Statistics for Environmental Engineers

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Y = в0 + в1 (x1 — x1) + в2 (x2 — x2)


The coefficients are:


во = Уо = У = f (x 1, x2),    в1



-У1—У0


Ax1



and в2



У20


Ax2


The linearization will be valid over a small region about the centering point, where “small” can be reasonably defined as about x ± 2xx .

Case Study: Reactor Kinetics


A lab-scale reactor is described by the model:


1 + kX( V /Q)


where S0 is influent BOD, S is effluent BOD, X is mixed-liquor suspended solids, V is reactor volume, Q is volumetric flow rate, and k is a reaction rate coefficient. Assume that the volume is fixed at V = 1.0. The independent variables Q, X, S0, and S will be measured for each experiment. Their expected measurement errors, given as standard deviations, are Oq = 0.02, aX = 100, aS = 2, oSq = 20. How precisely will k be estimated?


The sensitivity of k to changes in the other variables can be calculated analytically if we solve the model for k:


= (S0-S)Q SXV


The linearized model is:


k =    e0    +ds0(So —    So)    +    ds(s — S) + ex(x —    x)    +    eQ(q — q)


The 0’s are sensitivity coefficients that quantify how much a unit change in variable i will change the value of k from the centered value 00 = 0.0045. Because the 0’s are evaluated at particular values of the variables, they have meaning only near this particular condition. The coefficients are evaluated at the centered values: S0 = 200, X = 2000, Q = 1, and S = 20. For convenience, the overbars are omitted in the following equations.

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