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<br />II <br />I <br />I <br />I <br />I <br />I <br />I <br />I <br />I, <br />I <br />1 <br />1 <br />I <br />1 <br />I <br />I <br />I <br />I <br />I <br /> <br />TABLE 5-6: PUMPING HOURS REQUIRED TO MEET PEAK MONTHLY <br />AND SEASONAL CROP IRRIGATION REQUIREMENTS <br />FOR WHEAT FOR 122 ACRE PIVOT <br /> <br />WELL HOURS TO MEET PEAK HOURS REQUIRED TO MEET <br />CAPACITY, MONTHL Y DEMAND FOR SEASONAL WATER REQUIREMENT <br />GPM WHEAT' FOR WHEA T2 <br />500 534 1401 <br />550 486 1274 <br />600 445 1168 <br />650 411 1078 <br />700 382 1001 <br />750 356 934 <br />800 334 876 <br />850 314 824 <br />900 297 778 <br /> <br />Footnotes <br />1. The peak monthly demand for wheat as determined by using the Blaney-Criddle <br />method was determined to be 4.11 inches for May, see Table 1. The volume of <br />water required to meet that demand is estimated to be 4.84 inches during the <br />month of May, assuming an irrigation efficiency of 85 percent (4.11 inches/0.85). <br />This is equivalent to 49.2 acre-feet for a 122 acre center pivot (4.84 inches x 122 <br />acres/12 inches/ft.). <br /> <br />2. The number of hours required to operate an irrigation well to satisfy the seasonal <br />water requirement for wheat is determined by dividing the total volume, 1.06 <br />acre-feet per acre (0.9 acre-feet per acre, presented in Table 1 divided by an <br />irrigation efficiency of 85 percent). Based upon this seasonal requirement, a 122 <br />acre pivot will require 129 acre-feet per season (1.06 ac. ft.lacre x 122 acres). <br /> <br />V-10 <br />