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Water Resources Engineering

by: Aron Lehner

Water Resources Engineering CWR 4103

Aron Lehner
GPA 3.77

Mark Ross

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Mark Ross
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This 110 page Class Notes was uploaded by Aron Lehner on Wednesday September 23, 2015. The Class Notes belongs to CWR 4103 at University of South Florida taught by Mark Ross in Fall. Since its upload, it has received 24 views. For similar materials see /class/212655/cwr-4103-university-of-south-florida in Civil Water Resources at University of South Florida.

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Date Created: 09/23/15
SURFACE RUNOFF What factors in uence surface runoff Basin characteristics size shape slope land use cover soil type antecedent conditions Storm characteristics storm intensity it and total depth storm duration spatial variation movement Hydrography size shape condition of ow conveyance systems sometimes called the degree of development of the ow system or drainage density See Appendix Figure Relationships Between Flow Elements mrossw0rkcourseWkvwlresnotesnotesS ovh 1 HYDROGRAPH ANALYSIS De nitions Losses Hydrograph separations Unit Hydrographs Synthetic Unit Hydrographs Can we learn anything about the basin from a measured runoff event Stream ow Direct Runoff Base ow Direct Runoff rainfall excess is rainfall losses Losses interception in ltration depression storage etc sometimes called basin recharge Rainfall excess or direct runoff Overland Flow mrossw0rkcourseWkvwlresnotesnotesS ovh 2 Inter ow classically is runoff that in ltrates the top layers of soil and exits to stream prior to reaching zone of saturation In Florida sometimes call inter ow water which in ltrates to shallow water table and moves laterally to streams Inter ow is a rapid phenomenon base ow is a slo phenomenon varies by season Base ow entry of groundwater into stream Surface ow 1 Perennial all year 2 Ephemeral seasonal dry period wet period 3 Intermittent streams that go dry along their length that may come back up again downstream measure ow in perennial stream mrossw0rkcourseWkvwlresnotesnotesS ovh 3 HYDROGRAPH ANALYSIS basins with a lot of storage have a large recessional limb recessional limb typically has a long tail for Florida recession occurs exponentially for base ow hard to say what part is surface runoff and what part is base ow have to make assumptions mrossw0rkcourseWkvwlresnotesnotesS ovh 4 BASEFLOW SEPARATION have to separate base ow before we can construct hydrograph empirical and common practice methods Empirical Method N A0392 N is from the point of peak discharge to the point where ow is completely dominated by base ow mrossW0rkcourseWkvwlresnotesnotesS ovh Opoak Dlrect Runoff lntarl39low extender of ems k N 4 curve P Calculate N A N In days A In mllasz Common Practice Methods 1 ByEye Method 39 l mmmun nm mrossw0rkc0urseWkthresnotesnotesS ovh 6 2 In ection Point Method long hydrographs TB large basins Dlrect Runoff A lnllalttlon polnt 3 Exponential Point Method take advantage of exponential shape Q Qoe39kt ln 6 ln 60 kt 1 Plot on semilog paper 2 Draw straight lines through recessional data 3 Get slopes slope is same to tp since relates to physical conditions of basin slope relates to k riserun 4 Transfer back to arithmetic paper See Appendix Base ow Separation Bedient mrossw0rkcourseWkvwlresnotesnotesS ovh 7 PREDICTING RAINFALL RUNOFF Ilnear or nonllnear 1 Regression Analysis works best for long time periods works best when there is not appreciable lag storage in the system eg more urbanized catchments see paper by Diskin 1970 Better if basins haven39t experienced signi cant changes development mrossw0rkcourseWkvwlresnotesnotesS ovh 8 2 Coef cient Method Runoff volume CP DS Volume gt Depth inches or cm P Rainfall volume C RunoffCoef cient0 s C s 1 DS Depression Storage Initial Abstractions IA Ralnlall real but hard to measure volume lled prior to runoff dif cult to separate DS from other losses espin ltration For areas that are mostly impervious e g urban Chicago Study DS ll6 in for impervious 00625 in DS 14 in for pervious 025 in Los Angeles Soil DS in Sand 020 Loam 015 Clay 010 Viessman Data Chap 2 or 3 DS 013 00301 S linear function S slope in dp 00303 39 S39O3949 curvilinear mrossw0rkcourseWkvwlresnotesnotesS ovh 9 3 Rational Method Qp CiA for predicting peak ows also called Lloyd Davis Method Ireland 1850 Q1 peak ow cfs C runoff coef cient i rainfall intensity inhr A area ac cfs z inhr ac 1008 handy conversion For runoff coef cient many tables gt see Viessman Table 111 Bedient amp Huber Table 66 page 396 C 2 095 for paved areas C 2 010 for natural areas Correct application of Rational Method Obtain i from m curves with tr and T de ned Higher T for more important ood control T 5 2 3 yr for residential T 2 1025 yr for major highways mrossw0rkcourseWkvwlresnotesnotesS ovh 10 Set tr t0 basin response time or time of concentration time of wave not particle of water tr gt tc lower peak because i is too low tr lt tc also not correct whole basin not contributing Big criticism no catchment really behaves like this with exception of small metal roof tops because of DS For peak ow estimation OK IDF FDOT Drainage Manual currently being revised New manual Ch 6 see also KE Weldon FDOT Rainfall IDF quotCurve Generationquot mrossw0rkcourseWkvwlresnotesnotesS ovh 1 1 4 Index Flood Method Regression Technique Viessman p214524 For Florida USGS Barnes amp Golden How does it work 1 Region is diVided into homogeneous subareas for purposes of predicting peak ows a Regression Moan Annual Flow b Then take Flood Flow Mean Annual Flood Flow Mean Annual elves 0pm T return perlod Storage Impact effect of lakes and swamps mrossw0rkcourseWkvwlresnotesnotesS ovh 12 5 SCS Method prediction of runoff volume depth very popular based upon eld measurements valid for areas 5 s A s 2000 ac 3 sq miles massive database e g hydrologic classi cation ofa soil A B C D soil survey interpretation sheets De ne Pe P I2 F Q like rainfall excess Cumulative values inches mrossw0rkcourseWkvwlresnotesnotesS ovh 13 De ne P6 Precipitation excess inches F In ltration P6 Q inches S Storage maX available soil storage inches Q Runoff inches normalized by area P 8 F EP I E 9 5 P 2 P qr 2 Q PeS P IaS Ia Assumed by SCS 02 39 Storage 0239 S Therefore SCS Runoff Equation for Total Volume 2 Q SCS Runoff Equation P Precipitation total inches may be design storm or IDF value Q Runoff value inches gt Careful used for larger events only P gt 25quot mrossW0rkcourseWkvwlresnotesnotesS ovh What is S use SCS database Assumed fsoil S represents soil storage inches Each soil has a curve number CN a measure of soil storage If S 0 CN 100 implies 100 runoff only time it represents actual percent of runoff 100 90 s 1000 10 0 CN 70 CN39s Function of Land use Hydrologic condition good fair poor Hydrologic Soil Group ABCD 1000 10 5 Altemately CN mrossw0rkcourseWkvwlresnotesnotesS ovh 15 Hydrologic Soil Groups llin Infiltration Rate inhr A welldrained soil sand high S 03 045 B 015 030 C 005 015 D poorlydrained clay or muck low S 0 005 Dual classi cation AC or BD eg AC A arti cial drains to lower the water table urbanization or agric development C natural conditions Poorly drained Hydrologic Condition Land Use poor average good Can Adjust CN for Antecedent Moisture Condition AMC 3 over 2 inches in last 5 days AMC 2 normal condition 12 in last 5 days AMC 1 dry conditions lt 1 in last 5 days may need to look at longer periods than 5 days mrossw0rkc0urseWkthresnotesnotesS ovh 16 SCS Method Using Triangular Unit Hydrograph STEPS 1 Find CN for area use area weighted curve number if different land uses CNavg 2 AiCNi Z A but better to average storages san 2 Also2 Al CN 1000s 10 2 Choose rainfall need total depth P d i tr usually use IDF curve for speci ed T use maX 6 hrs to tr gt Weak point in methodology assumes a uniform intensity during storm 3 Obtain total runoff with P Q inches from graph or SCS equation 1 4 Choose family of dimensionless unit hydrographs UH see gure 2 a P 4 5 Get t1 ftc or SCS lag equation 5 on t 08 s107 P 1900 y 5 mrossw0rkcourseWkvwlresnotesnotesS ovh 17 Then 6 qp 484 Atp unit peak ow comes from triangular UH Ks 484 too high for Florida commonly use 256 qp in cfsinch of runoff 7 Multiply qp X Q to get storm peak ow cfs 8 Use Table 6 of hydrograph to get actual hydrograph In Viessman Ch 4 see SCS dimensionless UH Volume of runoff 808 Trllngullr UH lpplmdml on 12 qp tptr 1quotA com Iqu Q from graph in inches 1 m a tp 1Ins1npeak 11 6453 2 g t r t q p 1 p qlqp tP 484 comes from conversion must multiply by area in sq miles to obtain qp in cfs Q 2 Qpc ktphrAmz x 6453 Q in inches 6453 is conversion What is k mrossw0rkcourseWkvwlresnotesnotesS ovh 18 Earlier hydrograph has tr 167 tp k 21 ttp m k 21 53 34 075 7 KS k 6453 O756453 M Qp 484QAtp Constant 484 depends on shape of triangular hydrograph BUT in Florida longer recessions tails so tr is greater If trtp larger then k must be smaller qq mp Study found values of 64539 k lt 484 in fact lt 100 and as low as k 12 in Kiss River basin Capece Campbell Baldwin quotEstimating Runoff Peak Rates and Volumes from Flat HighWater Table Watershedsquot Finally SCS does not do very well in Florida unless take into account water table S porosity X d upper bound d s porost x d upper bound mrossw0rkcourseWkvwlresnotesnotesS ovh 19 ICA October 26 1999 Due November 4 1999 Homework 1 Find recession dimensions of a triangular unit hydrograph with a quotshape factorquot of 256 tr tp 1 Plot the storm hydrograph resulting from a 1000 acre basin in Hillsborough County for the 10yr 12hr Type IIM storm Use either a dimensionless curvilinear or triangular hydrograph 1000 ft hydraulic length 03 slope and a CN75 mrossworkcourseWkvwlresnotesnotesS ovh 20 RAINFALL RUNOFF Basin Urbanization Hydrograph Comparison Q A apurb quot 39 39 Urbanlzed basln t urb opnatquot39quotquot39 pm Natural basln vurb vnut t Result 1 Qpurb gtgt Qpnat 2 Vurb gtgt Vnat 3 tpurb ltlt tpnat 439 TBurbltlt TBnat Consequence Must provide attenuation reduce peak discharge delay time to peak mrossw0rkcourseWkvwlresnotesnotesS ovh 21 Stormwater Design Objectives 1 Provide for reduced ooding potential design for ood minimization within limitations 2 Provide increased storage and suf cient attenuation measures to minimize surface water and ground water impacts downstream 3 Provide for water quality treatment and stormwater BMP39s to reduce pollutant loading to recieving water systems How do we do this provide for increased storage frictional characteristics ie retention detention ponds swales exflltration pipes etc mrossw0rkcoursevvldekesnotesnoteSS ovh 22 RetentionDetention Ponds Typical RetentionDetention P ond Stormweter Dlecherge Emergency or H Deelgn Splllwey V Deelgn Detemlen Volume W eg 2524vDeelgn Storm Bleed down erlflee Retentlon Volume POND Retention long term storage volume removed from the urbanized discharge hydrograph quotpermanentlyquot stored on site Detention short term storage of hydrograph to be slowly released over time quottemporarilyquot stored on site Typical retention volume SWFWMDF DER design calls for rst 1 inch of rainfall excess for basin gt 20 acre 05 inch for lt 20 ac mrossw0rkcourseWkvwlresnotesnotesS ovh 23 Factors that In uence Direct Runoff Hydrograph l Basin characteristics slope soils shape land use etc Antecedent Moisture Conditions based on rainfall totals for preceding week AMCI dry lt O5quotweek AMC 11 med 05 to l5quotweek AMC III high gt l5quotweek In Florida z 52quotyear 52 weeks gt z lquotweek AMC 11 Usually tables based on AMC 11 can adjust tables up or down to I or III mrossw0rkcourseWkvwiresnotesnotesS ovh 24 2 Storm characteristics a b c d e m need separate UH for each duration in practice maybe have UH39s for durations of every 2 hrs or so 2 hr UH means 2 hr rainfall excess Time Variation of Intensity can39t make UH39s for all possible patterns might try to have separate UH39s for convective constant and cyclonic storms small basins show larger effect of variations than do larger basins Spatial Distribution large basins are effected more than small basins try to construct UH39s for various patterns practical matter not usually done can try to limit size of basin for application eg A lt 2000 miles2 topographic effects mush affect all storms in pm same way arm mmmom downltrolm W worst possible case storm and quot pquott 39 runoff peak in phase 3 Amount of Runoff assume linearity gt time base for all storms of same duration is the same but in fact larger storms have somewhat longer time bases better assumption for average storms than extremes mrossw0rkcourseWkvwlresnotesnotesS ovh 25 3 Hydrography condition of ow conveyance systems basin size and shape eg effect of basin shape on hydrography Badlal pattern type Elongated type Dlvldod subbasln type Nola falrly Note double peaked peaked Q A Q Q t t t mrossw0rkcourseWkvwlresnotesnotesS ovh 26 4 Consistency of Data a common check is the z variable Double Mass Analysis z variables X and y Plot 21 X VS 21 y eg 2 rain vs 3 runoff or 2 raingage 2 vs 2 raingag 31 How signi cant are the breaks 1 Use double mass curve to identify likely data time periods 2 Test for signi cance using analysis of covariance z Raln z Raln zRam 939 2 mrossw0rkcourseWkvwlresnotesnotesS ovh 27 UNIT HYDROGRAPHS Interception Depreeelon Storage 396 3916 ha equlllbrlum hydrogreph 01 AS SUTVIE 1 2 3 Identical or similar storms with identical antecedent conditions produce identical directly related discharge hydrographs Time base of all oods caused by rainfall excess of same duration is the same time base and time to peak for similar storms of the same duration are the same Time base duration of direct runoff If rainfall distribution in time and space for several storms is similar then the ordinates of each hydrograph will be proportional to volume of direct runoff ie lquothr 2quothr gt same shape hydrograph but one twice as big as the other Time Unit TU of the storm duration of rainfall excess eg If 25 hrs of rainfall and rst 12 hour goes to abstractions like in ltration the time unit is 2 hours mrossw0rkcourseWkthresnotesnotesS ovh 28 Unit Hydrograph UH is the direct runoff hydrograph resulting from a rainfall excess direct rainfall volume of 1 inch over the catchment area Strongly implied linearity different amounts same ordinates I39 1lnhl39 2hr1lnhr 2ln 2 hr 0 Unlt Hydrograph t TB eg From previous example TU 2 hr Runoff Volume fQdt 2quot Basin Area UH for this storm based on 1quot gt diVide by 2 gt 1quot 39 Area of basin UH is 12 the size but TB and TP are the same mrossw0rkcourseWkvwlresnotesnotesS ovh 29 Basin La TP time between centers of mass of rainfall excess and direct runoff Inltlll Wlona wetting Balmquot Rulnfall excess Hymn Con nuoua nbstracllona In ltratlon 11m Unlt Center of gmvlly of ralnl39all 0 DI Ru off In not n L Hydrogmph Cantor ravlty of runoff t Tlme Baa UH works best if time unit is approximately equal to 1A of basin lag Often done on rainfall and not rainfall excess for ease mrossw0rkc0urseWkthresnotesnotesS ovh 30 UNIT HYDROGRAPH PROCEDURES 90 rum ni1 Where Qnt is storm hydrograph ordinates P is incremental rainfall excess depth typically hourly QUHn is unit hydrograph ordinates n number of storm hydrograph ordinates Mathematical expression for displacement and scaling shown easily in handout How to construct UH from isolated storms 1 Separate base ow and compute volume of direct i Sum of am Vollme of direct runoff 0 mm mquot 2 Scale direct runoff ordinates by dividing by Tlme the direct runoff volume 39 expressed as inches over Volume1incthreaofbasin the drainage area this is 5 Unit Hydrograph the UH Tlme 3 Study the rainfall records to determine the duration of the rainfall excess time unit 4 Label UH with the time unit or duration e g 2 hr storm mrossw0rkcourseWkvwlresnotesnotesS ovh 31 Can generate a UH for durations other than those of the original data in the absence of storms of the desired duration eg can generate a 6hr UH by adding displaced ordinates of 3 2 hr UH39s and rescaling t E1 E2 m Sum Sum3 6hr UH hr 1000 cfs 0 0 0 0 0 0 2 25 0 0 25 08 4 60 25 0 85 28 6 102 60 25 187 62 8 164 102 60 326 109 10 253 164 102 519 17 3 but for 3quot of rain Can get any desired duration from Scurves mrossw0rkcourseWkvwlresnotesnotesS ovh 32 SCURVE S H dro ra h is the hydrograph produced by a continuous series of consecutive rainfalls of equal duration equal time units each have 1 inch of rainfall excess per time unit ie hydrograph produced by continuous rainfall excess of intensity lTU I W 311 1 t 09 Inlamlty Dnlnlae Area TU 1 Inrl39U A Q9 8Hvdroanph a 01 02 as 04 11me Obtain Scurve by adding UH39s Qe Intensity X drainage area l inTU X A But quotrealquot Scurves have oscillations must smooth can calculate asymptote mrossw0rkcourseWkvwlresnotesnotesS ovh 33 How do we construct UH39s from Scurves 1 Have Scurve from UH of time unit TU1 Want to nd UH of TU2 139ITU Area between Shydrogrephe Area under removed unit hydrogreph 080 Time Hypothesize that we could also have an Scurve caused by UH of time unit At 2 Assume that the Scurve for time unit At TUZ is geometrically similar to Scurve for time unit TU1 3 If displace Scurve by At it must include all runoff except that caused by rain during the rst At Hence difference in S curves is the hydrograph resulting from rainfall of duration At and intensity lTU 4 Take difference in Scurves and rescale by dividing by lTU39 At TUzTU1 Be sure and rescale Lysw gsu Am i 777 1U1 UH IA 1 Qs Q5A t At mrossw0rkcourseWkvwlresnotesnotesS ovh 34 SYNTHETIC UNIT HYDROGRAPHS Snyder empirical method Basins 10 10000 mil De ne Lag Time t1 time from center of mass of rainfall excess to peak runoff careful also called tp time to peak hrs t CJL39 La L main stream length outlet to diVide miles Lca length along channel to a point opposite watershed centroid miles oullut TYPICALLY 18 lt Ct lt 22 the steeper the slope the lower the value of Ct up to Ct z 80 along the Gulf of Mexico mrossw0rkcourseWkvwlresnotesnotesS ovh 3 5 Time Unit TU UH duration TR time of rain TRt55 TUO18tL if a time unit TM is desired then make adjustment to lag t1R tP O25TRd 39 TR Where th time lag of new UH hrs t1 time lag of original UH hrs TRd speci ed TU TR calculated TU tL 55 Please Note For this course tp Time to peak discharge from start of rainfall t1 Time to peak discharge from time of center of mass of rainfall excess mrossw0rkcourseWkvwlresnotesnotesS ovh 36 Peak Flow 21 SUH39s QPM39QP394Q tL A ow producing area milesz Qp peak ow cfs 04 lt Cp lt 08 large Cp with smaller Ct Q volume of runoff inches Unit Peak qpSCS 64039 Q A tL Actually 6453 ifuse Cp 075 qp 484 A39 QtL Time Base TB TB 3 tL8 TB days tL hrs For larger basins TB C 39 tL 3 lt C lt 5 Sketch in the Unit Hydrograph mrossw0rkcourseWkvwlresnotesnotesS ovh 37 INSTANTANEOUS UNIT HYDROGRAPH IUH In malhematlcs lhla Is a delta funcllon gamma functlon shine or exponential t IUH response runoff due to an instantaneous zero duration rainfall excess i of 1 inch over the catchment Rainfall oontlnuous ralnfall i pulse mathematical 39 quot direct delta function i tollizglztof a pulse height of pulse gt in nity at arm area of pulse gt 1 inch t break up into a series of direct delta pulses volume of a pulse it dt Convolution or superposition SEW integration Linear System Analysis hlt i v 0t Qt 0 tht 1i1d1 t mrossw0rkcourseWkvwlresnotesnotesS ovh 3 8 t RUNOFF and Stream Flow Surface Runoff l Overland Flow thin sheet ow especially urban areas and sat soil 2 Low Order Stream Flow streams fed by overland ow and smaller rivulets 3 Higher Order Stream Flow fed primarily by smaller streams Governing Equations Objectives De ne to predict ow as a function of space and time Flow fXt l dimension QXt gt cfs voltime q l cfsft of width volwidthtime yXt gt depth will have momentum and continuity equations for derivations of governing eq see Fluid Mechanics texts eg Eagleson Henderson Chow 11391 I ralnfall I In ltration V avg velocity 0 smallmtan Ocslno I l l l l qr B th overland lateral Inflow 1 m b mrossworkcourseWkvlmnresnotesnotesS ovh 39 Overland Flow Assume l Hydrostatic Pressure neg any quotover pressuresquot eg by raindrops 2 Wide Rectangular Channel F gtgt 1 yb ltlt 1 q vy 3 Small Bottom Slope 9 5 sin 9 z tan 9 4 No correction in momentum equation for nonuniform velocity distribution 5 Any lateral in ows are perpendicular to ow Momentum Eqn 2 v g i f v 2y To dt dx dx by 1 bpygsm where dydX surface slope i rainfall f in ltration 2qLb lateral in ows both sides bottom width 1 2yb bottom and side friction to p y wall shear stress g sin bottom slope 2 unknowns V and y mrossw0rkcoursevvldekesnotesnoteSS ovh 40 Continuity Eqn v y i f dt dx dx b Order of magnitude analysis on momentum equation yields rst four terms in equation small compared to the 5th and 6th terms Therefore the momentum eqn now becomes 1 0 gsi110 gt To pgysinG yysinO Py mrossw0rkcourseWkvwlresnotesnotesS ovh 41 How do we get V Now use empirical results to describe t0 Empirical results gt basic hydraulics 17 0 C5 p 39 VZZg Cf Friction coef cient Cf f4 where f DarcyWeisbach friction factor 21 v 0 2Yy5me cm Cfp Cfp where C Chezy coef cient 2gCf 50 150 sin 9 9 or slope for small 9 If assume C constant then C sin on Then V ocyl2 and gVy cayy2 mrossw0rkcourseWkvwlresnotesnotesS ovh 42 In general momentum equation for stagedischarge q ay Simpli ed Momentum Eq for steady uniform ow cc m ffriction loss ow regime One example for turbulent ow gt m 32 on 2 g39 sinQnyz and m 32 For laminar ow Cf 6R for wide rectangular ow down a wide plain where R Vyv v kinematic Viscosity this gives m 3 a g39 sin 3 v exact solution For turbulent ow Manning39s Eq V l49n yak 91 wide uniform ow Cf 09 g39 112 yquotquot V 14911 q a y53 a 612 m for turbulent n 3 ow mrossw0rkcourseWkvwlresnotesnotesS ovh 43 m What to use 53 turbulent lt m lt 3 laminar R gt turbulence ffriction factor R Vhv v 1X10396 mZs eg f 03 R 2103 m z 2 is sometimes observed for overland ow Momentum equation gt steady uniform ow Say m 2 from experiments what is o If have data get 0 from equation of the line From Manning39s Eq on 149n39 912 q 149n 9 yS3 y39Ayvquot l49n 8 y2 y gives a fy 149n39 9 y Then evaluate a at an average value of y linearization approximation mrossw0rkcourseWkvwlresnotesnotesS ovh 44 Manning39s n some estimates from Stanford Watershed Model SWlVI documentation obtained from model calibrations Surface L Smooth asphalt 0012 Concrete paving 0014 Packed clay 003 Light turf 020 Dense turf 03 5 Dense shrubbery and forest litter 040 gt overland ow for FL higher for vegetated surfaces Now need to solve for unknowns q and y m dt dx 2qL 0 i 4 y fb Solve for qXt yXt Momentum q my Continuity dydt dydX i f 2qLb i f rainfall excess Solve for qXt yXt Analytical Solution gt Method of Characteristics Two governing equations Kinematic Wave Solution mrossw0rkcourseWkvwlresnotesnotesS ovh 45 Kinematic Wave vs Dynamic Wave Kinematic Wave moves only downstream therefore m backwater effects retains only friction and gravity terms in momentum equation Dynamic Wave can simulate backwater and ow reversal that is can move upstream as well as downstream retains all terms in momentum equation Method of Characteristics follow pathway in time and space of a kinematic wave if can do this then the coupled partial differential equations are converted to uncoupled ordinary diff equations mrossw0rkcourseWkvwlresnotesnotesS ovh 46 TIME OF CONCENTRATION tc l The time of concentration is the travel time of a wave to move from the hydraulically most distant point in the catchment to the outlet 2 The tc is the time to equilibrium of the catchment under a steady rainfall excess ie when Qcatchment iA gt The tc is not the travel time of a parcel of water to move downstream through the catchment The tc is the time taken for the discharge at the outlet to re ect quotfeelquot the runoff contribution from all parts of the basin 1wavc lt tparcel wave speed gtgt parcel speed 5 Vavgybasin wave speed C xgy mrossw0rkcourseWkvwlresnotesnotesS ovh 47 Recall tCSCS Lag method Time lag L 1039839 S l 397 190039 Y0395 and tc 53 L tcSCSVelocity method tc 1 Vavg Vavg aVerage parcel velocity fps L thinematic wave method tc m1m cc 139 L length of overland ow plane Lavg ZmaX 39 Slopeavg is average rainfall excess intensity ftc on m hydraulic ow regime parameters cc l49 s n m 53 for turbulent ow What about quot11quot see FDOT Drainage Manual Overland Flow Manning39s 11 Values Separate into overland ow sheet ow channelized ow Overland ow maximum of 300 feet nationally 600 feet in Florida q dischargeunit width gt calculate ow depth of discharge and friction momentum equation mrossw0rkcourseWkthresnotesnotesS ovh 48 Miscellaneous Topics One more characteristic length average length of overland ow LS Total length of stream drainage collection network A Basin area tOF A 2 LS Drainage Density measure of the drainage structure density Dr total miles of streams sq mi of catchment Dr ELS Ar 1 lt Drlt 100 lmile High in deserts and areas of high relief low ratios of PET Low in humid areas and areas of low relief high ratios PET In Florida 1 lt Dr lt 10 lmile mrossw0rkcourseWkvwlresnotesnotesS ovh 49 Particular Methods for Predicting Runoff TIMEAREA METHODS 1 2 3 osswork Select a rainfall hyetograph real or synthetic design storm Compute losses most methods use Horton in ltration no special concern about case B could use SCS S really should consider case B could use curve shifting method or make f a funcF Remember depression storage if used is subtracted rst in time In ltratlon rainfall excess Construct isochrones of equal travel time water velocities would make more physical sense to use wave travel time NOTE Cmv if t AXW could adjust by multiplying velocities by m to get c eg if At 5 min should be same as hyetograph At 99 If A 5 mln coursewldelresnotesnoteSS ovh 50 Basin 4 5 6 Optional Construct a TimeArea Curve OOI IGGVB Lots of contributing area early in storm A1 MA3M M GOI39IVOX Construct TimeArea Concentration Curve Wt dAtdt derivative of time A Wt area Fine for continuous t t functions but for discrete time use At increment In discrete time simply plot A vs ti gt Time Area Concentration Curve actually use Runoff Computation by Convolution linear superposition and addition Qn 2 mm Amk ik ik rainfall excess Q1 A39 i1 Q2 Az39ii l39Ai39 i2 Q3ZA339i1A239 i2A1 i3 Q4A4i1A3i2A2i3A139i4 etc RRL only considers directly connected impervious surfaces mrossw0rkcourseWkvwlresnotesnotesS ovh 51 Comments If we work in the continuous time domain Qt j tWt1ie1d1 tdAt I f 713mm 0 Since Wt dAtdt Instantaneous Unit Hydrograph Qt f tW Iiet 1d1 0 or n Qn Zk1Akln kl No attenuation included in timearea method alone sometimes include reservoir routing at the lower end of basin to account for attenuation of hydrograph Simplest case use a linear reservoir Options a timearea curve linear reservoir could use Puls Method ILLUDAS Illinois Urban Drainage Simulator considers pervious areas with your choice of Horton or Holton in ltration also pipe routing by translation earlier model RRL Transport Road Research Lab Great Britain considers only directly connected impervious areas mrossw0rkcourseWkvwlresnotesnotesS ovh 52 Santa Barbara Urban Hydrograph Method Method SBUH 1 Compute runoff from three subareas A Directly Connected ImperVious Qjlc P Adc P precipitation at each time step B Other ImperVious Qimp mr P 39 Aimmher subtract losses C PerVious P A Qpervious perv Qtotal Qdc Qimpother Qperv f Aperv f in ltration by Horton method Qimp th r in model some losses from these 2 Route through Linear Reservoir with te hours t0 basin mrossw0rkcourseWkvwlresnotesnotesS ovh 53 FLOOD ROUTING Hydrologic vs Hydraulic Contintu ql qo dsdt V Ola AL Simplest methods are based on a form of the lumped continuity eq For large rivers aren39t concerned with rainfall or lateral in ows only change in Qi Assume Qi known solve for qo s Linear Reservoir Q0 k39s k39 ltime out ow linearly related to storage dQOdt k39Q0 k39Q0t can integrate analytically mrossw0rkcourseWkvwlresnotesnotesS ovh 54 INFILTRATION In US average of 70 of precipitation is abstracted primarily in ltration In Florida average of 90 of precipitation is abstracted primarily in ltration Depends on 1 Degree of soil moisture 2 Soil type 3 Vegetation type roots provide channelization m 2 Water can be T K b returned 0 ET water Rhizosphere aim by ET l 3quot 59quot zone of aeration unsaturated zone 1 Gravitational vadose zone water quot39V39quotquotI39 6ni r939za quott a39 zone of saturation movement water 32 Terminology Water moving into soil in ltration through soil percolation out of soil seepage Zone of aeration unsaturated zone vadose zone Percolation driving forces for movement of water 1 Gravity 2 Capillary suction in unsaturated zone Capillary fringe caused by capillary suction head function of soil moisture negative pressure effect more signi cant with lower moisture content capillary effect goes to zero as 9 goes to sat G soil moisture content F cumulative in ltration f in ltration rate actual net movement of water across the air soil horizon inhr f f in ltration capacity of the soil maximum rate of in ltration f initial in ltration capacity inhr finitial 9 ff ultimate in ltration capacity inhr Both f and ff are fsoil type vegetation or treatment at surface land use 6 Volume of waterTotal volume of soil 0 lt 399 lt z 50 upper limit is porosity Porosity n Volume of voidsTotal volume voids solids Typically sustained F rates 1 2 inhr but may be up to 5 or 6 inhr Can observe in ltration rate as long as water is at surface f must be less than the rainfall rate or supply rate K Kh are used interchangeably a lot K permeability xed intrinsic soil property relates to ability of soil to convey liquids Kh hydraulic conductivity refers to different uids moving through the soil ow of water in soil KhVERTICAL is less than KhHORIZONTAL due to layering over time increase 399 increase Kh less friction Measurement of In ltration usually measure ff 1 Ring in ltrometer Ring appmmmy 23 cm diameter Try to maintain constant l l l mgh ghga water level head in ring keep track of how much water added during time interval Problem with end effects gt gives much higher in ltration rates than actually occur during rainfall attempted solution is double ring in ltrometer 2 Double Ring In ltrometer Inmr I Std ASTM Procedure for I J l l l l l l DRI tests 135mg see handouts in notes Maintain water level in both rings but only keep track of water added to inner ring more representative of actual vertical in ltration rates Still problematic and often unrepresentative of rainfall in ltration 3 Sprinkler in ltrometer SprllldenmsMuubmnl ln1llrl on 39Bllnflll39 Rmofl39 Plot 6 x 12 lid sum sloped Runoff ASTM Standard Procedure slope is average of basin Know Q application ow rate ft3test time or ft3 sec quotRainfallquot Problem small scale try to relate as well as possible to actual basin conditions try to maintain constant ow rates eld procedures make it dif cult INFILTRATION CASES Case A excess water on surface i gt f Dashed lines represent 5 N m different initial 9 g 4 and Initial f is function of 52 m soilinitial 9 E 2 clay quotumquot Final f is NOT function E 1 k I 39 of soil initial 0 1 2 a 4 11m hr Case B rainfall intensity i lt f rainfall less than initial in ltration capacity Equations for predicting in ltration to t the observed data 1 Horton 1940 takes advantage of exponential behavior of in ltration f fmgtxlt f0gtxlt fmekt u k constant decay coef cient units of ltime fsoil properties initial 6 if f f soil properties initial 9 quot39 quotquotquot f0 in l capacity at t O ff in l capacity at t 0 a Best known b Widely used because good t to data c k fquot depend on initial 9 but hard to know how d Easy to obtain cumulative in l F Ff f0dt Ilnaar Increase L In the slope 11m hrs For case B can quotadjustquot equation to account for i lt f initially shift curve Since fl f0 need to determine equivalent 9 when i f F potential In ltratlon 39corrected39 A tatlon capaclty 2 Kostyakov Equation irrigation f atb ab constants 39 1 ltb lt O a finitial 9 a b c d Developed from Flood Irrigation Data 39 Simplicity is attractive Predictsf gt no att0 andf gt O att no not very realistic a b finitial 8 hard to predict in advance 3 Holtan Equation 1961 USDA faSF F volume of water already in ltrated I dt S volume of llable pore space in F soil a constant depending on soil and surface vegetation 025 lt a lt 08 N 1 4 from Holtan a f fF not time b Derived from hundreds of in ltrometer tests c As long as F lt S then get higher in ltration rate d Easy to use for case B because simply adjust F until f has fallen to i ie F I i dt until f lt i e Dif cult to determine S working depth sometimes used with SCS data Storage fcurve number 1 f fm F S a Soil Properties measure of pressure or suction 4 capillary pressure moisture content 0 measure of transmission K hydraulic conductivity See handouts 4 vs 9 1 vs K Freeze amp Chary 1979 Governing Equations z 1 Darcy39s Law 3 6H 2 72 k9 az qz average velocity QtotalAtotal speci c discharge H total head pressure head elevation head q z 2 Continuity Eq 1 g E o Tz W at 62 lqzldz 0 0 3 Richard39s Eq If substitute E i m j 0 at 62 62 a ak93ll12 at z 62 66 6 am 62 k6 k6 at 62 az 62 aKe at 62 62 az Let 9 dependent variable De ne D0 HMS lg Recall 6430 3 604 ten 3 aZD9 62 109 Richard39s Equation unsaturated ow equation Alternatively we can solve for 1 De ne CIII Coma ll k t 3123 at 1 9 form Solve for 1 continuous solution into saturated zone 1 can be or but doesn39t work for low soil moisture Solve for 9 limited to unsaturated zone Can only solve numerically for most real situations ie variable rainfall Often used to check solutions obtained by simpler methods Requires soil properties Examine several solutions Phillip Equation 1957 solved analytically for uniform initial moisture content 9i for case A only water at surface cannot handle case B if St39 2 A small terms both S and A are fporosity 9i zunsat soil in principle can predict S and A in advance rst eq that we can explicit eqns not empirical GreenAmpt Equation 191 1 rediscovered in 1957 derived independently by Phillip lots of current interest in literature 39url 2177 from Darcy39s Law 5 k surface dz wazsurface llJf L 0 W 739 ksTo quots l f F volume in ltrated 9 sat 9i X L e 9 f ks1 w Original form of GreenAmpt Equation Comments 1 Average suction 1f lt 0 f always gt KS f KS for sat soil 2 See graph ff vs F 3 Initial moisture content included explicitly in principle can predict in ltration in advance 4 Can handle Case B easily at moment of surface saturation f i FS volume of water in ltrated up to point of surface saturation Rearrange GreenAmpt Eq to solve for F3 esat ei llJf 1 F 5 L ks only works for i gt ks if i lt kS then all rainfall infiltrates GreenAmpt Eq 2 step process 1 fi i until reaching FS 2 for F gt FS then f GA Equation See handout Example Use of GreenAmpt Equation SECTION VI FLOOD ROUTING Lav A Basin Delineation 439 Flaw Lines Stream Reach subbasin Delineation Consider the watershed with 6 subbasins Q1 QA QB Runoff from A amp B hi onvh 1 Q2 2 QA QB2 QC QD Routed runoff from Q Direct runoff from C amp D What causes attenuation 1 Storage and 2 Friction What causes ow lag 1 Flood wave travel time celerity flength depth friction slope Consider a short reach or reservoir pond In ow Outflow Q Storage Release from Volume Storage As lt o S As gt Al At 0 t Continuity Eq Qin Qoutz hi oth 2 See Appendix Figure 42 Bedient Simple Hydrologic Flood Routing Simple methods based on lumped continuity equation quotlumpedquot not fX Vtz v11 At Qi gt S gt Q0 Qi Q0 ASAt Problem Qi is known need to solve for Q0 What about S hi onvh 3 To solve the ood routing problem a relationship between S and Q is needed Either MM See handout USGS rating curve 2 Q fS or S fQ directly Less common than 1 Q Q 3 Solve for Qy from momentum equation eg Q ky kinematic SA A crosssectional area hi Ith 4 RESERVOIR ROUTING simplest hydrologic method sometimes called quotPond Routingquot quotLinearquot Reservoir Q kS k ltime routing coef cient f channel geometry Q out ow linearly related to storage then Qj11 Qout A SAt Homework s 41 3 4 5 7 hi Ith 5 an Qout dk391Qoutdt onutdt Qout Solution k at ftime Can solve analytically exponential solution QM ehf thite dt constant Constant Qout at t0 Need functional form for Qi ie Qit so that it can be integrated over time For example could assume Qi fsine function 125 1 a g 075 039 E 05 025 0 0251 t O 025 05 075 1 t tB 125 1 TI39 sin t g 075 3 0 05 025 O o 25 o 2 W o25 o 025 05 075 1 125 ttB 0139 hi Ith could use numerical discrete step methods t0t LtQidtL1QidtLZQidtL3QidtnU Finite Difference Method of Continuity Equation Q Q dsdt Qi1Qi2 Q 1Q 2 S2Sl 2 2 At Subscript 1 implies beginning of a time step Known Q11 Q12 Q01 SI Subscript 2 implies end of a time step Unknown Q02 82 hi Ith 7 Modi ed Puls Method Finite Difference Continuity Eq also called Storage Indication Method book collect knowns and unknowns on opposite sides of the equation At At At Q02 TQQI 791 1 sz S 22 To solve use a relationship between S and Q om rating curve stage discharge relation weir equation uniform ow assumption or other information A t 202 3 Can construct a table or graph of Q fS A t2 Q0 Book has good example p 256 260 hi Ith 8 Consider a river reach shown Rising hydrograph S wsgt Sm v Noam now Sws Shed Falling hydrograph Sws lt Sm Sb Outflow Falling t Rising Consider uniform ow 149 R2 3S12A n eql Qa hi onvh 9 Manning39s Equation for uniform ow Squ Shed A crosssectional area L length of channel reach and S L A reach storage Then Lwa RMS nL R Hydraulic radius Area Wetted perimeter For rectangular channel y depth b bottom width V y b Areayb WP2yb Rh yb2y b 1 y2 for b 2y Q Constant y S hionvh 1 0 Muskingum Method For Flood Routing S prism storage wedge storage KQO Kx Qi Q0 Wedge Storage KXQi Qo Prism Storage Two Parameters K x x is not distance Can substitute into continuity i Q Q dt dtKQ KxQ Qo K timeE travel time through a reach hionvh 1 1 Can solve analytically as before x dimensionless shape parameter if x 0 linear reservoir In general 0 s x s 05 note x is higher for more regular quotimprovedquot channels x is lower for more natural channel eg natural irregular channels x 015 concrete lined trapezoidal channel 03 s x s 04 if x 05 pure translation X05 hionvh 1 2 Finite differences applied to continuity Equation ds Q Q Q A Qi39Qf MK Kx dt 2 2 At Qa Q 2 Q0 Co 2C1 7 2 a Muskingum Eq 05At Kx K1 x05 At 05 AtKx k 139 Mus i m K 1 05 At gquot 30 2 Ci 1390 Coe icients Kl x 05 At 2 K1 x05 At For a linear reservoir x 0 Must have storage and ow data to evaluate parameters Plot graph shown for trial values of x keep trying with different x until loop narrows approximates a line SlopeK SKin1XQol hionvh 1 3 Aa muo coon 0 Thousands 1 1 Amwamdo v 000th S Thousands 14 hi onvh in 1 xQo Thousands When storage is a maximum dSdt 0 Qi Q0 substitute S K Q KxQi QO Kan Kx in 612 0 dt dt dt dt 19 on xdtL 1xdtL hionvh 1 5 If we had these at the same time we could solve XQi1 XQO Use two slope values and solve for x hionvh 1 6 In the real river reach the parameters are a function of the ow Q For this case need parameter estimation for several ow rates ie variable parameter K ow x ow Muskingum Method Probably better to just go to full St Venant dynamic equations Another method is MuskingumCunge Method A better way for parameter estimation relate them to physical parameters of the channel ref Cunge 1969 Ref 66 Qi 00 Qi QoQxAx x x Ax distance let Muskingum parameter x now be 0 x in the following eqs reps distance dS QiQa Qx QxAx dt SK0 QxtKl 6 Qx Axt E Ke dQxt K1 a dQxAxt dt dt dt hionvh 1 7 Taylor Series Remember sz BZQ 2 atax x QxAx Qx AxZ Qlw x and 2 2 3 QxAxt m AxaQ w M 39 lx at at am 2 ataxZ Substitute for Qx Ax t and dQx Ax tdt in continuity equation neglect d3Qdt dx2 term 8Q 6Q 82Q at Iquot at Iquot axatlx 8Q sz 62Q xt xt Ax Q Q axlx 2 a lm Use continuity equation of the form aQ 3A 8x 8t De ne c 6 9 wave speed 8A MuskingumCunge hionvh 1 8 dQ0Qc forca t at ax Solve for aQ caQ 3 01 at 6x at 8x azQ cazQ atax at 6x 8x 8x 3x2 LHS 2 Ira Q cl BAx Ira Q RHS at 8x2 Combine dZQdx2 terms 2 2 Q 2 C1eAx Ax E 2k 6x2 at k 8x hionvh 1 9 Muskingum Cunge This is a form of the advection diffusion equation the bracketed term represents a sort of ow wave diffusivity D RHS gt Numerical DiffusiVity Likely to be curved eg c f ow Also approximate c by ood profiles Hydraulic Flood Routing Large quotDynamicquot Rivers quotDynamicquot subjected to rapid uctuations in ow requiring inclusion of acceleration terms in equations of ow Must use St Venant Eqns to adequately describe ow p 23 78 hionvh Surface Water Hydrology De nition of Hydrology As Hydrologists As Engineers As a reference Average Global Precipitation Average U S Precipitation Average Florida Precipitation mrossW0rkcourseWkvwlresnotesnotesl ovh Natural Science that deals with the transport and distribution of water liquid gas solid in the atmosphere on and beneath the earth39s surface Interested in forecasting means extremes and time histories of hydrologic events and processes Charged with the evaluation planning and design of facilities to best utilize mitigate and manage water resources including catasphic hydrologic events 1016 cmyr 4O inyr 762 cmyr 3O inyr 1397 cmyr 55 inyr In theory we use quot rst principlesquot to evaluate hydrologic events and processes ie 1 Conservation of Mass Continuity 2 Conservation of Momentum N ewton39s 2nd Law 3 Conservation of Energy 2nd Law of Thermodynamics 4 Equation of State PV nRT for Gases My ram Am 2 INPLOWS 2 OUTFLows m A STORAGE haoIm 31 3115 An 1nventory of all sources s1nks and storages is a Water Budget Transport Mechanisms 1 Energy a Potential b Kinetic c Heat 2 Mass 3 Momentum Introduction to hydraulic design must start with understanding of the hydrologic cycle Storages and Transport Processes mrossworkcourseWkvwlresnotesnotesl ovh 2 Hydrology De nitions Hi stog First dam built across the Nile 4000 BC Romans measured stream ow 97 AD Modern hydrology 19001930 quotPeriod of Empiricismquot Many of the present day agencies formed as a result Hydrology has evolved as an analytical science in the last half century 97593 Computers in hydrology Film about Florida Hydrology Hydrologic cycle terms de nitions mrossworkcourseWkvwlresnotesnotesl ovh 3 Water budgets Watersheds and watershed divides catchments etc PRGETDS Runoff coef cients C RD Water balance units inches ux units cfs acft 1 acft 12 cfs 1 ac 43560 ft2 MGD l MGD 155 cfs 1m3s 1 m3s 228 mgd 1 gal 1 ft3 748 gal 1 mile2 1 mile2 640 acres mrossworkcourseWkvwlresnotesnotesl ovh 4 Water balance for a pond problem 11 lake Bedient Homework Read Ch 1 p 10 p 11 12 Bedient Florida Water Story mrossworkcourseWkvwlresnotesnotesl ovh 5 Hydrologic C 39cle insulation Transpiration mummy Wmim v Evalmra ou Evaporation len 39 Con xfcd by Argcsj p Aquif Hydrologic systems are quite complex but behave according to 3 elemental laws of nature so called conservation laws 1 Conservation of Mass Continuity closed system 2 Qmass 21 anmass 2 QOutmass A Storagemass At Inventory of sources sinks and storages Water Budget 2 Conservation of Momentum Newton39s 2nd Law 2 Forces 2 F dMdt A momentum 3 Conservation of Energy 2nd Law of Thermodynamics Z Inenergywork Z Outum1 energy A Storage A E At Unfortunately lack of understanding necessitates lSt Order approach mrossworkcourseWkvwlresnotesnotesl ovh 7 Water Budget quotBlack Boxquot Model Conservation of Mass 21 2 O dSdt I In ows O Out ows S Mass Storage Preclplmlon Raln snow etc Sum Evapomtlron ani39isplra on Sum Mar Mm 4 quot Inflows l 39 Consumptlve Use 5 3 quot39 39 I 1soii mow quot3990rt 30m Plants amp Anlmals aemw39m quotmm m yawL 3Snow Com m sum mm39 Grouner aumiz Arllllclill GroundwatorL Lum In ows Out ow 39 Note Work is done in the process of water transport with the primary energy source being derived from the incoming radiation from the sun insolation The radiation that is received by the Earth at the edge of the atmosphere is the so called Solar Constant which is 5 20 lymin 1 1y l calcmz l lymin 1434 X 10393 wattm2 mrossWorkcourseWkthresnotesnotesl ovh Transport Mechanisms for mass momentum energy Source of energy is the sun overall driving force Types of energy Energy potential kinetic heat Transport ux of some quantity quantity area time ux eg Mass ux lb 2 2 cm sec ft hr cal watts Heat ux cm2 mm m2 mrossworkcourseWkthresnotesnotesl ovh 9 3 MECHANISMS of ener trans ort l Advecti on 2 Diffusion 3 Radiation I ADVECTION bodily transport due to motion of the uid Quantity moves with uid velocity Advection can be in any direction convection is only vertical transport Flux velocity X concentration Mass Heat Momentum Flux length v quantity quantity time volume area time Xulc w A s m 3 m 2 s w cal FquXuTpCp 8 cm3 9 DC cmz s Flux u p u Fickian mrossworkcourseWkvwlresnotesnotesl ovh 10 II DIFFUSION the process of transport of a quantity in the direction of decreasing concentration of the quantity proportional to the concentration gradient Diffusion of Mass Flux X X where k diffusivity areatime lengchtime c concentration quantitytime Flux is proportional to the gradient of concentration c of anything Positive ux is in the direction of decreasing concentration Turbulence Reynolds Stress Advection given by the time mean products Diffusion given by the uctuation products For example c 5 Temperature T U 39T is the advection of T sometimes called convection although meteorologists use that for vertical 0 T39 is diffusion ofheat mrossw0rkcoursewkvw esnotesnotesl ovh l l Fourier39s Law qX U k dTdx k diffusivity areatime lengchtime T Temperature Diffusion of Heat Fourier39s Law of Heat Conduction for heat diffusion k k39 thermal conductivity calOCcmsec k39 p Cp on qX heat ux calcm3s p density gcm3 Cp speci c heat calgmOC a thermal diffusivity cmZs S for turbulence eddy thermal diffusivity see HW2 prob 4 0x Diffusion of Momentum Newton39s Law of Viscosity quantity momentum m u m mass u velocity Fqu m m a fome area time area 1 coeff m uvolumeddy pressureshearstressT p mvol 1 coeff p uddy mrossworkcourseWkvwlresnotesnotesl ovh 12 u39c39 c39 u39 if density constant 1 coeffp dudy u dudy pV dudy Newton39s Law of Viscosity p coeff of dynamic Viscosity gmseccm uid property V kinematic Viscosity Given Concentration gradient in a stream mrossworkcourseWkvwlresnotesnotesl ovh 13 Turbulent ux how to measure U39C39 Measure From experiments we know U39C39 gradient of average concentration Turbulent Mass Flux U39C39 k didx k a diffusivity FluxGradient lengchtime velocity uctuation x mixing length velocity uctuation mixing lengthmixing time k in water z 01 1000 cmZs l ftZs k in atm w 1000 107 cmZs In turbulent ow heat mass momentum same order of magnitude k f ow conditions Schmidt No kmom kmass 5 07 Prandtt No kmom kheat w 07 cmZsec stoke 10394 mZsec c Zsec p01se 01N39secm lgmcm39 sec mrossworkcourseWkvwlresnotesnotesl ovh l4 III Radiation Stefan Boltzman Law Intensity Emissivity X Radiation Flux I e o T4 T dependent on absolute temp Water Surface Receiving Radiation Long wave albedo ofwater 16 003 R 003 3 re ection For atmospheric radiation Icloudy Iclear Careful Longwave radiation Short wave radiation measurement Pyranometer Thermocouple black white voltage For Total Radiation all wave lengths Radiometer Weather Service measures radiation of various locations around the state Langleysday 10 to 200800 range mrossworkcourseWkvwiresnotesnotesl ovh l 5 Atmospheric Moisture Ref Viessman Ch 2 Eagleson amp Meter Books eg S Hess Location Surface Area Water Volume Percentage of milez mile3 Total Water Surface water Freshwater lakes 330000 30000 0009 Saline lakes 270000 25000 0008 Stream Channels 300 00001 Subsurface water Groundwater lt 12 mi deep 50000000 1000000 031 Groundwater gt mi deep 50000000 1000000 031 Soil moisture etc 50000000 16000 0005 IcecaEs and glaciers 6900000 7000000 215 AtmosEhere at sea level 197000000 3100 0001 Oceans 139500000 317000000 972 approx totals 326000000 100 mrossW0rkcourseWkvwlresnotesnotesl ovh Equation of State Gas Law pVf7 Boyle 39s Law M PV nRT RT Universal Gas Law m equation which governs gases in the atmosphere applies to the constituents as well as the total R Universal Gas Constant 831 X 107 ergmole 0K erg unit of energy F X L g cms2 X cm ch5RT PpRT p a speci c volume volmass l p Rm gas constant for individual gases mrossworkcourseWkvwlresnotesnotesl ovh l7 For water MW 2H 1202 2 16 18 grn molecular wt 111 m X 107 462 X 106 erggm 0K Energy 1 erg 1 dyne cm 1 mb 1000 dynecm2 1 dyne 1 gm cms2 MT MW M0l For Dry Air Mixture of Nitrogen OXygen etc Md 2002 80N 02 32 08 28 288 gmmolecular wt By quotHarmonic Meanquot Md 2897 for dry air Rd mX 107 287 X 106 erggm 0K 2897 287 X 103 mb cm3gm 0K gas constant for dry air mrossworkcourseWkvwlresnotesnotesl ovh 1 8 Gas Law for Water Vapor PV e p VRVT e vapor pressure pV vapor density or absolute humidity Dry Pressure Pd p dRT Total Pressure Sum of Partial Pressures PutmPde p L i L LE V RVT TR Rdmd TRd md my my pv 06221 mrossw0rkcourseWkthresnotesnotesl ovh l9 Pressure Measurement Conversions Avg atm pressure 10132 mb 760 mm Hg 2992 in Hg 1 atm 147 psi pV u 910395 gcm3 pd u 910393 gcm3 Atmospheric density Patm e 6 Palm 9 1 pm p pquot RT RT RT atm De ne T Virtual Temperature Virtual temperature of moist air Temperature at which dry air would have same density 4 1 0378 3 atm T mrossworkcourseWkvwlresnotesnotesl ovh 20


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