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<br />CUO::.9G <br /> <br />. <br /> <br />First, determine the farm irrigation system efficiency, Eis' by the <br />method outlined, Then calculate: <br /> <br />R = <br /> <br />Du <br /> <br />Dd <br /> <br />where Du is the requirement at the time of irrigation and Dd is the amount <br /> <br /> <br />of water delivered. The units of requirement and water delivered are volume <br /> <br /> <br />or volume per unit area. Three cases are considered. <br /> <br />Case lEis :': R <br /> <br />This implies that an excess of water over that required for good irri- <br /> <br /> <br />gation is available, All in excess is lost, Therefore, <br /> <br />D~ = Dd - Du <br /> <br />where D is the water lost, in the same units as Dd and Du' <br /> <br />Case II Eis < R; Dd > Du <br /> <br />. <br /> <br />This implies that there is not sufficient water to meet the requirement, <br /> <br /> <br />although there could be enough if the irrigation efficiency were increased. <br /> <br />Assume that there is some adjustment by calculating a new value of field <br /> <br /> <br />irrigation system efficiency, <br /> <br /> <br />Eis' = Eis + O.2[(Du/Eis) - Dd1/[(Du/Eis) - Dul <br /> <br />where Eis' is the adjusted value of Eis' The losses are calculated as follows: <br /> <br />D = (1 - Eis') Dd <br /> <br />It is to be noted that the above calculation limits the value of Eis' <br />to a maximum of Eis + 0.2, <br /> <br />Case III Eis < R; Dd ~ Du <br /> <br />Water has become even more limiting. The maximum possible value of <br /> <br /> <br />irrigation efficiency is used, and this is the initial value plus 0,2. <br /> <br /> <br />Thus, <br /> <br />D~ (0.8 - Fis)Dd <br /> <br />. <br /> <br />-33- <br />