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Drip Irrigation Disposal Area Effluent disposal through drip irrigation is designed to take advantage of the warm and w7withan climatic conditions in this locale between six and eight months out of the year. Drip irri effluent disposal will comprise an area of approximately 19,500 ft' (195 ft x 100 ft) whi divided into three subareas of 6,500 ft'. The three subareas will be automatically switch indexing valve (or manually switched)to maintain sufficient head pressure and equal effluent distribution. Each tubing drip line will extend 100 feet through the disposal area in an east-west direction, and will have a pressure compensating emitter every two feet to deliver effluent at a rate of 0.53 gph at 20 psi. Tubing will be placed two feet apart. To compensate for the loss of plowed surface soil, engineered fill soil may have to be imported. Calculated flow: 50 emitters/100 ft of tubing. 195 ft N-S disposal area_ 3 subareas= 65 ft-2 ft spacing between drip lines=32 drip lines/subarea. 0.53 gph x 50 emitters/line=26.5 gph/line x 32 drip lines= 848 gph/subarea x 3 subareas=2,544 gallons. The estimated wastewater flow per day of 2,475 gallons can be theoretically distributed over the entire disposal area. Wastewater effluent drip irrigation technology is relatively new. Initial problems have been resolved, such as root clogging. However, the drip lines do have a life span which is indeterminable due to numerous factors, and typically range from two years to several years. Nitrogen Loading Through the Drip Irrigation System Determination of the nitrogen concentration available from the drip irrigation disposal area for grass groundcover uptake: It was determined from the Hantzsche/Finnemore Equation(most conservative) that 7.4 NO3 N (32.9 ppm NO3)=20.1 lbs N in an acre-foot of water. As determined on Page 11, 62.5 inches of domestic irrigation water must be applied to the grass groundcover over the drip irrigated disposal area, April through September. Water applied= 19,500 ft-43,560 ft'/ac= 0.45 ac x 27,154 gallons/ac-in. = 12,219 gallons applied per inch of water x 62.5 inches=763,687 gallons. Assuming that all 2,475 gallons of wastewater contributes to the domestic irrigation water applied per day during April through September: 2,475 gallons/day x 183 days =452,925 gallons of wastewater. 763,687 gallons minus 452,925 gallons =310,762 gallons of domestic well water needed to supplement irrigation. Wastewater Effluent Volume in ac ft=452,925 gallons_ 325,851 gallons/ac-ft= 1.39 ac-ft Domestic Well Water Volume in ac ft=310,762 gallons- 325,851 gallons/ac-ft=0.95 ac-ft. Assume the future domestic well has a nitrate-nitrogen(N0;N) concentration of 4.5 ppm: 4.5 ppm x 2.72= 12.2 lbs N/ac-ft. x 0.95 ac-ft= 11.6 lbs N. Assume the wastewater effluent will have 20.1 lbs N/ac-ft as calculated above x 1.39 ac-ft=27.9 lbs N: 11.6 + 27.9=39.5 lbs N in applied water. Kentucky blue grass requires approximately 124 lbs N/ac/yr for optimum growth: 19,500 ft' drip irrigation area-43,560 ft'/ac =0.45 ac x 124 lbs N/ac/yr= 55 lbs N/yr required for grass in the drip irrigated area. Therefore, a theoretical nitrogen deficit of: 55 lbs N/yr minus 39.5 lbs N/yr in applied water= 15.5 lbs N/yr may exist within the drip irrigated disposal area. Page -21- Chesney Consulting