Sam Maurer
10.28 Fermentation Problem Set #4
1
a. See attached spreadsheet.
b. For the three latest sampling points, the carbon mass balance is not completely satisfied.
Specifically, the mass of carbon produced as a result of cell growth and CO2 evolution exceeds
that of carbon consumed as a result of glucose in the medium degradation. The error is at most
29.3% (for the time point at 2.25 hours, the mass of carbon consumed, .369 g, is 29.3% lower
than that of carbon produced, 52.273 g) and at minimum 8.3% (for the 1.5 hour time point).
One likely explanation for this discrepancy is the fact that the glucose concentration was
approximated using the optical density. While the concentration of glucose should indeed vary
with optical density, the presence of waste products and other contaminants could certainly
change the optical density unpredictably. If, for example, the concentration of glucose at 2.25
hours were 7.2 g/L instead of 7.596 g/L, Mc,glu would become .496 g instead of .369 g, resulting
in an error of less than 5%. This would correspond to an unknown concentration of waste
products changing the optical density of the solution from 3.232 to 4.0, which is certainly
conceivable.
Another factor causing the apparent excess of produced carbon might be error in
measuring the dry cell weight. The measurement was carried out simply by weighing a cell
pellet and assuming that 55% of the cell pellet was comprised of carbon. If the cell pellet also
contains salts that were not completely washed away or other inorganic waste materials, the
overall cell pellet weight would increase, but the weight of carbon in the cell pellet would not,
causing the observed carbon production to be too high. Such an error could certainly affect the
experiment greatly, since Mc,cell has for all three data points over twice the magnitude of Mc,CO2.
Sam Maurer
10.28 Fermentation Problem Set #4
2
a. See attached spreadsheet.
b. See attached graphs.
c. The data for constant aeration are difficult to compare directly because the aeration used in
the first trial (rushton impeller) was 20% higher than the aeration used in the second trial
(pitched blade impeller). However, since the kLa curves are somewhat close even given the
pitched blade impeller’s inherent disadvantage, it can be concluded that the pitched blade
impeller is probably superior to the rushton impeller for affecting mass transfer of oxygen at
aeration levels in the range of 1 – 1.25 vvm. Certainly, it is superior for higher agitation rates
(above 600 rpm) and it seems to be not significantly worse for other agitation rates. The rushton
impeller is also a high-shear, high-mass-transfer impeller not suitable for mammalian cell
culture, a fact that is unimportant for this cell-free system or for a microbial culture, but this also
limits its applications relative to the pitched-blade impeller.
Conversely, the rushton impeller seems to be superior for most aeration rates at a
constant agitation rate of 500 rpm. In this case, the pitched blade impeller results in a higher kLa
only for the aeration values above 2.6 vvm, which is somewhat high for a typical experiment in a
bioreactor. Since 500 rpm is not a particularly high agitation rate, for a small-scale bioreactor
expected to run at such an agitation, using a Rushton impeller would be preferable to a pitched-
blade impeller. However, the pitched-blade impeller remains superior at higher agitation rates
and larger systems since it has a small diameter, yet still affects a kLa larger than the rushton
impeller at corresponding conditions.
Sam Maurer
10.28 Fermentation Problem Set #4
3
a. Sparge rates in excess of .1 vvm can be detrimental to mammalian cell culture. When an
aeration rate is high, cells are damaged by bubbles bursting at the top of the bioreactor.
Additionally, high aeration rates can cause foaming, which is also detrimental to mammalian
cells because it inhibits diffusion of glucose and other nutrients into the cell, inhibits mixing
patterns of feed solutions, and causes clogging of exhaust filters, leading to a possible risk of
contamination.
b. It has been shown experimentally that an impeller tip speed greater than 500 cm/sec
decreased viability. The tip speed in this case is
(3.14 * 80) cm/rot * 200 rot/min * 1/60 min/sec = 837 cm/sec
This causes damage to mammalian cells by exerting a large shear stress upon those cells which
pass by the impeller tip, which can tear apart their cell membranes.
c. The rushton turbine impeller indeed has a higher rate of mass transfer than the marine
impeller. However, because the marine impeller primarily causes mass transfer in the axial
direction, it exerts much lower shear stress upon the cells. The higher shear stresses created by
the rushton impeller could tear apart the cell membranes of the fragile mammalian cells. A
rushton impeller can be used for microbial cell culture because most microbes have cell walls
and are thus more sturdy than mammalian cells.
Sam Maurer
10.28 Fermentation Problem Set #4
1, calculations supplement
Note: In this and all other submissions, I have performed the calculations exactly as I did on the
original assignment submitted, even if I have now realized that they are incorrect.
Calculations were performed according to the Carbon mass balance described in Appendix D of
the 10.28 Fermentation Laboratory Manual. Here is a sample calculation for T = 1.5 hr:
First, calculate the mass of carbon from glucose that has been consumed. Assume constant
reactor volume of .8 L. As the YSI2700 analyzer was broken, glucose concentrations were
extrapolated from the optical density via a first-order linear approximation from known values.
MC,glu = (Cglu,o * Vo - Cglu,t * Vt) * .4
= (8.750 g/L * .8 L – 8.100 g/L * .8 L) * .4
= .20800 g
Next, calculate the mass of carbon from the cells that has been created. Assuming dried cells are
about 55 % carbon, the formula follows, where Mcell,t represents the dry cell weight at time t.
Mcell,t = (DCWsample,t – DCWo,t) * Vreactor / Vsample
= (.0007 g - .0001 g) * .8 L / .0015 L
= .32003 g
MC,cell = (Mcell,t – Mcell,o) * .55
= (.32003 g - .0533 g) * .55
= .14669 g
Finally, calculate the mass of carbon produced in evolved CO2 gas. First calculate the CER.
CER = 1 / Vreactor * (%N2,in / 100) * (Fg / 22.4 L/mol) * (%CO2,out / %N2,out - %CO2,in / %N2,in)
= 1 / .8 L * (79.83 / 100) * (.8 L/min / 22.4 L/mol) * (.33 / 80.12 - .02 / 79.83)
= 1.3786e-4 mol/L min
= .00827 mol/L hr
Use this CER and the two previously-calculated CER values, along with the trapezoidal
approximation from elementary calculus to approximate the area under the CER vs. t curve.
AUC = .5 * .75 hr * (.00401 mol / L hr + .00507 mol / L hr)
+ .5 * .75 hr * (.00507 mol / L hr + .00827 mol / L hr)
= .00841 mol / L
Finally, convert this value to MC,CO2
MC,CO2 = .00841 mol / L * 44.01 g CO2 / mol * .273 g C / g CO2 * .8 L in reactor
= .08083
Use these three masses to complete the mass balance. Note that on the shown graph, the second
MC,CO2 should actually MC,tot – the total carbon output from the reactor.
Sam Maurer
10.28 Fermentation Problem Set #4
2, calculations supplement
Calculations were performed according to the method described in Appendix A of the 10.28
Fermentation Laboratory Manual. A sample calculation will be performed for Day 2 data,
agitation 100 rpm, aeration 1.25 vvm.
First, calculate the OUR.
OUR = 1 / Vreactor * (%N2,in / 100) * (Fg / 22.4 L/mol) * (%O2,in / %N2,in - %O2,out / %N2,out)
= 1 / .8 L * (80.48 / 100) * (1 L/min / 22.4 L/mol) * (19.5 / 80.12 – 19.44 / 80.53)
= 4.022e-5 mol/L min
= 2.413 mmol/L hr
Next, find C* O2 . Use PO2 = .5 * (PO2,in + PO2,out)
C*O2 = kH * PO2
= .18 mmol/atm L * .5 * (19.5 + 19.44) / 100 * 1 atm
= .03505 mmol / L
Finally, use these two values to directly obtain kLa. Assume CO2,bulk is equal to 0.
kLa = OUR / (C*O2 - CO2,bulk)
= 2.413 mmol/L hr / (.03505 mmol/L – 0 mmol/L)
= 68.85 1/hr