Although some microbial systems can be represented as batches (Section 11.7.2), many others have flows of material entering and leaving, including streams, lakes, groundwater

Figure 11.25 Schematic of a continuous culture system.

aquifers, and most biological wastewater treatment plants. Such continuous culture systems receive inputs of new substrates, but also lose substrates and even organisms in the outflow. Thus, to model a continuous culture, in addition to knowing the reactions that are occurring, it is necessary to keep track of the inputs to and outputs from the system. This is usually done by means of a mass balance.

A general mass balance equation for a particular component, A, in a continuous flow system (Figure 11.25) can be written as rate of change of massA in system = (rate of massA input) โ (rate of massA output)

+ (rate of massA production through reactions)

or if Ma is used for the mass of A and RA for the reaction term, then dMA

For a continuous-flow (Q) system of volume V, in which concentration (C = M /V) of component A is measured in the influent and effluent (subscripts i and e, respectively) this can be expressed as d(VC = (QiCi) โ (QeCe) + (VRc ) (11.21)

Note that in a batch system, Qi = Qe = 0 and V = constant, so that dC

Thus, from equations (11.11) and (11.17), for biomass and substrate, respectively, we see that

Chemostat One particular type of laboratory continuous culture system in which certain constraints are imposed is known as a chemostat (Figure 11.26). It is of great utility as a research tool but is also of interest because many observations about its functioning can be generalized to other continuous culture systems, such as activated sludge wastewater treatment.

One requirement for a chemostat is that Q,- = Qe = Q = constant. If the influent and effluent flows are equal and constant, the chemostat volume will also be constant. It is also assumed that the substrate concentration in the influent (S,-) will be constant and that this flow will contain no biomass (X,- = 0).

Another requirement for a chemostat is that it must be completely mixed (indicated by the "propeller" in Figure 11.26), so that the concentration of each constituent in every drop of water in the reactor is equivalent to its concentration in every other drop. A consequence of complete mixing is that the concentration in the effluent will be equal to the concentration in the reactor (Xe = X, Se = S), since a drop leaving the reactor is equivalent to a drop inside the reactor. This means that if the influent concentration is 1000 mg/L of substrate, for example, and the effluent concentration is 5 mg/L, the concentration everywhere within the reactor is 5 mg/L. Thus, a microorganism in this reactor "sees" only 5 mg/L of substrate, never the 1000 mg/L being fed to it. This concept may seem difficult to grasp at first, but keep in mind that every drop entering the reactor (at 1000 mg/L) is instantaneously mixed and diluted throughout the reactor volume, and that further, the substrate in it is being consumed by the microorganisms present.

The general mass balance equation [equation (11.21)] can now be rewritten for biomass in the chemostat as d(VX) dt

where D = Q/V. D has units of (time)-1 and is referred to as the dilution rate. It represents the number of times the contents of the reactor is replaced per unit of time. The inverse of D, referred to as the hydraulic residence time (HRT), indicates the amount of time the water (and hence any soluble materials) spends in the reactor. It is widely used in environmental engineering and science for a variety of reactor types and systems and is often represented by the symbol 0, so that 0 = V/Q = 1/D.

For substrate, the mass balance equation is d(VS)

One characteristic that makes the chemostat so interesting is that once it is inoculated with microorganisms that can grow in the system and utilize the substrate, it will naturally move toward a stable condition referred to as steady state. Steady state occurs when the concentrations in the reactor no longer change, that is, when dX/dt = 0 and dS/dt = 0. If the microorganisms grow faster (dX/dt > 0), the substrate concentration drops (dS/dt < 0), and this slows the growth back down. If the growth is too slow (dX/dt < 0), organisms wash out of the system faster than they are growing; this drop in biomass allows the substrate to increase (dS/dt > 0), and this leads to faster growth. Thus, the biomass concentration grows to the point at which it uses up substrate at the same rate that it is added.

At steady state, dX

and X does not change. This can occur under two conditions: when mn โ D = 0 (mn = D) or when X = 0 (no biomass in the system). If X = 0, substituting for mn and using S for the steady state value of S, gives m- = ยป(jhd -" = D (1L27)

Thus, for the steady-state chemostat reactor, the net growth rate is equal to the dilution rate. Since D = QN, the dilution rate can be controlled merely by controlling Q for a given reactor volume. Thus, the biological growth rate can be controlled directly by the chemostat operator. A similar relationship holds for some wastewater treatment processes, such as activated sludge (Section 16.1.3), in which the net growth rate can be controlled by the amount of sludge wasting. This gives the process operator a means of direct control over the biology of the process.

Equation (11.27) can be solved for the effluent substrate concentration:

Note that since S is a function of only constants, it too must be a constant, confirming that dS/dt = 0:

Solving for gives

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