Excess Rainfall
description
Transcript of Excess Rainfall
Excess Rainfall Reading for today’s material:
Sections 5.3-5.7
Slides prepared by V.M. Merwade
Quote for today (contributed by Tyler Jantzen)
"How many times it thundered before Franklin took the hint! Nature is always hinting at us. It hints over and over again. And suddenly we take the hint.“
Robert Frost
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
-index
-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
-index method
M
mmd tRr
1
• Goal: pick t, and adjust value of M to satisfy the equation
• Steps1. Estimate baseflow2. DRH = streamflow
hydrograph – baseflow3. Compute rd, rd =
Vd/watershed area4. Adjust M until you get a
satisfactory value of 5. ERH = Rm - t
interval timerunoffdriecttongcontributi
rainfallofintervals#indexPhi
rainfall observedrunoffdirect ofdepth
t
M
Rr
m
d
ExampleTime Observed
Rain Flow
in cfs
8:30 203
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
1:30 2246
2:00 1802
2:30 1230
3:00 713
3:30 394
4:00 354
4:30 303
0
2000
4000
6000
8000
10000
12000
7:30 PM 9:00 PM 10:30 PM 12:00 AM 1:30 AM 3:00 AM 4:30 AM 6:00 AM
Time
Stre
amflo
w (c
fs)
0
0.5
1
1.5
2
2.5
No direct runoff until after 9:30And little precip after 11:00
Have precipitation and streamflow data, need to estimate losses
Basin area A = 7.03 mi2
Example (Cont.)
• Estimate baseflow (straight line method)– Constant = 400 cfs
0
2000
4000
6000
8000
10000
12000
7:30 PM 9:00 PM 10:30 PM 12:00 AM 1:30 AM 3:00 AM 4:30 AM 6:00 AM
Time
Stre
amflo
w (c
fs)
baseflow
Example (Cont.)
• Calculate Direct Runoff Hydrograph– Subtract 400 cfs
Total = 43,550 cfs
Example (Cont.)
• Compute volume of direct runoff
37
3
11
1
11
1
ft10*7.839
/sft 550,43*hr5.0*s/hr3600
n
nn
nd QttQV
• Compute depth of direct runoff
in80.4ft4.0
ft5280*mi03.7ft10*7.839
22
37
AV
r dd
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
)5.0*3*08.220.233.1(80.41
M
mmd tRr
in27.0t
in54.0
in80.4dr
Example (Cont.)
0
2000
4000
6000
8000
10000
12000
7:30 PM 9:00 PM 10:30 PM 12:00 AM 1:30 AM 3:00 AM 4:30 AM 6:00 AM
Time
Stre
amflo
w (c
fs)
0
0.5
1
1.5
2
2.5
t=0.27
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
PPee is runoff volume, P is is runoff volume, P is precipitation volume, Fprecipitation volume, Faa is continuing is continuing abstraction, and Iabstraction, and Iaa is the is the sum of initial losses sum of initial losses (depression storage, (depression storage, interception, ET)interception, ET)
Time
Prec
ipit
atio
n
pt
aI aF
eP
aae FIPP
Abstractions – SCS Method• In general
• After runoff begins
• Potential runoff
• SCS Assumption
• Combining SCS assumption with P=Pe+Ia+Fa
Time
Prec
ipit
atio
n
pt
aI aF
eP
aae FIPP
StorageMaximumPotentialSnAbstractioContinuing
nAbstractioInitialExcess Rainfall
Rainfall Total
a
a
e
FIPP
PPe
SFa
aIP
a
eaIPP
SF
SIP
IPP
a
ae
2
SCS Method (Cont.)
• Experiments showed
• So
SIa 2.0
SPSPPe 8.0
2.0 2
0
1
2
3
4
5
6
7
8
9
10
11
12
0 1 2 3 4 5 6 7 8 9 10 11 12Cumulative Rainfall, P, in
Cum
ulat
ive
Dir
ect R
unof
f, Pe
, in
10090807060402010
• Surface– Impervious: CN = 100– Natural: CN < 100
100)CN0Units;American(
101000
CN
S
100)CN30Units;SI(
25425400
CNCN
S
SCS Method (Cont.)
• S and CN depend on antecedent rainfall conditions
• Normal conditions, AMC(II)• Dry conditions, AMC(I)
• Wet conditions, AMC(III)
)(058.010)(2.4)(IICN
IICNICN
)(13.010)(23)(IICN
IICNIIICN
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 C
Land use % 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 43.40
8.8340.4338.40 CNCN values come from Table 5.5.2
Example (SCS Method – 1 Cont.)
• Average AMC
• Wet AMC3.92
8.83*13.0108.83*23
)(13.010)(23)(
IICNIICNIIICN
in25.393.1*8.05
93.1*2.058.0
2.0 22
SPSPPe
in93.1108.83
1000 S
8.83CN
in13.483.0*8.05
83.0*2.058.0
2.0 22
SPSPPe
in83.0103.92
1000S
101000 CN
S
Example (SCS Method – 2)• Given P, CN = 80, AMC(II)• Find: Cumulative abstractions and excess rainfall hyetograph
Time (hr)
Cumulative
Rainfall (in)
Cumulative Abstractions (in)
CumulativeExcess Rainfall
(in)
Excess RainfallHyetograph (in)
P Ia Fa Pe
0 0
1 0.2
2 0.9
3 1.27
4 2.31
5 4.65
6 5.29
7 5.36
Example (SCS Method – 2)
• Calculate storage• Calculate initial abstraction• Initial abstraction removes
– 0.2 in. in 1st period (all the precip)– 0.3 in. in the 2nd period (only part
of the precip)• Calculate continuing abstraction
in50.21080
1000101000
CNS
a
ea IP
PSF
in5.05.2*2.02.0 SIa
aae FIPP )0.2(
)5.0(5.2)(
)(
PP
SIPIPSF
a
aa
in34.0)0.29.0(
)5.09.0(5.2hr)(2 aF
Time (hr)
CumulativeRainfall (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
Example (SCS method – 2)• Cumulative abstractions can now be calculated
Time (hr)
Cumulative
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
)0.2()5.0(5.2
PPFa
Example (SCS method – 2)• Cumulative excess rainfall can now be calculated• Excess Rainfall Hyetograph can be calculated
Time (hr)
CumulativeRainfall
(in)
Cumulative Abstractions (in)
CumulativeExcess Rainfall (in)
Excess RainfallHyetograph (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
aae FIPP
Example (SCS method – 2)• Cumulative excess rainfall can now be calculated• Excess Rainfall Hyetograph can be calculated
Time (hr)
CumulativeRainfall
(in)
Cumulative Abstractions (in)
CumulativeExcess Rainfall (in)
Excess RainfallHyetograph (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
aae FIPP
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