110 Engineering Hycrology

Infiltration Indices
3.20
found convenient to use a cons.

it is
calculations involving floods, infltras
In hydrological duration of the
storm. The defined average ltration
rate for the in common use.
value of infiltration two types of
indices are
index and
rate is called infiltration

3.20.1 ndex
rainfall
the average
The pindex is Runoff
rainfall volume
above which the
volume. The
is equal to the runoff 2.5
index is derived from the rainfall 2.0
of
hyetograph with the knowledge 1.5
The 1.0
the resulting runoff volume,
initial loss is also considered as 0.5 Losses E Index
infiltration. The pvalue is found by 4 6 8 10 12 14
L U 2
infiltration Time (h)
treating it as a constant
capacity. If the rainfall intensity is Fig. 3.16 p-Index
less than p then the infiltration rate
is equal to the rainfall intensity; however, if the rainfall intensity is larger than p, the
difference between the rainfall and infiltration in an interval of time represents the
runoff volume as shown in Fig. 3.16. The amount ofrainfall inexcess of theindexis
called rainfall excess. In connection with runoff and flood studies, it is also known
aseffective rainfal, (details in Sec. 6.5, Chapter 6). The p-index thus accounts for
the total abstraction and enables magnitudes to be estimated for a given rainfall hye-
tograph. A detailed procedure for calculating -index for a given storm hyetograph
and resulting runoff volume is as follows.

Procedure for Calculation of p-index

Consider a rainfall hyetograph of event duration D hours and having N pulses of time
interval Ar such that
N At = D.
(In Fig. 3.16, N = 7)
Let, be the
intensity of rainfall ini" pulse and R= total direct runoff.
N
Total rainfall P E, A
Ifpis the value of p-index then P- p 1,=
where t,= duration of rainfall excess. Ra
If the rainfall
hyetograph and
storm can be determined total runoff depth Rg are given, the p-index of the
by trial-and-error
1. Assume that out
of given N procedure as given below.
pulses, M
number of
number of pulses in pulses
(Note that MSN). Select M have rainfall excess.
intesity decreasing order of rainfall
2. Find the value of
p that satisfies the relation

R E - p)Ar
 110Engineering Hydrology
3.20 Infiltration Indices
convenient to use a cono
floods, it is
found nstant
calculations involving The defined average infltro
ration
storm.
In hydrological
the duration of the
rate for in common use.
value of infiltration two types
of indices are
inder and
rate is called infltration

3.20.1 p-Index
the average
rainfall Runoff
The p-index is
rainfall volume
above which the
volume. The
isequal to the runoff the rainfall 2.
p-index is derived from 2.02.0
the knowledge of 1.5
hyetograph with
the resulting runoff
volume. The S1.0 Loses Index
considered as 0.5
initial loss is also 12 14
10
infiltration. The value
is found by 2 4 6 8
p infiltration Time (h)
constant
treating it as a
rainfall intensity is Fig. 3.16 p-Index
capacity. If the
less than pthen the infiltration rate is larger than p, the
however, if the rainfall intensity
is equal to the rainfall intensity; infiltration in an interval
of time represents the
rainfall and
difference between the indexis
shown in Fig. 3.16. The amount of rainfallin excess ofthe
flood studies, it is also known
runoff volume as
In connection with runoff and
called rainfall excess. thus accounts for
(details in Sec. 6.5, Chapter 6). The P-index
as effective rainfall magnitudes to be estimated for a given
rainfall hye-
the total abstraction and enables storm hyetograph
for calculating -index for a given
tograph. A detailed procedure follows.
and runoff volume is as
resulting
Procedure for Calculation of p-index
Consider a rainfall hyetograph of event
duration D hours and having N pulses of time
interval Ar such that
N At = D. (In Fig. 3.16, N= 7)
be the intensity of rainfall in " pulse and Ra = total direct runoff.
Let ,
Total rainfall P = E , A

If pis the value of p-index then P- p t , = Ra

where t, = duration of rainfall excess.

If the rainfall hyetograph and total runoff depth R, are given, the p-index of the
storm can be determined by trial-and-error procedure as given below.
1. Assume that out of given N pulses, M number of pulses have rainfall excess.
(Note that M M). Select M number of pulses in decreasing order of rainfal
intesity
2 Find the value of p that satisfies the relation
M
R E ( - p)Ar
 Abstractions from Precipitation 111
rainfall
. Using the value of pof Step 2, find the number of pulses (M,) which give
excess. (Thus M, = number of pulses with rainfall intensity , 2 P).
the
4, If M. M then P of Step 2 is the correct value of p-index. If not, repeat
=

be used
procedure Step I onwards with new value of M. Result of Step 3 can
guidance to the next trial.
as

Example 3.11 illustrates this procedure in detail.

Example 3.11
Example3.1A storm with 10 cm.of precipitation produced a direct runoff of 5.8 cm,The
duration of the rainfall was 16 hours and its time distribution is given below.
Estimate the g-index of the storm.

Timefromstart(h) 0246810 1214 16

16
6.9 8.5 9.5 10.0
Cumulative rainfall (cm) 0 0.4 1.3 2.8 5.1
Solution numbered
Pulses of uniform time duration At 2 h = are considered. The pulses are

shown below.
sequentiallyand intensity of rainfall in each pulse is calculated as
Table 3.12 Calculations for Example 3.11

3 5 6 8
Pulse nunmber

Time from start of 6 8

rain (h)| 2 10 12 14 16
4
Cumulative rainfall (cm)0.4 1.3 2.8 5.1 6.9:8.59.5 10.0
1.00 0.50
Incremental rain (cm)) 0.40 0.90 1.50 2.30 1.80 1.60
0.80 0.50 0.25
Intensity ofrain) incmih. 0.20 045 0.75 1.15 0.90 8.
16 h, Ar 2h and N =

Here, duration of rainfall D

= =

Trial 1:
hours.
and hence t, =M ArF16
Assume M =8, Ar =2hincluded.
Since M= N, all the pulses are
2G-pAt= 2x Ar-p(8 2)
x
Runoff R=5.8 cm =

x 2) + (1.15 x2) + (0.90 x 2)+ (0.80x

2)
2) + (0.45 x 2) + (0.75
5.8 {(0.20x
x 2)} 16 p= 10.0 14 p -

+ (0.50 x 2) + (0.25
-

p=4.2/14 = 0.263 cm/h

of pulses having 2 p, that is
of row 5 of Table 3.12, M. =number
By inspection
20.263 cm/h is 6. M is not correct. Try a new value of
M <8 in the
Hence assumed
Thus, M, 6 # M.=

next trial.
Trial 2: At= 14 hours.
Assume M 7, At = 2h and t, =M hence
omitted.
Select these 7 pules in decreasig
order of 7, Pulse 1 is
(U- 9) Al= 24Ar-p(7 x2)
Runoff R =5.8 cm

x + (1.15 x 2) + (0.90 x 2)
5.8 {(0.45 x 2) + (0.75 2) + 14 p
+ (0.50 x 2) (0.25 x 2)} 14 p=9.6-
-

+ (0.80 x 2)
cm/h
3.8/14 =0.271
=
112 EngineeringHydrology
Time since 8 10 12 14 16
6
start of rain (h)

1.50

1.25 Rainfall excess

1.00

0.75
Loses
"

0.50

0.25
0.200 0.275 0.275 0.275 0.275 0.275 0.275 0.250 p 0.275cm/
cm/h
0.00
2 3 4 5 6 7 8

Pulse number (pulse of 2-hour duration)

Initial loss

Fig. 3.17 Hyetograph and Rainfall Excess of the Storm-Example 3.11

By inspection of row 5 of Table 3.12, M. = number of pulses having l;2 p, that is

,20.271 cm/h is 6.
Thus, M. =6*M. Hence, assumed M is not O.K. Try a new value of M < 7 in the next
trial
Trial 3:
Assume M=6 Ar =2 h and hence t, = M At=12 hours.
Select these 6 pulses in
decreasing order of , Pulse 1 and 8 are omitted.
Runoff R 5.8 cm =a
5.8= {(0.45 x 2) + (0.75 x 2)
+(1.15 x 2) + (0.90 x 2) + (0.80 x2)
+ (0.50 x
2)) 12 p=9.1 12 p -
-

3.3/12 =0.275 cm/h

By inspection of row 5 of Table 3.12, M,
20.275 cm/h is 6.
=
number of pulses having 1, 2 9. that is
Thus, M=6= M.-Hence,
The -index assumed Mis OK.
of the storm is 0.275 cm/h and the
12 hours. The
hyetograph of the storm, duration of rainfall excess =

excess are shown in losses, the rainfall excess and the duration of
3.17. Fig. rainfall
3.20.2 W-Index
In an
attempt to renne the
p-index, the initial
abstracions and an
average value of infiltration losses are separated from the tota
rate, called W-index, is
P-R-I defined as
W
where P= total storm (3.29)
R total storm precipitation (cm)
=
runoff (cm)
=
initial losses
(cm)
, =
duration of the rainfall excess, 1.e.
the
intensity is greater than W (in hours) total and
time in which the rainta
 Abstractions from Precipitation 113

W defined average rate of infiltration (cm).

rates are difficult to obtain, the accurate estimation of W-index is rather
dince I,

value of the W-index obtained under very wet soil conditions,

difficult.

The minimum is known as

scnting the constant minimum rate of infiltration of the catchment,
storm.
that both the p-index and W-index vary from storm to
It is to be noted
Wi
Computation of W-index.
I and
W-index from a given hyetograph
storm with known values of
To Compute
is followed:
nunoff R, the following procedure
Deduct the initial loss I, from the
storm hyetograph pulses starting from thefirst
)
pulse. the procedure indicated
diagram and follow
i) Use the resulting hyetograph pulse
in Sec. 3.20 for -index. except for
same as in the determination of p-index
Thus, the procedure is exactly
hyetograph is appropriately
modified by deducting ,
the fact that the
storm

is given
Example3.12 Ihe
Thenmass of rainfall of duration 180minutes a catchment
curve
on

an initial loss of 0.5 cm. The.- indexof the catchment is

knowñto he
below. The catchmenthad
the catchment duetothis storm.
04 cm/hour Calculate the totalsurface runoff from
6080100 120 150A60
Time fromstart (min 030 3.3
0.6 1.3 2.6 2.8 3.0 3.2
Cumulative rainfall (cm)

Solution rainfall in that time step minus

time is equal to the incremental
Runoff in each step to or greater
infiltration loss calculated as (¢. t). This number is aliways equal
=
the Note that the initial loss has
runoff is always a positive quantity.
than zero, as the of o index.
for in the definition (and hence in the calculation)
accouted
already been redundant information in
this problem. Calculations are
the initial loss is
As such,
shown in Table 3.13.
3.12.
Table 3.13 of Runoff-Example
Calculation
in the data.
Time steps are as given
-index =0.4 cm/hour. Infiltration Runoff
Ineremental
Cumulative
Increment (cm)
Time from rainfall l o s s = .At
rainfall
start (Min) of time (At)
(cm) (cm) (cm)
hours
0.000 0.000 0.000
0.000 0.00
0 0.600 0.200 0.400
0.60
30 0.500 0.700 0.200 0.500
0.500 1.30
60 0.133 0.567
0.333 2.00 0.700
80 0.800 0.133 0.667
0.333 2.8 0.067
100
3.00
0.200 0.133
120 0.333 0.200 0.000
3.20 0.200
I50 0.500 0.200 0.000
0.100
0.500 3.30 2.200
180 Total=
the catchment is 2.20 cm.
stom from
due to the
ne total surface runoff
 114 Engineering Hydrology

Example 3.13
Details related to an isolated
6-hour stom that occurred over a
runoff from the catchment due to
the storm.
catchment are gíven. Eoi
Areal Extent of
mate the.
Sub sub area as % of Rainfall (cm)
area catchment
p-index First 2-hours Second
2-hours Third
A 25 0.3 0.80 2- hours
urs
1.60 1.30
0.45 0.5 0.90 1:40 1.10
0.30 0.4 0.86 1.30 0.90
Solution
The runoff from each sub-area is
calculated separately and converted
unoff over the whole catchment. These to depth of
of runoff the
depths are in turm added to get the total depth
over catchment. The calculations
indicated in Table 3.14.
are
performed in a tabular form as

Table 3.14 Calculation of

Runoff -

Example 3.13
Areal
Extant of Intensity Runoff Runoff
Sub- area 0-Index of Depth of depth depth
Sub (% of
RainfallDu- rainfall Rainfall in suh-
over total
of sub- intensity ration Excess excess area catchment
area catchment) area (em/h) (h (cm/h) (cm)
A (cm) area (em)
0.25 0.30 0.40 2.00 0.10 0.20 jeiTITo,
0.25 0.30 0.80 2.00 0.50
1.00 1.900.475
0.25 0.30
B 0.45
0.65 2.00 0.35 0.70 g6ro25
0.50 0.45 2.00 0.00 0.00
0.45 0.50 0.70 2.00 0.20 0.40 0.50 0.225
0.45 0.50 0.55 2.00 0.05 0.10
0.30 0.40 0.43 2.00 0.03 0.06
0.30 0.40 0.65 2.00 0.25 0.50 0.66 0.198
0.30 0.40 0.45 2.00 0.05 0.10
Total 0.898
The total depth of runoff over the catchment due to the storm = 0.898 cm

3.20.3 p-index for Practical use

The p-index for a catchment, during a storm, depends in general upon the soil type,
vegetal cover, initial moisture condition, storm duration and intensity. To oblain
complete information on the interrelationship between these factors, a detailed
extensive study of the catchment is necessary. As such, for practical use in tn
estimation of flood magnitudes due to critical storms, a sinnlified relationship tor
252 EngineeringHydrology
Table 6.2 Calculation for Example 6.3
ER Intensity
Time trom Time Accumulated Depth of
rainfall in (cm) (cm) of ER
interval ruinlall in
start of (em/h
time r (cm) Ar (em)
storm, e(hY (h)
4 6 7
3

0
0.6 0.8 0 0
0.6
2.2 0.8 14 0.7
4 2.8
2.4 0.8 1.6 0.8
5.2
1.5 0.8 0.7 0.35
(6.7
7.5 0.8 0.8 0
10
9.2 1.7 0.8 0.9 045
12
0.4 0.8 0 0
14 9.6

In a given time interval Ar, effective rainfali (ER) is given by

ER = (actual depth of rainfall o A)
or
ER = 0, whichever is larger.
The calculations are shown
0.8
in Table 6.2. For plotting the
hyetograph, the intensity of 0.7
effective rainfall is calculated
0.6
in Col. 7.
The effective rainfall 0.5
hyetograph is obtained by plotting
ER intensity (Col. 7) against time 0.4
from start of storm (Col. 1), and
0.3
is shown in Fig. 6.8.
Total effective rainfall = Direct 0.2
runoff due to storm = area of ER
0.1
hyetograph =
(0.7 + 0.8 +0.35 +
0.45) x 2 = 4.6 m 6 10 12 14 16
2 4
Volume of Direct runoff = Time from start of storm (h)

4.6
1000X5.0x(1000) = 23000 m3 Fig. 6.8 ERH
of Storm-Example 6.3
1000

6.6 Unit Hydrograph

6.6.1 Theory
The problem of predicting the flood hydrograph resulting from a known storm in a
catchment has received considerable attention. A large number of methods are proposeu
to solve this problem and of them probably the most popular and widely
used method
in 1932
is the unit hydrograph method. This method was first suggested by Sherman
and has many refinements since then.
undergone
 Hydrographs253
A Mnit hydrograph is defined as the hydrograph of direct runoff resulting frorD one2
nit depthLPm) of rainfalexcess ccurring uniformDover the basin.and
vetthe.basin auniform
andatauniform
Fate or aaEpecihed
rale durat+on (D
peciñed duration hours) T
(Dhours). The term unit here refers to a unit depth of
Tanfall excess whichh iS usually taken as l cm. The duration, being a very important
characteristic, is used as a prefix to a specific unit hydrograph. Thus, one has a6-h unit
hydrograph, 12-h unit hydrograph, etc. and in general a D-h unit hydrograph applicable
to a given catchment. The definition of a unit hydrograph implies the following:
.The unit hydrograph represents the lumped response of the catchment to a
unit rainfall excess of D-h duration to produce a direct-runoff hydrograph. It
relates only the direct runoff to the rainfall excess. Hence the volume of water
contained in the unit hydrograph must be equal to the rainfall excess. As 1 cm
depth of rainfall excess is considered, the area of the unit hydrograph is equal
to a volume given by 1 cm over the catchment.
The rainfall is considered to have an average
of 1/D cm/h for the duration D-h of the storm.
intensity of excess rainfall (ER)
The distribution of the storm is considered to be uniform all over the catchment.
Figure 6.9 shows a typical 6-h unit hydrograph. Here, the duration of the rainfall
excess is 6h.
Area under the unit hydrograph =12.92 10 m x

0 6h
- Rainfall excess
160
Catchment area
120
1292 km2

6-h unit
80 hydrograph

40

Direct runoff = 1 cmn

O 6 12 18 24 30 36 42 48 54 60 66
Time in hours
Fig. 6.9 Typical 6-h Unit Hydrograph
Hence
Catchment area of the basin = 1292 km2
Two basic assumptions constitute the foundations for the unit-hydrograph theory.
Theseare (i) the time invariance, and (i) the linear
response.
1. Time Invariance
This first basic assumption is that the direct-runoff response to a given effective
rainfall in a catchment is time-invariant. This implies that the DRH for a given ER in
a catchment isalwaysthe same irrespective of when it occurs.
2. Linear Response
The direct-runoff response to the rainfall excess is assumed to be linear. This is the