Unit Hydrograph 45
Unit Hydrograph 45
Unit Hydrograph 45
• 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).
1
Unit Hydrograph Theory
2
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
3
Graphical Representation
Duration of
excess
precipitation.
Lag time
Time of
concentration
Base flow
4
Methods of Developing UHG’s
5
Unit Hydrograph
6
0.
00
0
0.0000
100.0000
200.0000
300.0000
400.0000
500.0000
600.0000
700.0000
0. 0
16
00
0.
32
0
0. 0
48
00
0.
64
0
0. 0
80
00
0.
96
0
1. 0
12
0
1. 0
28
00
1.
44
0
1. 0
60
00
1.
76
0
1. 0
92
00
2.
Surface
08
0
Response
2. 0
24
00
2.
40
0
2. 0
56
0
2. 0
72
00
2.
88
0
3. 0
04
00
Baseflow
3.
20
0
3. 0
36
00
3.
52
0
3. 0
68
00
Derived Unit Hydrograph
7
Derived Unit Hydrograph
700.0000
600.0000 Total
Hydrograph
500.0000
Surface
400.0000 Response
300.0000
Baseflow
200.0000
100.0000
0.0000
0.0000 0.5000 1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000
8
Derived 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).
9
Deriving a UHG from a Storm
sample watershed = 450 mi2
25000 0,8
0,7
20000
0,6
Precipitation (inches)
0,5
15000
Flow (cfs)
0,4
10000
0,3
0,2
5000
0,1
0 0
0 8 16 24 32 40 48 56 64 72 80 88 96 10 4 11 2 12 0 12 8
Time (hrs.)
10
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 semi-
log fashion with flow being plotted on the log scale and noting the time
at which the recession side fits a straight line.
11
Semi-log Plot
100000
1000
Flow (cfs)
100
10
1
4
9
4
9
4
9
4
29
34
39
44
49
54
59
64
69
74
79
84
89
94
99
10
10
11
11
12
12
13
Time (hrs.)
12
Hydrograph & Baseflow
25000
20000
15000
Flow (cfs)
10000
5000
0
0
7
14
21
28
35
42
49
56
63
70
77
84
91
98
105
112
119
126
133
Time (hrs.)
13
Separate Baseflow
25000
20000
15000
Flow (cfs)
10000
5000
0
0 7 14 21 28 35 42 49 56 63 70 77 84 91 98 10 5 11 2 11 9 12 6 13 3
Time (hrs.)
14
Sample 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.
15
Obtain UHG Ordinates
16
Final UHG
25000
Storm #1 hydrograph
15000
Flow (cfs)
Storm # 1 unit
hydrograph
10000
Storm #1
baseflow
5000
0
0
7
14
21
28
35
42
49
56
63
70
77
84
91
98
105
112
119
126
133
Time (hrs.)
17
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 1.65 inches of
excess precipitation was determined to be approximately 0.2
inches per hour.
18
Estimating Excess Precip.
0.8
0.7
0.6
Precipitation (inches)
0.5
0.3
0.2
0.1
0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Time (hrs.)
19
Excess Precipitation
0,9
Small amounts of
0,6
excess precipitation at
beginning and end may
0,5
be omitted.
0,4
0,3
0,2
0,1
0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Time (hrs.)
20
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.
21
Develop S-Curve
60000,00
50000,00
40000,00
Flow (cfs)
30000,00
20000,00
10000,00
0,00
0
6
12
18
24
30
36
42
48
54
60
66
72
78
84
90
96
102
108
114
120
Time (hrs.)
22
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 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.
23
Final 1-hour UHG
14000,00 60000,00
12000,00
50000,00
Unit Hydrograph Flow (cfs/inch)
S-curves are
10000,00 lagged by 1 hour
and the difference 40000,00
is found.
Flow (cfs)
8000,00 1-hour unit
hydrograph resulting
30000,00
from lagging S-
6000,00 curves and
multiplying the
difference by 6. 20000,00
4000,00
10000,00
2000,00
0,00 0,00
Time (hrs.)
24
Shortcut Method
•There does exist a shortcut method for changing the duration of the
unit hydrograph if the two durations are multiples of one another.
•For example, if you had a two hour unit hydrograph and you
wanted to change it to a four hour unit hydrograph.
25
Shortcut Method Example
•First, a two hour unit hydrograph is given and a four hour unit
hydrograph is needed.
•There are two possiblities, develop the S - curve or since they are
multiples use the shortcut method.
Tim e (h r) Q
0 0
1 2
2 4
3 6
4 10
5 6
6 4
7 3
8 2
9 1
10 0
26
Shortcut Method Example
Tim e (h r) Q D is p la c e d U H G
0 0
1 2
2 4 0
3 6 2
4 10 4
5 6 6
6 4 10
7 3 6
8 2 4
9 1 3
10 0 2
11 1
12 0
27
Shortcut Method Example
Tim e (h r) Q D is p la c e d U H G S um
0 0 0
1 2 2
2 4 0 4
3 6 2 8
4 10 4 14
5 6 6 12
6 4 10 14
7 3 6 9
8 2 4 6
9 1 3 4
10 0 2 2
11 1 1
12 0 0
28
Shortcut Method Example
31
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.
32
Basic Relationships
t LAG
tduration
5.5
t LAG
tbase 3
8
640 AC p
q peak
t LAG
33
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:
34
SCS
Flow ratios
Cum. Mass
0.8
0.6
Q/Qpeak
0.4
0.2
0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
T/Tpeak
35
Dimensionless Ratios
Time Ratios Discharge Ratios Mass Curve Ratios
(t/tp) (q/qp) (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 .375
1.1 .990 .450
1.2 .930 .522
1.3 .860 .589
1.4 .780 .650
1.5 .680 .700
1.6 .560 .751
1.7 .460 .790
1.8 .390 .822
1.9 .330 .849
2.0 .280 .871
2.2 .207 .908
2.4 .147 .934
2.6 .107 .953
2.8 .077 .967
3.0 .055 .977
3.2 .040 .984
3.4 .029 .989
3.6 .021 .993
3.8 .015 .995
4.0 .011 .997
4.5 .005 .999
5.0 .000 1.000
36
Triangular Representation
D SCS Dimensionless UHG & Triangular Representation
1.2 Excess
Precipitation
Tlag
0.8
Flow ratios
Cum. Mass
Q/Qpeak
Triangular
0.6
Point of
Inflection
Tc
0.4
0.2
0
0.0 Tp 1.0 2.0 3.0 4.0 5.0
Tb
T/Tpeak
37
Triangular Representation
D SCS Dimensionless UHG & Triangular Representation
Tlag
Tr Tb - Tp 1.67 x Tp 0.8
Flow ratios
Cum. Mass
Q/Qpeak
Triangular
0.6
Point of
Inflection
Tc
0.4
qpT p qpT r qp
Q= + = (T p +T r )
2 2 2 0.2
0
0.0 Tp 1.0 2.0 3.0 4.0 5.0
Tb
T/Tpeak
2Q
qp=
T p +T r
654.33 x 2 x A x Q
qp= The 645.33 is the conversion used for
T p +T r
delivering 1-inch of runoff (the area
under the unit hydrograph) from 1-square
484 A Q
qp= mile in 1-hour (3600 seconds).
Tp
38
484 ?
484 A Q
qp=
Tp
39
Duration & Timing?
Again from the triangle
D
T p= +L
2
L = Lag time
L 0.6 * Tc
Tc D 1.7 T p
D
+ 0.6 T c = T p
2
40
Time of Concentration
• Regression Eqs.
• Segmental Approach
41
A Regression Equation
L0.8 (S 1) 0.7
Tlag
1900(% Slope) 0.5
42
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.
43
1
A Basic Approach V kS 2
Flow Type K
Small Tributary - Permanent or intermittent 2.1
streams which appear as solid or dashed
blue lines on USGS topographic maps.
Waterway - Any overland flow route which 1.2 Sorell & Hamilton, 1991
is a well defined swale by elevation
contours, but is not a stream section as
defined above.
Sheet Flow - Any other overland flow path 0.48
which does not conform to the definition of
a waterway.
44
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.
45
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
46
Time-Area
47
Time-Area
100%
Time
Q % Area of conc.
Time Time
48
Time-Area
49
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.
50
Rule of Thumb
51
Basin Breakdown
52
Incremental Area
40
35
25
20
15
10
0
1 2 3 4 5 6 7 8
Time Increment (hrs)
53
Cumulative Time-Area Curve
9
0
0 20 40 60 80 100 120 140 160 180 200
Time (hrs)
54
Trouble Getting a Time-Area
Curve?
55
Instantaneous UHG
IUH i cI i (1 c) IUH ( i 1)
2t
c
2 R t
56
Computations
Time Inc. Inc. Inst. IUHG 2-hr
(hrs) Area Translated UHG Lagged 2 UHG
(mi2) Flow (cfs) hours (cfs)
(1) (2) (3) (4) (5) (6)
0 0 0 0 0
2 14 4,515 1391 0 700
4 44 14,190 5333 1,391 3,360
6 53 17,093 8955 5,333 7,150
8 79 25,478 14043 8,955 11,500
10 0 0 9717 14,043 11,880
12 6724 9,717 8,220
14 4653 6,724 5,690
16 3220 4,653 3,940
18 2228 3,220 2,720
20 1542 2,228 1,890
22 1067 1,542 1,300
24 738 1,067 900
26 510 738 630
28 352 510 430
30 242 352 300
32 168 242 200
34 116 168 140
36 81 116 100
38 55 81 70
40 39 55 50
42 26 39 30
44 19 26 20
46 13 19 20
48 13
57
Incremental Areas
90
80
70
Area Increments (square miles)
60
50
40
30
20
10
0
0 2 4 6 8 10
Time Increments (2 hrs)
58
Incremental Flows
30000
25000
Translated Unit Hydrograph
20000
15000
10000
5000
0
1 2 3 4 5 6
Time Increments (2 hrs)
59
Instantaneous UHG
16000
14000
12000
10000
Flow (cfs/inch)
8000
6000
4000
2000
0
0 10 20 30 40 50 60
Time (hrs)
60
Lag & Average
16000
14000
12000
10000
Flow (cfs/inch)
8000
6000
4000
2000
0
0 10 20 30 40 50 60
Time (hrs)
61
Geomorphologic
62
Strahler Stream Ordering
1
1
1 1
2
2 1
1
2
63
Probability Concepts
64
GIUH Equations
• Channel-based, triangular instantaneous
unit hydrograph:
0.55
1.31 0.43 0.44 L W R B
qp RL V tp R L0.38
LW V RA
• LW in km, V in m/s, qp in hr-1, tp in hrs
65