unit hydrographs transforming the runoff. unit hydrograph theory moving water off of the...
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Unit Hydrographs
Transforming the Runoff
Unit Hydrograph Theory
Moving water off of the watershed… A mathematical concept Linear in nature Uses convolution to transform the
excess precipitation to streamflow….
The Basic Process
Excess Precip. Model
Excess Precip.
Excess Precip.
Runoff Hydrograph
Runoff Hydrograph
Stream and/or Reservoir “Routing”
Downstream Hydrograph
Basin “Routing” UHG Methods
Necessary for a single basin
Unit Hydrographs
Unit Hydrograph TheoryUnit Hydrograph Theory
Sherman - 1932 Horton - 1933 Wisler & Brater - 1949 - “the hydrograph of surface runoff
resulting from a relatively short, intense rain, called a unit storm.”
The runoff hydrograph may be “made up” of runoff that is generated as flow through the soil (Black, 1990).
Unit Hydrograph “Lingo”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
Graphical RepresentationGraphical Representation
Lag time
Time of concentration
Duration of excess precip.
Base flow
Methods of Developing Methods of Developing UHG’sUHG’s
From Streamflow Data Synthetically
Snyder SCS Time-Area (Clark, 1945)
“Fitted” Distributions
Unit HydrographUnit Hydrograph
The hydrograph that results from 1-inch of excess precipitation (or runoff) spread uniformly in space and time over a watershed for a given duration.
The key points : 1-inch of EXCESS precipitation Spread uniformly over space - evenly over the
watershed Uniformly in time - the excess rate is constant over
the time interval There is a given duration
Derived Unit HydrographDerived Unit Hydrograph
0.0000
100.0000
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400.0000
500.0000
600.0000
700.0000
0.00
00
0.16
00
0.32
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0.48
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0.64
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0.80
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0.96
00
1.12
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1.28
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1.44
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1.60
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1.76
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1.92
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2.08
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2.56
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2.72
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2.88
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3.04
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3.20
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3.36
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3.52
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3.68
00
Baseflow
Surface Response
Derived Unit HydrographDerived Unit Hydrograph
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0.0000 0.5000 1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000
Total Hydrograph
Surface Response
Baseflow
Derived Unit HydrographDerived Unit Hydrograph
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 of excess precipitation (the area under the hydrograph after
correcting for baseflow).
Deriving a UHG from a Deriving a UHG from a StormStormsample watershed = 450 mi2sample watershed = 450 mi2
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0 8 16 24 32 40 48 56 64 72 80 88 96 104
112
120
128
Time (hrs.)
Flo
w (
cfs)
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Pre
cip
itat
ion
(in
ches
)
Separation of BaseflowSeparation 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.
In 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.
Semi-log PlotSemi-log Plot
1
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100000
29 34 39 44 49 54 59 64 69 74 79 84 89 94 99 104
109
114
119
124
129
134
Time (hrs.)
Flo
w (
cfs)
Recession side of hydrograph becomes linear at approximately hour
64.
Hydrograph & BaseflowHydrograph & Baseflow
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0 7 14 21 28 35 42 49 56 63 70 77 84 91 98 105
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Time (hrs.)
Flo
w (
cfs)
Separate BaseflowSeparate Baseflow
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0 7 14 21 28 35 42 49 56 63 70 77 84 91 98 105
112
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Time (hrs.)
Flo
w (
cfs)
Sample CalculationsSample Calculations
In the present example (hourly time step), the flows are summed and then multiplied by 3600 seconds to determine the volume of runoff in cubic feet. If desired, this value may then be converted to acre-feet by dividing by 43,560 square feet per acre.
The depth of direct runoff in feet is found by dividing the total volume of excess precipitation (now in acre-feet) by the watershed area (450 mi2 converted to 288,000 acres).
In this example, the volume of excess precipitation or direct runoff for storm #1 was determined to be 39,692 acre-feet.
The depth of direct runoff is found to be 0.1378 feet after dividing by the watershed area of 288,000 acres.
Finally, the depth of direct runoff in inches is 0.1378 x 12 = 1.65 inches.
Again - Summing Flows
Continuous process represented with discrete time steps
Obtain UHG OrdinatesObtain UHG Ordinates
• The ordinates of the unit hydrograph are obtained by dividing each flow in the direct runoff hydrograph by the depth of excess precipitation.
• In this example, the units of the unit hydrograph would be cfs/inch (of excess precipitation).
Final UHGFinal UHG
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0 7 14 21 28 35 42 49 56 63 70 77 84 91 98 105
112
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126
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Time (hrs.)
Flo
w (
cfs)
Storm #1 hydrograph
Storm#1 direct runoff hydrograph
Storm # 1 unit hydrograph
Storm #1 baseflow
Determine Duration of UHGDetermine Duration of UHG
The duration of the derived unit hydrograph is found by examining the precipitation for the event and determining that precipitation which is in excess.
This is generally accomplished by plotting the precipitation in hyetograph form and drawing a horizontal line such that the precipitation above this line is equal to the depth of excess precipitation as previously determined.
This horizontal line is generally referred to as the -index and is based on the assumption of a constant or uniform infiltration rate.
The uniform infiltration necessary to cause 1.65 inches of excess precipitation was determined to be approximately 0.2 inches per hour.
Estimating Excess Precip.Estimating Excess Precip.
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0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Time (hrs.)
Pre
cip
itat
ion
(in
ches
)
Uniform loss rate of 0.2 inches per hour.
Excess PrecipitationExcess Precipitation
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0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Time (hrs.)
Exc
ess
Pre
c. (
inch
es)
Small amounts of excess precipitation at beginning and end may
be omitted.
Derived unit hydrograph is the result of approximately 6 hours
of excess precipitation.
Changing the DurationChanging the Duration Very often, it will be necessary to change the duration of the unit
hydrograph. If unit hydrographs are to be averaged, then they must be of the same
duration. Also, convolution of the unit hydrograph with a precipitation event
requires that the duration of the unit hydrograph be equal to the time step of the incremental precipitation.
The most common method of altering the duration of a unit hydrograph is by the S-curve method.
The S-curve method involves continually lagging a unit hydrograph by its duration and adding the ordinates.
For the present example, the 6-hour unit hydrograph is continually lagged by 6 hours and the ordinates are added.
Develop S-CurveDevelop S-Curve
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0 6 12 18 24 30 36 42 48 54 60 66 72 78 84 90 96 102
108
114
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Time (hrs.)
Flo
w (
cfs)
Continuous 6-hour bursts
Convert to 1-Hour DurationConvert to 1-Hour Duration
To arrive at a 1-hour unit hydrograph, the S-curve is lagged by 1 hour and the difference between the two lagged S-curves is found to be a 1 hour unit hydrograph.
However, because the S-curve was formulated from unit hydrographs having a 6 hour duration of uniformly distributed precipitation, the hydrograph resulting from the subtracting the two S-curves will be the result of 1/6 of an inch of precipitation.
Thus the ordinates of the newly created 1-hour unit hydrograph must be multiplied by 6 in order to be a true unit hydrograph.
The 1-hour unit hydrograph should have a higher peak which occurs earlier than the 6-hour unit hydrograph.
Final 1-hour UHGFinal 1-hour UHG
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Time (hrs.)
Un
it H
ydro
gra
ph
Flo
w (
cfs/
inch
)
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Flo
w (
cfs)
S-curves are lagged by 1 hour and the difference
is found.1-hour unit
hydrograph resulting from lagging S-
curves and multiplying the difference by 6.
Average Several UHG’sAverage Several UHG’s
It is recommend 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 of several unit hydrographs which are different “shapes” may result in an “unrepresentative” unit hydrograph.
It 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 1 inch of runoff for the basin.
Synthetic UHG’sSynthetic UHG’s
Snyder SCS Time-area
SnyderSnyder
Since peak flow and time of peak flow are two of the most important parameters characterizing a unit hydrograph, the Snyder method employs factors defining these parameters, which are then used in the synthesis of the unit graph (Snyder, 1938).
The parameters are Cp, the peak flow factor, and Ct, the lag factor.
The basic assumption in this method is that basins which have similar physiographic characteristics are located in the same area will have similar values of Ct and Cp.
Therefore, for ungaged basins, it is preferred that the basin be near or similar to gaged basins for which these coefficients can
be determined.
Basic RelationshipsBasic Relationships
3.0)(catLAG
LLCt
5.5LAG
duration
tt
)(25.0.. durationdurationaltLAGlagalt
tttt
83 LAG
base
tt
LAG
p
peak t
ACq
640
Final ShapeFinal Shape
The final shape of the Snyder unit hydrograph is controlled by the equations for width at 50% and 75% of the peak of the UHG:
SCSSCS
SCS Dimensionless UHG Features
0
0.2
0.4
0.6
0.8
1
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
T/Tpeak
Q/Q
pe
ak
Flow ratios
Cum. Mass
Dimensionless RatiosDimensionless RatiosTime Ratios
(t/tp)Discharge Ratios
(q/qp)Mass Curve Ratios
(Qa/Q)0 .000 .000.1 .030 .001.2 .100 .006.3 .190 .012.4 .310 .035.5 .470 .065.6 .660 .107.7 .820 .163.8 .930 .228.9 .990 .300
1.0 1.000 .3751.1 .990 .4501.2 .930 .5221.3 .860 .5891.4 .780 .6501.5 .680 .7001.6 .560 .7511.7 .460 .7901.8 .390 .8221.9 .330 .8492.0 .280 .8712.2 .207 .9082.4 .147 .9342.6 .107 .9532.8 .077 .9673.0 .055 .9773.2 .040 .9843.4 .029 .9893.6 .021 .9933.8 .015 .9954.0 .011 .9974.5 .005 .9995.0 .000 1.000
Triangular RepresentationTriangular RepresentationSCS Dimensionless UHG & Triangular Representation
0
0.2
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0.8
1
1.2
0.0 1.0 2.0 3.0 4.0 5.0
T/Tpeak
Q/Q
pea
k
Flow ratios
Cum. Mass
Triangular
Excess Precipitation
D
Tlag
Tc
TpTb
Point of Inflection
Triangular RepresentationTriangular Representation
pb T x 2.67 T
ppbr T x 1.67 T - T T
)T + T( 2
q =
2
Tq +
2
Tq = Q rp
prppp
T + T
2Q = q
rpp
T + T
Q x A x 2 x 654.33 = q
rpp The 645.33 is the conversion used for
delivering 1-inch of runoff (the area under the unit hydrograph) from 1-square
mile in 1-hour (3600 seconds). T
Q A 484 = q
pp
SCS Dimensionless UHG & Triangular Representation
0
0.2
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1
1.2
0.0 1.0 2.0 3.0 4.0 5.0
T/Tpeak
Q/Q
pea
k
Flow ratios
Cum. Mass
Triangular
Excess Precipitation
D
Tlag
Tc
TpTb
Point of Inflection
484 ?484 ?
Comes from the initial assumption that 3/8 of the volume under the UHG is under the rising limb and the remaining 5/8
is under the recession limb.
General Description Peaking Factor Limb Ratio (Recession to Rising)
Urban areas; steep slopes
575 1.25
Typical SCS 484 1.67 Mixed urban/rural 400 2.25 Rural, rolling hills 300 3.33 Rural, slight slopes 200 5.5
Rural, very flat 100 12.0
T
Q A 484 = q
pp
Duration & Timing?Duration & Timing?
L + 2
D = T p
cTL *6.0
L = Lag time
pT 1.7 D Tc
T = T 0.6 + 2
Dpc
For estimation purposes : cT 0.133 D
Again from the triangle
Time of ConcentrationTime of Concentration
Regression Eqs. Segmental Approach
A Regression EquationA Regression Equation
TlagL S
Slope
08 1 0 7
1900 05
. ( ) .
(% ) .
where : 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 %
Segmental ApproachSegmental Approach
More “hydraulic” in nature The parameter being estimated is essentially the time of
concentration or longest travel time within the basin. In general, the longest travel time corresponds to the longest
drainage path The flow path is broken into segments with the flow in each
segment being represented by some type of flow regime. The most common flow representations are overland, sheet, rill
and gully, and channel flow.
A Basic ApproachA Basic Approach 2
1
kSV
McCuen (1989) and SCS (1972) provide values of k for several flow situations
(slope in %)
K Land Use / Flow Regime 0.25 Forest with heavy ground litter, hay meadow
(overland flow) 0.5 Trash fallow or minimum tillage cultivation; contour
or strip cropped; woodland (overland flow) 0.7 Short grass pasture (overland flow) 0.9 Cultivated straight row (overland flow) 1.0 Nearly bare and untilled (overland flow); alluvial fans
in western mountain regions 1.5 Grassed waterway 2.0 Paved area (sheet flow); small upland gullies
Flow Type K Small Tributary - Permanent or
intermittent streams which appear as solid or dashed blue lines on USGS
topographic maps.
2.1
Waterway - Any overland flow route which is a well defined swale by
elevation contours, but is not a stream section as defined above.
1.2
Sheet Flow - Any other overland flow path which does not conform to the
definition of a waterway.
0.48
Sorell & Hamilton, 1991
Triangular ShapeTriangular Shape
In general, it can be said that the triangular version will not cause or introduce noticeable differences in the simulation of a storm event, particularly when one is concerned with the peak flow.
For long term simulations, the triangular unit hydrograph does have a potential impact, due to the shape of the recession limb.
The U.S. Army Corps of Engineers (HEC 1990) fits a Clark unit hydrograph to match the peak flows estimated by the Snyder unit hydrograph procedure.
It is also possible to fit a synthetic or mathematical function to the peak flow and timing parameters of the desired unit hydrograph.
Aron and White (1982) fitted a gamma probability distribution using peak flow and time to peak data.
Fitting a Gamma DistributionFitting a Gamma Distribution
)1(),;(
1
ab
etbatf
a
bta
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250.0000
300.0000
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500.0000
0.0000 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000
Time-AreaTime-Area
Time-AreaTime-Area
Time
Q % Area
Time
100%
Timeof conc.
Time-AreaTime-Area
Hypothetical ExampleHypothetical Example
A 190 mi2 watershed is divided into 8 isochrones of travel time. The linear reservoir routing coefficient, R, estimated as 5.5
hours. A time interval of 2.0 hours will be used for the computations.
WatershedBoundary
Isochrones
2
345
66
7
8
6
6
5
7
7
1
0
Rule of ThumbRule of Thumb
R - The linear reservoir routing coefficient can be estimated as approximately 0.75
times the time of concentration.
Basin BreakdownBasin Breakdown
Map Area #
Bounding Isochrones
Area (mi2)
Cumulative Area (mi2)
Cumulative Time (hrs)
1 0-1 5 5 1.0 2 1-2 9 14 2.0 3 2-3 23 37 3.0 4 3-4 19 58 4.0 5 4-5 27 85 5.0 6 5-6 26 111 6.0 7 6-7 39 150 7.0 8 7-8 40 190 8.0
TOTAL 190 190 8.0
WatershedBoundary
Isochrones
2
345
66
7
8
6
6
5
7
7
1
0
Incremental AreaIncremental Area
0
5
10
15
20
25
30
35
40
Incr
emen
tal
Are
a (s
qau
re m
iles
)
1 2 3 4 5 6 7 8
Time Increment (hrs)
WatershedBoundary
Isochrones
2
345
66
7
8
6
6
5
7
7
1
0
Cumulative Time-Area CurveCumulative Time-Area Curve
0
1
2
3
4
5
6
7
8
9
0 20 40 60 80 100 120 140 160 180 200
Time (hrs)
Cu
mu
lati
ve A
rea
(sq
aure
mil
es)
WatershedBoundary
Isochrones
2
345
66
7
8
6
6
5
7
7
1
0
Trouble Getting a Time-Area Trouble Getting a Time-Area Curve?Curve?
0.5) Ti (0for 414.1 5.1 ii
TTA
1.0) Ti (0.5for )1(414.11 5.1 ii
TTA
Synthetic time-area curve - The U.S. Army Corps of Engineers (HEC 1990)
Instantaneous UHGInstantaneous UHG
)1()1(
iiiIUHccIIUH
tR
tc
2
2
t = the time step used n the calculation of the translation unit hydrograph
The final unit hydrograph may be found by averaging 2 instantaneous unit hydrographs that are a t time step apart.
ComputationsComputationsTime(hrs)
(1)
Inc.Area(mi2)(2)
Inc.TranslatedFlow (cfs)
(3)
Inst. UHG
(4)
IUHGLagged 2
hours(5)
2-hrUHG(cfs)(6)
0 0 0 0 02 14 4,515 1391 0 7004 44 14,190 5333 1,391 3,3606 53 17,093 8955 5,333 7,1508 79 25,478 14043 8,955 11,50010 0 0 9717 14,043 11,88012 6724 9,717 8,22014 4653 6,724 5,69016 3220 4,653 3,94018 2228 3,220 2,72020 1542 2,228 1,89022 1067 1,542 1,30024 738 1,067 90026 510 738 63028 352 510 43030 242 352 30032 168 242 20034 116 168 14036 81 116 10038 55 81 7040 39 55 5042 26 39 3044 19 26 2046 13 19 2048 13
Incremental AreasIncremental Areas
0
10
20
30
40
50
60
70
80
90
0 2 4 6 8 10
Time Increments (2 hrs)
Are
a In
crem
ents
(sq
uar
e m
iles
)
Incremental FlowsIncremental Flows
0
5000
10000
15000
20000
25000
30000
1 2 3 4 5 6
Time Increments (2 hrs)
Tra
nsl
ated
Un
it H
ydro
gra
ph
Instantaneous UHGInstantaneous UHG
0
2000
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6000
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10000
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14000
16000
0 10 20 30 40 50 60
Time (hrs)
Flo
w (
cfs/
inch
)
Lag & AverageLag & Average
0
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14000
16000
0 10 20 30 40 50 60
Time (hrs)
Flo
w (
cfs/
inch
)
Convolution
Putting It All Together
The Basic Process….
Excess Precip. Model
Excess Precip.
Excess Precip.
Runoff Hydrograph
Runoff Hydrograph
Stream and/or Reservoir “Routing”
Downstream Hydrograph
Basin “Routing” UHG Methods
Necessary for a single basin
ConvolutionConvolution
CN = 68 S= 4.706 IA = 0.9412Time Ppt Cum Ppt.cum. ROInc. ROUHG(@360)RO1(UHG)RO2(UHG)RO3(UHG)RO4(UHG)
0 0 0 0.00 0.00 0.00 0.00 0.00 0.00 0.001 0.7 0.7 0.01 0.01 750.00 0.00 0.00 0.00 0.002 1.2 1.9 0.16 0.15 1500.00 0.00 9.77 0.00 0.003 0.6 2.5 0.39 0.23 2250.00 0.00 19.54 111.95 0.004 1.5 4 1.20 0.82 2850.00 0.00 29.31 223.90 169.195 6.3 10.3 6.23 5.02 3625.00 0.00 37.13 335.85 338.376 4 14.3 9.88 3.65 3400.00 0.00 47.23 425.41 507.567 0 14.3 9.88 0.00 3150.00 0.00 44.30 541.09 642.918 1.6 15.9 11.38 1.50 2725.00 0.00 41.04 507.50 817.749 0.4 16.3 11.76 0.38 2500.00 0.00 35.50 470.19 766.9810 0.2 16.5 11.95 0.19 2250.00 0.00 32.57 406.75 710.5911 0.6 17.1 12.51 0.57 1925.00 0.00 29.31 373.16 614.7112 0.1 17.2 12.61 0.09 1675.00 0.00 25.08 335.85 563.9613 0 17.2 12.61 0.00 1450.00 0.00 21.82 287.34 507.5614 0 17.2 12.61 0.00 1225.00 0.00 18.89 250.02 434.2515 0 17.2 12.61 0.00 1000.00 0.00 15.96 216.43 377.8516 0 17.2 12.61 0.00 775.00 0.00 13.03 182.85 327.1017 0 17.2 12.61 0.00 550.00 0.00 10.10 149.27 276.3418 0 17.2 12.61 0.00 325.00 0.00 7.17 115.68 225.5819 0 17.2 12.61 0.00 100.00 0.00 4.23 82.10 174.8320 0 17.2 12.61 0.00 0.00 0.00 1.30 48.51 124.0721 0 17.2 12.61 0.00 0.00 0.00 0.00 14.93 73.3122 0 17.2 12.61 0.00 0.00 0.00 0.00 0.00 22.5623 0 17.2 12.61 0.00 0.00 0.00 0.00 0.00 0.0024 0 17.2 12.61 0.00 0.00 0.00 0.00 0.00 0.0025 0 17.2 12.61 0.00 0.00 0.00 0.00 0.00 0.0026 0 17.2 12.61 0.00 0.00 0.00 0.00 0.00 0.0027 0 17.2 12.61 0.00 0.00 0.00 0.00 0.00 0.0028 0 17.2 12.61 0.00 0.00 0.00 0.00 0.00 0.0029 0 17.2 12.61 0.00 0.00 0.00 0.00 0.00 0.0030 0 17.2 12.61 0.00 0.00 0.00 0.00 0.00 0.0031 0 17.2 12.61 0.00 0.00 0.00 0.00 0.00 0.0032 0 17.2 12.61 0.00 0.00 0.00 0.00 0.00 0.0033 0 17.2 12.61 0.00 0.00 0.00 0.00 0.00 0.00
Individual ResponsesIndividual Responses
Individual Responses
-5000.00
0.00
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15000.00
20000.00
0 5 10 15 20 25 30 35 40 45
Time (hrs)
Flo
w (
cfs)
Overall ResponseOverall Response
Overall Response
-5000.00
0.00
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0 5 10 15 20 25 30 35 40 45
Time (hrs)
Flo
w (
cfs)
UHG Application
UHG’s in the NWSRFS – a few issues
The SAC-SMA UHG
The SAC-SMA model computes the following components:Surface runoff which occurs when the storage capacity of the upper zone free water is exceeded.Impervious runoff from impermeable surfaces (if the percent impervious is set to a value greater than zero.Direct runoff from additional impervious surfaces (if applicable).Interflow and baseflow contributions.
More on SAC-SMA UHG
When developing a unit hydrograph for the SAC-SMA model, the user should attempt to separate out both baseflow and interflow.
SAC-SMA – more…
The very nature of the unit hydrograph is that is time distributes the runoff or excess precipitation. Therefore, it accounts for lagging or delays. The SAC-SMA model also accounts for delays in the interflow and baseflow components; therefore, they should not be accounted for in the unit hydrograph that is to be used with the SAC-SMA.
Issues w/ UHG in Forecasting
Storm Size Moving Storms