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` Nitrate Mass Balance Calculation (7 people, 2.10 acres) Data Input.- Occupancy: 7 people Waste Loading(W): 2.24 inlyr Gallons Per Person Per Day: 50 gpd/person Site Area(A): 2.90 Acre Effluent Quantity(Q): 350 Gals Concentration Rain(Nb): 0.97 mg1L-N Effluent Stream(Nw): 45,00 mg/L-N Denitrification(d): 25% Deep Perc.of Rain(R): 6.79 inches Result: Mass Balance(N8.50 mg/L-N MCL Drinking Water Nitrate as N: 10.00 m N Percent of MCL Nitrate as Equations: 7 people x 50 gals/day/person =350 gals/day Find Effluent Quantity(Q) (W) Ln = (Q) gal x 1 ft? x 365 da x 1 acre x 12 in x 1 Find OM yr day 7.48 gal 1 year 43,560 fe 1 ft (A) 2.24 in/yr(site)=350 gallday x(9 cu-ft 17.48 gals)x(365 days/9 year)x(1 acre 143,660 sq-ft)x(12 in/i ft)x(1 site 12.1 acres) Nc= WN„(1-d)+RNy Hantzsche-Fennemore Equation(NO W+R 8.5 mgt-N=((2.24 inlyr x 45 mg1L-N x(1-0.25))+(6.79 in/yr x 0,17mg1L-N))1(2.24 in/yr+6.79 inlyr) Variables: Nc=Average nitrate-N concentration(mgA)of combined effluent and rainfall percolate(8.5 mg/L-N). W=Uniform waste water loading for study area(inlyr)(2.24 incheslyear). Nw=Total nitrate-N concentration of waste water percolate(45 mg/L-N). d=Fraction of nitrate-N loss due to denitrification in the soil(25%). R=Uniform deep percolation of rainfall(5.79 in/yr). Nb=Background nitrate-N concentration of rainfall(0,17 mg/L-N). Assumptions: 1. Nltrate-N concentration of waste stream based on estimate of 45 mg/L-N. Assumption based on work from Hantzsche and Finnemore,1992. 2. Fraction of nitrate-N loss due to denitrification in the soil is assumed to be 25%. Assumption based on work from Hantzsche and Finnemore,1992. 3. Estimated deep percolation of rainfall is 6.79 inlyr,see deep percolation of rain worksheet. i 4. Assume background nitrate-N concentration of rainfall is 0.17 mg/L-N based on range by Hantzsche and Fennemore,1992. pl R� NEIL O.ANDERSON&ASSOC.,INC. r °s ¢oioncK ernw as•v2rnyjpyaaaS W ]Z iknrs�anLnd L�kCh,cs�fx�e ClOel Al-Ao1 a pC� FLATS 4 }