An optimizing start-up strategy for a bio-methanator

Abstract

This paper presents an optimizing start-up strategy for a bio-methanator. The goal of the control strategy is to maximize the outflow rate of methane in anaerobic digestion processes, which can be described by a two-population model. The methodology relies on a thorough analysis of the system dynamics and involves the solution of two optimization problems: steady-state optimization for determining the optimal operating point and transient optimization. The latter is a classical optimal control problem, which can be solved using the maximum principle of Pontryagin. The proposed control law is of the bang–bang type. The process is driven from an initial state to a small neighborhood of the optimal steady state by switching the manipulated variable (dilution rate) from the minimum to the maximum value at a certain time instant. Then the dilution rate is set to the optimal value and the system settles down in the optimal steady state. This control law ensures the convergence of the system to the optimal steady state and substantially increases its stability region. The region of attraction of the steady state corresponding to maximum production of methane is considerably enlarged. In some cases, which are related to the possibility of selecting the minimum dilution rate below a certain level, the stability region of the optimal steady state equals the interior of the state space. Aside its efficiency, which is evaluated not only in terms of biogas production but also from the perspective of treatment of the organic load, the strategy is also characterized by simplicity, being thus appropriate for implementation in real-life systems. Another important advantage is its generality: this technique may be applied to any anaerobic digestion process, for which the acidogenesis and methanogenesis are, respectively, characterized by Monod and Haldane kinetics.

Appendix

If (t ₁, t ₂) is a singular interval, then by (42) and (39)

$$ s_{1}(x,p) \equiv 0 $$

(48)

$$ s_{2}(x,p) \equiv 0 $$

(49)

$$ \dot{s_{1}}(x,p) \equiv 0 $$

(50)

From (38) and (49) if follows that for t ₁ < t < t ₂

$$ (q-p_{4})x_{4} = p_{3}x_{3}\frac{\mu_{1}(\xi_{1})}{\mu_{2}(\xi_{2})} $$

(51)

$$ x_{4} = \frac{p_{3}x_{3}}{(q-p_{4})}\frac{\mu_{1}(\xi_{1})}{\mu_{2}(\xi_{2})} $$

(52)

while from (37) and (50) it follows that

$$ \dot{s_{1}}(x,p)= -p_{3}x_{3}\mu_{1}'(\xi_{1})(\xi_{{\rm in}_{1}}-\xi_{1}) +(q-p_{4})x_{4}\mu_{2}'(\xi_{2})(\xi_{{\rm in}_{2}}-\xi_{2}) - q\mu_{2}(\xi_{2})x_{4} \equiv 0 $$

(53)

Using (51) and (52) in (53) leads to

$$ p_{3}x_{3}\times \left[-\mu_{1}'(\xi_{1})(\xi_{{\rm in}_{1}}-\xi_{1}) + \frac{\mu_{1}(\xi_{1})}{\mu_{2}(\xi_{2})}\mu_{2}'(\xi_{2})(\xi_{{\rm in}_{2}} -\xi_{2}) - \frac{q}{q-p_{4}}\mu_{1}(\xi_{1})\right]\equiv 0 $$

(54)

Two possibilities result from (54):

Case 1

$$ p_{3}x_{3}\equiv0 $$

(55)

In this case, p ₃≡0 as x ₃ ≠ 0. Then, $\dot{p}_{3}\equiv0,$ which becomes using the costates equation (33)

$$ (q-p_{4})x_{4}\mu_{2}'(\xi_{2})c\equiv0 $$

(56)

As x ₄ ≠ 0, it follows that

$$ p_{4}=q $$

(57)

and consequently $\dot{p}_{4}=0$ since q is constant. However, using (57) in the costates equation (34) leads to

$$ \dot{p}_{4}=p_{4}u $$

(58)

which is in contradiction with $\dot{p}_{4}=0.$ Hence, (55) does not hold for t ₁ < t < t ₂.

Case 2

$$ \mu_{1}'(\xi_{1})(\xi_{{\rm in}_{1}}-\xi_{1}) - \frac{\mu_{1}(\xi_{1})}{\mu_{2}(\xi_{2})}\mu_{2}'(\xi_{2})(\xi_{{\rm in}_{2}} -\xi_{2}) + \frac{q}{q-p_{4}}\mu_{1}(\xi_{1})\equiv 0 $$

(59)

In this case, μ₁′(ξ₁) and μ₂′(ξ₂) are calculated from (8) and (9) as

$$ \mu_{1}'(\xi_{1}) =\mu_{m_{1}}\frac{K_{s_{1}}}{(K_{s_{1}}+\xi_{1})^{2}} $$

(60)

$$ \mu_{2}'(\xi_{2}) =\mu_{m_{2}}\frac{K_{s_{2}}-\frac{\xi_{2}^{2}}{K_{i_{2}}}} {\left(K_{s_{2}}+\xi_{2}+\frac{\xi_{2}^{2}}{K_{i_{2}}}\right)^{2}} $$

(61)

Using (60) and (61) in (59) leads to

$$ K_{s_{1}}\xi_{2}\left(K_{s_{2}}+\xi_{2} +\frac{\xi_{2}^{2}}{K_{i_{2}}}\right)(\xi_{{\rm in}_{1}}-\xi_{1}) -\xi_{1}(K_{s_{1}}+\xi_{1}) \left(K_{s_{2}} -\frac{\xi_{2}^{2}}{K_{i_{2}}}\right)(\xi_{{\rm in}_{2}}-\xi_{2})+ \frac{q}{q-p_{4}}\xi_{1}(K_{s_{1}}+\xi_{1})\xi_{2} \left(K_{s_{2}}+\xi_{2}+\frac{\xi_{2}^{2}}{K_{i_{2}}}\right)\equiv0 $$

(62)

Relationship (62) is a quadratic equation in ξ₁ with the solutions

$$ \xi_{1_{1}}=f_{1}(\xi_{2},p_{4}) $$

(63)

$$ \xi_{1_{2}}=f_{2}(\xi_{2},p_{4}) $$

(64)

Since ξ₁ = x ₁ − ax ₃, it follows that for t ₁ < t < t ₂

$$ x_{1_{1}}=f_{1}(\xi_{2},p_{4})+ax_{3} $$

(65)

$$ x_{1_{2}}=f_{2}(\xi_{2},p_{4})+ax_{3} $$

(66)

which is in contradiction with the fact that x ₁ is the solution of $\dot{x}_{1}=u(w_{1}-x_{1})$ for all t ≥ 0. Hence (59) does not hold.

Thus, it is concluded that singular intervals cannot occur.