# 02-24 - Assume that the rate of which organics are consumed...

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Assume that the rate of which organics are consumed is directly proportional to their conc. in solution. i.e . What type of kinetics? _____________________ then: where k 1 is a rate constant for the reaction (we've assumed the cell conc., X , remains approximately constant for the test) Note: k 1 depends (among other things) on - type of organics in the waste - type of bacteria used for test - physical and chemical test conditions (pH, temp., etc.) first order 1 k dL L dt Note that this is similar to our empirical model for substrate utilization @ the limiting case where S << K S s dS kXS dt K S so, k 1 = kX/K S

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To get L as a function of time integrate the rate equation. Use as boundary conditions L = L o @ t = 0 L = L @ time = t separate variables and integrate to get L =   o Lt 1 L0 ln L k t   1 kt o Le 1 o L L e L O t o 1 L o L dt k L dL plot of the result Note that the decay of BOD has the same exponential form as the decay of a radioisotope t L Instantaneous rate of change = Slope = dL/dt = -k 1 L
Since the organic concentration is conveniently expressed in units of mg O 2 demand (i.e. mg BOD L ), it is easy to extend the above L L results to describe O 2 consumption. Define: = O 2 consumed at time = t = BOD "exerted" at time = t y i.e., y L y = L O - L = BOD exerted But, 1 kt o L L e 1 o y L (1-e ) so time L L O @ time = 0, BOD exerted = 0

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plot y vs. time | | | y | | | time This model can (and has) be used to describe the time course of oxygen utilization in a natural stream, a reactor, or a BOD test bottle. 1 kt o y L (1-e ) Note: y = BOD L = L o at time = L L O =BOD L
Sources of Confusion : 1. Some texts use k 1 for a base 10 log system i.e.

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## This note was uploaded on 04/13/2010 for the course CEE 3510 taught by Professor Lion during the Spring '10 term at Cornell.

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02-24 - Assume that the rate of which organics are consumed...

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