civil depth notes for mar 17th-hydrolic

7/22/2019 Civil Depth Notes for Mar 17th-HYDROLIC

1/67

CIVILFEDEPTH

HYDRAULICSndHYDROLOGSYSTEMS

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2/67

FLUID PROPERTIESDENSITY:A fluid's massperunit volume.Also terrned .MassDensity".value:52.4lb/ft' @ room emperature (8.34 b/gar)(7aggal/#)Greeksymbol:p rho Units:Lb/ft3SPECIFICGRAVITY: Dimensionlessatioof a fluid's density o somestandard

reference ensity.For liquidsandsolids,hereference ensitys purewatesG liquid: prquia/pwater pwater:2.41b/# @roomtemperatureoils andhydrocarbonsavespecif,rcravitiesess han 1.Acids,caustics ndwastewater ludges avespecificgravitiesgreaterhanGreeksymbol: none units: none dimensionlessExampleProblem:Thespecificgravityof ferricchloride olution quals .How muchdoesonegallonof ferricchlorideweigh?Solution:prerricchroride/pwat.,1.3 - Conversions: .48 gallffoftquid

prerricchroride:G.3)(62.4.1b1ff) g34lb/galofwaie:81.Izlb/f t '(81.12 b/tr)lQ 48gaVft3) 10.84 b/gar oR (8.34b/gal)(l.3)

SPECIFICWEIGI{T: A fluid's weightperunitvolume.Value:62.4lb/ft3 or water@ roomtemperature,umericallyequal o densGreeksymbol:y gfinma Units: b/ft3VISCOSITY:A measure f a fluid's resistanceo flow whenacteduponby anexternal orcesuchaspressure r gravity.Also termed, AbsoluteViscositHeavyoils, sympsand ellies aremoreviscoushanwater.Increasing fluid's temperature ill decreasehe fluid's viscosity.Greeksymbol:p mu Units: b-sec/ft2KINEMATIC VISCOSITX:Ratioof a fluid'sViscosit5ro thesameluid's DensiGreeksymbol:v nuv:t/p v:AbsoluteViscosity/IvlassDensity Units: t2lsecVAPORPRESSURE: hepartialpressure xerted t thesurfaceof a liquid attheliquid/vapor nterface.The iquid molecules rereleasednto thevaporphawhen he iquid boils.Boiling occurswhen hevaporpressure f the iquidexceedshe vaporpressure bovehe iquid- usuallyatmosphericressureVaporpressures a functionof temperaturenly. Increasing liquid'stemperafurencreasests vaporpressure.Greeksymbol: noneUnits: lb/ftzIblinz: pli : (tulrt2)(ft21144n21I : (lb/in')(2.31 f/psi)

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3/67

PROPERTIES OF'WATERI (SI METRIC UNITS)Uoli

Lrq)

qJL/

q]

>J -ro > .E A2 a ?? 5f;

au)

) o0 9.805 999.8 0 .001781 0.00000785 0.615 9.807 1000.0 0 .00158 0.000005 1 8 0.87l 0 9.804 999.7 0.001307 0.00000306 t .231 5 9.798 999.1 0 . 0 0 1 1 3 9 0.0000039 r .1020 9.789 998.2 0.001002 0.0000000 3 z . J +25 9.777 997.0 0.000890 0.000000893 J . t I30 9.7 4 99s.7 0.000798 0.000000800 4.2440 9.730 992.2 0.000653 0.000000658 7.3850 9.689 988 . 0 0.000s47 0.000000553 ] Z . J J60 9.642 983.2 0.000466 0.000000474 19.9270 9 . 589 977.8 0.000404 0.000000413 31 t680 9.530 971.8 0.000354 0.000000364 47.3490 9.466 965.3 0.000315 0.000000326 70.10t00 9.399 958.4 0.000282 0.000000294 1 0 1 . 3 3

PROPERTIES OF WATER (ENGLISH UNITS)TemperahrrccF)

Spccific r'cightv "0b/fr)MassDensityGb.s5"7ft')

AbsolutcDynamicViscosityu(t 10 ' t lb . sec/ f t2 )KinematicViscosityt)(" lo-5 fflscc)

Vapor Pressurep, ,(psi)32 62 42 1 940 3', l46 I 9 3 1 0 0 940 62 43 I 940 22 9 L664 0 1 250 62.41 940 2 735 L 4 1 0 0 1 860 62 37 I 93 8 2 359 | 2 1 7 02 6'70 6230 I 936 05 0 I 059 0 3 680 62 22 | 93 4 | ' 7 9 9 0 930 0 5 t90 6 2 l I 9 3 1 I 595 0.826 07 010 0 6200 1927 42 4 0 139 0 9 51 1 0 61 86 | 923 | 284 0 667 t 24t20 6 l ;7 I 9 1 8 I 1 6 8 0 609 r.691 3 0 6 1 . 5 5 I 9 1 3 1 069 0 55 8 22214 0 6 13 8 1.908 0 9 8 1 0 5 1 4 28 91 5 0 61 20 902 0 905 o.476 3'7216 0 61 0 89 6 0 83 8 0 442 4 7 4170 60 80 l 89 0 0 780 0 4 t 3 5 9 9t 8 0 60 58 1 . 8 8 3 o.726 0 3tt5 1 .s lI 9 0 6036 l 87 6 0 67 8 0 362 9 3 4200 6 01 2 I 868 o 63't 0 341 | 5 2212 59 83 I 860 0 593 0 .3 1 9 t4 '70

0"Frorn"HydraulicModels,"ASCEManualofEngineeringpractice,No 25,ASCE,lg4|"Frorn J H Keenan and F G Keyes,T'hernodynanric Properties oJ,Sleaz, John Wiley & Sons, 1936rConipiled flom many sources including those indicated: J'Ia clbook of Chentistry ancl PlDtsic5,54tlt ed ,The CRC Press, 1973, and.Handbook of Tables for Applied Engineering Science,The Chemical Rubber Co , l970Vennard, J K ald Robert L S:f;eel,EleilxentaryFIuid Mechanics,6th ed , Wiley, New york, 1982

70 FLUID ECHANICSThis copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


4/67

F'LUIDSTATICS.._t .-Y_P_.8_9rSIAL g'F-F.ES.SIJ.BF

Hyilrostotic pre$1ffe is the preRsluea fluid exer-tson a.nirnAersed object or containervrdlb.4 Pressure is equalto-the force per unit a.ria of surface.

o Pressure is indep'endent of an sbitct's area arrddize'aad the weight (-*) of wajen above the ob.ject. Figr:re 15.6 illusbrates the hgdrostatic pat.E.-dor. The pressures at depth h a,re he same T' allfour columils becausepressure depeudson depth,not volume.

Hgure 15.6 h@iosEitc Parubx

FLUIP HEIGHT' EQUIVALENT TOP-.EES.F.-U--BFPressure va,ries inea,rly with dept'h. Tbe relatioqsbipbetween prsure and depth (i.e-, the hgilrcstotic h?d)for an incompressible fluid iBg\rili b5rEq..16.7-

PRESSURE NAIf_9..EI.tgt I-a_f:..pLaf.F..9_u-B.fa$.

Flgura 15.7 F/yckostdficFlessureona HorizontalptaneSwfacT1" tot"I """ti."f forceon the horizontatptaneofr{ is givenby Eq. 18.13. R:pA

PRESSUFEONAN .INGLTNEDPLANE SURFACETte averagepreds re and resultant force on sf inreetangular plane surface are calculated in musame ashion ae dr thd vertical plane surfac6.Thesurs varies liirearly with depth- Tlre resultant islated ftom the averagdpressure,which, in turn, a

Fp : a

p= psh tslln: 94 :.rh . tu.s.ln:'#:ui [r].sJ s.B(h)

height equivalente

ontJre a'rlerage depth

Flgura 15.10 Hydrostatlcpressurdonan tnclined.E&n@Plane&fiacaP: leo(hb* hr) sinO lstl

15.7(a)rs.4b).

waterwaterwatertl:70 ln/6n0.01S03/psf2.31ft/psi

. Af.SealevelPa = atmgsptreric ressure='I4.7 psiPg: gagepressure:0 ps i

0.03tt1 ri/io624rd.|fr,.0.432ei/ft

NcEEsuppLlED REFERENCEANDBooKAGE_64: -6-?

:f,t(h+h2): *t(h"* ha)'sing Ft.sR: FA

The following statements escribehe relationshipbetweenabsolutepressureandgagepressure,GagePressurs AbsolutePressure AtmosphericPressure



5/67

FLUID FLOW PARAMETERSBERNOULIEOUATION

DEVELOPEDFORENERGYWITHOUTFRICTION.TFM TOTAI ENERGYOF A FLTNDFLOWINGWITHOUTFRICTIONLOSSESS CONSTANT.THREECOMPONENTS:

PRESSUREENERGY:pIyI : specificwt of fluidunits: fir (tb/ft\ I (tb/ft3)

KINETIC ENERGY: v, / 29.Units: ft: (ff/sec)z l,1nlsecr1POTENTIALENERGY: ZUnits: ft

TOTAL ENERGY: PRES.+ KINETIC + PoT. ENERGIESCAN BE EXTENDED TO INCLUDEENERGYADDEDBY PUMPS, ENERGYEXTRACTEDBY TURBINESAND HEADLOSSDUE TO FRICTION.(p y) , + (rr ' Zg)r+ Z)r4ha: @ i l2+ (v , lZe)r+ Z)r+hE+hr

hAhBhf



6/67

FLUID FLOW PARAMETERS(continued)FTYDRAULICRADIUS

AREA IN FLOW DIVIDED BY THE WETTEDPERTMETER.USED IN OPENCI{ANNEL FLOWEX: ChBZY-MANNING EQUATION

EQUIVALENT DI,AMETER OR HYDRAULIC DIAI ffiTERFORA FULL FLOW PIPE: PIPEDIAMETER.

REYNOLDS NUMBERRATIO OF TI{E INERTIAL FORCESTO THEVISCOUSFORCES.USED TO DETERMINE FRICTION FACTOR"f'IN TIIE MOODY FRICTIONFACTORCIIART.EX: DARCY HEADLOSSEQUATION

LAMINARFLOWREYNOLDSNUMBERLESSTHAN 2TOO.VISCOUSFORCESARE DOMINANT.VELOCITY IS LOW, FLUID IS VISCOUS.

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7/67

FLUID FLOWPARAMETERS(continued)TURBULENT F'LOW

REYNOLDSNIIMBER IS GREATERTI{AN 4OOO.INERTIAL FORCESDOIVTTNi'iE.-THREEDIMENTIONAL F'LUIDMOVEMENT.CzuTICAL FLOW

REYNOLDSNTIMBER S BETWEEN2IOOAND4000.

ENERGY GRADELINE (EGL)GRAPHOFTHE TOTAI ENERGY.WHEN FRICTIONLESS, HE EGL IS CONSTANT.

HYDRAULIC GRADELINE (HGL)GRAPH OFTFIESI]M OFTFIEPRESSUREENERGYAND POTENTIALENERGY.DIFFERENCEBETWEENTFIEEGL AND HGL ISTFIEKINETIC ENERGY.

SPECIFICENERGYUSEDIN OPENCHANNELFLOW. IT IS TTIETOTAL ENERGYWITH RESPECTTO TI{ECHANNEL BOTTOM.SINCEPRESSURE NERGY: HYDROSTATICPRESSURE T FLUID DEPTH,TIIEN SPECIFICENERGY: DEPTHOFFLUID +KINETIC ENERGY.

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8/67

The lrsdraulic rodius.is defi:qedas the a^rea.inlow di-vided Ly th'e ruetted perimetlei'. (tne lyArlufic rpsliusis not the sEurres theradiru of a pipe.) The a.rea n howis the cross-sectionela.reaof ,the luid flowinE. When afluid is flowing under prpssrue n a pipe (ie., presstrefrot), thc area io $* will be the internd area of tbepipe. However, the fluid mayinot completely filI thepipe and may flow simply becauseof asloped:surface(i.e., grauity l,ow or open chanriel lou),The wetted perimeter is the length of.the line repre.senting the hterface bebween he fluid.and the pipe orchennel. It doesnot include the /ree surJoceength (i.e.,the interface between fluid and aLuosphere).

PAR.dVflETERS..F.H-I.YSHF.$..T..PIAH.FI.FYDRAULTCBADIUS

area in flswRu= - rh : wetted perim,eter

FLUID FLOW

t6.18

4t =D. : 4 rh 16.4.: - P

Consider + cirqular,pipe flowing completely fr:ll. Theareh in fl.orr s zrr2. The'wetted perimeter is the entirecircumferencq Znr. The hydraulic radius isr r 2 r Drh ,p ipa : znr : t = t f 16 .19

The hydraulic radius of a pipe flowing hnlf full is sIBorfZ, aiice tlie fl.owarea and wetted perimeter a,rebbthhalved.Fog;acirculss r.b+nngl lowi--nS ither full or half-full, tbobydraulic fadiue.is'pnefourth of the equivalent diaqclerr:D"14.

Determine the hydraulic radiusand equivalentdiameter or a rectangularbox culvertflowing full. The culvert hasdimensionsof 6 ft wide by 3 ft high.F l o w i n g F u l l H + = ( 6 x3 ) l e + 6 + 3 + O : 1 g f t z / t g f t : 1 f t' NOTE:Channels flowing ull, flow is touchinghe op.

EXAMPLE PROBLEMS

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9/67

FLUID FLOW PARAMETERSREYHOLDSNUHBER

The Reynolds number, Re, is a dimsnsisnlessnumberinterpreted as the ratio of inertial forces o viscous or-ces in the fluid.16.21

16.23

.The inertial forces a,reproportional to the flow di"-e-ter, velocity, and fluid densiby. (Increasing these va.ri-ables vrill increase the momentum of the fluid in flow.)The viscous force is representedby the fluid's absoluteviscosity, pr,.Thus, the Reynolds number is calculatedaszu: D",tf '- [sq ft.az(a)pnr: D=",tf p.s.l 16.p2(b)9elt

Since p,/p is defined as the kinematic uiscoaity, v,Eq. 16.22 ca.nbe simplified.

R": Dtov

.lAll.lltAB..Ftgt[Laminar flonr gets ts nnrne from the word,laminoe(lay-e,rs) If atl of the fluid parbides move n paths pa.ralld tothe overall flow direction.(i.e., in layers), he flow is sa.idto be lorninar: (The terms viscous lou a.nd sttEanilineflou ate also used.) This'occure in'pipeline flow whenthe R.eynoldsnr:mber is lse han (approrirnately) 2100.],n.rninss flow is tJTicaf w-hen he flow channel is small,the velocity is low, a.nd the fluid is viscou"i. Viscousforcee are domins,nt ia lnminnr flqrw.In lnminar'flow, a'strea.m of dye inserted in the flowwili continue from the aource in a continuous, unbrokenIine with very little mixing of the dyo and surrouadingliquid. The fluid partide paths coincide with imagi-, ns.ry streomlines. (Strearnlines aud velocity vectors sre'always lnngent to each other.) A 'bundle" of thesesftssa.rnlinsi.e., a Stteomtube) onstitutes a completefluid flow,

TUBBULEHT FLOW

"qnH g+Ltg-$'

FJ.F.-c_ Il9..FI FB.$JThe to -al head possessedby a fluid is given byBernoulli equation-

SpeclficEnergy,E, is defi-neda.s he total head withrespectci'the channel bottom. I.n' his case, :0 andplt = d'

p fpn+2n+'n v 24 + 2 n + '

ts4p.s.I

Equation19.66 s not mea.nt1q imply-that the poten-iial energy is an unimporta.nt factor in open e}la.nniflow problems. The concept of specific enersr is usedfor convenienceonly, a,nd it should be clear-that theBernoulliequation is still the valid enerry colseriationequationIn uniform floru, total head decreasesdue to the fric-tioualeffects,but specific errergr is constant. In nonr:-uform low, total head decreases,but specificenergymayincreaser decrease.Since : QlA, Eq. 19.66can be written as

E: d+ #t [general]

' .,2. E : d + i -' 2 s

u : 9 : QA u d

19.6

19.6For a rectangulsr chennsl, the velocity can be writtenin terms of the width and flow denth.

19.6

NCEESUPPLIEDREFERENCEANDBOOKAGE

Thespecificenergf equation for a rectangularchan''eis givenby ES. 19.69aad shswn n Fig. 10.fa.E: d. #ry [reciangular] ts.



10/67

, FLUrp byEJ i l iEE

. ' -I?) vatvp, itFng,orgfiefruqtion

Flgura17-9EGL N HGLor tufinoftosias

F.llF,E-q-Y.SBAP_F.H FThe cncrggyadr" ;"t EGL) is a graphof r,he oraleaergr-(totalspecific'e1qry)."lo"s .i tensrhof prpi

-In 1 tictig-nless ip with.irii pumpsoriuUink, inepd "pT'i$glqory is constaal,"oa tle ECl, witt tihgrilutat (This s arest+tenentof thcBirnouli equa.tiou.) elerationof EGL=.b + h,"+ h, . rO.4

eleve,tioirofGL# tb+h, fi.a1Tbe {iFer-en9e-berweenLhb_gI, ".aat[d ECL is thd.velbcity iead,-ft,;of the A{id.h,.+elwirtion'sf'EGl. eterra$bsgfEGT, rF.eFT. c.follo*lugruf applyto,rhesed rinrg n a ftip"tretug$Fquibnur'\tr lrl aplps ng.1eiiig.fug,(i..9.;rtffpiessuis),'tiithqpt niuirDs il'ttiiUh"s.-r Ths EGII is ahtayr odt-oesa.. T-lieSGJ,,rsalwayigquirl o or b4oS.tLUpGI,.' H-qI (v,=_0),fluid.ar:.ift* durfece; GI,,=nLrL (i.e.r tbe EGL coingide$"withthp fluids.uffa$iu a rFervpir).

'r \ilIhppthe0cq ere&dcrp+$ei,beSGr.i{Eerta$s.r llheq the4ffi dre+hcreas*, the $el,,t#Fps{:

k^1f,f; i"j (ie, esbe.ap{sr.arer. nm:,sosi),tu net llrieri&?tn rheetaparabolic .S+th-

refcrencqinp or aEGL anCiGLFlgurc 1 .1 EnergyandHydradb Gradi[ineEWiM.t Frlr;tion6 5

(blsqddehnlargemcm (clrVdCen'coh

NCEES1'PPUED.- EFERENCEANDBOOKAGEThis copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


11/67

F'LUIDDYNAMICS

M{55 IN FLUID SYSTEMSS ALWAYSCONSERVED:P A r V t : p A z v z

IF'TFIEFLUID DENSITYIS THE SAMEAT BOTHPOINTS N THE FLOW,THE CONTINUITYEQUATIONBECOMES:

A r V t : A z Y zENERGY CONSERVATIONPRTNCIPLE

BERNOTILIENERGYEQ_UATION / FRICTION@ y), + (,rt 2g),+ (Z\ :(p / r)z+ 1v2zg)2+(Z)z+ I7r

ALL UMTS ARE IN ft.}IEADLOSSDUE TO FRICTION S ALWAYSPLACED ON TI{E DOWNSTREAMPOINT OF TFIESECTIONBEING ANALY ZED

EXTENDED BERNOULI EQUATION w/ FRICTION@ Dr + (.r' 2g)r+ (Z)r* hn:(p / ilz + 7v2Zglr+ (Z)z+hE+hf



12/67

F'LUID DYIIAMICSBERNOULTEQUATION(CONTTNTTED)

Ep: ho= (P/r): "pressure ead"oR "statichead"oR,,hydrostatichead"Err: hu (v2/2g): velociqihead"oR "dynamichead', R,.kinetichead,Er:hr: (z): "potentialhead"oR "gravitationalhead,'oR.,elevationhea

H: Total Head: Ep* Eu * Er: Total Energywithout friction: ftP: TotalPressureyH: Ob/ft3xft) rb/# OR (Dl23t ftlpsi) psiExample Problem: Determine the total headand total pressureof waterflowing at a point in a pipe which has he following data:Flow:2500 gpmPressure40 psi; PipeDiameter 12 nch; Velocity: 7 ff/sec;Elevation30 ft; Twater 62.4lbffi3Solutions:Ep (a0psi)(2.31 l/psi) :92.4 ftnn: 6f tzg): 1tff1z*t2.2)0.76tE":30 ftH : TotalHead Ep+ Eu Er: 92.4 t + 0.76ft+ 30 t.: 123.2t.

P : TotalPressureTH (62.4 b/tr)OZ3.Z t): 76gg b/ff :53.3 p

DARCY.WEISBACH EQUATIONhr - fLvzlZDg

HAZEN-WILLIAMS EQUATIONTwo EQUATIONS whichto usedepends pon he unitsprovided

hr ((3.022)(vfps)"'(r))/(ct t'(Dpl:esyhr : ( I0.44)L)(Qgpm)' -*,)/((c) -ts(d iny+sess,

Nll lssuppLrED-REFERENcEHANDBoo[J4qE5This copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


13/67

FLUID DYNAMICS (continued)

DARCY EQUATION

FRICTIONFACTOR .f 'SPECIFICROUGHNESSRELATIVE ROUGHNESSREYNOLDSNUMBER

KINEMATIC VISCOSITYDIRECTLY RELATED TO FLUIDTEMPERATUREMOODY FRICTIONFACTORCFIARTMOODY FRTCTION ACTORTABLES

L: PIPELENGTHV: FLUID VELOCITYD : INSIDE PIPEDIAMETERg: ACCELERATIONDUE TO GRAVITY

NcEEsupptfEDREFERENcEANDBooTaar 6SThis copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


14/67

MOODY (STANTON)DIAGRAM

MaterialRivetedsteelConcreteCast ronGalvanizedronComrnelcialsteelor wrought ronDrawn tubing

e (ft)0.003-0.030.001-0.010.000850.00050.000150.000005

e (mm)0.9-9.00.3-3.00.250 .150.0460.0015

. 6 4Re

0.050.040.03o.o20.0150.0100.0080.0060-0040-0020.00100.00080.00060.00040.00020.000100.00005

bU'auJzT(tftrulF5I.IJE

E.oFC)trz.^c)E.Lz.Fz.sgo

o.o2

0 .0 1 5

1 0 3 2 3 5 1 0 4 2 3 5 2 3 5 2 3 5REYNOLDSUMBER,tg= DV Ptr

From ASHRAE (The American Society of Heating. Refrigerating and Air-Conclitioning Engineers,Inc )

0.000011 0 8

FLUID ECHANICSThis copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


15/67

FLUID DYN/MIC-g

Enarnple 77.350oF water is pumped through 1000 bof 4 in, schedule.40 welded steel.pipe at the rate of 800 gpm. Whatfriction loss (rgi 2 ., is predicted bf tne Darcyequation?

SolutionFirst, it is necessary o collect data on the pipe andwater. The fluid viscosjty,pipe dimensions,and otherpararneterscan be fou+d from the appendicei:

u : L . A L x 1 0 - 5 f t 2 / s e c r , , . 'e : 0 . 0 0 0 2 f i ' - - ' .D = 0,gg55 t ;, ':

A = 0.0884 i2 i -The flow quantrty is converted from ga,llonsper minuteto cubic feef per.second.

: 1.8x 106Therelative ougbaesss

e 0.00ti2D: gFE = o'oQo6trtoq the .fricrioh factoj ihble (ot'the Moftiitip.p factpscha,rt), / :0.0lg$

: 51.6 t ._ _

The Relmoldsmrmber is.56seT-E '

t usec

(

10-Re: D\r :v

(0.3355r)

The velocity is vt : 7 : : 7.56 t/sec

P R O F E . S S I O H A L .This copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


16/67

FLIIID DYNANflCSMINORLOSSES

METHOD OF'LOSSCOEFF'ICTENTS

pressed ia fractio.ns (or multipl*) of the velocity head-

h',n:Kh, -K ltftZglhr root: Sum h- * hrnipine

The losh coeffident for aayminonloss.caU.be calculatedlf +lre equivaledt length i.e ngwu l[trrenrer, the.rd is-noeCmntage t9 using ono.mrctlrod. crver he othep,exc@tfor consisteilcy in, calculations.

TABLE17.4Typical Loss Coefficients

ddvilce K.angle vat#bend, closeretrmbutbb.rfty valveb2 t O 8 i n10 to 14 in16 to.?Ain .ijlieck va(ve,sying,ruuy opell .cornlgated bends

52 2a5f;35f;25fi2.3.-

0.19.1-L5' 5 : 6 '?AlU

1-S q 1.6 times^ ^valiil for s:m0.90-6o.42_

$.4 o 10

I{:+ 1015. Ftpe eczJ: (frypjecfihg cxit, s.haif>erigedor rounded)

K : L . O .pipe'entrapcalssntra,nt: l(:0.78sharpeilg"d, 4 i 0.50.r r -bend radiusroUIloeo: .# I{D

0.02 0.28,0.04 0.24o.oo' '0"15'. 0.rq: q.0.e0.15 0.04NcEESuppuED REFERENCEANDBooKAGEG S t 66,

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17/67

FLUIDD-TNAMICSMINORLOSSES

METHODOFLOSSCOEFFICIENTS

0.50.91.8 EL7800.2

Lake

ExampleProblern12" Devices LossCoefficientsPipeEntance "A"STD90 DegElbowt tBrr, *prr, *grt

STDTee C"Full OpenGateValvetVlttr ttV2ttTurbineWheel Flow 6-2Metero'Mr"PipeExit "F" 1.0Lake water lows at 700 gpmby gravity rom a laketo a Raw WaterReservoirhrough6,000 t of straightpipe. Severalminor ossesshown n the Figureand isted n theTable), ropresentn the 12rrr'Schedule-40eldedsteelpipingsystemts0.018).1. Detennine the total headloss n the system.

hf to,"t: SUm h-* hfpipioesumK:0.5+(3 x 0.9)+ 1.8 (2x0.2)+ 6.2+ 1.0: 12.6sum hm: surn


18/67

HAZEN.W|LL|AMS ElgctTYTbe empirical Haaen-Willi^msequation was developed iroccasionally used in the Uity sewers. It is applicablehigh Reynolds numbers a.usional-analysis. Ilowwer, Iwere developed experiment

en-Witliayns mnsta.nt, C, toI of the channel. Since ilewater within "normal" am-

on the roughues, not on tlis also the method's mrinsional judgment -rqrequired

v: o.gb0ao.63go.satsll te.vR: HydraulicRadiusSoor S= Geometric lopen Unifo

NCEESUPPLIEDREFERENCEANDBOOKACTYIL ENGINEERTNC REFEN^ENqE MANUAI.

HAZEIY-unILrAMSPrPE LOWQ:0.285642'6tgos+where: Q = thc discharge ingallonsperminurcd = thepipediamctcrininchesSr = rheslopcoftheenerg:fgrade line

Example S-1 - plplng amtysls _ compuilng CProblem: The valuo:19p, wouldpredicta 3 ftrleecdischargcn arig_in.diam-eterpi1rclinc hat is 2,000 t rorrgandias a headloss f 56.g t.-is mostncarly:(A) r2o(E) tzs(c) 130(D) 13sSolutionr

step 1. convertprobrem ataor usen the lazen-williarnsonnuraQ= 3ftt / sec= (3X449) l347gal minSr= 56.8 ZffiO=0.0284ft/ t

step2. Thecalculatedarues re hensubstitutcdnto theHazen-williamsequation ndC solvcdon8 = 0.285Cd2.6t5;'x7347. 0.285C(g)2.6r0.02g410.r.C = 136 Sehct (D)

Environmenralnglneering.E.Examinationurde HandbookThis copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


19/67

EXAMPLE PROBLEM' circular plpglines1, 2, andB, each1000 t (s00m) Iongandof6,B,attd 12 n (0.3m) diarheier, bspectivgly,carrywater from reseiioir A "nioin pipeliae4 to reservoirB. pipeline4 is 2ft (0.6m) n diameterand b00ft(150m) Iong.Determinethepercentages f flowpassingpipelines1, z, andBusing the Hazen-williams equation.The Gvalue ror [ip"ti"e 4 is'rro ana, :for the other three, 100' ihe head oss n the three "-r]t"i pipelines is 1o t per 1000t.

S : 10 / 1000 :g . g1

Assume onduits1,2,3 &.4 we flowing till.

soLurloNStgp'l. Qalculateve]ocitf in pipelines1, p, and BEor a circuJ.arpipe, the hydraulic radius is

' R:+the4, from the Hazen-Williams equgtion,

vr : 1.3189l?0'6350.6a-j- .81s 1oo" (Ei)o'uu,o.orro.un

1o.eo(g{)o'ut\ 4 /2.96ft/s

rz1 10,e6fgig)tt': B.b4ftls\ 7 /

Va 10.96(tJo'ut: .bgft/s

Step. . Determine the total flow eaer :Ar v :f ,nlVr:0.?8b(o.b)22.9

:0.59 ftu , .Qz=AzYz * 0.7850.6602 3.54

: 1.23ts/sQa AaYs 0.?85 1)2 4.5S: 3.60 t3/s' Q+: Qt * Qz+Qe 0.59 L.2s+ 8.60: 5.42 t3/sStep3. Determine p.ercentage.o.flowi .fn pipes L, 2, a:rd.Bpipe1= (QJQDx 100= 0.89 xL00lE.4j : 1 0 . 9 %pipe2 = 1.29xLAO/8.42=22.7%pipe3': 3.60x LOO/8.42 66.4%

water andwastewater alculationsManuar,McGraw-HillThis copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


20/67

FLUIDDYNAMICS(continued)

TWO ORMOREPIPESPLACED INPARALLEL ORTGINATINGANDTERMINATING AT COMMON JUNCTIONS.FLOW DIVIDES TO MAKE THE HEAD LOSSIN EACH BRANCH AND BETWEEN TI#JLINCTIONST}IE SAME.TFM TOTAL FLOWRATEIS TFTESUM OFTIIE FLOWRATES N THE BRANCHS.COMMON PROBLEM:GIVEN DATA FOREACH BRANCH DETERMINE TFIEHEADLOSSDUE TO FRICTION:

USETIM FIEADLOSS UE TOFzuCTIONEQUATIONISOLATEAND SOLVEFORVELOCITYUSINGALL KNOWNFACTORSEXCEPTHEADLOSS UETO FRICTION(UNKNOWN).USETFIECONTINUITYEQUATIONAND SOLVEFOR IEADLOSSDUETOFRICTION SAME N EACHBRANCH)Q : A rv r

NcEEsupplrED REFERENCEANDBooKAGE6ZThis copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


21/67

10*q0:f 19' PVe;C=130

500'qf10"cast ron,C=100

WORKSHOPPROBLEMS

17 . ln the diogrom obove, which of fhe porollelbroncheswil lcorrymore flow?(A)(B)(c)(D)

bronchAbrqnchBneither,heywillcorry he some lownotenough nformqtiono solve

Flownporollel ipes ivides uch hol ihe frictionosssequol neoch pipe.Theres500 t morepipe ength n brqnchA; however, ecquse t iso smoothPpipemoteriol,he roughnesss esshon brqnchB. Thebesfwoy to quickly olvthisproblem s o setup on equolity.Hqzen-WilliomsEqn.17.30] t-',, $?I], r c l .BsDl . lTBydefinit ion f porollelpipes: hrn= hre3.o22vor'BuLo3.o22vrl'BsL*Col 'Bspot ' tz- Cut ' tuDrt ' t t

Thediometercqncelsbecouse he pipe diometer s he sqme n both brqnche(uk"hooot)_(ul'uXsoot)136t.es 1ggt'as0.l23vl85 O.l0Ovl'85In order for the two sidesof the equation o be equal,vB flust be greaterthanv6. Since he pipesare he samediameter 10 in), AreaB: AreaA'SinceQ : Ar, Qs s greater han Qa.The flow will be greater n branchB'



22/67

FLUID DYNAMICS

Eaarnple 77.93.0 fi3/sec of water enter. the paratlp .pipe ngtwork.shown. Aii piir'esare schedule 40steelwiJh t[e norniiraliizes shown. lnioot lossesa"b insigioifi.aot. Wnrt is'thetotal ftiction head loss between unctions AandB?

Solve the friction her(either Da.rcyor Elaatfy in each bra,nch.knourn, first convert i

t;D-"h,v: r / [D"..V]u t L0.bb0CDo.6B7ro.6av: -- .F;sa--r- [Haaen-Willia.m00 t

Soluti,onstep 1: Oollect the pipe dimensions.

2 in 4inflow area 0.0233 t2 0.0884 tz 0,2006,fu2dia,meter Q.L72g t 0.ffi35 fr 0.50b4 b

Since the Hazen-Williams ioss'coefficieuts are given for ea,chbraach, theIlaaen-Williat"s friction losSequation must be used.

6 i n,st'ep2:, The flow rdtes are

l z : A vi r . / ^.,v?r,; (o.o?$?.flrxQ.agr)hoiY,

: o.868oh?.64I/ao (o.og84r2)1z.zoo;1rg.sa: 0.2032hr54.

h : V i ' * % i . * V + i ,

I ttt /*C : q.01.94h3'5f.0.oe nor.l4.g.Z1tzhoThe fiiction head loss is the sa,men alatlel.bra,gchgs. -*o ,3 fta/sec (0.0194 0.8680+0.2032)h9.bh r : 6 ' 5 f t

step 3:

: Q.$1h?.54 :The velogities'in.the other two bianchesard, .ve'q= 4.32.7h;'54 .v+ii='2.299hoi64

l. Determinehevelocity n eachof the hreepipes?2. Determineheflow in eachof the hreepipes?



23/67

HYDRAULICMACHINES )..EXTENDEDBERNOULIEQUATIONwi FRICTION(p y),+ (ut Ze)r+ Z)rahr.=@ y)r+ (,0,Ze)r+ Z)z+ E+ rISOLATEENERGYADDEDBY PUMPhe:(p /y)r - (pf .y ) ,+ (v t tZe)r- u ' /2g)r+ (Z)r- (Z\+h*+hrHORSEPOWER OUATIONS

HYDRAULTCHP-ALSO WATER HP $HP).FIPAT PUMP,USEWHP EQUATIONS TABLEhA,Q, CONSTANT,SPECIFICGRAVITY.BRAKE I{P (BIIP): HP BETWEEN MOTOR & PUMPMOTOR I{P (EIIP): FIPTO TIIE MOTOR

EFFICIENCIESPUMPEFF:WHP/B I IPX IOOMOTOR EFF: BFIP EIIP x tooOVERALL EFF WIIP / EIIP x too



24/67

HYDRAULIC MACHINES (continued)

ITEADDECREASES S THE Q INCREASES.INCREASINGRPMAND IMPELLER SIZEINCREASESCIJRVEPOSITION.DECREASINGRPMAND IMPELLER SIZEDECREASESCURVEPOSITION.

SYSTEMCURVESIMAD INCREASESNONLINEAR AS THE QINCREASESBECAUSE 2RELATED TO Q.PLOT OF THE RESISTANCE O FLOW INTHE PIPINGSYSTEM:TDH - TSH+ hr+ v22s,f t zsusuAt,t,y A sMALLvALtrEASSUMEQr,' lUSEhtr/htz:CONSTANT *iADD TSH * v2 2g AND PLOT AT Q VALUES.

OPEBATTNGPOT.NTINTERSECTIONOFTHE CHARACTERISTICCURVEAND SYSTEMCURVEOPERATINGPOINTDATA IS USEDFORAFFINITY LAWS AND PUMP SIMILARITY

NcEEsuppllED REFERENcEANDBooKAGEa q /This copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


25/67

-tto,loE.Eo ,

HYDRAULIC MACHINES

PUMP PERFORMANCEGURVESl - L . , j - ; r . : 1 , ; . : , . , , .Fo; ;.* $w" 1. mp ellef;di a, a,et.{an{, constant: spegd,thgheadeddgdy.ill decreasg s h.e lorr ratg iqclpqfqg.

flow quentity (Q)Ilead ver-sus Flow Rate

' t40" go-: 120 160 '2obgal/rnin(a)variable peed

Hlurq 18.11 CentrifugalumpCharacteistics'urues

PROEESSIONAL PUBLICATIONS

120E i o g ,'=t s s o: eooF + oo'

20

tog.g'b 8 0 o oo. t r . -E, +ooP

80 120 160 200 240 280 32Ogal/min(b)variable mpeller diameter



26/67

HYDRAULICMACHINES

arid is disrega,rded.)hA : h , * h thr: h,(d) hz(")h1 : h1p1+.hr51t

18,37tF,:liit18.:39: Figure 18.'12 SystemCuNe

h t ,hf ,, ( * ) ' 18.

OPERATING POINT

hr4".yrt".

o"yo"-Flgure 18.13 Ertremo Operating poin

hA

II



27/67

HYDRAULICMACHINESExampleProblemOnepump lifls stormwater frorn a storageunnelwet well a constantverticaldistanceof 150 t. The equivalent engthL, ofthe 14-indiameterpiping is1,000 t. Pressure rop data or different low ratesare isted n the table.

1. Determine he flow rate f thepump s turnedat2,000{pm:Theoperatingpoint is atthe ntersectionof the 'oo.characteristic curveand /'othe system curve. z@

T OH

i l_r'l..-' ' - : I--'-r-Fl, - . . . . fi i l I i. ;+---t - t - -1 1--

l. L$t lr" r

.J1-#t -' F '

l r

' \44'i:Ttloo'.. FF

1 - ' i : -::-F-.---i -i-i--fT -I'k-1. {-t-+t-a,w', :_ JEpFt, r t I ' \.i-, r r tt-i--.i ) ffi |->lr : :r f-:-r.

+; lOO "Pft t -I

I n I lI 1J .t : t ' ' It . : t II ._-r--\ .-i--t > l ; Il l' - - F l t _ . : l

j?

: tI

I l-t---t :l - r ' I i t - . -I r Ht :.J ,4 . a ,V . t7 . . f . t t t * . * 3l

Theflow rate s 17cfs.

CIs

2. Determinehe flow rate f thepump s turnedat 100 pm:Thepurnpcannot ift theflow at 100 pm.

3. Determineheflow rate f thepump s turnedat2,700 pm:Theflow ruteat the operating oint s 23cfs.

cfs gpm Pressure rop(psi)per1,000t Head(ft) TDH(ft)TSH + Head4.45 2,000 2 . r l 4.9 154.96.68 3.000 4.45 10.3 160.38.91 4,000 7.92 i,8.3 168.31 1 . 1 4 5.000 1 1 . 6 26.8 176.813.36 6.000 15.4 35.6 185.615.60 7.000 21.0 48.5 /,98.s17.82 8.000 27.4 63.3 213.320.04 9,000 34.7 80.2 230.222.27 10,000 42.9 99.1 249.126.73 12,000 61.8 142.8 292.83 1 . 1 8 14,000 84.0 194.0 344.0



28/67

HYDRAULIC

If the speed is held constant and the impeller size svaried.

${ACHTNES

AEFf-$-Tlr_$wgMost parameters (impeller diameter, speed,and flowrate) determining a pump's performancecan vary, Ifthe impeller diameter is held'constant and the speed-i.gvaried, the following ratiob'are rnaintainedwith nochange n efficiency.

BUMP S MIIARITY

Qz - rizQr rL1h ,_ ( r z \ ' : ( e r \ 'n r - \ r z r l - \ d /P, _ / " r \ t : - f 'or \ 'P 1 \ " ' / - \ A ' l

P2

use"operatingPoint" datatocompare ande values.

The performanceof one pump can be used o pthe performanceof a dynimicolly similoi dr;;;;;;pump. This can be done.by u.singEqs.ntDt{htQt

nzDz{hz

Qz8mP2

p,fDf-;,EEhJSt - nztf6(hr;o.zs (hr1o.ts

Eranple 1-8.9A pump operati4g at 1770rpm delivers800 gal/minagainst_a otal head of 200 ft. Changesn the pipingsystemhave ncreasedhe total head o SZb . .q_t natspeed hould this pump be operated o achievehis newhead at the same effi.ciency?

SolutionFlom Eq. 18.42,

IETLz ItrLt/;: : L770 rpm. Y n t: Z424rpm

NCEESUPPLIEDREFERENCEANDBOOKAGEA1 IThis copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


29/67

NET POSTTIVESUGTION 1IEADLiquid is not sucked nto a pump. A positivehead (uor-mally atmospheric pressure) mrut push the liquid intothe impeller (i.e., "flood" the impeller)- tYet PositiucSuction Head RcEired (lVPSfIn) is the miniirrum fluiddirerry required at the inlet by the pump for satisfactoryoperation. NPSHR is usually specifiedby the Pumpmanufacturer.: Net Positiae Suction Hcad Aaailable(NPSIIA) is the actual fluid energyat the inlet-

GAVITATTONIf NPSIIA is less than NPSHR, the fluid will cavitate'iaaitation is the vaporization of 0uid within the cas'

tion, and possible structural damage o the pump'Cavitation may be causedby any of the following con-ditions:

1. Discharge heads far below the pump's calibratedhead at Peak efficiencY.2. Suction lift higher or suctionhead ower thnn themanufacturer's recommendation3. Speedshigher than the manufacturet's recourmen-

dation-4. Liquid temperatures (thus, vapor pressures)higher tha.u that for which the system was de'signed.NPSIIA cau be increased bY:a. increasing the heigbt of t'he free fluidlevel of the suPPlY ankb. reducing the distance 35d minoi lossesbetweeu the supply tank and the pump'or bY using a larger PiPesizec. reducing the temperature of the fluidd. pressurizing he suPPlYar*e. reducing the flow rate or velocityNPSHR canbe reducedbY:

b. using a double suctionPumP

HYDRAULIC MACHINES

NPSHA : harm h"(s)- ht1) - h* 18.30(a)NPSHA: hp(r)+ lr"G)- h* Il.nP)

Enarnple'78.8:2.0 ft3/sec (56 L/s) of 60"F (16"C) water a,repurfrom an elevated feed tenk to an openreservoir hro6 in (15.2 cm), schedule-40steel pipe, as shog4.friction loss f6r the pipiag and fittinp in the irucline is 2.6 ft (0.9 m). The friction loss for the piand fitti'g* in the discharge line ie 1-3 t (4.3 m)-atmospheric'presnueis 14.7psia (101kPa). WhthE NPSHA?

h9 gpecific weight of water is approximatelylbf/ft3. The atmospheric head is

for 60"F water, the vapor prehead s 0.59 t. Use 0.6 ft.FhomEq. 18.30(a), he NPSHA is

NISHA : hatm. h.G) - htq) - h"p:33.9 f t+5 f t+16 f t - I f t - 2.6 t -0.:50.7 ft



30/67

OPENCHANNELFLOW

CI#ZY-IyIANNING EOUATIOhIv: (t .49 n)Gf, (S)t"R: HYDRAULICRADIUSn : ROUGHNESS OEFFICIENTS: SLOFEFLOWCALCULATIONS OR:

RECTANGULARCHANNELS

TRAPEZOTDAL I{ANNELSNATIJRAL C}IANNELSCIRCULARCHANI\TELSLOWINGFULL

CIRCULAR CHANNELSFLOWING PARTIALLYFUCIRCI]LARCHANNELRATIOS

r " ,NCEESUPPLIEDREFERENCEANDBOOK AGE IO IThis copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


31/67

UNIFORMAND NONUNIFORMSTEADY FLOW ll"_:tor, ,9, s the. lopeof be energyine. In gentneslope anbecalculatedromtheBernorilliequas heenergyossperunit lengthofchannel.

If the flow is uniform, the slopeof the energy ine wparallel the water surfacea.ndchannelbottom, and thenergggrodient will equal the geometric slopelSs.

t9

,96 - LzT [unifoirn flow] tg

^ d EO : E

Any openchannelperformanceequationcanbe writteusing the geometricslope, ,gs, nstead of the hydraulslope, .9, but only under the condition of unifoim flowVARIED FT.OW

Accelerotetlltw gccurs n any channelwhere he actuaslope exceedshe friction loss per foot.So,E 19.

Figurc19.12SlopesUsednOpenChannel towUnder conditions of uniform flow, all of these threeslopes are equal since the flow quantity and flow depthare consta,ut along the length of flow. With nonuni-form flow, however, the flow velocity and depth'varyalong the length of channel and the three slopesare notnecessarily equal.

Retardedlou occurswhen the actual slope s less hathe unit fiction loss.t r aT

P r r oF E s s t o t { a L p u B L t c A T t o i t s . t ? | ,

__d:::']_**""

E4*tT

Wr)

hLItIvlYtIIZ +

- - - Refcrcncc lcvationFigurr-2:EnergyGrodients NCEESUPPLIEDREFERENCEANDBOOKAG

As shownn Figure -2, inuniform flow both te energy-grodeineond he hydroulic-grodeineore porollelo theboltomof thechonnrT.i o 6rr[yt : yz,ond hus he honnel ross eclionmusl lso econslonl.

vz - - - - -- _slope of energygradient ine2C

v12

llIElro,ol



32/67

WORKSHOPPROBLEMS

22. Calculate he veloclty and flolvrate of stormflow in an 8 ft wide rectangular concretechannel 10.015) if the depthof flonr is 2.4 ft.The chan:relslope 0.01.

(A) 8 ff/sec,155 t3/sec(B) 1 fVsec; 5 ft3lsec(C)4 ff/sec,85 t3/sec(D) 13 ft/sec,250 t3/secsolution: JseCheqy-ManningEquationsolve or v, then calculateQ : Avv: I.4g n ft;2/3 (S;1/zlz (1 4910.01sX1.5t)zr':10.0t;trz_ (ee.33)1.31)0.1)- .13t/ secA - (8 tX2.4): 1e.2tzQ: Av: (19.2 z)(13ft/ sec :250 ft3 sec



33/67

TRAPEZOIDAL CHANNELSExample roblerustormwater lows at a depthof 4 ft in thenaturalchannel n = 0.025)shown.Thechannel asa slopeof 0.01.

l. Determineheflow area:Area : Tana,:opp/adj adj:oppflanq adj_ 4/I.19:3.36f

(%)bh: (%)(3.36X4)6.7 ft2Area2: bh: (9X4) 36 ftzArea3:Tayg, - opp/adjadj opp/Tanq adj 410.g4:4.76t(%)bh: (%)(4.76)(4): .5 t2TotalArea:Area *Area2+ Area3:6.7 f+36f t2 +9.5 P =522. Determineheflow velocity:v - (1.a9lnXR)o'ez1g;o.s : Alp pro,a pA,"ur* par*z* par"uPar*r : hYPer"a sinq: opparuu/hypar* hypar" 1:4/sini: SPAr"u, hYPar.a: sinar: oppn"u: hypa."": hyp*"" 3: 4/slner: 6P: Par"u* PArquz*Ar"u= 5.2ft + t f t+ 6.2 t. :2A-4 t .R: A/P 52 r / 20.4t:2.5 ftv: (t 49/0.025x2.5)0.ut(0.01)o.t(59.6X1.88X0.1):1.2 t lsec3. Determinehe low rate:Q: vA: (112 tlsec)(szt\: 582 t3lsec

4. Determinehehydraulicgradeine (HGL) at this chalnelsection:Thehydraulicgradeine s thedepthof flow : 4 ft

5' Determineheenergygradeine (EGL)at this channel ection:p/r 4 ft z:0 ft (bonom) 1zg_ 11D2164.4:.95tEGL 4ft + 0ft + 1.95t= 5.95 t

Ae 4++ I



34/67

OPEN CHANNEL FLOWEXAMPI,E - TRAPEZOIDAL CHAIYNEI,

The rapezoidal hannelshown s aid at aslope ol.002.The depth of flow ls 2.0 ft. Daterminehe flowrate.Usen: .013. cIRcuL'ARCHANNEL LoWING ARTtAEzamplc2.5cF of water are in uniformflow in a 2d, se(n = 0.015,S = 0.001).Whatare hedepthanity? (Assnnen varieswith depth.)If the pipe were o flowfull, it wouldcarrye6rx

D =201t2 = t.d67ft_ t rR = iD = i(1,662) 0.412tenu: fr(r.ooz)zffi) (o.lrz;a/s: 3.8i1cfs3.8i1utuu:Iai3EF = 1'75t/sec9lQrnin=2.513.830.65.rtomAppendix d0.66 nd r/orou)== .92.So,

u= (0.92)(r.25) t.6t ft/secd = (0.G6)(20)rg.ZnchesPRO ESSIONAPUBLIfr'-T//ONS,NBelmont,U g1Mz

o Actuel: #:#= o.f,istrlom App. assumingthatn rrarie witb de" - 0.68vtul

(a) (0.6Exto.tl)l@lr t- ^(b) ft9;1 App,. the maxinumvuhu..o1.04 8tdlD * 0.9).

vn* = (1.04X16.2t)[ t6s;il1(c) Similarty,Q^,,lQnn=. l.O2 4t dl Der

Qae,j= (1.02X203.?)= 2O7.Ers/secc

o Rrll flowing:

e: (6xz)*3)(l9 -?t ftzp:6* 2rMW- 18.68r*: ft =L2uils: (24|ffi) O.?Bn2t:l 6mr1a5.6 tP/sec(cfs)

PNOFESS/NAL PUB IUTI ONS,NCBelmont,CAg4mz

f t :

EXAMPLE - CIRCULAR CHANNEL FLOWINGPARTIALLY FULL

I =t=toFlomEq./ 1 . 4 e \ ," = (ffi ) Q)2r\ffi = 16'21u6rc

e - Av= (;) (4)2(16.2r)- fr3.7e3/seccfs1This copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


35/67

MAXIMUM F .c Vvfr Fr E : i r o --E+{* + ::4:r_Eq FA IE D ,

^2\-

\s-4-a4, 1 M I N ' M LM F OW

+ $EIYAGEFLOW RATIO CURYES

2 3 4 5 6 7 8910 20 S0 40 60 80100 ?fra 400 6008001000PopulationnThouoandsP)

HYDRAULIGELEMTNT$GRAPHFORCIRCULARSAWERS

1 0I=

6o ' 5o,g 4dta 336a 2ctCDE{}z6 1'rt 0.8{lt* 0.6; 0,5E o.1E o'a=.b 0.2o(rE

0. 1

.rla:"pF(fi5o*i4oooo6E

1

o,90.8o.70.6o.59.40.3o.2o.1

0

.o 1.2 14 1.6 1.8 2.0 2,2 2.4 2"8 3.0 9.2 3.4 3.6

o o.1 0.2 0.3 Q.4 0.5 0.6 0.7 0-8 0.9 1.0 1.1 1.2 '1.3

ValueEof:t*O#,

Hydrauficrementff#r,*r, *r, *o E,) DedgnudfunslructlutofsanltaryaNslomfuwen,WaterPdlulionContelFedetaliEnandAmpdean$ocietyofCitfEoginwra,l9Tfi.160 clvrLEt{qttEERtf,{G

-n,f wtiable withdepth-. - n,f conslant I- lndependent f n,



36/67

OPENCHANNEL FLOWPracticeProblemCalculatehevelocity and low rate n a24 inch circularReinforcedConcretePipe(RCP)stormsewernstalledat a slopeof0.02andn : 0.013, ariablewith depth:

1. Flowing full

2. Flowing ata depthof flow of seventeen17) nches



37/67

NORMAL DEPTHEQUATIONTheequation s derived romtheChezy-Manning quation.As writtenbelow t solves or thediameterof a circularstormor sanitary ewerwhentheQ6,x,oughnessoefficient n) andslope s)are known.

D * 1.335 n erul/(s)o's; 'rzsRearranged,he equationcan determineQ1,1l hen the inside diameter,roughnesscoefficient and slopeareknown.

eru' 0'37s_ D (*) o'tt*)l(t.zz5 (n) 0.37s)Rearranged, he equationcan determine he slopeof a sewer nstallation ifQ611,oughnesscoefficient and diameterare known.

(r) o't (nen ,)/(Dt.tz5)z.etExampleProblemAn l8-inch circularreinforcedconcretepipe (RCP) stormsewer s designedto have a flowing full capacityof 4.4 cfs. The invert elevation at the inlet isEL = 345 ft. The engthof thesewer s 1700 t. Usen:0.014.1. What s the nvertelevation t thestormseweroutlet?

|o,lL335)2'67n Qruu(r) o't D: 1.5 t n:0.014 eru':4.4 cfs(r)o't: n Q* l@1t335)2'67r)o't (0.04*4.qle.slr33s)2'

n


38/67

Opnn cHANNEL Low

A ucir is an obstructionn iur openchennsloverwhichflowoccurs. weirsaredesignedfor lowmeasurphent.Theseweirsconsistof a vertical flat plate with ,h."fenededges. Becauseof their construction, hey arecalled ofrarp ctestcdueils.Sharpcrestedweirs arg most frequeutlyrectangular,consisting f astraight,horiz6ntalcrest.Uo*"ror'*ilmay alsohave rapezoidaland triangularopming;s.If a rectangularweir is constmcted' ith an openingwidth less hatrthechaln{ width, theoverfattindliq"iisheet (called thc nappc)decreasesn width as t falls.This contractionof the nappecausesheseweirs o becdled conhactcdoc#galthough t is tl" ""pp"-tn"tlictuallycontracted..

FLO'W MEASUREMENT WITH WEIRS

Flgure 70-6 QsalTa.ted aud SuppressedWeirs

If it is assunedttrat the contractionsupetreamverocitf"*ir;;'i"_ jr*r"ili#tllcrest, aappe prssureis 1ero, h; ;"pp" [s fidlyilated, sud viscosity,.J*brrt;;; ;;t"rf*" tegEectsareueglig.bederived oil'&. Bernoulri.q."J:lg equationo=I,,Fi[(".*)','_ 4)',J rIf u1 s negtigible, hen

o=?o,fttyf1zEquation q-*l b-e orrectedor all of the assumtions nade' rnis is gg" ty i"t .io.r"g " "o"ni.-Ttf1, to.accountprimarily fo.; ;;-d;"rm vetocityditribution. q =?r@r)tr,Eigyrz p.Cr-r[o.oors+o.oe*#).**FJ [,*o.T*,'r

If the contractlonr.arept suppresgedi.e, oueor bothidsouotrded ril o"Illiffii theuheactuidrh, ,should" reeracJGi li"?lrr"nr uidttLD"ficctt'r -- Decturt (g.lXJVXfi) B.s2ff is one if onesidc is contracted, and JV s turo if therere two end conhactiors. -' -s '

NcEEsupplrEDREFERENCEANDBooKAGE 7 ,

contraAed

EXAMPLE PROBLEM

H :4/12:0.333 ftN = Z andtheeffective idth is

Dcfccttvc { - (0.tX2X0.ggB) B.gg

The Rehbock coefrcient ftom equation/c,= (o.ooa'+o.oEr'Tf).ry) 0*ouH= 0,624

Frorncquation the flow isg= @.oz4xa.es)= 2.52 fs ,/tz to.3s1rThis copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


39/67

lvr=# F =

. O P E T f C H A H H E L F E . O W

ALTERHATEEEFTFfiSFor a gilres fsw rate, the two alternate de.pths avetbg eane energr. Onercpresents -highvelocitywithlcm deptfu the o.ther eprwuts a loct locity with highdepth. The former iEcaltd aryavrit*nl (tWiil) florqthe latter is cailed srabcritirr.ltrqrrqtil) flou.

GNIIIGAL FLOWAIID CRTTIGALDEIf$IHTbre tt onedepth, U*" o th"' *r;depffi that

dt=O.TLBde

F|glra i.18 Sper;ffrclergy taganNCEESUPPLIEDREFERENCEANDBOOKAGEtrbr couynnience,he Itoude nmoba can be rtitteoturuiof the flonr ate per uuritwiilth.

A b iB th nr&EerdFFh= -+F tre+sguled*n"a

(ll =.0.0_I$,s;-b.@2"aniluL $ ft.i$=n)f th.iB250.,f/sc (f no"/4 and ,ho epth i aA fl (L3(") Is tb6 f,ffi tEonquit, citicqt m radflf{"1 {f llE "4slinelflde iu a ftee or*fnlt, B}sr iEbd4kdepte?

(e) Iboa Eq.19.?4Ae c*ricatCrytUis

Since be actusldepth ercedshe qitical depthfcnr ir ksnqdl(c) FbonEq. 19.82, b brink denth s

4 = 0.7154= (0.n5x2.69t)Flgut'r I 9.16 FreeOutail



40/67

DISCTIARGE'ROMTANKORIF'ICES

Elgure17, O Dlschargi frmt a Tank v1:1/ffih: zr -.ze 17.Q'.17,6

V : C"vrA.: C6v2Aa: GaA"{ziE t7.zCa- CoC"actual discha.rgetheoretical Fd;,ge 17.7

NcEEsupplrEDREFERENcEANDBooKAGE i;8 , 63

illustration description

CTILYERTS

Figlvla17,14 SlmplePlpeCulvert'f : CaAt/Zfr

Ca '. Ca ..CABCDEFGH(none)

0.62 0.68r'0;0.98 1.00 0.0,61 1.00 0.60.92 .1,00: 0.gq.e? .0._e0 0.90i54 0;55' 0.90.72 1.00 '0.70.51 0.52 0.90.98 0.99 0.9

:PI l*L,(U"*u sepa.ratesromwalrs)snarplupe(ng separation)sharp ube vrith roundedentranceI*l""''ll fupe, ]engtl l* tha,none_half f pipe dia,meteryenfrant tube,.length2 to Bpipedia,rneters50roasmooth,well-taperednozzle.@Ashort tube hac a length letg then 2 to 3 dia.met+rs.This copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


41/67

HYDROLOGYHYDROLOGY: study of the circulation of water in nature.HYDROLOGIC CYCI,E: The movementof water on earth hrough its differentphases solid, liquid, vapor). Poweredby the energyof the *un. th" total volumof water is constant.EVAPORATION: The rate of water in the liquid state ransformed o vapor.Evaporationoccurs from openwater, baresoil or vegetationwith baresoil benethe vegetation.TRANSPIRATIQN: The part of evaporation n which watervapor enters heatmospherehrough the plants.The o'transpired"water originated in the soil.INFILTRATION: Water entering he soil from rainfall, snowmeltor irrigation. Aused o describeunwantedwaterwhich enterssewersystems hrough leaky oin

and cracks n the sewerpipe.EROSION: Removal of soil or otherchannelmaterial causedby the movementowater over the channelsurface.The removedmaterial is termedsediment.SEDIMENTATION: The filling in of lower velocity areas n channels,ponds,etcsedimentpreviously removedby erosion.

: Designedand installed to reduceandcontthe peakrunoff from developedareas, hus limiting the runoff rate to pre-development f possible.Also termed,equalizationbasins or capturingpeakfloentering a wastewater treatment plant.

RETURN PERIOD T or F: Also calledrecurrencenterval, returnperiod andfrequency.Measured n years.For Floods, t is the period of time, determinedbhistorical meteorological dataand statisticalanalyses, hat the streamdischargea result of randomrainfall) shouldequal he predicteddischarge.The return peris important to determine he designed evel of protection for the channelor damstructure o reducethe risk of flooding. Onceconstructed, he dischargequantitshouldnot be exceeded r floodinganddamagemost likely will occur.DESIGN DISCHARGE: Maximum capacitya channelor structure s designed ohandle. The smaller the designdischargeof a channelor structure, he lessexpensive t is to construct,however, he risk is greater hat flows may occur whwill exceedhe designdischarge. xamples:a 10-yrdesigndischargeissmallemagnitudecompared o a 5O-yeardesigndischarge.PROBABILITY OF OCCURRENCE: The reciprocal of the return period. 1/T = p



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RAINFALL INTENSITY DURATION FREQUENCYRECIJRRENCEINTERVALFREQUENCY F)RETURNPERIOD (T)UNITS:YEARS

PROBABTLTTYP) OFEVENTOCCI]RINGP= llF OR llTUNITS: DECIMAL OR

PERCENT

PROBABILITY (P) OF EVENTNOT OCCURINGIN THE NEXT YEAR(l-P)UMTS: DECIMAL ORPERCENT2 Yz:0.5 100 50Yo I _ 0.5 0.5 * 100:50%5 l / 5 : 0 . 2 * 100 : 20Vo | _0 . 2 : 0 . 9 * 100 : g 0%r0 l / 10 : 0 . 1 100 : l 0Yo 1 - 0 . 1 0 . 9 * 1 0 0 = 9 0 %25 l l 25=0.04*100:4Yo I _ 0.04 0.96 * 100 g6vo50 l/50 0.02 100 2Yo I -0 .02=0.98 100:98Vo100 l / 1 0 0 : 0 . 0 1 1 0 0 = l Y o 1_ 0.01 0.99 100 ggYo500 l/500= 0.002 100= 0.2Yo I - 0.002 0.998 100 99.8YoExarnple 7.1Determine the return period and intensity of a rainfall in the Columbus;Ohio area for a rain lasting two hours if the total rainfall depth wasmeasured o be 3 in.

Solulion;Frorn FigureT ^l a 2-hourduration rain at 3-in. rainfall depth correspondsoa 50-year return period.

ocE s. 8 4i 3dcE 2o

4$ rct

4n l--41zIT 4,z_4+I

5 10 15 30 60 lm 10OO Minutesl 2 3 6 1 2 2 4 H o u r s

0.5

0-1Duration

FIGURE ?.1 Rainfall depth, duration, and frequcncyat Coh.rmbus,Ohio.

By definition the rainfall intensity is the rainfall depth divided by itsduration in hours;hence,

I :1= 1.5n. /hourThis copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


43/67

IIYDROTOGYTOTALSTR.EAM SIJRT'ACE RUNOFF FROMI{YDROGR'APHS

the time frompeaknrnoE o cessation.

isthigrorrnfi. Thcraiawhich actuall{-appcan as inncdlstc ru:roE lr hrovt u wfuttruf, ourladfus, wfwfuq atd ndrai'r. Thc watcr sHci irabcorbcdby f,hesoil aud whlch doca ot contdbutc to tlc qnrfaccyaoEir loorra as h*floq grorarrretdct,tfuddo4 arrd&yutluflos.

o Ftgsre Ey&ograph SepalatimMethod1Oncc he bas lw ic scparatedout, the bydrograoverlaadlow will tslr dn tne "ppgi"rn"tfigule

UI.[T ITYDROGRAPHSFlgure A Streem Eydrogaph

In figure the portion of tbe hydmsraph to the leftof the d b h.own as the rhing Iimh. To the rigbt ofthe crest, the cume fu hown as the rcccesion fat;rgIimb.The strean dinchorge c asEumed'toconsist of oacrlarulfoo aud goantdwatct /ora Since cuherts do not haveto be designed o carry groirudwater, aprocedure cqledhg&ogmph sbparationot hgdrqra14h a,llallyshs nces.rary to oeparatesurfsce ant grorudwater.sThere are serrsal method.s of separating basflow frourovcrland flow. All of the methods are somewhst arbi-trary. Three of the methods easily ca,niedout aanuallyarc prescuted hetC.

Scthod J: (Straight Line Methoi): A horizontglline is dra.cm from the start of the ri+'ing limb to the fafling lirnh. 'fitl of theflow r:nder the horizontel line ie consid-ered base flow.

Onceh owrtendtrm lfdfogapb fq r stqra hs bdsntqc4 thc total rehhll mtunc csuL forudfi+-learceund:r t.haorrva ftrtt@cq sinccbc uethr aniuegc badb il fDffi4 tt erdgo frecipitacsEbeP = -Ad

J$ nre$$$nonrill bcs@numbcr f in&er (l.T. qr-0:{P, g{rc.).f arerypoiuro thehyarodtodividd br thi 8n"TqcGpit*im, I urit htdgraphwiUbGdcrivd.tit nra tydmgapf of astdrspltrDtz of r8in o thc ertir.bd".' Figurc+-t howr unit hydrogr?Dhonpsrs6 s dyatormdroppingm sverap otf.T oirah.

tFtgurc A Stonn ard Unit EydroefaphNCEESUPPLIEDREFERENCEANDBOOKAGE

blr. dmc brrt \

unit



44/67

HYDROGRAPHExampleProblemA drainage asinhasan areaof 12,000t x 8,500 t. A stream ollectsunofffromthe drainage asin.A recentstorm esultedn theflow rates ecordedn thstreamat the times isted n the ablebelow.StormHour Runoffcfs GroundWatercfs SurfaceWatercfs Unit HydrographcfsStart 0 200 200 0 04 29s 200 95 588 4t5 200 215 1 3 1t2 535 200 3 35 204t 6 425 200 22s 13720 300 200 100 6IEnd24 200 200 0 0

1. Determinehepeak unoff,and he ime of thepeakmeasuredromthe startof the runoff: 535cfs athour 122.Using he hydrograph eparation ethod, eterminehe otalvolumof surfacewater esulting romthestorm areaunder hehydrogra

From he able, he sumof the surfacewater data:95+21 +33 +225+100970cfs(970cfs)(dataecorded very4 hours)(3600econds/hour)13,968,000ubic eet - 104,480,640allons

3. Determinehe average recipitationhat ell during hestorm:x precipitation total vol. of surfacewater drainage asinarea: (13,968,000t)(rz irlft) l(12,000ft 8,500 t) : 1.64 n.4. Determineheunit hydrographlow rates:Surf,ace ater low rute T precipitation Unit flow rate l-inAnswersare n the astcolumnof the able.



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FTYDROGRAPHExampleProblem

"Stream"Hydrograph: SurfaceWater Overland low) * Groundwaterbase)low

"Storm" Hydrograph:'osurfaceWater"Hydrograph o'Overlandflow'' Hydrograph, rn-Ho-+ - i, \+t ATa

: l o oi i r-:l1 inch stormo'unit" Hydrograph

UNIT I{YDROG,RAPH:Hydrographof one-inchor l-cm of precipitationoveradrainagebasinarea watershed).t is consideredo be he hydrograph romwhichothehydrographs, nd stream low rates,canbe determined y ratio whenotherprecipitatioquantities < or > l-inch or l-cm) for the samedrainage asinareaareknown.4. Determineheunit hydrograph low rates:

Surfacewater low rate x precipitation Unit flow rate 1-inThis copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


46/67

WORKSHOFPROBLtrMS33. a unit hydrograph or a stream n a drainage asin s plottedon thefigure. A storm occurs overthe entire drainagebasin. Runoffimmediately ncreaseshe stream low (discharge)and welve hours

after the beginning of the storm the streamdischarge s measuredo be150m3/sec. etermine hepeakdischargemeasured uring he storm.

(A) 260 -'/s".(B) 86 m'/sec(C) 175 'lr.t(D) 118 m3/sec

18time (h )

70Q 6 0'L 50E +o-Itlk 3 0E r oE 10

A measuredpeak discharge *i/.qQu hydrograph peak discJmrge *-/=Q" measuredtscharge atl}h o"?/tQtr: hydtPg.?Ph disqharge at 12 h *3/tFioro- the illustration, at 12 h the dis^harge is 34 m3/santJ' he Saglag. station measures'l5O m.Js- The tnithy.drirgraph dhows the peak occurring at 6 h with a dis-cha,rgeof 59 *3/"-

a Q "Q n Q n -

A QnQ"V- - -= -/ rt'3 \l s g : - l_ \ E /

'aAt t -

: 260 */*

(ruo+)em"

;



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PEAK R UNOFF FRO\4 THE RATtOxlr,-Although otal nrno$data a.re equired or r*ervoh anddem design, the inqtaurareouli pealc unoff is ueededo-sizeculverts ald storm'6funinq.

eveutually equal the rate of rainfdl. Tbe time betweeotbe start of rainfall asd the time of pea.k lorr is loovraas the time of concentration, f". Tlpical naluesof t" forareas ess than 50 acres range from i to B0minutes.Tbe -rotional lormuldis based on the assumptionsgiven .and has been in widespread use for sinall areas(i.e.l hgstha,a several hundred actesor so) but is seldomus"a f*areas greater thal five square miles. Values of 6 arefound in appeudix

Qp: CIAa

AppendixRational Method Runoff Coeffcat_eorize_d y surfaceroresteclasphalUbrickconcreteshingle rooflawns, well drained (sa.udy oil)up to 2% slone296 o t% slojte- over 7% slopelawns,pog: drainage (claysoil)' up to 298 slope

l% ro Z% slopeover 7% slopedriveways, walkways

0,0590.7-00.7-0.0.8{r.0.75-0.0s40.r0{0.1H.20.13{.0.rE{.0.25{t0.7H.

lv"ragerrinfellIntcnsity(inlhrl I

Figureduretion or lime to concentrarion. rgIntensi ty-Duration-FrequencyCurves

pastureunimprovedpaslcs'cemeteriesrailrold yardIITflT*_(ercenr asphatrrconcrere).2+.sbusiuess istrictsueighborhoodcity (dorvutown)resideutialsingle familymulti-plexes, detachedmulti-plerces,attachedsuburbao. apdrtrrents, condominirrrnq

industriallishthearry

0.05{t.0.05-00-t-0.30.1-0.20.1-0.20-2-0-4

o5-o.70.7-0.90.3{t.50.4-0.60.H).750.25-0.40.5-0.70.5-0.80.H.9

The follorvingstepsconstitute the rational.method.step7: Estimate f.. This is the sumof the overlandflow and eonduit flow times. t, will changethe farther you get from the drainagearea-.slcp 2z Chgosea valueof C. If more than one areacontributes to the runoff. C is rveighted ythe areas.slep 9: Selecta frequencyor return period for thestorm.stepl: Caleulate or deternrine he average tormintensiy from int errsiy-du ation-fiequencyeurues.step 5: Useequation to calculate he peak lorv.step 6: Use open channel florv design echaiques'tosize he channelcarrying surfacewatei away.

NCEESSUPPLIED-REFERENCEHANDBOOKAGE



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EY-DROLOGYEzamplcTwo adjacent fieldscsntribute runoff to a collertorwhosecapacity s to bedetermined.The intensiqyoi a25 minutedurations 3.9 n/hr-

A1 . 2 rcrcrC1 0.35t l . 15min.A2.4 acrcrC2. 0.5512 l0 min.

stcp: The overland.low time is given for berea . Tbe time for water from the fartcotaer to reach te colleci,orstc= lS*10=2Sminutes

stcpEz The nrnoff coefrcieutsere given for earea.Sitrce 1*Tt to size he pipe carrythe totd runof, thecoefrcieutsareweishby their respective outributiugareas.c _ (2x0.35)t )(0.65)= 0.552 + 4

I

2

ExampleA dralnage area has the followiug dhdracteristics:

area &ze. overlandflow C(acres) time (min)

should be used o rize liuc

Tho hstaotaneous rain&ll iltDsity for the area (and aepeCf,ea$trm freqtiency) ic grl,EtrbyI t Er=;ifu" (tD/t')

r= _'11520TIE = 3.29iuAre = (0JXr0)(s.29) 9.87cfs

step & J: The intensity for a 2Eminute duratwasgiveuas3.9 nftr.stcp5: The total area e 4* 2 : 6 acres.The peflow is foundfrom equation6.2g.

qo: (9.58)(g.gxo)t2.ecfs

For areaB:At t : 5, J : S.?S DAE. The nrnoff from 2 acres(0.7)(2X5.25)r 8.05cfr. Hsweverr t r = (20+ r), tpeak ,ow from area.r{ nill rearh the secondma,nUAt t = 21,1= 3.rgh/hr.Thegun of CA valuess (0.gxt0) +(0.2)(A)= a.a

g -- (a.a)(3.19) 14.0cfrI_tievalueofQ shouldbeused o designsecdon2of thpipe. $t.he plpe is As$med o flown lt, tle regired ianeter is fourd from o,pm hannel ttw consideratiod=(1.3i)t{ffitg]"'= r.?efttouao2.Fpr areaC:Assnrni4gI &o s" the flow wlocity in the pip, t;ti'ne from the start of thestorm for waternoilpiato reach he 3rd rnrnhols s

2 0 + l * I = 2 2Siuee2$ is larger thao 22, the maxinum nrnofr woccur25minutesafter the start of the storm. ltre 2milut datumis not used.

. r=; . l l -=2.8TSThe1'" of CAvaluess(O.BX1O)+(0.?X2)+(0.= 10.4. Q- (r0.4X2.875')29.9c&

PNO ESSIOAL PUB CATIO NS, NC,Belmont,CAg@2

a r 0B 2c 1 520 0.35 0 . ? :23 0.4

Thir value of Q (9.8?cfs)#t.



49/67

\MCR.KSI{CF PROELtrh{S

34. A 500 t by 1,500 t site s presentlycoveredwith woodlands.Adeveloperhasproposedconstructionof single family and multi-familyresidentialuse.PlanrLingas50% of the site o be zonedand developed ssinglefamily residential.Another 25o/o f the sitewill be zonedanddevelopedasmulti-family residentiatr. unoff coefficientsfor the variouslandusesareas ollows:C : 0.15 or woodlands;C : 0.40 or singlefamily residential;C : 0.50 or multi-family residential.

3a.@)Determinehepre-deveiopmenteak unoff rateusing : 4.6'lnthr.

i = 4-6 in/hr500 ft x 1.500 tA = _:_ = '1..?.2 aCIg


50/67

35. Four 5-acreareasdrainstormwater nrnoffinto a 1200 t stormsewer n: 0.013, iope 0.005). :The runoff coefficient for eachareais 0.55.The time of concentrationacross ach5-acreareais15minutes. nlets to eachstormsewerconnectionare 300 feetapart.Theflow velocities n eachsectionofstorrnsewerare as ollows: Section1: 4.6 t/sec;Section : 5.3 l/sec;Section :5.14 ft/sec.A stormwithrainfall intensity according o theequationshown s to be used odesign he storm sewersystem.

WCRKSHCF FR.CELEV{S

300 i 30a t s00 t ' -sm tt

; - 1 0 0' - - [ + J oi is in incheslhrt" is in minutes

35.(a)Thediameterof thereinforcedconcretepipe (Appendix 16.F)to be installed or stormsewersection1 is closesto:(A) 18 n : 100e) z4n i : ffi:4 iqrlu D=1.33*\"- - t1.s3)g'413x(c) zT \Vs7 ' L Vo;o0(D) 21in 8. : CiAd : t-73 ft.Q= (0.ssx4xb)'11 n3/.seccrs) n = 21 in [standardpipe sizb]

35, (b)The flow rate in storm sewersection4 is closest o:(d) 39cfs To detennine he flow in section 4,the longest ime of concentration1El) 4i cf's is needed.The lengthof eachstorm sewersection s 300 ft (given) aniC) 11cfs the velocities n eachsectionof stonn sewer are alsogiven.(D) 37cfs

Stonn SewerSection Length(ft) Velocity(ff/sec) t": length/velocityI 300 4.6 1.09min2 300 5 .3 0.94min

'1J 300 5 . 1 0.97minSewer low t" 3.00min

fc : L

Total t": 15 min (time across irst 5-acre area)+ 3"0 min (sewer flow): 18 min. 100'i : ,5ff10 :3-57 in/iu q _- (0"55X3j5?)(20)'3e-?ne1**s

tII 2 v 3



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RE=SEBYOIRIZING.'[tc rcseryohyield probleu ic aa accouBtlng roble,n{ic., tecpitrg tncl of vhat cmer n aud whet goelo$). Thc Frupu of tbc rsrvoiryield analysir r todtGmine the proper eizc of l dem or rcswoirr c toeyllustr tb abillty of ea e*isting dsn to mect watcrdensndr. fhere are threc badc Eethodr of sotving hereremruhyiddproblen"

Fhttbe *on Aq n(ebobmnrat Hpf, ttiogrem)ontbs rc 6voir. Tbi i a eimlteie,ou pilot of thc.susulattvc darrrnd (luowl er drqfi)edcmraledw inflow. Thc masedlslfen i rrred o graDhtcdlf daauho the rerormir stffilg reqdiru dr (ilG.,drc)-Ar loug ar tbc clopccof thc cmulastvc deoiud adinfrorv tUpr atc eqttd, thri water Tsenf in the relsw6j' vill rot e&e.gE Wh6 tbe dope of itc lrf,ffi iIer than the 8lW of tbGdofl4 tbp inf,sr ern'rotb i{scf EefiEryba cmwiffa'weter rud+ 8hdthtcsEr.Tolrh dwtr dmra to m*b ql thc diEcaeoca dpcek trlowcit bv a trmsb ir, tberefm, e&uugb cbc'dds.lf tbF rdrqsbir i tb bc rizcd co thag thG eornrrnnit]r

In oadcr for thereaemoir o nrpS coougbwatcrdltrbga abouC 0mdidou, tbe reaewo& irut bG oll.Fitr toth6 $rst of ths druu$b Thb'&gt ig not te@edwhs tlic nnrr disgraa i drry ' bcnca hc adtodruw rpeeudtrdcntaud finc prraltd to tb psk

After a drorrghtgqal to thc capacityof tbe nessvoirthr reg*volr wiUbs eupty, At tbtrougb howe*er, heresreobwilt begb to filI up 8$b- Who tbc crurrlstiva mer cNcesdrha raervoArcafecitf, the reservojr will bseGo epill (ilc., relcrse)ruatc 1tir oceuswheutbc cunrtatin bfow Snccros6e he Frior gaktpaedo-AensDd ins a Fhffi infigpn 0.29.A f,nd-qaW 4o fu hdlt t0 bep wrter b anduuatbe sized o that i'Etr i ldt cpilcd. The uasr dtg,FaEcan*ill bc uss4 ht trls nra*irrmr Firaff@ b,tsetrougb -aod subec{IunfDGs& uot pGEbfollmed btaouds) becmee thc caplcitf.

.EG;t.5ET.ItEil

Ftgrlrc

E)hIVIPLE ,1. A ros?ryolrh ncedad o provlde20 acn-ft of wetersrch inonth.Thr Inllow for eachof 13repreaudatlvrnronflu lr ghenbelow. Slzc thr rescrvolrbyultdovcrmeehr you ehooFr.Asaumcthc retervolr t til futl.r N O N t f tM A M J J Aldov 30 6020 10 5 10 5(acrs-ttl

8 O ' V D J F10 20 90 8575 50 lf thc rcscit,oir hgd been futl in Atrf , tbc ruxinuEshort&fl rcutd hsrn bsn6{l ecr+& and thc reserrrirnilld be empty n O


52/67

o-i-+\ 8 \ \ 9r lvbf)aQAoN+20+irSenalroauir1,Iruttda

J_t -or

,q.49ay



53/67

WATERSTOR.AGE4 *rur diagEa,Bor a mr:.nicipalwater storage.an i.ssbcmm.Wbat ie the milimu:a requiredcapacityof thestorage taak?

PROBLEH 2b

50004500

(B)E}^. 4000E' o 3500I;e 3oooJ .I zsooo '* zooog5 .g .1SOO5o 100{tl2 16

tlme (hl

(A) tB00me6500m313000ms45000ns '

Eoi:utron zofls rnftn{mrrrnequired ank eaBacityudetermhed. E-i;q tbe slopo of the avenagelow. T1ris s.detecmiaedby drawin8 tbe eve'ragelow line oathe ilhrshation aadtheu dra;wlng iaeeparallel to the anreragelour io" thai.aaeaiso taogent to th6 manimurn'd,irviatipuntom theavera,gelow line, The verdcC separatioaoJthepare,lletlinee representshe drinlrnrul requbed storage oir:me.Passing a line throug . the'ordinate. o the eunmrtativevohrme at 24 h,drarrs the averagelow lihe.

4500. ^ 40.00dttr; 35oogE qooo. : Ip. zsoo

'E zooof rsoo:t(.it 1000

V storage volum.e mglhom the illustration" tb vertical separation oftheta,n-gent parallel lines iE

P: 13000mg' Theansw.prs (C),

. E S S I OThis copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


54/67

G R O U N D W A T E R

-D-+_Fglr9St+tMovement of groundwater through an aquifer is giveuby Darcg's lazr,Eq. 21.10.4The hydraulic gradientmaybe specified n either ft/ft (m/m) or ft/mi (m/km), de"pending on the units of area used.

Q: -I{iAgroo : -vur{g.o* 21.10Tbe specific discharge s the sa,rre as the effectiue ueloc-itg, v.. nc : = L : R i : v e 2 1 . 1 1AguDarcy's law is applicable only wheu the Reynolds num-ber is less than 1. Significant dwiations have beennoted when the Reynolds nurnber is even ss high 0s2. In Eg. 21.12, D-rro is the mean grain dia,meter.

T}le hydroulic Arodient, i, 'is the change in hydraulichearl over a particular distance. The.hydraulic herld ata poirrt is determined as he piezometricheadat obser-vation wells.

FTYDRAULIC RADIENT

. AI/, : TTBANSMISSIVITY

kansmhsiuity ("1"o known a.s he coeficiart of trans-missivity) ie an indoc for the rate of groundwatermorrement.The transmissivity of flow from asatuatedaquifer of thidrness If and width b s given by Eq. 21.13.The thidoess, If, of a confined aquifer is the difierencein elevations of the bottom and top of the spturatedfornation. For permeable soil, the thicknese, Y, is thedifference n elevations of the impermeable bottom andthe urater able.

2t.0

PgD^.o QD-".or - t E : _ p v 21. t2 T: KYCombiningEqs.21.10and 21.13gives

Q: bT i

21.13

21.14

For studies inrrolving the flow of water tbro'gh an aoui-fer, efiects of intrinsic permeabilitSr and the water a,rcombined into the hyilmalic coniluctiuity, also}nown asfhe m_eficient of permenbility ot qimdV the pnneabiLity, K.

PEBXIEABILITY

tabte2l.l TYPical eml,abillliesk(.-') k(darcys) K(cm/sec)k(-') K(gallday-fts)

gravelgravelly sandclean sandsa,ndstenedenseshele orlimestonegranite orquartziteclay

10-5-10-310-510-610-E10-e10-r l10-rr

10-r-1010:r10-210-81.0-510-710-7

103-1051081031 0 .10-r

50-50005050.50.0050.000050.00005

104-10610110 31021L0-2t 0-2

10-310-3

NCEESUPPLIEDREFERENCEANDBOOKAGE 51

P H 0 F C S S T O H A L P U L t C A ? t O t a 3 , t N C .This copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


55/67

WORKSHOPPROBI,EMS

36. Groundwatermonitoring wells areinstalled nto anunconfinedaquifer asshown n theplan view. Theelevationsof the monitoring wellcasing ops (at grade)are isted in thetable.Also isted or eachmonitoringwell is the vertical distanqe rom gradeto thewater table.The averagehydraulic conductivrtyfor the aquiferis 0.42 mlday and he effectiveporosity s 0.34.

casing top elevation . grou_ndwater depth(m above mea.n belorr casingwell sea level) top (m)

r MW-s/f\Ir- ' I IvM-z/T\

torthDetenninehe compass irectiothegroundwaters moving.(A)N 4soE,(B) N 4soW(c) s 4so(D)S 4sow

I'1T\YmonitoringwellMW-1MW-2MW-3MW-4MW-5MW-6lr[w-?

4s.n49.7449.5949.6049.094 1.3149.63

4"745-665.595.955.575.254-62

, . . . . - . - . . - . . . , . - , . . . i - . . r .

sa.qing top g:ou.Ldwater grdgndwir,terelevation depih ekirration

(m above meqn belorv casing (m abo-ve mba,nwell sea.level) top (*) sea level)

MW-1 49.77MW-2 49.7+MW-3 49.59M\H-4 49.60MW-s 49.09MW-6 49.31mrni-z .49.63

,t rfA

b.ob5.595.95u .u rt r qr, O -'Lr)4.62

45.0344.0844.0043.6543.5244-0645.01F'rom he water table gradient,the directionof the groundwaters S 45oW'



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oa

AQUIFERS

,IIANDBOOK OF ENVIRONMENTAJ.ENGiNEERING

tCI.EA E

E g5 aEE V



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FIGTTRE.I3lmmiscible lumccssdensehsn r, t

FSGURE .T4Dicsolvedeodtaminatioodume-

FXC{JR,ETsImmircible tumc cnscrielr*.rcs



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(a)FIG- 5.ll Pumping ffccron uquifcn (cl Elfectof pumping n conr:of drprcssirrn(3t.

FIG' 5J rcontiitueul (Er Effccr of aquifcr marcriaron coneot dcprcssinn(2r.A

FrG. 5.t (crrntinucrrrt.l EITcct f ovcrrappingierr n influcnce umJrcrrcilsr3r.

NCEESuppltED REFERENcEANDBooKAGE I 5 clThis copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


59/67

WELI, DRAWDOWNUNCONF'IN-EDQUIFER. DUPUT EQUATION

If the drawdown is small with respect to the aquiftirpbreatic zone hickness,Y, andthe well completelypen-etrates the aguifer, the equilibrirrm (steady-state) welldischarge s given by the Dupdt eqtation, Eq. 21.25"Equajion 2L.25 can only be used to determine the equi-librir:n flow rate a "long timen after pumplng has be-gul1.

21-27

21.?5

In Eq. 2l.25r gr and lz are the aquifen depths at radialdistanccls- 1 end ry, respectinelyr.from the well. gi.canalso be taken as thd original aquifr depth, Y, if 11 isthe well'g rcdhl's oJ influence, the distince at which thewell has go effgct on the water table levelNCEESUPPLIEDREFERENCEANDBOOK AGE | .-'r ;

water table/ , / / /. / ^ ' t r / AAuter .;-.' ./

/ Phrertic zone' , / / /

Factors for Both EquafionsQ = steady low rate (ft?day)K : coeffiicient f permeability (hydraulift3/day-#

v Vadosezone f1= somGdistance from well head ) 12

yz= aquifer thickness minus drawdown aY = aquifer thicluress

IN OTIIER REFERTNCES. iUNCONFINED AQTIIFER i--GradePumpingWell

Piezometris $urfa6s -'Z

Q:2rK1mffio Hydrology.Fourth Edition,

m: AquiferThickness

Radial flow to a well.in a confined aquifer.This copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


60/67

WELL DRAWDOWNUNCONFINEDAQUIFER

PRACTICEPROBLEMe- r$@7:rnT= fr'lday- h(rJr)K: fts/day:FQ = K(hl - h?\ Q in gpm,L055tos(ir/r)

Kin gpd/ft21055=

3.14 0.4343lnn= log. OR 2.3 ogr= ln ,An 18-indiamete_rPumping well penetratesan unconfinedaquifer that hasathicknessof 100 t Two observationwells ocated100ft ano igS ft from thepumpedwell havedrawdowns of 22.2 t and 2 .0 ft, respectively. f the flow issteadyand K (or K) - 1320gpdfif. What is thedischaige rom the pumpedwell?

K(h? h?)O'loe(r"/r')

h2hLo

1055lag(rJrr): tog(2Jf-_t"*oQ),,0,37.L07: tog- iI-:'79 fi,l: 100 2Z.z=li?;A:irg- tzzoi?g it.:i;,\105s 0.1:7i07': 634gpm

o : f t&7 :n i l


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HYDROLOGYNCEES UPPLIEDREFERENCEANDBOOKAGE ,3 J-ISEEPAGE AI{D FLOVf NETS

Gror:nd.water flows fron locatioas of bigh hydxaulichead to locations of lower hyilraulic head' Proble-sreqr:iriag tJreactual calculatioa of the flow quantity areUestsoGd with ibe flow nbt coucept,pa'rticularly whena manual solutioa is requiied'FIow netsare constructed from streaulines end equipo-i*UA fi"o. Str:rr,mhnes(fi'au lhesl sbow tbe pathtalen by the seepage.Equiqotentfal afle'eoanect.?oi-"ltof consta.nt.p-ressrre- he flosr net.colcept is linited toca.seshere the flow is steady, hn+'dimeu'siopal incom-p"*iUft, tLrough a homogeneorrsmedirun" and wbereinu Uqoia fia,ca constant viecosify'

.The flow uet-is conshrreted accordiag to the follorvingrules.o Strearn'llte enter and leavepervious suriacesper-.' i"rra"iiiil do'ihose urfa'ces'o Streamlines approarh the .lrne of seepage abor-ewhich there is no bydrostatic pressure) asymptot-icaIIY to tbat surface.o Streamline a.repasallel to but cennot touch im'perrious surfaceswhich are st ea'uline'o Stree'nlines are parallel to the flosr direction'e Equipotmtial tines are ilrem perpendicular tos'"reamlines, such t[at theresulting cellsare squareand the intersection's a're 90" angles'o Equipotentiat lines'etrter aod leeveimperviots sur-faces perpendicules to tihoseer:rfaces'

The size of tJre cells fom.ed by the intersection of t'h'estreamlines and equipotmtial lines is not important, al-though the more cells forned, the greater will be theacfltra,cJLTte object of a graphigal solutig is-to construet auehrork of flow paths (outlined by the streamlines)and equal pressure drops (bordered by equipotmtiaLliaes). No fluid flows ecro$t streenliles, and a consta'utt*ottot bf fuid flovrs between any f,gts5t'reqrnlines'

TableApprqiruate Coeficients of Permeability' Kpft/daY or

sataiat gsTday-ft2 g?f,lday-fr,2 -darcys

Once he flour net i6 drauro, t can be used o calcuthe seepage. First, tbe number of flow channefu,behar'eenhe streprnlines s cbunted. Thm, the uumof equipotential drops, tVp,behveenequipotential iis cgurted. The total hydraulic head is determineda ftaction of the water nrrface levels.Q=KpE#l)

f igure ll.3 TYpicatFlowNets

ClaySandGravelGravel/SandSaadstoaeDon.e shale& IirnestoneQuartzi'regranite

1.3EE-31.3EEr1.3EBI1.3EE31.3EEl1.3EFr-l1.3EF,-3

EF--zEEi}EE5EFlEEz

EE-3E,EzEBIEE3EE1EE.1E13

1.0EE-2

i'i'.t:-:il..).: " concrEt:=:'ii'F:ll+iii

surfacsof segpage

einh Uam

earrh dara



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SOLUTION,(a) The *5st irnPortant stap ia drawing the flcm net is#.;;;h"'diur *ad 'oil l*y"" to scale' Then' tbeflow and..equipotential lises cair be drarnn' While{argeamouata of time cau be takea to get the flow aet'"al-m*t f*fu*,o there ie uo iirstiflcatiol iil this problemf.- i.ii"g so' Likewiee, there is ao ueed to use a largenumber of flow Pat'hs-

HYDROLOGY

INC- s BeLuoat' CA

R,EOUXR,EH4ENT(a) Draw the seepagelorr net. Include both flow E*d liou, of equifotential piessuredrop'(b) Calculatean approximateseepageate per foodam width.(c) Discuss(qualitatively) the value of the cutoff w

(b) Wiih 5 flow chanaelsas.d 14 equipoteutialtU" Aott rate ie aPPrcximaielYno. flsw channelsq = k p E * @

I n o l \ / 5 \- {o.ooe# )(40ft);J\ day-fr,-l . \r?/- S.0?1gat/daY-ft

(e) 'While the cutoff wall disturbs ttie flow uet sih* efect past the second lorv liue is mininal' Tofieall doe: not sigaifcantly ' iach" downthesThe eff6ct of the oat i" ptobably lessthan 20%

40ft



63/67

WastewaterTreatmentCollection Systems9: -SUN9JP-+-l-W-A-q-r-Y-+IF-EMunici,pal wostewater is the general name given to theliquid collected in sanita.ry sewers and routed to muaic-ipal sewage treatment plants. Many older cities havecomb'ind sewei systems where storm water a.nd sani-tary wastewaters are collected. in the same lines. Thecombined flows are conireyed to the treatment plant forprocessing during dry weather. Dufing wet weather,when the combined flos' exceeds the plant's treatmentcapacity, the excess flon' often bypasses the piant andis discharged directly into the watercourse.

7. GRAVITY AND FORCECOLLEGTION SYSTEMSWastewater collectioa systems are made up of a network of discharge and flow lines, drains, inlets, valveworks, and comections for traasporting domestic andindustrial wastewater flows to f,reatment facilities-Flow through gradually sloping grouifu sewerss themost desirable mban.sof moving sewage.siuce t doesnot require pumping enersr. In some nstances, hough,it may be necessar5ro use pressurizedfore, mainstocaxry $ewageuphill or over long, flat distance.

4. UTIASTEWATER UANTITYApproxim.ately 70 to 80% of a commu-nit5r's domesticandindusbrialwater supply returns aswastewater- Thiswater is discharged into the sewer systems, which mayor may not also finction as storm drains. Therefore,the nature of the return system must be Inown beforesizing can occur-InfiItration, due to cracks a,nd poor joints in old orbroken lines, can increase the sewer flow significantly.InfiItration per mile (kilometer) per ir (mm) of pipediameter is limited by some municipal codqs to 500gpd/mi-in (46 Lpd/h.-mm). Modern piping materialsand joints easily reduce the iufiItratiou to 200 gpd/in-mi (18 Lpd/km'um) and below- I-ufiItration can alsobe roughly estimated as 3 to 5% of the peak hourlydomestic rate or a,s10% of the average rate-Inflowis another contributor to the flsw in sewers. Tnflow iswater discharged into a sewer system ftom such sourcesas roof down spouts, ya.rd and area flarins, parking areacatch basins, crub fuIets, and holes in rnanhole covers.Sanitary sewer sizing is commonly based on an sqsumedaverage of 100 to 125 gpcd (380 to 474 Lpcd). Therewill be va.riations in the flow over time, although thevariations are not as protrolrnced as they a,re for watersupply. Hourly variations are the most significant. Theflow rate pattern is essentially the sa.me ftom day today. Weekend flow patterns a,te not significantly dif-ferent ftom weekday flow patterns. Seasonal variationdepends on the location, local industries, and infiltra-tion.Table 28.1 Ests peaking Jactors (i.e., peak multiptiers)for treatment pla^nt iafluent volume.

Table28.1 Typical Vafiations in WastewaterFlows(based on average daily flow)flowdescription t5ryical time location

typicalvariation

daily averagedaily peakdaily minimrrmseasonalaverB,ge .May, Juneseasonal peak late nummerseasonalminimrrm late winter

LOO%treatment plant 225%outfall $Oyotreatment plant 40%LOOTor25%90Fd

1O o 12 a.m.12 a:rn-4 to 5 a-m.

carry the peak flow as a gravit;r flow. I-n' the a,bsenceof any studias or other justifiable m.ethods, the ratio ofpeak hourly flow to average flonr bhould be calculatedfr9m the following reLrationship in which p is the pop.ulation seryed in thousands'of people at a particula.rpoint in the network.Qp"*t 18+\/F_--.---:-4+ \/ P 28-1

POPI]LATION EQUTVALENTP in 1,000s

: (BoD: mg/lXQgpdX8.34b/gal)( I O"gaVMG)(000 person )(0.2 bBODeenon4ay)= (SSmy'lXQepdX8ia lbieal)@*o.21tb SSfienon-aay1

I_t:,ll,tatr_tlglEgMantroles along sewer lines should be provided at sewerline intersections and af, c.hanges n elevation, dirscf,iea,size, dia,meter, and slope. If a sewer line is too sma.llfor a person to enter, rnenholes should be placedever5r400 ft (120 m) to allow for cleaning. Recommendedmarcimum spacings are 400 ft (120 m) for pipes withdia,meters ess than 18 tu (460 mm), 500 ft (150 m) for18 to 48 in (460 to 1220 mm) pipes, and 600 to 200 ft(180 to 210 m) for larger pipe.

Qrt.



64/67

Water TreatrnentDlstribution Systeffxsg?: STQR+GFEy_DlsrnlHnll_@_E{Mlater is stored to provide water pressule, equalizepu-mping rates, equalize supply and demand overperiods of high consumption, provide surge relief, alrdfur .ish water duri'g firm and other emergencies whenpower is disrupted. Storage may also serve as parb ofthe treatment process, either by providing increased de.tention time or by blending water supplies to obtain adesired concentratioa.Several methods are used to distribute water depend-ing on terrain, econo.rics, and other local conditions-Graaity distri,buti,on is used when a lake or reservoir islocated significa,ntly higher ia elevatiou than the popu-lation.Distribution from pumped.storage is tlre most cornmonoption when gravity distribution cannot be used. Ex-cess water is pumped during periods of low hydraulica.nd electrical demands (usually at night) into elevatedstorage. During periods of high consumption, water isdrawn from the storage. With pumped storage, pumpsare able to operate at a uriform rate and nea,r theirrated capacity most of the time.Ifsing pumps without storage to force water directlyinto the ma.ins is the least desirable option. Withoutstorage, pumps and motors will not always be able tonrn in theh most efficient ranges since they must op.erate during low, average, and peak flows. In a poweroutage, all water supply will be lost unless a backuppo.vyer source comes online quickly or water ca,rebe ob-ta.ined by gravity flow.\ffater is commouly stored in surface and elevated tanks.The elevation of the water surface in the ta^rokdirectlyd-etermines the distribution pressrue- This elevation iscontrolled by an alti,tude aulae fhat operates on the dif-ferential in pressure between the heiglrt of the water anda-n adjustable spring-loaded pilot on the valve. A-ltitudevalves are insta^lled at ground level and, when properlyadjusted, can mairrtain the water levels io v/ithh 4 ilt.Ta*s must Lrevented to the atmosphere. Otherwise,a, rapid withfuawat of water will create a vacuum thatcould easily cause the tank to collapse inviard.The preferred location of a,rq. leyated tark is on theopposite side of the high-consrrmFtion district from thepumping station. During periods of high water use, thed-istrict will be fed from both sides, reducing the Iossof head in the mains below what would occur withoutetevated storage.Equalizing the pumping rate during the day ordina.riiyr


65/67

PRACTICEROBLEMSOLUTIONS



66/67

FLUID DYNAMICSPRACTICE PROBLEM SOLUTIOI\S

I . Determine he velocity in eachof the threepipes?h1:6.5 t

vz o=(0.83lxhf)o's4 =2,28 t/secve n (4.327xhf)0's4 =11.89 tlsecv+ n (2.299xhf)o's4 =6.32 tlsec

2 Determinehe flow in eachof the hreepipes?V: Av

Vz n:(0.02233t2x0.831Xhr)o'tn0.05 t3/secVe ,, (0.2006 t2x4.327xhr)0's42.38 ts/sectn,o :(0.088 t( )(2.:2gg)(hr)o'tn0.56 tt/sec

3.00 t3/sec



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OPENCHANNELFLOWPractice roblemSolutionCalculatehe velocityand low rate n a 24 inchcircularReinforcedConcrete ipe RCP)stormsewer nstalled t a slopeof0.02andn : 0.013, ariablewith depth:

1. Flowing uIlv-(1.a9ln)(R)o '0215;o 's:0.013 S:0.02R : D / 4 : 2 f t 1 4 : 0 . 5 f tvtu, e.4g10.01X0.5)0'021gz;O'svtur: 10 tlsecQn1: (vn XA) A - (0.7854XD)tQnrr: (10 t/sec)(34 ftz)Qrur31.4 ft3/sec

2. Flowingata depthof flow of seventeen17) nchesUseCircularChannelRatios igure. d/I) : 17124 0.77At dlD 0.71, Q/Qn :0.72,v/v61:0.96

9.6

civil depth notes for mar 17th-hydrolic

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