This problem focuses on the activated sludge process as shown in Fig 615 in your textbook and presented in class Lets assume that the activated sludge basin receives 03 m3s of primary treatment effluent with a BOD5 of 220 mgBOD5L The concentration of mixed liquor within the basin is 2,100 mgVSSL and the RAS concentration is 10,000 mgVSSL You calculate the FM ratio for the activated sludge basin to be 052 mgBOD5mgVSSd and the solids retention time (SRT) to be 9 days

Answer rating: 100% (QA)

Solution The data which is given in the question is Bod s Is 210 mgl Dis... View the full answer