MASS TRANSFER AND SCALE UP
A 50 m3 bioreactor (H/DT=2.5; working volume=60%) equipped with two sets of a standard flat blade
turbine is used for yeast growth, the bioreactor is operated continuously at a dilution rate of 0.3 hr-1. The organism
obeys the Monods equation (m=0.4 hr-1 and Ks=2 kg/m3). The inlet sugar feed concentration is 50 kg/m3. The
bioreactor is aerated and agitated at 0.5 vvm at 60 rpm. The yield of biomass based on glucose is 0.5 g cell (dry)
per gram glucose consumed. The density and viscosity of the broth are 1200 kg/m3 and 0.02 Pas.
Cell formula: CH1.8O0.5N0.2
State whether the system is mass transfer limited or biochemical reaction limited.
Given:
VT=50 m3 D=0.3 hr-1
H/DT=2.5 m=0.4 hr-1
working volume=60% Ks=2 kg/m3
two sets of a standard flat blade turbine; Pmo=2 Pmo CSO=50 kg/m3
Q=0.5 vvm 0.5 g cell (dry)
YX/S=
N=60 rpm
= 1200 kg/m3
=0.02 Pas
Cell formula: CH1.8O0.5N0.2
Required: Mass transfer limited or Biochemical reaction limited
Soln:
= 2.5
H = liquid height
DT = tank diameter
= 2 = 2 (2.5 ) = (2.5) ( 3 )
4 4 4
VT = 50 m3
50 3
50 3 = (2.5) ( 3 ) ; = 3
4 (2.5)
4
DT = 2.9420 m
0.02
= = 1200 /3 = 1.67 105 3 /
From Biochemical Engineering by James Lee: = 3 ; DI = impeller diameter
I
2.9420
I = = = 0.9807
3 3
For an air-electrolyte solution:
0.70
= 2.0 103 [ ] 0.2 (eq. 9.72, p. 267, James Lee)
kLa =volumetric mass- transfer coefficient
Pm = gassed power
VL = volume of the liquid or broth
Vs = superficial velocity
For the power number, Np,
1 2
2
60 ( ) (0.9807 ) (1200 )
, =
=
60 3
0.02
= 57706.3494
Re 10000; Np = 6 (p. 258, James Lee)
N = (60/60) rps
For a flat-blade turbine at Re 10000
60
= 3 5 = 6 (1200 3 ) (60 )3 (0.9807 )5
= 6531.5066 ( )
For the gassed power
0.115 1.96 ( )
4.38 2 2
log10 = 192 [ ] [ ] [ ] [ 3 ] (eq. 9.53, p. 258, James Lee)
0.5 3 2
= (50 3 0.60) = 0.25 3 /
3
0.9807
2 60
0.115 60 1.96 ( )
0.9807 4.38 (0.9807 ) (60 ) (0.9807 ) ( )2 2.9420
0.25 3 /
60
log10 2 (6531.5066 = 192 [2.9420 ] [ ] [ ] [ 60 ]
) 1.67 105 9.81 /2 ( ) (0.9807 )3
60
Pm = 6191. 1993 W
For computation of kLa, superficial velocity
1
0.5 3 ( )
60
3
= = (50 3 0.60) = 0.0368 /
(2.9420 )2
4
6191.1993 0.70
= 2.0 103 [ 50 3 (0.60) ] (0.0368 )0.20 = 0.0431 1
Assume: Solution of H2SO4 @ 2.0 mols/L, find the corresponding oxygen solubility from Table 9.2 of
Biochemical Engineering by James Lee, p. 262
CO2 = 1.02 mmol/L
Use the solubility of O2 from Table 9.2 to determine the Henrys constant and the equilibrium
concentration of O2 at that condition.
1
2 = 1.02 / = 0.9804
0.21
= = 0.2142
0.9804
For kLa,CL*, the volumetric mass-transfer coefficient at the equilibrium:
0.0431 3600 1 1 32 1000
, = [ ] (0.2142 ) [1000 ] [1000 ] [1 ] [ 1 3 ] = 1.0635
1 3
Biochemical reaction limited: Using Monod Equation
= = ( + )
Determine the Exit substrate concentration, Cs
0.3 0.4
= [ ] ; = 6.0 /3
2 3 +
Then, determine the cell concentration that was produced
= = ( ) = 0.5 (50 6 ), 3 = 22 3
For ZO2, grams of oxygen required per gram dry cell produced (Maleles, 1971)
32+816
2 = + 0.01 0.0276 + 0.01714 0.08
Where:
C, H, O is the number of atoms present in the carbon source
C, H, O, N, - is the percent carbon, hydrogen, oxygen and nitrogen respectively in the cell
*From C6H12O6 (carbon source) *From CH1.8O0.5N0.2 (cell formula)
C= 6 C= 1(12) C= (1(12)/ 24.6) x100 = 48.78
H=12 H= 1.8(1) H= (1.8(1)/ 24.6) x100 = 7.32
O=6 O= 0.5(16) O= (0.5(16)/ 24.6) x100 = 32.52
MW= 6(12) + 12(1) + 6(16) N= 0.2(14) N= (0.2(14)/ 24.6) x100 = 11.38
= 24.6
32(6)+8(12)16(16)
2 = + 0.01(32.52) 0.0276(48.78) + 0.01714(11.38) 0.08(7.32)
0.5 (180)
2
2 = 0.7656
0.4
2 = 0.7658 2 (22 3 ) ( ) = 6.73904 3
ZO2(x)(m) > kLa, Cl*
Biochemical reaction > Mass Transfer
Thus, it is Mass transfer limited because theres very little oxygen that is being transferred or dissolved
in the broth.
In an oxygen absorption study, the following data were obtained on the oxygen transfer capacity of an air diffusion
unit.
Q= 9.439 m3/s Air bubble diameter: 2.5 x 10-3 m
T= 12C Air bubble velocity: 0.3 m/s
HL= 4.27 m DT= 99.5 m
The dissolved oxygen concentration measurement was tabulated as:
Time, min Cs, kg/m3
3 0.6 x10-3
6 1.6 x10-3
9 3.1 x10-3
12 4.3 x10-3
15 5.4 x10-3
18 6.0 x10-3
21 7.0 x10-3
From the information, compute the:
a. KLa and kL
b. Mass of oxygen per hour transferred per 28.317 m3 and zero dissolved oxygen concentration and the
oxygen transfer efficiency
c. How much oxygen will be transferred to waste with =0.80 at temperature of 32C and an operating
dissolved oxygen of 15 x10-3 kg/m3? Assume the saturation concentration of oxygen in the liquid at 12C
to be 1.08 x10-2 kg/m3, at 20C to be 1.07 x10-2 kg/m3 and at 32C to be 8.673 x10-2 kg/m3
Given:
Cs= 1.08 x10-2 kg/m3
Solution:
A. = [29.4
+ 42 ]
Pb - absolute pressure at the depth of air release, psi
Ot concentration of O2 in air leaving the tank, %
Cs - saturation concentration of O2 @ experimental temperature
Csm mean oxygen saturation concentration
@ 10% absorption
2 21 (10.1)
= 100 = 100 = 19.3 %
21 (10.1)+79
For Csm,
20.2 19.3
= 1.08 x 102 3 [ 29.4 + ] = 0.01238 3
42
Time, min Csm Cs, kg/m3
3 0.01238 - 0.6 x10-3 0.01178
6 0.01238 - 1.6 x10-3 0.01078
9 0.01238 - 3.1 x10-3 9.28 x10-3
12 0.01238 - 4.3 x10-3 8.08 x10-3
15 0.01238 - 5.4 x10-3 6.98 x10-3
18 0.01238 - 6.0 x10-3 6.38 x10-3
21 0.01238 - 7.0 x10-3 5.38 x10-3
Plot Csm Cs vs. Time
60
= = 3.6071 104 3 [ 1 ] = 0.0216 3
Determine the kLa
0.0216 3
= =
= 2.7 1
2 (08103 ) 3
Determine the interfacial area for the calculation of the mass transfer coefficient, kL
6
= ; = 4 2
Where :
A/V is the interfacial area
dB is the air bubble diameter
VB - is the air bubble velocity
= (99.5 )2 (4.27) = 33,201.97 3
4
6 (9.439 3 )(4.27 )
= = 9.7114 1
(2.5 103 )(0.3 )(33,201.97 3 )
For kL,
2.7/
= / = = 0.2677
9.7114/
B. () = (20) (1.02)20
(12) 2.6/ 3.046
(20) = = =
1.021220 1.021220
( ) = (20) (,20 ,20 ) (eq.1)
(20)
( ) = (32) (,32 ,32 ) (eq.2)
(32)
(32) = (20) (1.02)20
Determine the amount of oxygen transferred per hour at 20C.
3.046
( ) = (20) ( ,20 ) = (1.07 102 0) 3
(20)
3.046
(20) = (1.07 102 0) (28.317 3 ) = 0.923
3
C. From eq.1 and 2, in terms of equal working volume,
(32) (,32 ,32 )
(32) =
(20) (,20 ,20 ) (20)
(20) (1.02)3220 (,32 ,32 )
(32)@ 80% = 0.80 [ ] (20)
(20) (,20 ,20 )
(0.923 )(8.673102 1.5102 ) 3 (1.02)3220
(32)@ 80% = 0.80 [ ] = 6.2778
(1.07102 0)
SCALE UP: FERMENTER
The usual procedure of scale up of fermenter is to fix one of several criteria involving Reynolds
Number, power consumption per unit volume of liquid, tip velocity of an impeller, the liquid circulation
time and the volumetric oxygen transfer coefficient. The choice of criterion will depend on the
fermentation being studied. Estimate, using two methods, the required speed of an impeller and the power
requirements of a production scale fermenter of 60 m3, to match the volumetric mass transfer coefficient.
Following optimum conditions were obtained with a 0.03 m3 fermenter:
Density of broth: 1200 kg/m3 Liquid volume: 0.018 m3
Aeration rate: 1vvm Oxygen transfer rate: 0.25 kmol/m3hr
Liquid Height: 1.2 DT
Two sets of standard flat blade turbine impellers were installed.
Given:
L= 1200 kg/m3 HL= 1.2 DT VL= 0.018 m3
Q= 1.0 vvm OTR= 0.25 kmol/m3hr
V1= 0.03 m3 V2= 60 m3
Solution:
Determine the tank diameter, DT, and impeller diameter, DI, for each conditions.
1 = 1 2 = (1 )2 (1.2 1 ) = 0.31 3
4 4
1 1
1 3 0.0183 3
1 = (0.3) = [ ] = 0.267
0.3
1 1
1 = 3 1 = 3 (0.267 ) = 0.089 (Biochemical Engineering by James Lee, pg.274)
1 = 1.2 1 = 1.2 (0.267 ) = 0.32
1
1 0.018 3
2 3 60 3
0.03
2 = (0.3) = [ ] = 3.36
0.3
1 1
2 = 3 2 = 3 (3.36 ) = 1.12
2 = 1.2 2 = 1.2 (3.36 ) = 4.03
Aeration flow rate:
1 3 1
2
= 3 (0.018 3 ) [ ] = 3 104 3 /
60
For superficial velocity
3 3600
3104 [ ]
1
= = = 19.3
(0.267)2
4
Determine the partial pressure of oxygen in the system
0.32 2
1 +(1 + ) 1 +[1+ ]
10.3 2 10.3 2
2 = (0.21) = (0.21) = 0.213
2 2
Then determine the kLa
0.25 3
= = = 1.17
2 0.213 3
For the determination of the power requirement
Cooper et. al. (1944) correlated kLa, gassed power per unit volume (Pm/V)and superficial velocity (Vs) for
Varied Disk Impellers as:
0.95
= 0.0635 ( ) ( )0.67 , 3
For Standard-Flat Blade Turbine, Aiba et.al. (1965)
0.95
= 0.0318 ( ) ( )0.67 , 3
For Turbine Type Impellers, Vant Riet (1979)
0.4
= 0.026 ( ) ( )0.5 , 3
For the gassed power,Pm
0.95
1.17 3 = 0.0318 (0.0183 ) (19.3 )0.67 = 0.1
Then determine the ungassed power, Pmo, for two sets of impeller, Np= 2x6=12
3 5
= 3 5 ; =
3 5 12 (1200 3 ) ( )3 (0.089 )5
= =
= 8.197 103 3 , 0.108 103 3
9.81 2
Using Michaelis correlation:
2
( ) 3
1 = 0.5 [ 10.56 1 ]
1
2
(0.108 103 3 ) (0.089 )3
0.1 = 0.5 [ ]
(3104 3 /)0.56
60
= 18 [ 1 ] = 1080
1 = 0.108 103 3 , = 0.108 103 (1080)3 = 0.630
0.10
[ ] = 0.630 = 0.1587
1
For Constant Power Input in a Baffled Vessel
3 5 = 3 5 ; = = /
1 1 2 2
1 1
[ 3 5
] = [ ]
1 1 2 3 2 5
1 2
For constant power in geometrically similar vessels
1 3 1 5 2 3 2 5
=
1 2
*Accdg. To Ghose *Accdg to James Lee; Pm/V= constant
1 5 2 2/3
1 3 1 5 1 3
3
2 = 1
2 3
; 2 = 1 [ ] [ ]
3
[1 ] = [ 2 ] ; 2 = 1 [ 2 ]
5 1 2 2 1 1
2 2 2/3
1 5 2 0.089 2/3
60 3 0.089 3 2 = 1 [ ] = 720 [ 1.12 ]
2 = 1080 [0.03] [ 1.12 ] , 1
2 = 199.8575 2 = 3895.39
*for constant impeller tip velocity *for constant impeller speed
1 1 = 2 2 2/3 2/3
0.089
1 = 2 [ 2 ] ; = [ ]
2 = 1 [ 1 ] = 720 [ 1.12 ] 1
2 2 = 3895.39
2 = 57.2143 Or
( ) = 1.7 ( )
