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. Thetopic 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, andmomentum equations for steady uniform flow inopen 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. Foruniformfiow thevelocity is uniform throughout thecontrol volume, sothat VI = Vl for thecontrolvolume in Figure 5.1.1. Hencefor a uniform flow, QI = Ql,A I = Al, VI = Vl, andYI = Yl. Thedepth of flow in uniform open-channel flow is also referred to as the normal depth. Figure 5.1.2shows an open-channel flow, in an aqueduct of the Central Arizona Project.

5.1.1 Energy

Theenergy equation foropen-channel flow canbederived inasimilar manner astheenergy equationforpipeflow (equation 4.2.13) using thecontrol volume approach. Insection 3.4,thegeneral energyequation forsteady fluid flow wasderived asequation (3.4.20). Considering open-channel flow inthecontrol 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 thedefinition of a. The shaftwork termis dWs/dt = 0 because no pump or turbine exists. Becausehydrostatic conditions prevail, the terms (p / p+eu +gz) can be taken outside the integral inequation (5.1.1):

(5.1.2)

113

114 Chapter 5 Hydraulic Processes: Open-Channel Flow

CD ---- ................ ..................... .........................................

Uniform flow

CD------ ................---- ................-------

I-

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

®

The terms JPVd4 = m are the mass rate of flow at sections 1 and 2 and the terms

J(pV3/2)d4 = (pV3/2)A = m ~2, so that equation (5.1.2) becomes

A

dH (P2 ).. Vi (P1 ).. Vfdi= p+eUz+ gz2 m+mT- p+eU1+ gz1 m-mT

Dividing through by mg and rearranging yields

P1 Vf P2 Vi [euz - eUj 1 dH]-+Zl+-=-+Z2+-+ --.---Y 2g Y 2g g mg dt

(5.1.3)

(5.1.4)

(5.1.5)

Similar to equation (4.2.10), the terms in square brackets represent the headloss hL due to viscousstress (friction). This energy lossdue to friction effects per unit weight of fluid is denoted as hL .

The energy equation for one-dimensional flow in an open-channel is

P1 Vf P2 Vi- +Zl +a1 - = - +Z2 +a2- +hLY 2g Y 2g

5.1 Steady Uniform Flow 115

Figure 5.1.2 Hayden-Rhodes Aqueduct, Central Arizona Project. (Courtesy of the U.S. BureauofReclamation (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 isexpressed 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 slopeSo = tan e, where e is the angle of inclination. If e is small « 100), then tan e~ sin e=(ZI - Z2)/L.


5.1.2 Momentum

Theforces acting uponthefluid control volume in Figure5.1.1 arefriction, gravity, andhydrostaticpressure. Thefriction force, Ff , is the product of the wallshearstress 'toand the area overwhich itacts, PL, where P is the wetted perimeter of the cross-section, thus

Ff = -'toPL (5.1.9)

where the negative signindicates that the friction force acts opposite to the direction of flow. Thegravity force Fg relates to the weight of the fluid yAL, where y is the specific weightof the fluid(weight perunitvolume). Thegravity force on thefluid 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 formof the integral momentum equation (3.5.6) in the x

direction is

(5.1.11)cs

or

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

yALsine - 'toPL = 0

For esmall, So ~ sin eso

yALSo = 'toPL

(5.1.12)

(5.1.13)

(5.1.14)

(5.1.15)

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

yALSo'to=--

PL

or

'to = yRSo = yRSj (5.1.16)

where R = AlP is the hydraulic radius. Equation (5.1.16) expresses theeffects offriction throughthe wallshearstress 'toas represented from a momentum viewpoint andthrough the rate of energydissipation Sfrepresented froman energy viewpoint. Consequently, equation (5.1.16) expresses alinkage between the momentum and energy principles.

Theshearstress 'to forfullyturbulent flow canbeexpressed asa function ofdensity, velocity, andresistance coefficient Cf as

(5.1.17)

5.1 Steady UniformFlow 117

Equating (5.1.16) and (5.1.17) yields

(5.1.18)

and solving for the velocity gives

v= (2g~ (5.1.19)yc;Defining C = V2g/Cf, then equation (5.1.19) can be simplified to the well-known Chezyequation

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)

(5.1.22)

wheren is the Manning roughness coefficient. Values of n are listed in Table5.1.1.Values of n fornatural 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/ 3S~/2n

which is valid for SI units and So = Sf.

Table 5.1.1 Values of theRoughness Coefficient n(Boldface figures are values generally recommended in design)

Type of channel anddescription

A. Closed conduits flowing partly fullA-I. Metal

a. Brass, smoothb. Steel

1. Lockbar and welded2. Riveted andspiral

c. Castiron1. Coated2. Uncoated

d. Wrought iron1. Black2. Galvanized

e. Corrugated metal1. Subdrain2. Storm drain

A -2. Nonmetala. Luciteb. Glassc. Cement

1. Neat, surface2. Mortar

Minimum Normal Maximum

0.009 0.010 0.013

0.010 0.012 0.0140.013 0.016 0.017

0.010 0.013 0.0140.011 0.014 0.016

0.012 0.014 0.0150.013 0.016 0.017

0.017 0.019 0.0210.021 0.024 0.030

0.008 0.009 0.0100.009 0.010 0.013

0.010 0.011 0.0130.011 0.013 0.015

(Continued)

Table 5.1.1 (Continued)

Type of channel and description Minimum Normal Maximum

d. Concrete1. Culvert, straight and free of debris 0.010 0.011 0.0132. Culvertwith bends, connections, and some debris 0.011 0.013 0.0143. Finished 0.011 0.012 0.0144. Sewerwith manholes, inlet, etc., straight 0.013 0.015 0.0175. Unfinished, steel form 0.012 0.013 0.0146. Unfinished, smooth woodform 0.012 0.014 0.0167. Unfinished, rough woodform 0.015 0.017 0.020

e. Wood1.Stave 0.010 0.012 0.0142. Laminated, treated 0.015 0.017 0.020

f Clay1. Common drainage title 0.011 0.013 0.0172. Vitrified sewer 0.011 0.014 0.0173. Vitrified sewerwith manholes, inlet, etc. 0.013 0.Ql5 0.0174. Vitrified subdrainwith openjoint 0.014 0.016 0.018

g. Brickwork1. Glazed 0.011 0.013 0.Ql52. Lined with cement mortar 0.012 0.015 0.017

h. Sanitary sewers coated with sewage slimes, 0.012 0.013 0.016with bendsand connections

i. Pavedinvert, sewer, smoothbottom 0.016 0.019 0.020j. Rubblemasonry, cemented 0.018 0.025 0.030

B. Lined or built-up channelsB-1. Metal

a. Smooth steel surface1. Unpainted 0.011 0.012 0.0142. Painted 0.012 0.013 0.017

b. Corrugated 0.021 0.025 0.030B-2. Nonmetal

a. Cement1. Neat, surface 0.010 0.011 0.0132. Mortar 0.011 0.013 0.Ql5

b. Wood1. Planed, untreated 0.010 0.012 0.0142. Planed, creosoted 0.011 0.012 0.Ql53. Unplaned 0.011 0.013 0.0154. Plank with battens 0.012 0.015 0.0185. Lined with roofing paper 0.010 0.014 0.017

c. Concrete1. Trowel finish 0.011 0.013 0.0152. Float finish 0.013 0.015 0.0163. Finished, with gravelon bottom 0.015 0.017 0.0204. Unfinished 0.014 0.017 0.0205. Gunite, good section 0.016 0.019 0.0236. Gunite, wavy section 0.018 0.022 0.0257. On good excavated rock 0.017 0.0208. On irregular excavated rock 0.022 0.027

d. Concrete bottomfloat finished with sides of1. Dressed stone in mortar 0.015 0.017 0.0202. Random stone in mortar 0.Q17 0.020 0.024

118



3. Cement rubble masonry, plastered 0.016 0.020 0.0244. Cement rubble masonry 0.020 0.Q25 0.0305. Dryrubble or riprap 0.020 0.030 0.035

e. Gravel bottom with sides ofI. Formed concrete 0.017 0.020 0.Q252. Random stone in mortar 0.020 0.023 0.0263. Dry rubble or riprap 0.023 0.033 0.036

f BrickI. Glazed 0.011 0.013 0.0152. In cement mortar 0.012 0.015 0.018

g. MasonryI. Cemented rubble 0.017 0.025 0.0302. Dryrubble 0.023 0.032 0.035

h. Dressed ashlar 0.013 0.015 0.017i. Asphalt

I. Smooth 0.013 0.0132. Rough 0.016 0.016

j. Vegetal lining 0.030 0.500C. Excavated or dredged

a. Earth, straight and uniformI. Clean, recently completed 0.016 0.018 0.0202. Clean, after weathering 0.018 0.022 0.Q25

3. Gravel, uniform section, clean 0.022 0.025 0.0304. Withshortgrass, few weeds 0.022 0.027 0.033

b. Earth, winding and sluggishI. No vegetation 0.023 0.Q25 0.0302. Grass, someweeds 0.Q25 0.030 0.0333. Dense weeds or aquatic plants in deepchannels 0.030 0.035 0.0404. Earthbottom and rubble sides 0.Q28 0.030 0.0355. Stony bottom and weedy banks 0.025 0.035 0.0406. Cobble bottom and cleansides 0.030 0.040 0.050

c. Dragline-excavated or dredgedI. No vegetation 0.Q25 0.028 0.0332. Lightbrush on banks 0.035 0.050 0.060

d. RockcutsI. Smooth and uniform 0.Q25 0.035 0.0402. Jagged and irregular 0.035 0.040 0.050

c. Channels not maintained, weeds and brush uncutI. Dense weeds, high as flow depth 0.050 0.080 0.1202. Cleanbottom, brush on sides 0.040 0.050 0.0803. Same, highest stage of flow 0.045 0.070 0.1104. Dense brush, high stage 0.080 0.100 0.140

D. Natural streamsD-l. Minor streams (top width at flood stage <100 ft)

a. Streams on plain1. Clean, straight, full stage, no riftsor deep pools 0.025 0.030 0.0332. Sameas above, but more stones and weeds 0.030 0.035 0.0403. Clean, winding, some pools and shoals 0.033 0.040 0.0454. Same as above, but some weeds and stones 0.035 0.045 0.0505. Sameas above, lowerstages, more ineffective 0.040 0.048 0.055

slopes and sections(Continued)

119

120 Chapter 5 HydraulicProcesses: Open-Channel Flow



0.030 0.040 0.0500.040 0.050 0.070

0.025 0.030 0.0350.030 0.035 0.050

0.020 0.030 0.0400.025 0.035 0.0450.030 0.040 0.050

0.035 0.050 0.0700.035 0.050 0.0600.040 0.060 0.0800.045 0.070 0.1100.070 0.100 0.160

0.110 0.150 0.2000.030 0.040 0.0500.050 0.060 0.0800.080 0.100 0.120

0.100 0.120 0.100

6. Same as 4, but more stones7. Sluggish reaches, weedy, deep pools8. Very weedy reaches, deeppools, or floodways

with heavy stand of timber and underbrushb. Mountain streams, no vegetation in channel,

banks usually steep, trees and brush along bankssubmerged at highstages1. Bottom: gravels, cobbles, and few boulders2. Bottom: cobbles with large boulders

D-2. Flood plainsa. Pasture, no brush

1. Short grass2. High grass

b. Cultivated areas1. No crop2. Mature row crops3. Mature field crops

c. Brush1. Scattered brush, heavy weeds2. Lightbrush and trees, in winter3. Lightbrush and trees, in summer4. Medium to dense brush, in winter5. Medium to dense brush, in summer

d. Trees1. Dense willows, summer, straight2. Cleared landwith tree stumps, no sprouts3. Same as above, but with heavy growth of sprouts4. Heavy stand of timber, a few down trees, little

undergrowth, flood stage below branches5. Same as above, but with flood stage reaching branches

D-3. Major streams (top width at flood stage> 100ft).Then value is less than that for minor streams of similardescription, because banks offerless effective resistance.a. Regular section with no boulders or brushb. Irregular and rough section

Source: Chow (1959).

Manning's equation in SI units can also be expressedas

Q= ! AR2/3S6/2

n

0.0450.0500.075

0.Q250.035

0.0500.0700.100

0.0600.0800.150

0.0600.100

(5.1.23)

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

V = 1.49R2/3S6/

2 (5.1.24)n

and equation (5.1.23) can be written as

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

n(5.1.25)

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


Table 5.1.2 Geometric Functions for Channel Elements

Section: Rectangle Trapezoid Triangle

r-B-;1 r-B--jI ~ II U 1z L.l

f-Bw-J f-Bw-l

AreaA BwY (Bw+zy)y zl1 .g(8-sm 8)d~

Wetted Bw+2y Bw+2y~ 2Y\h +z2118do

perimeter P

HydraulicBwY (Bw+zy)y zy ~ (1- sin 8)d

radius R Bw+2y Bw+2YVl +z2 2V1 +z2 4 8 0

Top width B[sin(~) ]do

Bw Bw+2zy 2zyor

(Bw+ 2zy)(SBw+ 6yVl + z2)2Jy(do-Y)

4(2 sin 8 + 38 - secas 6)

2dR 1M SBw+6y \-4z/V1 +Z2 8 3do8(8 - sin e)sin (8/2)

3Rdy + Ady 3y(Bw+2y) 3y(Bw+ zy)(Bw+2yVl + z2)3y

where 8 = 2 cas - 1 ( 1 - ~)

Source: Chow (1959) (with additions).

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

n6JRSf ? 1.9 x 10 -13 (R in feet) (S.1.26a)

n6JRSf ? 1.1 x 10 -13 (R in meters) (S.1.26b)

EXAMPLE 5.1.1 An 8-ft widerectangular channel witha bed slopeof 0.0004 ftlft has a depthof flow of 2 ft. Assumingsteady uniform flow, determine the discharge in the channel. The Manning roughness coefficient isn=0.015.

SOLUTION Fromequation (5.1.25), the discharge is

Q = 1.49AR2/3S~/2n

= 1.49 (8)(2)[ (8)(2) ]2/3(00004)1/20.015 8 + 2(2) .

= 38.5 ft3/s


EXAMPLE 5.1.2

SOLUTION

EXAMPLE 5.1.3

SOLUTION

Solve example 5.1.1 using SI units.

Thechannel width is 2.438 m,with a depthof flow of0.610 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)

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;Q--S

J - n 0 (Bw

+ 2y;) 2/3

5/3 (8 )5/3Q. = 1.49 (0.0004)1/2 (8y;) = 1.987 Y;

J 0.Q15 (8+2y;) 2/3 (8+2yi/3

Using a numerical method such as Newton's method (see Appendix A), the normal depthis 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 adecrease in the wetted perimeter. Consequently, the channel section with the smallest wettedperimeterfor a givenchannel section area will havemaximumconveyance, referredto as the besthydraulic section or the cross-section of greatest hydraulic efficiency. Table 5.1.3 presents thegeometricelementsof the besthydraulicsectionsfor six cross-section shapes. These sectionsmaynot alwaysbe practical because of difficulties in construction and use of material. The concept of

Table 5.1.3 BestHydraulic Sections

Wetted Hydraulic HydraulicCross-section Area perimeter radius Top width depth

A p R T D

Trapezoid, halfof a hexagon 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

Parabola, T = 2V2y 13/21 'l3/2y Y2Y 2V2y %Y

Hydrostatic catenary 1.39586i 2.9836y 0.46784y 1.917532y 0.72795y


EXAMPLE 5.1.4

SOLUTION


besthydraulic section is onlyfornonerodible channels. Eventhough thebesthydraulic section givesthe minimum area for a given discharge, it may not necessarily have the minimum excavation.

Determine the cross-section of greatest hydraulic efficiency for a trapezoidal channel if the designdischarge is 10.0 m3/sec, thechannel slope is 0.00052, and Manning's n = 0.025.

From Table 5.1.3, thehydraulic radius should be R = y/2, so that thewidth B and area A are

2V3y 1B =-3- = 1.155y (becauseB = Jpfor halfof a hexagon)

A = v'3l = 1.732l

Manning's equation (5.1.23) is used to determine thedepth:

1 1 ( ) 2/3Q= -;:,AR2/3S~/2 = 0.025 (1.732y2) ~ (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 riverreach of length L1x with known cross-sectional areas of flow at the upstream, Aw and downstream, Ad, endsof the reach.Theuse of high-water marksfrom a flood and a survey of the crosssections allow computation of the cross-sectional areasof flow. Manning's equation (5.1.25) canbe expressed as

Q=KVSo (5.1.27)

1.49 2/3whereK is the conveyance factorexpressed as K = --AR . Conveyance is a measure of then

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

The discharge is then expressed as

K = VKuKd (5.1.28)

(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 endsof 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)

(5.2.1)


Thedifference in water surface elevations is referred to as thefall. Thek is a factor to account foracontraction and expansion of a reach. For a contracting reach Vu < Vd so k = 1.0 and for anexpanding reach (Vu> Vd) sok = 0.5. Thefirst approximation would compute thedischarge usingQ= K/Sf with the friction slope computed ignoring the velocity heads. Using the first approximation of Q, theupstream anddownstream velocity heads arecomputed forthenextapproximationof the friction slope, which is used to compute the second approximation of the discharge. Theprocedure continues computing the new friction slope using the last discharge approximation tocompute the new discharge. This process continues until the discharges approximations do notchange significantly.

5.2 SPECIFIC ENERGY, MOMENTUM, AND SPECIFIC FORCE

5.2.1 Specific Energy

The total heador 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)andthepressure distribution ishydrostatic (i.e., p = yy).Using thechannel bottom as thedatum (i.e., z = 0) thendefine the totalheadabove the channel bottom as the specific energy

V2

E=y+2g

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

Q2E=Y+2gA2 (5.2.2)

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

y

ConstantQ

Q2E=y+ --2

2gA

Figure5.2.1 Specific energy.

(5.2.3)

5.2 Specific Energy, Momentum, and Specific Force 125

y

QlE=y+

2gA2

Figure 5.2.2 Specific energy showing subcritical and supercritical flow ranges.

Criticalflow occurswhenthe specific energy is minimum fora givendischarge (i.e.,dEIdy = 0),

so that

dE=l_~dA=Ody gA3 dy

Referring toFigure5.2.1,thetop-width is defined asT = dAIdysoequation (5.2.3)canbeexpressedas

(5.2.4)

or

(5.2.5)

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

(5.2.6)

(5.2.7)

or

V~ Ae

g t:

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

The hydraulic depth is defined as D = AIT so equation (5.2.7) becomes

V~ = 1gDc

(5.2.8)

(5.2.9)


or

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

{

< 1subcritical flow

F, =~ = 1critical flowgD >1supercritical flow

(5.2.10)

(5.2.11)

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

£1 = £2 or

V2 V21 2

Y1 + -=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)

Ifwelet qbe the flow rate perunitwidthof channelfor a rectangularchannel, i.e., q = Q/BwhereT = 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

EXAMPLE 5.2.1

SOLUTION

_ (q2) 1/3Ye -

g

Compute thecritical depth for thechannel 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)2/3y~/2 = 8/8 or Yc = 8/8 = 1.69ft

Alternatively, using equation (5.2.14) yields

((100/ 8)2) 1/3

Ye = = 1.69ftg

(5.2.14)

EXAMPLE 5.2.2 Fora rectangular channel of 20ft width, construct a family of specific energy curves forQ = 0,50, 100,and300cfs. Draw thelocus of thecritical depth points on these curves. Foreach flow rate, what is theminimum specific energy found from these curves?

SOLUTION

5.2 Specific Energy, Momentum, and SpecificForce 127

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:

Q = 100ds:

Q=300ds:

Emin = 0.868

Emin = 1.379

Emin = 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 0.50 0.89 2.05 14.860.6 0.60 0.87 1.68 10.570.8 0.80 0.95 1.41 6.411.0 1.00 1.10 1.39 4.591.2 1.20 1.27 1.47 3.691.4 1.40 1.45 1.60 3.231.6 1.60 1.64 1.75 3.001.8 1.80 1.83 1.92 2.912.0 2.00 2.02 2.10 2.902.2 2.20 2.22 2.28 2.942.4 2.40 2.42 2.47 3.022.6 2.60 2.61 2.66 3.132.8 2.80 2.81 2.85 3.263.0 3.00 3.01 3.04 3.403.5 3.50 3.51 3.53 3.794.0 4.00 4.01 4.02 4.224.5 4.50 4.50 4.52 4.685.0 5.00 5.00 5.02 5.14


y(ft)

7

7 E(ft)

EXAMPLE 5.2.3

SOLUTION

Figure 5.2.3 Specific energy curves for example 5.2.2.

Arectangular channel 2mwide hasaflow of2.4m3/s atadepth of1.0m.Determine whether critical depthoccurs at (a)a section where a hump of ilz = 20 em high is installedacrossthe channelbed, (b) a sidewall constriction (with no humps)reducingthe channel width to 1.7 m, and (c) both the hump andside wall constrictions combined. Neglect headlosses of the hump and constriction caused byfriction, 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 +L1zctit orL1zctit = E - Emin, where E is the specific energy of thechannel flow andE rnin is the minimum specificenergy, which isatcritical depth bydefinition. If<1zctit :::: L1z, thencritical depthwilloccur. Using equation(5.2.2) yields

Q2 q2E = y+ 2gA2 = y+ 2gy2

which can be solved for q:

Differentiating this equation with respect to y because maximum q and minimum E are equivalent(see Figure 5.2.4) yields

dq =!!.- [ /2g(y2E _ y3)] = 0dy dy V

2 3Yc = "3 Emin or Emin = :??cTocompute specific energy, use

q2 (2.4/2.0)2E =Y+ -22 =1.0+ 2 =1.073 m

gy 2(9.81)(1)


Y

2Ye=gEmin

qconstant

Emin E

Y

Subcritical

Supercritical

Econstant

q

5.2.2 Momentum

Figure 5.2.4 Specific energy curve andY versus q for constant E.

Next compute Emin using Emin = 3/2Ye, where Ye = (l /g) 1/3 (equation (5.2.14)):

[

2] 1/3= (2.4/2.0) = 0 28Ye 9.81 .5 m

So Emin = 3/2(0.528 m) = 0.792 m. Then ~Zerit = E - Emin = 1.073 - 0.792 = 0.281 m. In thiscaseM = 20 em = 20/100 m = 0.2 m<Merit = 0.281 m. Therefore, Ye does not occurat the hump.

(b) The critical depth at the side wall constriction is

= [(2.4/1.7)2] 1/: 0 588Ye 9.81 . m

Thus Emin = (3/2)Ye = (3/2)(0.588) = 0.882 m. E is computed above as E = 1.073 m. BecauseEmin = 0.882 m<E = 1.073 m,critical depthdoes notoccurat theconstriction. Remember thatenergylosses arenegligible sothatthespecific energy intheconstriction andupstream oftheconstriction mustbeequal. For critical flow to occur, the constriction width can be computed as follows: Emin =E = 1.073 m = (3/2)Ye, so that Ye = 0.715 m. Then using equation (5.2.14), 0.715 = [(2.4/Be)2/

]1/39.81 and Be = 1.267 m.

(c) With both the hump and the side wall constriction,Yc is 0.588 m, so Emin = 0.882 m. ThenMerit = E - Emin = 1.073 - 0.882 = 0.191 m.

Because ~z = 20/100 m = 0.20 m>Merit = 0.191 m, critical depth will occur at the hump with aconstriction.

Applyingthemomentumprinciple(equation3.5.6) to a shorthorizontalreachofchannelwithsteadyflow (Figure 5.2.5), we get

(2.5.6)

where

(5.2.15)


Control volume

w--------~Fj ----------

9 sin9-SoZ1 rc--------dx Z2

--------------------------- --Dawm

Figure5.2.5 Application of momentum principle.

The momentum entering from the upstream is - P~IVIQand the momentum leaving the controlvolume is P~2V2Q, where ~ is called the momentum correction factor that accounts for thenonuniformity of velocity (equation 3.7.8), so that

(5.2.16)

The forces are(5.2.17)

The hydrostatic forces are

(5.2.18)

and

(5.2.19)

where YI andY2 are the distances to thecentroid. The gravity force Fg due to the weight Wof thewater is W sin e= pgAdx sin e, where W = pgAdx. Because the channel slope is small,So ~ sin e, and the force due to gravity is

r, = pgAdxSo (5.2.20)

where A = (AI +A2)/2 is the average cross-sectional area of flow. The external force due tofriction created by shearbetween thechannel bottom andsidesof thecontrol volume is - 'toPdxwhere 'to is the bed shear stress and P is the wetted perimeter. From equation (5.1.6),'to = yRSf = pg(A/P)Sf. So the friction force is then

For our purposes here we will continue to use Fg and F;.Substituting equations (5.2.15) through (5.2.21) into (3.5.6) gives

yAIYl - yA2Y2 +Wsin e- F; = - P~IVIQ+P~2V2Q

(5.2.21 )

(5.2.22)

which is the momentum equation for steadystate open-channel flow.It should beemphasized thatin theenergy equation theFf(lossduetofriction) is a measure of the

internal energy dissipated in the entire mass of water in the control volume, whereas F; in the

5.2.3 Specific Force

5.2 Specific Energy, Momentum, and Specific Force 131

momentum equation measures the losses dueto external forces exerted on the waterby the wettedperimeter of thecontrolvolume. Ignoring thesmalldifference between theenergy coefficient a andthe momentum coefficient ~ in gradually varied flow, the internal energy losses are practicallyidentical with the losses due to external forces (Chow, 1959). For uniform flow, Fg = Fj.

Application of the energyand momentum principles in open-channel flow can be confusing atfirst. It is important to understand the basic differences, even though the two principles mayproduce identical or very similar results. Keep in mind that energy is a scalar quantity andmomentum is a vector quantity and that energy considers internal losses in the energyequationand momentum considers external resistance in the momentum equation. The energyprinciple is simpler and clearer than the momentum principle; however, the momentumprinciple has certain advantages in application to problems involving high internal-energychanges, such as the hydraulic jump (Chow, 1959), which is discussedin section 5.5 on rapidlyvaried flow.

Forashorthorizontal reach(control volume) withe= 0 andthegravity forceFg = W sine= 0, theexternal force of friction Fj, can be neglected so F; = 0 andFg = O. Alsoassuming ~I = ~2' themomentum equation (5.2.22) reduces to

(5.2.23)

Substituting VI = Q/A I and V2 = Q/A2, dividing through by 'Y and substituting l/g = ph andthen rearranging yields

The specific force F (Figure 5.2.6) is defined as

F=Q2 +AygA

(5.2.24)

(5.2.25)

Y

----T----+1

1-- -----YlC Y

j2

__1__ _ ~ ---.F

Figure 5.2.6 Specific force curves.


which has units of ft3 or m3. The minimum value of the specific forcewith respectto the depthis

determined using

which results in

dF = _ TQ2 +A = 0dy gA2

(5.2.26)

(5.2.27)

Referto Chow (1959) or Chaudhry (1993) for theproofandfurther explanation of d (Ay) /dy = A.Equation (5.2.27) reduces to - V2/ g +A/T = 0 where the hydraulic depth D = A/T, so

V2 V2

-=D or -=1 (5.2.28)g gD

which we have already shown is the criterion for critical flow (equation (5.2.9) or (5.2.10».Therefore, at critical flow the specific force is a minimum for a given discharge.

Summarizing, critical flow is characterized by the following conditions:

• Specific energy is minimum for a given discharge.• Specific force is minimum for a givendischarge.• Velocity head is equal to half the hydraulic depth.• Froude number is equal to unity.

Two additional conditions that are not proven here are (Chow, 1959):

• The discharge is maximum for a givenspecific energy.• Thevelocity offlow inachannel ofsmall slopewithuniform velocity distribution isequaltothe

celerity of smallgravity waves in shallow watercaused by local disturbances.

Whenflow is ator nearthecritical state, minorchanges inspecific energy nearcritical flow causemajorchanges in depth (see Figures 5.2.1 or 5.2.2), causing the flow to be unstable. Figure5.2.7illustrates examples of locations of critical flow.

(a)

--(c)

l nnft Hydrauic jump

------- Yc

I(d)

Figure 5.2.7 Example locations ofcritical flow. (a) Critical depth near free overfal!; (b) Change ingradeof channel bottom; (c) Flow over a broad-crested weir; (d) Flow through hydraulic jump.

·EXAMPLE 5.2.4

SOLUTION


Compute the specific force curves for the channel and flow rates usedin example 5.2.2.

Thespecific force values arecomputed using equation (5.2.25) withthevalues presented in Table 5.2.2.The curves are plotted in Figure 5.2.8.

Table 5.2.2 Computed Specific Force Curve Values for Example 5.2.4

Specific force, F (fr')

Depth, y (ft) Q=O Q = 50 Q = 100 Q = 300

0.1 0.10 38.92 155.38 1397.620.2 0.40 19.81 78.04 699.160.4 1.60 11.30 40.42 350.980.6 3.60 10.07 29.48 236.520.8 6.40 11.25 25.81 181.091.0 10.00 13.88 25.53 149.751.2 14.40 17.63 27.34 130.861.4 19.60 22.37 30.69 119.421.6 25.60 28.03 35.30 112.941.8 32.40 34.56 41.03 110.042.0 40.00 41.94 47.76 109.882.2 48.40 50.16 55.46 111.922.4 57.60 59.22 64.07 115.832.6 67.60 69.09 73.57 121.35

2.8 78.40 79.79 83.95 128.31

3.0 90.00 91.29 95.18 136.58

3.5 122.50 123.61 126.94 162.43

4.0 160.00 160.97 163.88 194.94

4.5 202.50 203.36 205.95 233.56

5.0 250.00 250.78 253.11 277.95

y(ft)

4.5

4

3.5

3

2.5

2

1.5

20 40 60 80 100 120 140 160 180 F(tt3)

Figure 5.2.8 Specific force curves.


5.3 STEADY, GRADUALLY VARIED FLOW

5.3.1 Gradually Varied Flow Equations

Several types of open-channel flow problems can be solved in hydraulic engineering practiceusing the concepts of nonuniform flow. The first to be discussed are gradually varied flowproblems in which the change in the water surface profile is small enough that it is possible tointegrate the relevant differential equation from one section to an adjacent section for thechange in depth or change in water surface elevation. Consider the energy equation (5.1.6)previously derived for nonuniform flow (Figure 5.1.1) using the control volume approach (withal = a2 = 1):

(5.1.6)

Because hi. = SfL = Sftu lettingAy = Y2 - YI and liz = Zl - Z2 = SoAx, then equation (5.1.6)can be expressed as

(5.3.1)

Rearranging yields

and then dividing through by tu results in

Ay = So _ Sf_ (Vi _Vl) 2tu 2g 2g tu

Taking the limit as tu~ 0, we get

lim (AY) = dyLlx-->O tu dx

and

lim (Vi _Vl) (2-) _~ (V2)

Llx-->O 2g 2g tu - dx 2g

Substituting these into equation (5.3.3) and rearranging yields

(5.3.2)

(5.3.3)

(5.3.4)

(5.3.5)

(5.3.6)dy d (V2

)dx + dx 2g = So - Sf

[(V2)]d V2 d 2 d

The second term dx (2) can be expressed as --g- 2, so that equation (5.3.6) can besimplified to g dy dx

dy [1 + d(~;)] =S-Sdx dy 0 'f (5.3.7)

5.3 Steady, Gradually Varied Flow 135

'-dA= Tdy

.J.------T-----+l,01dy

T 1,,-----"_'I(a) (b)

Figure5.3.1 Definition of top width (T = dAldy). (a) Natural channel; (b) Rectangular channel.

ordydx

So -Sf(5.3.8)

Equations (5.3.7) and(5.3.8) aretwoexpressions of thedifferential equation for gradually variedflow. Equation (5.3.8) can also be expressed in terms of the Froude number. First observe that

(5.3.9)

By definition, the incremental increase in cross-sectional area of flow dA, due to an incremental increase in thedepthdy, is dA = Tdy, where Tis the top widthof flow (seeFigure5.3.1).AlsoAIT = D, which is the hydraulic depth. Equation (5.3.9) can nowbe expressed as

(5.3.l0a)

= _F2r (5.3.l0b)

where F; = ~ = ~. Substituting equation (5.3.lOb) into (5.3.8) and simplifying, we findygD AygD

that the gradually varied flow equation in terms of the Froude number is

dy

dxSo -Sfl-F2r

(5.3.11)

EXAMPLE 5.3.1 Consider a vertical sluice gate in a wide rectangular channel (R = AlP = ByI (B+2Y) ~ y becauseB » 2y). The flow downstream of a sluice gate is basically a jet that possesses a venacontracta(seeFigure 5.3.2). Thedistance from thesluice gatetothevenacontracta asaruleisapproximated asthesameasthesluice gateopening (Chow, 1959). Thecoefficients ofcontraction forvertical sluicegates are approximately 0.6,ranging from 0.598 to 0.611 (Henderson, 1966). Theobjective of thisproblem is to determine the distance from the venacontracta to a pointb downstream where thedepthofflow isknown tobe0.5mdeep. Thedepth offlow at thevenacontracta is0.457 mfora flowrate of 4.646m3/s per meter of width. Thechannel bed slope is 0.0003 andManning's roughnessfactor is n = 0.020.

O.6h


T-;~__~h

h L----+-t

SOLUTION

Figure 5.3.2 Flow downstream of a sluice gate in a wide rectangular channel.

To compute the distance, Axfrom Ya to Yb, thegradually varied flow equation (5.3.1) can be used,

where Lly = Y2 - Yl. Solving for Ax, weget

Thefriction slope is computed using Manning's equation (5.1.22) with average values of thehydraulicradius

_ ~ 2/3 1/2Vave - Rave Sf

n

so

S = [nVave] 2'f R2/3

ave

Let us use the following values for thisexample:

Now we get

Location

ab

Y (m)

0.4570.500

R = Y (m)

0.4570.500

V (mlS)

10.179.292

V2/2g (m)

5.274.40

_ 10.17+9.292 _ 9 73 rnfVave - 2 -. s

_ 0.457+0.500 _ 0 479Rave - 2 -. m

S = [0.020(9.73)]2 = 0.101rnfm'f 0.4792/3

5.3 Steady, Gradually Varied Flow 137

The distance from a to b, Ax, is

_ (0.500 - 0.457) +(4.40 - 5.27) _ - 0.827 _ 821Ax - 0.0003 _ 0.101 - _ 0.101 -. m

The distance from the sluice gate to b is (0~~~7) +8.21 == 8.97 m.

5.3.2 Water Surface Profile Classification

Channelbedslopesmaybeclassified asmild(M),steep(S),critical(C),horizontal (H) (So = 0),andadverse(A) (So <0). Todefinethevarious typesof slopesfor the mild, steep,andcritical slopes,thenormal depth Yn and critical depth Yc are used:

Mild: Yn>Yc Yn 1 (S.3.12a)or ->Yc

Steep: Yn <» or Yn 1 (5.3.12b)-<Yc

Critical: Yn = Yc or Yn = 1 (S.3.12c)Yc

Thehorizontal andadverseslopesarespecialcasesbecausethenormaldepthdoesnotexistfor them.Table5.3.1lists the typesandcharacteristics of the varioustypesof profiles andFigure5.3.3 showsthe classification of gradually varied flow profiles.

Table 5.3.1 Types of Flow Profiles in Prismatic Channels

ChannelDesignation Relation of Y to Yn and Yc

slope Zone I Zone 2 Zone 3 Zone I Zone 2 Zone3 General type of curve Type of flow

Horizontal None Y > Yn > Yc None NoneSo == 0 H2 Yn >Y >Yc Drawdown Subcritical

H3 Yn > Yc > Y Backwater Supercritical

Mild MI Y > Yn > Yc Backwater SubcriticalO<So «s, M2 Yn >Y >Yc Drawdown Subcritical

M3 Yn > Yc > Y Backwater Supercritical

Critical CI Y > Yc Yn Backwater SubcriticalSo == Sc>O C2 Yn ==y == Yc Parallel to channel bottom Uniform-critical

C3 Yc Yn > Y Backwater Supercritical

Steep SI Y > Yc > Yn Backwater SubcriticalSo »s. >0 S2 Yc >Y > Yn Drawdown Supercritical

S3 Yc > Yn > Y Backwater Supercritical

Adverse None Y> (Yn)* > Yc None NoneSo < 0 A2 (Yn)* >Y >Yc Drawdown Subcritical

A3 (Yn)* > Yc > Y Backwater Supercritical

*Yn in parentheses isassumed a positive value.


I

III

II

I

II

I

II

II

I

II

I

IIII

III

III

IIIII

III

III

II

I

II

I

II

II

I

II

I

IIII

I

I1

1

¥j

\

-- \

{In)

FIll"'" 5.3.3 1'\OW "",Jiles (fiOmChOW (\9

59)).

dyldx "'+

--_ ...'---~~- "':'--

5.3 Steady, Gradually VariedFlow 139

The three zones for mild slopes are definedas

Zone 1:

Zone 2:

Zone 3:

Y>Yn>Ye

Yn>Y>Yc

Yn>Ye>Y

The energygrade line, water surface, and channel bottomare all parallelfor uniformflow, i.e.,Sf =So = slopeofwatersurfacewheny = Yn.FromManning'sequationforagivendischarge,Sf<SoifY>Yn'

Now consider the qualitative characteristics using the three zones.

Zone 1 (Ml profile):

Zone2 (M2profile):

Zone3 (M3 profile):

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--2 =- = +

x -F, +thenY increases withx so thatY --+ Yn

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--2 = - = -

x -F, +thenY decreases with x so thatY --+ Yc

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--2 = - = +x -F,

thenY increases with x so thatY --+ Yc

.EXAMPLE 5.3.2

SOLUTION

This analysis can be made of the other profiles. The results are summarized in Table 5.3.1.

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

In example 5.1.3, thenormal depth is computed asYn = 3.97ft, andinexample 5.2.1, thecriticaldepthiscomputed as Ye = 1.69ft. Because Yn > Ye' this is a mildchannel bed slope.

For a flow depth of 5.0 ft, 5.0 > Yn > Ye, so that an Ml profile with a backwater curveexists (refer toTable 5.3.1 and Figure 5.3.3). The flow is subcritical,

Fora flow depth of 2.0 ft, Yn > 2.0> Ye' so that an M2profile witha drawdown curveexists. The flow issubcritical.

Fora flow depthof 1.0ft, Yn > Ye > 1.0,so that an M3profile witha backwater curveexists. The flow issupercritical.

Prismatic Channels with Changes in Slope

Consider a channel that changes slope from a mild slope to a steep slope. The critical depth is thesame for each slope;however, the normal depth changes, for the upstream mild slope,Ynl >Ye, andfor the downstreamsteep slope,Yn2 <Ye' The only control is the critical depth at the break in slopeswhere flow transitions from a subcritical flow to a supercritical flow. The flow profiles are an M2


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

profile for theupstream reach andan S2profile for thedownstream reachas shown in Figure 5.3.4.Thespecific energy forthereaches, En l andEn2, andthespecific energy forcriticalconditions canbecomputed. Achannel thatchanges slope from asteep slope toa mildslope ismorecomplicated inthata hydraulic jumpforms. Thisjumpcould form ontheupper 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 energyequation. Thisprocedure canbeextended tocompute reachlengths forspecified depths at eachendof a reach. Computations areperformed stepby stepfrom oneendof thechannel reachby reach tothe otherend. Thisprocedure is calledthe direct stepmethod and is applicable only to prismaticchannels.

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)

(5.3.14)

EXAMPLE

Thedirectstepmethod isbasedonthisequation. Manning's equation is usedtocompute thefrictionslope at the upstream and downstream endsof each reachfor the specified depths using

n2 y2Sf = -2.-22-R-2-j3

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

Atrapezoidal channel has theflowing characteristics: slope = 0.0016; bottom width = 20ft; and sideslopes = 1vertical to2horizontal (z = 2).Water atadownstream location isanembankment where thewater depth is5.0ftjustupstream oftheembankment and the discharge is400cfs.Determine thedistanceupstream to where the flow depth is 4.60 ft. (Adapted from Chow, 1959.) (Assume a = 1.10.)

SOLUTION

5.4 Gradually Varied Flow for Natural Channels 141

Thecritical depth isYc = 2.22 ftandthenormal depth isYn = 3.36 ft, sotheflow issubcritical. Because thedepth of5.0ft isgreater than thenormal depth of3.36 ft,thisisanMl profile. Then thedepth will decreaseproceeding 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 upstreamto a depth of 5.0to 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,andatY = 4.80ft then A = 142.1 ftz, R = 3.43 ft, V = 2.82 ft/sec,Sf = 0.00043, andE = 4.936 ft. Using equation 5.3.14,

- 1where Sf = 2: (0.00037+0.00043) = 0.00040,

Ez -E1 (5.123-4.938)then~x = So _ Sf 0.0016_ 0.00040 = 156ft.

Second reach: Aty = 4.60ft,A = 134.3 ftZ,R = 3.3lft, V = 2.98 ftIsec,Sf= 0.00051,andE = 4.752ft.- 1Sf = 2: (0.00043 +0.00051) = 0.00047, and~x = 163 ft.

So thedistance upstream to a depth of 4.60 ft is 156 + 163 = 319ft. Theprocedure canbe continuedupstream to where the normal depth occurs.

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 beexpressed in termsof thewatersurfaceelevation for application to naturalchannelsby consideringw = z+y where w is the water surfaceelevation abovea datumsuch as mean sea level.The totalenergy H at a sectionis

V2 V2

H = z+y+a- = w+a- (5.4.1)2g 2g

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

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

dH-= -Sf-Sedx

(5.4.2)

(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 Sfcan be expressed using Manning's equation (5.1.23) or (5.1.25):

whereK is defined as the conveyance in SI units

K = !AR2/3

n

(5.4.5)

(5.4.6a)


or in U.S. customary units as

(5.4.6b)

(5.4.8)

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

Q2 Q2[1 1]Sf = K2 =T Kr + Ki (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 bedetermined using an average conveyance, i.e., X = (Kl+K2) /2 and Q= XS}/2; then

Q2Sf=

X2

The contraction-expansion loss term S; can be expressed for a contraction loss as

(V2) (V

2V

2)for d a- = a2~ -a1.-l >0

2g 2g 2g (5.4.9)

and for an expansion loss as

(5.4.10)

Thegradually variedflow equation for anatural channel isdefined bysubstituting equation (5.4.7)into equation (5.4.4):

or

_ Q2 [~ +~] _s, = dw + !!..- (a V2)

2 Kr Ki dx dx 2g(5.4.11a)

(5.4.11b)

Rearranging yields

(5.4.12)

EXAMPLE 5.4.1

SOLUTION

Derive anexpression forthechange in water depth asa 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):


with

Then

dydx

So -Sf

(1- Q2T)gA3

EXAMPLE 5.4.2

SOLUTION

Todetermine a gradually varied flow profile, this equation is integrated.

Forariversection witha subcritical discharge of6500ft3/S, thewatersurface elevation atthedownstreamsection is5710.5 ft witha velocity headon.72 ft.Thenextsection is500ft upstream witha velocity headof 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 thewater 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 termis

Q2[~ ~]Lix=65002[ 1 + 1 ]500=2.79ft2 Ki + KI 2 76,1402 104,3002

Because ~[a(v2 /2g)] > 0, a contraction exists for flow fromcross-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)


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 acompound channel, sothatthevelocity head fortheentire channel isu(V2/2g) where Vis themeanvelocity in the compound channel. Equation (5.4.13) can be expressed as

N

EV/A;;=1

U;:::j--N-

V3 E A;;=1

(5.4.14)

where N is thenumber of sections (subareas) of thechannel (e.g., in Figure 5.4.1,N = 3),Vis themean velocity in eachsection (subarea), andAi is the cross-sectional area of flow in each section(subarea).

The mean velocity can be expressed as

N

EViAiV=~

N

EA ii=l

(5.4.15)

Substituting Vi = Q;/Ai andequation (5.4.15) for V into (5.4.14) and simplifying yields

(5.4.16)

N (Q )3E -.!. Aii=l Ai

[

N 1

3

'" Q;Z:: iii A; N;~1 EA-t.', C,)

U=----...,,-----

5.4 Gradually Varied Flowfor Natural Channels 145

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

1/2Qi =«s;and solving for S)./2 yields

(5.4.17)

(5.4.18)

Assuming thatthefriction slopeis thesameforall sections, Sf; = SfU = 1, ... ,N), thenaccordingto equation (5.4.18)

(5.4.19)

This leads to

(5.4.20)

and the total discharge is

(5.4.21)

Substituting the aboveexpression for L Qi and

into equation (5.4.16) and simplifying, we get

N(K3)(N )2~ At ~Ai

a = (tK;)31=1

or

whereAt and K, are the totals.The friction slope for the reach is

(5.4.22)

(5.4.23)

(5.4.24)

S = (2:Qi)2'f, 2:Ki

(5.4.25)

EXAMPLE 5.4.3

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

Forthecompound cross-section at river mile 1.0shown inFigure 5.4.2, determine theenergy correctionfactor a. Thedischarge is Q = 11,000 cfs andthewater surface elevation is 125 ft.


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

50' I 115' 60'.1 150'

n=.0451 n = .025 In=.045

70'

n=.04

Cross-section at river mile 1.0Q=11,000cfsWS=125ftC,=.3Cc=.1

-------~------- 125'

......- ... 110'

100'

Elev.---l

70'

n=.04

AM 1.0L= 2640ft

AM 1.5

aV2 Vhf--- --2C u

-f-a2

- ..L

r----1115'

105'

Elev.---l

I

Cross-section atriver mile 1.5Q=10,500 cfsWS=? ftC,=.3Cc =.1

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

SOLUTION Step1 Compute thecross-sectional areas offlow fortheleftoverbank (L), channel (C), andrightoverbank (R):

AL = 1050ft2,Ac = 3000fr,AR = 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 ()( )2/3K = 1.486ALRL = 1.486 1050 12.35 = 208 700L n 0.04 '

K = 1.486(3000)(21.4)2/3 = 1 716300c 0.02 "

KR = 208,700(KR = Kd

Step4 Compute totals At andK;

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

At = AL +Ac +AR = 5100 ft2

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

Step6 Useequation (5.4.24) to compute a:

a = E(%) (At? = (578.3 x 109)(5100)2 = 1.55

(Kt)3 (2,133,700)3

ExAMPLE5.4.4 ForthedatainFigure 5.4.2, startwith theknown water surface elevation atrivermile1.0anddetermine thewater surface at rivermile 1.5 (adapted from Hoggan, 1997).

(5.4.2)


SOLUTION Computations are presented in Table 5.4.1.

Table 5.4.1 Standard Step Backwater Computation

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16)Water Average Energy

River surface Hydraulic Manning Average friction Friction correctionmile elevation Area radius roughness Conveyance conveyance slope loss factor Velocity v2 Ll(*)Wk A R n K K Sf hL K3/A2 a V

a"2gh ** Wk***0

(ft2) (ft) (£1) (109) (ftlsec) (ftlsec) (ftlsec) (ft) (ft)

1.0 125.0 1050 12.35 0.040 208,700 8.25

3000 21.43 0.020 1,716,300 561.80

1050 12.35 0.040 208.700 8.25

5100 2,133,700 578.30 1.55 2.16 0.11

1.5 126.1 666.0 9.37 0.045 97,650 2.10

2426.5 17.97 0.025 989,400 164.50

666.0 9.37 0.045 97,650 2.10

3758.5 1,184,700 1,659,200 0.000042 0.111 168.70 1.43 2.79 0.17 -0.060.02 125.07

1.5 125.0 600 8.57 0.045 83,000 1.59

2300 17.04 0.025 905,300 140.22

600 8.57 0.045 83,000 1.59

3500 1,071,300 1,602,350 0.000043 0.113 143.30 1.43 3.00 0.20 -0.09 0.03 125.05

(A )z" K3/A2*a= t w J I

(Kt )3

"ho = 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 channelcontraction).

'''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

dH dw d ( V2

)dx = dx + dx a 2g

Thetotalenergy losstermis dHjdx = - Sf - Se, where Sfis thefriction slope defined by equation(5.4.7) orequation (5.4.8) andS, is theslope of thecontraction orexpansion loss. Thedifferentialsdwandd[a(V 2 j2g)] aredefined overthechannel reachas dw = Wk - Wk+ 1 andd[a(V 2j2g)] =

V2 V2

ak+1 ~; 1 - ak 2;' where wedefine k + 1at thedownstream andk at theupstream. River cross-

sections are normally defined from downstream to upstream for gradually varied flow. Equation(5.4.2) is now expressed as

Vf Vf+lWk +ak 2g = Wk+l +ak+l~ +Sfdx+Sedx (5.4.26)

The standard step procedure for water surface computations is described in the following

steps:

a. Startata pointinthechannel where thewatersurface isknown orcanbeapproximated. Thisisthedownstream boundary condition forsubcritical flow andtheupstream boundary condition


forsupercritical flow. Computation proceeds upstream forsubcritical flow anddownstream for

supercritical flow. Why?b. Choose a water surface elevation Wk at the upstream endof the reach for subcritical flow or

Wk +1 at thedownstream endof thereach forsupercritical flow. Thiswatersurface elevationwill be slightly lower or higher depending upon the type of profile (see Chow (1959);Henderson (1966); French (1985); or Chaudhry (1993)).

c. Next compute the conveyance, corresponding friction slope, andexpansion and contractionloss terms in equation (5.4.26) using the assumed water surface elevation.

d. Solve equation (5.4.26) for wk+ 1 (supercritical flow) or Wk (subcritical flow).e. Compare thecalculated water surface elevation Wwiththeassumed water surface elevation w'.

If thecalculated andassumed elevations donotagree within anacceptable tolerance (e.g.,0.01ft), thensetwk+l = Wk+l (forsupercritical flow) andWk = Wk (forsubcriticalflow)andretumto step (c).

Computer models for determining water surface profiles using the standard step procedureinclude 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 ofany cross-section (HEC, 1997a-c). HEC-RAS can simulate flow through a single channel, adendritic system ofchannels, or a fullnetwork ofopenchannels (sometimes calleda fullyloopedsystem).

100 0 100 200I"II! I I

Scale (ft)

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


HEC-RAS can model sub- or supercritical flow, or a mixture of each within the same analysis.A graphicaluser interface providesinput data entry,data modifications, and plots of stream crosssections, profiles, and other data. Program options include inserting trapezoidal excavations oncross-sections, and analyzing the potential for bridge scour. The water surface profile throughstructuressuchasbridges,culverts,weirs,andgatescan becomputed.TheWorldWideWebaddressto obtain the HEC-RAS model is www.hec.usace.army.mil.

,EXAMPLE 5.4.5 A planview of theRedFoxRiver in California is shown in Figure 5.4.3, along with the location of fourcross-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, andc 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 CorpsofEngineers frommaterial developed bytheU.S. Bureau ofReclamation (1957). Distance between crosssections 1and2 is 500ft, between cross-sections 2 and3 is400 ft, andbetween cross-sections 3 and4 is400 ft.

SOLUTION The computations for this example are illustrated in Table 5.4.2.

17001500600(690, 0) 800 1000

Station (It)Cross-section No. 1

(a)

400200

(20,25) f I I I (1635,25)

1:\ A. A3 A A, AI,

'"n 0.10 n f 0.05 n 0.05 n 0.10

-f+ 1--1= 0.03(110,18)(415,17)

(650,14)(1020,14) (1590,14)(710,13)

(675,1) \ (71°11)

5710

5700o

5705

6 5715

~m

5725

5720

5730

5725

g 5720c.2 5715~ijj 5710

5705

I A, A, A A3 v,L"J,-I n =0.10 n =0.05 I n =0.10

~,25) (200~ 20) ~ I- Jn =0.03

(110,20)

~7)'\ (1195,18)(640,18)

r\.(575,9.5)

I(58,,4) (611,4)

o 200 400 600 800 1000Station (It)

Cross-section No.2

(b)

1200 1400

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

ISO Chapter 5 Hydraulic Processes: Open-Channel Flow

5710

1000400 600 800

Station (tt)Cross-section No.3

(c)

200

I A3 A, 1 A21 ~=0.11 " =0.03 ,,=o.OS

(40.25) I (875,25)-

"""~ 1--,,=0.10 WI I_I...- (850,22

(260,22) .....-l370,18.7)\ f< (600.20)

I 1\ (560,17:3)

(420.15)1 \

J \ I) (s+.7.5)(500,7.1)

5715

5720

57050

~ 5725~

~~w

5730

1000400 600 800

Station (tt)Cross-section No.4

(Ii)

200

I AJ A3 1 A, A2 TJ" .. 0.10I,,- "i o

.os ,,= "io

.os I-..0.10 0.036 (700.26)..,..,

1~0,22)30,26) (130,24) 7(460.22330,23) I--

""ri1/

(400'1 10)

5715

5730

57050

5710

~ 5725~

,g 5720

~W

Figure 5.4.4 (Continued)

572118 16

5720

~ 5716.91ii~m

5710

o

~~ ~ 400 ~ ~ 1~ 1~ 1400 1~ 1~Area(tt2)

(a)

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 ofReclamation(1957)).

o1657211r-8---T----r--=~..-_Tjr--r-v~:::;::;;;;p,_{_;~5720

5710

57080 200 400 600 800 1000 1200 1400 1600 1800

Area (tt2)(b)

A2 14 12 10 8 4572118 16

5720 A1

5718

g5716c

0

il>

5714CDiii

5712

5710

57080 200 400 600 800 1000 1200 1400 1600 1800

Area (fI2)(c)

572118 16 14 4 2 0

5720

5718

g5716c

0..~ 5714iii

~~ ~ 400 ~ ~ 1~ 1~ 1400 1~ 1~Area (tt2)

(d)

Figure 5.4.5 (Continued)

151


Table 5.4.2 Standard Step Backwater Computation for Red Fox River

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16)

Water Average Energy

Cross- surface Hydraulic Manning Average friction Friction correction

section elevation Area radius roughness Conveyance conveyance slope loss KJ factor Velocity E~( E)Wk + ! A R n K K S/1O-3

) hLA2 a V u2g U 2g ho Wk

(ftl) (ft) (ft/ft) (ft) (106) (ft/sec) (ft) (ft) (ft) (ft)

1 5710.5* 420 7.0 0.03 76,100 1.0 15.5 3.72

2 5714.7 470 7.6 0.03 90,100 3311.1

260 2.5 0.05 14,200 42.0

730 104,300 90,200 5.19 2.60 3353.5 1.58 8.90 1.95 + 1.77 0.18 5715.07**

5715.0

500 7.85 0.03 97,800 3741.8

300 2.7 0.05 17,300 57.5

800 115,100 95,600 4.62 2.31 3799.3 1.59 8.13 1.63 +2.09 0.21 5715.1

3 5718.0 1145 5.85 0.03 184,100 149,600 1.89 0.76 1.0 5.68 0.50 + 1.13 0.115717.1

5717.1 970 5.6 0.03 151,500 133,300 2.38 0.95 1.0 6.70 0.70 +.93 0.095717.1

*Known starting watersurface elevation.**Wk+1 = 5710.7+1.77+2.60 +0.18= 5715.07 = 5715.1; a.= (A,/ ,£Kl!Af /(K,)3; h, = eel (a.V2/2g)Ifor~(aV22g) <0 (lossdue to

channel expansion); ho = Ccl~(aV2 /2g) I for~(aV2 /2g) >0 (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 overaveryshortdistance. Thefollowing are characteristic features of rapidly varied flow (Chow, 1959):

• Curvature of the flow is pronounced, so that pressure distribution cannot be assumed to behydrostatic.

• The rapid variation occurs over a relatively short distance so that boundary friction iscomparatively small and usually insignificant.

• Rapid changes of water area occur in rapidly varied flow, causing the velocity distributioncoefficients a andpto be much greater than 1.0.

Examples of rapidly varied flow are hydraulic jumps, transitions in channels, flow overspillways, flow in channels of nonlinear alignment, and flow through nonprismatic channelsections such as flow in channel junctions, flow through trash racks, and flow between bridgepiers.

The discussion presented in this chapter is limited to the hydraulic jump. The hydraulicjump occurs when a rapid change in flow depth occurs from a small depth to a large depthsuch 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 downstreamof regulating sluices, at the foot of spillways, or where a steepchannel slopesuddenly becomesflat.

Figure 5.5.1 illustrates a hydraulic jumpalong with thespecific energy andspecific force curves.Thedepths of flow upstream anddownstream of thejumpare called sequent depths or conjugatedepths. Because hydraulic jumps are typically short in length, the losses due to shear along thewetted perimeter aresmall compared to thepressure forces. Neglecting these forces andassuming ahorizontal channel (Fg = 0), themomentum principle can be applied as in section 5.2.3 to derive

y

Figure 5.5.1 Hydraulic jump.

equation (5.2.24):

Hydraulic jump

y

5.5 Rapidly Varied Flow 153

Specific force curve

Q2 _ Q2 _- +AlYl = - +A2Y2 (5.2.24)gAl gA2

Consider a rectangular channel of width B > 0, so Q = Al VI = A2V2,Al = BYl,A2 = BY2,)11 = yI!2, and )12 = y2/2:

(5.5.1)

Simplifying yields

Q2 (~_~) = !B2(y~ -yi)g Yl Y2 2

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 = Y2 (Y2 +1)gYl Yl Yl

(5.5.2)

The Froude number for a rectangular channel is Fr 1 = VI! JgDl = VI!Viii; thereforeequation (5.5.2) reduces to

or(5.5.3)

(5.5.4)

using the quadratic formula and discarding the negative roots.


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 sequentdepths 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

EXAMPLE 5.5.3

SOLUTION

Consider the8-ftwiderectangular channel usedinexamples 5.1.1,5.1.3,and5.3.1 withadischarge of 100cfs.If a weirwereplacedin thechannel and thedepth upstream of the weirwere5 ft, would a hydraulicjump form upstream of the weir?

Forthedischarge of 100cfs,thenormal depthisYn = 3.97ft from example 5.1.3,andthecriticaldepthisYe = 1.69. Because Ye <Yn < 5 ft, a hydraulic jump would not form. As a resultof Yn >Ye, a mild slopeexists. For a jump to form, Yn <Ye, whichis a steepslope.

For example 5.3.1, determine whether a hydraulic jump will occur.

Thenormal depthandcriticaldepthmustbecomputed andcompared. Usingequation (5.2.14) withq =4.646 m3/s per meterof width, we get

_ (q2) 1/3 _ (4.6462) 1/3 _

Ye - g - 9.81 -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 depthis computed using Manning's equation (5.1.23):

Q = ! AR2f3 56/ 2

nor

4.646 = 0.~20y~f3(0.0003)1/2

ThusYn = (5.365)3 /5 = 2.74 m. Undertheseconditions Yn > Ye > 0.5 m, an M3 water surfaceprofileexistsand a hydraulic jump occurs. If normal depthoccurs downstream of the jump, what is the depthbefore the jump? Usingequation (5.5.5) withY2 = 2.74 m, we find that Yl is 0.5 m.

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 ftdeep.Will a jump form upstream from the obstruction? If so, howfar upstream? What type of curvewill be present?

Firsta determination mustbemadewhether ajumpwillformbycomparing thenormal depthandcriticaldepth. Usingequation (5.2.14), we find

(q2) 1/3 ( 402

) 1/3Ye = g = 32.2 = 3.68 ft


andYc>Yn = 3.0ft, therefore thechannel issteep. Because Yn <Yc<8 ft, asubcritical flow existsonasteepchannel, a hydraulic jumpforms upstream of theobstruction. If thedepthbefore thejumpisconsidered tobenormal depth.y, = Yl = 3,thentheconjugate depthy, canbecomputed using equation (5.5.4), wherethe Froude number is

so

Yl (Y2 = 2 -1+

~~(-1+= 4.45 ft

1+ 8(40)2)

g(3)3

Nextthe distance Ax fromthe depth of 4.45ft to the depth of 8 ft is determined using equation (5.3.1):

y2 y2SoAx + 2~ = i1y + 2~ + Sfi1x

which can be rearranged to yield

y2 y2(So -Sf)i1x = Y3 +-1. -Y2 _---.l

2g 2g

= E3-E2so

E3-E2 MAx=--=--

So -Sf So -Sf

To solvefor i1x, first compute E2 and E3:

Depth(ft)

84.45

8044.5

R(ft)

3.082.35

Y(ft/s)

5.008.99

0.3881.25

E(ft)

8.3885.70

Rave

(ft)

2.72

Yave

(ft/s)

7.00

Now compute SffromManning's equation using equation (5.1.24) with Yave andRave, and rearrange toyield

S = n2Y;ve = 0.017

2x 7

2= 0.00168

'f 2.22R~~~ 2.22 x 2.724/3

Compute So using Manning's equation with the normal depth:

1.49 (30)2/3400 = 0.017 x 3 x 10 x 16 jS;


M 8.388 - 5.71 322 f h di f hThus So = 00100 Now using fix =--= = t, t e istance rom t e, .., So - Sf 0.0100 - 0.00168

conjugatedepthof thejumpYz = 4.45ft downstream to thedepthY3 of 8 ft (locationof the obstruction) is322 ft. In otherwords,the hydraulic jump occursapproximately 322 ft upstreamof the obstruction. Thetype of water surface profile after the jump is an S1 profile.

EXAMPLE 5.5.4

SOLUTION

A hydraulic jump occursin a rectangular channel3.0 m wide.The waterdepth before the jump is 0.6 m,andafterthe jump is 1.6 m.Compute(a)the flow rate in thechannel,(b) thecriticaldepth,and(c) theheadloss in the jump.

(a) To compute the flow rate knowing Yl = 0.6 m and yz = 1.6m, equation (5.5.4) can be used:

Yl [Y2 =2 -1+ 8q2 ]1+-gyj

EXAMPLE 5.5.5

in which Fr 1 = q/ ;;;i has been substituted:

0.6[ 81 ]1.6 = T - 1 + 1 + 39.81(0.6)

6.33 = VI + 3.775q2

40.07 = 3.7751

q = 3.26 m3/s per meter width of channel

and

Q = 3q = 9.78 m3/s.

(b) Critical depth is computedusing equation (5.2.14):

_ (q2) 1/3 _ (3.262) 1/3 _

Yc - g - 9.81 - 1.03 m

(c) The headloss in the jump is the change in specific energy before and after the jump

hi. = M = £1-£2

so

V2 V2

h: = YI + ---l - Y2 - ~2g 2g

so

VI = Q/A I = 9.78/(3 x 0.6) = 5.43 mls

V2 = Q/A2 = 9.78/(3 x 1.6) = 2.04rn1s

h = 0 6 5.432

_ _ 2.042

L . + 2(9.81) 1.6 2(9.81)

= 0.6 + 1.5- 1.6 - 0.21

h: = 0.29m.

Derive an equation to approximate the headloss (energy loss) of a hydraulic jump in a horizontalrectangularchannel in terms of the depths before and after the jump, Y, and Yz, respectively.

SOLUTION


Theenergy loss canbe approximated by

ht: = £1 -£2

= (Yl + ~:) - (Y2+ ~!)Thevelocities canbe expressed as VI = Q/A1 = q/Yl and V2 = q/Y2, so

1 q2 1 q2b: = Yl +-- - Y2 - --

2gyi 2gy~

Thebalance between hydrostatic forces andthe momentum flux per unit width of the channel canbeexpressed using equation (5.2.23):

where A = (1)(yl) andy, = Yl /2, so

Ii- Ii = pq(!L _!i)2 1 2 2 Y2 Yl

Solving, weget

l (Yl +Y2)-=Y1Y2 --g 2

Substituting this equation into the above equation for the headloss andsimplifying gives

Prismatic Channels with Change in Slope

Consider the channel that changes slopefroma steepslope to a mild slope shown in Figure5.5.2.In thiscaseYnz >Yn! .Theconjugate depthofYn! isy' computed usingequation (5.5.4). Ify' <Ynz. thehydraulic jumpoccurs upstream of theslopebreakwithcontrolat A,andif y' >Ynz. thejumpoccursdownstream of the break in slopewith control at B. If (En! - tilli!) >Enz• then a hydraulic jumpoccurson the downstream mildslope. If Enz> (En! - tilliI ) . then thehydraulic jump occurson theupstream steepslope. tilliI is the energy lossin thejumpin reach 1. Another wayto lookat thisis ifEy' >Enz controlis at B because the energy loss tilliI in the hydraulic jump in reach 1 is not largeenoughto decrease the energy from En! to Enz.

Mildslope

Yn,

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


5.6 DISCHARGE MEASUREMENT

5.6.1 Weir

A weir is a device (oroverflow structure) that is placed normal to the direction of flow. The weiressentially backs up water so that in flowing overthe weir, the water goes through critical depth.Weirs have been usedforthemeasurement ofwater flow inopenchannels formany years. Weirs cangenerally beclassified assharp-crested weirs andbroad-crested weirs. Weirs arediscussed indetailin Bas et al. (1984), Brater et al. (1996), andReplogle et al. (1999).

Asharp-crested weir isbasically athinplatemounted perpendicular totheflow withthetopoftheplate having a beveled, sharp edge, which makes the nappe spring clear from the plate (seeFigure 5.6.1). Therateof flow is determined bymeasuring thehead, typically in a stilling well(seeFigure 5.6.2) at a distance upstream from the crest. The headH is measured using a gauge.

Suppressed Rectangular Weir

These sharp-crested weirs areaswide asthechannel, andthewidth ofthenappe is thesamelengthasthecrest. Referring toFigure 5.6.1, consider anelemental areadA = Bdhandassume thevelocity isJ2ih; then the elemental flow is

(5.6.1)

Thedischarge isexpressed byintegrating equation (5.6.1) overtheareaabove thetopof theweircrest:

H H

Q = JdQ = ViiBJh1/2dh = ~ViiBH3

/2

o 0

(5.6.2)

v2a

2g Drawdown

Figure5.6.1 Flow over sharp-crested weir.

tr11

·1--------B-------+II·

u u~

•Stilling-

well tdh

C ~Crest

Figure5.6.2 Rectangular sharp-crested weir without endcontraction.

5.6 Discharge Measurement 159

Friction effects havebeenneglected inderivation ofequation (5.6.2). Thedrawdown effectshown inFigure 5.6.1 and thecrestcontraction indicate that the streamlines are not parallel or normal to thearea in the plane. To account for these effects a coefficient of discharge Cdis used, so that

Q= Cd~ ~gBH3/2 (563)3 v-« ..

where Cdis approximately 0.62. Thisis thebasicequation fora suppressed rectangular weir, whichcan be expressed more generally as

Q = CwBH3/2 (5.6.4)2

w~ere c, is theweircoefficient, c, = c, "3 /li. ForU.S. customary units,c, ::::; 3.33, andforSIunits Cw ::::; 1.84.

If thevelocity of approach Va where H is measured is appreciable, thenthe integration limitsare

(5.6.5a)

(5.6.5b)

v2

H +J* [( 2) 3/2 (2) 3/2]Q= /liB h

1/2dh = CwB H + ~; - ~;

v2...!L2g

(V2) 3/2

When 2a ::::; 0, equation (5.6.5a) can be simplified tog V2 3/2

Q= CwB ( H + 2;)

Contracted Rectangular Weirs

Acontractedrectangular weirisanother sharp-crested weirwithacrestthatisshorter thanthewidthof the channel and one or two beveled end sections so that watercontracts both horizontally andvertically. Thisforces the nappe width to be less than B. The effective crest length is

B' = B - 0.1 nH (5.6.6)

where n = 1if theweiris placed against onesidewallof thechannel so that thecontraction on oneside is suppressed and n= 2 if the weir is positioned so that it is not placed against a sidewall.

Triangular Weir

Triangular or V-notch weirs aresharp-crested weirs thatareusedforrelatively smallflows, buthavetheadvantage thattheycanalsofunction forreasonably largeflows aswell. Referring toFigure5.6.3,

Figure5.6.3 Triangular sharp-crested weir.


the rate of discharge through an elemental area, dA, is

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]

andH

Q = Cd2y1Lgtan(~)J(H-h)hl/2dho

= CdC85) fig tan(~)HS/2

(5.6.7)

(5.6.8)

(5.6.9)

= C",HS/ 2

Thevalue of Cw for a valueof e= 90° (themostcommon) is Cw = 2.50for U.S. customary unitsand 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 weirsare high enough, critical depth occurs on the crest of the weir. For critical flow conditions

2 1/3 3Ye = (q jg) and E = 2Ye for rectangular channels:

~2 ) 3 (2)3/2

Q=B.q=Bj;i=Byg\3E) =B 3 y'gE3/2

or, assuming the approach velocity is negligible:

(2)3/2

Q= B 3 y'gH3/2

Q= CwBH3/2

Figure5.6.5 illustrates a broad-crested weir installation in a concrete-lined canal.

(5.6.10)

__1__~ _i Y1P

y2c

28

Figure 5.6.4 Broad-crested weir.


//

/ Survey point forestablishing gauge zero/ / reference. L/4 toLl3 from end. onweir

/ / center line/

//

Cross-section

rrY2 d

Top ofcanal

l_J

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

Recorder1----'L..., I

-* 1 IF : l

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

EXAMPLE 5.6.1 A rectangular, sharp-crested suppressed weir3 m longis 1.0 m high. Determine thedischarge when thehead is 150 mm.

SOLUTION Using equation (5.6.4), Q = 1.84BH1.5, the discharge is

(150 ) 1.5Q= 1.84(3) 1000 = 0.321 m3/s

5.6.2 Flumes

Boset al. (1984) providean excellentdiscussionof flumes. Aweir is a controlsectionthat is formedby raising the channel bottom, whereasaflume is formed by narrowing a channel.When a controlsectionis formedbyraisingboththechannelbottomandnarrowingit, thestructureisusuallycalledaflume. Figure 5.6.6 shows a distinction between weirs and flumes.

Weirscanresult in relativelylargeheadlossesand,if the waterhas suspendedsediment,can causedeposition upstream of the weir, resulting in a gradual change of the weir coefficient. Thesedisadvantages can be overcome in many situationsby the use of flumes.

Figure5.6.7 illustratesthe generallayoutof a flow measuringstructure.Most flow measurementand flow regulating structures consist of (a) a converging transition, where subcritical flow isaccelerated and guided into the throat without flow separation; (b) a throat where the water


Broad-crested weir

Long-throated flume

Usually flume

Cross-sections are through controlatweir crest orflume throat

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

Tailwater. Converging.. channel

Distance to transition Diverginggauging station'\ ~. C transition

~ Throat I

: ~ "L;i'~~~"e-..Approach channel I I I

IUpstream channel

entrance

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

accelerates to supercritical flow so that the discharge is controlled; and (c) a diverging transitionwhere flow velocity is gradually reduced to subcritical flow and the potential energy is recovered.

Oneof the morewidely usedflumes 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 flumeswith widths of 1 ft (0.31 m) to 8 ft (2.4 m) is

(5.6.11)

Extend wingwallintocanal bank~

as required A.;'<.. ... I I'" I

P

------------Flow

ThroatrDivergingSection Section

Hb

PLAN

PROFILE

Water surface

-------

W A gA B C 0 E F G M N P R FREE·FLOW CAPACITY3 MINIMUM MAXIMUM

FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. CFS CFS

0 6 2 7 1 4~ 2 0 1 31 1 3§. 2 0 1 0 2 0 1 0 0 41 2 111 1 4 .05 3.9iii 2 8 2 2

9 2 10£ 1 11 1 2 10 1 3 1 10§. 2 6 1 0 1 6 1 0 41 3 61 1 4 .09 8.98 8 8 2 2

1 0 4 6 3 0 4 41 2 0 2 91 3 0 2 0 3 0 1 3 9 4 10~ 1 8 .11 16.18 4 4

1 6 4 9 3 2 4 71 2 6 3 4~ 3 0 2 0 3 0 1 3 9 5 6 1 8 .15 24.68 8

2 0 5 0 3 4 4 101 3 0 3 111 3 0 2 0 3 0 1 3 9 6 1 1 8 .42 33.18 2

3 0 5 6 3 8 5 41 4 0 5 11 3 0 2 0 3 0 1 3 9 7 31 1 8 .61 50.44 8 2

4 0 6 0 4 0 5 10§. 5 0 6 41 3 0 2 0 3 0 1 6 9 8 10~ 2 0 1.3 67.98 4 4

5 0 6 6 4 4 6 41 6 0 7 6§. 3 0 2 0 3 0 1 6 9 10 11 2 0 1.6 85.62 8 4

6 0 7 0 4 8 6 10~ 7 0 8 9 3 0 2 0 3 0 1 6 9 11 31 2 0 2.6 103.58 2

7 0 7 6 5 0 7 41 8 0 9 111. 3 0 2 0 3 0 1 6 9 12 6 2 0 3.0 121.44 8

8 0 8 0 5 4 7 101 9 0 11 1~ 3 0 2 0 3 0 1 6 9 13 81 2 0 3.5 139.58 4 4

Figure 5.6.8 Standard Parshall flume dimensions. 103-D-1225 (fromU.S. Bureau of Reclamation (1978)).

163


whereQis thedischarge in fe Is, Wis thewidthof the flume throat,andHa is the upstreamheadinft.

For smaller flumes, e.g., 6-inch flumes,

(5.6.12)

and for 9-inch flumes(5.6.13)

EXAMPLE 5.6.2

SOLUTION

Water flows through aParshall flume with athroat width of4.0ft atadepth of2.0 ft.Whatis theflow rate?

Using equation (5.6.11).

Q= 4.0(4.0)(2.0)(1.522)(4.0r 26= 47.8 ft3/s

Figure 5.6.9 illustrates a flow-measuring structure for unlined (earthen) channels that are longer, andconsequently more expensive, than structures forconcrete-lined channels. Forconcrete-lined channels,the approach channel and sides of the control section are already available.

5.6.3 Stream Flow Measurement: Velocity-Area-Integration Method

As for weirsand flumes, streamflow is not directly measured. Instead, water level is measured andstream flow is determined from a rating curve, which is the relationship between water surfaceelevation and discharge.

Upstreamchannel

Figure 5.6.9 Flow-measuring structure forearthen channel with rectangular control section (from Boset al. (1984)).

Battery


Gas cylinder

Figure5.6.10 Water level measurement using a bubble gauge recorder. Thewater level is measured astheback pressure onthebubbling stream ofgas byusing amercury manometer (from Rantz etal.(1982)).

Water surface level(elevation) can be measured manually or automatically. Crest stagegaugesareusedtomeasure flood crests. Theyconsist ofawooden staffgaugeplacedinsideapipewithsmallholesfor water entry. Corkin thepipefloats as the waterrises andadheres to the staff(scale) at thehighest water level. Bubble gauges (shown in Figure 5.6.10) sense the water surface level bybubbling acontinuous streamofgas(usually carbondioxide) intothewater. Thepressure required tocontinuously force the gas streamout beneath the watersurface is a measure of thedepthof waterover the nozzle of the bubble stream. The pressure is measured with a manometer assembly toprovide a continuous record of water level in the stream (gauge height).

Rating curves are developed using a set of measurements of discharge and gaugeheightin thestream. Thedischarge is Q= AVwhere V is themeanvelocity normal to thecross-sectional areaofflow A,which is a function of thegaugeheight. Soinorderto measure discharge thevelocity andthegaugeheightmustbe determined. In a streamorriver, thevelocity varieswithdepth, asdiscussed inChapter3 (Figure 3.7.1). Therefore, the velocity mustbe recorded at various locations and depthsacross the stream.

Referring toFigure5.6.11, thetotaldischarge iscomputed by summing theincremental dischargecalculated from each measurement i, i = 1, 2, ... , n, of velocity Vi, and depth Yi- These

Vi = mean of velocitiesat 0.2and0.8 depth

Figure 5.6.11 Computation of discharge from stream gauging data.


.....-- Wading rod

(a)

Revolution counter -~-lr:!;. J

Anemometer

Support cable

Rudder with adjustablebalancing weight

(h)

Figure 5.6.12 (a) Propeller- and (b) Price-type current meters (from James (1988)).

measurements represent average values over the width L1wi of the stream. The total discharge iscomputed using

n

Q= l: ViYiL1wii=l

(5.6.14)

Boththeory andexperimental evidence indicate thatthemeanvelocityinavertical section canbeclosely approximated by the average of the velocities at 0.2 depth and 0.8 depth below the watersurface, as shown in Figure 5.6.11. If the streamis shallow, it may be possible to take a singlemeasurement of velocity at a 0.6 depth.

Tomeasure thevelocity ina stream, a current meter, which isanimpellordevice, canbeused.Thespeed at which the impellor rotates is proportional to the flow velocity. Figure 5.6.12a shows apropeller typecurrent meterona wading rodandFigure5.6.l2b shows a Pricecurrentmeter, whichis the most commonly used velocity meter in the United States. Refer to Wahl et al. (1995) fordetailed descriptions on the stream-gauging program of the U.S. Geological Survey.

PROBLEMS

5.1.1 Compute the hydraulic radius and hydraulic depth for atrapezoidal flood control channel with abottom width of20ft,sideslopes 2:1 (h:v), and a top width of 40 ft.

5.1.2 Compute the hydraulic radius and hydraulic depth for atrapezoidal flood control channel with a bottom width of4 m,sideslopes 2:1 (h:v), and a topwidth of 8 ft.

5.1.3 Compute the hydraulic radius and hydraulic depth for a36-inch diameter culvert witha depthof flow of 24 in.

5.1.4 Compute the hydraulic radius and hydraulic depth for a1.5-m diameter culvertwitha depth of flow of 1.24m.

5.1.5 A 2-m wide rectangular channel with a bed slope of0.0005 has a depth of flow of 1.5 m. Manning's roughnesscoefficient is 0.015. Determine the steadyuniform discharge inthe channel.

5.1.6 Determine theuniform flow depth in a rectangular channel2.5 m wide with a discharge of 3 m3/s. The slope is 0.0004 andManning's roughness factor is 0.015.

5.1.7 Determine the uniform flow depth in a trapezoidal channelwith a bottom width of 8 ft and side slopes of 1 vertical to2 horizontal. The discharge is 100 ft3/s. Manning's roughnessfactor is 0.015 and the channel bottom slope is 0.0004.

5.1.8 Determine the uniform flow depth in a trapezoidal channelwith a bottom width of 2.5 m and side slopes of 1 vertical to2 horizontal with a discharge of 3 m3/s. The slope is 0.0004 andManning's roughness factor is 0.015.

5.1.9 Determine the cross-section of the greatest hydraulic efficiency for a trapezoidal channel with side slope of 1 vertical to2horizontal if thedesigndischarge is 10m3Is. Thechannel slopeis0.001 and Manning's roughness factor is 0.020.

5.1.10 Fora trapezoidal-shaped channel (n = 0.014andslopeSoof0.0002 witha 20-ftbottomwidthandsideslopes of 1vertical to1.5 horizontal), determine the normal depth for a discharge of1000 cfs.

5.1.11 Show thatthebesthydraulic trapezoidal section isone-halfof a hexagon.

5.1.12 Atrapezoidal channel hasa bottom width of 10ft andsideslopes of 2:1 (h:v). The channel has a slope of 0.0001 and aManning's roughness of 0.018. If the uniform flow depth is 4 ft,what is the discharge in the channel?

5.1.13 Compute the normal depth of flow in a 36-in diameterculvert with a slope of 0.0016 and Manning's n of 0.015 for adischarge of 20 cfs.

5.1.14 A 6-ft diameter concrete-lined sewer has a bottomslope of 1.5 ft/mi Find the depth of flow for a discharge of20 cfs.

5.1.15 Compute the uniform flow depthin a trapezoidal channelwith a bottom width of 20 ft, slope of 0.0016, Manning's n of0.025, andsideslopes of2:1 (h:v) fora discharge of500cfs.Whatis the velocity of flow?

5.1.16 Design a trapezoidal concrete-lined channel (n = 0.015) toconvey 100 cfs on a slope of 0.001. Assume the use of a besthydraulic section for the design.

5.1.17 Design atrapezoidal concrete-lined channel (n = 0.015) toconvey 20 m3/s on a slopeof 0.0001. Assume the use of a besthydraulic section for the design.

5.1.18 Phillips and Ingersoll (1998) presented equations fordetermining the Manning's roughness factor for gravel-bed

Problems 167

streams using the relative roughness defined as (R/dso) whereR is the hydraulic radius and dso is the median grain size. Theequation was verified for Arizona (for the range in dso 0.28 to0.36 ft) is n = (0.0926RI/6)/[1.46 + 2.2310g(R/dso)]. Usethis equation to develop a graph illustrating the relation of n, R,and dso.

5.1.19 Using the slope-area method, compute the flood discharge through a river reach of 800 ft having knownvalues ofwater areas, conveyances, and energy coefficients of the upstream and downstream end sections. The fall of the watersurface is 1.0 ft.

Au = 11,070 ftl

Ad = 10,990 ftl

t; = 3.034 X 106

Kd = 3.103 X 106

au = 1.134

ad = 1.177

5.2.1 Solveexample 5.2.2 for discharges of 0, 25, 75, 125, and200 fe/s.

5.2.2 Rework example 5.2.3 for a 30-cmhigh hump and a sidewall constriction that reduces the channel widthto 1.6 m.

5.2.3 Compute thecriticaldepthforthechannel inproblem 5.1.5.

5.2.4 Compute thecritical depthforthechannel inproblem 5.1.6.

5.2.5 Rework example 5.2.4withdischarges of0,25,75,125,and200 cfs.

5.2.6 Compute thecriticaldepthina36-indiameter culvertwithaslopeof 0.0016 for a discharge of 20 cfs.

5.2.7 Compute the critical flow depth in a trapezoidal channelwith a bottom width of 20 ft a slope of 0.0016, Manning'sn of 0.025, and side slopes of 2:1 (h:v) for a discharge of 500cfs.

5.2.8 A trapezoidal channel has a bottomwidthof 10ft and sideslopes of 2:1 (h:v). Manning's roughness factor is 0.018. For auniform flow depth of2.9ft,whatis thenormal slope(correspondingtouniform flow depth) andthecriticalslopeofthechannel foradischarge of 200 cfs?

5.3.1 Resolve example 5.3.1 for a channel bed slopeof 0.003.

5.3.2 A 2.45-m wide rectangular channel has a bed slope of0.0004 andManning's roughness factorof 0.015. Fora dischargeof 2.83 m3/sec, determine the type of water surface profile fordepths of 1.52m, 0.61 m, and 0.30 m.

5.3.3 Rework problem 5.3.2 with a bed slopeof 0.004.

5.3.4 If thechannel ofproblem 5.1.10ispreceded byasteepslopeand followed by a mild slope and a sluice gate as shown inFigure P5.3.4, sketch a possible water surface profile with theelevations to a scaleof 1 in to 10ft. Consider a discharge of 1500cfs. For this discharge, the normal depthfor a slopeof 0.0003 is8.18ft and for a slopeof 0.0002 is 9.13 ft.


So= 0.00023'

So= 0.0003

Figure P5.3.4

M M

(a)M

----1------H

(b)

-- ~---..........--------- ----. .--.~:t-- ---~i------

H------- Ye• critical depth

_._.- Yn• normal depth

(e)

Figure P5.3.5

.~.--------

5.3.5 Sketch possible water surface profiles for the channel inFigure P5.3.5. Firstlocate andmarkthecontrol points, thensketchtheprofiles, marking eachprofile withtheappropriate designation.Show any hydraulic jumps that occur.

5.3.6 Show thatfordepths lessthanthenormal depth (y <Yn) thatSf> So and that for Y > Yn thenSf< So.

5.3.7 Show that dyfdx = + for the Sl profile; dyldx = - forthe 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 andforuniformflow.

5.3.9 A trapezoidal channel with a bottom width of 20 ft, aslope 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 channelcauses a backwater profile with a depth of 6.5 ft just upstreamof the obstruction. What would be the depth of flow 200 ft

and 400 ft upstream of the obstruction? How far upstreamdoes the normal depth occur? Assume an energy coefficientof 1.1.

5.3.10 Consider a concrete (n=0.013) wide rectangular channelthatdischarges 2.0m3Isperunitwidth offlow. Thechannel bottomslope is 0.0001. There is a stepriseof 0.2 m. Determine the flowdepth downstream of the step assuming no transition loses. Doesthe water rise or fall at the step?

5.3.11 Consider a 5-mwide rectangular channel with a dischargeof 12.5 m3/s. Thedepthofflow upstream ofthestepis2.5m.Thereis a stepriseof0.25 minthebottom ofthechannel. Determine theflow depthdownstream of the stepassuming no transition losses.Does the water riseor fall at the step?

5.3.12 Consider the trapezoidal channel with a bottom width of20ft, a slope of 0.0016, Manning's n of0.025, andsideslopes of2:1(h:v) having a discharge of500cfs.Now a stepriseof 1.0ft isplaced in the channel bottom. Determine the flow depth downstream of thebottom stepassuming notransition losses. Doesthewater surface rise or fall at the step?

5.3.13 A 5-m wide rectangular channel with two reaches, eachwithadifferent slope, conveys 50m3Isofwater. Thechannel slopefor the first reach is 0.001 andthena sudden change to a slope of0.010. The Manning's n for the channel is 0.Q15. Perform thenecessary computations to sketch the water surface profile anddefine the typeof profiles.

5.3.14 Consider a concrete (n=0.013) wide rectangular channel(R= y) thatdischarges 2.0m3/sec perunitwidth offlow. Theslopeof the channel is 0.001. A low damcauses a backwater depth of2.0 m immediately behind (upstream of) the dam. Compute thedistance upstream of the damto where the normal depthoccurs.

5.3.15 Consider a concrete (n =0.013) wide rectangular channel (R = y) thatdischarges 2.0m3/sec perunitwidthof flow. Theslopeof the channel is 0.0001. A low dam causes a backwaterdepth of 2.0 m immediately behind (upstream of) the dam.Compute the distance upstream of the damto where the normaldepth occurs.

5.3.16 Consider aconcrete trapezoidal channel with a4-mbottomwidth, 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 ameasured depthof 2.0 m.

5.3.17 Awide rectangular channel changes in slopefrom0.002 to0.025. Sketch thewater surface profile fora discharge of 1.7m3/s1

m andManning's n=0.025.

5.3.18 A 5-m wide rectangular channel with two reaches, eachwith adifferent slope, conveys 40m3Is ofwater. Thechannel slopeforthefirst reach is 0.0005 andthena sudden change to a slope of0.015 so thatcritical flow occurs at the transition. Determine thedepths of flow at locations 10m, 20 m and30 m upstream of thecritical 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

Problems 169

culvert is0.02. Theoutletof theculvert isplaced atanelevation sothatit willnotbesubmerged. Fora discharge of 230cfs,computethe water surface profile. Develop a spreadsheet to perform thecomputations.

5.3.20 A 500-ft, 6-ft diameter reinforced concrete pipe culvert(n= 0.012) isused toconvey stormwater from adetention reservoirto a downstream flood control channel. Theslope of theculvert is0.001.Theoutlet oftheculvert isplaced atanelevation sothatitwillnot be submerged, and the flow falls freely into the flood controlchannel. Foradischarge of?5cfs,compute thewater surfaceprofile.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 cfsandadownstream water surface elevation of 123.5 ft.

5.4.3 Rework example 5.4.4using a discharge at rivermile1.0of8000cfs anda discharge of7500 cfs at rivermile 1.5. Thewatersurface elevation at rivermile1.0is 123.5 ft. Allotherdataarethesame.

5.4.4 Consider a starting (assumed) water surface elevation of5719.5 ft at cross-section 1 for example 5.4.5 and determine thewater surface elevation at cross-section number 4.

5.4.5 Consider a starting (assumed) water surface elevation of5717.6 ft at cross-section 1 for example 5.4.5 and determine thecomputed water surface elevation at cross-section 4.

5.4.6 Perform the backwater computations at mile 2.0 for thesituation in example 5.4.4. TheManning's n values at mile2 are0.02 for the main channel and 0.04 for the overbanks. Use anassumed trialwatersurface elevation of 125.5 ft.Cross-sections atmiles 2.0 and 1.5 are the same.

5.4.7 Formostnatural channels andmany designed channels, theroughness varies along thewettedperimeterofthechannel. Inordertoperform normal flow computations forthesecomposite channelsit is necessary to compute thecomposite (equivalent or effective)roughness factor. Forthecomposite channel inFigure 5.4.2, computetheeffective roughness factor (ne) atrivermile1.0fora watersurface elevation of 125 ft using thefollowing equations:

and

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

Manning's n foreachsubsection of thechannel; P = L, Pi is thei=l

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

5.5.11 A hydraulic jump is formed in a lO-ft wide channel justdownstream ofasluicegatefora discharge of450cfs.If thedepthsof flow are 30 ft and 2 ft just upstream and downstream of thegate, respectively, determine the depth of flow downstream of thejump. What is the energy loss in the jump? What is the thrust[Fgate = y(M)] on the gate? Illustrate the thruston the specificforce and specific energy diagrams for this problem.

5.5.12 Consider a 40-ft widehorizontal rectangular channel witha discharge of400cfs.Determine theinitialandsequent depths ofa hydraulic jump, if the energy loss is 5 ft.

5.6.1 A rectangular, sharp-crested weir with end contraction is1.6m long. How highshould it beplaced in a channel to maintainan upstream depth of 2.5 m for 0.5 m3/s flow rate?

5.6.2 Fora sharp-crested suppressed weir(Cw = 3.33) of lengthB = 8.0 ft, P = 2.0ft, andH = 1.0ft, determine the dischargeover 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 5.6.2using equation (5.6.5b).

5.6.5 A rectangular sharp-crested weir with end contractions is1.5m long. How highshould theweircrestbeplacedin a channelto maintain an upstream depth of 2.5 m for 0.5 m3/s flow rate?

5.6.6 Determine theheadona60°V-notch weirfora discharge of150 lis. Take Cd = 0.58.

5.6.7 The head on a 90° V-notch weir is 1.5 ft. Determine thedischarge.

5.6.8 Determine the weircoefficient of a 90° V-notch weirfor ahead of 180mmfor a flow rate of 20 lis.

5.6.9 Determine the required head for a flow of 3.0 m3/s overabroad-crested weir 1.5 m highand 3 m long witha well-roundedupstream comer (Cw=1.67).

5.6.10 Water flows through aParshall flume witha throatwidth of4.0 ft at a depth of 7.5 ft. Determine the flow rate.

5.6.11 Water flows through aParshall flume witha throatwidth of5.0 ft at a depth of 3.4 ft. Determine the flow rate.

5.6.12 Thefollowing information wasobtained froma dischargemeasurement on a stream. Determine the discharge.


5.4.9 UsetheU.S. ArmyCorps ofEngineers HEC-RAS computercode to solveexample 5.4.5.

5.5.1 Consider a 2.45-m wide rectangular channel with a bedslopeof0.0004 anda Manning's roughness factor of0.015. Aweiris placed in the channel and the depth upstream of the weir is1.52mforadischarge of5.66m3Is. Determine whether ahydraulicjump forms upstream of the weir.

5.5.2 A hydraulic jump occurs in a rectangular channel 4.0 mwide. Thewaterdepthbefore thejumpis0.4mandafterthejumpis 1.7m. Compute the flow rate in thechannel, thecriticaldepth,and the headloss in the jump.

5.5.3 Rework example 5.5.3 withaflow rateof450cfsatanormaldepth of 3.2 ft. All otherdata remain the same.

5.5.4 Rework example 5.5.4if thedepth before thejumpis0.8mand all otherdata remain the same.

5.5.5 A rectangular channel is 3.0 m wide (n = 0.018) with adischarge of 14 m3/s at a normal depth of 1.0m. An obstructioncauses thedepthjust upstream ofthe obstruction tobe2.7mdeep.Will a jump form upstream of the obstruction? If the jump doesform, howfar upstream is it located?

5.5.6 Rework example 5.5.4if the depth after thejump is 1.8mand all otherdata 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. Asluicegateat thedownstream endof thechannel controls theflowdepth just upstream of thegateto a depthz.Determine thedepth zsothata hydraulic jumpis formed just upstream of thegate. Whatis thechannel bottom slope? Whatis theheadloss (energy loss) inthe hydraulic jump?

5.5.8 A 3-mwiderectangular channel (n = 0.02)hasa dischargeof 10m3Isat a uniform flow (normal) depthof0.8m.A sluice gateat the downstream endof thechannel controls the flow depth justupstream of the gate to a depth z. Determine thedepth z so thatahydraulic depthis formed just upstream of the gate. Whatis thechannel bottom slope? What is the headloss (energy loss)in thehydraulic jump?

5.5.9 A 5-m wide rectangular channel with two reaches, eachwith a different slope, conveys 40 m3Is of water. The channelslope for the first reach is 0.015 and then a sudden change to aslope of 0.0005. The Manning's n for the channel is 0.Ql5.Does a hydraulic jump occur in the channel? If there is ahydraulic jump, where does it occur: on the first reach or thesecond reach?

5.5.10 A 5-m wide rectangular channel with two reaches, eachwith a different slope, conveys 80 m3/s of water. The channelslope for the first reach is 0.01 and then a sudden change to aslope of 0.001. The Manning's n for the channel is 0.Ql5.Does a hydraulic jump occur in the channel? If there is ahydraulic jump, where does it occur: on the first reach or thesecond reach?

Distance frombank (ft)

o1232527292

100

Depth (ft)

0.00.14.44.65.74.30.0

Mean velocity (ft)

0.000.370.871.091.340.710.00

References 171

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