Biological Reactors: Material Balances and Kinetics

TERZA PARTE

1. Fed-batch biological reactor: material balances with Monod kinetics and operating methods:

• it maintains some operational characteristics of the batch, while ensuring constant biomass production such as a CSTR
• Therefore, there is an inlet supply with V˙e and [substrate] Cse and negligible Cxe
• since it is not a continuous reactor, it will never reach the steady state condition, there will always be an accumulation
• at zero time there will be: a certain initial V_R with an initial [substrate], initial [biomass] and subsequently the volume will increase over time.
• Total biomass balance:
• dmtot / dt = me, with me = ρe V ,e,
• from this relationship we obtain that
  1. dVr / dt = V˙e.
• At this point two feeding strategies can be considered,
  1. 1. one with constant V˙e
  2. 2. one with Vr / V˙e = constant = λ.
• 1. integral is solved with first strategy and boundary conditions of @ t = 0 Vr = Vrin:
  1. Vr = Vrin + V˙et.
  2. à proceeding this way we have a volume that increases linearly over time
• 2. In the second case, under the same boundary conditions, the volume increases exponentially:
  1. Vr = Vrin * and ^ λt.
  2. Still referring to this second case, the material balances for biomass and substrate are made:
  3. dms/dt= V˙eCse+rsVR
  4. dmx/dt=+rxVR
  5. dms/dt=Cs V˙e+VRdCs/dt and dmx/dt=Cx V˙e+VRdCx/dt
  6. by replacing in the balance sheets
  7. rs=-1/Yxs rx=-1/Yxs*KCxCs/Cs+km
  8. the equations are obtained
     YxsCse-(YxsCs+Cx)=[YxsCse-(YxsCsin+Cxin)]e^-λt
     Cx=Yxs(Cse-Kmλ/k-λ)
• What can be seen from the concentration profile over time for substrate and biomass is that initially the two vary and then stabilize and tend asymptotically towards constant values
• This is the property of this type of reactor, i.e. producing biomass at constant concentration like a CSTR reactor.

3. Perfectly mixed biological reactor: material balance for components, biomass growth kinetics, substrate consumption and product formation:

• In the case of a perfectly mixed biological reactor, the material balances concern the three components, the substrate, the product and the biomass:
  S) Cse-Csu+rsuθ=0
  P) Cpe-Cpu+rpuθ=0
  X)Cxe-Cxu+Rxθ=0
• it is assumed that the incoming product concentration is not there, because it has not yet been produced, therefore Cpe = 0.
• Furthermore, it is noted that Rx biomass reaction term is = rx-kdCxu, where Kd is the death constant, because biomass grows but also dies.
• The concept of yield Y is introduced, in order to express the reaction rates: yield of substrate in biomass Yxs = rx / -rs and yield of substrate in product Yps = rp / -rs and refer to the fact that the substrate is consumed both both in the product and in biomass.
• There are two kinetic models to describe the growth of x-ray biomass which are the law of Malthus and that of Monod.
• In the first rx = μmaxCx, where mu max represents the exponential growth rate, this model is not complete, in fact it only describes the exponential growth and there is no reference to the Cs which sooner or later becomes limiting.
• The second Monod kinetic law was therefore introduced which is rx / Cx = μmaxCs / (Cs + Km). This is formally the same as Michaelis Menten's kinetic law and therefore can be rewritten as rx = KCx * Cs / Cs + Km.
• From this relationship and referring to the previous equations, the growth kinetics for substrate and product can be rewritten: rs = -1 / Yxs * kCx * Cs / Cs + km, obviously negative because the substrate wears out; rp = Yps / yxS KCx * * Cs / Cs + Km.

4. Biological reactor perfectly mixed with biomass concentration and recycling: setting of balances for components and blocks, yield concept

Reattore biologico perfettamente miscelato con concentrazione e riciclo della biomassa: impostazione dei bilanci per i componenti e per i blocchi, concetto di resa:

• In a biological reactor perfectly mixed with biomass concentration and recycling, the biomass is recovered which would be removed from the outgoing current and circulated in the reactor, thus increasing productivity and eliminating the wash out problem.
• In this system there are 16 budgets to be written as we have 3 components and 4 blocks.
• To write the total balances, it is based on the assumption that the densities are always the same and that the flow rates must be balanced in all the blocks, so:

at the mixer: V˙0+V˙r-V˙e=0
the settler: V˙e-V˙s-V˙u=0
the splitter: V˙u-V˙r-V˙w=0
At the mixer the material balances for the components are:
S) V˙0Cs0+V˙rCsu-V˙eCse=0
X) V˙rCxu-V˙eCxe=0
P) V˙rCpu-V˙eCpe=0
from these the respective values of Cpe, Cxe e Cse, are obtained, considering that at the mixer V˙e=V˙0+V˙r and that V˙r=RV˙0, recycling ration.
Cse=Cs0+RCsu/(1+R)
Cpe=RCpu/(1+R)
Cxe=RCxu/(1+R)
to the settler:
S) V˙eCsu-V˙sCsu-V˙uCsu=0
X) V˙eCxu-V˙uCxr=0, it is assumed that the settler completely separates the biomass from the suspension, therefore Cxs=0
P) V˙eCpu-V˙sCpu-V˙uCpu=0
to the splitter:
S) V˙uCsu-V˙rCsu-V˙wCsu=0
X) V˙uCxr-V˙rCxr-V˙wCxr=0
P) V˙uCpu-V˙rCpu-V˙wCpu=0
Infine i bilanci al reattore:
S) V˙eCse-V˙eCsu+rsuVR=0
X) V˙eCxe-V˙eCxu+RxuVR=0
P) V˙eCpe-V˙eCpu+rpuVr=0

from these equations we can express the θ which is given by VR / V˙0 and we can rewrite the rs in terms of yields Y.

• The yield concept derives from the fact that the available substrate S is consumed both in the product, according to the yield of substrate in product Yps = rp / -rs, both is consumed in biomass, therefore to support cellular metabolism, according to the yield of substrate in biomass Yxs = rx / -rs.
• We then rewrite the equations for the substrate and biomass, also including the Monod kinetics:
  S) Cs0-Csu-1/Yxs*kCxu*Csu/Csu+Km*θ=0
  X) RCxr-(1+R)Cxu+(kCxu*Csu/Csu+km-KdCxu)θ=0
  P) -Cpu+Yps/YxskCxuCsu/Km+Csu*θ=0

7. Biological reactor perfectly mixed with recycling: material balances with Monod kinetics. Dependence of substrate concentration leaving the SRT and wash-out conditions:

• The material balances at the reactor in a bilogical reactor perfectly mixed with recycling can be written as:
  V˙eCse-V˙eCsu+rsuVR=0
  V˙eCpe-V˙eCpu+rpuVR=0
  V˙eCxe-V˙eCxu+RxuVR=0
  
• In these the values of Cse=Cs0+RCsu/1+R, Cxe=CxrR/1+R e Cpe=CpuR/1+R and the kinetics of Monod are replaced:
  Cs0+RCsu-Csu-RCsu+[-1/Yxs*kCxu*Csu/Km+Csu]θ=0
  RCpu-Cpu-Rcpu+[Yps/Yxs*KCxu*Csu/Csu+Km]θ=0
  Cxr-(1+R)Cxu+[kCxu*Csu/Csu+Km-kdCxu]θ=0
  A new operating variable, Solid retention TIme, θc, is introduced in th:
  θc= VRCxu/V˙wCxr
  
• If we consider the steady state, then we can conclude that the biomass that is removed is the one that is produced, because the accumulation is zero. This means that V˙wCxr=VRRxu which can be replaced in the equation of θc:
  
• θc= VRCxu/VRRxu, namely (ossia) θc= Cxu/rxu-kdCxu
• If we consider rxu according to Monod´s kinetics, we can replace rxu=KCxu*Csu/Csu+Km:
  θc= Cxu/ KCxu*Csu/Csu+Km -kdCxu, from which you can simplify Cxu and derive the value of Csu
  Csu=Km(1+kd θc)/(k-kd) θc-1, which represents the equation of the substrate concentration at the output in dependence of the θc.
• In theory, this type of reactors with recycling solve the wash out problem, but in reality this is not true; in fact, even the new operating variable SRT is associated with the flow rate and in particular with the purge flow rate V˙w, it is therefore possible as for biological reactors without recycling, that wash out conditions occur.
• What can happen is that at a value of bassoc too low (θc wo), too much biomass is removed compared to its formation capacity in the reactor, in practice it is removed before it is completely formed.
• The graph with Csu as a function of θc is very similar to the graph with θ, and does not always mean for values ​​θc <θc wo and for csu> Cs0. θc>
• Furthermore, when the temperature drops, the bacteria reproduce more slowly and the wash out situation is reached very early.