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'+ ' Attachment 1 (Letter from B. R. Sabey) Nitrogen Calculations - Sludge and Manure SLUDGE Total N in the sludge 1500 mg/liter NH4 -N in the sludge 1275 mg/liter 1275 = 85% of total N is NH4+-N 1500 Total N=1.5gN=15gN N 1000 g 50 g dry sludge x 100 = 3% N on a dry weight basis 15 Tons/acre = 30,000 #/A @ 3% = 900# Total N 900 x .85 = 765# NH -N 900 - 765 = 135# Or$-N 765# NH -N x 20% left after volatilization = 151# NH -N 13.5# Or$-N x 20% mineralized or less = 27# inorganic4N Total inorganic N - 178# Most of the NH -N will nitrify rapidly to form N03-N. Much of the 178# of N03 will denitrify especially being buried which excludes considerable air. ~,.,,; Assume 60-80% denitrification - this leaves ~ for plant growth during the first year from the sludg . MANURE One ton of animal manure is equivalent to a 10-5-10 fertilizer, which means it contains 10# of Total N. 30 tons x 10 = 300# ~f Total N. + 300 x 50% = 150# NH4 -N. If we assume that 80% of the NH4 -N volatilizes that leaves 30# NH4-N for the plant. 300# Total N - 150# NH4+-N = 150# organic N. If 40% of the organic N is mineralized the first year, 60# of inorganic N will be produced from the organic N. Total inorganic N is 30# + 60# = 90# from the manure - this will be nitrified to form N03 - N. If 50% of the N03 - N is denitrified, it will leave 45# of inorganic N for the plants to use from the manure. Total inorganic N for plant use = 72# from the sludge + 45# from the manure = 117#N/acre. Some N will be made available from the soil. If the soil has 1% organic matter, it will contain 1000# N. Less than 1% will become available or about 10#/acre. ~I Total N_av~i a~l~e-the-f-i-rst year-should-approach 117C~~`10=127#-N._- ~ ~nr ~Y'ri ated r_as es -the N re uirements could ran e from 5.0-.125# 9 ~J ~_ q 9 J ~ PJ/a 4r-e~epending on-the Species. ~a~