Hydrology
“Course notes”
Dr. Amr A. El-Sayed
1- Basic Hydrologic cycle
The main processes in the hydrologic cycle are:
123456-
Precipitation (rainfall)
Runoff (surface)
Transpiration (from plants)
Evaporation
Infiltration
Groundwater flow
P
R
T
E
F
G
1-1 Water budget equation
For any hydrologic system, a water budget can be developed to account for various flow
pathways and storage components. The simplest system is an impervious inclined plan, confined
on all four sides with a single outlet. The hydrologic continuity equation for any system is:
Qin − Qout =
dS
dt
………. (1-1)
Where:
Qin
Qout
inflow rate (volume / time)
outflow rate (volume / time)
dS
dt
Change in storage in (volume / time)
You can think of inflow in the form of (Precipitation),
And the outflow (Runoff, Groundwater flow, Evaporation, and Transpiration)
P–R–G–E–T=∆S
………. (1-2)
Note:
When dealing with the surface hydrology, infiltration “I” is considered as “loss”, but when
dealing with subsurface hydrology, infiltration “I” is considered as a gain to the ground water
surface, which is called “recharge”.
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Prepared by Amr A. El Sayed, aelsayed@vt.edu
1-3 Watershed boundaries
Watershed
A watershed is defined as an area of land that drains to a single outlet and is separated
from other watersheds by a watershed divide.
Base Flow
The channel or the stream may contain a certain amount of “Base Flow” coming from
groundwater and soil contribution even if there is no rainfall, and that amount of water appeared
in the gage devices when estimating the runoff into channels or streams. Discharge from rainfall
excess, after losses have been subtracted, makes up the direct runoff hydrograph. The total runoff
hydrograph is direct runoff + Base flow.
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Prepared by Amr A. El Sayed, aelsayed@vt.edu
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Prepared by Amr A. El Sayed, aelsayed@vt.edu
Water balance unit conversion:
Each process in the hydrologic cycle may be presented in the same flow rate units
(Volume / time), and when this volume is spread over a certain area (like watershed), one can use
depth units. The depth units represent a volume of water when multiplied by the surface area of
the watershed.
Length units:
1.0 inch = 2.54 cm
1 mile = 1.6093 km
Area units:
1 acre = 4047 square meters
1 hectare (ha) = 2.471 acre
1- Volume:
1 ft = 12 inches
1 ft = 30.48 cm
1 yard = 3 feet
1 mile = 5280 ft
1 acre = 43560 ft2
1 ha = 10000 m2
1 mile2 = 640 acre
1 cfs-hr = 3600 ft3
1 cfs.day = 1 * 24 * 60 * 60 = 86400 ft3
2- Depth:
The volume is expressed as a depth of water over a certain area, thus:
Volume = Depth of water * Area.
1 cfs.hr = 3600 ft3, this volume is spread over an area of 1 acre.
1 acre
12inches
= 0.9917 in ≅ 1.0 inch over 1.0 acre
depth = 3600 ft 3 *
*
2
ft
43560 ft
1 cfs.hr = 1.0 acre-inch
1 cfs.day = 86400 ft3 (over an area of 1.0 acre = 43560 ft2)
depth =
86400
= 1.9834 ft ≅ 2.0 ft over 1.0 acre
43560
1 cfs.day = 2.0 acre-ft
4
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Example:
For a given month, a 300 acre lake has 15.0 cfs of inflow, 13 cfs of outflow, and a total
storage increase of 16 acre-ft. A USGS “United States Geological Service” gage next to the
lake recorded a total of 1.3 in. precipitation for the lake for the month. Assuming that
infiltration is insignificant for the lake, determine the evaporation loss in inches over the lake.
Solution:
Putting all the income and outcome flow into depth units (inches):
1- inflow = 15 cubic feet / sec. (cfs)
a- Converting the flow rate into volume = flow rate * time (one month).
= 15.0 “ft3/s * 30 “days” * 24 “hours” * 3600 “seconds” = 38880000.0 ft3.
b- Expressing the volume as a depth over unit area (acre-ft):
38880000 ft 3 *
acre
43560 ft 2
= 892.562 acre-ft.
c- Converting the volume into depth over the watershed area:
= 38880000 ft 3 *
acre
1
= 2.975 ft = 2.975 * 12 = 35.7024 in.
*
300 acre 43560 ft 2
2- outflow = 13 cfs.
a- Converting flow rate into volume:
= 13 * 30 * 24 * 3600 = 33696000 ft3.
b- Expressing the volume as a depth over unit area (acre-ft):
33696000 ft 3 *
acre
43560 ft 2
= 773.554 acre-ft.
c- converting the volume to depth over the watershed area:
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Prepared by Amr A. El Sayed, aelsayed@vt.edu
=
33696000
= 2.573 ft = 2.573 * 12 = 30.9421 inches.
300 * 43650
3- Precipitation “P” = 1.30 in
4- change in storage = 16.0 acre-ft
= 16 acre. ft *
12 in
1
= 0.64 inches.
*
ft 300 acre
The water budget equation:
Inflow – Outflow = Change in Storage.
Inflow
I
Precipitation
P
+
-
Evaporation
E
-
outflow
O
=
Change in storage
∆S
E = I − O + P − ∆S = 35.7024 - 30.9421 + 1.30 – 0.64
E = 5.4203 inches
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Prepared by Amr A. El Sayed, aelsayed@vt.edu
Example (2)
In a given year, a watershed with an area of 2500 km2 received 130 cm of precipitation.
The average rate of flow measured in a river draining the watershed was 30 m3/s. Estimate the
amount of water lost due to the combined effects of evaporation, transpiration, and infiltration
to groundwater. How much runoff reached the river for the year (in cm)? What is the runoff
coefficient?
1- converting the output rate of flow into water depth (cm)
m 3 365 * 24 * 60 * 60s
a- Volume of outflow = 30
*
*1 year = 94608*104 m3.
s
year
b- depth of the flow volume over the area:
= 94608 *10 4 m 3 *
1km 2
(1000)2 m 2
*
1
2500km 2
= 0.3784 m
= 37.8432 cm
In flow – out flow = change in storage
The losses are:
Surface runoff “R” + Groundwater flow “G” + Evaporation “E” + Transpiration “T”
P–R–G–E–T=∆S
Assume that the water levels are the same at (time = 0.0, and at time = 1.0 year)
∆ S = 0.0
The water reached the river as an out flow is from the surface runoff. Drained water to the
river = 37.8432 cm.
P – R = G + ET,
where “ET” is evaporation + transpiration.
130 – 37.8432 = G + ET “LOSSES”
G + ET = 92.1568 cm
The runoff coefficient, defined as runoff divided by Precipitation:
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Prepared by Amr A. El Sayed, aelsayed@vt.edu
The Runoff Coefficient =
=
Runoff
Precipitation
37.8432
= 0.2911
130
2. Principal Water Resources Agencies and Data Sources
2-1 Water Resources Data, Virginia
Water-resources data for the 2005 water year for Virginia includes records of stage, discharge,
and water quality of streams and stage, contents, and water quality of lakes and reservoirs. This
volume contains records for water discharge at 172 gaging stations; stage only at 2 gaging
stations; elevation at 2 reservoirs and 2 tide gages; contents at 1 reservoir, and water quality at 25
gaging stations. Also included are data for 50 crest-stage partial-record stations. Locations of
these sites are shown on figures 4A-B and 5A-B. Miscellaneous hydrologic data were collected at
128 measuring sites and 19 water-quality sampling sites not involved in the systematic datacollection program. The data in this report represent that part of the National Water Data System
collected by the U.S. Geological Survey and cooperating State and Federal agencies in Virginia.
http://pubs.usgs.gov/wdr/2005/wdr-va-05-1/
3. Precipitation Mechanisms and Meteorology Definitions
Precipitation is the primary mechanism for transporting water from the atmosphere to the surface
of the earth. There are several forms of precipitation, the most common of which for the United
States is rain.
3-1 Rainfall Mechanism
3-2 Saturation vapor pressure
Is the partial pressure of water vapor when the air is completely saturated ( no further
evaporation occurs) and is a function of temperature.
3-3 Relative humidity “H”
is approximately the ratio of water vapor pressure to that which would prevail under
saturated conditions at the same temperature.
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H=
e
*100
es
Where:
e
es
vapor pressure in (mb)
Saturated vapor pressure in (mb),
Thus, a 50% relative humidity means that the atmosphere contains 50% of the maximum
moisture it could hold under saturated conditions at that temperature.
3-4 Dew point temperature Td
Is the value at which an air mass just becomes saturated (e = es), when cooled at constant
pressure and moisture content.
An approximation relationship for saturated vapor pressure over water es as a function of
dew point Td is:
⎛
4278.6 ⎞
⎟⎟
es = 2.7489 *108 * exp⎜⎜ −
+
242
.
79
T
⎝ d
⎠
Where, es is in (mb), and Td is in oC.
3-5 Example:
Use Penman equation (combined with Meyers equation) to estimate monthly evaporation from a
lake surface in July, given the following data:
Temperature of water at the lake surface “Tw”
Relative humidity
Temperature of air at 25 ft above water surface
Wind velocity “W” at 25 ft above water surface
=
=
=
=
70o F
60%
80o F
15 mph
Solution:
Tair = 80o F = 26.70o C @ 25 ft above the lake surface.
Eh =
∆
γ
.Q N +
.E A
∆ +γ
∆ +γ
Eh
Flux of latent heat due to evaporation =
Where:
energy
− time
area
E h = q.Le .E , with E in units of L/T (Length / Time).
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Latent heat of vaporization in cal/g:
Le = 597.30 − 0.57 * T oC
Le
∆
Slope of “es” (saturated vapor pressure) vs. T air (temperature) curve.
∆ in units of
mb
oC
γ
Psychometric constant from equation:
P
Atmospheric pressure (mb).
QN
net radiation absorbed by the water body.
γ = 0.66 *
P
1000
⎛
des 2.7489 *108 * 4278.6
4278.6 ⎞
⎟⎟
∆=
=
. exp⎜⎜ −
dT
(T + 242.79)2
⎝ Tair + 242.79 ⎠
air
Where:
Tair in that equation is in Co. 80 Fo = 26.7 Co.
∆ = 2.058 mb/ Co.
P
γ = 0.66 *
, assume P (atmospheric) = 1000 mb,
1000
1000
then γ = 0.66 *
= 0.66
1000
According to Meyers
E a = 0.0106(1 + 0.1 *W )(es − ea ) * ρ .Le
Le = 597.30 − 0.57 * 26.7 oC = 582.1 cal/g
ρ = 1.0 gm/cm3
At the level of (25.0 ft), the relative humidity was 60%, and the temperature was 80 F0
The saturation pressure “es” at 80 F0 “26.7 oC” = 35.0 m bar ???
0.60 = ea / 35.0
Actual vapor pressure ( ea ) = 0.60 * 35 = 21.0 m bar
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Prepared by Amr A. El Sayed, aelsayed@vt.edu
E a = 0.0106(1 + 0.1 *15)(25 − 21)mb *1.0
⎛ cal ⎞
⎟⎟
* 582.1⎜⎜
⎝ g ⎠
cm3
g
Ea = 0.27
Q N = Qθ − Qv + Qh + Qe
4. Rainfall Intensity-Duration-Frequency Curves
http://hdsc.nws.noaa.gov/hdsc/pfds/
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4-1 Design storm Hyetograph
4-2 Example
5. Rainfall Abstraction by Phi-index and SCS method
5-1 Rainfall Runoff analysis:
5-1-1 Precipitation:
1- Gross precipitation - ∑ Losses = Pe
Pe
excess rain,
net rain
effective precipitation
direct runoff
∑ Losses
= Hydrologic Abstractions
pond ing / evaporation
surface detention storage
infiltration
Methods:
1- Pabt Horton
2- Ф index
3- USDA soil conservation surface
4- NRCA soil cover complex method.
5-2 Pabt Horton Method.
Ft = Fc + (Fo − Fc ).e − kt
where:
Fo
Fc
k
slope constant
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5-2-1 Example
Problem 1-14 “text”
Data: 5 hour storm, 5 in rainfall
Time
hours
Rain rate
in/hr
0
0.0
1
Ft
Ft = Fc + (Fo − Fc ).e
1.20
− kt
average
Ft
Pe
0.66
0.54
0.35
1.75
0.25
0.65
0.22
0.18
0.21
0.18
0.90
0.43
2
2.10
0.28
3
0.90
0.23
4
0.40
0.21
5
0.40
Σ = 5.0
0.20
apply Horton equation:
k = 1.10 1/hr
Fc = 0.20 in/hr
Fo = 0.90 in/hr
Ft = Fc + (Fo − Fc ).e − kt
Ft = 0.20 + (0.90 − 0.20).e −1.10t
5-3 Phi-index
5-3-1 Example
5-4
Development of SCS Rainfall-Runoff Equation and Runoff Curve No.
(Soil-cover-complex method)
5-4-1
5-4-2 Drive SCS Rainfall-Runoff equation
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6. Unit Hydrograph Derivation and Application
Hydrograph of excess runoff:
we can’t measure “Q” directly
This step depends on hydraulic control, and geometry. The control establish a critical
depth condition.
Types of stream gauges:
1- Non-recording gages: ex (Staff gauges)
2- Recording gauges:
* Stilling well
* Gas bulbar (air or nitrogen gas)
* Electromagnetic devices.
* A costic Doppler devices.
Site characteristics for good gauging location:
1- section control : overall fall weir, bridge pad, or constriction, or anything forces critical depth
to occur.
2- Channel control:
Q vs. H curve controlled by geometry, roughness, slope, as according to Manning’s
equation:
1
Q = .R 2 3 .S 1 2 . A
n
3- Avoid backwater:
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7.30
9
10
11 12
0.0
8.60
8
1.30
7.50
7
2.00
6
3.20
5
4.40
4
5.40
3
4.10
2
1.20
Depth
1
0.0
∆w
Method of measuring discharge:
Measurement
Station
Distance
across
stream (ft)
Width
(ft)
Measured
Depth
(ft)
Mean
velocity
ft/s
Area
(ft2)
Discharge
ft3/s
1
2
3
4
5
6
7
8
9
10
11
12
0
2
4
6
8
10
12
14
16
18
20
22
1
2
2
2
2
2
2
2
2
2
2
1
0.0
1.2
4.1
7.5
8.6
7.3
5.4
4.4
3.2
2.0
1.3
0.0
0.0
1.20
1.80
2.45
2.75
2.55
2.25
2.05
1.90
1.60
1.40
0.0
0.0
2.40
8.20
15.0
17.20
14.60
10.80
8.80
6.40
4.00
2.60
0.0
90.0
0.0
2.88
14.76
36.75
47.30
37.23
24.30
18.04
12.16
6.4
3.64
0.0
203.46
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Prepared by Amr A. El Sayed, aelsayed@vt.edu
The first Hydrology test:
1- You are the engineer representing a group of tobacco farmers in the Danville, VA area who
want to construct a small irrigation reservoir on the Dan River for storage of flood flows in order
to irrigate their fields during the summer growing season. The farmers will need to withdraw
1500 acre-feet of water each year for irrigation and processing purposes. Average surface area of
the lake is 500 acre (constant for year).
Lake evaporation is estimated to be 57 inches/year; while reservoir seepage losses are
estimated to be approximately 4.0 inches/month. Average annual rainfall in the area is 44 inches.
Average daily runoff diverted into the reservoir from the Dan river is estimated to be 2 cfs.
Treated waste-water from the city’s wastewater treatment plant will also flow into the reservoir at
the rate of 1.0 mgd (million gallons/day; 1 mgd = 1.547 cfs.
In your opinion, is reservoir construction feasible?, assuming the irrigation demand is the
only outflow from the reservoir?. Work in acre.feet units.
Solution:
Water budget equation over the watershed:
Qin − Qout =
dS
dt
Where:
Qin
Qout
inflow rate (volume / time)
outflow rate (volume / time)
dS
dt
Change in storage in (volume / time)
Qin, (+) sign means input:
(+) Precipitation (Rainfall) over a period of one year = 44.0 inches
=
Volume of Rainfall (over one year)
Surface area of the lake
Volume of Rainfall = 44.0 inch * 500 acre *
16
ft
= 1833.33 acre-feet
12 inch
Prepared by Amr A. El Sayed, aelsayed@vt.edu
(+) Runoff = 2 cfs per day
Volume of Runoff over one year = 2.0 * 24 * 60 * 60 * 365 = 63072000 ft3
acre
Depth of Runoff over the area of 1.0 acre = 63072000 ft 3 *
43560 ft 2
= 1447.93 acre.ft
(+) Rin from treated wastewater = 1.0 mgd
Volume = 1.547 cfs.day
Volume over one year = 1.547 * 24 * 60 * 60 * 365 = 48786192.0 ft3.
acre
Depth of Rin over the area of 1.0 acre = 48786192 ft 3 *
43560 ft 2
= 1119.97 acre.ft
Qout, (-) sign means output:
(-) Rout Irrigation demand = 1500 acre.ft (for one year).
(-) Evaporation from lake = 57.0 inches/year
Volume of Evaporation (over one year)
= 57.0 inches.
Surface area of the lake
Volume of Evaporation from lake over one year = 57.0 inch * 500 acre *
ft
12 inch
= 2375.0 acre.ft
(-) Seepage loss = 4.0 inches per month = 4*12 = 48.0 inches/year
Volume of seepage losses over one year = 48.0 inch * 500 acre *
ft
12 inch
= 2000 acre.ft
Applying water budget equation:
Qin − Qout =
dS
dt
Change in storage = (1833.33 + 1447.93 + 1119.97) – ( 1500 + 2000 + 2375 ) =
= -1473.77 acre.ft.
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So, actually with this income flow and outflow the construction of a reservoir in this area
is not feasible.
2- Match the definitions on the right with items on the left by placing the letter of the best
definition in space provided. There may be up to two definitions per item. Not all are used.
G
I,M
E,P
Method for potential
evaporation estimate.
Dewpoint.
Basin-wide rainfall
Ф-index
H
Greenhouse effect
E
P
L
N
F
G
H
E
Precipitation excess
TP No. 40
Stable channel with section
control
Raindrop growth
VDOT IDF charts
Effective precipitation
SCS rainfall-runoff equation
N
E
Critical depth
Green-Ampt equation
M
N
I,M
Mean-areal precipitation.
O
J
O
D
lake
A
B
C
D
I
J
K
L
P
18
Housing unit on a Navajo
Reservation.
Snyder method
Iso-average method
Regional rainfall frequency
method
Alternative
to
Horton
infiltration equation
Isotropic averaging
Temperature of saturated air.
Helps compensate for radiation
imbalance.
Iso-hyetal method.
Pan measurement
Fraternity house drop-out rate.
National rainfall frequency
atlas.
Thiessen method
Desirable site for stream gage
location.
Ice-crystal
and
raindrop
coalescence.
Depth of direct runoff.
Prepared by Amr A. El Sayed, aelsayed@vt.edu
3- The direct runoff from a 2.20 sq. mile watershed is measured by digitizing the area beneath the
direct runoff hydrograph, which has been separated from baseflow. This volume is 5040 cfshours. The hydrograph was produced by a 3-hour rainfall with a constant intensity of 2.0 inches
per hour. Determine:
A- The net rain or excess rainfall (or depth of direct runoff) in inches.
B- The Ф-index in inches per hour.
C- The SCS runoff curve number, CN (only to the nearest whole number; consider using Fig.
2.14, page 130 text).
D- S, and Ia in the SCS rainfall-runoff equation.
Solution:
Volume of direct runoff = 5040.0 * 60 *60 = 18144000 ft3
Depth of DRO = Volume / Area
=
18144000 ft 3
2.2 mi 2
*
mi 2
12 in
= 3.55 inches.
(5280)2 ft 2 ft
*
b1.0
0.8
Rainfall in (inch/hr)
The Ф-index line is taken that
the volume of the hatched area
equals to the depth of direct
runoff.
0.6
0.4
0.2
0.0
Φ-index line
0
6
12
18
24
Time in Hours
In our case rainfall intensity is constant, and equals to 2.0 inches per hour.
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4.0
Area = 3.55 inches
2.0
1.0
0.0
Φ-index line
0
Φ-index
Rainfall in (inch/hr)
3.0
1.0
2.0
3.0
4.0
Time in Hours
(2 - Ф-index) * 3 = 3.55
Ф-index = 2 – 3.55 / 3 = 0.8167 inch / hr
C- Total rainfall depth = Rain intensity (in / hr) * Rainfall duration time (hr).
P = 2.0 (in/hr) * 3.0 (hrs) = 6.0 inches.
Pe (rainfall excess “DRO”) = 3.55 inches.
Q=
(P − 0.20 * S )2
P + 0.80 * S
Where:
Q
P
3.55 =
Total accumulative runoff depth, inches
Total accumulative precipitation at time t
(6 − 0.20 * S )2
6 + 0.80 * S
21.30 + 2.84 S = 36 – 2.40 S + 0.040 S2
S2 – 131 S + 367.50 = 0.0
Which is a second degree equation, and the solution will be on the form:
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S=
− b ± b 2 − 4ac 2
2a
131 − 1312 − 4 *1 * 367.50
S=
2 *1
CN =
D-
= 2.868
1000
= 77.71
S + 10
S = 2.868
Ia = 0.20 S = 0.20 * 2.868 = 0.5736
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Prepared by Amr A. El Sayed, aelsayed@vt.edu
Unit Hydrograph
Looking for a method of forecasting a storm hydrograph, given the time distribution of
rainfall excess.
The unit hydrograph is defined as the direct runoff hydrograph (DRH), we get from
exactly 1.0 inch of rain excess over a T. hr period, or Unit hydrograph is a hydrograph of direct
runoff produced by 1.0 inch of rain excess.
The unit hydrograph once defined becomes a property of a watershed, and it can’t be
applied to any future storms, provided that no significant land-use change in the watershed since
it was established.
The direct runoff hydrograph ordinates for a future storm are directly proportional to
ordinates of the T. hr unit hydrograph when it is applied to rainfall excess distribution in T. hr
intervals.
Basic steps in Unit Hydrograph (UH):
1- Derivation of a (UH) from an isolated storm rainfall and stream flow observation are required.
(Gauged rainfall hydrograph plus gauged stream flow hydrograph).
2- Separate base flow from total runoff hydrograph to get direct runoff or (rainfall excess). This is
done by extending the portion (A) of the runoff hydrograph to point (1) “under the peak”, and
then construct a line from point (1) to point (2), where point (2) lies on the rainfall hydrograph at
a distance :
N = A 0 .2
where:
N
A
number of days beyond peak where base flow becomes 100% of discharge.
drainage area in sq. miles.
(The period of the Unit Hydrograph “UH” is equal to the period of Direct runoff)
3- construct the direct runoff hydrograph, by subtracting base flow from total runoff:
Direct runoff = Total runoff – Base flow.
4- Find the area under the direct runoff hydrograph, and this is done by adding the vertical
ordinates of the DRH, and multiply the summation by the time interval.
n
Volume of DR = ∑ H i * time interval
i =1
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5- Find the direct runoff as a depth over the drainage area:
Depth of direct runoff =
Total DRH volume
Drainage area
6- Divide each ordinate of the DRH by the depth of direct runoff, thus obtaining the ordinates of
the unit hydrograph.
Example:
Design a unit hydrograph for an isolated storm (has only one peak), the rainfall, and total runoff
gauges are given.
It is noticed from the rainfall data that the time period that caused the direct runoff
(rainfall excess) is 6.0 hours, so, the resulted unit hydrograph is called “a 6-hr UH”.
Date
16-Feb
17-Feb
23
Time
600
800
1000
1200
1400
1600
1800
2000
2200
2400
200
400
600
800
1000
1200
1400
1600
Total
flow
500
5600
9200
10100
7800
6600
5550
4700
4000
3300
2700
2300
1950
1650
1400
1200
1000
800
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Steps:
1- Estimate the base flow:
- Extend the Recession Limb (1) to a point just under the peak “point B”
- Draw a vertical line parallel to line “AB”, and to a distance: N = A0.2 (convert N from days
to hours).
- The vertical line will intersect with the right recession limb in point “C”
- Construct a line between “B”, and “C”
- The dashed line will be the base flow line, and the ordinates from that line to the X-axis will be
subtracted from the total flow ordinates to give the DRO ordinates.
Total Flow
12
A
Discahrge (1000) cfs
10
8
6
4
Base Flow
2
1
N (hr)
B
0
0
2
4
6
8
10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40
Time (Hours)
Volume of direct runoff
= Area under the DRO Hydrograph
= ∑ of DRO Ordinates * Time Interval
Volume of DRO = 59850
3600 sec .
ft 3
* 2.0hr *
= 430920000 ft3
sec .
hr
Covering the volume of DRO into depth over area:
mi 2
430920000 ft 3
Depth of DRO =
= 0.3864 ft
*
40.0 mi 2
(5280)2 * ft 2
24
Prepared by Amr A. El Sayed, aelsayed@vt.edu
C
Depth of DRO = 0.3864 * 12 = 4.637 inches
All the ordinates of the DRO are to be divided on (4.637) to give the ordinates of the UH.
A
Date
16-Feb
17-Feb
B
Time
600
800
1000
1200
1400
1600
1800
2000
2200
2400
200
400
600
800
1000
1200
1400
1600
C
Total
flow
500
5600
9200
10100
7800
6600
5550
4700
4000
3300
2700
2300
1950
1650
1400
1200
1000
800
D
Base
flow
500
450
400
400
450
450
500
550
600
600
600
650
650
700
700
750
750
800
E
Direct
F
UH
runoff
0
5150
8800
9700
7350
6150
5050
4150
3400
2700
2100
1650
1300
950
700
450
250
0
Ordinate
s
0
0
1111
2
1898
4
2092
6
1585
8
1326
10
1089
12
895
14
733
16
582
18
453
20
356
22
280
24
205
26
151
28
97
30
54
32
0
34
59850
12907
Hours
To check and see if the results are correct, find the volume under the UH, and divide that volume
over the area to get the water depth, the result must be (1 inch) according to the definition of the
Unit Hydrograph:
Depth of water of UH = 12907 * 2 * 3600
25
1
40 * 5280 2
*12 = 1.000025
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Total Flow
Unit Hydrograph
12
Discahrge (1000) cfs
10
8
6
4
2
0
0
2
4
6
8
10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40
Time (Hours)
Example (Application of the Unit Hydrograph)
Apply the Unit Hydrograph “of the previous example” to the storm:
Time (Hours)
Rainfall
in/hr
Total Rainfall
(inch)
0-6
0.317
6-12
0.383
12-18
0.55
18-24
0.20
24-30
0.35
0.317*6
= 1.9
0.383*6
= 2.3
0.55*6
= 3.3
0.20*6
= 1.2
0.35*6
= 2.1
The Ф-index = 0.30 in/hr.
Steps:
1- break up the storm into (time-hour) period, each period equals to the UH period (6-hr).
2- Apply the loss rate and determine the precipitation excess in each time-hour period.
Rainfall excess (DRO) = (Total rainfall depth - Ф-index) * time
(inches)
for example:
DRO for the period (0-6) = (0.317 – 0.30) * 6 = 0.102 inch.
The DRO volumes are presented as hatched area in the following chart.
26
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Rainfall Hyetograaph
Phi-index
0.6
0.5
Rainfall depth (in/hr)
0.4
0.3
0.2
0.1
0
6
12
18
24
30
Time in Hours
Time (Hours)
Rainfall
in/hr
Total Rainfall
(inch)
DRO (inch)
0-6
0.317
6-12
0.383
12-18
0.55
18-24
0.20
24-30
0.35
1.9
2.3
3.3
1.2
2.1
0.102
0.498
1.5
0
0.30
3- Apply UH to each 6-hr intervals by multiplying interval rain excess by each ordinate in (timehour) UH.
27
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Individual DRO Hydrographs
Day Cloc Elapse Unit
k
d
Hyd
Time Hours Ord.
(0 - 6) (6 - 12) (12 18)
0.1
0.5
1.5
inch inch
inch
(18 24)
0.0
inch
(24 30)
0.3
inch
(cfs)
0
1111
1898
2092
1585
1326
1089
895
733
582
453
356
280
205
151
97
54
0
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
52
54
56
58
28
0
111
190
209
159
133
109
89
73
58
45
36
28
20
15
10
5
0
0
555
949
1046
793
663
545
447
367
291
226
178
140
102
75
49
27
0
0
1666
2847
3138
2378
1989
1634
1342
1100
873
679
534
421
307
226
146
81
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
333
569
628
476
398
327
268
220
175
136
107
84
61
45
29
16
0
Base
Storm
Flow
Hyd.
cfs
Assum cfs
ed
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500
Prepared by Amr A. El Sayed, aelsayed@vt.edu
500
611
690
709
1214
1582
1655
3048
4083
4241
3370
2892
2453
2423
2362
2151
1763
1507
1296
1103
946
820
717
607
584
561
545
529
516
500
Storm Hydrograph cfs
1.5 in cfs
0.1 in cfs
0.0 in cfs
0.5 in cfs
0.3 in cfs
5000
4500
4000
Rainfall (cfs)
3500
3000
2500
2000
1500
1000
500
0
0
6
12
18
24
30
36
42
48
Time hr
29
Prepared by Amr A. El Sayed, aelsayed@vt.edu
54
60
7. Lagging and S-curves for Unit Hydrograph
The main problem is that we have a t-hr unit Hydrograph, and we want to get another t-hr
Unit Hydrograph from the original UH.
What if the rainfall comes in 30-min intervals, and the UH is a 60-min UH?, there are two
options:
1- Add the rainfall data (every 0.5-hr) to form 1.0-hr UH intervals and continue to use 1-hr Unit
Hydrograph.
2- Use 0.5-hr UH rainfall record and redefine 1.0-hr to be a 0.5-hr UH, and apply it to rainfall
data.
The method that is used to convert a D-hr UH into a t-hr UH is called (S-Hydrograph) which
means how to develop a new t-hr UH from an original D-hr UH:
where:
t-hr
D-hr
(new UH we want)
(Original UH we have).
Example:
The original Unit Hydrograph is a 12-hr UH, and we want to get a 6-hr UH from that 12hr UH.
Steps:
1- Build D-hr (S-Hydrograph) from D-hr UH, use alternative addition at D-hr intervals. (A
curve).
2- Locate a second D-hr (S-Hydrograph) behind the original at a distance of (t-hr) (B curve).
t: the label of the new UH(time).
3- Subtract (A-curve) from (B-curve).
4- Multiply differences (A-B) by
30
D
, and this will produce the new t-hr UH.
t
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Time
in hours
12-hr UH
in cfs
D-hr UH
6
12
18
24
30
36
42
48
54
60
66
72
78
84
90
96
102
108
114
120
126
132
138
144
900
3400
6900
10100
12300
13600
13900
13200
11800
10300
8950
7650
6400
5250
4200
3200
2280
1580
1100
750
500
300
150
50
31
Curve A
12-hr
S-Hyd.
900
3400
7800
13500
20100
27100
34000
40300
45800
50600
54750
58250
61150
63500
65350
66700
67630
68280
68730
69030
69230
69330
69380
69380
Curve B
12-hr SHyd. shifted
6-hr
900
3400
7800
13500
20100
27100
34000
40300
45800
50600
54750
58250
61150
63500
65350
66700
67630
68280
68730
69030
69230
69330
69380
69380
A-B
6-hr UH
( A − B)* D
t
900
2500
4400
5700
6600
7000
6900
6300
5500
4800
4150
3500
2900
2350
1850
1350
930
650
450
300
200
100
50
0
1800
5000
8800
11400
13200
14000
13800
12600
11000
9600
8300
7000
5800
4700
3700
2700
1860
1300
900
600
400
200
100
0
Prepared by Amr A. El Sayed, aelsayed@vt.edu
12-hr S-Hyd.
12-hr S-Hyd. shifted 6-hr
new 6-hr UH
70
50
40
30
20
10
0
0
6
12
18
24
30
36
42
48
54
60
66
72
78
84
90
96
102
108
114
120
126
132
138
144
Discharge in thouthanss cfs
60
Time in (Hours)
32
Prepared by Amr A. El Sayed, aelsayed@vt.edu
6-hr UH
Time in hr
33
Prepared by Amr A. El Sayed, aelsayed@vt.edu
144
138
132
126
120
114
108
102
96
90
84
78
72
66
54
60
48
42
36
30
24
18
6
12
0
Rainfall in thousand cfs
12-hr UH
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Example (2)
It is required to design a 3-hr UH from a given 2-hr UH by using S-Hydrograph
Procedure.
The given 2-hr UH
Time
Total
Discharge
cfs
0
0
1
2
3
4
5
6
7
8
9 10 11 12
50 150 300 600 750 650 550 450 350 250 150 50
13
0
Solution:
Time
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
2-hr UH
in cfs
D-hr UH
0
50
150
300
600
750
650
550
450
350
250
150
50
0
0
34
Curve A
2-hr
S-Hyd.
Curve B
2-hr S-Hyd.
shifted 3-hr
0
50
150
350
750
1100
1400
1650
1850
2000
2100
2150
2150
2150
2150
0
0
0
0
50
150
350
750
1100
1400
1650
1850
2000
2100
2150
A-B
3-hr UH
( A − B)* 2
3
0
50
150
350
700
950
1050
900
750
600
450
300
150
50
0
0
33
100
233
467
633
700
600
500
400
300
200
100
33
0
Prepared by Amr A. El Sayed, aelsayed@vt.edu
14
0
If the 2-hr UH is tabulated in 3-hr intervals,
then plot the UH, and interpret for the values
each 2-hr intervals
Rainfall in thousand cfs
Note:
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
A 2-hr UH tabulated in 3-hr period
0
3
6
9
12
15
18
21
24
27
30
33
36
Time in hr
2500
2-hr S-Hydr.
2-hr S-Hydr. shifted 3-hr
3-hr UH
2000
Discharge in cfs
1500
1000
500
0 0
1
2
3
4
5
6
7
8
9
10
11
12
Time in Hours
35
Prepared by Amr A. El Sayed, aelsayed@vt.edu
13
14
It is an interesting exercise to get the original 2-hr UH, from the resulted 3-hr UH
considering it as a given data:
Example:
It is required to design a 2-hr UH from a given 3-hr UH
The given 2-hr UH
Time
Total
Discharge
cfs
0
0
1
2
3
4
5
6
7
8
9 10 11 12 13
33 100 233 467 633 700 600 500 400 300 200 100 33
Solution:
Time
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
3-hr UH
in cfs
D-hr UH
0
33
100
233
467
633
700
600
500
400
300
200
100
33
0
Curve A
3-hr
S-Hyd.
Curve B
3-hr S-Hyd.
shifted 2-hr
A-B
2-hr UH
( A − B)* 3
2
0
33
100
233
500
733
933
1100
1233
1333
1400
1433
1433
1433
1433
0
33
100
233
500
733
933
1100
1233
1333
1400
1433
1433
0
33
100
200
400
500
433
367
300
233
167
100
33
0
0
0
50
150
300
600
750
650
550
450
350
250
150
50
0
0
It is obvious that we got the same original 2-hr UH.
36
Prepared by Amr A. El Sayed, aelsayed@vt.edu
14
0
8. Dimensionless Unit Hydrograph by SCS
This method is used for ungaged watersheds, and there are mainly two methods:
1- Snyder Synthetic UH (1940): for large watershed sometimes (1000 mi2)
2- Soil Conservation Service “SCS” Dimensionless UH : for smaller areas ≤ 50 sqm.
Water drop
path in the stream
path on the ground
Drainage area
( Watershed Boundaries )
Target Point
No gages at this point
The required data are:
1- A full topographic map of the watershed, from which we can determine:
a- The drainage area (A)
b- How long it takes for a drop of water from the moment of hitting the ground surface untill
reaching the target point, which is called (Concentration time), or (Traveling time) TC.
The Concentration Time “TC” is defined as the time of travailing of the most distant point
in the watershed to the target point (over the ground, and in the streams).
We have to check every possible path of a water drop to know approximately the most
distant point in the watershed. TC value is affected very much by the nature of streams, and land
development over the watershed.
c- Rainfall time distribution.
37
Prepared by Amr A. El Sayed, aelsayed@vt.edu
d- Infiltration loss curve or SCS runoff curve number CN.
Equations:
Peak Discharge Q p (cfs ) =
Time to Peak T p (hrs ) =
( )
484 * A mi 2 * Q(in )
T p (hrs )
∆D
+ Tl
2
Where:
484
A
Q
Tp
Tl
constant refers to the shape factor
Drainage area in square miles.
Depth of rainfall excess (1.0 inch) for UH.
Time to hydrograph peak in hours, referenced from origin.
Basin lag time in hours = time laps from mid-point of rain excess period to
peak.
Rainfall in cfs
∆D
T
Tp
Time in hr
An approximate equation to calculate “ Tl ” :
Basin lag time Tl (hrs ) =
38
L0.8 * (1000 − 9 * CN )0.7
1900 * CN 0.7 * Y 0.5
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Where:
L
Y
CN
Length along main channel to divide, ft.
Average land slope (%).
SCS runoff curve number.
Example:
There is a storm over a watershed area of 40.0 mi2, has a CN of 75, and the mean slope of
the land = 1.20 %. The total length along main channel to divide = 6.92 miles. The period of
rainfall excess equals 6.0 hours. Construct the 6-hr UH for that watershed.
Solution:
Tl (hrs ) =
L0.8 * (1000 − 9 * CN )0.7
1900 * CN 0.7 * Y 0.5
L = 6.92 mi = 6.92*5280 = 36537.6 ft
(
36537.6 )0.8 * (1000 − 9 * 75)0.7
= 5.9927 ≈ 6.0 hours
Tl (hrs ) =
1900 * 750.7 * (1.2 )0.5
Time to peak T p =
Q p (cfs ) =
Note:
∆D
+ Tl = 6/2 + 6 = 9.0 hr.
2
484 * A(mi 2 ) * Q(in ) 484 * 40 * (1)
=
= 2151.11 cfs
T p (hrs )
9
if we have perfect data about the watershed, ∆D ≅ 0.133Tc , and 0.20 ≤ ∆D/Tp ≤ 0.25
∆D = 0.133 *10 = 1.33 hr, but ∆D = 6.0 hrs, so, the resulted UH will not be optimal.
39
Prepared by Amr A. El Sayed, aelsayed@vt.edu
The SCS dimensionless UH ordinates:
t
tp
Q
Qp
t
0.00
0.20
0.40
0.60
0.80
1.00
0.00
0.10
0.31
0.66
0.93
1.00
tp
Q
Qp
1.20
1.40
1.60
1.80
2.00
0.93
0.78
0.56
0.39
0.28
t
tp
Q
Qp
2.20
2.40
2.60
2.80
3.00
0.207
0.147
0.107
0.077
0.055
t
tp
Q
Qp
t
tp
Q
Qp
3.20
3.40
3.60
3.80
4.00
0.040
0.029
0.021
0.015
0.011
4.20
4.40
4.60
4.80
5.00
0.0100
0.0070
0.0030
0.0015
0.000
1.0
0.9
0.8
0.7
Q / Qp
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0
1
2
3
4
5
t / tp
From the above table, results for t, and Q can be obtained, this is done in the following
table:
t (of SCS UH ) =
t
tp
Q(of SCS UH ) =
(from table) * t p (calculated)
Q
(from table) * Q p (calculated)
Qp
40
Prepared by Amr A. El Sayed, aelsayed@vt.edu
A
B
t
tp
Q
Qp
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
2.20
2.40
0.00
0.10
0.31
0.66
0.93
1.00
0.93
0.78
0.56
0.39
0.28
0.207
0.147
tp = 9
t
A * tp
Qp=2151
Q
B * Qp
0.0
1.80
3.60
5.40
7.20
9.00
10.80
12.60
14.40
16.20
18.0
19.80
21.60
0
215
667
1420
2000
2151
2000
1678
1205
839
602
445
316
A
B
t
Q
Qp
tp
2.60
2.80
3.00
3.20
3.40
3.60
3.80
4.00
4.20
4.40
4.60
4.80
5.00
0.107
0.077
0.055
0.040
0.029
0.021
0.015
0.011
0.0100
0.0070
0.0030
0.0015
0.000
tp = 9
t
A * tp
23.4
25.2
27
28.8
30.6
32.4
34.2
36
37.8
39.6
41.4
43.2
45
Qp=2151
Q
B * Qp
230.157
165.627
118.305
86.04
62.379
45.171
32.265
23.661
21.51
15.057
6.453
3.2265
0
2400
UH by SCS UH obs.
2200
2000
1800
Q cfs
1600
1400
1200
1000
800
600
400
200
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
0
Time (hours)
The UH produced by the SCS method comes with non-steady time intervals, so, after
plotting the UH, interpret for points of (Q) at suitable fixed time intervals (every 2, 3 … hours).
41
Prepared by Amr A. El Sayed, aelsayed@vt.edu
2200
UH by SCS
2000
1800
1600
1400
1200
1000
800
600
400
45
42
39
36
33
30
27
24
21
18
15
12
9
6
0
3
200
0
Q cfs
time UH
Ordinat
es
0
0
3
470
6
1708
9
2153
12 1818
15 1063
18 603
21 352
24 206
27 118
30 69
33 41
36 24
39 18
42 5
45 0
8648
Time (hours)
To check the resulted 6-hr UH, the height of DRO over the watershed area must equal
(1.0 inch):
Depth of water of UH = 8648 * 3 * 3600
42
1
40.0 * 5280 2
*12 = 1.04 inch
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Application to the SCS UH:
Example:
Given a Hyetograph of a storm over periods of 6.0 hours.
Given data: Curve number CN = 75, find the direct runoff, by using the SCS curve number
method, and then apply the resulted UH from the previous example to that storm to produce the
storm Hydrograph.
Time (Hours)
Rainfall
in/hr
Total Rainfall
(inch)
0-6
0.317
6-12
0.383
12-18
0.55
18-24
0.20
24-30
0.35
0.317*6
= 1.9
0.383*6
= 2.3
0.55*6
= 3.3
0.20*6
= 1.2
0.35*6
= 2.1
18-24
24 - 30
Rainfall Hyetograaph
0.6
Rainfall depth (in/hr)
0.5
0.4
0.3
0.2
0.1
0
0-06
6-12
12-18
Time in Hours
43
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Solution:
Q=
(P − 0.20 * S )2
P + 0.80 * S
S=
1000
− 10
CN
where:
Q
S=
(depth of Direct Runoff in inches).
1000
− 10 = 3.33
75
The calculations of (Q) are tabulated in the following table.
Q=
Time
in hours
P (total rainfall)
inches
P (cumulative)
inches
0–6
6 – 12
12 – 18
18 – 24
24 – 30
1.90
2.30
3.30
1.20
2.10
1.90
4.20
7.50
8.70
10.80
44
(P − 0.20 * 3.33)2
P + 0.80 * 3.33
Direct runoff
∆Q
Q (cumulative)
inches
0.333
1.818
4.593
5.678
7.625
inches
0.333
1.485
2.775
1.085
1.948
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Applying the UH to the direct runoff results to get the actual DRO hydrograph:
time UH**
0
3
6
9
12
15
18
21
24
27
30
33
36
39
42
45
Pe=0.33 Pe=1.48 Pe=2.77 Pe=1.08 Pe=1.94 Base
3
5
5
5
8
Flow
0
500
156
500
569
0
500
717
697
500
605
2536
0
500
354
3197
1303
500
201
2700
4740
0
500
117
1579
5974
510
500
68
895
5045
1853
0
500
39
522
2951
2336
915
500
23
305
1672
1973
3327
500
14
176
976
1154
4193
500
8
103
571
654
3542
500
6
61
328
382
2071
500
2
35
192
223
1174
500
0
26
113
128
685
500
7
66
75
401
500
0
49
44
231
500
14
26
135
500
0
19
80
500
5
46
500
0
34
500
10
500
0
500
0
470
1708
2153
1818
1063
603
352
206
118
69
41
24
18
5
0
45
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Storm Hyd.
cfs
500
656
1069
1914
3642
5354
8140
8679
8362
7263
7801
7012
5377
3348
2126
1453
1049
824
674
599
552
534
510
500
Storm Hyd.
UH
10000
Total Rainfall cfs
8000
6000
4000
0
0
3
6
9
12
15
18
21
24
27
30
33
36
39
42
45
48
51
54
57
60
63
66
69
72
2000
Time in Hours
Snyder method for ungaged watersheds:
The Snyder method to construct a unit hydrograph for watersheds with areas ≤ 10000 mi2,
Steps of the method:
a-
t p = Ct (L * Lc )0.3
where:
tp
Ct
Basin lag time in hours
Empirical slope coefficient. ( 0.40 ≤ Ct ≤ 2.20 )
If the value of Ct = 0.40 it indicates the watershed surface slope is steep “Y > 1.60
%, and if Ct = 2.20 it indicates the watershed surface slope is flat (Y < 0.50 %).
L
Lc
Main channel length in miles.
Main channel length to a point nearest to the watershed centroid.
The time to peak is referred from
46
∆D
in units of hours.
2
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Water drop
A
path on the ground
C
D
Drainage area
( Watershed Boundaries )
B
Target Point
No gages at this point
L = distance AB
Point C is the centroid point of the watershed, from that point draw a perpendicular line to the
main channel, will intersect in D.
Lc = distance DB.
b-
Qp =
640 * C p * A
tp
Where:
Qp
A
Cp
Peak discharge in
Drainage area of the watershed in square miles.
Storage coefficient, 0.40 ≤ Cp ≤ 0.80,
If the value of Cp = 0.40 it indicates that watershed surface is swampy, and if the
value of Cp = 0.80 it indicates that there is no storage in the form of flood plains
cover bank areas.
Note: Large Ct and small Cp go together.
c-
Tb = ( 4 ~ 5 ) tp
47
Prepared by Amr A. El Sayed, aelsayed@vt.edu
The form of equation Tb = 3 days + tp / 8 is suitable only for ***????
d-
⎛ Qp ⎞
⎟
W75 = 440 * ⎜⎜
⎟
A
⎝
⎠
⎛ Qp ⎞
⎟
W50 = 770 * ⎜⎜
⎟
A
⎠
⎝
−1.08
“place
W75
to the left of vertical line through Qp”
3
−1.08
W75, and W50 are in hours.
Tp
∆D
5
W 75
3
W 75
2
6
W 50
3
Qp
3
3 Qp
4
Rainfall in cfs
4
Qp
2
W 50
7
1
Tb
Time in hr
Adjustment equation for tp, and Qp:
t`p is the working value of tp:
t ` p = t p + 0.25(D`− D )
Where:
D = tr =
D`
tp
5.50
is the working duration of rainfall excess.
48
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Apply the value of t`p to get the value of Q`p:
Q` p =
640 * C p * A
t`p
Example:
Develop a 6-hr Unit Hydrograph for the watershed of 40.0 mi2, by using Snyder method,
knowing that:
L = 6.92 mi, Lc = 5.0 mi
∆D = D`= t `r = 6.0 hours
Ct = 1.90 “semi-flat watershed”
Cp = 0.54
Solution:
t p = Ct (L * Lc )0.3
t p = 1.90 * (6.92 * 5.0 )0.3 = 5.5 hours
D = tr =
tp
5.50
“Snyder saturated rain excess duration”
D = 5.50 / 5.50 = 1.0 hour
but, we want a 6-hr UH, so, adjust D to 6-hr by applying the equation:
t ` p = t p + 0.25(D`− D ) , where D` = 6.0 hours
t ` p = 5.50 + 0.25(6.0 − 1.0 ) = 6.75 hrs
Q` p =
640 * C p * A
t`p
=
640 * 0.54 * 40
6.75
Q`p = 2048 cfs
Tb = (4~5) t`p = (27~33.75) hours
⎛ Qp ⎞
⎟
W75 = 440 * ⎜⎜
⎟
A
⎝
⎠
49
−1.08
⎛ 2048 ⎞
= 440 * ⎜
⎟
⎝ 40 ⎠
−1.08
= 6.27 hours
Prepared by Amr A. El Sayed, aelsayed@vt.edu
⎛ 2048 ⎞
= 770 * ⎜
⎟
⎝ 40 ⎠
= 11.0 hours
3.67
4.2
7.33
2048 cfs
6.75
2.1
Q cfs
−1.08
1536 cfs
3
−1.08
1024 cfs
⎛ Qp ⎞
⎟
W50 = 770 * ⎜⎜
⎟
A
⎝
⎠
33.75 hours
Time in Hours
50
Prepared by Amr A. El Sayed, aelsayed@vt.edu
9- Reservoir and Channel Routing
Tank Analogy for reservoir routing:
I1
t1
2
Inflow Hydrograph
I 1+ I 2
Inflow
I2
t2
Time in hr
Fig. (9-1)
The inflow hydrograph can be estimated before the actual storm by Unit Hydrograph.
Storage Equation:
I +I
O + O2
∆S
= I −O = 1 2 − 1
∆t
2
2
………. (9-1)
∆S = S 2 − S1 = (h2 − h1 ) * At
Substituting for ∆S in equation (9-1)
I +I
O + O2
⎛ h2 − h1 ⎞
⎜
⎟ * At = 1 2 − 1
2
2
⎝ ∆t ⎠
………. (9-2)
I1, and I2 are always known from the inflow hydrograph.
h1 is the water level in the reservoir at time t = t1, and is known too.
51
Prepared by Amr A. El Sayed, aelsayed@vt.edu
The two unknowns in equation (9-2) are h2, and O2.
t2
h1 - h 2
t1
Tank
Fig. (9-2)
The outflow rating equation, considering the reservoir as a tank, and the outlet as an
orifice:
O = CAo 2 gh
As there is two unknowns, and only one equation (9-2), the second equation needed for
solution comes from Storage vs. Elevation table, and Outflow vs. Elevation table.
Rearranging equation 9-1
I +I
O + O2
∆S
= I −O = 1 2 − 1
∆t
2
2
I1 + I 2 O1 + O2 S 2 − S1
−
=
2
2
∆t
………. (9-1)
Multiply by 2
S −S
I1 + I 2 = O1 + O2 + 2 2 1
∆t
52
I1 + I 2 = O1 + O2 +
2 S 2 2 S1
−
∆t
∆t
Prepared by Amr A. El Sayed, aelsayed@vt.edu
⎞
⎛ 2S
⎞ ⎛ 2S
I1 + I 2 = ⎜ − 1 + O1 ⎟ + ⎜ 2 + O2 ⎟
⎝ ∆t
⎠ ⎝ ∆t
⎠
⎛ 2S
⎞ ⎛ 2S
⎞
I1 + I 2 + ⎜ 1 − O1 ⎟ = ⎜ 2 + O2 ⎟
⎝ ∆t
⎠ ⎝ ∆t
⎠
⎛ 2S
⎞
⎛ 2S
⎞
− O⎟
=⎜
+ O⎟
I1 + I 2 + ⎜
⎝ ∆t
⎠ at step 1 ⎝ ∆t
⎠ at step 2
………. (9-3)
Equation (9-3) represents the basic procedure to reservoir routing, which is done step by
step:
Steps in reservoir routing:
1- the inflow hydrograph is known. Select ∆t , use (5~6) points on the rising side of
inflow hydrograph to the peak point.
2- The water surface elevation of the pool is known for different values of storage S, and
output O, from that curve design a table relating
between
2S
+ O and O, and draw a curve
∆t
2S
+ O , and outflow O
∆t
3- At the initial step “step (1)”, Inflow (I1, and I2) are known from the inflow hydrograph.
4- At the same initial step (1), the storage S1, and output O1 are known from the (water
surface elevation of the reservoir vs storage, and outflow curve).
⎛ 2S 2
⎞
+ O2 ⎟ is determined.
⎝ ∆t
⎠
5- from 3, and 4, the right hand side of equation 9-3 ⎜
6- Go to the curve constructed in step 1, and find the outflow O2 corresponding to the
⎛ 2S 2
⎞
+ O2 ⎟ .
⎝ ∆t
⎠
value ⎜
⎞
⎞
⎛ 2S
⎛ 2S
=⎜
+ O⎟ − 2*O
− O⎟
⎠
⎠ at next step ⎝ ∆t
⎝ ∆t
7- ⎜
53
Prepared by Amr A. El Sayed, aelsayed@vt.edu
The work is continuos till the table is completed, then the outflow hydrograph is
determined. It is preferable to plot both the input, and output flow hydrograph together to find out
if the reservoir is doing its job efficiently. The area between the inflow, and outflow curves
represents the maximum storage capacity required for the reservoir.
Example:
The design inflow hydrograph shown in fig. (9-3 ), developed for a commercial area, is to
be routed through a reservoir. Assume that initially the reservoir is empty (S0 = 0 “storage at time
zero = 0.0), and there is no initial outflow (O0 = 0.0). Using the water surface elevation, Storage
and outflow relationships given in table (9-1), rout the hydrograph through the reservoir. What is
the maximum height reached in the reservoir for this inflow? Use ∆t = 10 min.
Table (9-1) Water surface elevation, Storage and outflow
54
Water surface
elevation “ft”
Storage
(acre-ft)
Outflow
(cfs)
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
11.0
12.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.50
10.50
12.0
13.50
20.0
22.0
0
15
32
55
90
125
158
185
210
230
250
270
290
Prepared by Amr A. El Sayed, aelsayed@vt.edu
400
Inflow Hydrograph cfs
350
300
250
200
150
100
50
0
0
10
20
30
40
50
60
70
80
90 100 110 120 130 140 150 160
Time in min
Fig. (9-3) Inflow Hydrograph
Outflow cfs
300
250
250
200
200
150
150
100
100
50
50
0
0
1
2
3
4
5
6
7
8
9
10
11
12
Outflow cfs
Storage cfs.hr
Storage cfs.hr
300
0
Water surface elev. ft
Fig. (9-4) Water surface elevation, Storage and outflow
Note: The water surface elevation in the reservoir, and the corresponding storage is determined
by using the contour map of the reservoir to find the volume of water in the reservoir
corresponding to each water elevation.
55
Prepared by Amr A. El Sayed, aelsayed@vt.edu
For example:
∆S (storage between contour lines h1, and h2) =
h2
A1 + A2
* (h2 − h1 )
2
h 2 - h1
h1
A2
A1
Steps:
1- ∆t = time to peak / (5~6) = 10.0 min
2- table relating
2S
+ O and O:
∆t
∆t = 10.0 min = 1 / 6 hr
Table (9.)
Reservoir Storage Storage
Outflow
Elev. Ft
acre-ft
cfs-hr
conduit cfs
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
11.0
12.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.50
10.50
12.0
13.50
20.0
22.0
To convert from acre.ft to cfs.hr:
43560 ft 2
hr
1.0acre. ft *
acre
56
*
60 * 60 sec
0
12
24
36
48
61
73
91
127
145
163
242
266
0
15
32
55
90
125
158
185
210
230
250
270
290
2S*∆t+O
cfs
0
160
322
491
671
851
1029
1274
1735
1972
2210
3174
3484
= 12.10 cfs.hr
Prepared by Amr A. El Sayed, aelsayed@vt.edu
300
250
Outflow cfs
200
150
100
4000
3800
3600
3400
3200
3000
2800
2600
2400
2200
2000
1800
1600
1400
1200
1000
800
600
400
200
0
0
50
2*S / delta t + O (cfs)
Fig. (9-5)
2S
+ O vs. Outflow
∆t
4- Initial conditions:
At t = 0.0:
S0 = 0.0
O0 = 0.0
2* S
− O = 0.0
∆t
“at time = 0.0”
1- add the number in the column (I1 + I2) for example “420” to the number in the column
(
2* S
2* S
− O ) “378”, that will give the number in the column (
+ O ) “798”.
∆t
∆t
2- Use that number
2* S
+ O “798” and by the curve produced before “Fig. (9.5)”to determine
∆t
the corresponding value of “O” “115”.
3- Apply the equation
2* S
2* S
+ O - (2*O) to determine
− O of the next step
∆t
∆t
“798 – 2*115 = 568”.
4- Use the values of O to determine both Storage, and water surface elevation by using the curve
in Fig. (9.4).
57
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Table (9.3) Routing table.
Time
Inflow
Hyd.
cfs
I1+I2
0
0
In +
In+1
60
10
20
30
60
120
180
180
300
420
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
240
300
360
320
280
240
200
160
120
80
40
0
540
660
680
600
520
440
360
280
200
120
40
0
0
0
0
0
0
in Min
58
2S/∆t O
cfs
+
0
50
186
+
378
568
774
1046
1306
1458
1516
1498
1418
1282
1090
854
628
464
362
288
232
186
2S/∆t + Outflow Storage Elevatio
n
O
cfs
O (cfs) cfs.hr
ft
0
0.0
60
230
486
5
22
54
4.2
17.4
35.6
0.27
1.45
2.95
798
1108
1434
1726
1906
1978
1956
1858
1698
1482
1210
894
628
464
362
288
232
115
167
194
210
224
231
229
220
208
196
178
133
82
51
37
28
23
57.2
77.1
99.8
127.0
139.8
145.9
144.2
112.5
124.7
105.3
84.6
63.7
45.3
34.3
27.0
21.5
18.2
4.7
6.3
7.25
8
8.7
9.05
8.95
7.6
7.9
7.4
6.7
5.2
3.8
2.85
2.25
1.8
1.5
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Inflow Hyd.
Outflow Hyd.
20
60
400
350
300
Q cfs
250
200
150
100
50
0
0
40
80
100
120
140
160
180
Time in Min
Fig. (9- ), Inflow, and Outflow hydrographs.
59
Prepared by Amr A. El Sayed, aelsayed@vt.edu
200
Muskingum Channel Routing
The routing techniques can be used for design systems when the target outflow
hydrograph is given.
We should be able to develop sub area out flow hydrograph given are rainfall and
watershed properties using one of the unit hydrograph procedures.
SA 1
SA 2
SA 3
SA 4
Reservoir
Channel routing methods:
1- Hydraulic routing method
Mathematically and physically correct. Based on Partial Deferential Equations PDE of
mass and momentum conservation. They are generally varied unsteady flow equations, and they
are difficult to solve.
2- Hydrologic routing procedures (like Muskingum), based only on mass conservation + a simple
storage inflow-outflow relationship.
60
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Inflow Hydrograph
Rainfall in cfs
L
k = vL
v
Time in hr
v
k
Average velocity through the channel.
Travel time.
Channels, like reservoirs, have storage associated with it causing inflow hydrograph to
attenuate (stretch out), and lag.
So, flow in channel will affect the shape of inflow hydrograph at the end of traveling in
the channel.
Fundamental Equations:
1- Mass conservation equation:
I1 + I 2 O1 + O2 S 2 − S1
−
=
2
2
∆t
(9-1)
I1, I2, O1, and S1 are known.
O2, and S2 are unknown.
Bring in second relationship between O2, and S2 besides equation (9-1):
S = k {xI + (1 − x )O}
Where:
k
x
Travel time through the reach in hours or days.
Wedge coefficient. ( 0.0 ≤ x ≤ 0.50 )
if x = 0.0, it will indicated that there is only the prism, and water surface is parallel to the
channel bed.
61
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Total storage beneath flood wave:
S = prism + wedge
= k .O + k .x(I − O )
S = k {x.I + (1 − x )O}
Wedge
Storage= k.x (
I-O)
Inflow
Prism
Storage= k.O
Outflow
L=k.v
If x = 0.0 means that reservoir exists.
if x = 0.50 means very steep wave, or storm sewer, and it means also that there will be very little
storage in the channel. The outflow will be like inflow but shifted in time.
S1 = k {x.I1 + (1 − x )O1}
S 2 = k {x.I 2 + (1 − x )O2 }
S 2 − S1 = k {x(I 2 − I1 ) + (1 − x )(O2 − O1 )}
………. (
Substitute into equation (9-1), combine terms, and solve for O2
O2 = C0 .I 2 + C1.I1 + C 2 .O1
62
………. (9-
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Where:
C0 =
− k .x + 0.50 * ∆t
k (1 − x ) + 0.50 * ∆t
C1 =
k .x + 0.50 * ∆t
k (1 − x ) + 0.50 * ∆t
C2 =
k − k .x − 0.50 * ∆t
k (1 − x ) + 0.50 * ∆t
3
∑ Ci = 1 or –1, and this is a check if the values are correct.
i =1
Steps in channel routing:
1- set ∆t =
1
* t p “on the rising side of inflow hydrograph”
5~6
2- Estimate or compute x, and k such that:
2.k .x ≤ ∆t ≤ k
“Compatibility condition”
2
3- Compute C0, C1, and C2,
∑ Ci = 1
i =0
4- Organize routing table in ∆t intervals.
63
Prepared by Amr A. El Sayed, aelsayed@vt.edu
If the compatibility condition is not satisfied, the output hydrograph will be of strange
shape, and to overcome that problem one can divide the total channel length into two or more
reaches, and by that decrease the travel time “k”. The output hydrograph from the first portion of
the channel is considered inflow hydrograph for the second portion, and so on.
Inflow Hydrograph
Outflow Hydrograph
at the end of portion 1
Outflow Hydrograph
Portion 1
Portion 2
Example:
Perform the flood routing for a reach of river given:
x = 0.2, and k = 2 days.
The inflow hydrograph with ∆t = 1 day is given, assume equal inflow and outflow rates
on day 1.
Time
Days
Inflow
cfs
Time
Days
Inflow
cfs
1
2
3
4
5
6
7
8
9
10
11
12
13
4260 7646 1116 1673 2159 2095 2657 4600 5996 5774 4789 3446 2166
7
0
0
0
0
0
0
0
0
0
0
14
15
16
17
18
19
20
21
22
23
24
25
3468 4518 4914 4129 3383 2051 1472 1143 9294 7831 6228 6083
0
0
0
0
0
0
0
6
Solution:
Choose ∆t = 1 day, note that k, and ∆t must be in the same time units (min, hr, or day).
64
Prepared by Amr A. El Sayed, aelsayed@vt.edu
2.k .x ≤ ∆t ≤ k
2*2*0.2 ≤ 1 ≤ 2
o.k
− k .x + 0.50 * ∆t
0.10
− 2 * 0.20 + 0.5 *1
C0 =
=
=
= 0.04762
k (1 − x ) + 0.50∆t 2(1 − 0.20 ) + 0.50 *1 2.10
C1 =
k .x + 0.50 * ∆t
0.90
=
= 0.42857
k (1 − x ) + 0.50 * ∆t 2.10
C2 =
k (1 − x ) − 0.50 * ∆t 1.10
=
= 0.52381
k (1 − x ) + 0.50 * ∆t 2.10
2
∑ Ci = 0.04762 + 0.42857 + 0.52381 = 1.0
i =0
Time
Days
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Inflow
cfs
4260
7646
11167
16730
21590
20950
26570
46000
59960
57740
47890
34460
21660
34680
45180
49140
41290
33830
20510
14720
11436
9294
7831
6228
6083
65
C0 * I2
C1 * I1
C2 * O1
364
532
797
1028
998
1265
2190
2855
2750
2280
1641
1031
1651
2151
2340
1966
1611
977
701
545
443
373
297
290
1826
3277
4786
7170
9253
8979
11387
19714
25697
24746
20524
14769
9283
14863
19363
21060
17696
14499
8790
6309
4901
3983
3356
2669
2231
2316
3208
4605
6706
8882
10018
12360
18296
24484
26982
25744
21761
17126
17883
20735
22923
22120
19693
15287
11597
8874
6930
5543
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Outflow
cfs
4260
4421
6125
8791
12803
16957
19126
23596
34929
46743
51511
49147
41544
32695
34140
39586
43762
42229
37595
29184
22140
16941
13230
10583
8502
Inflow
Outflow
70
60
Discharge (1000 cfs)
50
40
30
20
10
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 25
Time in days
66
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Task / Hydrologic
Operation
Data
Input
Output
1- Define Rainfall time distribution for some TP40
.
National Rainfall Hydrograph.
specified storm
weather services and a
known distribution
2- Define Sub-area boundaries, and measure USGS map, or
drainage areas.
30 m DEM
Sub-area
and areas.
boundaries,
3- Land use and soil data
Montgomery
Arial photography, long term zoning map, or soil map.
comprehensive plan for the area.
County HSG type and CN
4- Estimate time of concentration from the Land slope, length of Tc for each sub-area.
channels
(flow
most distant point Tc.
elements).
5- Unit Hydrograph for each sub-area
SCS Unit Hydrograph
Lag for Tc,
drainage area.
and UH for all sub-areas for
specified time duration.
6- Apply Uh in each sub-area to distribution
of rain excess using ∆t intervals = time
labels of UH.
7- Combine SA1 + SA2 storm hydrograph.
8- Main channel routing
x, L, k
+combined
inflow
hydrograph from step 7
9- Combine outflow from step 8 with SA3.
10- Reservoir routing of step 9 inflow inflow from step 9 +
hydrograph
elevation vs. outflow,
elevation vs. storage
67
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Step (1) in problem set “8”
Summarizing the data for each sub-area:
Sub area
No.
1
2
3
Drainage
area m2
Curve No
CN
58
79
90
TC
min
211
Depth Dn
5.0 ft
5.0
9.0 AB
12.0 BC
Main Channel
VBF
L ft
4.56
37488
TCh.
2.28
Where:
CN
TC
VBF
L
curve number of each sub-area according to the land use.
Time of concentration (the time required for a water drop at the most
distant point in the sub-area to reach the target point).
the mean velocity of water in the channel.
Length of the channel.
TCh.
Time of traveling water through the channel =
L
VBF
Methods of calculating Tc
By definition, the time of concentration is the time of travel for a drop of water leaving
the most distant point in the contributing area and arriving at the downstream point of interest.
The time of concentration includes:
1- Overland or sheet flow time.
2- Ditch or swale flow time.
3- Flow in channel or pipe.
An example calculation of the total flow time in the upper reach sub-area of the Calder
Ally watershed in downtown Salem, VA is shown by different methods:
68
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Method
and Date
1- Kirpich
1940
Formulas for TC
minutes
Developed from SCS data for
7 rural basins in Tennessee
with well-defined channel and
steep slopes (3% to 10%). For
overland flow on concrete or
Length of channel/ditch asphalt surface multiply TC by
from head water to outlet.
0.40. For concrete channels
Average gully slope, ft / ft multiply
by
0.20.
No
adjustment for overland flow
on bare soil or flow in roadside
ditches.
Reference: Civil Engineering,
Vol. 10, No. 6, June 1940.
⎛ L2 ⎞
Tc = 0.0078 * ⎜ ⎟
⎜ S ⎟
⎝ ⎠
L
S
2- Izzard
1946
Tc =
i
c
L
S
69
Remarks
0.385
41.025(0.0007i + c )L0.33
S 0.333 .i 0.667
Rainfall intensity, in/hr
Retardance coefficient
Length of flow path, feet
Slope of flow path, ft/ft
Developed
in
laboratory
experiments by Bureau of
public Roads for overland flow
on roadway and turf surface.
Values of the retardance
coefficient range from 0.007
for very smooth pavement, C =
0.012 for concrete pavement;
C = 0.06 for dense turf.
Solution is extremely tedious
and requires iteration. Product
(i times L) should be < 500.
Reference: Proc. Highway
Research Board Vol. 26, pp.
129-146, 1946
Prepared by Amr A. El Sayed, aelsayed@vt.edu
L:
L
3- Federal Aviation Agency
1970
4- Kinamatic wave formula
Morgali 1965
5- Old SCS lag equation
1975
6- New SCS sheet flow eqn.
70
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Flood Frequency Analysis:
Example:
Water year Date
1928-29
1929-30
1930-31
1931-32
1932-33
1933-34
1934-35
1934-35
1934-35
1935-36
1935-36
1936-37
1937-38
1937-38
1937-38
1937-38
1938-39
1939-40
1939-40
1939-40
1940-41
1940-41
1940-41
1940-41
1941-42
1941-42
1941-42
1941-42
1942-43
1942-43
71
3 Feb
15 Dec
23 Jan
24 Dec
16 Mar
29 Dec
4 Jan
28 Feb
8 Apr
11 Jan
21 Feb
14 Feb
20 Nov
11 Dec
2 Feb
23 Mar
8 Mar
2 Jan
28 Feb
30 Mar
24 Dec
10 Feb
1 Mar
4 Apr
3 Dec
16 Dec
27 Jan
6 Feb
21 Jan
8 Mar
Flow cfs Max. flow in
a single year
1520
1520
6000
6000
1500
1500
5440
5440
1080
1080
2630
2630
4010
4010
3190
3040
3930
4380
4380
3310
3310
4700
23000
23000
5050
4950
1260
1260
4600
11400
11400
9360
6240
12200
12200
4250
7260
4130
6910
5450
11000
11000
6450
6970
6970
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Water year Date
1943-44
1944-45
1945-46
1945-46
1946-47
1947-48
1947-48
1948-49
1949-50
1950-51
1950-51
1950-51
1951-52
1951-52
1951-52
1952-53
1952-53
1953-54
1953-54
1953-54
1954-55
1955-56
1955-56
1955-56
1956-57
1957-58
1957-58
1957-58
1957-58
1957-58
72
4 Mar
5 Feb
4 Dec
21 Dec
12 Feb
23 Mar
28 Apr
11 Mar
4 Feb
16 Nov
22 Jan
11 Feb
1 Dec
26 Dec
1 Feb
9 Jan
27 Apr
17 Jan
13 Feb
4 Apr
11 Nov
22 Dec
7 Jan
22 Feb
24 Feb
26 Jan
12 Feb
24 Feb
21 Mar
2 Apr
Flow cfs Max. flow in
a single year
3220
3220
3230
3230
3660
6180
6180
4070
4070
7320
7320
3380
3870
3870
4430
4430
3870
3510
3870
3660
4930
5280
5280
4650
7710
7710
3070
4910
4910
3300
4240
2480
2480
9180
9180
5020
6480
6140
6140
3060
4330
6880
6880
4540
3970
Prepared by Amr A. El Sayed, aelsayed@vt.edu
No. Max. flow
in descending order
1
23000
2
12200
3
11400
4
11000
5
9180
6
7710
7
7320
8
6970
9
6880
10 6180
11 6140
12 6000
13 5440
14 5280
15 4910
16 4430
17 4380
18 4070
19 4010
20 3870
21 3870
22 3310
23 3230
24 3220
25 2630
26 2480
27 1520
28 1500
29 1260
30 1080
174470
Plotting
Position
1.67
5.00
8.33
11.67
15.00
18.33
21.67
25.00
28.33
31.67
35.00
38.33
41.67
45.00
48.33
51.67
55.00
58.33
61.67
65.00
68.33
71.67
75.00
78.33
81.67
85.00
88.33
91.67
95.00
98.33
PE (The expedience probability):
1- According to Weibll
2- According to Hazen
Plotting Position =
73
2m − 1
2N
m
N +1
2m − 1
PE =
2N
PE =
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Statistical Operations
1- Mean value:
Q=
1 N
∑ Qi
N i =1
Where:
N
No. of years in the sample.
2- Standard deviation:
SQ =
(
)
2
1 N
Qi − Q
∑
N − 1 i =1
Another simpler form of standard deviation:
SQ =
2⎞
1 ⎛N 2
⎜⎜ ∑ Qi − N .Q ⎟⎟
N − 1 ⎝ i =1
⎠
3- Mean log:
( )
x = “mean log10 of Qi = log10 (Qi ) ≠ log10 Q
1
x = * [log10 (Q1 ) + log10 (Q2 ) + log10 (Q3 ) + .................. + log10 (QN )]
N
or:
xi = log10 (Qi ),
1 N
x = ∑ xi
N i =1
and
4- Standard deviation of logs of Q:
Sx =
2⎞
1 ⎛N 2
⎜⎜ ∑ xi − N x ⎟⎟
N − 1 ⎝ i =1
⎠
5- Skew coefficient of xi
gx =
N
(N − 1)(N − 2 )(Sx )
74
N
(
* ∑ xi − x
3
i =1
)
3
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Example:
For the previous table:
Q=
1
* 174470 = 5815.67 ≅ 5816 cfs
30
SQ =
2⎞
1 ⎛N 2
⎜⎜ ∑ Qi − N .Q ⎟⎟
N − 1 ⎝ i =1
⎠
SQ =
1
2
1569033500 − 30 * (5815.67 ) = 4372.225 cfs
30 − 1
(
)
1 N
x = ∑ xi
N i =1
x=
1
* 109.9673 = 3.67
30
4- Standard deviation of logs of Q:
Sx =
2⎞
1 ⎛N 2
⎜⎜ ∑ xi − N x ⎟⎟
N − 1 ⎝ i =1
⎠
Sx =
1
(405.7572 − 30 * 3.67 ) = 0.303
30 − 1
5- Skew coefficient of xi
gx =
gx =
N
(N − 1)(N − 2 )(Sx )
30
N
i =1
(30 − 1)(30 − 2 )(0.303)3
75
(
* ∑ xi − x
3
)
3
* (− 0.12425) = -0.165
Prepared by Amr A. El Sayed, aelsayed@vt.edu
1- Normal Distribution to flood flows:
Fit standard normal distribution to annual series, use Cumulative Distribution Function CDF
table (table D.3.1)
F(z)
8
z
8
-
z
1
−z2 2
F (z ) =
e
.dz
2π −∫∞
QTR = Q + K TR * SQ
“General frequency factor method for QTR”
Where:
flood discharge in a return period of (TR).
factor determined from standard normal table “SND”.
QTR
KTR
Ex:
Q = 5816 cfs
SQ = 4372 cfs
x = 3.67
Sx = 0.303
Find:
1- Q2 (the flood discharge of a return period of 2 years)
2- Q
Solution:
QTR = Q + K TR * SQ
Q2 = Q + K 2 * SQ
To find K2 use (cumulative probability of the standard normal distribution table), and note
that the value of (z) in the table will equal to the non-exceeding probability of (K).
76
Prepared by Amr A. El Sayed, aelsayed@vt.edu
For example, if (K = 2),
PE =
1
1
, non-exceeding probability = 1 −
= 1- 0.50 = 0.50
K
K
You will find that the value of (z) corresponding to (0.50) is (0.0), so K2 = 0.0
Q2 = Q + K 2 * SQ
Q2 = 5816 + 0.0 * 4372 = 5816 cfs.
K=5
Non-exceeding probability = 1 – 1/5 = 0.80
You will find that the value of (z) corresponding to (0.80) is between (0.84, and 0.85)
You can find K5 by linear interpolation between (0.7995, and 0.8023)
0.80
0.04 0.001785714 0.05
0.7995
0.8023
0.8
K5 = 0.8+0.04+0.001785714 = 0.8418
Q5 = 5816 + 0.8418 * 4372 = 9496 cfs.
K = 10
K10
Non-exceeding probability = 1 – 1/10 = 0.90
You will find that the value of (z) corresponding to (0.90) is between (1.28, and 1.29)
you can find K10 by linear interpolation between (0.8997, and 0.9015)
1.20
0.08
0.8997
0.001666667
0.9
0.09
0.9015
K10 = 1.20 + 0.80 + 0.0016667 = 1.2816
Q10 = 5816 + 1.2816 * 4372 = 11,419 cfs.
77
Prepared by Amr A. El Sayed, aelsayed@vt.edu
z
0.00
0.0
0.1
0.2
0.3
0.4
0.500
0.01
0.02
0.03
0.04
0.5
0.6
0.7
0.8
0.9
0.05
0.06
0.07
0.08
0.09
0.7995 0.8023
1.0
1.1
1.2
1.3
1.4
0.8997 0.9015
1.5
1.6
1.7
1.8
1.9
0.9798 0.9803
2.0
2.1
2.2
2.3
2.4
0.9898 0.9901
K = 50
K50
Non-exceeding probability = 1 – 1/50 = 0.98
You will find that the value of (z) corresponding to (0.98) is between (2.05, and 2.06)
You can find K50 by linear interpolation between (0.9798, and 0.9803)
2.0
0.05
0.9798
0.004
0.98
0.06
0.9803
K50 = 2.0 + 0.05 + 0.004 = 2.054
Q50 = 5816 + 2.054 * 4372 = 14,796 cfs
78
Prepared by Amr A. El Sayed, aelsayed@vt.edu
K = 100
K100
Non-exceeding probability = 1 – 1/50 = 0.99
You will find that the value of (z) corresponding to (0.98) is between (2.32, and 2.33)
You can find K100 by linear interpolation between (0.9898, and 0.9901)
2.3
0.02
0.9898
0.0067
0.98
0.03
0.9901
K100 = 2.30 + 0.02 + 0.0067 = 2.3267
Q50 = 5816 + 2.3267 * 4372 = 15,988 cfs
Return period
TR (years)
2
5
10
50
100
Q
5816
KTR
QTR = Q + K TR * SQ
0.0
0.8418
1.2816
2.054
2.3267
5816
9496
11,419
14,796
15,988
Plot the data on a special probability graph.
Example:
Calculate Q1.5 by using the Cumulative Distribution Function, and the following results from a
sample size = 30 years:
Q = 5816 cfs
SQ = 4372 cfs
x = 3.67
Sx = 0.303
Solution:
For K1.5, PE = 1 / 1.50 = 66.667 %
Non-exceeding probability = 1 – 0.66667 = 0.333
The smallest value in the CDF table is 0.50
79
Prepared by Amr A. El Sayed, aelsayed@vt.edu
2- Log Normal Distribution
The lognormal probability density function (PDF) analysis of flood flow:
xTR = x + K TR * Sx
Where:
xTR
x
Mean log10 of Qi =
KTR
Sx
Standard deviation of logs of Q
Example:
From a field data of flood peaks of a sample equals to 30 years, the following results were
obtained:
Q = 5816 cfs
SQ = 4372 cfs
x = 3.67
Sx = 0.303
gx = -0.165
By using the lognormal distribution, find Q2, Q5, Q10, Q50, and Q100.
Solution:
TR = 2
K2 = 0.0 “As calculated before by using CDF table”
xTR = x + K TR * Sx
x2 = 3.67 + 0.0 * 0.303 = 3.67
Q2 = 103.67 = 4677 cfs
TR = 5
K5 = 0.8418
x5 = 3.67 + 0.8418 * 0.303 = 3.925
Q5 = 103.925 = 8415 cfs
TR = 10
K10 = 1.2816
x10 = 3.67 + 1.2816 * 0.303 = 4.058
Q10 = 104.058 = 11,437 cfs
80
Prepared by Amr A. El Sayed, aelsayed@vt.edu
TR = 50
K50 = 2.054
x50 = 3.67 + 2.054 * 0.303 = 4.29
Q50 = 104.29 = 19,605 cfs
TR = 100
K100 = 2.3267
x100 = 3.67 + 2.3267 * 0.303 = 4.375
Q100 = 104.375 = 23,713 cfs
x
3.67
Return period
TR (years)
2
5
10
50
100
KTR
xTR = x + K TR * Sx QTR = 10 xTR
0.0
0.8418
1.2816
2.054
2.3267
3.67
3.925
4.058
4.29
4.375
4677
8415
11,437
19,605
23,713
3- The Log-Pearson type III Distribution:
xTR = x + K TR * Sx
where:
Factor depends on the skew coefficient (gx), and the exceeding probability
(PE).
The values of KTR are placed in table (3.4) in the text book.
KTR
Example:
From a field data of flood peaks of a sample equals to 30 years, the following results were
obtained:
Q = 5816 cfs
SQ = 4372 cfs
x = 3.67
Sx = 0.303
gx = -0.165
By using the Log-Pearson type III Distribution find Q2, Q5, Q10, Q50, and Q100.
81
Prepared by Amr A. El Sayed, aelsayed@vt.edu
Solution:
The skew coefficient (gx) is rounded to one decimal digit.
gx = -0.165 ≅ -0.20
TR = 2
PE = 1 / 2 = 50 %
From the table: PE (Percent change) = 50, and Skew coefficient = -0.20, KTR = 0.033
x2 = 3.67 + 0.033 * 0.303 = 3.68
Q2 = 103.68 = 4786 cfs.
TR = 5
PE = 1 / 5 = 20 %
From the table: PE (Percent change) = 20, and Skew coefficient = -0.20, KTR = 0.850
x5 = 3.67 + 0.850 * 0.303 = 3.93
Q5 = 103.93 = 8464 cfs.
TR = 10
PE = 1 / 10 = 10 %
From the table: PE (Percent change) = 10, and Skew coefficient = -0.20, KTR = 1.258
x10 = 3.67 + 1.258 * 0.303 = 4.05
Q10 = 104.05 = 11,251 cfs.
TR = 50
PE = 1 / 50 = 2 %
From the table: PE (Percent change) = 2, and Skew coefficient = -0.20, KTR = 1.945
x50 = 3.67 + 1.945 * 0.303 = 4.26
Q50 = 104.26 = 18,169 cfs.
TR = 100
PE = 1 / 100 = 1 %
82
Prepared by Amr A. El Sayed, aelsayed@vt.edu
From the table: PE (Percent change) = 1, and Skew coefficient = -0.20, KTR = 2.178
x100 = 3.67 + 2.178 * 0.303 = 4.33
Q100 = 104.33 = 21,376 cfs.
x
3.67
Return period
TR (years)
2
5
10
50
100
KTR
xTR = x + K TR * Sx QTR = 10 xTR
0.033
0.850
1.258
1.945
2.178
3.68
3.93
4.05
4.26
4.33
4786
8464
11,251
18,169
21,376
4- The Gumbel (Extreme Value Type I) Distribution:
Sample
Size
15
20
25
30
40
50
60
70
75
100
10
1.703
1.625
1.575
1.541
1.495
1.466
1.446
1.430
1.423
1.401
1.305
83
Gumbel Extreme-Value Frequency Factors
Recurrence Intervals
20
25
50
75
2.410
2.632
3.321
3.721
2.302
2.517
3.179
3.583
2.235
2.444
3.068
3.463
2.188
2.393
3.026
3.3393
2.126
2.326
2.943
3.301
2.086
2.283
2.889
3.241
2.059
2.253
2.852
3.200
2.038
2.230
2.824
3.169
2.029
2.220
2.812
3.155
1.998
2.187
2.770
3.109
1.866
2.044
2.592
2.911
100
4.005
3.836
3.729
3.653
3.554
3.491
3.446
3.413
3.400
3.394
3.137
Prepared by Amr A. El Sayed, aelsayed@vt.edu