 We have already discussed the theories in this water requirements of crops and we discussed about the base period delta and duty. We also discussed about different efficiencies in the irrigation system. Also, we have discussed about the soil moisture relationship. Now, we will solve some problems, some numerical problems, so that the things will be more clear. The first problem in this series is a crop requires 1 meter of water for its base period of 120 days, find the duty of water. So, delta is equal to 1 meter. I can write this as 100 centimeter and B is 120 days and D is what? So, we know the formula for delta is 864 B over D. So, if I exchange the terms D will be equal to 864 into B over delta which will be equal to 864, B is 120 in our case and delta is 100. So, you calculate and find out whatever is the number and this unit will be in hectare per cubic. Let us go to the second example. Water is released at the rate of 10 meter cube per second. So, this is the cube given at the head of a canal. Estimate the area that can be irrigated if duty at field is 1220 hectare per cubic and transit loss of water is 20 percent. So, here cube is given which is 10 meter cube per second and area is given which is 12 sorry this is duty is 1220 hectare per cubic and the question is what is the area that can be irrigated if duty is this. So, as we know D is equal to A over Q. So, this says A will be equal to D times Q which is equal to 1220 into 10. Please remember this Q mech and meter cube per second they are same thing. So, this becomes hectare whatever is the number 1, 2, 2, 0, 0 hectare. Now there is a term transit loss of water is 20 percent. So, when the loss is this much we cannot say that this is the area because there is some loss. That means in place of the Q given we should have used the actual Q available to the field. That means in place of 10 we should have multiplied 80 percent of that because 20 percent will be loss. So, multiply a factor of 0.8 which is 80 percent. So, this number will be changed and 0.8 and the answer will be in hectare. Now let us go to problem number 3. The gross commanded area for an irrigation canal is 10000 hectare and 80 percent of this land is cultural commanded area. Intensity of irrigation is 40 percent. Calculate the outlet discharge and delta if base period is 60 days. The outlet factor is 1500 hectare per Q mech. So, this is a kind of duty. Although it is written outlet factor you can think that this is a duty and also you can verify from the unit this is hectare per Q mech. So, 1500 is the duty. The gross commanded area for an irrigation canal is 10000 hectare and 80 percent of this land is culture able. So, the area that will be culture able will be 80 percent of this that means 8000. Then intensity of irrigation is 40 percent. So, out of this 8000 hectare we have irrigation at 8000. So, irrigated land will be 8000 into 40 percent. So, 0.4 which will give me 3200 hectare. Now for the calculation I should use this as the A and duty is given. Now the question is calculate the outlet discharge. So, what will be Q? Q is what? So, if this is area and duty is 1500 hectare per Q mech. So, Q will be duty multiplied by A because duty is area divided by Q. So, Q duty is area divided by Q. So, Q will be equal to A divided by D which is equal to in this case 3200 hectare divided by 1500 whatever is the. So, to whatever is the number 20 you use your calculator to calculate this number 2. something this much Q mech. So, the outlet discharge will be 2.1 or 2 Q mech and also it is asking about delta because base period is given. So, if I have to calculate delta what should I do? I should use the relationship between base period, duty and delta which is delta is equal to 864 times B over D and here B is 60 days. So, 60 divided by D is 1500. So, use your calculator and find out this number whatever is the. So, for delta we should use this formula and whatever is the answer will be in centimeter because B is in days and D is in hectare per Q mech. So, let us go to the next problem. Problem number 4, a water course commands an irrigated area of 1000 hectare. Intensity of irrigation for this area is 75 percent. Considering the first 15 days for a crop total depth requirement is 50 centimeter. During this period useful rainfall is 15 centimeter. Calculate the duty for the crop on the field for this period. In addition find out the duty at the head of the water course assuming a loss of 20 percent. So, in any problem what you should do is first try to note down whatever data are given in the problem then also note down what is required. So, in this problem we have the area is 1000 hectare and intensity of irrigation is 75 percent. That means area to be irrigated will be equal to 7500 sorry 750 hectare 75 percent of 1000. Considering the first 15 days for a crop. So, here though the base period is not 15 days, but the problem is asked for the first 15 days. So, the time here is 15. So, we can take B as 15 days. Total depth requirement is 50 centimeter. So, delta is 50 centimeter and you can see during this period useful rainfall is 15 centimeter. That means if the requirement by the crop is 15 centimeter and rainfall is 15 centimeter. That means through irrigation how much water we have to give this will be 50 minus 15 which is 35 centimeter. So, for the purpose of calculation we should use delta is equal to 35 centimeter and not 50 centimeter. Calculate the duty for the crop on the field. So, now B is given delta is given and duty is asked. So, duty will be what? You should use the formula delta is equal to 864 B over D and all are in consistent units. So, here 35 centimeter will be equal to 864 here B will be 15 and D is what? So, D will be equal to 864 times 15 divided by 35 and this will be whatever is the number the unit will be vector per cubic. This is the duty. Now, this is the duty on the field. The next question is find out the duty at the head of water course and the loss is 20 percent. So, remember that this is the duty whatever number we got here is the duty on the field. I am just writing it through a subscript F. So, at the head of the water course the duty whether it will increase or decrease in the discussion on duty it varies with the downstream movement of the canal system. So, as we go down the duty increases that means here this duty is on the field. So, as we go up duty will decrease. So, this loss is 20 percent that means 80 percent is available we should multiply 0.8 to this number. So, D duty on the head of water course will be equal to whatever number you got here multiplied by 0.8. 5.8 because 20 percent is loss while water is moving downstream there will be loss of 20 percent. So, duty here will be less. So, this problem is over. Let us go to the next problem. Question number 5 calculate the discharge required at the head of the canal and the design discharge if the capacity factor is 0.8 and time factor for the canal is 0.55 use the following data. This is an interesting problem because you are asked to design your canal system at different period of a year the water requirement might be different, but while designing what we should do we should look for the maximum water requirement and it is generally in the summer days. So, we should look for that and depending on that maximum value we should design the canal system. Now, let us solve this problem here in the table 1, 2, 3, 4, 5 different crops are given one is sugarcane and the base period is 320. So, it almost covers most part of the year and similarly this is sugarcane overlap in hot weather which is for 90 days and also we have a typical Robby crop, a typical Kharib crop and a typical hot weather vegetables with different base periods and the area irrigated will be respectively 900 hectare, 150 hectare, 750 hectare, 600 hectare and 320 hectare and the duties are respectively 580, 580, 1600, 2600. Why duties are different? We are considering the same land for duty may be different depending on the base period, depending on the water requirement of the crop. So, let us consider individual crops and let us try to find out the discharge for individual crops. If I designate 1, 2, 3, 4, 5 for 5 different crops then I can find out the required discharge for each one of these crops. For example, here for sugarcane I have duty equal to 580 and area is equal to 900 hectare. So, for sugarcane discharge will be what we know duty is A divided by Q. So, Q will be A divided by D. So, this area divided by the duty I can get. So, for the first thing it will be 900 divided by 580 whatever is the number. Similarly, for other crops I can find out 150 divided by 580, 750 divided by 1600, then 600 divided by 2000, 320 divided by 600. So, whatever number you get here you write down. Please remember in the discussion on these problems I am trying to give you only the clues. The detailed calculations are to be done by you. Now, we get Q, 5 Q values or discharge values for the individual crops. Now, out of these crops I should categorize them for Kharif, Rabi and hot weather and try to see what is the water requirement. For example, in case of Kharif what should I consider? I should consider number 1 and number 4 which is Bajra. So, for calculating the water discharge required in Kharif season I should use number 1 and number 4. Similarly, for Rabi what should I use? I should use number 1 and number 3 and for hot weather what should I use? I should use 1. Why 1 is used everywhere? Because sugarcane is almost spanning throughout the year. So, you should use the water required for sugarcane in all these calculations. So, for hot season I should use number 1 then also number 2 and number 3. Common sense says that this will give you the maximum number because during hot weather you need more water. But you verify these numbers whatever number you are getting and choose the maximum out of these. So, you choose the maximum discharge Q Kharif, Q Rabi and Q hot season. Then that Q will be your design discharge. But wait a minute because we have to use these factors also. The capacity factor is 0.8 and time factor for the canal is 0.55. So, you have to use judiciously these factors to find out what is the design discharge. So, let us say Q H is the maximum. So, whatever number you got by calculating this discharge you have to again take into account these factors. So, Q H you should divide by 0.8 for finding out I am writing here 0.8 and also you should divide by 0.55 for the canal. Now, this is the design discharge for your canal. So, if you are planning to construct a canal system your canal system has to be designed for this much discharge. Let us go to the next problem, problem 6. Here the data are field capacity of soil is equal to 30 percent. Permanent wilting percentage is equal to 10 percent. Density of soil is equal to 1300 kg per meter cube. Effective depth use root zone is equal to 70 centimeter. Daily consumptive use of water is equal to 1.5 centimeter. Moisture content must not fall below 25 percent of the water holding capacity between the field capacity and the permanent wilting point. Find out the frequency of irrigation. Let us say this is the end of the ground water. This is the field capacity and this is the permanent wilting point and this is the optimum water content. So, this is 30 percent and permanent wilting point is 10 percent. So, this is 0.2 or 20 percent and here it says moisture content must not fall below 25 percent of the water holding capacity. So, this is the water holding capacity. That means we should allow only 75 percent of this because it should not fall below 25 percent. So, what is the water required by the irrigation water for this depth? It will be this moisture content which is 20 percent and again 75 percent which is this. Why 75 percent? I am using this information. Then I should multiply the density of soil which is 1300 and I should divide it by water which is 1000. Now, with this I should multiply the root zone depth which is in our case 70 centimeter. So, my answer this will be in terms of centimeter whatever number I got. This is 102030 and maybe I can delete this 0. So, basically 2 into 75 into 13 into 7 divided by 1000. This is D of water in centimeter because this number 70 is in centimeter. Now, the question is Delhi conjunctive use of water is 1.5 centimeter and this is the centimeter we have to provide. So, when Delhi conjunctive use is 1.5 centimeter. So, the number of days in which we should repeat irrigation will be this divided by 1.5. So, this number will be the number of days we should repeat because first water we will try to in one day it will calm down to 1.5 centimeter and again we have to supply this water. Again it will in one day it may calm down to 1.5 centimeter. So, in this way you calculate this number whatever is the number it will be around this is 150, this is 91 let us say 100 divided by 1.5 into 1000. So, you get this is 100 and 30. So, around 10 days you get. So, here the frequency of irrigation is 10 days. There is a precaution here suppose your answer comes out to be let us say 10.2 or may be it is 10.8. So, if it is 10.8 let us say you should not feel that 10.8 is more than 10.5. So, it should be equal to 11 because you cannot fix your irrigation at 10.8 days it has to be either 10 days or 11 days. So, to be in this same side 10.8 should be taken as 10 days not 11 days. Similarly, if it is 10.2 it will also be 10. So, the nearest whole number or the nearest integer number smaller than the number you get here that will be your number of days after which you should repeat irrigation. This problem is over now I will switch over to the next problem. 7. The discharge available from a tube well is 150 meter cube per hour. Assuming 10 hour of pumping per day for 330 days in a year find out the cultured area that this tube well can command. The intensity of irrigation is 40 percent and average water depth required for the crops is 50 centimeter. Let us write the data given in this problem and then we will solve this is 150 meter cube per hour and it is supplied for 10 hour in a day. So, if I want to calculate the volume of water in a year then it will be 150 which is the discharge multiplied by 10. So, this will be the water volume in a day, but throughout the year I am supplying the water for 330 days. So, multiply this whatever is the number you get here this will be in meter cube. Now, it says find out the cultured area. So, this is the water volume. Now, if I divide this through the depth then I get area and here the depth is the water depth required for the crops is 50 centimeter. So, whatever is your volume so area will be volume divided by depth and in this case depth is 50 centimeter. If I write this in meter this will be 0.5. So, this will be in meter square. Please remember that you have to calculate this and place that here then you find the final number here in terms of meter square this is the area and this will be generally a very large number and we express area in hectares. So, for that matter whatever is the number here you have to divide that by 10 to the power 4 this will give you in hectare. Now, we must take this into account the intensity of irrigation is 40 percent. So, let us say you get here 100 hectare then this should be multiplied by 40 percent because your intensity of irrigation is 40 percent it can actually serve 40 percent of this area. So, the final answer will be this number whatever number you get at this step multiplied by 40 percent that much hectare. So, in these calculations we must take into account the losses or if there is intensity of irrigation those factors are to be considered to find out the final answer. Let us go to the next problem. So, problem number 8 following particulars we are recorded from an agricultural land. Field capacity is 20 percent, wilting point is 10 percent, permissible depletion of available moisture content is 50 percent, root zone depth is 2 meter, dry unit weight of soil is 1400 kg per meter cube, effective rainfall is 2.5 centimeter, daily consumptive use is 0.6 centimeter. Find the net and field irrigation requirement if loss in field application is 10 percent. What is the frequency of irrigation? Now, let us see one by one this is field capacity which is 20 percent and this is permanent wilting point which is 10 percent and 50 percent is permissible. I am trying to explain you here something sometimes this data will be given in terms of moisture content like this optimum level sometimes it will be given 14 percent. That means that number is in between these two but here it is 50 percent and as you are seeing clearly this 50 is not in between 10 and 20 what is the meaning of this? This means we can use 50 percent of this moisture. So, we should allow to drop down by 50 percent of this moisture that means if this is 10 this is 20 we can go up to 15. So, what is the water required? The water requirement in terms of depth will be 20 minus 15, 20 is the upper limit 15 up to which we can go and in percentage so divide by 100 multiply the dry unit weight which is 1400 and divide by the unit weight of water which is 1000 and root zone depth and root zone depth is 2 meter. So, the answer here will be in terms of meter this will be the water we have to supply through irrigation for the crop and here it will be 22 this is 5, 10, 10 divided by 1000 that means 10 millimeter of water. We have to supply 10 millimeter of water, water required by the plant in this case will be 10 millimeter of water. Out of this something is given by rainfall 2.5 centimeter. So, here is some error I have not included this 14 so it will not be 10 millimeter. I was wondering because this 2.5 centimeter is higher than this if suppose this is the case if the water requirement is 10 millimeter and the rainfall is 2.5 centimeter then there is no need for irrigation. So, in this case it will not be 10 millimeter it will be something else this is 5 this is 14 this is 2. So, it will be 14 into 5 into 2 divided by 1000 which is 5 to 10, 140 by 1000. So, this is 140 by 1000. So, 140 millimeter or 14 centimeter. So, out of 14 centimeter in this case we are supplied with 2.5 centimeter by the rain requirement is 14 centimeter and 2.5 centimeter is by the rain. So, by irrigation we need 11.5. Now, let us see what are the other data. So, total water requirement by irrigation is 11.5 centimeter and here the Delhi consumptive use is 0.6 centimeter. So, the irrigation frequency will be decided on that, but let us first calculate this one find the net and field irrigation requirement if loss in field application is 10 percent. So, to find that loss is 10 percent. So, net irrigation requirement we call it NIR will be equal to 11.5 centimeter and field irrigation requirement will be more than that because losses should be taken into account. And as in this case the application the field application loss is 10 percent. So, you can say 0.9 will be available. So, requirement in the field will be 11.5 divided by 0.9 whatever number you get. So, it will be 125. So, around 13 centimeter. Now, Delhi consumptive use is let us say 0.6 centimeter which is a very small number that is why the irrigation frequency will be this 13 divided by 0.6 whatever is the number around 20. Please remember while finding out the frequency of irrigation what we must do we should not take into account NIR net irrigation requirement we should take into account the field irrigation requirement FIR. So, in our case the net irrigation requirement is 11.5 centimeter and field irrigation requirement is 13 centimeter. So, based on 13 centimeter value we should find out the frequency of irrigation and in this case this is 13 divided by 0.6 which is approximately 20. So, let us go to the next problem. Problem number 9 find the Delhi consumptive use and discharge required in the canal feeding a certain crop on an area of 5000 hectares. Other data for the canal system are field capacity is 30 percent, optimum moisture content is 14 percent, permanent wilting point is 10 percent, effective depth of root zone is 0.8 meter, apparent relative density of soil is equal to 1.4, frequency of irrigation is 10 days, overall irrigation efficiency is 25 percent. So, we have to find out Delhi consumptive use. So, let us do one by one let us say this is field capacity which is equal to 30 percent then this is permanent wilting point which is 10 percent and optimum moisture is 14 percent here. So, this is 16. So, if I have to find out what is the water required for this then it will be 16 percent which is 16 by 100 multiplied by the density, dry density for soil which is in this case not given but it is given indirectly. This factor 1.4 the apparent relative density is basically when you divide the density of soil with the density of water. So, it is 1.4 remember that this 1.4 takes into account both and multiplied by the depth and here the effective depth of root zone is 0.8 meter. So, 0.8 meter and the number will be in meter we need this much of water and this will be this is. So, here this is the total water requirement in meter and we know that the frequency of irrigation is 10 days. So, if I have to find out the daily consumptive use then what should I do? So, this much water is consumed in 10 days. So, this should be divided by 10 days. So, whatever answer I got here this divided by 10 days will give me the daily consumptive use and of course the answer will be in meter which will be a very small number. So, what you have to do is you must express that in terms of centimeter or millimeter to have some meaningful number. So, with this we conclude the series of problems and I am sure the theory is clear now and if you practice more problems then the concept will be more clear to you. Thank you.