Rainfall - Runoff: Unit Hydrograph

Rainfall - runoff:Unit HydrographManuel Gómez ValentínE.T.S. Ing. Caminos, Canales yPuertos de Barcelona

Options in many commercialcodes, HMS and others HMS Menu Transformmethod,Userspecified,SCS, etc

Rainfall - runoff Different options Unit hydrograph (most popular)Lumped (global) modelHydrological response at the basinoutlet from an effective rainfall, 1 mm,duration D minutes, uniformlydistributed all over the basin

UH Hipothesis Linear response Time invariant (rain event)Q1 mm effective rainfallEffective rainfallUnit hydrograph,Unit hydrograph,duration d minutesDuration tdTb tTiempo

Problems in the application Lack of real data to obtain it We need to use synthetic UH

The Basic ProcessNecessary for asingle basinUnitHydrographsExcess Precip.ModelExcess Precip.Excess Precip.Basin “Routing”UHG MethodsRunoffHydrographRunoffHydrographStream and/orReservoir“Routing”DownstreamHydrograph

Proposal of the HU Sherman - 1932 Horton - 1933 Wisler & Brater - 1949 - “the hydrograph ofsurface runoff resulting from a relatively short,intense rain, called a unit storm” Standardly used in most professional codes forrural basins

Unit Hydrograph “Lingo” DurationLag TimeTime of ConcentrationRising LimbRecession Limb (fallinglimb)Peak FlowTime to Peak (rise time)Recession CurveSeparationBase flow

Graphical Representation DurationLag TimeTime ofConcentrationRising LimbRecession Limb(falling limb)Peak FlowTime to Peak (risetime)Recession CurveSeparationBase flowDuration ofexcess precip.Lag timeTime ofconcentrationBase flow

How to get the UH Field data measurements I(t), Q(t) Approach with synthetic unithydrographs SCS (NRCS)Time - area curve (Clark, 1945)

Unit Hydrograph Hydrological response at the basin outlet from aneffective rainfall, 1 mm, duration D minutes,uniformly distributed all over the basin Puntos capitales: 1-inch 1-mm of effective rainfall Uniformly distributed in space and time (durationD minutes) Different hydrographs for different durations

How to use the UH Graphical process Hyetograph definedwith time steps “d” Use the unithydrograph forduration “d” Addition of differentsub-hydrographs

How to use the UH Matrix approach Prepare matrix Pand U Direct operationI1 0 0I 2 I1 000u1I 3 I 2 I1 0u20I 3 I 2 I1u3I3 I 20 I3u40 00 0 Q1 Q2 Q3 Q4 Q5 Q6

UH obtention We need field measurements RainfallRunoff hydrograph at the basin outletWe measure the total rainfall (effectiverainfall is “estimated”)SOURCE OFUNCERTAINTYOther problems (errors, spatial distrib.)

UH obtentionRules of Thumb : the storm should be fairly uniform in nature and the excessprecipitation should be equally as uniform throughout the basin.This may require the initial conditions throughout the basin tobe spatially similar. Second, the storm should be relatively constant in time,meaning that there should be no breaks or periods of noprecipitation. Finally, the storm should produce at least an inch (1 mm)of excess precipitation (the area under the hydrograph aftercorrecting for baseflow).

Deriving a UHG from aStorm250000.80.7200000.5Flow (cfs)150000.4100000.30.250000.1Time itation (inches)0.6

Derived Unit 003.50004.0000

UH obtentionsurface andgroundwaterresponse We want justthe 0.0000. 016000.3200. 048000.6400. 080000.9601. 01201. 028001.4401. 060001.7601. 092002.0802. 024002.4002. 05602. 072002.8803. 004003.2003. 036003.5203. 06800 Measured Q(t),

Separation of Baseflow . generally accepted that the inflection point on the recessionlimb of a hydrograph is the result of a change in the controllingphysical processes of the excess precipitation flowing to the basinoutlet. In this example, baseflow is considered to be a straight lineconnecting that point at which the hydrograph begins to riserapidly and the inflection point on the recession side of thehydrograph. the inflection point may be found by plotting the hydrograph insemi-log fashion with flow being plotted on the log scale andnoting the time at which the recession side fits a straight line.

Semi-log Plot100000 Groundwaterresponse,exponentialRecession side of hydrographbecomes linear at approximately hour64.1000010Time 3944to determine theseparation point1002934 Use log paperFlow (cfs)1000

Hydrograph & Baseflow2500020000Flow (cfs)15000100005000Time (hrs.)13312611911210598918477706356494235282171400

Separate Baseflow2500020000Flow (cfs)15000100005000Time (hrs.)13312611911298105918477706356494235282171400

Separation of Baseflow If no significant contribution fromgroundwaters, use a horizontal straight line Constant base flow

UH from field data Matrix approachI1000I2I100I3 I 2I100I 3 I 2 I1u1u2u300I3 I2u4000I3 Q1 Q 2 Q3 Q4 Q5 Q6 PU Q

UH from field data Considering a matrix algebra, we can obtainthe vector U, from vectos Q and matrix PPU QPU QP T PU P T Q U P PT 1TP Q

UH of D’ from D minutes UH Sometimes you have UH for duration D, butyou need the D’ UH duration Use of S-curve Just for real UH, not synthetic ones

S hydrograph60000.00 Consider a40000.00Flow (cfs)very long,constant rainevent50000.00Addition ofUH, duration 8423630241854Time (hrs.)102 Hydrological120.000mm/h10000.006 Intensity 1/D

S curve

UH of D’ from D minutes UH From D minutes UH, establish S curve Move D’ minutes the S-curve Sustract both hydrographs Convert to unit rainfall, the obtainedhydrograph (1 mm rainfall)

Problems of the UH obtention Rain event selection Errors in rainfall or runoff measurements Non-uniform rain events Use of more than 1 event to obtain UH Make an average of the UH Optimization methods to obtain the UH fromseveral rain event at the same time

Average Several UHG’s It is suggested that several unit hydrographs be derived and averaged.The unit hydrographs must be of the same duration in order to beproperly averaged.It is often not sufficient to simply average the ordinates of the unithydrographs in order to obtain the final unit hydrograph. A numericalaverage of several unit hydrographs which are different “shapes” mayresult in an “unrepresentative” unit hydrograph.It is often recommended to plot the unit hydrographs that are to beaveraged. Then an average or representative unit hydrograph shouldbe sketched or fitted to the plotted unit hydrographs.Finally, the average unit hydrograph must have a volume of 1 mm ofrunoff for the basin.

Synthetic UHG’s SCS Clark (Time-area method)

SCS SUH SCS proposal SimpleSCS Dimensionless UHG Features1 Basin of regularFlow ratiosCum. Mass0.8shapes Single peakQ/Qpeak0.60.40.2000.511.522.5T/Tpeak33.544.55

Triangular SHUDSCS Dimensionless UHG & Triangular RepresentationExcessPrecipitation1.2Tlag10.8Flow ratiosQ/QpeakCum. MassTriangular0.6Point ofInflectionTc0.40.200.0Tp1.02.0Tb3.0T/Tpeak4.05.0

Dimensionless RatiosTime Discharge 7.147.107.077.055.040.029.021.015.011.005.000Mass Curve .934.953.967.977.984.989.993.995.997.9991.000

Triangular RepresentationDSCS Dimensionless UHG & Triangular RepresentationExcessPrecipitation1.2Tb 2.67 x TpTlag1Tr Tb - Tp 1.67 x Tp0.8Flow ratiosQ qpT p2 qpT r2 qp22Qqp T p T rqp 654.33 x 2 x A x QT p T rqp 484 A QTpQ/QpeakCum. MassTriangular0.6(T p T r )Point The 645.33 is the conversion used for delivering 1inch of runoff (the area under the unit hydrograph)from 1-square mile in 1-hour (3600 seconds).

Duration & Timing?Again from the triangleDT p L2L Lag timeL 0.6 * TcTc D 1.7 T pD 0.6 T c T p2For estimation purposes should be around : D 0.133 TcTo be used with SCS concept and expressions

Time of Concentration Regression Eqs. Segmental Approach

Time of Concentration In Spain, we use the Témez’s formula Different concept for Tc than the SCS Modify expressions for SHU, SCS

A Regression EquationL0.8 (S 1) 0.7Tlag 1900(% Slope) 0.5where : Tlag lag time in hoursL Length of the longest drainage path in feetS (1000/CN) - 10 (CN curve number)%Slope The average watershed slope in %Tc 0.3(LJ0.76)0.25

Tc, is it always the same?L0.8 (S 1) 0.7Tlag 1900(% Slope) 0.5Tc 0.3(LJ0.25)0.76 Tc for SCS and Tc for other expressions are not the same T’c for SCS, time to inflexion point of the UH Tc as time needed to exit the basin from the farthest pointTlag 0.6 T 0.35 Tc'c

Clark, synthetic UH Propossed by Clark Considering the basin shape, not just thetotal area Consider delays attributed to sub-surfacerunoff Need to be applied in non regular shapebasins

Clark - Time-Area

Time-Area Synthetic UH, equal form as the time-area curve It can show more than one peak100%QTimeof conc.% AreaTimeTime

Additional delay Presence of sub-surface runoff Runoff shows an additional delay, that cannot be explained just for surface runoff

Reservoir model for thedelay Conceptual model Assume that additional delay is equal to theproduced by a water reservoirdSI Q dtI (UH Clark)KQ

Storage description General approachS K1 Q K 2 Q K n Q2n Simplified to a linear reservoir modelS KQ K has dimensions of timeK T

Mathematical description If K is contant in timedQI Q Kdt We can solve the diff. Equation as:tQ (t ) 0I ( )eK tKd

Practical application A finite difference scheme can be usedI1 I 2Q1 Q 2Q 2 Q1 K22 t From an initial condition Q1, and from thevalues of the previous hydrograph we canproceed as:2 tQ2 2K tQ1Q1 I1 2 K t

K values We need to estimate the K value Best approach, field data (I , Q) From correlations obtained in other Proposed K 0.75 Tc

Basin application Basin 190 Km2 , Tc 8 hours K 5.5 hours Time step, 1 or 2 hours87776665240315

Basin geometryIsocronesArea -8Area(km2)59231927263940190Accum. Area(km2)514375885111150190190AccumTime (hrs)1.02.03.04.05.06.07.08.08.0

Time area curve40Area (Km2)Incremental Area (sqaure miles)358302520151057707616650315345Time Increment (hrs)242678

Time accumulated area9Cumulative Area (sqaure miles)8Isochrone43207720406080100Time 0200

No time / area curve ?TAi 1.414Ti 1.51 TAi 1.414(1 Ti )1.5for (0 Ti 0.5)for (0.5 Ti 1.0)U.S. Army Corps ofEngineers (HEC 1990) What about if the synthetic curve does not matchthe real one?

SUH, comments SUH is an approach to the real UH, could begood or not SHU Clark, problems to estimate K SHU SCS, one peak value, can only beapplied to basins with regular shape You must make your choice according thebasin characteristics

Average Several UHG’s It is suggested that several unit hydrographs be derived and averaged. The unit hydrographs must be of the same duration in order to be properly averaged. It is often not sufficient to simply average the ordinates of the unit hydrographs in order to obtain the final unit hydrograph. A numerical average o