 So, we have looked at various abstractions from precipitation because as we have already said our interest is in most of the times in the runoff which results from rainfall. And for that to find 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 so on and then we have some infiltration which goes underground. We have evaporation and some transpiration. So, we would look at all these abstractions in terms of taking some examples and looking at what is 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 us start with the initial abstraction which is when the precipitation occurs there would be some small depressions on 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. So, the initial abstraction is the one which occurs immediately after the rainfall and that means that without satisfying this initial abstraction we will not have runoff. So, 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. So, in the initial abstraction typically we would be including interception and depression storage. We have already discussed the mechanism of these and the estimation of these typically the soil conservation service of the US SCS has proposed a technique because this is a very difficult to estimate quantity. Initial abstraction is depends on a lot of factors which are changing from place to place and then again from storm event to different storm event it may be different. So, a rough estimate of this can be made by this SCS method in which the initial abstraction is taken as 20 percent of S where S is a potential maximum retention. Naturally, if we have a surface which is let us say paved then there would be very small retention on the surface, but if we have a land which is agricultural then there will be 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 C n and typically this C n is popularly known as the SCS curve number. The curve number depends on the type of soil and the land use of that area. It also depends on what is the existing moisture condition. For example, if the soil is already moist then less infiltration will occur. If the soil is initially dry then more infiltration will occur. So, initial abstraction will be high if the soil is dry initially. So, this relation between the curve number and the value of S which is the potential maximum retention is given by 25.4. Now, this factor converts the inches because originally the formula was an inches. So, to convert this inches into millimeter we have this 25.4 factor then 1000 over C n minus 10. So, as you can see from here C n can be a maximum of 100 because when C n is 100 then S will be 0. 100 means there is no storage. So, perfectly everything which falls down on the surface will go as runoff there will be no infiltration evaporation. The curve number can be shown as in this figure in this table for normal existing soil moisture conditions. If the soil is dry then the normal conditions or if it is wetter than the normal conditions then the curve number will change. But for normal conditions depending on the land use and the type of 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. But for example, if we have a paved area then naturally the soil type will not matter because the 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 for example, it may be 70 percent covered or 80 percent 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 we have a cultivated area then the curve numbers are even smaller compared to the industrial area. So, paved area will have very high curve number and high curve number as you can see from this equation high curve number will indicate low S and low S means low value of initial abstraction. So, for paved area initial abstraction will be very small. Some of the values may be for example, paved area using a curve number of 98 we can guess we can get S equal to 5.18 millimeter and initial abstraction of 1.04 millimeter. This indicates that if the rainfall is less than about 1 millimeter 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 millimeter 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 millimeter and initial abstraction of 0.78 millimeter. Curve number can be as small as 25 for example, if we have a forest area and soil is sandy and a good cover. So, that means the foliage density is very high sandy soil will cause more infiltration curve number of 25 will give us initial abstraction which will be 20 percent of this. So, 6 millimeter which is quite high and that means that your initial rainfall of 6 millimeter would be completely absorbed by the catchment and there will be no runoff. So, for forest area with permeable soils, sandy soil with a very good cover the initial abstraction is quite high and it also depends on the climate. So, for example, 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 a little more evaporation and we can reduce the curve number to 95. So, this method gives us an idea of how to find out initial abstraction from the precipitation. It is 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 anti-sident 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 type of soil, but we will not go into details of that. The other part which we have after initial abstraction then we have other abstractions like evaporation and infiltration and transpiration. Typically the evaporation from a water surface can be described as changing with time of the year for example, 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 figure in which January has very small evaporation, December has small evaporation, but as we can see here may June, July even may have high evaporation as high as may be up to about 170 millimeters this is monthly evaporation. So, 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 to 20 millimeters per month or so. So, the question is how to find 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. So, evaporation will vary a lot from month to month and for a typical month or for a typical year if you want to find 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. So, 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 which we have seen for example, for evaporation we have the Meyers equation and the Rovers equation. Rovers equation and Meyers equation both are based on the fact that the evaporation will be directly proportional to the vapor pressure deficit. So, E w here as we have seen is the saturation vapor pressure and E a is the actual vapor pressure. So, E w minus E a 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 is once it becomes saturated it is moved away then a drier air will come and then evaporation can continue. So, wind speed larger wind speed will mean larger rate of evaporation. So, most of the equations have one factor which depends on the saturation deficit and another factor which depends on the wind velocity. In the Rovers we have an additional factor which accounts for atmospheric pressure also at that location. So, we would need to know what is E w saturation vapor pressure saturation vapor pressure is generally a function of temperature and therefore, we can find out if we know the temperature we can find out what will be the value of the E w. So, some of the tables which are easily available in literature for example, one of the tables is shown here in which with temperature the variation of vapor pressure and there is a slope of vapor pressure curve this we will need in some other equations later on, but for now this is what is important how the vapor pressure saturation vapor pressure E w changes with temperature. There are equations also given for this and there are figures available like this which show for any temperature what is the saturation vapor pressure in terms of millimeter of mercury. So, for example, if we take the example of a lake we want to find out the evaporation from the lake given some data for example, surface area of the lake is given as 4 square kilometers water temperature is 25 degree Celsius. Now, this water temperature mean that we have to find out E w at 25 degrees. So, we can use the graph or some equations to find out what is the E w corresponding to this temperature. And as you can see from here it is obtained as 23.77 millimeter of mercury which we can get from the figure or the table also. So, for a temperature of 25 degrees Celsius vapor pressure of 23.77 is obtained the actual vapor pressure will be obtained by multiplying this saturation vapor pressure with the relative humidity. And the data which is given for this problem is a relative humidity of 45 percent. So, this will indicate that the actual vapor pressure would be 45 percent of the saturation vapor pressure which is obtained as 10.70 millimeter of mercury. In the Myers equation there is a an empirical constant k here which depends on the kind of body water body which we are dealing with. If we have large and deep body then k is taken as 0.36. So, since this is a large body 4 kilometer square area we can take k as 0.36. So, knowing k E w and E a we have we are now left with obtaining the value of U 9 which 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. So, 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 kilometers per hour at 0.6 meter. What we need in the Myers equation is U at 9 meters. So, we need to use some correlation for example, we can use the 1 7 power law which tells that the velocity varies as 1 7 power of the distance from the surface. So, using that relationship we can estimate the velocity at 9 meters. So, the wind velocity profile would be like this at 0.6 meter the velocity is given as 14 and our interest is in finding out 9 meters what is this velocity. So, if we assume 1 7 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 kilometers per hour. So, looking back at the Myers equation now we know everything we have k, we have the E w and E a, we have U 9 and we can obtain the evaporation rate in terms of millimeters per day 0.36 is the value of k saturation deficit 20.6 is the velocity at 9 meters. So, in this equation 1 plus U 9 by 16 we have 20.6 by 16. So, this turns out to be 10.77 millimeters per day. So, we have evaporation from the lake surface equal to almost 1 centimeter in a day and we can also find out in a month how much volume of water will evaporate from that lake. So, the surface area is 4 square kilometers we convert into meter square and this is in millimeter per day convert into meters and 30 days assuming in a month. We have 1.29 million meter cube of water evaporated from the lake in a month. Now, we can also use the robust equation and see how much estimate of the evaporation we get. Since all these are empirical equations they would naturally give different values because they are based on different data. So, robust equation has this additional parameter P a and we have this given information mean barometric pressure is let us say 768 millimeters of mercury. So, this is in millimeters of mercury the velocity wind velocity which is used in robust equation is u naught which is at the surface. But, since the velocity at the surface it is very difficult to measure it ideally would be 0 if we say no slip condition therefore, u 0 is typically taken as u 0.6 and that is what we have measured here u at 0.6 meter above ground as 14 kilometer per hour. So, using the robust equation we have this constant 0.771 barometric pressure term, wind velocity term and saturation deficit term and this u 0 we will be using as u 0.6. So, using this data we get a value of 13.33 millimeters per day you can see that here we have 10.77 millimeters per day here we have obtained a value of just 13.33 millimeters per day. So, they are a little different and not exactly the same thing. The other thing which we do this is only evaporation. Now, most of our times we would be interested in combining evaporation and transpiration. So, we need to know what is the evaporation from a water body as well as what is transpiration from plants or vegetation in that area and as we have already discussed we combine these two and call that evapotranspiration. Evapotranspiration will depend on a lot of factors for example, what is the solar radiation available at that point, what is the number of daylight hours because if we have cloud cover then naturally evapotranspiration will be smaller. So, 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. So, we would look at some of these techniques and see how to compute evapotranspiration based on some of the empirical and theoretical approaches. For example, we have seen the Blaney-Cradle equation which says that the evapotranspiration is equal to 25.4 again this 25.4 comes because we convert inches into millimeter and there is some factor k here which will be function of the crop type. So, 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 it and there is another method which is commonly used is the thorn to it equation which again tells that E t is equal to 16 L a 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. So, let us take this data we have 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-Cradle equation is taken as 0.65 for rise it is 1.1. Now, let us look at two months months of November and December for which the mean monthly temperatures are 15 for November and 12 for December and what we need to find out is evapotranspiration for this period of two months November and December. In the Blaney-Cradle equation which is written as evapotranspiration is 25.4 k f 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 for example some of the values are rise 1.1, maize 0.65, sugar k in 0.9 natural vegetation if it is dense then 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 p h p h is the monthly percent of annual daylight and this will depend on latitude and of course the time of year which we say as month. T average is given for November and December as 15 and 12 degrees in the Blaney-Cradle equation which was based on FPS system of units the temperature was in degrees for an height. So, we can convert this 15 and 12 degrees Celsius to 59 and 53.6 degrees Fahrenheit. The p h 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 for example 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 you are talking about equator 0 degrees north in January there would be 8.5 percent daylight hours compared to the number of daylight hours for the whole year. So, 8.5 percent of the year daylight hours will occur in January. Similarly if we look at February the number of daylight hours in this case you would see is the smallest at 7.66. If number of daylight hours are same in every month then the percentage would be 100 by 12 or 8.33 percent, 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. So, this depends on the latitude as well as the season and therefore from this table we can find out the percentage which is required in the Blenie Cradle equation. So, for example the latitude which is given if we look at the data given for a 30 degree north latitude and the year the month has November and December we can look at the percentage values 7.19 percent for November and 7.15 percent for December. So, using these two values of percentage and using the mean temperature we can estimate the value of F which is summation of P H monthly percent we have already seen this as 7.19 and 7.15 percent. T average is 59 for November 53.6 for December. So, the summation of these two sigma P H T average for these two months comes out to be 8.07. So, knowing F and K we can find out the evapotranspiration 25.4 conversion from inch to millimeter K for wheat and F for November and December months for 30 degrees north latitude. It gives us 133.3 millimeters of evapotranspiration over the two months of November and December. Similarly, if we use the thorned weight equation we can write E T equal to 16 L A 10 T average over I T and exponent A. The exponent A as you can see from here is given by another equation which relates it with I T. I T is the heat index and it is given as for the whole year summation from 1 to 12 T bar over 5 1.514 which is this equation. So, T bar is the average monthly temperature. The heat index can be obtained if average temperature are given for each month. So, we have this additional data which should be given to us. This is the mean monthly temperature and of course, this is in degree Celsius. So, in thorned weight we are using Celsius. So, for January you can see that the mean monthly temperature is 10 degree centigrade. Number and December of course, these were 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 I T and it turns out to be 96.1. So, T average by 5 for each month to the power 1.514 summing over all the month. So, this summation goes from I equal to 1 to 12 as written here. So, for each month we need to add up all these factors and when we do that we get I T equal to 96.1. A is a function of I T. So, some constant into I T cube again I T square I T and a constant term which gives us 96.1 sorry 2.1. For I T equal to 96.1, A is given as 2.1 and then the value of L A is the one which is required for different is given in table for different latitudes and different months. So, this is the adjustment factor L A and for the given latitude of 30 degrees for months of November and December we can see that the correction factor of the adjustment factor is 0.89 and 0.88. So, using this for these two months we can find out the value of E T as 36.3 for November. So, this is the December and this is November. So, 36.3 millimeters in November and 22.7 millimeters 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. So, 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. So, 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 equations other tables which show behavior with different latitudes and different times of the months. For example, in this case we can look at mean solar radiation in terms of millimeter of evaporable water per day. Again for different latitudes and different times of months, 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. So, the mean solar radiation is given in terms of millimeters of evaporable water per day because evaporation of water requires some energy. So, this millimeter of evaporable water per day indicates what is the energy coming in at the atmospheric level. And as you can see for different at for example, this 30 degrees north latitude in November the solar radiation will be able to evaporate 9.1 millimeter of water per day in December on the 7.9 millimeter per day. At 20 degrees north it would be able to evaporate 11.2 and 10.3 for the months of November and December. So, the other thing which we need for most of the theoretical equations is what is the sunshine hours per day. The mean potential sunshine hours per day actual sunshine hours of course, will be different depending on the cloud cover on that day. But this is the potential sunshine hours per day and of course, this will depend on the latitude and the time of the year. At the equator of course, this value does not change with time of the year. So, it is 12.1 hours of sunshine every day expected on the equator throughout the year. For 10 degrees north latitude, again Wednesday winter months since we are talking about northern hemisphere, January, February, November, December these are the winter months. So, the sunshine hours are smaller during this time and then they become larger during the summer months. So, that trend you can see here also a smaller number of daylight hours, sunshine hours in the winter months and large number during the summers. So, 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 finding out the evapotranspiration involves solar radiation. The equation which is given by Penman is potential evapotranspiration would be some factor A times h n this is the radiation. Again another factor gamma E A this is the evaporation and weighted by. So, these are the weights A and gamma and then divide by total value A plus gamma. The data which is given in this example is that for the month of November at a latitude of 30 degrees north mean monthly temperature is given as 15 degrees centigrade. In this case let us take the mean relative humidity as 75 percent and actual mean daily sunshine hours for the month of November are taken as 9. Potential the sunshine hours are larger than this, but actual daily sunshine hours are only 9 because sometimes there may be cloud covers. Wind velocity is measured in this case at 2 meters above the surface and was found as 4 kilometers per hour, but in Penman's equation we need the velocity in kilometers per day. So, we converted into 96 kilometers per day. There is an albedo required in the Penman's equation which is the reflectance of the surface. It is given that this area has very close to ground crops which have albedo of 0.2. This is the reflectance 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. So, 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 typically taken as 0.49. Then there is one more term which we require the slope of the saturation vapor pressure curve. This also depends on a temperature. So, if we take a temperature of 15 degrees centigrade the saturation vapor pressure is 12.8 and the slope of the curve is 0.82. So, for example, this shows the saturation vapor pressure and the next curve shows the slope of the saturation vapor pressure curve in terms of millimeter per degree centigrade. And for different temperatures we can find out its value. So, in this case for a temperature of 15 degrees Celsius the values which we will be using will be 12.8 millimeter of mercury pressure saturation vapor pressure E w 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 the mean solar radiation. So, this tells us that using these tables and the values of the coefficients we can estimate or find out the values of different parameters in the Penman's equation. So, we will look at the example. The month is November that is 30 degrees north. So, using these data we can obtain from that table which we have shown that h a is 9.1 millimeter of evaporable water per day. This we have seen in the previous equation function of latitude and month. So, the table gives us the available radiation as 9.1 millimeter. E w also we have seen for a temperature of 15 degrees was 12.8 millimeters of mercury. The relative humidity given in this case is 75 percent. So, the actual vapor pressure will be 75 percent of E w which is obtained as 9.6. So, E w and E a 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 deficit term u 2 means the wind velocity at 2 meters. In this case the velocity is given at 2 meters. So, we do not need to use the height correction and therefore, we will be using 96 kilometers per day as the velocity. So, using 1 plus u 2 by 160 the wind velocity term saturation deficit term multiplied by 0.35 we get E a as 1.79 millimeters per day. The slow slope of the saturation vapor pressure curve again we have obtained from a temperature of 15 degrees and it was obtained as 0.82. So, that is the value of A. A is the slope of E w versus T curve and this is in millimeters of mercury per degree Celsius. The value is obtained from the table as 0.82. So, A is also obtained a psychometric constant gamma is 0.49 millimeters of mercury per degree Celsius that value is almost a constant. In the Penman's equation we use the St. Pans constant and the value which is obtained is 2.01 10 to the power minus 9 millimeters per day. The Stefan Boltzmann constant is used here sigma T to the power 4 T is the temperature in degree Kelvin. So, T as we have written here for a temperature of 15 degrees Celsius we have to add to 73.3 and we get a 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 30 degrees north can be obtained from the table and the value which is obtained is 10.6 hours. So, 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. So, this may happen due to number of reasons some days may be cloudy. So, sunshine hours are limited. So, 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 degrees and 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 seen the value as 9.1. So, in this equation all the terms are known and therefore, we can find out the net radiation in terms of evaporable water per day as 1.74 millimeter per day. So, as we can see here E A and H n both have been obtained in this case they are very close H n is 1.74 and E A is 1.79. So, in this case the relative magnitude of A and gamma will not really matter because these two values are very close. So, 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 millimeters per day. So, 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 percent. Then we can expect for event velocity of 4 kilometers per hour and for an albedo of 0.2 that the potential evapotranspiration from that area would be 1.76 millimeters per day. So, using either the theoretical equation where spendments is the most common or some empirical equations for example, thorns width we can find out what is the potential evapotranspiration from that area. So, knowing the initial abstraction and the potential evapotranspiration the only other abstraction which we need to look at is the infiltration. Infiltration depends on the soil type and as we have seen there are some curves which relate the temporal variation of infiltration rate. For example, infiltration 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. So, generally the infiltration rate will decrease with time. So, there will be some rate here at time 0 there will be some rate here at infinity and then typically an exponential decrease is specified as the change of infiltration with time, but this becomes very complicated to estimate these parameters f 0, f infinity and there is some rate here which typically is like 8 power minus 8 power minus k t. So, this rate finding out this k also becomes complicated. So, there are some simple methods which can be used to estimate the infiltration loss. The infiltration indices in a way represents 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. So, if we are given a storm hydrograph for example, a storm may look like this intensity may be millimeter per hour and time in hours. So, we may be given that rain occurs with this intensity and this temporal variation and we may be given runoff. So, 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 expressed in terms of phi index. So, let us take an example in which the precipitation of 88 millimeters occurs in 8 hours as shown in this table that from 0 to 1 hour the intensity is 4 millimeters per hour. Similarly, 1 to 2 is 8 and 7 to 8 intensity is 5. So, if we add all these up we get 88 millimeters of precipitation. It is also given that because of this rain there is a 46 millimeter of runoff and therefore, a total loss of 42 millimeter 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 find out the phi index what we need to do is draw a line here such that the area above this represents runoff and the area below this represent the loss. Then this value is called the phi index. So, 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 the loss. So, in this case for example, we can say that the average loss is in 8 hours we have loss of 42 millimeters. Therefore, the average loss is 5.25 millimeters 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. So, first period is only 4, last period it is 5. So, in the first period in the last period the losses cannot be 5.25 because the losses are limited to the 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 one more iteration in which we assume that in the first hour the loss is 4 millimeters, in the 8th hour the loss is 5 millimeters. And in other 6 hours we have to find out what will be the loss. So, we can say that a loss of 42 minus 4 minus 5 will be in 6 hours, which means that a loss of 33 millimeters in 6 hours giving us average loss of 5.5 millimeters per hour. And all the other values are higher than 5.5 therefore, we can say that phi index is 5.5 millimeter per hour. So, in this way we can obtain the value of an average loss over the entire period. T e we can also if we look at the hydrograph and let us say that the phi index is somewhere here. Then we can say that an effective duration. So, this T e will be 5.5 millimeter per hour duration of effective rainfall or excess rainfall. So, this rainfall is what is contributing to the runoff and the rest is lost to infiltration, evaporation and abstraction. So, the 6 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. For example, there is a W index, which is given as precipitation minus runoff minus runoff minus runoff minus and initial abstraction divided by the duration of rainfall excess. But phi index is little more common. There are some other methods also of estimating phi index. So, 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. So, this 4 is completely lost. Out of this 8, we have a loss of 5.5 millimeters. So, all the other values we subtract 5.5 except the first in the loss will become 0 and sum of all these will turn out to be 46 millimeters. Central water commission CWC of India has given an equation by correlating the runoff with rainfall. So, if rainfall of i centimeter per day occurs in 24 hours, the runoff can be correlated by alpha i to the power 1.2. Alpha is depending on the soil type 0.2 for sandy 0.45 for clay. So, 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 excess rainfall duration. So, 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 infiltration. We have looked at various empirical and theoretical methods to obtain an estimate of these parameters which can be used to find 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 event.