Tutorial (Chapter 3)
ERT 317 Biochemical Engineering
Problem:
A new strain of yeast is being considered for biomass production. The following data were
obtained using a chemostat. An influent substrate concentration of 800 mg/l (S0=800 mg/l)
and an excess of oxygen were used at a pH of 5.5 and T=35°C. Using the following data
calculate µm, Ks, YMX/S, kd, and ms.
Dilution rate, D (h-1)
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Carbon concentration, S (mg/l)
16.7
33.5
59.4
101
169
298
702
Cell concentration, X (mg/l)
366
407
408
404
371
299
59
Practise to solve the problem! Otherwise
you will face difficulty during the exam. If
you have any problem then please do see
me. Solution for this tutorial is provided.
Solution:
1. Plot 1/YAPX/S versus 1/D, where YAPX/S is calculated from X/(S0-S)
YAPX/S = X/(S0-S)
366/(800-16.7)= 0.467
407/(800-33.5)= 0.531
408/(800-59.4)= 0.551
404/(800-101)= 0.578
371/(800-169)= 0.588
299/(800-298)= 0.596
59/(800-702)= 0.602
1/YAPX/S
2.14
1.88
1.82
1.73
1.70
1.68
1.66
D
0.1
0.2
0.3
0.4
0.5
0.6
0.7
1/D
10.00
5.00
3.33
2.50
2.00
1.67
1.43
y-intercept = 1/YMX/S =1.58
2.20
2.10
1/Y APX/S
n=7
n
1
2
3
4
5
6
7
2.00
1.90
1.80
1.70
1.60
1.50
0.00
2.00
4.00
1/D
6.00
8.00
10.00
2. The y-intercept is 1/YMX/S= 1.58 or YMX/S = 0.633 gX/gS
3. Calculate the slope using the following equation to determine ms:
y = mx + c, where m = slope of the straight line:
∑
∑ ∑
∑
∑
x = 1/D, y = 1/YAPX/S and n = no. of samples or data = 7
x=1/D
10.00
5.00
3.33
2.50
2.00
1.67
1.43
∑x =25.93
y=1/YAPX/S
2.14
1.88
1.82
1.73
1.70
1.68
1.66
∑y =12.61
x2
100.00
25.00
11.11
6.25
4.00
2.78
2.04
2
∑x =151.18
xy
21.40
9.42
6.05
4.33
3.40
2.80
2.37
∑xy =49.77
Thus m = 49.77 – [(25.93)(12.61) / 7]
151.18 – [(25.93)2 / 7]
= 49.77 – 46.71
151.18 – 96.05
= 0.056 ≈ 0.06 = ms =0.06 gS/gX-h
.
4. Recall ms = kd/YMX/S, thus kd = ms YMX/S = (0.06 gS/gX-h)(0.633 gX/gS) = 0.038 h-1
5. Plot 1/(D + kd) versus 1/S
n
1
2
3
4
5
6
7
D + kd
(0.1 + 0.038)= 0.138
(0.2 + 0.038)= 0.238
(0.3 + 0.038)= 0.338
(0.4 + 0.038)= 0.438
(0.5 + 0.038)= 0.538
(0.6 + 0.038)= 0.638
(0.7 + 0.038)= 0.738
1/ (D + kd)
7.25
4.20
2.96
2.28
1.86
1.57
1.36
S
16.7
33.5
59.4
101
169
298
702
1/S
0.0599
0.0299
0.0168
0.0099
0.0059
0.0034
0.0014
y-intercept = 1/µm = 1.25
8
7
1/(D + kd)
6
5
4
3
2
1
0
0.0000
0.0100
0.0200
0.0300
0.0400
0.0500
1/S
6. The y-intercept is 1/µm = 1.25 or µm= 0.8 h-1
7. Calculate the slope using the following equation to determine Ks:
y = mx + c, where m = slope of the straight line:
∑
∑
∑ ∑
∑
x = 1/S, y = 1/ (D + kd) and n = no. of samples or data = 7
0.0600
n=7
x=1/S
0.0599
0.0299
0.0168
0.0099
0.0059
0.0034
0.0014
∑x =0.1272
y=1/(D + kd)
7.25
4.20
2.96
2.28
1.86
1.57
1.36
∑y =21.48
x2
0.00359
0.00089
0.00028
9.8 x 10-5
3.5 x 10-5
1.13 x 10-5
2.03 x 10-6
∑x2 =0.00491
Thus m = 0.6496 – [(0.1272)(21.48) / 7]
0.00491 – [(0.1272)2 / 7]
= 0.6496 – 0.3903
0.00491 – 0.00231
= 99.7 ≈ 100
Since m = Ks/µm = 100 or Ks = m. µm = (100)(0.8) = 80 mg/l
Thus:
YMX/S = 0.633 gX/gS
ms = 0.06 gS/gX-h
kd = 0.038 h-1
µm = 0.8 h-1
Ks = 80 mg/l
xy
0.434
0.125
0.0498
0.0226
0.0110
0.00526
0.00193
∑xy =0.6496