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Nitrate Mass Balance Calculation <br /> F. <br /> Data Input: <br /> I <br /> Effluent Quantity(Q): 730 gals/day Concentration Rain (Nb): 1.00 mg/L-N <br /> t Effluent Stream Nw : 56.50 m <br /> ( ) g/L-N Denitrification (d): 30.0% <br /> Site Area(A): 5.00 Acres Deep Perc. of Rain (R): 5.76 in/yr <br /> Result: <br /> Mass Balance(Nj: 10.8mg/L-N <br /> MCL Drinking Water Nitrate as N: 10.0 mg/L-N Waste Loading (W): 1.96 in/yr <br /> Percent of MCL Nitrate as N 108% <br /> Equations: <br /> q:�D <br /> (W) in = (Q) gal x 1 ft3 x 365 day x 1 acre x 12 in x 1 Find <br /> yr / day 7.48 gal 1 year 43,560 ftZ 1 ft (A) acre <br /> ✓ 0• -0002.z1g 12-- <br /> 1.96 in/yr(site) =730 gal/day x(1 cu-ft/7.48 gals)x(365 days/i year)x(1 acre/43,560 sq-ft)x(12 in/1 ft)x(1 site/5 acres) <br /> Nr= WN-(1-d)+RN„ Hantzsche-Fennemore Equation Nc <br /> W+R <br /> n_'a- -+ 4-'�-Z <br /> 10-8 mg/L-N=((1.96 in/yr x 56.5 mg/L-N x (1-0.3))+(5.76 in/yr x 1mg/L-N))/(1.96 in/yr+5.76 in/yr) <br /> Variables: G� {/ 7,92— = /6.SSI- �,- /p,g <br /> Nc=Average nitrate-N concentration (mg/1)of combined effluent and rainfall percolate(10.8 mg/L-N). <br /> W= Uniform waste water loading for study area(in/yr) (1.96 inches/year). <br /> Nw=Total nitrate-N concentration of waste water percolate (56.5 mg/L-N). <br /> d=Fraction of nitrate-N loss due to denitrification in the soil (30%). <br /> R=Uniform deep percolation of rainfall (5.76 in/yr). j <br /> Nb= Background nitrate-N concentration of rainfall (1 mg/L-N), <br /> Assumptions: <br /> 1. Total nitrogen concentration of waste stream based on estimate of 56.5 mg/L-N. <br /> 2. Fraction of nitrate-N loss due to denitrification in the soil is assumed to be 30%. <br /> 3. Estimated deep percolation of rainfall is 5.76 inlyr, see deep percolation of rain worksheet. <br /> 4. Assume background nitrate-N concentration of rainfall is 1 mg/L-N. <br /> Fj' <br /> F., <br /> k <br /> NEIL O. ANDERSON <br /> Ff" "44 AN D ASSOCIATES } <br /> i <br /> Plate 10 <br />