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# CHAPTER 9: AGITATION AND AERATION

9.1 Derive the relationship between the overall mass transfer coefficient for gas phase K G and the
individual mass-transfer coefficients, kL and kG. How can this relationship be simplified for
sparingly soluble gases?
Given:
a. Overall mass transfer coefficient for gas phase, KG
b. Individual mass transfer coefficients, kL and kG
Required:
a. Derive the relationship between KG, and kG and kL
b. How can the relationship be simplified for sparingly soluble gases?
Solution:
Expressing the molar flux in terms of overall driving force for mass transfer and an
overall mass transfer coefficient,
P A
P A , G
N A =K G
c Ac A , L

N A =K L
where KG is termed the overall mass transfer coefficient based on gas phase driving
forces. In an equivalent manner, we also can write
Using Henrys Law,
mc A
P A , G=
where m is the Henrys law constant.
We can rewrite the first two equations as,
N
P A , GP A ,i= A
kG

c A , ic A , L =

NA
kL

## Substituting the equilibrium relationships,

mNA
P A=
kL
P A , i
Adding the partial pressure differences we obtain
N mN A
1 m
P A= A +
=N A ( + )
kG
kL
kG kL
P A=P A ,G
P A , GP A ,i + P A , i
Simplifying further,
1
1 m
= +
K G kG k L
For sparingly soluble gases,
1
=0
kG
Therefore,
1
m
=
K G kL

## CHAPTER 9: AGITATION AND AERATION

9.2 Prove that Eq. (9.25) is the same with Eq. (9.26), and Eq. (9.27) is the same with Eq. (9.28)
Given:
a) Eq. (9.25) and Eq. (9.26)
b) Eq. (9.27) and Eq. (9.28)
Required:
a) Prove that
2

c g

Eq. (9.25)

k L =0.31 N Sc 3 (

Eq. (9.26)

N Sh=0.31 N Sc 1/ 3 N Gr 1/3

1 /3

is the same as

b) Prove that
2
1 /3
2 D AB
3 c g
Eq. (9.27) k L = D + 0.31 N Sc ( 2 )
32

Eq. (9.28)

is the same as

## N Sh=2.0+ 0.31 N Sc1 /3 N Gr1/ 3

Solution:
a.)
N Sh=0.31 N Sc

1/ 3

N Gr

1/3

2
3

k L =0.31 N Sc (

c g

1 /3

Substituting Eq. (9.21), Eq. (9.22) and Eq. (9.23) into Eq. (9.26):
k L D32
c 1/ 3 D332 c g 1 /3
=0.31(
) (
)
D AB
c D AB
2c
c 1 /3
c 1 /3 g 1/3 1 /3
k L =0.31 1/ 3
c D AB2 /3
2c /3
k L =0.31

1/ 3

1/3
c

D AB

1 /3

1/3

1 /3

D32
D AB
Combining similar
terms

2 /3

1/3

1 /3

## Dividing both sides by

k L =0.31 1/ 3 c 2 /3 c
c D AB

1 /3

1/3

Multiplying

1/ 3

c 2 /3
c 1/ 3 g1 /3 1 /3
k L =0.31 2/ 3
c D AB2 /3
2/ 3c

1/3

)( c2/ 3 )
2/ 3
c
c

c
c D AB

k L =0.31
2

k L =0.31 N Sc 3 (

c g 1 /3
)
2

k L =0.31 N Sc 3 (

c g 1 /3
)
2

b.)
N Sh=

2 D AB
+0.31 N Sc 1/ 3 N Gr1 /3
D 32
2

k L =2.0+0.31 N Sc 3 (

c g

1/ 3

Substituting Eq. (9.21), Eq. (9.22) and Eq. (9.23) into Eq. (9.28):
k L D32
c 1/ 3 D 332 c g 1/3
=2.0+0.31(
) (
)
D AB
c D AB
2c

## Dividing both sides by

D32
D AB

1/ 3

1/ 3

1 /3
c

1/ 3

k L =2 AB + 0.31 1/ 3 c 2 /3 c
D32
c D AB
k L =2

D AB

c
+ 0.31 1/ 3
2 /3
D32
c D AB

1 /3

1 /3

1 /3

1 /3

g
2 /3
c
g

D AB
c1 /3
c 1/ 3 g1 /3 1 /3
k L =2
+ 0.31 1/ 3
D32
1
c D AB2 /3
2/ 3

k L =2

D AB

c
+ 0.31 2/ 3c
2 /3
D32
c D AB

Combining similar
terms

1/ 3

c
c D AB

D AB
k L =2
+ 0.31
D32
2
1 /3
D AB
3 c g
k L =2
+ 0.31 N Sc (
)
2
D 32

2
1 /3
D AB
3 c g
k L =2
+ 0.31 N Sc (
)
D32
2

1 /3

g
2/ 3
c

1 /3

Multiplying

3
1/
1/3
c
c
)(
)
2/ 3
c
2/c 3

## CHAPTER 9: AGITATION AND AERATION

9.4 A cylindrical tank (1.22m diameter) is filled with water to an operating level equal to the tank
diameter. The tank is equipped with four equally spaced baffles, the width of which is one tenth
of the tank diameter. The tank is agitated with a 0.36 m diameter, flat-blade disk turbine. The
impeller rotational speed is 4.43 rps. The air enters through an open ended tube situated below
the impeller and its volmetric flow rate is 0.0217 m3/s at 1.08 atm and 25 deg C.
density= 997.08 kg/m3
Calculate:
a.
b.
c.
d.
e.

Power Requirement
Gas Hold-up
Sauter-mean Diameter
Interfacial area
Volumetric mass-transfer coefficient

Given:
DT = 1.22 m
W = 1/10 DT
DI = 0.36 m
N = 4.43 rps
Q = 0.0217 m3/s

Required:
a.
b.
c.
d.
e.

Pm
H
D32
a
kla

## viscosity = 8.904 x 10^-4 kg/m-s

Solution:
a. Nre = (997.08)(4.43)(.36)^2
8.904 x 10^-4
Nre = 642,915.034 > 10,000

d. Interfacial Area:
a =6H/ D32
a= 98.5447 /m

= 3144.8862 W

## Using equation 9.53 of James Lee

Kl = 4.58x10^-4 m/s

Pm= 1341.3218 W

## Kla = 4.58x10^-4 (98.5447)

b. v= (/4)(1.22)(1.22)^2
v= 1.43 m3
Vs= (4x .0217) / (1.22)^2
= 0.0186 m/s
<
0.02 m/s
Using equation 9.48 of James Lee
H = 0.0790
c. Using equation 9.42
D32= 4.8132 x 10^-3 m
D32= 4.8132 mm

## CHAPTER 9: AGITATION AND AERATION

Kla = 0.0451 /s

9.5 Estimate the volumetric mass-transfer coefficient kLa for the gas-liquid contractor described
in Problem 9.4 by using a correlation for kLa and compare the result with the experimental value.
Given:
Reactor volume, v= 1.43m3
Vs= 0.0186 m/s
PM= 1342 Watts
Required:
kLa (using equation 9.71 by James Lee)
% kLa compared with experimental value
Solution:
1342
kLa = 0.026( 1.43 )0.4 (0.0186)0.5 = 0.0548 s-1

Experimental

## Estimated value= 0.0451 s-1

% for volumetric mass-transfer coefficient =

( 0.05480.0451 ) x 100
= 17.7007% error
(0.0548)

## CHAPTER 9: AGITATION AND AERATION

9.6 The power consumption by an agitator in an unbaffled vessel can be expressed as
P mo
ND2I
=f

N 3 D5I

( )

Can you determine the power consumption and impeller speed of a 1,000-gallon fermenter
based on findings of the optimum condition from a one-gallon vessel by using the same fluid
system? Is your conclusion reasonable? Why or why not?
Given:
P mo
ND2I
=f

N 3 D5I

( )

VP =1000 gallons
Vm=1 gallon
Required:
Can the power consumption and impeller speed of VP be determined on findings of the
optimum condition from Vm by using the same fluid system? Why?
Solution:
VP
=1000
Vm
The scale ratio is
1
DI , P
=1000 3 =10
DI,m
To achieve dynamic similarity, the three numbers for the prototype and model must be
equal
Pmo
P mo
=
3
5
N DI P N 3 D5I m

] [

[ ] [ ]
ND 2I
ND 2I
=
P

## Using the same fluid for model and prototype, P =m ;

5

( Pmo )p =10 [ P mo ] m

P =m

[ ]
NP
Nm

## The equality of Reynolds number requires

N P =0.01 N m
while the equality of Froude number requires
1
N P=
Nm
10
which shows two conflicting concepts.
If

P m ;

[]

P m ;

[]

1
=
m 31.6

## Therefore, if kinematic viscosity of prototype is similar to water, the kinematic viscosity

of the fluid which needs to be employed for the model should be 1/31.6 of the kinematic
viscosity of water. It is impossible to find the fluid whose kinematic viscosity is that
small. As a conclusion, if all three dimensionless groups are important, it is impossible to
satisfy the dynamic similarity.

## CHAPTER 9: AGITATION AND AERATION

The specific oxygen demands and critical oxygen concentrations for typical microbial plants and
animal cell cultures are listed below
Cell culture

qo

Concentration, mmol/L

Escherichia Coli

0.5 mmol(gdw)-1h-1

0.0082

Vitisvinifera (grape)

0.6 mmol(gdw)-1h-1

0.055

## Chinese Hamster Ovary

3.0x10-10 mmol(gdw)-1h-1

0.020

a.Estimate the kLa requirement to achieve cell concentrations of 25 gram dry weight/L for E.
Coli and V. Vinifera and 3.0x109 for CHO cells, while maintaining dissolved oxygen
concentration above critical. The oxygen solubility in the media used for the cultures is 7.2x10 -3
kg/m3
b. The relationship between kLa and the power input to a 1-m3 stirred bioreactor is kLa
Pt

Vl

0.5

, compare the bioreactor power requirements for culture of the three different cell types

Solution:
a.)

## for e coli: q= 8.5mmol/h (25/L)=292.65 mmol/L.h

v.vinifera: q= 0.6 mmol/h (25/L)=15 mmol/L.h
CHO: q= 3x10-10mmol/h (3x109/L)=0.9 mmol/L.h

## Oxygen Solubilty: 7.3x10-3 kg/m3

292.5 mmol/L.h= [7.2 x10-3 kg/m3 (1kmol/32kg)(1x106 mmol/kmol)-0.0002)kla
kLa= 980.1661/hr

## 15 mmol/L.h= [7.2 x10-3 kg/m3 (1kmol/32kg)(1x106 mmol/kmol)-0.055)kla

kLa= 88.2353/hr
0.9 mmol/L.h= [7.2 x10-3 kg/m3 (1kmol/32kg)(1x106 mmol/kmol)-0.02)kla
kLa= 4.3902/hr

Pt
b.) kLa= k ( Vl )

0.5

## PT E.C.= 960725.5836 m3/hr2

K2
PT v.v. = 7765.4682 m3/hr2
K2
PT CHO= 19.2739 m3/hr2
K2
PT E.C.> PT v.v.> PT CHO

## CHAPTER 9: AGITATION AND AERATION

ESTIMATING KLa USING THE SIMPLE DYNAMIC METHOD
A stirred fermenter is used to haematopoietic cells isolated from umbilical cord blood. The liquid
volume is 15 Liters. The simple dynamic method is used to determine K La. The air flow is shut
off for a few minutes and the dissolved oxygen level drops; the air supply is then reconnected at
a flow rate of 0.25 L/s. The following results are obtained at a stirrer speed of 50 rpm.
Time, s
5
20
Oxygen tension (%air saturation)
50
66
When steady state is established, the dissolved oxygen tension is 78% air saturation. In
separate test experiments, the electrode response to a step change in oxygen tension did not vary
with stirrer speed above 40 rpm. The probe response time under these conditions was 2.8
seconds. When the KLa measurement was repeated using nitrogen to deoxygenate the mixture,
the results for oxygen tension as a function of time were similar to those listed. Estimate KLa.
Solution:

KLa =

C C AL1
ln ( AL
)
C ALC AL2
t 2t 1
7850
)
7866
( 205 ) s

ln (
KLa =

= 0.056/s

a)

## CHAPTER 9: AGITATION AND AERATION

A 20-L stirred fermenter containing Bacillus thuringiensis is used to produce a microbial
insecticide. The oxygen balance method is applied to determine KLa. The fermenter operating
pressure is 150 KPa and the culture temperature is 30oC. The oxygen tension in the broth is
measured as 82% using a probe calibrated to 100% in situ using H 2O and air at 30oC and
150KPa. The solubility of oxygen in the culture fluid is the same as in H 20. Air is sparged into
the vessel; the inlet gas flow rate measured outside the fermenter at 1 atm pressure and 22 oC is
0.23/s. The exit gas from the fermenter contains 20.1% oxygen and has a flowrate of 8.91 min-1.
a. Calculate the volumetric rate of oxygen uptake by the culture
b. What is the value of KLa.
Solution:
NA =

1
RV L

([ F TP ) i( F TP ) o]
g

Ag

Ag

0.201 x 1.48

1
8.91 x
( ( 30+273 ) K )
60

0.23 L
(0.2099 atm)
s

( 22+273 ) K
1

0.08205 L. atm
(20 L)
mol . K

((

1
( 1.6357 x 104 1.4572 x 104 ) =1.0878 x 105 mol
L.s
0.08205 (20)

Because of steady state the rate of oxygen transfer is equal to the rate of oxygen uptake
by the cells; the volumetric rate of oxygen uptake by the culture is 1.0878x10-5 mol/ L.s
b) Assume that the gas phase is well-mixed so that the oxygen concentration in the bubbles
containing the liquid is the same as the outlet gas, that is 20.1%. As the difference in the
composition of the gas phase to be constant throughout the fermenter.
The solubility of oxygen in H 20 @30oC and 1atm air pressure is 8.05x10 -3 kg/m3.
Determine the solubility at the fermenter operating pressure of 1.48 atm and gas phase
oxygen mole fraction of 0.201.

C*AL,2

PT 2 Y AG 1
PT 1 Y AG 2

C*AL,1

## 1.48 atm ( 0.201 ) 8.05 x 103 g

(
)
=
= 0.0114g/L
L
1 atm ( 0.2099 )

CAL in the fermenter is 82% of the oxygen solubility at 30oC and 1.48atm air pressure
CAL= 0.820

## 1.48 atm 8.05 x 10 g

(
) = 9.77x10-3 g/L
1 atm
L

32 g
)
mol
3
0.0114 g 9.77 x 10 g

L
L
5

1.1 x 1 0 mol / L . s (

KLa =

## CHAPTER 9: AGITATION AND AERATION

DYNAMIC TECHNIQUE

## KLa = 0.22 s-1

A strain of Azobacter vinelandii is cultured in a 15m 3 stirred fermenter for alginate production.
Under current operating conditions, KLa is 0.17 s-1. The solubility of oxygen in the broth is
approximately 8x10-3 kg/m3.
a. The specific rate of oxygen uptake is 12.5 mmol/g-hr. What is the maximum cell
concentration supported by oxygen transfer in the fermenter?
b. The bacteria suffer growth inhibition after copper sulphate is accidentally added to
the fermentation broth just after the start of the culture. This causes a reaction in the
oxygen uptake rate to 3 mmol/g-hr. What maximum cell concentration can now be
supported by oxygen transfer in the fermenter?
Solution:

a) Xmax =

0.17 8 x 1 0 kg
(
)
3
s
m
mmol 1 hr
1 mol
32 g
12.5
g . hr 3600 s 1000 mol 1 mol

( )
( )(

)(

1 kg
)( 1000
g)

12240 g
3
m

b) Assume that addition of copper sulphate does not affect C*AL of KLa
0.17 8 x 1 03 kg
(
)
s
m3
Xmax=
mmol 1 hr
1 mol
32 g
3
g . hr 3600 s 1000 mol 1mol

( )
( )(

)(

1 kg
)( 1000
g)

51000 g
3
m

## CHAPTER 9: AGITATION AND AERATION

GAS HANDLING WITH RUSHTON TURBINE
A fermenter of diameter and liquid height 1.4m is fitted with a Rushton impeller of diameter
0.5m and off-bottom clearance 0.35m operated at 75 rpm. The fermentation broth is sparged with
air at a volumetric flow rate of 0.28m3/min. Half-way through the culture some bearings in the
stirrer drive begin to fail and stirrer speed must be reduced to a maximum of 45 rpm for the
remainder of the process.
a. Under the normal operating conditions, is the gas completely dispersed?
b. After the stirrer speed is reduced, is the impeller flooded or loaded?
Solution:
1 min
a) Ni = 75/min 60 s

Fr =

2 Di
g

= 1.25s-1

1.25 s1 2 (0.5 m)

= 0.0796

Flg = 0.2

0.5

DI
DT

( )

Fr 0.5 = 0.2

0.5 m
1.4 m

0.5

## Fg = FlgNiDi3 = (0.0337)(1.25s-1) (0.5m)3 = 5.27x10 -3 m3/s > 0.28 m3/min

*Fg, volumetric flowrate of gas greater than the operating flow rate, we can conclude that
the air provided is completely dispersed under normal conditions.
b)

1 min
Ni = 45 /min ( 60 s ) = 0.75s-1
1 2

0.75 s (0.5 m)

Fr =

Di
g

= 0.0287

Flg = 30

DI
DT

3.5

( )

Fr

= 30

0.5 m
1.4 m

3.5

= 0.0234

## Fg = FlgNiDi3 = (0.0234)(0.75s-1) (0.5m)3 = 0.00219 m3/s

*At reduced stirrer speed, maximum air flow rate can be handled without impeller
flooding as operating flow rate (0.28m3/min) is greater than this. The impeller is
FLOODED.

## CHAPTER 9: AGITATION AND AERATION

Clostridium acetobutylicum carries out anaerobic fermentation and converts glucose into
acetone, butanol along with smaller concentrations of butyrate, acetate, etc. In fermentation the
following products were obtained from 100 moles of glucose and 11.2 moles of NH 3, as nitrogen
source. Products formed:
Cells = 13moles
Butanol= 56moles
Acetone= 22moles
Butyric acid = 0.4 moles

## acetic acid= 14 moles

CO2= 221 moles
H2= 135moles
Ethanol= 0.7 moles

By performing a carbon, nitrogen, hydrogen, and oxygen balance, determine the chemical
composition of the cells.
Solution:
By performing a carbon, hydrogen, nitrogen and oxygen balance, determine the element
composition of the cells.
100C6H12O6 + 11.2 NH3 13CaHbOcNd + 56C4H10O (butanol) + 22C3H6O(acetone) +
0.4C4H8O2(butyrate) + 14C2H14O2 (acetic acid) + 221CO2 +135H2 + 0.7 C2H6O(ethanol)
* where CaHbOcNd represents elemental composition of clostridium cells
Carbon Balance:
100(6) + 11.2(0) = 13(a) + 56(4) + 22(3) + 0.4(4) + 14(2) + 221(1) + 0.7(2) : a = 4.46
Hydrogen Balance:
100(12) + 11.2(3) = 13(b) + 56(10) + 22(6) + 0.4(8) + 14(14) + 135(2) + 0.7(6) : b = 16.02
Oxygen Balance:
100(6) + 11.2(0) = 13(c) + 56(1) + 22(1) + 0.4(2) + 14(2) + 221(2) + 0.7(1) : c = 3.88
Nitrogen Balance:
11.2(1) = 13(d) : d = 0.86

## Aerobic Culture of Saccharomyces cerevisae in a synthetic medium produced the following:

= 0.2/hr
RQ =1.0
Yy/s = 90 grams dry cell/gmol glucose
Vglutamate = 0.043 grams glutamate/ g cell-hr
Assuming that carbon content of the cell is 45%, check carbon and oxygen balance with
respect to given culture.
aC6H12O6 + bC5H9O4N + cO2 dCHOO + e CO2 + fH2O
glutamic acid
Solution:
RQ = 1.0 = Q02 = QCO2 = c = e
@ = 0.2 hr -1 and 45% C in biomass ; 1mol cell biomass = 12gC

d=

0.2
(0.45)
hr
12 g/mol

45% =

g cell biomass
g Carbon
(100)
MW cell biomass
C

MWcell =

12(100)
=26.67
45

## MW = 26.67 g/mol = CHON

26.67= 1(12) + (1) + (16) + (14)
14.67= + 16 + 14 (Equation 1)

a= Qglucose =

Y X /S

0.2/hr
g dry cell
90
g glucose

0.043
b= Qglutamate =

147

g glutamate
g cell . hr

g
glutamic acid
mol

## = 2.925x10-4 gmol glutamate/ g dry cell. Hr

Elemental Balance:
C: 6a + 5b = d + e
H: 12a + 9b = d + 2f
O: 6a + 4b + 2c = d + 2e + f : 2c=2e : 6a + 4b = d + f
N: b = d
=

b
d

2.925 x 104
7.5 x 1 03

Using H and O:
12a + 9b = d + 2f
-2
(6a + 4b = d + f)
b= d - 2d

= 0.0390

## 2.925x10-4 = (7.5x10-3) - 2 (7.5x10-3)

2.925 x 1 04 +0.015
=
7.5 x 1 03
from equation 1:
14.67 = + 16 + 14 : 14.67 = +16 + 14(0.0390)
14.124 = + 16
Substituting :
4

14.124 =

2.925 x 10 +0.015
3
7.5 x 10

+ 16 : = 0.7825, = 1.604

Using equation of O:
6a + 4b = d + f ; f = 6a + 4b d
f= 6(2.22x10-3) + 4(2.925x10-4) 0.7825(7.5x10-3)
f = 8.6213x10-3
Using equation of C:
6a + 5b = d + e ; e = 6a + 5b d
e = 6( 2.22x10-3) + 5( 2.925x10-4) 7.5x10-3
e = c = 7.2825x10-3
* aC6H12O6 + bC5H9O4N + cO2 dCHOO + e CO2 + fH2O

2.22 C6H12O6 + 2.925C5H9O4N + 7.2825 O27.5 CH1.604O0.7825O0.0390 + 7.2828 CO + 8.6213H2O

## CHAPTER 9: AGITATION AND AERATION

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
H/DT=2.5
working volume=60%
two sets of a standard flat blade turbine; Pmo=2 Pmo
Q=0.5 vvm

D=0.3 hr-1
m=0.4 hr-1
Ks=2 kg/m3
CSO=50 kg/m3
0.5 g cell (dry )
YX/S=
g glucose

N=60 rpm
= 1200 kg/m3
=0.02 Pas
Cell formula: CH1.8O0.5N0.2
Required: Mass transfer limited or Biochemical reaction limited

Soln:
H
=2.5
DT

H = liquid height
DT = tank diameter

2
2
3
V T = D T H = DT ( 2.5 D T )= ( 2.5 ) ( DT )
4
4
4

VT = 50 m3

50 m = (2.5 ) (D T 3 ) ;
4
3

DT = 3

50 m3

( 2.5 )
4

DT = 2.9420 m

0.02 Pa s
= =
=1.67 105 m3 / s
3
1200 kg /m

## From Biochemical Engineering by James Lee:

DI =

DT
=3 ; DI = impeller diameter
DI

DT 2.9420 m
=
=0.9807 m
3
3

0.70
3 P m
k L a=2.0 10
V s0.2
VL

[ ]

## kLa =volumetric mass- transfer coefficient

Pm = gassed power
VL = volume of the liquid or broth
Vs = superficial velocity
For the power number, Np,

## (eq. 9.72, p. 267, James Lee)

N D I2
'
Reynol d s Number , Re =
=

60 rpm

## ( 601minsec )(0.9807 m) (1200 kgm )

2

0.02 Pa s

57706.3494
Re 10000; Np = 6 (p. 258, James Lee)
N = (60/60) rps
For a flat-blade turbine at Re 10000
3
kg 60
Pmo=N P N 3 DI 5=6 1200 3 ( rps) (0.9807 m)5
m 60

## Pmo=6531.5066 W (ungassed power )

For the gassed power
4.38

0.115

DI
DT

( )

[ ] [ ] [ ] [ ]

P
D
log 10 m =192 I
Pmo
DT

D I2 N

DI N 2
g

1.96

Q
N D I3

## (eq. 9.53, p. 258,

James Lee)
3

Q=

0.5 m O 2
3

m broth min

( 50 m3 0.60 ) =0.25 m 3 / s
60 2

60

1.96

(0.9807)

( 0.9807
2.9420 )

0.25 m3 /s
60
( )(0.9807)3
60

Pm
0.9807 m
=192
2.9420 m
2 ( 6531.5066 W )
log 10

4.38

60
rps )
60
1.67 105

(0.9807)2 (

Pm = 6191. 1993 W
For computation of kLa, superficial velocity
1 min
0.5 m3 air
60 sec

V s=

Q
=
S

m3 media min

(2.9420 m)2
4

( 50 m3 0.60 )=0.0368 m/ s

0.115

k L a=2.0 10

6191.1993 W
3
50 m (0.60)

0.70

m 0.20
(0.0368 ) =0.0431 s1
s

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 atm
atm L
HO =
=0.9804
1.02 mmol/L
mmol
2

C L =

0.21 atm
mmol
=0.2142
atm L
L
0.9804
mmol

## For kLa,CL*, the volumetric mass-transfer coefficient at the equilibrium:

0.0431 3600 s
mmol
1 mol
1 kmol
32 kg
k L a ,C =
0.2142
s
1 hr
L
1000 mmol 1000 mol 1 kmol

](

)[

][

][

][

1000 L
kg
=1.0635 3
3
1m
m hr

Cs
D==m
K s +C s

## Determine the Exit substrate concentration, Cs

Cs
0.3 0.4
=
; C s=6.0 kg /m 3
hr hr
kg
2 3 +Cs
m

[ ]

## Then, determine the cell concentration that was produced

g dry cell
kg
kg
C x C xo =x= x ( C soC s )=0.5
( 506 ) , 3 =22 3
g glucose consumed
m
m
s
For ZO2, grams of oxygen required per gram dry cell produced (Maleles, 1971)
32 C+ 8 H 16 C
'
'
'
'
ZO =
+0.01 O 0.0276 C +0.01714 N 0.08 H
x MW
2

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)

C= 6
H=12
O=6
MW= 6(12) + 12(1) + 6(16)

ZO =
2

32 ( 6 ) +8 (12)16(16)
+ 0.01(32.52)0.0276(48.78)+0.01714 (11.38)0.08(7.32)
0.5(180)

Z O =0.7656
2

gO2
g dry cell

Z O x m=0.7658
2

## *From CH1.8O0.5N0.2 (cell formula)

C= 1(12)
C= (1(12)/ 24.6) x100 = 48.78
H= 1.8(1)
H= (1.8(1)/ 24.6) x100 = 7.32
O= 0.5(16)
O= (0.5(16)/ 24.6) x100 = 32.52
N= 0.2(14)
N= (0.2(14)/ 24.6) x100 = 11.38
= 24.6

g O2
kg
22 3
g dry cell
m

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
broth.

## CHAPTER 9: AGITATION AND AERATION

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
T= 12C
HL= 4.27 m

## Air bubble diameter: 2.5 x 10-3 m

Air bubble velocity: 0.3 m/s
DT= 99.5 m

Time, min
3
6
9
12
15
18
21

Cs, kg/m3
0.6 x10-3
1.6 x10-3
3.1 x10-3
4.3 x10-3
5.4 x10-3
6.0 x10-3
7.0 x10-3

## From the information, compute the:

a. KLa and kL
b. Mass of oxygen per hour transferred per 28.317 m 3 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.

C sm=C s

Pb O t
+
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
moles of unabsorbed O2
21(10.1)
Ot=
100=
100=19.3
total moles of leaving air
21 ( 10.1 ) +79

For Csm,
C sm=1.08 x 102

kg
+
=0.01238 3
3
29.4
42
m
m

Time, min
3
6
9
12
15
18
21

## 0.01238 - 0.6 x10-3

0.01238 - 1.6 x10-3
0.01238 - 3.1 x10-3
0.01238 - 4.3 x10-3
0.01238 - 5.4 x10-3
0.01238 - 6.0 x10-3
0.01238 - 7.0 x10-3

0.01178
0.01078
9.28 x10-3
8.08 x10-3
6.98 x10-3
6.38 x10-3
5.38 x10-3

## Plot Csm Cs vs. Time

m=slope=3.6071 104

kg
60 min
kg
=0.0216 3
m min 1 hr
m hr
3

## Determine the kLa

kg
slope
m3 hr
k L a=
=
=2.7 hr 1
mean concentration of O2 (
kg
08 103) 3
m
0.0216

Determine the interfacial area for the calculation of the mass transfer coefficient, kL
A 6Q HL

=
; V T = DT 2 H L
V dBV B V T
4
Where :
A/V is the interfacial area
dB is the air bubble diameter
VB - is the air bubble velocity

## V T = ( 99.5m )2 ( 4.27 m )=33,201.97 m3

4
A
=
V

6 ( 9.439 m3 ) (4.27 m)
=9.7114 m1
( 2.5 103 m ) 0.3 m (33,201.97 m3)
s

For kL,
k L=

kLa
2.7 /hr
m
=
=0.2677
A /V 9.7114 /m
hr

B.

k L a(20 )=

( qv )
( qv )

k L a(12 )
1220

1.02

2.6/ hr
3.046
=
1220
hr
1.02

## =k L a(20 ) (C s ,20 C L,20 )

(eq.1)

(20 )

=k L a(32 ) (C s ,32 C L, 32 )

(eq.2)

(32 )

## k L a(32) =k L a(20 ) (1.02)T 20

Determine the amount of oxygen transferred per hour at 20C.
q
3.046
kg

2
=k a
( C C L,20 ) = hr ( 1.07 10 0 ) 3
v (20 ) L (20 ) s
m

()

q(20 )=

3.046
( 1.07 1020 ) kg3 ( 28.317 m 3 )=0.923 kg
hr
hr
m

## C. From eq.1 and 2, in terms of equal working volume,

k L a(32) ( Cs , 32 C L, 32 )
q(32 )=
q
k L a(20 ) (C s , 20 C L, 20 ) (20 )
q ( 32 ) @80 eff =0.80

## k L a(20 ) (1.02)3220 (C s ,32 C L ,32 )

q(20 )
k L a(20 ) (C s ,20 C L ,20 )
1.02 3220

## (0.923 kgh ) (8.673 10

1.5 102 )

q( 32 )@ 80 eff =0.80

kg
kg
=6.2778
3
hr
m

## CHAPTER 9: AGITATION AND AERATION

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 m 3, to match the
volumetric mass transfer coefficient. Following optimum conditions were obtained with a 0.03
m3 fermenter:
Density of broth: 1200 kg/m3
Aeration rate: 1vvm
Liquid Height: 1.2 DT

## Liquid volume: 0.018 m3

Oxygen transfer rate: 0.25 kmol/m3hr

## Two sets of standard flat blade turbine impellers were installed.

Given:
L= 1200 kg/m3
Q= 1.0 vvm
V1= 0.03 m3
Solution:

HL= 1.2 DT
OTR= 0.25 kmol/m3hr
V2= 60 m3

VL= 0.018 m3

Determine the tank diameter, DT, and impeller diameter, DI, for each conditions.
V 1=

2
2
3
DT H L = ( DT ) ( 1.2 DT )=0.3 DT
4
4
1

1
3

( ) [

V1
DT =
0.3
1

0.018 m
=
0.3

3 1
3

=0.267 m

1
1
D I = DT = ( 0.267 m) =0.089 m
3
3
1

## (Biochemical Engineering by James Lee,

pg.274)
H L =1.2 DT =1.2 ( 0.267 m )=0.32 m
1

0.018
60 m
V 2 13
0.03
DT =
=
0.3
0.3
2

( )

1
3

=3.36 m

1
1
D I = DT = ( 3.36 m )=1.12 m
3
3
2

2

## Aeration flow rate:

1m3 O2
( 0.018 m3 ) 1min =3 104 m3 /sec
Q= 3
60 sec
m broth min

0.267 m2

4
m3 3600 sec
3 1 04
sec
1hr
Q
V s= =

1 atm+ 1 atm+
PO =

HL
10.3 m H 2 O

2 atm

1 atm+ 1 atm+

( 0.21 )=

0.32m H 2 O
10.3m H 2 O

2 atm

kmol
0.25 3
OTR
m hr
kmol
k L a=
=
=1.17 3
PO
0.213 atm
m hr atm
2

## 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:
kmol
V s 0.67 , 3
m hr atm
Pm 0.95
k L a=0.0635

( )

kmol
V s 0.67 , 3
m hr atm
0.95
Pm
k L a=0.0318

( )

kg
V s 0.5 , 3
m hr
0.4
Pm
k L a=0.026

( )

## For the gassed power,Pm

m 0.67
19.3 =0.1 hP
hr
Pm
kmol
1.17 3
=0.0318
m hr atm
0.018 m3

0.95

Then determine the ungassed power, Pmo, for two sets of impeller, Np= 2x6=12
P g
N N i 3 D i5
N p= mo3 c 5 ; Pmo = p
gc
N i Di

0.089 m

N i 3
kg
12 1200 3
m
N p N i3 Di5
Pmo=
=
gc

## Using Michaelis correlation:

2
3
( Pmo ) Di
Pm =0.5
0.56
Q1

0.1 hP=0.5
N i=18 rps

( 0.108 103 N i3 )

( 0.089 m )3

(3 104 m3 / sec)0.56

60 sec
=1080 rpm
1 min

1

[ ]

Pm
0.10
=
=0.1587
Pmo 1 0.630

## For Constant Power Input in a Baffled Vessel

Pmo
Pmo
=
; where P mo=constant =M /V
3
5
N 1 D I
N 23 D I 5
1

1
M
N 3 DI 5
V1 1
1

] [
=

model

1
M
N 3 DI 5
V2 2
2

prototype

## For constant power in geometrically similar vessels

N 13 D I 5 N 23 D I 5
=
V1
V2
1

*Accdg. To Ghose

N 13 D I 5
V
N2 =
; N 2=N 1 2
V1
V1
DI 5
V2
3

1
3

[ ][ ]

DI
DI

5
3

3
DI
N1
=
N2
DI

[ ][ ]
2

[ ]

DI
; N 2=N 1
DI

2/ 3

N 2=1080

1
3

[ ][
60
0.03

0.089
1.12

5
3

,rpm

N 2=199.8575 rpm

N 1 D I =N 2 DI
1

[ ] [ ]

DI
0.089
N 2=N 1
=720
DI
1.12
1

N 2=57.2143 rpm

2/ 3

[ ]

DI
N 2=N 1
DI

=720

0.089
1.12

2/ 3

N 2=3895.39 rpm

## *for constant impeller speed

2/ 3
DI
DI
N 1=N 2
; N p =N m
DI
DI

[ ]
2

2 /3

[ ]
m

N 2=3895.39 rpm
Or
(N DI ) p=1.7( N D I )m

## CHAPTER 9: AGITATION AND AERATION ; ADDITIONAL PROBLEM

A strain of azobacter vinelandii is cultured in a 15 m3 stirred fermenter for production of alginate.
Under conditions the mass transfer coefficient, K La, is 0.25 s-1. Oxygen solubility in a fermenter
3

## broth is approximately 8.5

max cell density in the broth?
REQD: Xmax
Soln:
X max=

K La C al
qo
2

10 kg
3
. The specific oxygen uptake is 15 mmol g/h. What is the
m

## ( 0.25s )(8.510 mkg )

3

1
1
32 g
1 kg
(
)(
)(
)
(15 mmol
)(
)
gh 3600 s 1000 mmol 1 mol 1000 g

Xmax 15937.5

g
g
=15.94
3
L
m

## CHAPTER 9: AGITATION AND AERATION ; ADDITIONAL PROBLEM

Biochemical Engineering and Biotechnology by Ghasem D. Najafpour
Calculate the gas hold-up for an agitated and aerated system with power input of 18 hp in an 80
m3 vessel with gas superficial velocity of 2.6 m.min-1
Given:
P = 18 hp

V = 80 m3
Vs = 2.6 m/min
Required:
Gas Hold-up, Ho
Solution:
From equation;
P
V
Vs
(

0.4
Where; P = power in hp
V = ungassed liquid volume in m3
Vs = gas superficial velocity in m/h

Solving for H:
18
80

2.6 x 60 min /h

H = 0.6 m
The gas hold up can be defined by the above definition using the gas height per volume,
where H = 0.6 m for aeration
Ho =

Vg
V g+V L

Ho = 8.5%

0.6
0.6+6.5

= 0.085

## CHAPTER 9: AGITATION AND AERATION ; ADDITIONAL PROBLEM

Biochemical Engineering and Biotechnology by Ghasem D. Najafpour
Calculate the speed of an impeller and the power requirements of a production-scale bioreactor
with 60m3 using two different methods. Also, match the volumetric mass transfer coefficient.
The following optimum conditions were obtained with a small scale fermenter of volume 0.3m 3
and 60% of the vessel working. The density of the broth, broth, 1200 kg.m-3, working volume
0.18m3, aeration rate of one volume of gas per volume of liquid (vvm), oxygen transfer rate 0.25
kmol.m-3.h-1, liquid height inside the vessel, H L, 1.2Dt. two sets of standard, flat-blade turbine
impellers were installed.
Given:
@ production-scale, V2 = 60m3
@ small-scale, V = 0.3m3
60% of vessel working
broth, 1200 kg.m-3
working volume, V1 = 0.18m3
Q = 1vvm
OTR = 0.25 kmol.m-3.h-1
HL = 1.2Dt
Two sets of standard flat-blade turbine impellers
Required: impeller speed, power requirement, kla
Solution:
V1 = (/4)(Dt)2(1.2Dt) = 0.3Dt3
Diameter of the vessel:
Dt = (V1/0.3)1/3 = (0.18/3)1/3 = 0.576m
Diameter of the impeller:
Di,1 = (1/3)(Dt) = 0.576/3 = 0.192m
Height of liquid media was assumed to be 1.2 times the diameter of the fermenter vessel.
HL,1 = 1.2 x 0.576 = 0.691m

## Diameter of the larger vessel:

Dt2 = (V1/0.3)1/3 = (60/0.3)1/3 = 3.36m
The impeller size for the larger vessel is:
Di2 = 3.36/3 = 1.12m
And the liquid media height in the second fermenter is:
HL2 = 1.2 x 3.36 = 4.03m
Assume the fermentation broth has the same viscosity as water:
1 = 1cp
Aeration rate for 1vvm is:
Us = (0.18 x 60)/(

## Let us take average values for the partial pressure of oxygen

PO2 = ({1 atm + [1 + (HL m/10.3m) x atm]}/2) x 0.21 = 0.213 atm
The oxygen transfer rate is
OTR = 0.25 kmol/m3.h
The mass transfer coefficient is
kL = 0.25/0.213 = 1.174 kmol/m3.h.atm
Use imperial correlation based on the following equation for mass transfer in the bioreactors. The
general equation for the evaluation of Kla is
Kla = x(Pg/V)Y (Us)Z
where x,y, and z are empirical constants. For Newtonian fluids, mnon-coalescing broth and gas
bubbles, the following correlation is valid for a working volumeof less than 4 m3 and a power per
unit volume of 500 10,000 W/m3

kL = 0.002(Pg/V)0.7(Us)0.5
1.174 = 0.002 (Pg/V)0.7(41.45)0.5
For the gassed power per unit volume (Pg/V) is 630.7 W; thegpassed power, Pg, was 0.15hp.
Since the flow regime is turbulent, the power number obtained from figure 6.6, Power number
versus Reynolds number, Re, reads Pno = 6. For two sets of impellers, Np = 2(6) = 12
`

Np = (P1 gc)/(N13Di5)
P1 = (12 x 1200 x N13 x 0.1925)/9.81 = 0.383 N13 W = 5x10-4 N13 hp

## Using Michel and Millers correction factor for power calculation:

Pg1 = 0.5(P121Di13/10.56)0.45
Knowing the power input, we can calculate the rotational speed:
0.15 = 0.5(5x10-4 x N13 x 0.1923 / (3x10-3)0.56)0.45
N1 = 10 rps
N1 = 600 rpm
The power input for ungassed system is
P1 = 5x10-4 N13 = 5x10-4 (10)3 = 0.5hp
Pg/P = 0.15/0.5 = 0.3
For constant power input based on geometric similarity of the vessels, agitation rate is calculated.
(N13Di15)/(V1) = (N23Di25)/(V2)
(N2/N1)3 = (V2/V1)(Di1/Di2)5
N2 = N1 (V2/V1)1/3(Di1/Di2)5/3 = 600(60/0.3)1/3(0.192/1.12)5/3 = 185 rpm
For constant input velocity for a large system:
N2 = N1(Di1/Di2) = 600(0.192/1.12) = 103 rpm
The content of the provided text is a series of problems and solutions related to biochemical engineering, specifically focusing on agitation and aeration in bioreactors. The problems involve deriving relationships, proving equations, and calculating various parameters such as mass transfer coefficients, power requirements, and gas hold-up.

### Analysis:
multiple_questions_with_answers

1. **multiple questions**: the text contains multiple distinct questions, each with its own set of given data and required calculations.
2. **answers provided**: for each question, detailed solutions are provided, including step-by-step calculations and explanations.

### summary of questions and answers:
1. **derive the relationship between overall mass transfer coefficient for gas phase \( k_g \) and individual mass-transfer coefficients \( k_l \) and \( k_g \)**:
   - **solution**: the relationship is derived and simplified for sparingly soluble gases.

2. **prove that eq. (9.25) is the same as eq. (9.26), and eq. (9.27) is the same as eq. (9.28)**:
   - **solution**: detailed algebraic manipulations are provided to show the equivalence of the equations.

3. **calculate power requirement,
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
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# CHAPTER 9: AGITATION AND AERATION

9.1 Derive the relationship between the overall mass transfer coefficient for gas phase K G and the
individual mass-transfer coefficients, kL and kG. How can this relationship be simplified for
sparingly soluble gases?
Given:
a. Overall mass transfer coefficient for gas phase, KG
b. Individual mass transfer coefficients, kL and kG
Required:
a. Derive the relationship between KG, and kG and kL
b. How can the relationship be simplified for sparingly soluble gases?
Solution:
Expressing the molar flux in terms of overall driving force for mass transfer and an
overall mass transfer coefficient,
P A
P A , G
N A =K G
c Ac A , L

N A =K L
where KG is termed the overall mass transfer coefficient based on gas phase driving
forces. In an equivalent manner, we also can write
Using Henrys Law,
mc A
P A , G=
where m is the Henrys law constant.
We can rewrite the first two equations as,
N
P A , GP A ,i= A
kG

c A , ic A , L =

NA
kL

## Substituting the equilibrium relationships,

mNA
P A=
kL
P A , i
Adding the partial pressure differences we obtain
N mN A
1 m
P A= A +
=N A ( + )
kG
kL
kG kL
P A=P A ,G
P A , GP A ,i + P A , i
Simplifying further,
1
1 m
= +
K G kG k L
For sparingly soluble gases,
1
=0
kG
Therefore,
1
m
=
K G kL

## CHAPTER 9: AGITATION AND AERATION

9.2 Prove that Eq. (9.25) is the same with Eq. (9.26), and Eq. (9.27) is the same with Eq. (9.28)
Given:
a) Eq. (9.25) and Eq. (9.26)
b) Eq. (9.27) and Eq. (9.28)
Required:
a) Prove that
2

c g

Eq. (9.25)

k L =0.31 N Sc 3 (

Eq. (9.26)

N Sh=0.31 N Sc 1/ 3 N Gr 1/3

1 /3

is the same as

b) Prove that
2
1 /3
2 D AB
3 c g
Eq. (9.27) k L = D + 0.31 N Sc ( 2 )
32

Eq. (9.28)

is the same as

## N Sh=2.0+ 0.31 N Sc1 /3 N Gr1/ 3

Solution:
a.)
N Sh=0.31 N Sc

1/ 3

N Gr

1/3

2
3

k L =0.31 N Sc (

c g

1 /3

Substituting Eq. (9.21), Eq. (9.22) and Eq. (9.23) into Eq. (9.26):
k L D32
c 1/ 3 D332 c g 1 /3
=0.31(
) (
)
D AB
c D AB
2c
c 1 /3
c 1 /3 g 1/3 1 /3
k L =0.31 1/ 3
c D AB2 /3
2c /3
k L =0.31

1/ 3

1/3
c

D AB

1 /3

1/3

1 /3

D32
D AB
Combining similar
terms

2 /3

1/3

1 /3

## Dividing both sides by

k L =0.31 1/ 3 c 2 /3 c
c D AB

1 /3

1/3

Multiplying

1/ 3

c 2 /3
c 1/ 3 g1 /3 1 /3
k L =0.31 2/ 3
c D AB2 /3
2/ 3c

1/3

)( c2/ 3 )
2/ 3
c
c

c
c D AB

k L =0.31
2

k L =0.31 N Sc 3 (

c g 1 /3
)
2

k L =0.31 N Sc 3 (

c g 1 /3
)
2

b.)
N Sh=

2 D AB
+0.31 N Sc 1/ 3 N Gr1 /3
D 32
2

k L =2.0+0.31 N Sc 3 (

c g

1/ 3

Substituting Eq. (9.21), Eq. (9.22) and Eq. (9.23) into Eq. (9.28):
k L D32
c 1/ 3 D 332 c g 1/3
=2.0+0.31(
) (
)
D AB
c D AB
2c

## Dividing both sides by

D32
D AB

1/ 3

1/ 3

1 /3
c

1/ 3

k L =2 AB + 0.31 1/ 3 c 2 /3 c
D32
c D AB
k L =2

D AB

c
+ 0.31 1/ 3
2 /3
D32
c D AB

1 /3

1 /3

1 /3

1 /3

g
2 /3
c
g

D AB
c1 /3
c 1/ 3 g1 /3 1 /3
k L =2
+ 0.31 1/ 3
D32
1
c D AB2 /3
2/ 3

k L =2

D AB

c
+ 0.31 2/ 3c
2 /3
D32
c D AB

Combining similar
terms

1/ 3

c
c D AB

D AB
k L =2
+ 0.31
D32
2
1 /3
D AB
3 c g
k L =2
+ 0.31 N Sc (
)
2
D 32

2
1 /3
D AB
3 c g
k L =2
+ 0.31 N Sc (
)
D32
2

1 /3

g
2/ 3
c

1 /3

Multiplying

3
1/
1/3
c
c
)(
)
2/ 3
c
2/c 3

## CHAPTER 9: AGITATION AND AERATION

9.4 A cylindrical tank (1.22m diameter) is filled with water to an operating level equal to the tank
diameter. The tank is equipped with four equally spaced baffles, the width of which is one tenth
of the tank diameter. The tank is agitated with a 0.36 m diameter, flat-blade disk turbine. The
impeller rotational speed is 4.43 rps. The air enters through an open ended tube situated below
the impeller and its volmetric flow rate is 0.0217 m3/s at 1.08 atm and 25 deg C.
density= 997.08 kg/m3
Calculate:
a.
b.
c.
d.
e.

Power Requirement
Gas Hold-up
Sauter-mean Diameter
Interfacial area
Volumetric mass-transfer coefficient

Given:
DT = 1.22 m
W = 1/10 DT
DI = 0.36 m
N = 4.43 rps
Q = 0.0217 m3/s

Required:
a.
b.
c.
d.
e.

Pm
H
D32
a
kla

## viscosity = 8.904 x 10^-4 kg/m-s

Solution:
a. Nre = (997.08)(4.43)(.36)^2
8.904 x 10^-4
Nre = 642,915.034 > 10,000

d. Interfacial Area:
a =6H/ D32
a= 98.5447 /m

= 3144.8862 W

## Using equation 9.53 of James Lee

Kl = 4.58x10^-4 m/s

Pm= 1341.3218 W

## Kla = 4.58x10^-4 (98.5447)

b. v= (/4)(1.22)(1.22)^2
v= 1.43 m3
Vs= (4x .0217) / (1.22)^2
= 0.0186 m/s
<
0.02 m/s
Using equation 9.48 of James Lee
H = 0.0790
c. Using equation 9.42
D32= 4.8132 x 10^-3 m
D32= 4.8132 mm

## CHAPTER 9: AGITATION AND AERATION

Kla = 0.0451 /s

9.5 Estimate the volumetric mass-transfer coefficient kLa for the gas-liquid contractor described
in Problem 9.4 by using a correlation for kLa and compare the result with the experimental value.
Given:
Reactor volume, v= 1.43m3
Vs= 0.0186 m/s
PM= 1342 Watts
Required:
kLa (using equation 9.71 by James Lee)
% kLa compared with experimental value
Solution:
1342
kLa = 0.026( 1.43 )0.4 (0.0186)0.5 = 0.0548 s-1

Experimental

## Estimated value= 0.0451 s-1

% for volumetric mass-transfer coefficient =

( 0.05480.0451 ) x 100
= 17.7007% error
(0.0548)

## CHAPTER 9: AGITATION AND AERATION

9.6 The power consumption by an agitator in an unbaffled vessel can be expressed as
P mo
ND2I
=f

N 3 D5I

( )

Can you determine the power consumption and impeller speed of a 1,000-gallon fermenter
based on findings of the optimum condition from a one-gallon vessel by using the same fluid
system? Is your conclusion reasonable? Why or why not?
Given:
P mo
ND2I
=f

N 3 D5I

( )

VP =1000 gallons
Vm=1 gallon
Required:
Can the power consumption and impeller speed of VP be determined on findings of the
optimum condition from Vm by using the same fluid system? Why?
Solution:
VP
=1000
Vm
The scale ratio is
1
DI , P
=1000 3 =10
DI,m
To achieve dynamic similarity, the three numbers for the prototype and model must be
equal
Pmo
P mo
=
3
5
N DI P N 3 D5I m

] [

[ ] [ ]
ND 2I
ND 2I
=
P

## Using the same fluid for model and prototype, P =m ;

5

( Pmo )p =10 [ P mo ] m

P =m

[ ]
NP
Nm

## The equality of Reynolds number requires

N P =0.01 N m
while the equality of Froude number requires
1
N P=
Nm
10
which shows two conflicting concepts.
If

P m ;

[]

P m ;

[]

1
=
m 31.6

## Therefore, if kinematic viscosity of prototype is similar to water, the kinematic viscosity

of the fluid which needs to be employed for the model should be 1/31.6 of the kinematic
viscosity of water. It is impossible to find the fluid whose kinematic viscosity is that
small. As a conclusion, if all three dimensionless groups are important, it is impossible to
satisfy the dynamic similarity.

## CHAPTER 9: AGITATION AND AERATION

The specific oxygen demands and critical oxygen concentrations for typical microbial plants and
animal cell cultures are listed below
Cell culture

qo

Concentration, mmol/L

Escherichia Coli

0.5 mmol(gdw)-1h-1

0.0082

Vitisvinifera (grape)

0.6 mmol(gdw)-1h-1

0.055

## Chinese Hamster Ovary

3.0x10-10 mmol(gdw)-1h-1

0.020

a.Estimate the kLa requirement to achieve cell concentrations of 25 gram dry weight/L for E.
Coli and V. Vinifera and 3.0x109 for CHO cells, while maintaining dissolved oxygen
concentration above critical. The oxygen solubility in the media used for the cultures is 7.2x10 -3
kg/m3
b. The relationship between kLa and the power input to a 1-m3 stirred bioreactor is kLa
Pt

Vl

0.5

, compare the bioreactor power requirements for culture of the three different cell types

Solution:
a.)

## for e coli: q= 8.5mmol/h (25/L)=292.65 mmol/L.h

v.vinifera: q= 0.6 mmol/h (25/L)=15 mmol/L.h
CHO: q= 3x10-10mmol/h (3x109/L)=0.9 mmol/L.h

## Oxygen Solubilty: 7.3x10-3 kg/m3

292.5 mmol/L.h= [7.2 x10-3 kg/m3 (1kmol/32kg)(1x106 mmol/kmol)-0.0002)kla
kLa= 980.1661/hr

## 15 mmol/L.h= [7.2 x10-3 kg/m3 (1kmol/32kg)(1x106 mmol/kmol)-0.055)kla

kLa= 88.2353/hr
0.9 mmol/L.h= [7.2 x10-3 kg/m3 (1kmol/32kg)(1x106 mmol/kmol)-0.02)kla
kLa= 4.3902/hr

Pt
b.) kLa= k ( Vl )

0.5

## PT E.C.= 960725.5836 m3/hr2

K2
PT v.v. = 7765.4682 m3/hr2
K2
PT CHO= 19.2739 m3/hr2
K2
PT E.C.> PT v.v.> PT CHO

## CHAPTER 9: AGITATION AND AERATION

ESTIMATING KLa USING THE SIMPLE DYNAMIC METHOD
A stirred fermenter is used to haematopoietic cells isolated from umbilical cord blood. The liquid
volume is 15 Liters. The simple dynamic method is used to determine K La. The air flow is shut
off for a few minutes and the dissolved oxygen level drops; the air supply is then reconnected at
a flow rate of 0.25 L/s. The following results are obtained at a stirrer speed of 50 rpm.
Time, s
5
20
Oxygen tension (%air saturation)
50
66
When steady state is established, the dissolved oxygen tension is 78% air saturation. In
separate test experiments, the electrode response to a step change in oxygen tension did not vary
with stirrer speed above 40 rpm. The probe response time under these conditions was 2.8
seconds. When the KLa measurement was repeated using nitrogen to deoxygenate the mixture,
the results for oxygen tension as a function of time were similar to those listed. Estimate KLa.
Solution:

KLa =

C C AL1
ln ( AL
)
C ALC AL2
t 2t 1
7850
)
7866
( 205 ) s

ln (
KLa =

= 0.056/s

a)

## CHAPTER 9: AGITATION AND AERATION

A 20-L stirred fermenter containing Bacillus thuringiensis is used to produce a microbial
insecticide. The oxygen balance method is applied to determine KLa. The fermenter operating
pressure is 150 KPa and the culture temperature is 30oC. The oxygen tension in the broth is
measured as 82% using a probe calibrated to 100% in situ using H 2O and air at 30oC and
150KPa. The solubility of oxygen in the culture fluid is the same as in H 20. Air is sparged into
the vessel; the inlet gas flow rate measured outside the fermenter at 1 atm pressure and 22 oC is
0.23/s. The exit gas from the fermenter contains 20.1% oxygen and has a flowrate of 8.91 min-1.
a. Calculate the volumetric rate of oxygen uptake by the culture
b. What is the value of KLa.
Solution:
NA =

1
RV L

([ F TP ) i( F TP ) o]
g

Ag

Ag

0.201 x 1.48

1
8.91 x
( ( 30+273 ) K )
60

0.23 L
(0.2099 atm)
s

( 22+273 ) K
1

0.08205 L. atm
(20 L)
mol . K

((

1
( 1.6357 x 104 1.4572 x 104 ) =1.0878 x 105 mol
L.s
0.08205 (20)

Because of steady state the rate of oxygen transfer is equal to the rate of oxygen uptake
by the cells; the volumetric rate of oxygen uptake by the culture is 1.0878x10-5 mol/ L.s
b) Assume that the gas phase is well-mixed so that the oxygen concentration in the bubbles
containing the liquid is the same as the outlet gas, that is 20.1%. As the difference in the
composition of the gas phase to be constant throughout the fermenter.
The solubility of oxygen in H 20 @30oC and 1atm air pressure is 8.05x10 -3 kg/m3.
Determine the solubility at the fermenter operating pressure of 1.48 atm and gas phase
oxygen mole fraction of 0.201.

C*AL,2

PT 2 Y AG 1
PT 1 Y AG 2

C*AL,1

## 1.48 atm ( 0.201 ) 8.05 x 103 g

(
)
=
= 0.0114g/L
L
1 atm ( 0.2099 )

CAL in the fermenter is 82% of the oxygen solubility at 30oC and 1.48atm air pressure
CAL= 0.820

## 1.48 atm 8.05 x 10 g

(
) = 9.77x10-3 g/L
1 atm
L

32 g
)
mol
3
0.0114 g 9.77 x 10 g

L
L
5

1.1 x 1 0 mol / L . s (

KLa =

## CHAPTER 9: AGITATION AND AERATION

DYNAMIC TECHNIQUE

## KLa = 0.22 s-1

A strain of Azobacter vinelandii is cultured in a 15m 3 stirred fermenter for alginate production.
Under current operating conditions, KLa is 0.17 s-1. The solubility of oxygen in the broth is
approximately 8x10-3 kg/m3.
a. The specific rate of oxygen uptake is 12.5 mmol/g-hr. What is the maximum cell
concentration supported by oxygen transfer in the fermenter?
b. The bacteria suffer growth inhibition after copper sulphate is accidentally added to
the fermentation broth just after the start of the culture. This causes a reaction in the
oxygen uptake rate to 3 mmol/g-hr. What maximum cell concentration can now be
supported by oxygen transfer in the fermenter?
Solution:

a) Xmax =

0.17 8 x 1 0 kg
(
)
3
s
m
mmol 1 hr
1 mol
32 g
12.5
g . hr 3600 s 1000 mol 1 mol

( )
( )(

)(

1 kg
)( 1000
g)

12240 g
3
m

b) Assume that addition of copper sulphate does not affect C*AL of KLa
0.17 8 x 1 03 kg
(
)
s
m3
Xmax=
mmol 1 hr
1 mol
32 g
3
g . hr 3600 s 1000 mol 1mol

( )
( )(

)(

1 kg
)( 1000
g)

51000 g
3
m

## CHAPTER 9: AGITATION AND AERATION

GAS HANDLING WITH RUSHTON TURBINE
A fermenter of diameter and liquid height 1.4m is fitted with a Rushton impeller of diameter
0.5m and off-bottom clearance 0.35m operated at 75 rpm. The fermentation broth is sparged with
air at a volumetric flow rate of 0.28m3/min. Half-way through the culture some bearings in the
stirrer drive begin to fail and stirrer speed must be reduced to a maximum of 45 rpm for the
remainder of the process.
a. Under the normal operating conditions, is the gas completely dispersed?
b. After the stirrer speed is reduced, is the impeller flooded or loaded?
Solution:
1 min
a) Ni = 75/min 60 s

Fr =

2 Di
g

= 1.25s-1

1.25 s1 2 (0.5 m)

= 0.0796

Flg = 0.2

0.5

DI
DT

( )

Fr 0.5 = 0.2

0.5 m
1.4 m

0.5

## Fg = FlgNiDi3 = (0.0337)(1.25s-1) (0.5m)3 = 5.27x10 -3 m3/s > 0.28 m3/min

*Fg, volumetric flowrate of gas greater than the operating flow rate, we can conclude that
the air provided is completely dispersed under normal conditions.
b)

1 min
Ni = 45 /min ( 60 s ) = 0.75s-1
1 2

0.75 s (0.5 m)

Fr =

Di
g

= 0.0287

Flg = 30

DI
DT

3.5

( )

Fr

= 30

0.5 m
1.4 m

3.5

= 0.0234

## Fg = FlgNiDi3 = (0.0234)(0.75s-1) (0.5m)3 = 0.00219 m3/s

*At reduced stirrer speed, maximum air flow rate can be handled without impeller
flooding as operating flow rate (0.28m3/min) is greater than this. The impeller is
FLOODED.

## CHAPTER 9: AGITATION AND AERATION

Clostridium acetobutylicum carries out anaerobic fermentation and converts glucose into
acetone, butanol along with smaller concentrations of butyrate, acetate, etc. In fermentation the
following products were obtained from 100 moles of glucose and 11.2 moles of NH 3, as nitrogen
source. Products formed:
Cells = 13moles
Butanol= 56moles
Acetone= 22moles
Butyric acid = 0.4 moles

## acetic acid= 14 moles

CO2= 221 moles
H2= 135moles
Ethanol= 0.7 moles

By performing a carbon, nitrogen, hydrogen, and oxygen balance, determine the chemical
composition of the cells.
Solution:
By performing a carbon, hydrogen, nitrogen and oxygen balance, determine the element
composition of the cells.
100C6H12O6 + 11.2 NH3 13CaHbOcNd + 56C4H10O (butanol) + 22C3H6O(acetone) +
0.4C4H8O2(butyrate) + 14C2H14O2 (acetic acid) + 221CO2 +135H2 + 0.7 C2H6O(ethanol)
* where CaHbOcNd represents elemental composition of clostridium cells
Carbon Balance:
100(6) + 11.2(0) = 13(a) + 56(4) + 22(3) + 0.4(4) + 14(2) + 221(1) + 0.7(2) : a = 4.46
Hydrogen Balance:
100(12) + 11.2(3) = 13(b) + 56(10) + 22(6) + 0.4(8) + 14(14) + 135(2) + 0.7(6) : b = 16.02
Oxygen Balance:
100(6) + 11.2(0) = 13(c) + 56(1) + 22(1) + 0.4(2) + 14(2) + 221(2) + 0.7(1) : c = 3.88
Nitrogen Balance:
11.2(1) = 13(d) : d = 0.86

## Aerobic Culture of Saccharomyces cerevisae in a synthetic medium produced the following:

= 0.2/hr
RQ =1.0
Yy/s = 90 grams dry cell/gmol glucose
Vglutamate = 0.043 grams glutamate/ g cell-hr
Assuming that carbon content of the cell is 45%, check carbon and oxygen balance with
respect to given culture.
aC6H12O6 + bC5H9O4N + cO2 dCHOO + e CO2 + fH2O
glutamic acid
Solution:
RQ = 1.0 = Q02 = QCO2 = c = e
@ = 0.2 hr -1 and 45% C in biomass ; 1mol cell biomass = 12gC

d=

0.2
(0.45)
hr
12 g/mol

45% =

g cell biomass
g Carbon
(100)
MW cell biomass
C

MWcell =

12(100)
=26.67
45

## MW = 26.67 g/mol = CHON

26.67= 1(12) + (1) + (16) + (14)
14.67= + 16 + 14 (Equation 1)

a= Qglucose =

Y X /S

0.2/hr
g dry cell
90
g glucose

0.043
b= Qglutamate =

147

g glutamate
g cell . hr

g
glutamic acid
mol

## = 2.925x10-4 gmol glutamate/ g dry cell. Hr

Elemental Balance:
C: 6a + 5b = d + e
H: 12a + 9b = d + 2f
O: 6a + 4b + 2c = d + 2e + f : 2c=2e : 6a + 4b = d + f
N: b = d
=

b
d

2.925 x 104
7.5 x 1 03

Using H and O:
12a + 9b = d + 2f
-2
(6a + 4b = d + f)
b= d - 2d

= 0.0390

## 2.925x10-4 = (7.5x10-3) - 2 (7.5x10-3)

2.925 x 1 04 +0.015
=
7.5 x 1 03
from equation 1:
14.67 = + 16 + 14 : 14.67 = +16 + 14(0.0390)
14.124 = + 16
Substituting :
4

14.124 =

2.925 x 10 +0.015
3
7.5 x 10

+ 16 : = 0.7825, = 1.604

Using equation of O:
6a + 4b = d + f ; f = 6a + 4b d
f= 6(2.22x10-3) + 4(2.925x10-4) 0.7825(7.5x10-3)
f = 8.6213x10-3
Using equation of C:
6a + 5b = d + e ; e = 6a + 5b d
e = 6( 2.22x10-3) + 5( 2.925x10-4) 7.5x10-3
e = c = 7.2825x10-3
* aC6H12O6 + bC5H9O4N + cO2 dCHOO + e CO2 + fH2O

2.22 C6H12O6 + 2.925C5H9O4N + 7.2825 O27.5 CH1.604O0.7825O0.0390 + 7.2828 CO + 8.6213H2O

## CHAPTER 9: AGITATION AND AERATION

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
H/DT=2.5
working volume=60%
two sets of a standard flat blade turbine; Pmo=2 Pmo
Q=0.5 vvm

D=0.3 hr-1
m=0.4 hr-1
Ks=2 kg/m3
CSO=50 kg/m3
0.5 g cell (dry )
YX/S=
g glucose

N=60 rpm
= 1200 kg/m3
=0.02 Pas
Cell formula: CH1.8O0.5N0.2
Required: Mass transfer limited or Biochemical reaction limited

Soln:
H
=2.5
DT

H = liquid height
DT = tank diameter

2
2
3
V T = D T H = DT ( 2.5 D T )= ( 2.5 ) ( DT )
4
4
4

VT = 50 m3

50 m = (2.5 ) (D T 3 ) ;
4
3

DT = 3

50 m3

( 2.5 )
4

DT = 2.9420 m

0.02 Pa s
= =
=1.67 105 m3 / s
3
1200 kg /m

## From Biochemical Engineering by James Lee:

DI =

DT
=3 ; DI = impeller diameter
DI

DT 2.9420 m
=
=0.9807 m
3
3

0.70
3 P m
k L a=2.0 10
V s0.2
VL

[ ]

## kLa =volumetric mass- transfer coefficient

Pm = gassed power
VL = volume of the liquid or broth
Vs = superficial velocity
For the power number, Np,

## (eq. 9.72, p. 267, James Lee)

N D I2
'
Reynol d s Number , Re =
=

60 rpm

## ( 601minsec )(0.9807 m) (1200 kgm )

2

0.02 Pa s

57706.3494
Re 10000; Np = 6 (p. 258, James Lee)
N = (60/60) rps
For a flat-blade turbine at Re 10000
3
kg 60
Pmo=N P N 3 DI 5=6 1200 3 ( rps) (0.9807 m)5
m 60

## Pmo=6531.5066 W (ungassed power )

For the gassed power
4.38

0.115

DI
DT

( )

[ ] [ ] [ ] [ ]

P
D
log 10 m =192 I
Pmo
DT

D I2 N

DI N 2
g

1.96

Q
N D I3

## (eq. 9.53, p. 258,

James Lee)
3

Q=

0.5 m O 2
3

m broth min

( 50 m3 0.60 ) =0.25 m 3 / s
60 2

60

1.96

(0.9807)

( 0.9807
2.9420 )

0.25 m3 /s
60
( )(0.9807)3
60

Pm
0.9807 m
=192
2.9420 m
2 ( 6531.5066 W )
log 10

4.38

60
rps )
60
1.67 105

(0.9807)2 (

Pm = 6191. 1993 W
For computation of kLa, superficial velocity
1 min
0.5 m3 air
60 sec

V s=

Q
=
S

m3 media min

(2.9420 m)2
4

( 50 m3 0.60 )=0.0368 m/ s

0.115

k L a=2.0 10

6191.1993 W
3
50 m (0.60)

0.70

m 0.20
(0.0368 ) =0.0431 s1
s

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 atm
atm L
HO =
=0.9804
1.02 mmol/L
mmol
2

C L =

0.21 atm
mmol
=0.2142
atm L
L
0.9804
mmol

## For kLa,CL*, the volumetric mass-transfer coefficient at the equilibrium:

0.0431 3600 s
mmol
1 mol
1 kmol
32 kg
k L a ,C =
0.2142
s
1 hr
L
1000 mmol 1000 mol 1 kmol

](

)[

][

][

][

1000 L
kg
=1.0635 3
3
1m
m hr

Cs
D==m
K s +C s

## Determine the Exit substrate concentration, Cs

Cs
0.3 0.4
=
; C s=6.0 kg /m 3
hr hr
kg
2 3 +Cs
m

[ ]

## Then, determine the cell concentration that was produced

g dry cell
kg
kg
C x C xo =x= x ( C soC s )=0.5
( 506 ) , 3 =22 3
g glucose consumed
m
m
s
For ZO2, grams of oxygen required per gram dry cell produced (Maleles, 1971)
32 C+ 8 H 16 C
'
'
'
'
ZO =
+0.01 O 0.0276 C +0.01714 N 0.08 H
x MW
2

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)

C= 6
H=12
O=6
MW= 6(12) + 12(1) + 6(16)

ZO =
2

32 ( 6 ) +8 (12)16(16)
+ 0.01(32.52)0.0276(48.78)+0.01714 (11.38)0.08(7.32)
0.5(180)

Z O =0.7656
2

gO2
g dry cell

Z O x m=0.7658
2

## *From CH1.8O0.5N0.2 (cell formula)

C= 1(12)
C= (1(12)/ 24.6) x100 = 48.78
H= 1.8(1)
H= (1.8(1)/ 24.6) x100 = 7.32
O= 0.5(16)
O= (0.5(16)/ 24.6) x100 = 32.52
N= 0.2(14)
N= (0.2(14)/ 24.6) x100 = 11.38
= 24.6

g O2
kg
22 3
g dry cell
m

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
broth.

## CHAPTER 9: AGITATION AND AERATION

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
T= 12C
HL= 4.27 m

## Air bubble diameter: 2.5 x 10-3 m

Air bubble velocity: 0.3 m/s
DT= 99.5 m

Time, min
3
6
9
12
15
18
21

Cs, kg/m3
0.6 x10-3
1.6 x10-3
3.1 x10-3
4.3 x10-3
5.4 x10-3
6.0 x10-3
7.0 x10-3

## From the information, compute the:

a. KLa and kL
b. Mass of oxygen per hour transferred per 28.317 m 3 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.

C sm=C s

Pb O t
+
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
moles of unabsorbed O2
21(10.1)
Ot=
100=
100=19.3
total moles of leaving air
21 ( 10.1 ) +79

For Csm,
C sm=1.08 x 102

kg
+
=0.01238 3
3
29.4
42
m
m

Time, min
3
6
9
12
15
18
21

## 0.01238 - 0.6 x10-3

0.01238 - 1.6 x10-3
0.01238 - 3.1 x10-3
0.01238 - 4.3 x10-3
0.01238 - 5.4 x10-3
0.01238 - 6.0 x10-3
0.01238 - 7.0 x10-3

0.01178
0.01078
9.28 x10-3
8.08 x10-3
6.98 x10-3
6.38 x10-3
5.38 x10-3

## Plot Csm Cs vs. Time

m=slope=3.6071 104

kg
60 min
kg
=0.0216 3
m min 1 hr
m hr
3

## Determine the kLa

kg
slope
m3 hr
k L a=
=
=2.7 hr 1
mean concentration of O2 (
kg
08 103) 3
m
0.0216

Determine the interfacial area for the calculation of the mass transfer coefficient, kL
A 6Q HL

=
; V T = DT 2 H L
V dBV B V T
4
Where :
A/V is the interfacial area
dB is the air bubble diameter
VB - is the air bubble velocity

## V T = ( 99.5m )2 ( 4.27 m )=33,201.97 m3

4
A
=
V

6 ( 9.439 m3 ) (4.27 m)
=9.7114 m1
( 2.5 103 m ) 0.3 m (33,201.97 m3)
s

For kL,
k L=

kLa
2.7 /hr
m
=
=0.2677
A /V 9.7114 /m
hr

B.

k L a(20 )=

( qv )
( qv )

k L a(12 )
1220

1.02

2.6/ hr
3.046
=
1220
hr
1.02

## =k L a(20 ) (C s ,20 C L,20 )

(eq.1)

(20 )

=k L a(32 ) (C s ,32 C L, 32 )

(eq.2)

(32 )

## k L a(32) =k L a(20 ) (1.02)T 20

Determine the amount of oxygen transferred per hour at 20C.
q
3.046
kg

2
=k a
( C C L,20 ) = hr ( 1.07 10 0 ) 3
v (20 ) L (20 ) s
m

()

q(20 )=

3.046
( 1.07 1020 ) kg3 ( 28.317 m 3 )=0.923 kg
hr
hr
m

## C. From eq.1 and 2, in terms of equal working volume,

k L a(32) ( Cs , 32 C L, 32 )
q(32 )=
q
k L a(20 ) (C s , 20 C L, 20 ) (20 )
q ( 32 ) @80 eff =0.80

## k L a(20 ) (1.02)3220 (C s ,32 C L ,32 )

q(20 )
k L a(20 ) (C s ,20 C L ,20 )
1.02 3220

## (0.923 kgh ) (8.673 10

1.5 102 )

q( 32 )@ 80 eff =0.80

kg
kg
=6.2778
3
hr
m

## CHAPTER 9: AGITATION AND AERATION

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 m 3, to match the
volumetric mass transfer coefficient. Following optimum conditions were obtained with a 0.03
m3 fermenter:
Density of broth: 1200 kg/m3
Aeration rate: 1vvm
Liquid Height: 1.2 DT

## Liquid volume: 0.018 m3

Oxygen transfer rate: 0.25 kmol/m3hr

## Two sets of standard flat blade turbine impellers were installed.

Given:
L= 1200 kg/m3
Q= 1.0 vvm
V1= 0.03 m3
Solution:

HL= 1.2 DT
OTR= 0.25 kmol/m3hr
V2= 60 m3

VL= 0.018 m3

Determine the tank diameter, DT, and impeller diameter, DI, for each conditions.
V 1=

2
2
3
DT H L = ( DT ) ( 1.2 DT )=0.3 DT
4
4
1

1
3

( ) [

V1
DT =
0.3
1

0.018 m
=
0.3

3 1
3

=0.267 m

1
1
D I = DT = ( 0.267 m) =0.089 m
3
3
1

## (Biochemical Engineering by James Lee,

pg.274)
H L =1.2 DT =1.2 ( 0.267 m )=0.32 m
1

0.018
60 m
V 2 13
0.03
DT =
=
0.3
0.3
2

( )

1
3

=3.36 m

1
1
D I = DT = ( 3.36 m )=1.12 m
3
3
2

2

## Aeration flow rate:

1m3 O2
( 0.018 m3 ) 1min =3 104 m3 /sec
Q= 3
60 sec
m broth min

0.267 m2

4
m3 3600 sec
3 1 04
sec
1hr
Q
V s= =

1 atm+ 1 atm+
PO =

HL
10.3 m H 2 O

2 atm

1 atm+ 1 atm+

( 0.21 )=

0.32m H 2 O
10.3m H 2 O

2 atm

kmol
0.25 3
OTR
m hr
kmol
k L a=
=
=1.17 3
PO
0.213 atm
m hr atm
2

## 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:
kmol
V s 0.67 , 3
m hr atm
Pm 0.95
k L a=0.0635

( )

kmol
V s 0.67 , 3
m hr atm
0.95
Pm
k L a=0.0318

( )

kg
V s 0.5 , 3
m hr
0.4
Pm
k L a=0.026

( )

## For the gassed power,Pm

m 0.67
19.3 =0.1 hP
hr
Pm
kmol
1.17 3
=0.0318
m hr atm
0.018 m3

0.95

Then determine the ungassed power, Pmo, for two sets of impeller, Np= 2x6=12
P g
N N i 3 D i5
N p= mo3 c 5 ; Pmo = p
gc
N i Di

0.089 m

N i 3
kg
12 1200 3
m
N p N i3 Di5
Pmo=
=
gc

## Using Michaelis correlation:

2
3
( Pmo ) Di
Pm =0.5
0.56
Q1

0.1 hP=0.5
N i=18 rps

( 0.108 103 N i3 )

( 0.089 m )3

(3 104 m3 / sec)0.56

60 sec
=1080 rpm
1 min

1

[ ]

Pm
0.10
=
=0.1587
Pmo 1 0.630

## For Constant Power Input in a Baffled Vessel

Pmo
Pmo
=
; where P mo=constant =M /V
3
5
N 1 D I
N 23 D I 5
1

1
M
N 3 DI 5
V1 1
1

] [
=

model

1
M
N 3 DI 5
V2 2
2

prototype

## For constant power in geometrically similar vessels

N 13 D I 5 N 23 D I 5
=
V1
V2
1

*Accdg. To Ghose

N 13 D I 5
V
N2 =
; N 2=N 1 2
V1
V1
DI 5
V2
3

1
3

[ ][ ]

DI
DI

5
3

3
DI
N1
=
N2
DI

[ ][ ]
2

[ ]

DI
; N 2=N 1
DI

2/ 3

N 2=1080

1
3

[ ][
60
0.03

0.089
1.12

5
3

,rpm

N 2=199.8575 rpm

N 1 D I =N 2 DI
1

[ ] [ ]

DI
0.089
N 2=N 1
=720
DI
1.12
1

N 2=57.2143 rpm

2/ 3

[ ]

DI
N 2=N 1
DI

=720

0.089
1.12

2/ 3

N 2=3895.39 rpm

## *for constant impeller speed

2/ 3
DI
DI
N 1=N 2
; N p =N m
DI
DI

[ ]
2

2 /3

[ ]
m

N 2=3895.39 rpm
Or
(N DI ) p=1.7( N D I )m

## CHAPTER 9: AGITATION AND AERATION ; ADDITIONAL PROBLEM

A strain of azobacter vinelandii is cultured in a 15 m3 stirred fermenter for production of alginate.
Under conditions the mass transfer coefficient, K La, is 0.25 s-1. Oxygen solubility in a fermenter
3

## broth is approximately 8.5

max cell density in the broth?
REQD: Xmax
Soln:
X max=

K La C al
qo
2

10 kg
3
. The specific oxygen uptake is 15 mmol g/h. What is the
m

## ( 0.25s )(8.510 mkg )

3

1
1
32 g
1 kg
(
)(
)(
)
(15 mmol
)(
)
gh 3600 s 1000 mmol 1 mol 1000 g

Xmax 15937.5

g
g
=15.94
3
L
m

## CHAPTER 9: AGITATION AND AERATION ; ADDITIONAL PROBLEM

Biochemical Engineering and Biotechnology by Ghasem D. Najafpour
Calculate the gas hold-up for an agitated and aerated system with power input of 18 hp in an 80
m3 vessel with gas superficial velocity of 2.6 m.min-1
Given:
P = 18 hp

V = 80 m3
Vs = 2.6 m/min
Required:
Gas Hold-up, Ho
Solution:
From equation;
P
V
Vs
(

0.4
Where; P = power in hp
V = ungassed liquid volume in m3
Vs = gas superficial velocity in m/h

Solving for H:
18
80

2.6 x 60 min /h

H = 0.6 m
The gas hold up can be defined by the above definition using the gas height per volume,
where H = 0.6 m for aeration
Ho =

Vg
V g+V L

Ho = 8.5%

0.6
0.6+6.5

= 0.085

## CHAPTER 9: AGITATION AND AERATION ; ADDITIONAL PROBLEM

Biochemical Engineering and Biotechnology by Ghasem D. Najafpour
Calculate the speed of an impeller and the power requirements of a production-scale bioreactor
with 60m3 using two different methods. Also, match the volumetric mass transfer coefficient.
The following optimum conditions were obtained with a small scale fermenter of volume 0.3m 3
and 60% of the vessel working. The density of the broth, broth, 1200 kg.m-3, working volume
0.18m3, aeration rate of one volume of gas per volume of liquid (vvm), oxygen transfer rate 0.25
kmol.m-3.h-1, liquid height inside the vessel, H L, 1.2Dt. two sets of standard, flat-blade turbine
impellers were installed.
Given:
@ production-scale, V2 = 60m3
@ small-scale, V = 0.3m3
60% of vessel working
broth, 1200 kg.m-3
working volume, V1 = 0.18m3
Q = 1vvm
OTR = 0.25 kmol.m-3.h-1
HL = 1.2Dt
Two sets of standard flat-blade turbine impellers
Required: impeller speed, power requirement, kla
Solution:
V1 = (/4)(Dt)2(1.2Dt) = 0.3Dt3
Diameter of the vessel:
Dt = (V1/0.3)1/3 = (0.18/3)1/3 = 0.576m
Diameter of the impeller:
Di,1 = (1/3)(Dt) = 0.576/3 = 0.192m
Height of liquid media was assumed to be 1.2 times the diameter of the fermenter vessel.
HL,1 = 1.2 x 0.576 = 0.691m

## Diameter of the larger vessel:

Dt2 = (V1/0.3)1/3 = (60/0.3)1/3 = 3.36m
The impeller size for the larger vessel is:
Di2 = 3.36/3 = 1.12m
And the liquid media height in the second fermenter is:
HL2 = 1.2 x 3.36 = 4.03m
Assume the fermentation broth has the same viscosity as water:
1 = 1cp
Aeration rate for 1vvm is:
Us = (0.18 x 60)/(

## Let us take average values for the partial pressure of oxygen

PO2 = ({1 atm + [1 + (HL m/10.3m) x atm]}/2) x 0.21 = 0.213 atm
The oxygen transfer rate is
OTR = 0.25 kmol/m3.h
The mass transfer coefficient is
kL = 0.25/0.213 = 1.174 kmol/m3.h.atm
Use imperial correlation based on the following equation for mass transfer in the bioreactors. The
general equation for the evaluation of Kla is
Kla = x(Pg/V)Y (Us)Z
where x,y, and z are empirical constants. For Newtonian fluids, mnon-coalescing broth and gas
bubbles, the following correlation is valid for a working volumeof less than 4 m3 and a power per
unit volume of 500 10,000 W/m3

kL = 0.002(Pg/V)0.7(Us)0.5
1.174 = 0.002 (Pg/V)0.7(41.45)0.5
For the gassed power per unit volume (Pg/V) is 630.7 W; thegpassed power, Pg, was 0.15hp.
Since the flow regime is turbulent, the power number obtained from figure 6.6, Power number
versus Reynolds number, Re, reads Pno = 6. For two sets of impellers, Np = 2(6) = 12
`

Np = (P1 gc)/(N13Di5)
P1 = (12 x 1200 x N13 x 0.1925)/9.81 = 0.383 N13 W = 5x10-4 N13 hp

## Using Michel and Millers correction factor for power calculation:

Pg1 = 0.5(P121Di13/10.56)0.45
Knowing the power input, we can calculate the rotational speed:
0.15 = 0.5(5x10-4 x N13 x 0.1923 / (3x10-3)0.56)0.45
N1 = 10 rps
N1 = 600 rpm
The power input for ungassed system is
P1 = 5x10-4 N13 = 5x10-4 (10)3 = 0.5hp
Pg/P = 0.15/0.5 = 0.3
For constant power input based on geometric similarity of the vessels, agitation rate is calculated.
(N13Di15)/(V1) = (N23Di25)/(V2)
(N2/N1)3 = (V2/V1)(Di1/Di2)5
N2 = N1 (V2/V1)1/3(Di1/Di2)5/3 = 600(60/0.3)1/3(0.192/1.12)5/3 = 185 rpm
For constant input velocity for a large system:
N2 = N1(Di1/Di2) = 600(0.192/1.12) = 103 rpm
user
The content of the provided text is a series of problems and solutions related to biochemical engineering, specifically focusing on agitation and aeration in bioreactors. The problems involve deriving relationships, proving equations, and calculating various parameters such as mass transfer coefficients, power requirements, and gas hold-up.

### Analysis:
TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS

1. **Multiple Questions**: The text contains multiple distinct questions, each with its own set of given data and required calculations.
2. **Answers Provided**: For each question, detailed solutions are provided, including step-by-step calculations and explanations.

### Summary of Questions and Answers:
1. **Derive the relationship between overall mass transfer coefficient for gas phase \( K_G \) and individual mass-transfer coefficients \( k_L \) and \( k_G \)**:
   - **Solution**: The relationship is derived and simplified for sparingly soluble gases.

2. **Prove that Eq. (9.25) is the same as Eq. (9.26), and Eq. (9.27) is the same as Eq. (9.28)**:
   - **Solution**: Detailed algebraic manipulations are provided to show the equivalence of the equations.

3. **Calculate power requirement,