CVEN3502 - Workshop SOLUTIONS - Biological Treatment Processes.pdf

CVEN3502 - Workshop SOLUTIONS - Biological Treatment Processes.pdf

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1 CVEN3502 Workshop Biological Treatment Processes SOLUTIONS QUESTION 1 (Adapted from Mines, Chapter 7, Q9). A conventional activated sludge process treats 3,785 m 3 /day of wastewater, containing 250 g/m 3 BOD 5 , and produces and effluent containing 20 g/m 3 BOD 5 . The nominal detention time in the aeration basin excluding the RAS flow is six hours and the MLSS concentration is 3,000 g.m 3 . Determine the following: a) The aeration basin volume (m 3 ) b) F/M ratio (d -1 ) c) Specific substrate utilisation rate (d -1 ) d) Substrate removal efficiency (%) Solution:
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2 QUESTION 2 Settled sewage at a flow rate of 2.5 ML/d and 200 mg/L BOD is treated in an activated sludge plant which is equipped with two aeration tanks each 25m long x 5m wide x 4m deep, and with a mixed liquor volatile SS (MLVSS) concentration of 2000 mg/L. Calculate the detention time and the F/M ratio. Solution: Detention time, ߬ ൌ ଶൈଶହൈହൈସ ଶହ଴଴ ൌ 0.4 ݀ܽݕݏ ൌ 9.6 ݄݋ݑݎݏ F/M ratio ൌ ܯܽݏݏ ܤܱܦ݈݋ܽ݀/݀ܽݕ
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