Rainfall - runoff: Unit Hydrograph. Manuel Gómez Valentín E.T.S. Ing. Caminos, Canales y Puertos de Barcelona

Transcription

1 Rainfall - runoff: Unit Hydrograph Manuel Gómez Valentín E.T.S. ng. Caminos, Canales y Puertos de Barcelona

2 Options in many commercial codes,, HMS and others HMS Menu Transform method, User specified, SCS, etc

3 Rainfall - runoff Different options Unit hydrograph (most popular) Lumped (global) model Hydrological response at the basin outlet from an effective rainfall, mm, duration D minutes, uniformly distributed all over the basin

4 UH Hipothesis Linear response Time invariant (rain event) Effective rainfall mm effective rainfall Unit hydrograph, duration d minutes Unit hydrograph, Duration t d Tb t Tiempo

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

6 The Basic Process Unit Hydrographs Necessary for a single basin Excess Precip. Model Excess Precip. Excess Precip. Basin Routing UHG Methods Runoff Hydrograph Runoff Hydrograph Stream and/or Reservoir Routing Downstream Hydrograph

7 Proposal of the HU Sherman - 93 Horton Wisler & Brater the hydrograph of surface runoff resulting from a relatively short, intense rain, called a unit storm Standardly used in most professional codes for rural basins

8 Unit Hydrograph Lingo Duration Lag Time Time of Concentration Rising Limb Recession Limb (falling limb) Peak Flow Time to Peak (rise time) Recession Curve Separation Base flow

9 Graphical Representation Duration Lag Time Time of Concentration Rising Limb Recession Limb (falling limb) Peak Flow Time to Peak (rise time) Recession Curve Separation Base flow Duration of excess precip. Lag time Base flow Time of concentration

10 How to get the UH Field data measurements (t), (t) Approach with synthetic unit hydrographs SCS (NRCS) Time - area curve (Clark, 945)

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

12 How to use the UH Graphical process Hyetograph defined with time steps d Use the unit hydrograph for duration d Addition of different sub-hydrographs

13 Matrix approach Prepare matrix P and U Direct operation u u u u u u u u How How to to use use the the UH UH

14 UH obtention We need field measurements Rainfall Runoff hydrograph at the basin outlet We measure the total rainfall (effective rainfall is estimated ) SOURCE OF UNCERTANTY Other problems (errors, spatial distrib.)

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

16 Deriving a UHG from a Deriving a UHG from a Storm Flow (cfs) Precipitation (inches) Time (hrs.)

17 Derived Unit Hydrograph Total Hydrograph Surface Response 3.. Baseflow

18 UH obtention Measured (t), surface and groundwater response We want just the surface response Surface Response Baseflow

19 Separation of Baseflow... generally accepted that the inflection point on the recession limb of a hydrograph is the result of a change in the controlling physical processes of the excess precipitation flowing to the basin outlet. n this example, baseflow is considered to be a straight line connecting that point at which the hydrograph begins to rise rapidly and the inflection point on the recession side of the hydrograph. the inflection point may be found by plotting the hydrograph in semi-log fashion with flow being plotted on the log scale and noting the time at which the recession side fits a straight line.

20 Semi-log Plot Groundwater response, exponential Use log paper to determine the separation point Flow (cfs) Recession side of hydrograph becomes linear at approximately hour Time (hrs.)

21 Hydrograph & Baseflow Flow (cfs) Time (hrs.)

22 33 Separate Baseflow 5 5 Flow (cfs) Time (hrs.)

23 Separation of Baseflow f no significant contribution from groundwaters, use a horizontal straight line Constant base flow

24 UH UH from from field field data data Matrix approach u u u u PU

25 UH from field data Considering a matrix algebra, we can obtain the vector U, from vectos and matrix P PU PU P T PU P T U T T P P P

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

27 S hydrograph 6. Consider a 5. very long, constant rain event Flow (cfs) Addition of UH, duration D ntensity /D mm/h.. Hydrological Time (hrs.) response?

28 S curve

29 UH of D 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 obtained hydrograph ( mm rainfall)

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

31 Average Several UHG s t 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. t is often not sufficient to simply average the ordinates of the unit hydrographs in order to obtain the final unit hydrograph. A numerical average of several unit hydrographs which are different shapes may result in an unrepresentative unit hydrograph. t is often recommended to plot the unit hydrographs that are to be averaged. Then an average or representative unit hydrograph should be sketched or fitted to the plotted unit hydrographs. Finally, the average unit hydrograph must have a volume of mm of runoff for the basin.

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

33 SCS SUH SCS proposal Simple Basin of regular shapes Single peak /peak SCS Dimensionless UHG Features Flow ratios Cum. Mass T/Tpeak

34 Triangular SHU D SCS Dimensionless UHG & Triangular Representation. Excess Precipitation T lag /peak.8.6 Point of nflection Flow ratios Cum. Mass Triangular T c.4.. T p. T b T/Tpeak

35 Dimensionless Ratios Discharge Ratios (q/q p ) Time Ratios (t/t p ) Mass Curve Ratios ( a /)

36 Triangular Representation D SCS Dimensionless UHG & Triangular Representation Tb.67 x T p. T lag Excess Precipitation = q q p T q p p T + r = T q T p T b r - T = +T p r p q x x A x = T p +T r.67 x p (T p +T T r p ) p The is the conversion used for delivering - inch of runoff (the area under the unit hydrograph) q p 484 A = T p /peak T c Point of nflection Flow ratios Cum. Mass Triangular. T p T b T/Tpeak from -square mile in -hour (36 seconds).

37 Duration & Timing? Again from the triangle T p = D + L L.6*T c L = Lag time T c D.7 T p D +.6 T c =T For estimation purposes should be around : p D.33 Tc To be used with SCS concept and expressions

38 Time of Concentration Regression Eqs. Segmental Approach

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

40 A Regression Equation L 8. ( S ) 7. T lag 9(% Slope) 5. where : T lag = lag time in hours L = Length of the longest drainage path in feet S = (/CN) - (CN=curve number) %Slope = The average watershed slope in % T c.3( J L.76 ).5

41 Tc,, is it always the same? Tlag L 8. ( S ) 7. 9(% Slope) 5. T c.3( J L.76 ).5 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 point lag ' c T.6 T. 35 T c

42 Clark, synthetic UH Propossed by Clark Considering the basin shape, not just the total area Consider delays attributed to sub-surface runoff Need to be applied in non regular shape basins

43 Clark - Time-Area

44 Time-Area Synthetic UH, equal form as the time-area curve t can show more than one peak % % Area Time of conc. Time Time

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

46 Reservoir model for the delay Conceptual model Assume that additional delay is equal to the produced by a water reservoir (UH Clark) ds dt K

47 Storage description General approach S K K Simplified to a linear reservoir model K has dimensions of time S K K n n K T

48 Mathematical description f K is contant in time K d dt We can solve the diff. Equation as: ( t ) t ( ) t K e K d

49 Practical Practical application application A finite difference scheme can be used From an initial condition, and from the values of the previous hydrograph we can proceed as: t K t K t K t

50 K values We need to estimate the K value Best approach, field data (, ) From correlations obtained in other Proposed K =.75 Tc

51 Basin application Basin 9 Km, Tc 8 hours K= 5.5 hours Time step, or hours

52 Basin geometry 8 7 socrones Area Accum. Area Accum Area # (km) (km ) Time (hrs) TOTAL

53 Time area curve Area (Km) ncremental Area (sqaure miles) Time ncrement (hrs) 5

54 Time accumulated area 9 8 Cumulative Area (sqaure miles) Watershed Boundary sochrone Time (hrs) 5

55 No time / area curve? TA i.44t i.5 for ( Ti.5) TA i.44( T ) i.5 for (.5 Ti.) U.S. Army Corps of Engineers (HEC 99) What about if the synthetic curve does not match the real one?

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