EXAMPLE 4.14 Problem In wastewater treatment the digestion of sludge produces a

Question

EXAMPLE 4.14 Problem In wastewater treatment the digestion of sludge produces a useful gas (carbon dioxide and methane), and in the course of digestion the sludge becomes less odoriferous. Raw sludges can contain pathogenic microorganisms, and digestion to some extent dis- infects the sludge and may make it easier to dewater. Typically, digestion is carried out in a two-stage system the primary digester, is the main reactor where the anaerobic (without oxygen) microor- ganisms convert the high-energy organic solids (called volatile solids) to methane gas, carbon dioxide, and lower-energy solids that make up the digested sludge. The second tank, the secondary digester, is used as both a gas storage tank and a solids separator where the heavier solids settle to the bottom and a supernatant is drawn off the sludge and reintroduced to the treatment process. The first digester, called Assuming steady state conditions, let the feed solids be 4%, of which 70% are volatile solids, and the feed flow rate be 0.1 m2/s. The gas, methane and carbon dioxide, contains no solids. The supernatant solids are 2%, with 50% being volatile, and the digested sludge is at a solids concentration of 6%, with 50% volatile. Find the flow rate of the supernatant and digested sludge.

Explanation / Answer

Non volatile solid sludge rate = 0.3*0.1*0.04 = 1.2*10-3 m3/s.

Non volatile Digested sludge rate = 0.06*0.5*1.2*10-3 = 3.6*10-5 m3/s.

Rate of supernatant (non volatile) = (0.96*0.1)-(0.01*0.1*0.96) = 0.09504 m3/s.

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