Transcript Unit Hydrograph Theory

Simulating the Hydrologic
Response
“Moving Water to the Basin Outlet”
Presented By
Dennis Johnson
For
COMET Hydrometeorology 99-2
Friday, 25 June 1999
Unit 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”
•
•
•
•
•
•
•
•
•
•
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 Representation
Duration of
excess
precipitation.
Lag time
Time of
concentration
Base flow
Methods of Developing UHG’s
• From Streamflow Data
• Synthetically
– Snyder
– SCS
– Time-Area (Clark, 1945)
• “Fitted” Distributions
Unit 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 Hydrograph
Streamflow & Assumed Baseflow
800
700
600
cfs
500
400
300
200
100
0
0
5
10
15
Time (hrs)
20
25
30
Derived Unit Hydrograph
Rules of Thumb :
æ… 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.
æ… Second, the storm should be relatively constant in time,
meaning that there should be no breaks or periods of no
precipitation.
æ… 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 Storm
sample watershed = 5.4 mi2
800
2
1.8
700
1.6
600
1.4
500
1.2
400
1
0.8
300
0.6
200
0.4
100
0.2
0
0
1
3
5
7
9
11
13
15
17
19
21
23
25
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.
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 semilog fashion with flow being plotted on the log scale and noting the time
at which the recession side fits a straight line.
Semi-log Plot
1000
Assuming inflection
@ ~ hour 17 (160 cfs)
100
10
1
0
5
10
15
20
25
30
Hydrograph & Baseflow
800
700
600
500
400
300
200
100
0
0
5
10
15
20
25
30
Separate Baseflow
800
700
600
500
400
300
200
100
0
0
-100
5
10
15
20
25
30
Represent Surface Flow as Discrete Blocks
(by averaging)
700
600
500
400
300
200
100
0
0
-100
2
4
6
8
10
12
14
16
18
20
22
24
Your Spreadsheet :
Time
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
Flow
32
33
32
32
89
223
450
620
680
691
685
640
550
465
370
290
210
160
123
95
77
58
46
39
33
Baseflow
32
33
32
32
41.142857
50.285714
59.428571
68.571429
77.714286
86.857143
96
105.14286
114.28571
123.42857
132.57143
141.71429
150.85714
160
123
95
77
58
46
39
33
Surface
Runoff
0
0
0
0
47.85714
172.7143
390.5714
551.4286
602.2857
604.1429
589
534.8571
435.7143
341.5714
237.4286
148.2857
59.14286
0
0
0
0
0
0
0
0
Average
Flow
0
0
0
23.9286
110.286
281.643
471
576.857
603.214
596.571
561.929
485.286
388.643
289.5
192.857
103.714
29.5714
0
0
0
0
0
0
0
0
Cubic
Feet of
Runoff
0
0
0
86142.86
397028.6
1013914
1695600
2076686
2171571
2147657
2022943
1747029
1399114
1042200
694285.7
373371.4
106457.1
0
0
0
0
0
0
0
0
16974000
Precipitation
0.5
1.2
1.8
0.3
0.6
0.1
Sample Calculations
• The total volume of runoff was calculated as
16,974,000 cubic feet.
• The basin average would now be :
runoff 
volume
16,974,000

*12  1.35inches
area
(5.4 * 640 * 43560)
Obtain 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 UHG
800
Flow
700
Surface Runoff
UHG
600
500
400
300
200
100
0
0
-100
5
10
15
20
25
30
Determine 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 F-index and is based
on the assumption of a constant or uniform infiltration rate.
• The uniform infiltration necessary to cause 1.35 inches of excess
precipitation was determined to be approximately 0.82 inches per hour.
Estimating Excess Precip.
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
1
2
3
4
5
6
Changing 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-Curve
4000
3500
3000
2500
2000
1500
1000
500
0
-500 0
10
20
30
40
50
Convert 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 2 hour duration of uniformly distributed precipitation, the
hydrograph resulting from the subtracting the two S-curves will be the
result of 1/2 of an inch of precipitation.
• Thus the ordinates of the newly created 1-hour unit hydrograph must
be multiplied by 2 in order to be a true unit hydrograph.
• The 1-hour unit hydrograph should have a higher peak which occurs
earlier than the 2-hour unit hydrograph.
Final 1-hour UHG
500
450
400
350
300
250
200
150
100
50
0
0
2
4
6
8
10
12
14
16
1, 2, 6, &12 Hr UHG’s
500
450
400
350
300
250
200
150
100
50
0
0
5
10
15
20
25
30
Average 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’s
•
•
•
•
Snyder
SCS
Time-area
IHABBS Implementation Plan :
NOHRSC Homepage
http://www.nohrsc.nws.gov/
http://www.nohrsc.nws.gov/98/html/uhg/index.html
Snyder
• 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 Relationships
tLAG  Ct ( L  Lca )0.3
tduration 
tLAG
5.5
talt.lag  tLAG  0.25(talt.duration  tduration)
tbase  3 
q peak 
t LAG
8
640 AC p
t LAG
Final 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:
SCS
SCS Dimensionless UHG Features
1
Flow ratios
Cum. Mass
0.8
Q/Qpeak
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
T/Tpeak
3
3.5
4
4.5
5
Dimensionless Ratios
Time Ratios
(t/tp)
0
.1
.2
.3
.4
.5
.6
.7
.8
.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
3.6
3.8
4.0
4.5
5.0
Discharge Ratios
(q/qp)
.000
.030
.100
.190
.310
.470
.660
.820
.930
.990
1.000
.990
.930
.860
.780
.680
.560
.460
.390
.330
.280
.207
.147
.107
.077
.055
.040
.029
.021
.015
.011
.005
.000
Mass Curve Ratios
(Qa/Q)
.000
.001
.006
.012
.035
.065
.107
.163
.228
.300
.375
.450
.522
.589
.650
.700
.751
.790
.822
.849
.871
.908
.934
.953
.967
.977
.984
.989
.993
.995
.997
.999
1.000
Triangular Representation
D
SCS Dimensionless UHG & Triangular Representation
Excess
Precipitation
1.2
Tlag
1
0.8
Flow ratios
Q/Qpeak
Cum. Mass
Triangular
0.6
Point of
Inflection
Tc
0.4
0.2
0
0.0
Tp
1.0
2.0
Tb
3.0
T/Tpeak
4.0
5.0
Triangular Representation
Tb  2.67 x Tp
D
SCS Dimensionless UHG & Triangular Representation
Excess
Precipitation
1.2
Tlag
1
Tr  Tb - Tp  1.67 x Tp
0.8
Flow ratios
Q/Qpeak
Cum. Mass
Triangular
0.6
Point of
Inflection
Tc
Q=
qpT p
2
+
qpT r
2
=
qp
2
0.4
( T p +T r )
0.2
0
0.0
2Q
qp=
T p +T r
qp=
654.33x 2 x A x Q
T p +T r
qp=
484 A Q
Tp
Tp
1.0
2.0
Tb
3.0
4.0
5.0
T/Tpeak
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).
484 ?
qp=
484 A Q
Tp
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
Urban areas; steep slopes
Typical SCS
Mixed urban/rural
Rural, rolling hills
Rural, slight slopes
Rural, very flat
575
484
400
300
200
100
Limb Ratio
(Recession to Rising)
1.25
1.67
2.25
3.33
5.5
12.0
Duration & Timing?
Again from the triangle
T p=
D
+L
2
L = Lag time
L  0.6 * Tc
Tc  D  1.7 Tp
D
+ 0.6 T c = T p
2
For estimation purposes :
D  0.133Tc
Time of Concentration
• Regression Eqs.
• Segmental Approach
A Regression Equation
. (S  1)0.7
L08
Tlag 
.
1900(%Slope)05
where : Tlag = lag time in hours
L = Length of the longest drainage path in feet
S = (1000/CN) - 10 (CN=curve number)
%Slope = The average watershed slope in %
Segmental 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 Approach
K
0.25
0.5
0.7
0.9
1.0
1.5
2.0
Land Use / Flow Regime
Forest with heavy ground litter, hay meadow (overland flow)
Trash fallow or minimum tillage cultivation; contour or strip
cropped; woodland (overland flow)
Short grass pasture (overland flow)
Cultivated straight row (overland flow)
Nearly bare and untilled (overland flow); alluvial fans in
western mountain regions
Grassed waterway
Paved area (sheet flow); small upland gullies
Flow Type
Small Tributary - Permanent or intermittent
streams which appear as solid or dashed
blue lines on USGS topographic maps.
Waterway - Any overland flow route which
is a well defined swale by elevation
contours, but is not a stream section as
defined above.
Sheet Flow - Any other overland flow path
which does not conform to the definition of
a waterway.
V  kS
1
2
McCuen (1989) and SCS
(1972) provide values of k
for several flow situations
(slope in %)
K
2.1
1.2
0.48
Sorell & Hamilton, 1991
Triangular 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 Distribution
t a e t b
f (t ; a, b)  a1
b (a  1)
500.0000
450.0000
400.0000
350.0000
300.0000
250.0000
200.0000
150.0000
100.0000
50.0000
0.0000
0.0000
1.0000
2.0000
3.0000
4.0000
5.0000
6.0000
Time-Area
Time-Area
100%
Q
Time
of conc.
% Area
Time
Time
Time-Area
Hypothetical 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.
Rule of Thumb
R - The linear reservoir routing coefficient
can be estimated as approximately 0.75
times the time of concentration.
Basin Breakdown
Map
Area #
1
2
3
4
5
6
7
8
TOTAL
Bounding
Isochrones
0-1
1-2
2-3
3-4
4-5
5-6
6-7
7-8
Area
(mi2)
5
9
23
19
27
26
39
40
190
Cumulative
Area (mi2)
5
14
37
58
85
111
150
190
190
Cumulative
Time (hrs)
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
8.0
Incremental Area
40
Incremental Area (sqaure miles)
35
30
25
20
15
10
5
0
1
2
3
4
5
Time Increment (hrs)
6
7
8
Cumulative Time-Area Curve
9
Cumulative Area (sqaure miles)
8
7
6
5
4
3
2
1
0
0
20
40
60
80
100
Time (hrs)
120
140
160
180
200
Trouble Getting a Time-Area Curve?
TAi  1.414Ti1.5
1  TAi  1.414(1  Ti )1.5
for (0  Ti  0.5)
for (0.5  Ti  1.0)
Synthetic time-area curve The U.S. Army Corps of
Engineers (HEC 1990)
Instantaneous UHG
IUH i  cI i  (1  c) IUH ( i 1)
c
2 t
2 R  t
ü t = the time step used n the
calculation of the translation unit
hydrograph
ü The final unit hydrograph may be
found by averaging 2
instantaneous unit hydrographs
that are a t time step apart.
Computations
Time
(hrs)
(1)
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
Inc.
Area
(mi2)
(2)
0
14
44
53
79
0
Inc.
Translated
Flow (cfs)
(3)
0
4,515
14,190
17,093
25,478
0
Inst.
UHG
(4)
0
1391
5333
8955
14043
9717
6724
4653
3220
2228
1542
1067
738
510
352
242
168
116
81
55
39
26
19
13
IUHG
Lagged 2
hours
(5)
0
1,391
5,333
8,955
14,043
9,717
6,724
4,653
3,220
2,228
1,542
1,067
738
510
352
242
168
116
81
55
39
26
19
13
2-hr
UHG
(cfs)
(6)
0
700
3,360
7,150
11,500
11,880
8,220
5,690
3,940
2,720
1,890
1,300
900
630
430
300
200
140
100
70
50
30
20
20
IUH i  cI i  (1  c) IUH ( i 1)
c
2 t
2 R  t
Incremental Areas
90
Area Increments (square miles)
80
70
60
50
40
30
20
10
0
0
2
4
6
Time Increments (2 hrs)
8
10
Incremental Flows
30000
Translated Unit Hydrograph
25000
20000
15000
10000
5000
0
1
2
3
4
Time Increments (2 hrs)
5
6
Instantaneous UHG
16000
14000
Flow (cfs/inch)
12000
10000
8000
6000
4000
2000
0
0
10
20
30
Time (hrs)
40
50
60
Lag & Average
16000
14000
Flow (cfs/inch)
12000
10000
8000
6000
4000
2000
0
0
10
20
30
Time (hrs)
40
50
60
Convolution
Current Modeling Techniques in NWS
• NWSRFS Models use spatially and
temporally averaged quantities.
• NWSRFS models are parametric(API) and
conceptual (SAC-SMA).
• Spatial & Temporal “lumping” may lead to
model errors.
Distributed Models - Basic Concept(s)
• Break basin into smaller “pieces” to
lessen the effect of averaging.
• Allow for the horizontal and vertical
interaction of water movement
throughout the basin (horizontal is
the greatest challenge).
Distributed Modeling Advantages
• The greatest advantage of distributed
modeling is the ability to overcome the
“loss” of data due to spatial and temporal
averaging.
Distributed Modeling DISAdvantage(s)
• Data requirements (disk space) - NOT
• Computational requirements - STILL?
• Parameterization
• Calibration
Development of a Hydrologic Model
with both
Spatially Distributed and Physically Based
Capabilities -
Could we use this time-area concept?
How About a Distributed UHG?
IHABBS Implementation Plan :
NOHRSC Homepage
http://www.nohrsc.nws.gov/
http://www.nohrsc.nws.gov/98/html/uhg/index.html
Each Grid Cell Contributes
Volume
TIME
Now Follow Time-Area Method
How About a “Dynamic UHG”
Only certain cells
contribute - based on
storm location therefore only route the
contributing cells using
the time-area method.
What are the Challenges?
• How much runoff from each cell?
• Lumped calibration being distributed!
• How about that “reservoiring” function?
• See USACE : HEC-HMS
• Many other models - no offense intended!!