Water Resources Engineering
Prof.

R Srivastava

Department of Civil Engineering
Indian Institute of Technology, Kanpur
Lecture

No.

21

We have looked at various abstractions from precipitation because as we have already
said our interest

is most

of the times

in the runoff

which

results

from

rainfall

and for that

to nd out the runoff we should know how much is being abstracted from the
precipitation. We have already discussed a few types of abstractions for example the
initial abstraction which consists of the depression, storage interception and then we have
some inltration which goes underground. We have our evaporation and some
transpiration. We would look at all these abstractions in terms of taking some examples
and looking at the magnitude, rough estimate of how to estimate these quantities. We
have looked at some of the equations and today we will look at some of the numerical
examples to understand these equations better.
Let's
start with the initial abstraction which is when the precipitation occurs there would
be some small depressionson the storage. There would be some water on the tree leaves
and then these may either evaporate directly from the surface or some of it will infiltrate.
The initial abstraction is the one which occurs immediately after the rainfall and that
means that without such . .0}.this initial abstraction we will not have runoff. Initial
abstraction is what initial amount is taken from the precipitation and after that the rest of
it will go as runoff or part of it will infiltrate and evaporate. In the initial abstraction it
would be including interception and depression storage. We have already discussed the
mechanism

of these

and the estimation

of these.

The

soil

conservation

service

of the US

SCS has proposed a technique becausethis is a very difficult to estimate quantity. Initial
abstraction depends on a lot of factors which are changing from place to place and then
again from a storm event to different storm event it may be different. A rough estimate of
this can be made by this SCS method in which the initial abstraction is taken as 20% of S
where S is a potential maximum retention.
Naturally if we have a surface which is paved then there would be very small retention on
the surface. But if we have a land which is agricultural then there will be a lot of retention
possible. Therefore S will depend on the land use and also what kind of soil it is. The
SCS has related S with an empirical constant called curve number or CN and this CN is
popularly known as the SCS curve number. The curve number dependson the type of soil
and the land use of that area. It also dependson the existing moisture condition. If the soil
is already moist then less inltration will occur. If the soil is initially dry then more
inltration will occur. Initial abstraction will be high if the soil is dry initially. This
relation between the curve number and the value of S which is the potential maximum
retention is given by 25.4. This factor converts the inches becauseoriginally the formula
was

in inches.

To

over

CN minus

10.

convert

this

inch

into

millimeter

we

have

this

25.4

factor

then

1000

As you can see from here CN can be a maximum of 100 becausewhen CN is 100 then S
will be zero. 100 means there is no storage; perfectly everything which falls down on the
surface will go as runoff. There will be no inltration, evaporation. The curve number
can be as shown in this table for normal existing soil moisture conditions. If the soil is
dry than the normal conditions or if it is better than the normal conditions then the curve
number will change but for normal conditions depending on the land use and the type of
the

soil.

We

can find

out

the curve

number

from

this

table.

A lot

of other

land

use and

soil combinations are not shown in this table. For example if you have paved area then
naturally the soil type will not matter becausethe area is paved. So a curve number of 98
can be used for all types of soils. For industrial area which is not completely covered it
may be 70% covered or 80% covered then the curve number will also depend on the type
of soil. For sandy soil we have 81 curve number. For clay we have 93 curve number
indicating that for the clay S will be small and the runoff will be larger; for sand S will be
large and the runoff will be smaller.
Similarly if you have a cultivated area then the curve numbers are even smaller compared
to the industrial area. Paved area will have very high curve number and high curve
number, as you can see from this equation, will indicate low S and low S means low
value

of initial

abstraction.

(Refer Slide Time: 6:44)

For paved area initial abstractions will be very small. For example paved area using a
curve number of 98 we can S equal to 5.18 millimetre and initial abstraction of 1.04
millimetre.

This

indicates

that

if the rainfall

is less than

about

1 millimetre

then

we can

expect that all of it will be absorbed by the catchment area and there will be no runoff.
Once the rainfall exceeds 1 millimetre, only then the runoff will occur. Similarly if we
have a cultivated sandy soil we can look at the curve number which is 72. Using that

curve number of 72 we can get S equal to 3.89 millimetre and initial abstraction of 0.78
millimetre. Curve number can be as small as 25. For example if we have a forest area and
soil is sandy and good cover; that means the foliage density is very high. The sandy soil
will cause more inltration. Curve number of 25 will give us initial abstraction which
will be 20% of this; 6 millimetre which is quite high and that means that your initial
rainfall of 6 millimetre would be completely absorbed by the catchment and there will be
no runoff. For forest area with permeable soil, sandy soil with a very good cover the
initial abstraction is quite high and it also depends on the climate. If we have a warmer
climate we can reduce the curve number. For paved area we have already seen that the
curve number is 98 for all types of soils. But if the climate is warmer then we would
expect little more evaporation and we can reduce the curve number to 95.
(Refer Slide Time: 9:02)

This method gives us an idea of how to nd out initial abstraction from the precipitation.
Its an empirical method. Tables of curve numbers are available for different types of
land uses, different types of soils and different types of existing moisture conditions what
is known

as the

antecedent

moisture

conditions

and

under

normal

conditions

we

have

shown the table. This is under normal conditions. If dry or wet then there are other tables
or other equations which relate the curve number with the land use and the type of soil
but we will not go into details of that.
After initial abstraction then we have other abstractions like evaporation, infiltration and
transpiration. The evaporation from a water surface can be described as changing with
time of the year. In summer months we would expect a very high evaporation. For winter
it would be lower. A typical variation of evaporation can be seen in this gure in which
January has very small evaporation, December has small evaporation but as we can see
here May, June and July may have high evaporation as high as may be up to about 170

millimetres. This is a monthly evaporation. Evaporation for the whole month may be
about 175, 170. This is for a typical station. Sometimes evaporation may be as high as
200 or even higher than that. In most of the dry areas in India it would be higher than
200; 220 millimetres per month.
The question is how to nd out this evaporation which varies over a wide range? If you
look at this value it is close to about 40, 50 and if you look at this value it is close to 170,
175.

(Refer Slide Time: 11:09)

Evaporation will vary a lot from month to month and for a typical month or for a typical
year if you want to nd out evaporation then there are empirical equations which are
given. We have already seen that we can measure the evaporation by using evaporation
pan and then using a pan coefficient. That measurement can be done but there are some
empirical equations which can be used if data is not available from evaporation pans.
Some of the empirical equations we have seen for evaporation. We have the Meyer's
equation and the Rohwer's
equation.
Rohwer's
equation and Meyer's
equation both are based on the fact that the evaporation
will be directly proportional to the vapour pressure decit. So ew here as we have seenis
the saturation vapour pressure and ea is the actual vapour pressure. ew-ea denotes the
deficit from the saturation value to the actual value and this will govern the amount of
evaporation from a water surface. We have also seen that not only the deficit but also the
wind velocity will affect the evaporation because the wind velocity u will affect how
fast

the

saturated

air

from

here

is removed

from

over

the

surface

and

then

another

air

packet which once becomes saturated is moved away. Then a dryer air will come and
then evaporation can continue. Larger wind speed will mean larger rate of evaporation.

Most of the equations have one factor which dependson the saturation deficit and another
factor which dependson the wind velocity.
In the Rohwer's
we have an additional factor which accounts for atmospheric pressure
also at that location. We would need to know ew, the saturation vapour pressure.
(Refer Slide Time: 13:49)

Saturation vapour pressure is generally a function of temperature and if we know the
temperature we can find out the value of the ew. Some of the tables are easily available in
literature. One of the tables is shown here in which with temperature there is variation of
vapour pressure and there is slope of vapour pressure curve. This we will need in some
other equations later on but for now this is what is important; how the saturation vapour
pressure ew changeswith temperature.

(Refer Slide Time: 14:31)

There are equations also given for this and there are figures available like this which
show for any temperature what is the saturation vapour pressure in terms of millimetre of
mercury. If we take the example of a lake we want to nd out the evaporation from the
lake given some data. For example surface area of the lake is given as 4 square
kilometres; water temperature is 25 degree Celsius. This water temperature means that
we have to find out ew at 25 degrees. 1514
(Refer Slide Time: 15: 14)

We can use the graph or some equations to nd out the ew corresponding to this
temperature and as you can see from here it is obtained as 23.77 millimetre of mercury
which we can get from the figure or the table also. For a temperature of 25 degree Celsius
vapour pressure of 23.77 is obtained. The actual vapour pressure will be obtained by
multiplying this saturation vapour pressure with the relative humidity and the data which
is given for this problem is a relative humidity of 45%. This will indicate that the actual
vapour pressure would be 45% of the saturation vapour pressure which is obtained as
10.70 millimetre of mercury.
In the Meyer's
equation there is an empirical constant K here which depends on the kind
of water body which we are dealing with. If we have a large and deep body then K is
taken as 0.36. Since this is large body 4 kilometre square area we can take K as 0.36.
Knowing K, ew and ea we are now left with obtaining the value of 119which is the
velocity at 9 meters above the ground level. The data which is given is for wind velocity
at 0.6 meter above ground. Wind velocity will be given based on where is this velocity
measuring point and in this case we have data which is available at 0.6 meter above
ground and the wind velocity is 14 kilometres per hour at 0.6 meter. What we need in the
Meyer's
equation is u at 9 meters.
We need to use some correlation. We can use the one seventh power law which tells that
the velocity varies as one seventh power of the distance from the surface. Using that
relationship we can estimate the velocity at 9 meters. The wind velocity profile would be
like this. At 0.6 meter the velocity is given as 14 and our interest is in nding out at 9
meters what this velocity is. If you assume one seventh power law then the velocity at 9
meters can be obtained as 14 into 9 divided by 0.6 which is the depth ratio to the power 1
by 7 and it turns out to be 20.6 kilometres per hour. Looking back at the Meyer's
equation, now we know everything. We have K, we have the ew and ea; we have 119and
we can obtain the evaporation rate in terms of millimetres per day. 0.36 is the value of K,
saturation decit, 20.6 is the velocity at 9 metres. In this equation 1 plus 119by 16 we
have 20.6 by 16. This turns out to be 10.77 millimetres per day.
We have evaporation from the lake surface equal to almost 1 centimetre in a day and we
can also nd out in a month how much volume of water will evaporate from that lake.
The surface area is 4 square kilometres. We convert into metre square. This is millimetre
per day converted into metres and 30 days assuming in a month we have 1.29 million
metre cube of water evaporated from the lake in a month. We can also use the Rohwer's
equation and see how much estimate of the evaporation we get. Since all these are
empirical equations they would naturally give different values becausethey are based on
different data. Rohwer's
equation has this additional parameter Pa and we have this given
information mean barometric pressure is let's
say 768 millimetres of mercury. This is in
millimetres of mercury.
The wind velocity which is used in Rohwer's
equation is U0 which is at the surface. But
since the velocity at the surface is very difcult to measure, ideally would be zero if we
say no slip condition. U0 is taken as U06 and that is what we have measuredhere U at 0.6
above ground as 14 kilometre per hour. Using the Rohwer's
equation we have this

constant 0.771 barometric pressure term, wind velocity term and saturation deficit term
and this U0 we will be using as U05. Using this data we get a value of 13.33 millimetres
per day. You can seethat here we have 10.77 millimetres per day. Here we have obtained
a value of just 13.33 millimetre per day. They are a little different and not exactly the
samething.
(Refer Slide Time: 21 :24)

This is only evaporation; most of the times we would be interested in combining
evaporation and transpiration. We need to know what is the evaporation from a water
body as well as water transpiration from plants or vegetation in that area and as we have
already discussed we combine these two and called that evapotranspiration.
Evapotranspiration will depend on a lot of factors. For example the solar radiation
available at that point and the number of day light hours becauseif we have cloud cover
then naturally evapotranspiration will be smaller. We need a lot of information. There are
again empirical methods which use some equations and there are some theoretical
methods which are a little more based on theory rather than only empirical coefficients.
We would look at some of these techniques and see how to compute evapotranspiration
based on some of the empirical and theoretical approaches.
We have seenthe Blaney Criddle equation which says that the evapotranspiration is equal
to 25.4; again this 25.4 comes because we convert inches into millimetre and there is
some factor K here which will be function of the crop type. This will affect the
transpiration, the crop type. Therefore this factor K will come and F is a consumptive use
factor, monthly consumptive use factor, which we will see later on how to compute and
another method which is commonly used is the Thornthwaite equation which again tells
that ET is equal to 16 La is a factor which depends on the latitude and the month, T
average is the average temperature and using these two equations we can find out the

evapotranspiration at a particular latitude for a given month where average temperature
will be known. Let's
take this data. We have a location where the latitude is 30 degrees
north, crop is given as wheat and for wheat we know that the value of K in the Blaney
Criddle equation is taken as 0.65; for rice it is 1.1. Let's
look at 2 months, months of
November and December for which the mean monthly temperature are 15 for November
and 12 for December and what we need to find out is evapotranspiration for this period of
2 months

November

and December.

(Refer Slide Time: 24:42)

In the Blaney Criddle equation in which is written as evapotranspiration is 25.4 KF we
have already discussed that for wheat K is given as 0.65. There is a table of values of K
for different types of crops. Some of the values are rice 1.1, maize 0.65, sugarcane 0.9,
natural vegetation if it is dense then the value can be taken as 1.2 to 1.3 and if it is light
0.8 to 1. Here since wheat is given we will take K equal to 0.65. The factor F is the
summation of Ph. P1,is the monthly percent of annual daylight hours and this will depend
on latitude and the time of year which we say as month. T average is given for November
and December as 15 and 12 degree in the Blaney Criddle equation which was based on
FBA system of units. The temperature was in degree Fahrenheit. We can convert this 15
and 12 degreesCelsius to 59 and 53.6 degreesFahrenheit. 2623
The Ph value is obtained from a table of values which relate the monthly sunshine hour
percent with the latitude and the time of the year and if we look at this table this gives us
the daytime hour percentage for each month, for different latitude and different months
we are given these percentages.For example if we are talking about equator zero degrees
north in January there would be 8.5% percent daylight hours compared to the number of
daylight hours for the whole year. 8.5% of the year daylight hours will occur in January.
Similarly if we look at February the number of daylight hours in this case we would seeit

as smallest at 7.66. If the number of daylight hours is same in every month then the
percentage would be 100 by 12 or 8.33%. But since February is a smaller month also and
colder also the percentage of daylight hours is 7.66. As we move to the north you can see
that in January the percentage is decreasing while in May it is increasing. This depends
on the

latitude

as well

as the

season

and

therefore

from

this

table

we

can

find

out

the

percentagewhich is required in the Blaney Criddle equation.
If we look at the data given for 30 degree north latitude and the month as November and
December we can look at the percentage values; 7.19% for November and 7.15% for
December. Using these two values of percentage and using the mean temperature we can
estimate the value of F which is summation of Ph monthly percent. We have already seen
this as 7.19 and 7.15%. T average is 59 for November, 53.6 for December. The
summation of these two sigma Ph T average for these two months comes out to be 8.07.
Knowing F and K we can find out the evapotranspiration; 25.4 is conversion from inch to
millimetre, K for weight and F for November and December months for 30 degreesnorth
latitude.

(Refer Slide Time: 29:38)

It gives us 133.3 millimetres of evapotranspiration over the two months of November and
December.

Similarly if we use the Thornthwaite equation we can write ET equal to 16 La 10 T
average over It and exponent a. The exponent a as you can see from here is given by
another equation which relates it with It. It is the heat index and it is given as for the
whole year summation from 1 to 12 T bar over 5 to the power 1.514 which is this
equation. T bar is the average monthly temperature. The heat index can be obtained if
average temperatures are given for each month. We have this additional data which
10

should be given to us. This is the mean monthly temperature and this is in degree Celsius;
Thornthwaite we are using Celsius. For January you can see that the mean monthly
temperature is 10 degree centigrade. November and December are already given to us.
But for this method we need to have this whole year data available to us and using this
entire data set we can estimate the value of It and it turns out to be 96.1; T average by 5
for each month to the power 1.514 summing over all the month. This summation goes
from I equal to 1 to 12 as written here. For each month we need to add up all these factors
and when we do that we get It equal to 96.1. a is a function of It. Some constant into Itq;
again It square, It and a constant term which gives us 2.1.
(Refer Slide Time: 32:06)

For It equal to 96.1 a is given as 2.1 and then the value of La is given in table for different
latitudes and different months and this is the adjustment factor La and for the given
latitude of 30 degrees for months of November and December we can see that the
correction factor or the adjustment factor is 0.89 and 0.88.

11

(Refer Slide Time: 32:43)

Using this for these two months we can find out the value of ET as 36.3 for November;
this is December

and this is November;

36.3 millimetres

in November

and 22.7

millimetres in December. These are empirical equations but there are some theoretical
equations also available to find out the evapotranspiration. The most commonly used
theoretical equation is the Penman's
equation.
(Refer Slide Time: 33:20)

12

If we look at the Penman's
equation it is based on the incoming solar radiation at that
particular location and for that particular period of the year. So we need to have some
data about what will be the solar radiation coming at different latitudes at different times
of the year. There are some tables which are available for this purpose and we will look at
a few of those tables which are needed in the Penman's
equation which are similar to the
other
months.

tables
In

which
this

show

case

we

behaviour
can

look

with

different

at mean

solar

latitudes

and

radiation

in

different
terms

of

times
millimetre

of

the
of

evaporable water per day; again for different latitudes and different times of the year we
can find out what is the mean solar radiation which occurs at the atmosphere; how much
water it will be able to evaporate?
The mean solar radiation is given in terms of millimetres of evaporable water per day.
Because evaporation of water requires some energy this millimetre of evaporable water

per day indicateswhat is the energycoming in at the atmosphericlevel and as you can

seefor this 30 degreesnorth latitudein Novemberthe solarradiationwill be ableto
evaporate 9.1 millimetre of water per day, in December only 7.9 millimetre per day. At
20 degreesnorth it would be able to evaporate 11.2 and 10.3 for the months of November
and December.

(Refer Slide Time: 35:24)

The other thing which we need for most of the theoretical equations is the sunshine hours
per day, the mean potential sunshine hours per day. The actual sunshine hours will be
different depending on the cloud cover on that day but this is the potential sunshine hours
per day and this will depend on the latitude and the time of the year. At the equator this
value does not change with the time of the year. It is 12.1 hours of sunshine every day
expected on the equator throughout the year. For 10 degrees north latitude again winter
months; since we are talking about northern hemisphere January, February, November
13

and December are the winter months. The sunshine hours are smaller during this time and
then they become larger during the summer months. That trend you can see here also
smaller number of daylight hours, sunshine hours, in the winter months and large number
during the summers.
(Refer Slide Time: 36:40)

These two tables mean potential sunshine hours and mean solar radiation. They would be
helpful in obtaining estimates of evapotranspiration using the theoretical equations. In the
Penman's
equation the equation which is given for nding out the evapotranspiration
involves solar radiation. In the equation which is given by Penman the potential
evapotranspiration would be some factor A times Hn; this is the radiation. Again another
factor gamma, Ea is the evaporation and weighted by mean of the weights A and gamma
and then divided by total value A plus gamma.

14

(Refer Slide Time: 37:41)

The data which is given in this example is that for the month of November at latitude of
30 degreesnorth. Mean monthly temperature is given as 15 degreescentigrade and in this
case let's
take the mean relative humidity as 75% and actual mean daily sunshine hours
for the month of November are taken as 9. Potential sunshine hours are larger than this.
But actual daily sunshine hours are only 9 becausesometimes there may be cloud covers.
Wind velocity is measured in this case at 2 metres above the surface and was found as 4
kilometres per hour. But in Penman's
equation we need the velocity in kilometres per
day. So we converted into 96 kilometres per day.
There is an albedo required in the Penman's
equation which is the reectance of the
surface. It is given that this area is very close to ground crops which have albedo of 0.2. It
is the reectance

of the surface.

So 0.2 is taken

as the albedo.

If we have

water

we have

0.05 albedo. For bare land it can vary from 0.05 to 0.45 and for snow it can go as high as
0.95. But in this case albedo is given as 0.2.

15

(Refer Slide Time: 39: 15)

In the Penman's
equation we would need the quantities; as we have seen the albedo is
given as 0.2, net radiation we want at the atmospheric level, we also want a psychometric
constant which is taken as 0.49. Then there is one more term which we require, the slope
of the saturation vapour pressure curve. This also dependson the temperature. If you take
a temperature of 15 degrees centigrade the saturation vapour pressure is 12.8 and the
slope of the curve is 0.82.
(Refer Slide Time: 39:57)

16

This shows the saturation vapour pressure and the next curve shows the slope of the
saturation vapour pressure curve in terms of millimetre per degree centigrade and for
different temperatureswe can nd out its value.
(Refer Slide Time: 40: 19)

In this case for a temperature of 15 degrees Celsius the values which we will be using
will be 12.8 millimetre of mercury pressure, saturation vapour pressure ew and 0.82 as the
slope of the curve. The other thing which we need is the amount of solar radiation
available

and if we look

at the month

of November

for

different

latitudes

we have

these

values of mean solar radiation. This tells us that using these tables and the values of the
coefficients we can estimate or find out the values of different parametersin the Penman's
equation. We will look at the example. The month is November and latitude is 30 degrees
north. Using these data we can obtain from the table which we have shown that Ha is 9.1
millimetre of evaporable water per day. This we have seen function of latitude and
month. The table gives us the available radiation as 9.1 millimetre. ew also we have seen
for the temperature of 15 degrees was 12.8 millimetres of mercury. The relative humidity
given in this case is 75% percent. The actual vapour pressure will be 75% of the ew
which

is obtained

as 9.6.

ew and ea can be obtained from the temperature and the relative humidity data. The
evaporation for the Penman's
equation is given by this equation 0.35, wind velocity term
and the saturation decit term. 112means the wind velocity at 2 meters. In this case the
velocity is given as 2 meters. We don't
need to use the height correction and therefore we
will be using 96 kilometres per day as the velocity. Using 1 plus 112by 160 the wind
velocity term, saturation decit term multiplied by 0.35 we get ea as 1.79 millimetres per
day. The slope of the saturation vapour pressure curve again we have obtained from a
temperature of 15 degrees and it was obtained as 0.82. That is the value of A. A is the
17

slope of ew versus t curve and this is in millimetres of mercury per degree Celsius. The
value is obtained from the table as 0.82. So A is also obtained. Psychometric constant
gamma is 0.49 millimetres of mercury per degree Celsius. That value is almost a
constant.

In the Penman's
equation, i.7.~...e
(00:43:58) constant and the value which is taken is 2.01
into 10 to the power -9 millimetres per day. The Stefan-Boltzmann constant is used here
sigma T to the power 4; T is the temperature in degree Kelvin. T as we have written here
for a temperature of 15 degree Celsius we have to add 73.3 and we get the T of 288.3
degree Kelvin. There are some correction factors here a plus b n by N and 0.1 plus 0.9 n
by N. These correct for the actual number of sunshine hours compared to the potential
sunshine hours. In this case the potential number of sunshine hours for the month of
November at latitude of 36 degrees North can be obtained from the table and the value
which

is obtained

is 10.6 hours.

This

means

that

we can have

a maximum

of 10.6

hours

of daylight in that period on that latitude but the actual number n is only 9. This may
happen due to a number of reasons. Some days may be cloudy so sunshine hours are
liIr1ited. Because of this number of actual sunshine hours being less than potential hours
we have these correction factors a plus b n by N here and 0.1 plus 0.9 n by N here in
which a is related to the latitude as 0.29 cosine of the latitude, in this case 30 degreesand
it comes out to be 0.25. b is constant equal to 0.52; r is the albedo given as 0.2. Ha
available solar radiation that we have seen will depend on the latitude and the month and
from the table we have seenthe value as 9.1. In this equation all the terms are known and
we can nd out the net radiation in terms of evaporable water per day as 1.74 millimetre
per day.
As we can see here Ea and Hn both have been obtained. In this case they are very close.
Hn is 1.74 and ea is 1.79. In this case the relative magnitude of A and gamma will not
really matter because these two values are very close. Whatever weight we assign to
individual values will not affect the calculation very much but still if we use A and
gamma as weights we get potential evapotranspiration as 1.76 millimetres per day. For
the month of November at latitude of 30 degrees North we can expect if the actual
sunshine hours are 9 per day on an average and relative humidity is 75% then we can
expect for a wind velocity of 4 kilometres per hour and for an albedo of 0.2 that the
potential evapotranspiration from that area would be 1.76 millimetres per day.

18

(Refer Slide Time: 47:40)

Using either the theoretical equation where Penman's
is the most common or some
empirical equations for example Thomthwaite we can find out what is the potential
evapotranspiration from that area. Knowing the initial abstraction and the potential
evapotranspiration the only other abstraction which we need to look at is the inltration.
Inltration depends on the soil type and as we have seen there are some curves which
relate the temporal variation of infiltration rate. Inltration rate will be very high at the
beginning of the storm and as we move further in time infiltration rate will decrease
because the soil has become more saturated. The soil may also become compacted
because of the impact of the raindrops. Generally the inltration rate will decreasewith
time. There will be some rate here at time zero, there will be some rate here at infinity
and then an exponential decreaseis specified as the change of inltration with time. But
it becomes very complicated to estimate these parameters fo, f infinity and there is some
rate here which is like e to the power kt; nding out this K also becomes complicated.

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(Refer Slide Time: 49: 15)

There are some simple methods which can be used to estimate the infiltration loss. The
inltration indices in a way represent an average value of the loss. The most commonly
used index is the phi index and what it means is some kind of average loss from the
precipitation. If we are given a storm hyetograph, for example a storm may look like this,
intensity may be millimetre per hour and time in hours. We may be given that rain occurs
with this intensity and this temporal variation and we may be given runoff. Knowing the
amount of rain and the amount of runoff we can estimate what are the losses and using
those losses we can estimate what is the average loss expressedin terms of phi index.
Let's
take an example in which the precipitation of 88 millimetre occurs in 8 hours as
shown in this table. From 0 to 1 hour the intensity is 4 millimetres per hour. Similarly I
to 2 is 8 and 7 to 8 intensity is 5. If we add all these up we get 88 millimetres of
precipitation. It is also given that because of this rain there is a 46 millimetre of runoff
and a total loss of 42 millimetre takes place for this particular event and when we say this
loss this includes all losses. So everything else goes as runoff. Now in order to nd out
the phi index what we need to do is draw a line here such that the area above this
representsrunoff and the area below this representsthe loss. Then this value is called the
phi index.
How to compute this phi index? Basically we want the total precipitation and runoff so
that we can find out the losses and then we have to adjust this line in such a way that the
area below that is equal to loss. So in this case we can say that the average loss is in 8
hours; we have loss of 42 millimetres. The average loss is 5.25 millimetres per hour but if
we look at the table we can see that for the first period and the last period, rainfall
intensity is smaller than 5.25. First period is only 4, last period it is 5. So in the first
period and the last period the losses cannot be 5.25 becausethe losses are liIr1ited to the
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amount of rain. Whatever rain is falling all of it can be lost but no more than that. So
what

we have

to do now

is one more

iteration

in which

we assume

that

in the rst

hour

the loss is 4 millimetres in the 8th hour the loss is 5 millimetre and in other 6 hours we
have to find out what will be the loss. We can say that a loss of 42-4-5 will be in 6 hours
which means that a loss of 33 millimetres in 6 hours giving us average loss of 5.5
millimetres per hour and all the other values are higher than 5.5. Therefore we can say
that phi index is 5.5 millimetre per hour.
(Refer Slide Time: 53:46)

In this way we can obtain the value of an average loss over the entire period. If we look at
the hyetograph and let's
say that the phi index is somewhere here then we can say that
this te will be the duration of effective rainfall or excess rainfall. This rainfall is what is

contributing to the runoff and the rest is lost to infiltration, evaporation and abstraction.

21

(Refer Slide Time: 55:02)

The six here represents the effective t or the time of excess rainfall. So phi index is very
commonly used to estimate the average loss. There are some other indices also. There is a
W index which is given as precipitation minus runoff minus initial abstraction divided by
the duration of rainfall excessbut phi index is little more common. There are some other
methods also of estimating phi index. After getting the phi index we can prepare a table
like

this

which

shows

for

different

times

what

is the

excess

rainfall?

That

means

after

taking care of the losses what is the additional amount and in this case if we look at this
table this 4 is completely lost; out of this 8 we have a loss of 5.5 millimetres. All the
other values we subtract 5.5 except the first and the last that becomes zero and a sum of
all these

will

turn

out to be 46 millimetres.

Central water commission, CWC of India has given an equation by correlating the runoff
with rainfall. If rainfall of I centimetre per day occurs in 24 hours, the runoff can be
correlated by alpha I to the power of 1.2. Alpha depends on the soil type; 0.2 for sandy
and 0.45 for clay.

22

(Refer Slide Time: 56:55)

We have a higher runoff for clay soils and then phi index can be defined as I minus R
divided by 24 and can be used to estimate the effective or excessrainfall duration.
In today's
lecture we have seen various abstractions and how to compute them. We have
looked at the initial abstraction, the evaporation, evapotranspiration and inltration. We
have looked at various empirical and theoretical methods to obtain an estimate of these
parameters which can be used to nd out how much percentage of rainfall will be lost
and how much will go as runoff which is our main part of interest as to how much runoff
will occur due to a particular storm effect.

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