APSC 150 Case 2 Assignment 5 Solution

# APSC 150 Case 2 Assignment 5 Solution - APSC 150 Biological...

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APSC 150 Biological Engineering Tutorial Solutions APSC 150: Engineering Case Studies Assignment 5 Solution #1. Note that the problem is to design an activated sludge reactor for the same wastewater as in Assignment 4. So the input flowrate is the same as Asst 4 (question 4-2) and so are the kinetics. a) Volume of aeration basin: First job is to figure out the SRT that will be required to achieve the desired effluent BOD. Since the equation used in question 4-2 can be used for activated sludge as well, we have already done this calculation. The required SRT is 6.94 days. C0 dC Y(S S) X (1 k ) θ− = θ+ θ substituting V/Q for HRT ( θ ) and rearranging () 3 1 3 m 5703 )) 935 . 6 )( 1 ( 1 )( 2000 ( ) 30 900 )( 75 . 0 )( 20000 )( 935 . 6 ( 1 = + = + + = d d L mgSS L mgBOD mgBOD mgSS d m d V k X S S QY V s d O c θ b) HRT = V/Q = 5703/20000= 0.285 d = 6.8 h c) You can get the amount of excess sludge (biomass or bacteria) generated each day in two different ways: i) By material balance around the clarifier:

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APSC 150 Biological Engineering Tutorial Solutions
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## This note was uploaded on 04/10/2010 for the course APSC 150 taught by Professor Hatzidizakes during the Spring '09 term at UBC.

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APSC 150 Case 2 Assignment 5 Solution - APSC 150 Biological...

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