Excess Rainfall
Reading for today’s material:
Sections 5.3-5.7
Slides prepared by V.M. Merwade
Excess rainfall
• Rainfall that is neither retained on the land
surface nor infiltrated into the soil
• Graph of excess rainfall versus time is
called excess rainfall hyetograph
• Direct runoff = observed streamflow baseflow
• Excess rainfall = observed rainfall abstractions
• Abstractions/losses – difference between
total rainfall hyetograph and excess rainfall
hyetograph
f-index
 f-index: Constant rate of abstraction
yielding excess rainfall hyetograph with
depth equal to depth of direct runoff
• Used to compute excess rainfall
hyetograph when observed rainfall and
streamflow data are available
f-index method
•
•
Goal: pick t, and
adjust value of M to
satisfy the equation
Steps
1. Estimate baseflow
2. DRH = streamflow
hydrograph – baseflow
3. Compute rd, rd =
Vd/watershed area
4. Adjust M until you get a
satisfactory value of f
5. ERH = Rm - ft
M
rd   Rm  ft 
m 1
rd  depth of direct runoff
Rm  observed rainfall
f  Phi index
M  # intervals of rainfall
contributing to driect runoff
t  time interval
Example
Have precipitation and streamflow data, need to estimate losses
Observed
Rain
Flow
in
cfs
8:30
12000
203
9:00
0.15
246
9:30
0.26
283
10:00
1.33
828
10:30
2.2
2323
11:00
0.2
5697
11:30
0.09
9531
12:00
11025
12:30
8234
1:00
4321
0
0.5
10000
1
1.5
2
Streamflow (cfs)
Time
8000
2.5
6000
4000
2000
1:30
2246
2:00
1802
2:30
1230
3:00
713
3:30
394
4:00
354
4:30
303
0
7:30 PM
9:00 PM
10:30 PM
12:00 AM
1:30 AM
3:00 AM
Time
No direct runoff until after 9:30
And little precip after 11:00
Basin area A = 7.03 mi2
4:30 AM
6:00 AM
Example (Cont.)
• Estimate baseflow (straight line method)
– Constant = 400 cfs
12000
Streamflow (cfs)
10000
8000
6000
4000
2000
0
7:30 PM
9:00 PM
10:30 PM
12:00 AM
Time
1:30 AM
3:00 AM
4:30 AM
baseflow
6:00 AM
Example (Cont.)
• Calculate Direct
Runoff Hydrograph
– Subtract 400 cfs
Time
8:30
9:00
9:30
10:00
10:30
11:00
11:30
12:00
12:30
1:00
1:30
2:00
2:30
3:00
3:30
4:00
4:30
Observed
Rain
Flow
in
cfs
0.15
203
0.26
246
1.33
283
2.2
828
2.08
2323
0.2
5697
0.09
9531
11025
8234
4321
2246
1802
1230
713
394
354
303
Direct
Runoff
cfs
428
1923
5297
9131
10625
7834
3921
1846
1402
830
313
43550
Total = 43,550 cfs
Example (Cont.)
• Compute volume of direct runoff
11
11
n 1
n 1
Vd   Qn t  t  Qn
 3600s/hr * 0.5 hr * 43,550 ft 3 /s
 7.839*107 ft 3
• Compute depth of direct runoff
V
rd  d
A

7.839*107 ft 3
7.03 mi * 52802 ft 2
 0.4 ft
 4.80 in
Example (Cont.)
• Neglect all precipitation intervals that occur
before the onset of direct runoff (before 9:30)
• Select Rm as the precipitation values in the 1.5
hour period from 10:00 – 11:30
M
rd   Rm  ft 
m 1
4.80  (1.33  2.20  2.08  f * 3 * 0.5)
f  0.54 in
ft  0.27 in
rd  4.80 in
Example (Cont.)
8:30
9:00
9:30
10:00
10:30
11:00
11:30
12:00
12:30
1:00
1:30
2:00
2:30
3:00
3:30
4:00
4:30
Observed
Rain
Flow
in
cfs
0.15
203
0.26
246
1.33
283
2.2
828
2.08
2323
0.2
5697
0.09
9531
11025
8234
4321
2246
1802
1230
713
394
354
303
Direct
Runoff
Excess
Rainfall
cfs
in
12000
0
ft=0.27
0.5
428
1923
5297
9131
10625
7834
3921
1846
1402
830
313
43550
10000
1.06
1.93
1.81
1
1.5
2
Streamflow (cfs)
Time
8000
2.5
6000
4000
2000
0
7:30 PM
9:00 PM
10:30 PM
12:00 AM
Time
1:30 AM
3:00 AM
4:30 AM
6:00 AM
SCS method
• Soil conservation service (SCS) method is an
experimentally derived method to determine
rainfall excess using information about soils,
vegetative cover, hydrologic condition and
antecedent moisture conditions
• The method is based on the simple relationship
that Pe = P - Fa – Ia
Precipitation
Pe is runoff volume, P is
precipitation volume, Fa is
continuing abstraction, and Ia is
the sum of initial losses
(depression storage,
interception, ET)
P  Pe  I a  Fa
Pe
Ia
Fa
tp
Time
Abstractions – SCS Method
• In general
Pe  P
• After runoff begins
• Potential runoff
Precipitation
Fa  S
P  Pe  I a  Fa
Pe
P  Ia
• SCS Assumption
Fa
Pe

S
P  Ia
• Combining SCS
assumption with
P=Pe+Ia+Fa
P  I a 2
Pe 
P  Ia  S
Ia
Fa
tp
Time
P  Total Rainfall
Pe  Rainfall Excess
I a  Initial Abstraction
Fa  Continuing Abstraction
S  Potential Maximum Storage
SCS Method (Cont.)
•
• Experiments showed
Surface
– Impervious: CN =
100
– Natural: CN < 100
I a  0.2S
• So
P  0.2S 2
P  0.8S
1000
S
 10
CN
(American Units; 0  CN  100)
25400
 254CN
CN
(SI Units; 30  CN  100)
S
100
90
80
70
11
Cumulative Direct Runoff, Pe, in
Pe 
12
10
9
60
40
20
10
8
7
6
5
4
3
2
1
0
0
1
2
3
4
5
6
7
Cumulative Rainfall, P, in
8
9
10
11
12
SCS Method (Cont.)
• S and CN depend on antecedent rainfall
conditions
• Normal conditions, AMC(II)
4.2CN ( II )
CN
(
I
)

• Dry conditions, AMC(I)
10  0.058CN ( II )
• Wet conditions, AMC(III)
CN ( III ) 
23CN ( II )
10  0.13CN ( II )
SCS Method (Cont.)
• SCS Curve Numbers depend on soil conditions
Group
Minimum Infiltration
Rate (in/hr)
Soil type
A
0.3 – 0.45
High infiltration rates. Deep, well
drained sands and gravels
B
0.15 – 0.30
Moderate infiltration rates. Moderately
deep, moderately well drained soils
with moderately coarse textures (silt,
silt loam)
C
0.05 – 0.15
Slow infiltration rates. Soils with layers,
or soils with moderately fine textures
(clay loams)
D
0.00 – 0.05
Very slow infiltration rates. Clayey
soils, high water table, or shallow
impervious layer
Example - SCS Method - 1
• Rainfall: 5 in.
• Area: 1000-ac
• Soils:
– Class B: 50%
– Class C: 50%
• Antecedent moisture: AMC(II)
• Land use
– Residential
• 40% with 30% impervious cover
• 12% with 65% impervious cover
– Paved roads: 18% with curbs and storm sewers
– Open land: 16%
• 50% fair grass cover
• 50% good grass cover
– Parking lots, etc.: 14%
Example (SCS Method – 1,
Cont.)
Hydrologic Soil Group
B
Land use
C
%
CN
Product
%
CN
Product
Residential (30% imp
cover)
20
72
14.40
20
81
16.20
Residential (65% imp
cover)
6
85
5.10
6
90
5.40
Roads
9
98
8.82
9
98
8.82
Open land: good cover
4
61
2.44
4
74
2.96
Open land: Fair cover
4
69
2.76
4
79
3.16
Parking lots, etc
7
98
6.86
7
98
6.86
Total
50
40.38
50
CN values come from Table 5.5.2
CN  40.38  43.40  83.8
43.40
Example (SCS Method – 1
Cont.)
• Average AMC
S
CN  83.8
S
1000
 10
CN
1000
 10  1.93 in
83.8
Pe 
P  0.2S 2 5  0.2 *1.932
P  0.8S

5  0.8 *1.93
 3.25 in
• Wet AMC
CN ( III ) 
S
23CN ( II )
23 * 83.8

 92.3
10  0.13CN ( II ) 10  0.13 * 83.8
1000
 10  0.83 in
92.3
Pe 
P  0.2S 2 5  0.2 * 0.832
P  0.8S

5  0.8 * 0.83
 4.13 in
Example (SCS Method – 2)
• Given P, CN = 80, AMC(II)
• Find: Cumulative abstractions and excess rainfall hyetograph
Time
(hr)
Cumulativ
e
Rainfall
(in)
P
0
0
1
0.2
2
0.9
3
1.27
4
2.31
5
4.65
6
5.29
7
5.36
Cumulative
Abstractions (in)
Ia
Fa
Cumulative
Excess Rainfall
(in)
Pe
Excess Rainfall
Hyetograph (in)
Example (SCS Method – 2)
• Calculate storage
• Calculate initial abstraction
• Initial abstraction removes
1000
1000
 10 
 10  2.50 in
CN
80
I a  0.2S  0.2 * 2.5  0.5 in
S
– 0.2 in. in 1st period (all the precip)
– 0.3 in. in the 2nd period (only part
of the precip)
• Calculate continuing abstraction
Pe
Fa  S
P  Ia
P  Pe  I a  Fa
Time
(hr)
Cumulative
Rainfall (in)
P
0
0
1
0.2
2
0.9
S (P  I a )
2.5( P  0.5)
Fa 

(P  I a  S )
( P  2.0)
3
1.27
4
2.31
5
4.65
2.5(0.9  0.5)
Fa (2 hr) 
 0.34 in
(0.9  2.0)
6
5.29
7
5.36
Example (SCS method – 2)
• Cumulative abstractions can now be calculated
Time
(hr)
Cumulati
ve
Rainfall
(in)
Cumulative
Abstractions (in)
P
Ia
Fa
0
0
0
-
1
0.2
0.2
-
2
0.9
0.5
0.34
3
1.27
0.5
0.59
4
2.31
0.5
1.05
5
4.65
0.5
1.56
6
5.29
0.5
1.64
7
5.36
0.5
1.65
2.5( P  0.5)
Fa 
( P  2 .0 )
Example (SCS method – 2)
• Cumulative excess rainfall can now be calculated Pe  P  I a  Fa
• Excess Rainfall Hyetograph can be calculated
Time
(hr)
Cumulative
Rainfall
(in)
Cumulative
Abstractions (in)
Cumulative
Excess Rainfall (in)
Excess Rainfall
Hyetograph (in)
P
Ia
Fa
Pe
0
0
0
-
0
0
1
0.2
0.2
-
0
0
2
0.9
0.5
0.34
0.06
0.06
3
1.27
0.5
0.59
0.18
0.12
4
2.31
0.5
1.05
0.76
0.58
5
4.65
0.5
1.56
2.59
1.83
6
5.29
0.5
1.64
3.15
0.56
7
5.36
0.5
1.65
3.21
0.06
Example (SCS method – 2)
• Cumulative excess rainfall can now be calculated Pe  P  I a  Fa
• Excess Rainfall Hyetograph can be calculated
R a i nf a l l ( i n)
Time
(hr)
Cumulative
Rainfall
(in)
P
0
0
1
0.2
2
0.9
3
1.27
Cumulative
2. 5
Abstractions (in)
2
1. 5
1
R a i nf a l l H y e t ogr a phs
Cumulative
Excess Rainfall (in)
Excess Rainfall
Hyetograph (in)
Ia
Fa
Pe
0
-
0
0
0.2
-
0
0
0.5
0.34
0.06
0.06
0.5
0.59
0.18
0.12
0.5
1.05
0.76
0.58
1.56
2.59
1.83
3.15
0.56
0. 5
4
2.31
5
4.65
6
7
5.29
5.36
0
0.5
0.5
0.5
0
1
1.64 2
1.65T i m e
3
( hour )
4
3.21
5
6
0.067
R a i nf a l l
E x c e s s R a i nf a l l
Time of Concentration
• Different areas of a
watershed contribute to
runoff at different times after
precipitation begins
• Time of concentration
– Time at which all parts of
the watershed begin
contributing to the runoff
from the basin
– Time of flow from the
farthest point in the
watershed
Isochrones: boundaries of
contributing areas with equal time of
flow to the watershed outlet
Stream ordering
• Quantitative way of studying
streams. Developed by Horton
and then modified by Strahler.
• Each headwater stream is
designated as first order stream
• When two first order stream
combine, they produce second
order stream
• Only when two streams of the
same order combine, the stream
order increases by one
• When a lower order stream
combines with a higher order
stream, the higher order is
retained in the combined stream