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
Solution:
viscosity = 8.904 x 10^-4 kg/m-s
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
Pmo = 6 (997.08) (4.43)^3 (.36) ^5
= 3144.8862 W
e. Volumetric mass-trans coeff.
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
Kla = 0.0451 /s
CHAPTER 9: AGITATION AND AERATION
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
3
5
N DI P N DI 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
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
Di
g
= 1.25s-1
1.25 s1 2 (0.5 m)
= 0.0796
Fr =
For complete gas dispersion
Flg = 0.2
DI
DT
0.5
( )
Fr 0.5 = 0.2
Flg: Flooding- loading transition
0.5 m
1.4 m
0.5
( 0.0796 )0.5 = 0.0337
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
0.75 s1 2(0.5 m)
Fr =
Di
g
= 0.0287
*Flooding-loading transition
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
Chemical Composition of the cell = C4.46H16.02O3.88N0.86
