrete-lined canal.

y2 c

__1__ ~ i

Y1

P

Figure 5.6.4 Broad-crested weir.

_

28

(5.6.10)

5.6 Discharge Measurement 161

/ // // /

/

/

/

Survey point forestablishing gauge zero reference. L/4 to Ll3 from end. onweir center line

/

Recorder

-*

F

1----' L..., I 1

:

I

Top of canal

l

rr

Y2 d

l_J

t=,_L_b~~~~=:I+---L----+l Section along center line

Cross-section

Figure 5.6.5 Broad-crested weirin concrete-lined canal (from Bos et al. (1984)).

EXAMPLE 5.6.1

SOLUTION

A rectangular, sharp-crested suppressed weir3 m longis 1.0 m high. Determine the discharge when the head is 150 mm. Using equation (5.6.4), Q = 1.84BH1.5, the discharge is 150 ) 1.5

Q = 1.84(3) ( 1000

= 0.321 m3/s

5.6.2 Flumes Bos et al. (1984) providean excellentdiscussionof flumes. Aweir is a control sectionthat is formed by raising the channel bottom, whereas aflume is formed by narrowing a channel. When a control sectionis formedbyraisingboththechannelbottomandnarrowingit, the structureis usuallycalled a flume. Figure 5.6.6 shows a distinction between weirs and flumes. Weirscan resultin relativelylargeheadlossesand,if the water has suspendedsediment,can cause deposition upstream of the weir, resulting in a gradual change of the weir coefficient. These disadvantages can be overcome in many situations by the use of flumes. Figure 5.6.7 illustratesthe generallayout of a flow measuringstructure.Most flow measurement and flow regulating structures consist of (a) a converging transition, where subcritical flow is accelerated and guided into the throat without flow separation; (b) a throat where the water

162 Chapter 5 Hydraulic Processes: Open-Channel Flow

Broad-crested weir

Long-throated flume

Usually flume

Cross-sections are through control atweir crest orflume throat

Figure 5.6.6 Distinction between a weir and a flume (from Bas et al. (1984)).

. Converging.. Distance to transition Diverging gauging station' \ ~. transition ~ Throat I

Tailwater channel

C

: Approach channel

~

I I

"L;i'~~~"e-..

I

I

Upstream channel entrance

Figure 5.6.7 General layout of a flow-measuring structure (from Bas et al. (1984)).

accelerates to supercritical flow so that the discharge is controlled; and (c) a diverging transition where flow velocity is gradually reduced to subcritical flow and the potential energy is recovered. One of the morewidely used flumes is the Parshall flume (U.S. Bureau of Reclamation, 1981), which is a Venturi-type flume illustrated in Figure5.6.8. The discharge equation for these flumes with widths of 1 ft (0.31 m) to 8 ft (2.4 m) is

(5.6.11)

Extend wingwall intocanal bank ~ as required

A.;'I I

So and that for Y > Yn then Sf < So. 5.3.7 Show that dyfdx = + for the Sl profile; dyldx = - for the S2 profile; and that dy/ dx = + for the S3 profile.

5.3.8 Using the gradually varied flow equation dyldx = (So - Sf)/ (1 - F;) define dy/ dx forcriticalflow andforuniform flow. 5.3.9 A trapezoidal channel with a bottom width of 20 ft, a slope of 0.0016, Manning's n of 0.025, and side slopes of 2:1 (h:v)has a discharge of 500 cfs. An obstruction in this channel causes a backwater profile with a depth of 6.5 ft just upstream of the obstruction. What would be the depth of flow 200 ft

Problems 169 and 400 ft upstream of the obstruction? How far upstream does the normal depth occur? Assume an energy coefficient of 1.1. 5.3.10 Consider a concrete (n = 0.013) wide rectangular channel thatdischarges 2.0m3Isperunitwidth offlow. Thechannel bottom slope is 0.0001. There is a stepriseof 0.2 m. Determine the flow depth downstream of the step assuming no transition loses. Does the water rise or fall at the step? 5.3.11 Consider a 5-mwide rectangular channel with a discharge of 12.5 m3/s. Thedepthofflow upstream ofthestepis2.5m.There is a stepriseof0.25 m in thebottom ofthechannel. Determine the flow depthdownstream of the stepassuming no transition losses. Does the water rise or fall at the step? 5.3.12 Consider the trapezoidal channel with a bottom width of 20 ft, a slope of 0.0016, Manning's n of 0.025, and sideslopes of 2:1 (h:v) having a discharge of 500cfs.Now a stepriseof 1.0ft is placed in the channel bottom. Determine the flow depth downstream of the bottom stepassuming no transition losses. Doesthe water surface rise or fall at the step? 5.3.13 A 5-m wide rectangular channel with two reaches, each withadifferent slope, conveys 50m3Isof water. Thechannel slope for the first reach is 0.001 and thena sudden change to a slope of 0.010. The Manning's n for the channel is 0.Q15. Perform the necessary computations to sketch the water surface profile and define the type of profiles.

culvert is 0.02. Theoutletof theculvert is placed at anelevation so thatit willnotbe submerged. Fora discharge of 230cfs,compute the water surface profile. Develop a spreadsheet to perform the computations. 5.3.20 A 500-ft, 6-ft diameter reinforced concrete pipe culvert (n = 0.012) is used toconvey stormwater from a detention reservoir to a downstream flood control channel. The slope of the culvert is 0.001.Theoutlet oftheculvert isplaced atanelevation sothatit will not be submerged, and the flow falls freely into the flood control channel. Foradischarge of?5cfs,compute thewater surface profile. Develop a spreadsheet to perform thecomputations. 5.4.1 Rework example 5.4.2using equation (5.4.8), Sf = (jl /7(2. 5.4.2 Resolve example 5.4.3witha discharge of 10,000 cfs anda downstream water surface elevation of 123.5 ft. 5.4.3 Rework example 5.4.4using a discharge at rivermile1.0of 8000cfs anda discharge of7500 cfs at rivermile 1.5. Thewater surface elevation at rivermile1.0is 123.5 ft. Allotherdataarethe same. 5.4.4 Consider a starting (assumed) water surface elevation of 5719.5 ft at cross-section 1 for example 5.4.5 and determine the water surface elevation at cross-section number 4. 5.4.5 Consider a starting (assumed) water surface elevation of 5717.6 ft at cross-section 1 for example 5.4.5 and determine the computed water surface elevation at cross-section 4.

5.3.14 Consider a concrete (n = 0.013) wide rectangular channel 3/sec (R = y) thatdischarges 2.0m perunitwidth offlow. Theslope of the channel is 0.001. A low dam causes a backwater depth of 2.0 m immediately behind (upstream of) the dam. Compute the distance upstream of the damto where the normal depthoccurs.

5.4.6 Perform the backwater computations at mile 2.0 for the situation in example 5.4.4. The Manning's n values at mile 2 are 0.02 for the main channel and 0.04 for the overbanks. Use an assumed trialwatersurface elevation of 125.5 ft.Cross-sections at miles 2.0 and 1.5 are the same.

5.3.15 Consider a concrete (n = 0.013) wide rectangular channel (R = y) that discharges 2.0 m3/sec perunit widthof flow. The slope of the channel is 0.0001. A low dam causes a backwater depth of 2.0 m immediately behind (upstream of) the dam. Compute the distance upstream of the dam to where the normal depth occurs.

5.4.7 Formostnatural channels andmany designed channels, the roughness varies along thewetted perimeterofthechannel. Inorder toperform normal flow computations forthesecomposite channels it is necessary to compute the composite (equivalent or effective) roughness factor. Forthecomposite channel in Figure 5.4.2, computetheeffective roughness factor (n e ) atrivermile1.0fora water surface elevation of 125 ft using thefollowing equations:

5.3.16 Consider aconcrete trapezoidal channel with a4-mbottom width, side slopes of 2:1 (h:v), and a bottom slope of 0.005. Determine the depth 150 m upstream from a section that has a measured depthof 2.0 m. 5.3.17 Awide rectangular channel changes in slopefrom0.002 to 0.025. Sketch thewater surface profile fora discharge of 1.7m3/s1 m and Manning's n = 0.025. 5.3.18 A 5-m wide rectangular channel with two reaches, each with adifferent slope, conveys 40m3Is of water. Thechannel slope forthefirst reach is 0.0005 andthena sudden change to a slope of 0.015 so that critical flow occurs at the transition. Determine the depths of flow at locations 10m, 20 m and 30 m upstream of the critical depth. The Manning's n for the channel is 0.015. 5.3.19 A 500-ft, 6-ft diameter reinforced concrete pipe culvert (n = 0.012) is used to convey stormwater from a detention reservoir to a downstream flood control channel. The slope of the

and

where Pi and n, are, respectively, the wetted perimeter and N

Manning's n for eachsubsection of the channel; P = L, Pi is the i=l

wetted perimeter of the complete channel section, and N is the number of subsections of the channel. 5.4.8 UsetheU.S. Army Corps ofEngineers HEC-RAS computer code to solve Example 5.4.4.

170 Chapter 5 Hydraulic Processes: Open-Channel Flow 5.4.9 UsetheU.S. ArmyCorps ofEngineers HEC-RAS computer code to solveexample 5.4.5. 5.5.1 Consider a 2.45-m wide rectangular channel with a bed slopeof0.0004 anda Manning's roughness factor of0.015. Aweir is placed in the channel and the depth upstream of the weir is 1.52mfora discharge of5.66m3Is. Determine whether a hydraulic jump forms upstream of the weir. 5.5.2 A hydraulic jump occurs in a rectangular channel 4.0 m wide. Thewaterdepthbefore thejumpis 0.4m andafterthejump is 1.7m. Compute the flow rate in the channel, the criticaldepth, and the headloss in the jump. 5.5.3 Rework example 5.5.3 witha flow rateof450cfsata normal depth of 3.2 ft. All other data remain the same. 5.5.4 Rework example 5.5.4if thedepth before thejumpis 0.8 m and all other data remain the same. 5.5.5 A rectangular channel is 3.0 m wide (n = 0.018) with a discharge of 14 m3/s at a normal depth of 1.0m. An obstruction causes thedepthjust upstream ofthe obstruction tobe 2.7m deep. Will a jump form upstream of the obstruction? If the jump does form, how far upstream is it located? 5.5.6 Rework example 5.5.4if the depth after thejump is 1.8m and all other data remain the same. 5.5.7 A lO-ft wide rectangular channel (n=0.015) has a discharge of 251.5 cfs at a uniform flow (normal) depth of 2.5 ft. A sluicegateat the downstream endof thechannel controls the flow depth just upstream of the gateto a depthz. Determine thedepth z sothata hydraulic jumpis formed just upstream of thegate. What is thechannel bottom slope? Whatis theheadloss (energy loss) in the hydraulic jump? 5.5.8 A 3-m widerectangular channel (n = 0.02)hasa discharge of 10m3Isat a uniform flow (normal) depthof0.8m.A sluice gate at the downstream end of the channel controls the flow depth just upstream of the gate to a depth z. Determine the depth z so that a hydraulic depth is formed just upstream of the gate. Whatis the channel bottom slope? What is the headloss (energy loss)in the hydraulic jump? 5.5.9 A 5-m wide rectangular channel with two reaches, each with a different slope, conveys 40 m3Is of water. The channel slope for the first reach is 0.015 and then a sudden change to a slope of 0.0005. The Manning's n for the channel is 0.Ql5. Does a hydraulic jump occur in the channel? If there is a hydraulic jump, where does it occur: on the first reach or the second reach? 5.5.10 A 5-m wide rectangular channel with two reaches, each with a different slope, conveys 80 m3/s of water. The channel slope for the first reach is 0.01 and then a sudden change to a slope of 0.001. The Manning's n for the channel is 0.Ql5. Does a hydraulic jump occur in the channel? If there is a hydraulic jump, where does it occur: on the first reach or the second reach?

5.5.11 A hydraulic jump is formed in a lO-ft wide channel just downstream ofa sluicegatefora discharge of450cfs.If thedepths of flow are 30 ft and 2 ft just upstream and downstream of the gate, respectively, determine the depth of flow downstream of the jump. What is the energy loss in the jump? What is the thrust [Fgate = y(M)] on the gate? Illustrate the thrust on the specific force and specific energy diagrams for this problem. 5.5.12 Consider a 40-ft widehorizontal rectangular channel with a discharge of 400cfs.Determine theinitialandsequent depths of a hydraulic jump, if the energy loss is 5 ft. 5.6.1 A rectangular, sharp-crested weir with end contraction is 1.6m long. How highshould it be placed in a channel to maintain an upstream depth of 2.5 m for 0.5 m3/s flow rate? 5.6.2 Fora sharp-crested suppressed weir (C w = 3.33) of length B = 8.0 ft, P = 2.0 ft, and H = 1.0 ft, determine the discharge over the weir. Neglect the velocity of approach head. 5.6.3 Rework problem 5.6.2 incorporating the velocity of approach head (equation (5.6.5a)). 5.6.4 Rework example 5.6.2 using equation (5.6.5b). 5.6.5 A rectangular sharp-crested weir with end contractions is 1.5m long. How highshould theweircrestbe placedin a channel to maintain an upstream depth of 2.5 m for 0.5 m3/s flow rate? 5.6.6 Determine theheadon a 60°V-notch weirfora discharge of 150 lis. Take Cd = 0.58. 5.6.7 The head on a 90° V-notch weir is 1.5 ft. Determine the discharge. 5.6.8 Determine the weircoefficient of a 90° V-notch weirfor a head of 180mm for a flow rate of 20 lis. 5.6.9 Determine the required head for a flow of 3.0 m3/s over a broad-crested weir 1.5 m highand 3 m long witha well-rounded upstream comer (C w = 1.67). 5.6.10 Water flows through a Parshall flume witha throatwidth of 4.0 ft at a depth of 7.5 ft. Determine the flow rate. 5.6.11 Water flows through a Parshall flume witha throatwidth of 5.0 ft at a depth of 3.4 ft. Determine the flow rate. 5.6.12 Thefollowing information wasobtained froma discharge measurement on a stream. Determine the discharge.

Distance from bank (ft)

o 12 32 52 72 92 100

Depth (ft)

Mean velocity (ft)

0.0 0.1 4.4 4.6 5.7 4.3 0.0

0.00 0.37 0.87 1.09 1.34 0.71 0.00

References

171

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