Rainfall - runoff: Unit Hydrograph

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Rainfall - runoff: Unit Hydrograph
Rainfall - runoff:
Unit Hydrograph
Manuel Gómez Valentín
E.T.S. Ing. Caminos, Canales y
Puertos de Barcelona
Options in many commercial
codes, HMS and others
 HMS Menu
 Transform
method,
User
specified,
SCS, etc
Rainfall - runoff
 Different options
 Unit


hydrograph (most popular)
Lumped (global) model
Hydrological response at the basin
outlet from an effective rainfall, 1 mm,
duration D minutes, uniformly
distributed all over the basin
UH Hipothesis
 Linear response
 Time invariant (rain event)
Q
1 mm effective rainfall
Effective rainfall
Unit hydrograph,
Unit hydrograph,
duration d minutes
Duration t
d
Tb
t
Tiempo
Problems in the application
 Lack of real data to obtain it
 We need to use synthetic UH
The Basic Process
Necessary for a
single basin
Unit
Hydrographs
Excess Precip.
Model
Excess Precip.
Excess Precip.
Basin “Routing”
UHG Methods
Runoff
Hydrograph
Runoff
Hydrograph
Stream and/or
Reservoir
“Routing”
Downstream
Hydrograph
Proposal of the HU

Sherman - 1932
 Horton - 1933
 Wisler & Brater - 1949 - “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
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
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
Time of
concentration
Base flow
How to get the UH
 Field data measurements I(t), Q(t)
 Approach with synthetic unit
hydrographs


SCS (NRCS)
Time - area curve (Clark, 1945)
Unit Hydrograph

Hydrological response at the basin outlet from an
effective 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 (duration
D minutes)
 Different hydrographs for different durations
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
How to use the UH
 Matrix approach
 Prepare matrix P
and U
 Direct operation
I1 0 0
I 2 I1 0
0
0
u1
I 3 I 2 I1 0
u2
0
I 3 I 2 I1
u3
I3 I 2
0 I3
u4
0 0
0 0
Q1
Q2
 
Q3
 
Q4
 
Q5
Q6
 
UH obtention
 We need field measurements




Rainfall
Runoff hydrograph at the basin outlet
We measure the total rainfall (effective
rainfall is “estimated”)
SOURCE OF
UNCERTAINTY
Other problems (errors, spatial distrib.)
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 (1 mm)
of excess precipitation (the area under the hydrograph after
correcting for baseflow).
Deriving a UHG from a
Storm
25000
0.8
0.7
20000
0.5
Flow (cfs)
15000
0.4
10000
0.3
0.2
5000
0.1
Time (hrs.)
12
8
12
0
11
2
96
10
4
88
80
72
64
56
48
40
32
24
8
0
16
0
0
Precipitation (inches)
0.6
Derived Unit Hydrograph
700.0000
Total
Hydrograph
600.0000
500.0000
Surface
Response
400.0000
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
UH obtention
surface and
groundwater
response
 We want just
the surface
response
700.0000
600.0000
500.0000
Surface
Response
400.0000
300.0000
Baseflow
200.0000
100.0000
0.0000
0.
00
0
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.
08
0
2. 0
24
00
2.
40
0
2. 0
56
0
2. 0
72
00
2.
88
0
3. 0
04
00
3.
20
0
3. 0
36
00
3.
52
0
3. 0
68
00
 Measured Q(t),
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.
Semi-log Plot
100000
 Groundwater
response,
exponential
Recession side of hydrograph
becomes linear at approximately hour
64.
10000
10
Time (hrs.)
99
10
4
10
9
11
4
11
9
12
4
12
9
13
4
89
94
79
84
69
74
59
64
49
54
1
39
44
to determine the
separation point
100
29
34
 Use log paper
Flow (cfs)
1000
Hydrograph & Baseflow
25000
20000
Flow (cfs)
15000
10000
5000
Time (hrs.)
133
126
119
112
105
98
91
84
77
70
63
56
49
42
35
28
21
7
14
0
0
Separate Baseflow
25000
20000
Flow (cfs)
15000
10000
5000
Time (hrs.)
13
3
12
6
11
9
11
2
98
10
5
91
84
77
70
63
56
49
42
35
28
21
7
14
0
0
Separation of Baseflow
 If no significant contribution from
groundwaters, use a horizontal straight line

Constant base flow
UH from field data
 Matrix approach
I1
0
0
0
I2
I1
0
0
I3 I 2
I1
0
0
I 3 I 2 I1
u1
u2
u3
0
0
I3 I2
u4
0
0
0
I3
 Q1 
Q 2
 
 Q3
 
Q4
 
Q5 
Q6 
 
PU  Q
UH from field data
 Considering a matrix algebra, we can obtain
the vector U, from vectos Q and matrix P
PU  Q
PU  Q
P T PU  P T Q

U P P
T

1
T
P Q
UH of 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
S hydrograph
60000.00
 Consider a
40000.00
Flow (cfs)
very long,
constant rain
event
50000.00
Addition of
UH, duration D
30000.00
20000.00
response?
120
114
108
96
90
84
78
72
66
60
48
42
36
30
24
18
54
Time (hrs.)
102
 Hydrological
12
0.00
0
mm/h
10000.00
6
 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 obtained
hydrograph (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 from
several 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 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 mm of
runoff for the basin.
Synthetic UHG’s
 SCS
 Clark (Time-area method)
SCS SUH
 SCS proposal
 Simple
SCS Dimensionless UHG Features
1
 Basin of regular
Flow ratios
Cum. Mass
0.8
shapes
 Single peak
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
Triangular SHU
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
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
Tb  2.67 x Tp
Tlag
1
Tr  Tb - Tp  1.67 x Tp
0.8
Flow ratios
Q=
qpT p
2
+
qpT r
2
=
qp
2
2Q
qp=
T p +T r
qp=
654.33 x 2 x A x Q
T p +T r
qp=
484 A Q
Tp
Q/Qpeak
Cum. Mass
Triangular
0.6
(T p +T r )
Point of
Inflection
Tc
0.4
0.2
0
0.0
Tp
1.0
2.0
Tb
3.0
4.0
5.0
T/Tpeak
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 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
For estimation purposes should be around : D  0.133 Tc
To 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 Equation
L0.8 (S  1) 0.7
Tlag 
1900(% Slope) 0.5
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 %
Tc  0.3(
L
J
0.76
)
0.25
Tc, is it always the same?
L0.8 (S  1) 0.7
Tlag 
1900(% Slope) 0.5
Tc  0.3(
L
J
0.25
)
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 point
Tlag  0.6 T  0.35 Tc
'
c
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
Clark - Time-Area
Time-Area
 Synthetic UH, equal form as the time-area curve
 It can show more than one peak
100%
Q
Time
of conc.
% Area
Time
Time
Additional delay
 Presence of sub-surface runoff
 Runoff shows an additional delay, that can
not be explained just for surface runoff
Reservoir model for the
delay
 Conceptual model
 Assume that additional delay is equal to the
produced by a water reservoir
dS
I Q 
dt
I (UH Clark)
K
Q
Storage description
 General approach
S  K1 Q  K 2 Q   K n Q
2
n
 Simplified to a linear reservoir model
SKQ
 K has dimensions of time
K  T 
Mathematical description
 If K is contant in time
dQ
I  QK
dt
 We can solve the diff. Equation as:
t
Q (t ) 

0
I ( )
e
K
 t
K
d
Practical application
 A finite difference scheme can be used
I1  I 2
Q1  Q 2
Q 2  Q1

 K
2
2
t
 From an initial condition Q1, and from the
values of the previous hydrograph we can
proceed as:
2  t
Q2 
2K  t
Q1
Q1 

 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 hours
8
7
7
7
6
6
6
5
2
4
0
3
1
5
Basin geometry
Isocrones
Area #
1
2
3
4
5
6
7
8
TOTAL
8
7
7
7
6
6
6
5
2
4
0
3
1
5
0-1
1-2
2-3
3-4
4-5
5-6
6-7
7-8
Area
(km2)
5
9
23
19
27
26
39
40
190
Accum. Area
(km2)
5
14
37
58
85
111
150
190
190
Accum
Time (hrs)
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
8.0
Time area curve
40
Area (Km2)
Incremental Area (sqaure miles)
35
8
30
25
20
15
10
5
7
7
0
7
6
1
6
6
5
0
3
1
5
3
4
5
Time Increment (hrs)
2
4
2
6
7
8
Time accumulated area
9
Cumulative Area (sqaure miles)
8
Isochrone
4
3
2
0
7
7
20
40
60
80
100
Time (hrs)
6
5
2
4
0
3
1
5
5
0
7
6
6
6
1
Watershed
Boundary
8
7
120
140
160
180
200
No time / area curve ?
TAi  1.414Ti 1.5
1  TAi  1.414(1  Ti )1.5
for (0  Ti  0.5)
for (0.5  Ti  1.0)
U.S. Army Corps of
Engineers (HEC 1990)
 What about if the synthetic curve does not match
the real one?
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

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