FACULTY OF ENGINEERING CHEMICAL ENGINEERING PROGRAMME SEMESTER 2 2014/2015 BIOPROCESS PRINCIPLES TITLE: SOLUTION FOR MID-SEMEST MID -SEMESTER ER EXAMINATION
Group Members: NAME
MATRIC. NO.
ANGEL KUOK MEI ERH
BK12110025
MASZIAH BINTI MANSUR
BK12110180
MOHD MIRZUAN BIN MUSLIN
BK12160438
RICCO IRZWIN SYAZRYN BIN MOHD IRFAN
BK12110303
SAUFI AZRA’EI BIN RAYMIE
BK12110425
Lecturer’s Name:
Prof. Pogaku Ravindra
Date of Submission:
27th April 2015
Question 1 Aiba et al. (1968) reported the results of a chemo stat study on the growth of a specific strain of baker s yeast as shown in the following Table 1. The inlet stream of the chemo stat ’
did not contain any cells or products. Table 1
Dilution rate, D, hr-1 0.084 0.100 0.160 0.198 0.242
Inlet conc. (gL-1) glucose Csi 21.5 10.9 21.2 20.7 10.8
Steady-state concentration (gL-1) Glucose, Cs Ethanol, Cp Cells, Cx 0.054 7.97 2.00 0.079 4.70 1.20 0.138 8.57 2.40 0.186 8.44 2.33 0.226 4.51 1.25
a) Find the rate equation for cell growth. Assume the cell growth follows Monod equation, and no endogenous material, k d=0, thus
= g
net,
=
For a chemostat system,
+
=D, g
∴ =
1
+
+
=
1
1
=
1
+
A graph of 1/D against 1/Cs is plotted. 1/D , hr 11.905 10.000 6.250 5.051 4.132
1/Cs, g-1L 18.519 12.658 7.246 5.376 4.425
From the graph, 1
= − = 2.0 = 0.500 ℎ− = = − = 0.5714 − . ℎ − 1
11.8 4.6
1
17.2 4.6
∴ = 0.5714− . ℎ × 0.500ℎ− = 0.2857 − = = + . ∴ = . + 1
1
1
b) Find the rate equation for product (ethanol) formation. Assume it is growth-associated product,
= = = = + 1
/
/
is obtained from the slope of graph of C /
P against
Cx.
From the graph,
= ∆ ∆ = − 2.5 = 3.5625 . ( )− = 8.2 2.2 − 0.6
1
. = . . +
Question 2 A simple, batch fermentation of an aerobic bacterium growing on methanol gave the results shown in the table. Calculate: a) b) c) d) e)
Maximum growth rate (µ max) Yield on substrate (Y X/S) Mass doubling time (td) Saturation constant (K s) Specific growth rate (µnet) at t = 10h Time (h) 0 2 4 8 10 12 14 16 18
X (g/l) 0.2 0.211 0.305 0.98 1.77 3.2 5.6 6.15 6.2
S(g/l) 9.23 9.21 9.07 8.03 6.8 4.6 0.92 0.077 0
Graph of X,S against t 7
10 9
6
8 5
7 6
) 4 l / g ( X3
5
g
(
S
4 3
2
2 1
1
0
0 0
2
4
6
8
10
Time,t(h)
12
14
16
18
20
)
l
/
a) Maximum grow rate (µmax) Maximum growth rate is between t = 10 and t = 12. X2 = X1eµ ∆ t
(1)
ln X2 = (ln X1)(ln eµ ∆ t) ln ln
= µ∆t
3.2 1.77
µ=(
= µ(12-10)
0.592
)
2
µmax = 0.296 hr-1
b) Yield on substrate (Y x/s) From the table, XO = 0.2 hr-1
Xf = 6.2 hr-1
SO =9.23 hr-1
XO = 0 hr-1
Y x/s
= =
− − 6.2−0.2 9.23−0
= 0.65
c) Mass doubling time (td) Mass doubling time is only applicable during the exponential growth phase -
Use the maximum growth rate period. Rearranging
td = td =
2 µ 2 0.296
td = 2.34 hr Mass doubling time, td = 2.34 hr
d) Saturation constant (K s) The value is in between 3 last points where the substrate is about to exhausted. It is between S = 0.0 gl-1 and S= 0.92 gl-1 K s
= (0 + 0.92)/2 = 0.46
e) Specific growth rate (µ net) at t = 10h Specific growth rate is in midway into the fermentation run. It is the same as μmax = 0.292 hr-1
3. The growth rate of E. coli can be expressed by monod kinetics with the parameters of µmax = 0.935 hr-1 and k s = 0.71 g/L. Assume that the cell yield Y x/s is 0.6 g dry cells per g substrate. If Cx0 is 1 g/L and Cs = 10 g/L when the cells start to grow exponentially, at t 0 = 0, show how ln Cx, Cx, Cs, dln Cx/dt and dCx/dt, change with respect to time. Solution:
µ = 1 = = − Net specific growth rate = growth 0f cells K d = 0, cells start to grow exponentially.
= . Assume that a single chemical species is in a growth rate limiting, Monod equation,
= . = . Based on the question, S = Cs, x = Cx
= . + + = .
Thus,
∫ = ∫ . [] = [] .
Cx increased at step size = 0.1. Thus, Cs I calculated.
−0 = 1 − / Value of t is calculated from integration value of Cs and Cx. For Cx = 1.0, Step size = 0.1
0.71 10 (ln1− ln1) = 0.935(10) t=0
= 10 − 10.−61 Cs = 10
= − 0 = 1 − 1 = 0 0 = −0 = 1 −1 = 0 0 For Cx = 1.1,
0.71 10 (ln1.1 −ln1) = 0.935(10) t = 0.1092 h
= 1.1 −1 = 0.9158 0.1092 = 1.1 − 1 = 0.8728 0.1092 Then, we continue to iterate the value of t by using Microsoft excel. The table below only shows some of the iterations values.
Cx
Cs
t
ln Cx
dCx
dln Cx
1.0
10 9.833 9.5 9.0 8.333 7.5 6.5 5.333 4.0 2.5 0.833
0 0.1092 0.2091 0.3016 0.3883 0.4706 0.5503 0.6295 0.7123 0.8083 0.9519
0 0.0953 0.1823 0.2624 0.3365 0.4055 0.4700 0.5306 0.5878 0.6419 0.6932
0 0.9158 0.9564 0.9947 1.0301 1.0625 1.0903 1.1120 1.1231 1.1134 1.0505
0 0.8728 0.8719 0.8699 0.8665 0.8616 0.8541 0.8429 0.8252 0.7941 0.7282
1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0
Based on the iterations, we will plot the graphs that show the relation of ln Cx, Cx, Cs, dln Cx/dt and dCx/dt against time.
Graph Cx,Cs against time 12
Cx,Cs against time
10 8 s C , x C
6
Cx Cs
4 2 0 0
1
2
3
4 time
5
6
7
Graph ln Cx against time
ln Cx against time 2.5 2
x C n l
1.5
1 0.5 0 0
1
2
3
4
5
6
7
time
Graph dCx/dt against time
dCx/dt against time 3 2.5 2 t d / x C d
1.5 1 0.5 0 0
1
2
3
4 time
5
6
7
Graph d(ln Cx)/dt against time
d(lnCx)/dt against time 1 0.9 0.8 0.7 t d / ) x C n l ( d
0.6 0.5
0.4 0.3 0.2 0.1 0 0
1
2
3
4 time
5
6
7
4. Penicillin is produced by P.chrysogenum in a fed-batch culture with the intermittent addition of glucose solution to the culture medium. The initial culture volume at quasi-steady state is V0 = 500L, and glucose-containing nutrient solution is added with a flow rate of F =
50 . Glucose concentration in the feed solution and initial cell concentration are S 0 = 300
. h-1, K s = 0.5 , and = 0.3
and X0 = 20 , respectively. The kinetic and yield coefficient of the organisms are u m = 0.2
(a) Determine the culture volume at t=10h.
= + =500+50 ℎ10ℎ =1000 # (b) Determine the concentration of glucose at t=10h at quasi steady state.
= = = 0.05 h-1 . = . = 0.1667 = − . − . # (c) Determine the concentration and total amount of cells at quasi-steady state when t =10h.
Amount of glucose
= + 300 10 ℎ = 20 500+50 0.3 = 10 000 g glucose + 45 000 g glucose = 55 000 Concentration of glucose
[ ] = 55 0001000 [ ] =55 #
(d) If qp = 0.05
and P =0.1, determine the product concentration in the vessel o .
at t=10h.
= + ( + 2)
x ) + (0.05 55 )( + . 10ℎ .
=0.1 =0.05 +2.75 0.7510 = 20.675 # ≈ 21
