Question 1 Concentration of Hydrogen At standard state is given: 10^-7MAt a ph=7.6 is: 10^-7.6Therefore concentration ratio of hydrogen is :( 10^-7.6/10^-7)=0.2511886The concentration ratio of Oxygen is:=0.13158Q==481.8lnQ=6.1775(given)Therefore:

Question 2

Figure 1 Typical Chemostat Reactor (http://2008.igem.org/Team:ETH_Zurich/Modeling/Chemostat_Selection)

Mass balance on the substrate concentration around the chemostat reactor in figure 1;Accumulation= In-Out+Generated-Consumed ..1Monod EquationAssuming:no extracellular product formation,,and steady state operation, dSdt=0.Equation 1 reduces to:)1aBut Therefore..1bBut dilution rate is equal to the cell growth rate for a batch process at steady state,i.e. .Thus, ..1cIn terms of rearranging Equation 1c ,

..1dMonod equation

Substitute monod equation into equation 1d..1eAssuming there is endogenous metabolism, ..(2)Substitute eqution 2 into 1b,..3Rearranging eqution 3..3aSubstitute equation 1d into 3a

..(4)Therefore, ..(5)Maintenance coefficient based on substrate concentration..(6)Substitute equation 6 into 5..(7)Rearranging equation (7) gives and proves that indeed:..(8)

2b)

Figure 2Used microsoft excel to plot figure 2 above,from the plot I got a linear line of best fit for the points and an R squared value.From the line of best fit equation y-intercept is 1.5958,i.e The gradient is 0.0555,which means .But thus;

Figure 3Used microsoft excel to plot figure 3 above,from the plot I got a linear line of best fit for the points and an R squared value.From the line of best fit equation y-intercept is 1.248 which means The gradient is 100.27h/mg/L.But .Therefore

Question 33A)Cone shapeVolume is

Surface area,

Ratio of H(height) to D(diameter) is 1/1Therefore

And

Cylinder shape

And

Ratio of H(height) to D(diameter) is 2/1i.e. H=2Dthereforeand

Total Volume is;

Total surface area is;

Small fermenter:volume is 0.1m^3Therefore from

And

Big fermenter(scale up):volume 25m^3

And

Mass produced on surface of scale up fermenter is;

Mass produced on volume of scale up fermenter is;

Therefore Total mass produced is :The volume increased by a factor of 250,whereas the production rate increased by ( . If there was no growth on the surface,i.e there was growth only in the liquid phase the overall yield would be 70000g which implies that surface growththe yield in a system when we scale up.3B)A batch reactor is filled only once up to full capacity, i.e. no more substrate addition takes place as the reaction progress until the end, where the product is removed. A fed batch reactor is partly filled in the beginning of the reaction and certain amounts of substrate are intermittently added as the reaction progresses. The product is removed periodically and this type of operation helps in preventing the inhibition of substrate thus increases production rate.A chemostat reactor with a recycle is a continuous process where you put feed once, get product periodically whilst its concentration is increasing due to the recycle stream.

AdvantagesDisadvantages

Batch reactorAllows for flexibility in the operation in that one reactor is able to manufacture one product one time and a another product the next time.Does not allow for the control of the growth limiting substrate, hence theirs is an over production of by-products.

Has a high conversion per volume in a single passGives product that has a quality with a great degree of variation.

Fed-Batch reactorAllows for the control of the growth limiting substrate, hence allows control over production of by-products.

There is a need an understanding of the physiology microorganism used , and what it requires.

Does not require any additional parts of equipment in comparison to the batch reactor.Caution must be taken in the design stage to ensure that toxins do not accumulate to levels that lead to inhibitions.

Chemostat with recycleHelps keep the cell concentration higher than the normal level.

Greatly inflexible, for it cannot be used in various processes without major retrofitting.

Increases the productivity of biomass.Requires that processing be adjusted to the flow through the bioreactor by fitting holding tanks, thus high costs.

Question 4A) Mass balance around the reactor is:..1Steady state and sterile feed ( assumptions reduce equation 1 into:..1b

B) Rearranging equation 1a and writing it in terms of :

(given)=1.5(given)Because this is a continuous system problem , D=Dopt. Essentially, dilution rate is not equal to the growth rate.Therefore;)

By definition

This means,

Monod equation

Therefore (Concentration of substrate in reactor effluent)

A mass balance on the substrate concentration around reactor is:

Assuming steady state,,thus: (Concentration of biomass in reactor effluent)

A mass balance on the cell concentration around reactor is:

(given)=1.5(given)Therefore

Then (Concentration of biomass in the effluent from cell separator)

Concentration of biomass in the recycle stream is: