Difference between revisions of "Team:UChile-OpenBio/Modelling"

Bioreactor
What is a bioreactor?
Is an equipment which keeps an active biological environment and is used to carry out a transformation due to the performance of a biocatalyst (1). In this project, these biocatalysts are bacteria.
Why is important a bioreactor?
“The heart of a typical bioprocess is the reactor or fermenter. Flanked by unit operations which carry out physical changes for medium preparation and recovery of products, the reactor is where the major chemical and biochemical transformations occur. In many bioprocesses, characteristics of the reaction determine to a large extent the economic feasibility of the project.”[2] How works a bioreactor? Para preguntar A bioreactor is an equipment in which some substrates, from the culture, are transformed to a product through the performance of bacteria. A bioreactor gives all necessary conditions for the culture, like mixing, temperature regulation, oxygen supply, substrates ports, sampling ports, pH control, etc. Then, within bioreactor is carrying out a bio-reaction, because this equipment gives operation conditions for a complete reaction. There are three ways to carry out the reaction after charging the bioreactor with cell y substrate: giving a continuous feed (substrate), giving a semi-continuous feed and without feed, they are called a continuous, fed-batch and batch bioreactor respectively [3].
In this project, a continuous bioreactor was used. Thus, a continuous feed rate is equal to an outflow rate, maintaining a constant volume within bioreactor. Once PLA is produced, it is got from the outflow and then will be purified to use it to make any PLA product.

Which are the goals of a bioreactor?
These are the goals of a bioreactor[1]
Keep cells distributed uniformly.
Keep temperature constant and homogeneous
Minimize nutrient concentration gradients
Preventing sedimentation and flocculation
Allow gas diffusion
Aseptic environment culture
Sampling ports
Foam control
Maximize performance and production
Reduce production costs
Optimize volume
Reduce the reaction time

What is the advantage of this project’s bioreactor?
Due to it uses genetically modified bacteria to transform glucose to PLA, which is released to the culture, it is not necessary to kill these bacteria to get PLA, because in this project, PLA is purified and characterized after its production within bacteria. That’s why the most important advantage of this bioreactor, over chemical bioreactor, is an environmental friendly reactor, because it does not pollute a lot the air, water neither land .
Besides, bacteria do not need to grow in an environment with high temperatures(between 28-30ºC [3]) or pressures, then the bioreactor does not use a great amount of energy (electrical energy or heat, comparing with the temperatures of chemical reactor that are about 130ºC [4]), being a low cost reactor.
In conclusion, this bioreactor is a sustainable equipment.

Equations related to bioreactor
Before writing the equations, it’s important to understand what is happening inside the bioreactor. That’s why a mass balance for the system is showed [2]:

$\left\{\begin{array}{c}Mass accumulated\\ within \\ system\end{array}\right\}=\left\{\begin{array}{c}Mass in \\ through the \\ system \\ boundaries\end{array}\right\}-\left\{\begin{array}{c}Mass out \\ through the\\ system\\ boundaries\end{array}\right\}+\left\{\begin{array}{c}Mass generated\\ within \\ system\end{array}\right\}-\left\{\begin{array}{c}Mass consumed\\ within \\ system\end{array}\right\}$

An example of a bioreactor, is a CSTR (continuous stirred-tank reactor) where the feed rate is equal to outflow, maintaining a constant culture volume. In figure 1 is showed a diagram of a CSTR.

Figure 1. Diagram of a CSTR. Where $F_i = Feed rate; C_{Ai}= Feed concentration ofthe component A; p_i= feed density; F_o = outflow concentration; C_{Ao}=outflow concentration; p_o= outflow density$

Then, considering the mass balance, we can write the following equations [5]:

Global mass balance
$\frac{d({\rho }_o*V)}{dt}=F_i*{\rho }_i-F_o*{\rho }_o$ (1)
Where:
$\begin{array}{l}{\rho }_o = culture density (kg/m^3 )\\ V = culture volume (m^3)\\ F_i = feed rate (m^3 / h)\\ {\rho }_o = feed density (kg/m^3 )\\ F_o = outflow (m^3 / h)\\ {\rho }_o = outflow density (kg/m^3 )\end{array}$ Supposed:
Constant density all the time $({\rho }_o={\rho }_{i })$
Constant volume all the time
Steady state

In conclusion, we have:
$F_i=F_o=F$ (2)
For next equations, it’s supposed the following scheme showed in the figure 2:


Figure 2. Scheme of the bioreactor. The substrate is glucose and the product is PLA. Within bioreactor are the two genetically modified bacteria.

Cell balance

$F*X_i-F*X_o+ \mu *X_o*V-\alpha *X_o*V=\frac{d\left(X_o*V\right)}{dt}=X_o*\frac{dV}{dt}+V*\frac{d{X}_o}{dt}$ (3)
Where:
$\begin{array}{l}{X}_{i,o} = cell concentration (kg/m^3) (feed rate and outflow)\\ \mu = specific growth rate of the bacteria (1/h)\\ \alpha = death rate of the bacteria (1/h)\end{array}$

Supposed:
Constant volume all the time $(\frac{dV}{dt} = 0)$
Steady state $( \frac{dx}{dt}=0)$
There are not cells in the feed $(x_o)$
Death rate(α) is less than the specific growth rate(µ)

Then we have:
$-F*X_o+\mu *X_o*V=0 F*X=\mu *X *V$ (4)
$\frac{F}{V}=\mu$ (5)
And we know that the dilution is $D =F/V D=\mu$

Substrate balance
$F*G_i-F*G_o-\frac{\mu *X_o*V}{Y_{\frac{X}{G}}}-m_S*X_o*V-\frac{q_P*X_o*V}{Y_{\frac{P}{G}}} =\frac{d\left(G_o*V\right)}{dt}=G_o*\frac{d\left(V\right)}{dt}+V*\frac{d\left(G_o\right)}{dt}$ (6)
Where:
$\begin{array}{l}{G}_{i,o} = glucose concentration (feed rate and outflow)(kg/m^3)\\ \mu = specific growth rate (1/h)\\ X_o = cell concentration ( kg/m^3)\\ Y_{\frac{X}{G}} = conversion of cell referred to consumed glucose ( kg cell / kg glucose )\\ m_S = maintenance coefficient(1/h) \\ q_P= specific product formation rate (kg product/ kg cell / h) \\ Y_{\frac{P}{G}}= conversion of product referred to consumed glucose ( kg product/ kg glucose )\end{array}$

Supposed:
1. $m_S* X_o << \mu *X_o$
2. Steady state $$( \frac{d\left(V\right)}{dt}=0 y \frac{d\left(G_o\right)}{dt}=0 )$$

Then we have:
$F*G_i-F*G_o-\frac{\mu *X_o*V}{Y_{\frac{X}{G}}}-\frac{q_P*X_o*V}{Y_{\frac{P}{G}}} =0$ (7)
Dividing by:
$\frac{F}{V}*(G_i-G_o)-X_o * ( \frac{\mu }{Y_{\frac{X}{G}}}-\frac{q_P}{Y_{\frac{P}{G}}} ) =0$ (8)
Using equation 5:
$X_o=\frac{\mu *(G_i-G_o)}{ ( \frac{\mu }{Y_{\frac{X}{G}}}-\frac{q_P}{Y_{\frac{P}{G}}})}$ (9)