Growth of a pathogenic bacterium Brucella abortus

So Eqs. (8.258) and (8.256) represent the growth kinetics. At stationary states, they become 

To satisfy Eq. (8.259), either. S = 0, or the contents of the bracket equals zero. As the amount of.S in the integral in Eq. (8.259) will increase with time, there is no fixed value of S that makes the contents of the bracket zero. Therefore, the only solution for Eq. (8.259) would be. S = 0. Now, Eq. (8.260) becomes 

Consider a microbe coupling catabolism, which consists of processes that liberate free energy, and anabolism, which is biosynthesis. In the vicinity of global equilibrium (and for some far-from equilibrium systems) linear relations exist between thermodynamic forces, which are the corresponding Gibbs free energy differences AG, and flows. Therefore, a microbial growth may be represented by 

The dissipation function P is always positive, as the dissipation of the Gibbs free energy should occur because of the irreversibility of the system 

Overall, the thermodynamic efficiency of growth 17 is the ratio of the Gibbs free energy output and the Gibbs free energy input 

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