Question

Determine the minimum volume of a first-stage, high-rate anaerobic digester based on a maximum volatile solids loading rate of 9.5 kg VS/(m3.d) and a detention time of 4.5 days. 500 cubic meters per day of combined primary sludge and waste activated sludge at 6.0% solids and a volatile content of 85% are fed to the digester.

Answer 1

therefore minimum volume of digester is 0.27 cubic.m.

day Q Given Detention time = 4.5 days discharge of combined primary sludge = 500 mo and waste activated sludge = 6% which means 6 mg 1.- img now solids content going to digester is = 6 mg x 500 x 10² lit x 4.5 days day - 13.5kg of solids also given that 85% solids arie Volatile » Volatile solids = 1305x0.85 = 11.475 kg from given dara maximum volatile solidare of digester = 9.5Kg/m3 I day . for Derention time of 4.5 days maximum solids that can be Looded in digester is = 9.5 kg x 4.5days = 42.75 kg Loading YSI mid i minimum volume of digestor = 11.475 kg (VS) 42.75 Kg (Y6) Jo. 268m/ T m3

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