No inflow of autocatalyst

In the especially simple case of a reactor fed only with the reactant A, so b0 = 0, the stationary states can be obtained explicitly. One solution of eqn (8.3) is ass = a0, corresponding to no reaction. The other solutions are given by the roots of a quadratic 

For the no reaction state ass = a0, the relaxation time given by eqn (8.10) is simply equal to the residence time. In terms of the eigenvalue, we have A = - l/tres, which is negative. The stationary state is always stable, irrespective of a0 and kl. Chemistry makes no contribution (formally we have l/tch,ss = 0, so the chemical time goes to infinity) the perturbation of a does not introduce any B to the system, so no reaction is initiated. The recovery of the stationary state is achieved only by the inflow and outflow. 

For the non-zero states, substitution of ass from eqn (8.12) into (8.10) leads to the expression 

Thus the upper root of eqn (8.10), which gives the middle branch of stationary-state solutions and requires the minus sign above, has a negative value for trelax. It then follows that the eigenvalue A for this branch is positive, so perturbations grow. This is an unstable state. 

For the other non-zero solution, with the lowest value for ass (highest extent of conversion), the relaxation time has the plus sign in eqn (8.13) and so is positive. The corresponding eigenvalue A is negative, so the solutions along this branch are stable. 

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As the residence time tends to infinity, the upper root of this equation tends to the equilibrium state 1 — ass = 1 — aeq. Multistability exists over all residence times satisfying inequality (6.33), so there is no upper limit corresponding to an ignition point because there is no inflow of autocatalyst. 

Equation (8.13) gives the appropriate form for l/treiax( = — X) for the cubic autocatalysis model with no inflow of autocatalyst. The condition for the turning point in the stationary-state locus (there is only one) is = 4. 

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