Kc nTLTy213

dR = 0 = dL(kLc-ky-y'ï = L2(KLC - kcy - y')(y) (2.14)

In the previous equations, kcLc and kc are the parameters of the partial differentiation. Thus, in Equation (2.14), where the differentiation is with respect to kc, the partial derivative of kcLc with respect to kc is zero, since the whole expression kcLc is taken as a parameter.

From Eqs. (2.13) and (2.15), Lc, the ultimate oxygen demand, may finally be solved producing

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