Hydrologic Design and Design
Storms
Reading: Applied Hydrology Sections
13-1, 13-2
14-1 to 14-4
Hydrologic design
• Water control
– Peak flows, erosion, pollution, etc.
• Water management
– Domestic and industrial use, irrigation, instream flows, etc
• Tasks
– Determine design inflow
– Route the design inflow
– Find the output
• check if it is sufficient to meet the demands (for management)
• Check if the outflow is at safe level (for control)
2
Hydrologic design scale
• Hydrologic design scale – range in magnitude of the
design variable within which a value must be
selected
• Design considerations
– Safety
– Cost
• Do not design small structures for large peak values
(not cost effective)
• Do not design large structures for small peak values
(unsafe)
• Balance between safety and cost.
3
Estimated Limiting Value (ELV)
• Lower limit on design value – 0
• Upper limit on design value – ELV
• ELV – largest magnitude possible for a hydrologic
event at a given location, based on the best available
hydrologic information.
– Length of record
– Reliability of information
– Accuracy of analysis
• Probable Maximum Precipitation (PMP) / Probable
Maximum Flood (PMF)
4
Probable Maximum Precipitation
Most recent
report 1999
http://www.nws.noaa.gov/oh/hdsc/studies/pmp.html
6
TxDOT Recommendations
Recommended Design Frequencies (years)
Design
-
Functional Classification and Structure Type
Check
Flood
100
2 5 10 25 50
Freeways (main lanes):
- -
-
-
 culverts
- -
-
-
X
X
 bridges
- -
-
-
X
X
Principal arterials:
- -
-
 culverts
- -
X (X) X
X
 small bridges
- -
X (X) X
X
 major river crossings
- -
-
-
(X)
Minor arterials and collectors (including frontage roads):
- -
-
-
-
 culverts
-
 small bridges
- -
 major river crossings
- -
-
Local roads and streets (off-system projects):
- -
-
 culverts
-
-
-
-
-
X
-
X (X) X -
X
X (X) X
X
X (X)
X
-
-
X X
X -
-
X
 small bridges
X X
X -
-
X
Storm drain systems on interstate and controlled access
highways (main lanes):
- -
-
-
 inlets and drain pipe
- -
X -
-
 inlets for depressed roadways*
- -
-
-
Storm drain systems on other highways and frontage:
- -
-
-
-
 inlets and drain pipe
X (X) -
-
-
 inlets for depressed roadways*
- -
(X) X
-
-
-
X
X
X
-
Notes.
* A depressed roadway provides nowhere for water to drain even when the curb height is
exceeded.
( ) Parentheses indicate desirable frequency.
7
X
X
Hydrologic design level
• Hydrologic design level – magnitude of the
hydrologic event to be considered for the
design or a structure or project.
• Three approaches for determining design level
– Empirical/probabilistic
– Risk analysis
– Hydroeconomic analysis
8
Empirical/Probabilitic
• P(most extreme event of last N years will be
exceeded once in next n years) P( N , n)  n
N n
• P(largest flood of last N years will be exceeded in
n=N years) = 0.5
• Drought lasting m years is worst in N year record.
What is the probability that a worse drought will
occur in next n years?
– # sequences of length m in N years = N-m+1
– # sequences of length m in n years = n-m+1
P ( N , n, m ) 
n  m 1
( N  m  1)  (n  m  1)
9
Example 13.2.1
• If the critical drought of the record, as
determined from 40 yrs of data, lasted 5 yrs,
what is the chance that a more severe drought
will occur during the next 20 yrs?
• Solution:
N = 40, m = 5 and n = 20
20  5  1
P(40,5,20) 
 0.308
40  20  2  5  2
10
Risk Analysis
• Uncertainty in hydrology
– Inherent - stochastic nature of hydrologic phenomena
– Model – approximations in equations
– Parameter – estimation of coefficients in equations
• Consideration of Risk
– Structure may fail if event exceeds T–year design
magnitude
n
R  1  1  P( X  xT )
n
P( X  xT ) 
1
T
 1
R  1  1  
 T
– R = P(event occurs at least once in n years)
• Natural inherent risk of failure
11
Example 13.2.2
• Expected life of culvert = 10 yrs
• Acceptable risk of 10 % for the culvert
capacity
• Find the design return period

 1
R  1  1  
 T
n
10
 1
0.10  1  1  
 T
T  95 yrs
What is the chance that the culvert designed for an event of
95 yr return period will have its capacity exceeded at least
once in 50 yrs?
1

R  1  1  
 95 
R  0.41
50
The chance that the capacity will not be exceeded during the next 50 yrs is 10.41 = 0.59
12
Hydroeconomic Analysis
• Probability distribution of hydrologic event
and damage associated with its occurrence
are known
• As the design period increases, capital cost
increases, but the cost associated with
expected damages decreases.
• In hydroeconomic analysis, find return period
that has minimum total (capital + damage)
cost.
13
14
Design Storms
• Get Depth, Duration, Frequency Data for the
required location
• Select a return period
• Convert Depth-Duration data to a design
hyetograph.
Depth Duration Data to Rainfall Hyetograph
http://hdsc.nws.noaa.gov/hdsc/pfds/index.html
An example of precipitation frequency estimates for a location in California
37.4349 N
120.6062 W
Results of Precip Frequency Query
TP 40
• Hershfield (1961) developed isohyetal maps of
design rainfall and published in TP 40.
• TP 40 – U. S. Weather Bureau technical paper no. 40.
Also called precipitation frequency atlas maps or
precipitation atlas of the United States.
– 30mins to 24hr maps for T = 1 to 100
• Web resources for TP 40 and rainfall frequency maps
– http://www.tucson.ars.ag.gov/agwa/rainfall_frequency.ht
ml
– http://www.erh.noaa.gov/er/hq/Tp40s.htm
– http://hdsc.nws.noaa.gov/hdsc/pfds/
20
2yr-60min precipitation GIS map
21
2yr-60min precipitation map
This map is from
HYDRO 35 (another
publication from
NWS) which
supersedes TP 40
22
Design aerial precipitation
• Point precipitation estimates are extended to
develop an average precipitation depth over
an area
• Depth-area-duration analysis
– Prepare isohyetal maps from point precipitation
for different durations
– Determine area contained within each isohyet
– Plot average precipitation depth vs. area for each
duration
23
Depth-area curve
(World Meteorological24Organization, 1983)
Depth (intensity)-duration-frequency
• DDF/IDF – graph of depth (intensity) versus
duration for different frequencies
– TP 40 or HYDRO 35 gives spatial distribution of
rainfall depths for a given duration and frequency
– DDF/IDF curve gives depths for different durations
and frequencies at a particular location
– TP 40 or HYDRO 35 can be used to develop
DDF/IDF curves
• Depth (P) = intensity (i) x duration (Td)
P  iTd
25
IDF curve
26
Example 14.2.1
•
Determine i and P for a 20-min duration storm with 5-yr return period in
Chicago
From the IDF curve for Chicago,
i = 3.5 in/hr for Td = 20 min and T
= 5yr
P = i x Td = 3.5 x 20/60 = 1.17 in
27
Equations for IDF curves
IDF curves can also be expressed as equations to avoid reading
from graphs
c
i e
Td  f
cT m
i e
Td  f
i is precipitation intensity, Td is the duration, and c, e, f are
coefficients that vary for locations and different return
periods
This equation includes return period (T) and has an extra coefficient
(m)
28
Example 14.2.4
Using IDF curve equation, determine 10-yr 20-min design
rainfall intensities for Denver
c
i e
Td  f
From Table 14.2.3 in the text, c = 96.6, e = 0.97, and f = 13.9
96.6
i  0.97
 3.002 in / hr
20  13.9
Similarly, i = 4.158 and 2.357 in/hr for Td = 10 and 30 min,
respectively
29
IDF curves for Austin
i
i  design rainfall intensity
t  Duration of storm
a
t  b c
a, b, c  coefficients
Storm Frequency
a
b
c
16
2-year
106.29
16.81
0.9076
14
5-year
99.75
16.74
0.8327
2-yr
5-yr
10-yr
25-yr
50-yr
100-yr
500-yr
10-year
96.84
15.88
0.7952
25-year
111.07
17.23
0.7815
50-year
119.51
17.32
0.7705
100-year
129.03
17.83
0.7625
Intensity (in/hr)
12
10
8
6
4
2
0
1
500-year
160.57
19.64
0.7449
Source: City of Austin, Watershed Management Division
10
100
1000
Duration (min)
30
Design Precipitation Hyetographs
•
•
Most often hydrologists are interested in
precipitation hyetographs and not just the
peak estimates.
Techniques for developing design
precipitation hyetographs
1. SCS method
2. Triangular hyetograph method
3. Using IDF relationships (Alternating block method)
31
SCS Method
SCS
(1973) adopted method similar to DDF to develop dimensionless rainfall
temporal patterns called type curves for four different regions in the US.
SCS type curves are in the form of percentage mass (cumulative) curves based on
24-hr rainfall of the desired frequency.
If a single precipitation depth of desired frequency is known, the SCS type curve is
rescaled (multiplied by the known number) to get the time distribution.
For durations less than 24 hr, the steepest part of the type curve for required
duraction is used
32
SCS type curves for Texas (II&III)
SCS 24-Hour Rainfall Distributions
T (hrs)
SCS 24-Hour Rainfall Distributions
Fraction of 24-hr rainfall
Type II
T (hrs)
Type III
Fraction of 24-hr rainfall
Type II
Type III
0.0
0.000
0.000
11.5
0.283
0.298
1.0
0.011
0.010
11.8
0.357
0.339
2.0
0.022
0.020
12.0
0.663
0.500
3.0
0.034
0.031
12.5
0.735
0.702
4.0
0.048
0.043
13.0
0.772
0.751
5.0
0.063
0.057
13.5
0.799
0.785
6.0
0.080
0.072
14.0
0.820
0.811
7.0
0.098
0.089
15.0
0.854
0.854
8.0
0.120
0.115
16.0
0.880
0.886
8.5
0.133
0.130
17.0
0.903
0.910
9.0
0.147
0.148
18.0
0.922
0.928
9.5
0.163
0.167
19.0
0.938
0.943
9.8
0.172
0.178
20.0
0.952
0.957
10.0
0.181
0.189
21.0
0.964
0.969
10.5
0.204
0.216
22.0
0.976
0.981
11.0
0.235
0.250
23.0
0.988
0.991
24.0
1.000
1.000
33
SCS Method Steps
•
Given Td and frequency/T, find the design
hyetograph
1. Compute P/i (from DDF/IDF curves or equations)
2. Pick a SCS type curve based on the location
3. If Td = 24 hour, multiply (rescale) the type curve with P to
get the design mass curve
1.
If Td is less than 24 hr, pick the steepest part of the type curve
for rescaling
4. Get the incremental precipitation from the rescaled
mass curve to develop the design hyetograph
34
Example – SCS Method
• Find - rainfall hyetograph for a 25-year, 24-hour duration SCS
Type-III storm in Harris County using a one-hour time
increment
• a = 81, b = 7.7, c = 0.724 (from Tx-DOT hydraulic manual)
i
a
81

 0.417 in / hr
c
t  b  24 * 60  7.7 0.724
P  i *Td  0.417 in / hr * 24 hr  10.01in
• Find
– Cumulative fraction - interpolate SCS table
– Cumulative rainfall = product of cumulative fraction * total 24-hour
rainfall (10.01 in)
– Incremental rainfall = difference between current and preceding
cumulative rainfall
TxDOT hydraulic manual is available at:
http://manuals.dot.state.tx.us/docs/colbridg/forms/hyd.pdf
35
SCS – Example (Cont.)
(hours)
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
Cumulative
Fraction
Cumulative
Precipitation
Incremental
Precipitation
Pt/P24
Pt (in)
(in)
0.000
0.010
0.020
0.032
0.043
0.058
0.072
0.089
0.115
0.148
0.189
0.250
0.500
0.751
0.811
0.849
0.886
0.904
0.922
0.939
0.957
0.968
0.979
0.989
1.000
0.00
0.10
0.20
0.32
0.43
0.58
0.72
0.89
1.15
1.48
1.89
2.50
5.01
7.52
8.12
8.49
8.87
9.05
9.22
9.40
9.58
9.69
9.79
9.90
10.01
0.00
0.10
0.10
0.12
0.12
0.15
0.15
0.17
0.26
0.33
0.41
0.61
2.50
2.51
0.60
0.38
0.38
0.18
0.18
0.18
0.18
0.11
0.11
0.11
0.11
3.00
2.50
Precipitation (in)
Time
2.00
1.50
1.00
0.50
0.00
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
Time (hours)
If a hyetograph for less than 24 needs to be prepared,
pick time intervals that include the steepest part of the
type curve (to capture peak rainfall). For 3-hr pick 11 to
13, 6-hr pick 9 to 14 and so on.
36
Triangular Hyetograph Method
Rainfall intensity, i
ta
Td: hyetograph base length = precipitation duration
tb
ta: time before the peak
r
ta
Td
h
Td
Time
r: storm advancement coefficient = ta/Td
tb: recession time = Td – ta = (1-r)Td
1
P  Td h
2
2P
h
Td
• Given Td and frequency/T, find the design hyetograph
1. Compute P/i (from DDF/IDF curves or equations)
2. Use above equations to get ta, tb, Td and h (r is available for
various locations)
37
Triangular hyetograph - example
• Find - rainfall hyetograph for a 25-year, 6-hour duration in
Harris County. Use storm advancement coefficient of 0.5.
• a = 81, b = 7.7, c = 0.724 (from Tx-DOT hydraulic manual)
a
81

 1.12 in / hr
c
t  b  6 * 60  7.7 0.724
h
2 P 2  6.72 13.44


 2.24 in / hr
Td
6
6
t a  rTd  0.5  6  3 hr
tb  Td  ta  6  3  3 hr
P  i * 6  1.12 in / hr * 6 hr  6.72 in
Rainfall intensity, in/hr
i
38
3 hr
3 hr
2.24
6 hr
Time
Alternating block method
• Given Td and T/frequency, develop a hyetograph in
Dt increments
1. Using T, find i for Dt, 2Dt, 3Dt,…nDt using the IDF curve for
the specified location
2. Using i compute P for Dt, 2Dt, 3Dt,…nDt. This gives
cumulative P.
3. Compute incremental precipitation from cumulative P.
4. Pick the highest incremental precipitation (maximum
block) and place it in the middle of the hyetograph. Pick
the second highest block and place it to the right of the
maximum block, pick the third highest block and place it
to the left of the maximum block, pick the fourth highest
block and place it to the right of the maximum block (after
second block), and so on until the last block.
39
Example: Alternating Block Method
Find: Design precipitation hyetograph for a 2-hour storm (in 10
minute increments) in Denver with a 10-year return period 10minute
Duration
(min)
10
20
30
40
50
60
70
80
90
100
110
120
Td 
e
Intensity
(in/hr)
4.158
3.002
2.357
1.943
1.655
1.443
1.279
1.149
1.044
0.956
0.883
0.820
f

96.6
Td 0.97  13.90
Cumulative
Depth
(in)
0.693
1.001
1.178
1.296
1.379
1.443
1.492
1.533
1.566
1.594
1.618
1.639
Incremental
Depth
(in)
0.693
0.308
0.178
0.117
0.084
0.063
0.050
0.040
0.033
0.028
0.024
0.021
i  design rainfall intensity
Td  Duration of storm
c, e, f  coefficients
0.8
Time
(min)
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
80-90
90-100
100-110
110-120
Precip
(in)
0.024
0.033
0.050
0.084
0.178
0.693
0.308
0.117
0.063
0.040
0.028
0.021
0.7
0.6
Precipitation (in)
i
c
0.5
0.4
0.3
0.2
0.1
0.0
0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90
40
Time (min)
90100
100110
110120