Quantitative Hydrology
Basin Recharge and Runoff
As rain fall towards the earth, a portion of it is
intercepted by the leafs and stems of
vegetation.
The water so retained, interception, together
with depression storage and soil moisture,
constitutes basin recharge.
Interception
Interception is that portion of precipitation
which, while falling, is intercepted by aerial
portion of vegetation, buildings and other
objects above the surface of earth and
evaporates back to the atmosphere.
The adhesive force between the water drops
and the vegetation holds back the drops of
water against gravity until they grow in size to
over weigh and slip down.
Interception
Factors on which interception depends are:
(a) Intensity and duration of the storm,
(b) Density of trees,
(c) Type of trees (i.e. long or short, coniferous or
deciduous, area of canopy) and other
obstructions,
(d) Season of the year and
(e) Wind velocity at the time of precipitation.
Interception
The part of precipitation retained by the aerial
portion of vegetation and other objects and is
either absorbed by them or evaporates back is
called interception loss.
Depending on the vegetal cover and precipitation,
a general form of equation for interception loss
can be written as:
IL = aP + b (1 – e (-p / b)) where
Interception
a and b are constants which depend on the
factors of infiltration loss, a varies between
0.01 and 0.2 and b between 2.5 and 38% of
rainfall
P is the precipitation depth in mm.
Interception
Example: Calculate the interception loss
for a tropical forest area in India (a = 0.11
and b = 15% of P). The intensity of
precipitation is low but uniform over the
area. Total depth of storm rainfall is 4 cm.
Interception
Solution: Here a = 0.11, b = 0.15P, P = 4.0
cm
Interception loss IL = aP + b (1 – e (-p / b))
IL = 0.11P + 0.15 P (1 – e (-p/0.15P)) = 0.2598
P = 1.039 cm = 10.4 mm.
Depression Storage
Any natural ground surface generally has
numerous shallow depressions of varying
size, shape and depth.
When precipitation fall, these depressions
form miniature reservoirs detaining water
temporarily. Water from these storages either
evaporates or infiltrates into the ground
charging the ground water reservoir.
Depression Storage
After filling all the small depressions,
overland flow from the area takes place. Due
to change in land use pattern, depression
storages also change with time, which makes
it almost a non-measurable quantity.
Depression Storage
The important factors affecting depression
storage are:
(a) land form,
(b) soil characteristics,
(c) topography,
(d) antecedent precipitation index and
(e) man made disturbances, like terrace
farming.
Depression Storage
Depression storage helps to reduce soil
erosion and increase soil moisture content.
Therefore, farmers are encouraged to go for
terrace farming to conserve soil and rain
water for beneficial uses.
The sum of infiltration and depression
storage can vary from 10 to 50 mm per storm
depending on their intensity, duration and
other characteristics.
Infiltration and Its Estimation
Infiltration is as the entry or the passage of
water into the soil through soil surface. It is
a major loss of precipitation affecting runoff
of a basin.
Infiltration and Its Estimation
This term should be properly understood and
quantified. Losses like interception,
depression storage and evaporation during
precipitation are small, which cannot change
the runoff of a basin significantly during
major floods, but infiltration is a major
process continuously affecting the magnitude,
timing and distribution of surface runoff at
any measured outlet of a basin. It is
responsible for the growth and nourishment
of life on earth.
Infiltration and Its Estimation
Infiltration process is initiated by creation of
hydrogen bond between soil particles and the
water. The adhesive force of attraction
between soil and water, the surface tension,
capillarity and gravitational forces help to
force more water between the pores of soil
particles as more water is added to the system
due to rain.
Infiltration and Its Estimation
Infiltration capacity is the maximum rate at
which a given soil can absorb water under a
given set of conditions at a given time.
At any instant the actual infiltration, can be
equal to infiltration capacity only when the
rainfall intensity is greater than infiltration
capacity, otherwise actual infiltration will be
equal to the rate of rainfall. This can be
observed during low intensity rainfall when
there is no surface runoff produced due to
precipitation.
Infiltration and Its Estimation
Once water enters into the soil, the process of
transmission of water within the soil known
as percolation takes place, thus removing the
water from near the surface to down below,
charging the ground water reservoir.
Infiltration and Its Estimation
Infiltration and percolation are directly
interrelated. When percolation stops,
infiltration also stops. During any storm,
infiltration is the maximum at the beginning
of the storm, decays exponentially and
attains a constant value as the storm
progresses.
Infiltration and Its Estimation
The effect of infiltration is to:
1. reduce flood magnitude
2. delay the time of arrival of water to the
channel
3. reduce the soil erosion
4. recharge to the ground water reservoir
5. fill the soil pores with water to its fieldcapacity, which subsequently supply
water to the plants
6. avail the ground water during the nonrain periods in the channels
7. help to supply water to plants
Factors Affecting Infiltration
Factors affecting infiltration depend on both
meteorological and soil medium
characteristics. These are:
Surface Entry: If a soil surface is bare, the
impact of raindrops causes in-washing of
finer particles and clogs the surface. This
retards infiltration. An area covered by grass
and other bushy plants has better infiltration
capacity than a barren land.
Factors Affecting Infiltration
Percolation: For infiltration to continue,
water that has entered the soil must be
transmitted down by the force of gravity and
capillary actions. When percolation rate is
slow, the infiltration rate is bounded by the
rate of percolation. This depends on the
factors like type of soil, its composition,
permeability, porosity, stratification,
presence of organic matter and presence of
salts.
Factors Affecting Infiltration
Antecedent Moisture Condition: Infiltration
depends on the presence of moisture in the
soil. For the second storm in succession, the
soil will have lesser rate of infiltration than
the first maiden storm of the season. Except
sandy soil most other soils have swelling
ingredients, which swells in presence of
water and reduce infiltration rate to the
extent of their presence
Factors Affecting Infiltration
Rainfall Intensity and Duration: During
heavy rainfall, the top soil is affected by
mechanical compaction and by the in-wash
of finer materials. This leads to faster
decrease in the rate of infiltration than with
low intensities of rainfall. Duration of rain
affects to the extent that when the same
quantity of rain falls in n number of isolated
storms instead of a continuous one, the
infiltration will be higher in the former case
Factors Affecting Infiltration
Human Activities: When crops are grown or
grass covers a barren land, the rate of
infiltration is increased. On the other hand
construction of roads, houses, overgrazing of
pastures and playgrounds reduce infiltration
capacity of an area considerably
Factors Affecting Infiltration
Depletion of Ground Water Table: Position
of ground water table should not be very
close to the surface for infiltration to
continue. The quantity of infiltrated water
entering into the soil should be drained out
fully from the top soil zone so that there is
some space available for infiltrated water to
store during the next rain
Factors Affecting Infiltration
Quality of Water: Water containing silt, salts
and other impurities affect the infiltration to
the extent they are present. Salts present
affect the viscosity of water and may also
react chemically with soil to form complexes
which obstruct the porosity of soil, thereby
affecting infiltration. Silts clog the pore
spaces retarding infiltration rate
considerably
Field Measurement Using Infiltrometers
Two types of infiltrometers used are:
1. Single cylindrical and
2. Concentric double cylindrical.
Single Tube lnfiltrometer
It consists of a hallow metal cylinder 30 cm
in diameter and 60 cm long driven into the
ground such that 10 cm of it projects above
ground level. Water is poured at the top such
that a head of 7 cm within the infiltrometer
is maintained above ground level. A
graduated jar or burette is used to add water,
to give directly the volume of water added
over time.
Single Tube lnfiltrometer
A plot of time in abscissa against rate of water
added in mm/h gives an infiltration capacity
curve for the area. The setup resembles to the
flooding type of irrigation situation in the field
which can be a possible representation of real
local conditions. Sufficient precautions should
be taken to drive the cylinder into the ground
with minimum disturbance to the soil structure.
In a single infiltrometer, the major criticism is
that water spreads out immediately beyond the
bottom of the cylinder which does not represent
a true infiltration condition of the field
Double Tube lnfiltrometer
To overcome the objections of a single ring
infiltrometer a set two concentric hollow
cylinders of same length are used. Water is
added to both the rings to maintain the same
height. Reading of the burette for the inner
cylinder is taken as infiltration capacity of the
soil. The outer cylinder is maintained to
prevent spreading of water from the inner one.
Double Tube lnfiltrometer
The important disadvantages still prevalent in
these types of infiltrometers are:
1. The size effect: Larger diameter
infiltrometers give more accurate and
always lesser value of infiltration than
smaller diameter type.
2. Boundary effect.
3. Disturbance of original soil due to driving
of the rings.
Field Measurement Using Infiltrometers
Example: From a double tube infiltrometer
with inside ring diameter of 30 cm, the
following observations were taken. Plot the
infiltrations capacity curve and find the
constant rate of infiltration that the
experimental field have towards the end.
Time (min)
0
2
5
10
20
30
50
80
120
150
Cumulative volume
(ml)
0
130
280
510
680
900
1040
1190
1280
1343
Solution
Area of the infiltrometer = (/4) 302 = 706.5 cm2
Rainfall-Runoff Analysis
A network of rain gauges gives the average
storm precipitation over a basin. The runoff
from the basin can be measured at the outlet or
at the desired location by gauging the stream.
The difference between precipitation and the
corresponding runoff averaged over the basin
over time should give the total loss. This should
form the basis for the estimation of infiltration.
Rainfall-Runoff Analysis
During storms, evapotranspiration is
negligible. Depression storage and
interception can be estimated as discussed
earlier. The balance gives a true
representation of the basin infiltration losses..
Rainfall-Runoff Analysis
The infiltration capacity of a soil at any time is
the maximum rate at which water will enter the
soil.
The infiltration rate is the rate at which water
actually enters the soil during a storm; and it
must equal the infiltration capacity or the
rainfall rate, whichever is less.
An ideal infiltration capacity curve proposed by
Horton (1933) is
ft = fc + (fo - fc) e-kt
Rainfall-Runoff Analysis
where
ft = the infiltration capacity at any time t from
the beginning of the storm in mm/h.
fc = the infiltration rate in mm/h at the final
steady stage when the soil profile becomes fully
saturated
fo = the maximum initial value when t = 0 in
mm/h at the beginning of the storm
k = empirical constant depending on soil cover
complex, vegetation and other factors
t = time lapse from the onset of the storm
A typical curve of ft separating the rainfall
intensity histogram between infiltration and
surface runoff is
Rainfall-Runoff Analysis
Since we are interested in determining the
infiltration rate for storms of large basins, an
average infiltration rate known as infiltration
index is assumed. This type of assumption
underestimates infiltration rate during the initial
part of the storm and slightly overestimates
towards the end of it.
Rainfall-Runoff Analysis
The assumption holds fairly good for analysis of
high intensity and long duration storms or in a
situation, when the area is fairly saturated
before the onset of the storm.
Difficulties with the theoretical approach to
infiltration and practical difficulties to
determine the values of fo and k, led to the use
of infiltration indices based on an empirical
approach.
Infiltration Indices
The following indices are mostly used for
computation of infiltration rate from rainfall
runoff data:
1.-Index
2.W-Index
Infiltration Indices/ -Index
The average rate of rainfall above which the
rainfall volume equals to runoff volume is
called -Index.
=I–R
in which
R = a I1.2
I = the rainfall intensity in mm/h on a daily (24
h) basis
R = the runoff in mm.
Infiltration Indices/ -Index
For computation of R, take I in mm/day and for
computation of , divide R by 24 to convert it
to mm/h. Values of coefficient a vary from 0.17
to 0.50. For sandy soils it may be taken as 0.20,
coastal alluvium 0.25 to 0.30, silt 0.35, red and
clayey soil 0.40, black cotton soil 0.45 and hilly
soil 0.50.
Infiltration Indices/ -Index
Example: Calculate -index of a storm from the
following data. Plot the rainfall histogram and mark
the -index on the plot. Assume losses due to
interception, depression storage and evaporation is 2
mm.
Volume of direct-runoff after separation of base flow
= 10.75 M.m3
Runoff started at 3.00 pm on 17/08/98
Time of
15 18 21 24 03
rainfall (h)
Depth of
1.2 1.5 0.9 2.2 0.2
rainfall
(cm)
Infiltration Indices/ -Index
Solution :
.
(10.75 X 106 x 100)
Depth of direct runoff = ------------------------ = 2.5 cm = 25 mm
(430 x 10 6 )
Rainfall has occurred for 15 h between 15th hour of 17.8.98 to 6th hour of 18.8.98.
Period of rainfall excess is from 15 h to next day 3 AM = 12 h.
Second trial
Rainfall excess
Precipitation Precipitation
Time
 = 2.75 mm/h
depth
intensity
for  = 2.8mm/h
Period
Rainfall excess
(mm)
(mm/h)
col. (3) - 2.8*3= mm
=col. (3) - 2.75 * 3= mm
(1)
(2)
(3)
(4)
(5)
15-18
12
4
3.6
3.75
18-21
15
5
6.6
6.75
21-24
9
3
0.6
0.75
0-3
22
7.33
13.6
13.75
3-6
2
0.67
Total
60 mm
24.4 mm
25 mm
So, -index = 2.75 mm/h
Total loss (12 x 2.75 + 2) = 35 mm of rainfall during the storm, which is the sum of all
losses due to infiltration, depression storage, interception and evaporation during 12 h of
storm period.
Infiltration Indices/ -Index
Storm histogram and  -index.
Infiltration Indices/ W-Index
This index is considered as an improvement
over -index in the sense that surface storage
and interception losses are considered in its
computation. It is defined as the average rate of
infiltration which equals to the rate of
precipitation minus surface runoff and retention
during time t and is expressed as:
W = (P - SRo - DR) / t
Infiltration Indices/ W-Index
where
P = the total depth of storm rainfall in mm
SRo = the total depth of surface runoff in mm
DR = the total depth of surface retention
(depth of depression storage plus
interception loss) mm and
t = the time in hour during which rainfall rate
exceeds infiltration rate.
Obviously when DR = 0, W-index and index become the same.
Infiltration Indices/ W-Index
Example: the following are the rates of
rainfall for successive 20 minutes period of a
140 minutes storm: 2.5, 2.5, 10.0, 7.5, 1.25,
1.25, 5.0 cm/hr. taking the value of -index
as 3.2 cm/hr, find out the runoff in cm, the
total rainfall, and the value of W-index.
Runoff = (10-3.2)20/60 + (7.5-3.2)20/60 + (5-3.2)20/60 =
(6.8+4.3+1.8)20/60 = 4.3 cm
Total precipitation = (2.5*20/60) + (2.5*20/60) + (10*20/60) +
(7.2*20/60) + (1.25*20/60) + (1.25*20/60) + (5*20/60)
= (2.5+2.5+10+7.2+1.25+1.25+5)20/60 = 10 cm
W-index = (P - SRo - DR) / t
(10-4.3)140/60 = 2.44 cm/hr
Hydrograph Analysis
The characteristics of direct and groundwater
runoff differ so greatly that they must be
treated separately in problems involving
short-period, or storm, runoff. There is no
practical means of differentiating between
groundwater flow and direct runoff after they
have been intermixed in the stream, and the
techniques of hydrograph analysis are
arbitrary.
Hydrograph Analysis
The typical hydro graph resulting from a
single storm consists of a rising limb, peak,
and recession. The recession represents the
withdrawal of water stored in the stream
channel during the period of rise. Double
peaks are sometimes caused by the geography
of the basin but more often result from two or
more periods of rainfall separated by a period
of little or no rain.
Hydrograph Analysis
Numerous methods of hydrograph separation
have been used. The method illustrated by
ABC in Figure is simple and as easily
justified as any other. The recession of flow
existing prior to the storm is extended to
point B under the crest of the hydrograph.
The straight line BC is then drawn to intersect
the recession limb of the hydrograph N days
after the peak.
Hydrograph Analysis
The value of N is not critical and may be selected
arbitrarily by inspection of several hydrographs
from the catchment. The selected value should,
however, be used for all storm events analyzed to
conform to the unit hydrograph concept. The
time N will increase with size of drainage basin
since a longer time is required for water to drain
from a large basin than from a small one. A
rough guide to the selection of N (in days) is
N = Ad0.2
Where Ad is the drainage area in square miles.
With Ad in square kilometers, computed values
of N should be reduced by about 20 percent.
Unit Hydrographs
If two identical rainstorms could occur over
a catchment with identical conditions prior
to the rain, the hydrographs of runoff from
the two storms would be expected to be the
same. This is the basis of the unit
hydrograph concept.
Unit Hydrographs
Actually the occurrence of identical storms
is very rare. Storms may vary in duration,
amount, and aerial distribution of rainfall. A
unit hydrograph is a hydrograph with a
volume of 1 in. (25 mm) of direct runoff
resulting from a rainstorm of specified
duration and aerial pattern. Hydrographs
from other storms of like duration and
pattern are assumed to have the same time
base, but with ordinates of direct runoff in
proportion to the runoff volumes.
Unit Hydrographs
A unit hydrograph may be constructed from
the rainfall and stream flow data of a storm
with reasonably uniform rainfall intensity
and without complications from preceding or
subsequent rainfall. The first step in the
derivation is the separation of groundwater
flow from direct runoff. The volume of
direct runoff (area ABCD, Figure) is
determined and the ordinates of the unit
hydrograph are found by dividing the
ordinates of the direct runoff by the volume
of direct runoff in inches (or cm). The
resulting unit hydrograph should represent a
unit volume (1 in.) of runoff, or 1 cm in
metric units.
Unit Hydrographs
Example: Derive a unit hydrograph from the
flows indicated by the previous Figure.
Solution: Since a unit hydro graph applies to
direct runoff, the first step is the estimation
of this quantity. The groundwater
hydrograph (i.e., base flow) is assumed to
follow the recession occurring before the
storm to a point under the peak of the total
hydrograph (point A to point B). From B the
base flow is assumed to increase slowly to
point C. In this case the location of point C
was chosen arbitrarily.
Unit Hydrographs
The table that follows illustrates the other
steps in the process. After entering the date
and time, the total flow is tabulated in the
third column and the corresponding base
flow is entered in column 4. Subtracting
base flow from total flow gives the direct
runoff values (column 5). Summing the
direct runoff ordinates gives the total direct
runoff, which must be converted to inches of
depth over the 25-mi2 catchment:
11,970 x (3/24) = 1496 cfs-days
Unit Hydrographs
In this calculation it is assumed that each
entry in the table represents an average flow
for 3 hr, that is, 3/24 day. Summing these
flows and multiplying by 3/24 gives the
volume of runoff for the storm in cfs-days.
There are 26.9 cfs-days in 1 in. of runoff
from 1 mi2. Hence the volume of direct
runoff in inches over the 25-mi2 catchment is:
1496/(25 x 26.9) = 2.22 in.
Unit Hydrographs
Dividing each ordinate of direct runoff by
2.22 gives the ordinates of the unit
hydrograph (column 6). The final step is the
assignment of an effective storm duration
from a study of the rainfall records.
Base
flow
Direct
runoff
Ordinates of
unit
hydrograph
(derived)
Hours
after
start
Date
Hour
Total
flow
(given)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
14
0500
470
470
0
0
0
0800
1200
440
760
342
3
1100
2250
410
1840
829
6
1400
2920
380
2540
1145
9
1700
2670
400
2270
1022
12
2000
2060
410
1650
743
15
2300
1430
420
1010
455
18
0200
1100
430
670
302
21
0500
910
440
470
212
24
0800
780
450
330
149
27
1100
680
460
220
99
30
1400
600
470
130
59
33
1700
540
480
60
27
36
2000
510
490
20
9
39
2300
500
500
0
0
42
15
Total
11 ,970
Unit Hydrographs
Data from at least one recording rain gage is
necessary. Periods of low-intensity rain at
the beginning and end of the storm should be
ignored if they did not contribute
substantially to the total runoff. In this case
the effective storm duration is 6 hr, and the
unit hydrograph is called a 6 hr unit
hydrograph.
The use of a unit hydrograph to estimate the
hydrograph of a storm of like duration is
illustrated in the following example.
Unit Hydrographs
Example: A storm occurs between 0400 and
1000 hours. The estimated depth of direct
runoff is 1.5 in. Construct the hydrograph to
be expected from this storm.
Solution: Assume the initial flow in the
stream is 600 cfs. All flows in the table that
follows are in cubic feet per second.
.
Date
Hour
Unit
hydrograph
Direct
runoff
Base
flow
Total
flow
9
0400
0
0
600
600
0700
342
513
570
1083
1000
829
1244
530
1774
1300
1145
1718
500
2218
1600
1023
1535
500
2035
1900
743
1115
500
1615
2200
455
682
510
1192
0100
302
453
520
973
0400
212
318
530
848
0700
149
224
540
764
1000
99
148
550
698
1300
59
88
560
648
1600
27
40
570
610
1900
9
14
580
594
10
Unit Hydrographs
The flow occurring prior to the storm serves as
a starting point for the line ABC representing
the base flow or estimated groundwater flow.
In this example it has been assumed to
decrease slowly to the time of peak and then to
rise slowly to meet the estimated direct runoff
33 hr after the peak. The ordinates of the unit
hydrograph are taken from previous example
and multiplied by the estimated depth of direct
runoff to generate the hydrograph of direct
runoff. The direct runoff is added to the ground
water flow to obtain the total hydrograph
(ADC). The direct runoff was estimated by one
of the methods discussed earlier in this course.
Unit Hydrographs
The number of unit hydrographs for a given
catchment is theoretically infinite since there
could be one for every possible duration of
rainfall and every possible distribution pattern.
Practically there need be only a few relatively
short durations considered, since these short
durations can be used to build a hydrograph
for a longer duration .
Unit Hydrographs
The effect of varying aerial patterns of rainfall
can be minimized by restricting the use of unit
hydrographs to relatively small catchments.
An area of 2000 mi2 (5000 km2) is often taken
as an upper limit. The effect of exceeding this
limit will decrease the accuracy of computed
hydrographs. Where rainfall is typically in the
form of showers or thunderstorms covering
small areas, the unit hydrograph is applicable
only to very small catchments.
Unit Hydrographs
Hydrographs for larger catchments can be
estimated by dividing the area into subcatchments and summing the flows from these
sub-catchments using routing techniques.
The application of a 3-hr unit hydrograph to a
storm of 12 hr duration 6 illustrated in the
following example.
Unit Hydrographs
Example: Develop the hydrograph of direct
runoff from a 12-hr storm on a given
catchment whose 3-hr unit hydrograph is
given in the first two columns of the following
table. The 12-hr storm occurs in four 3-hr
periods having estimated runoffs of 0.15, 0.50,
1.25, and 1.75 in.
Solution: The computations are illustrated in
the following table. Base flow is ignored.
Flows are in 1000 cfs.
0
3-hr
unit
hydrograph
0
0.15
0
1
2.5
0.4
0
2
9.0
1.4
3
10.0
4
Time
hr
Runoff per period
0.50
1.25
0
0
Total
1.75
0
0
0
0
0.4
0
0
0
1.4
1.5
0
0
0
1.5
8.5
1.3
1.2
0
0
2.5
5
5.2
0.8
4.5
0
0
5.3
6
2.4
0.4
5.0
0
0
5.4
7
1.2
0.2
4.2
3.1
0
7.5
8
0.6
0.1
2.6
11.2
0
13.9
9
0.3
0
1.2
12.5
0
13.7
10
0.1
0
0.6
10.6
4.4
15.6
11
0
0
0.3
6.5
15.8
22.6
12
0
0
0.2
3.0
17.5
20.7
13
0
0
0
1.5
14.9
16.4
14
0
0
0
0.8
9.1
9.9
15
0
0
0
0.4
4.2
4.6
16
0
0
0
0.1
2.1
2.2
17
0
0
0
0
1.0
1.0
18
0
0
0
0
0.5
0.5
19
0
0
0
0
0.2
0.2
Unit Hydrographs
The 3-hr unit hydro graph was developed
through analysis of several 3-hr storms using
the procedure illustrated in the previous
example. The depth of direct runoff for each
3-hr period of the storm is estimated by
subtracting an estimate of the infiltration from
the rainfall during the period. The unit
hydrograph ordinates are multiplied by 0.15,
0.50, 1.25, and 1.75, respectively, each lagged
by 3 hr from the previous increment. The
resulting values are summed to give the total
hydrograph of composite storm.
Example: A 4 h hydrograph for a Basin is given below.
Calculate (i) A 12-h unit hydrograph and (ii) 2-h UH by
S-hydrograph approach.
Solution: Required S- hydrograph from the given 4-h.UH
is calculated in the following table.
Synthetic Unit Hydrograph
The unit hydrograph from rainfall and stream
flow data on a watershed applies only for that
watershed and for the point on the stream
where the stream flow data were measured.
Synthetic unit hydro graph procedures are used
to develop unit hydrographs for other locations
on the stream in the same watershed or for
nearby watersheds of a similar character.
Synthetic Unit Hydrograph
There are three types of synthetic unit hydrographs:
(1) Those relating hydro graph characteristics
(peak flow rate, base time, etc.) to watershed
characteristics (Snyder, 1938; Gray, 1961),
(2) Those based on a dimensionless unit
hydrograph (Soil Conservation Service, 1972), and
(3) Those based on models of watershed storage
(Clark, 1943).
Snyder's Synthetic Unit Hydrograph
In a study of watersheds located mainly in the
Appalachian highlands of the United States,
and varying in size from about 10 to 10,000
mi2 (30 to 30,000 km2), Snyder (1938) found
synthetic relations for some characteristics of
a standard unit hydro graph [Figure below].
Snyder's Synthetic Unit Hydrograph
From the relations, five characteristics of a required
unit hydro graph [Figure below] for a given excess
rainfall duration may be calculated:
•the peak discharge per unit of watershed area, qpR,
•the basin lag tpR (time difference between the
centroid of the excess rainfall hyetograph and the
unit hydro graph peak),
•the base time tb, and
•the widths W (in time units) of the unit hydro graph
at 50 and 75 percent of the peak discharge.
Using these characteristics the required unit hydro
graph may be drawn. The variables are illustrated
in Figure below.
Snyder's Synthetic Unit Hydrograph
Snyder defined a standard unit hydrograph as
one whose rainfall duration tr, is related to the
basin lag tp by
tp = 5.5 tr
For a standard unit hydrograph he found that:
1. The basin lag is
tp = C1Ct (L Lc )0.3
Snyder's Synthetic Unit Hydrograph
where tp is in hours, L is the length of the
main stream in kilometers (or miles) from the
outlet to the upstream divide, Lc is the
distance in kilometers (miles) from the outlet
to a point on the stream nearest the centroid
of the watershed area, Cl = 0.75 (1.0 for the
English system), and Ct is a coefficient
derived from gauged watersheds in the same
region.
Snyder's Synthetic Unit Hydrograph
2. The peak discharge per unit drainage area in m
3/s'km2 (cfs/mi 2) of the standard unit
hydrograph is
qp = C2Cp / tp
where C2 = 2.75 (640 for the English system)
and Cp is a coefficient derived from gauged
watersheds in the same region.
Snyder's Synthetic Unit Hydrograph
To compute Ct and Cp for a gauged watershed,
the values of L and Lc are measured from the
basin map. From a derived unit hydrograph of
the watershed are obtained values of its effective
duration tR in hours, its basin lag tpR in hours,
and its peak discharge per unit drainage area, qpR,
in m3/s.km2.cm (cfs/mi2.in for the English
system). If tpR = 5.5tR, then tR = tr, tpR = tp, and
qpR = qp, and Ct and Cp are computed by
equations mentioned earlier.
Snyder's Synthetic Unit Hydrograph
If tpR is quite different from 5.5tR, the standard
basin lag is
tp = tpR + (tr – tR) / 4
3. The relationship between qp and the peak
discharge per unit drainage area qpR of the required
unit hydrograph is
qpR = (qp tp) / tpR
Snyder's Synthetic Unit Hydrograph
4. The base time tb in hours of the unit
hydrograph can be determined using the fact
that the area under the unit hydrograph is
equivalent to a direct runoff of I cm (1 inch in
the English system). Assuming a triangular
shape for the unit hydrograph, the base time
may be estimated by
tb = C3 / qpR
where C3 = 5.56 (1290 for the English system).
Snyder's Synthetic Unit Hydrograph
5. The width in hours of a unit hydrograph at a
discharge equal to a certain percent of the peak
discharge qpR is given by
W = Cw qpR-1.08
where Cw = 1.22 (440 for English system) for
the 75-percent width and 2.14 (770, English
system) for the 50-percent width. Usually onethird of this width is distributed before the unit
hydrograph peak time and two-thirds after the
peak.
Snyder's Synthetic Unit Hydrograph
Example: From the basin map of a given
watershed, the following quantities are
measured:
L = 150 km, Lc = 75 km, and drainage area =
3500 km2.
From the unit hydrograph derived for the
watershed, the following are determined:
tR = 12 h, tpR = 34 h, and peak discharge =
157.5 m3/s.cm.
Determine the coefficients Ct and Cp for the
synthetic unit hydrograph of the watershed.
Snyder's Synthetic Unit Hydrograph
Solution: From the given data, 5.5tR = 66 h,
which is quite different from tpR (34 h).
tp=tpR + (tr – tR) / 4
=34 +( tr-12) / 4
tr = 5.9 h and tp = 32.5 h
tp = C1 Ct(L Lc)0.3
32.5=0.75Ct(150 x 75)0.3
Ct=2.65
Snyder's Synthetic Unit Hydrograph
The peak discharge per unit area is qpR =
157.5/3500 = 0.045 m3/s.km2.cm.
The coefficient Cp is calculated with qp,qpR, and
tp = tpR:
qpR = (C2Cp) / tpR
0.045 = 2.75Cp / 34.0
Cp=0.56
Snyder's Synthetic Unit Hydrograph
Example: Compute the six-hour synthetic
unit hydrograph of a watershed having a
drainage area of 2500 km2 with L = 100 km
and Lc = 50 km. This watershed is a subdrainage area of the watershed in previous
example.
Snyder's Synthetic Unit Hydrograph
Solution: The values Ct = 2.64 and Cp = 0.56
determined in previous example can also be
used for this watershed. Thus,
tp = 0.75 x 2.64 x (100 x 50)0.3 = 25.5 h
tr = 25.5/5.5 = 4.64 h
Snyder's Synthetic Unit Hydrograph
For a six-hour unit hydrograph, tR = 6 h,
tpR = tp - (tr - tR) / 4 = 25.5 (4.64-6)/4 = 25.8
h.
qp = 2.75 x 0.56/25.5 = 0.0604 m3/s.km2.cm
qpR = 0.0604 x 25.5/25.8 = 0.0597
m3/s.km2.cm
the peak discharge is 0.0597 x 2500 = 149.2
m3/s.cm.
Snyder's Synthetic Unit Hydrograph
The widths of the unit hydrograph are:
At 75 percent of peak discharge, W = 1.22qpR1.08 = 1.22 X 0.0597-l.08 = 25.6 h.
A similar computation gives a W = 44.9 h at
50 percent of peak.
The base time tb = 5.56/qpR = 5.56/0.0597 =
93h.
The hydrograph is drawn, as in Figure below
and checked to ensure that it represents a
depth of direct runoff of 1 cm.
SCS Dimensionless Hydrograph
The SCS dimensionless hydrograph is a
synthetic unit hydrograph in which the
discharge is expressed by the ratio of
discharge q to peak discharge qp and the time
by the ratio of time t to the time of rise of the
unit hydro graph, Tp.
SCS Dimensionless Hydrograph
Given the peak discharge and lag time for the
duration of excess rainfall, the unit hydrograph
can be estimated from the synthetic
dimensionless hydrograph for the given basin.
Figure below shows such a dimensionless
hydrograph, prepared from the unit hydrographs
of a variety of watersheds. The values of qp and
Tp may be estimated using a simplified model
of a triangular unit hydrograph as shown in
Figure below, where the time is in hours and the
discharge in m3/s.cm (or cfs/in) (Soil
Conservation Service, 1972).
SCS Dimensionless Hydrograph
From a review of a large number of unit
hydrographs, the Soil Conservation Service
suggests the time of recession may be
approximated as 1.67 Tp. As the area under the
unit hydrograph should be equal to a direct
runoff of 1 cm (or 1 in), it can be shown that
qp = CA / Tp
where C = 2.08 (483.4 in the English system)
and A is the drainage area in square kilometers
(square miles).
SCS Dimensionless Hydrograph
Further, a study of unit hydrographs of many
large and small rural watersheds indicates that
the basin lag tp  0.6Tc, where Tc is the time of
concentration of the watershed. As shown in
previous Figure, time of rise Tp can be
expressed in terms of lag time tp and the
duration of effective rainfall tr
Tp = (tr/2) + tp
SCS Dimensionless Hydrograph
Example: Construct a 10-minute SCS unit
hydrograph for a basin of area 3.0 km2 and time
of concentration 1.25 h.
Solution: The duration tr = 10 min =0.166 h,
lag time tp = 0.6Tc = 0.6 X1.25 = 0.75 h,
and rise time Tp = (tr / 2) + tp = (0.166/2) +
0.75 = 0.833 h.
qp = CA / Tp = 2.08 x 3.0/0.833 = 7.49
m3/s.cm.
SCS Dimensionless Hydrograph
The dimensionless hydrograph in the previous
figure may be converted to the required
dimensions by multiplying the values on the
horizontal axis by Tp and those on the vertical
axis by qp .
Alternatively, the triangular unit hydrograph
can be drawn with tp = 2.67 Tp = 2.22 h. The
depth of direct runoff is checked to equal 1 cm.