CN5172 Biochemical Engineering Problem Set 3

3.1 In a two-stage chemostat system, the volumes of the first and second reactors

are V1=500 L and V2=300 L, respectively. The first reactor is used for biomass production and the second reactor is for a secondary metabolite formation. The feed flow rate to the first reactor is F=100 L/h, and the glucose concentration in the feed is S=5.0 g/L. Use the following constants for the cells:

m=0.3 h-1, Ks=0.1 g/L, Yx/S=0.4 g dcw/g glucose (a) Determine the cell and glucose concentrations in the effluent of the first

stage. (b) Assume that growth is negligible in the second stage and the specific rate

of product formation is qp=0.02 g P/g cell-h, and YP/S=0.6 gP/g S. Determine the product and substrate concentration in the effluent of the second reactor.

3.2 Predict the outcomes of the competition between bacterium A and bacterium

B when grown in a chemostat operated at dilution rates between 0.2h-1 to 1.0h-

1.

3.3 The chemostat growth of a microorganism on glucose gave a cell production

rate of:

hm

cellsgx S

Sxq 3

43

4

and cell yield coefficient of YX/S = 0.1 gg-1.

For a chemostat volume of 1m3 and feed glucose concentration of 60 gm-3, (a) calculate the feed flow rate for which maximum cell production rate is

obtained. What is this maximum cell production rate? (b) calculate the largest feed flow which can be handled in the given reactor. (c) A centrifuge is installed, and a fraction, � of the cells is returned to the

reactor. Determine the recirculation rate, R as a function of if the specific growth rate of the cells is the same as that found in (a) is to be maintained, and the feed flow rate is 2.5 m3h-1. Hence determine the minimum .

3.4 A chemostat study was conducted to investigate the effect of the feed limiting

substrate concentration on the steady state cell concentration. The results are graphically represented in Figure 3.4. Explain the results obtained assuming a

0

0 .2

0 .4

0 .6

0 .8

1

1 .2

0 2 0 40 6 0 80

S u b strate C o n c. (g /l)

Spec

ific

Gro

wth

Rat

e (/

h)

B

A

Monod model for cell growth. Comment on the relative magnitude of the affinity constant in the cell growth model.

Figure 3.4

0

1

2

3

4

5

6

7

8

0.0 0.2 0.4 0.6 0.8 1.0 1.2

Dilution rate (1/h)

Ce

ll c

on

ce

ntr

ati

on

(g

/l)

0

2

4

6

8

10

12

14

16

Su

bs

tra

te c

on

ce

ntr

ati

on

(g

/l)Feed concentration = 14g/l

Feed concentration = 6g/l

Feed concentration = 2g/l