arc mg biomass COD formed/mg subslrate COD removed. The intercept is equal to bn/Y,,; consequently, b,, has a value of 0.006 hr '.
b. The second task is to estimate the value of f,,. This is done by plotting 1 l\ versus ("), as indicated by Eq. 8.9. The resulting plot is shown in Figure E8.2. Application of linear least squares analysis to the data reveals that the slope of the line is 0.00109 hr The slope is equal to f,,-h,,- Since bn has a value of 0.006 hr ', f|, has a value of 0.18. It is dimensionless.
c. The final task is to determine the values of |iM and Ks. Before this can be done, it is necessary to calculate the biodegradable COD concentration. Ss. This is done by subtracting the inert soluble COD, S, from the soluble COD values given in Table E8.1, as indicated by Eq. 8.4. The Hanes linearization will be used because it does not require taking the reciprocal of Ss. Because the values of Ss are generally small, taking their reciprocal can greatly amplify the error in them, making it difficult to see the trends. The Hanes linearization requires plotting Ss/(l/("\ + bn) versus Ss as indicated in Eq. 8.10. The plot is shown in Figure E8.3. It suggests that |iM has a value of 0.10 hr ' and Ks a value of 25 mg COD/L. One thing that is evident from the plot is that small errors in the estimated values of Ss have a large effect on the estimates of the slope and the intercept. This is because the substrate concentration was low at all SRTs studied. While this suggests that it would be good to collect data at shorter SRTs, where the values of S., would be larger, this would be difficult to do because bioflocculation would be poor (See Section 10.2.1). If the scatter in the data is so great that no confidence can be placed in the resulting parameter values, then it would be better to determine k,, the mean reaction rate coefficient, which combines jiM, Ks. and Y|, into a single lirst order coefficient. This is done by plotting Ss versus (1 W. + b„) as indicated by Eq. 5.41. Figure E8.4 shows the result of doing
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