Centrifugal problem calculation

1. Centrifugation calculation
Solution of Examples

2. V= 50L
l = 1.5d
Example 1. Separation of Cells Growing on Support
m
d
cm
g
s
μ
ρ
150
/
02
.
1 2
=
=
sec)
/
011
.
0
(
1
.
1
/
00
.
1 3
cm
g
cp
cm
g
=
=
μ
ρ
(a) Settling time?
assumption : Beads quickly reach their maximum terminal velocity
time
time
lenght
length
t
g
=
= )
/
(
υ

Dextran
bead

3. Dextran
bead
FD FB
FG
Force balance
D
B
G F
F
F +
=
D
B
G F
F
F =
−
g
R
F S
G ρ
π 3
3
4
=
g
R
FB ρ
π 3
3
4
=
]
)
4
/
)[(
2
1
(
3
2
2
d
F
d
F
D
D
π
ρυ
κ
μυ
π
=
= 1
Re <
1
Re >
μυ
π
ρ
π
ρ
π d
g
R
g
R S 3
3
4
3
4 3
3
+
=
μυ
π
ρ
π
ρ
π d
g
R
g
R S 3
6
1
6
1 3
3
+
=
μυ
π
ρ
ρ
π d
g
d S 3
)
(
6
1 3
=
−
g
d
S )
(
18
2
ρ
ρ
μ
υ −
=

4. sec)
/
011
.
0
(
18
sec)
/
980
(
/
)
00
.
1
02
.
1
(
)
015
.
0
( 2
3
2
cm
g
cm
cm
g
cm
g
−
=
υ
cm
cm
V
d
3
.
52
10
50
)
5
.
1
/
(
4
4
3
3
2
2
=
×
=
=




π
π
sec
/
022
.
0
sec)
60
(min/
3
.
52
cm
cm
t
t
g
=
=
υ

min
6
.
39
=
sec
/
022
.
0 cm
g =
υ
μ
υρ
d
=
Re
1
03
.
0 <
=
sec)
/
011
.
0
(
/
1
sec)
/
022
.
0
)(
015
.
0
( 3
cm
g
cm
g
cm
cm
=

2
r
V π
=
• check

5. D
B
G
D
B
G F
F
F
F
F
F −
−
=
−
− '
'
'
D
B
G
B
G F
F
F
F
F −
−
=
− '
'
(b) Estimate the time to reach this velocity
Force balance
μν
π
ρ
ρ
π
ρ
ρ
π d
g
d
dt
dv
d S
g
S 3
)
(
6
1
)
(
6
1 3
3
−
−
=
−
)
(
6
/
3
3
ρ
ρ
π
μν
π
−
−
=
S
g
d
d
g
dt
dv
(Initial condition : t=0, v=0) c
t
g g +
=
−
− )
ln(
1
αυ
α
Step 1
Step 2
Terminal velocity
(υg=constant)
Initial position
(υg= 0)
dt
dv
g
g
g
=
−αυ
1
μυ
π
ρ
π
ρ
π
ρ
π
ρ
π d
g
R
g
R
R
g
R S
S 3
3
4
3
4
3
4
3
4 3
3
3
3
+
−
=
−
)
/(
18
ρ
ρ
μ
α
−
=
s
d
•

6. [ ]
))
(
/
18
exp(
1
18
)
( 2
2
ρ
ρ
μ
μ
ρ
ρ
υ −
−
−





 −
= s
s
g d
t
gd
1
))
((
18
2
>>
−
t
d s ρ
ρ
μ
sec)
/
01
.
0
(
18
)
/
02
.
0
(
)
015
.
0
(
18
)
( 3
2
2
cm
g
cm
g
cm
d
t s
=
−
>>
μ
ρ
ρ
sec
10
2 5
−
×
>>
g
c ln
2
1
−
=
g
t
v
g g ln
)
ln(
1
+
=
−
− α
α

7. Example 2 Centrifugation of Yeast Cells.
3cm
10cm
m
d
cm
g
s
μ
ρ
0
.
8
/
05
.
1 2
=
=
Spin =500r/min
Water
(a) How long does it take to have
a complete separation ?
ω
v
dt
dr
= r
d
s
2
2
)
(
18
ω
ρ
ρ
μ
−
=
t
d
cm
cm
s
2
2
)
(
18
3
10
ln ω
ρ
ρ
μ
−
=
dt
d
dr
r
s )
(
18
1 2
ρ
ρ
μ
−
=
t
d
r s )
(
18
]
[ln
2
10
3 ρ
ρ
μ
−
=
⋅
×
=
−
sec)
/
01
.
0
(
18
)
10
8
(
3
10
ln
2
4
cm
g
cm
cm
cm
sec
2500
=
t
t
cm
g sec)
60
/
2
500
)(
/
05
.
0
( 2
3
π
×
min
6
.
41
=
g
d
S )
(
18
2
ρ
ρ
μ
υ −
=
2
/
1 cm
g
=
ρ
sec)
/
01
.
0
(
1 cm
g
cp
=
μ

8. Example 3 Tubular Centrifugation of E.coli.
length
= 73.0cm
r = 12.7cm
Q = 200 liter/hr
Bowl speed
16,000 r/min
(a) Calculate the settling velocity υg
for the cells.
r
z
2
2
2 ω
π R
Qg
vg

=
2
2
2
3
3
min)
min)
(sec/
60
/
)]
000
,
16
(
2
([
)
7
.
12
)(
73
(
2
sec)
3600
/
)(
sec
/
980
(
/
10
200
π
π cm
cm
hr
cm
hr
cm
vg
×
=
sec
/
10
6
.
2 7
cm
−
×
=






=
g
R
v
Q g
2
2
2 ω
π

9. Tubular Bowl Centrifuge Theory
Av
Q =
)
( 2
1
2
0 R
R
Q
dt
dz
−
=
π
ω
v
dt
dr
= r
d
s
2
2
)
(
18
ω
ρ
ρ
μ
−
= 







=
g
r
v
dt
dr
g
2
ϖ
dt
dz
dt
dr
dz
dr
/
/
=
Q
R
R
g
r
vg
)
( 2
1
2
0
2
−








=
π
ω
dz
Q
R
R
g
v
dr
r
g
)
(
1 2
1
2
0
2
−








=
π
ω
)
/
ln(
)
(
1
0
2
1
2
0
2
R
R
R
R
g
v
Q
g −








=
π
ω

)
/
ln( 1
0
1
0
R
R
R
R
R
−
=
Logarithmic mean radius
2
1
0
1
0
1
0
1
0
2
1
2
0
2
)
/
ln(
)
)(
(
)
/
ln(
R
R
R
R
R
R
R
R
R
R
R
=
−
+
=
−






=
g
R
v
Q g
2
2
2 ω
π
]
[
= g
v
Q

10. (b) After disruption the diameter of debris is about one-half of the
original cell diameter and the viscosity is increased four times
Estimate the volumetric capacity of this same centrifuge operating
under these new conditions.
1/2d
d
Original cell
(μ, Qo)
Disrupted cell
( 4μ, Qd)
]
2
[
2
2
g
R
v
Q g
ω
π
=
)
2
(
)
(
18
2
2
2
g
R
g
d
Q s
ω
π
ρ
ρ
μ

−
=
o
o
d
d
o
d
d
d
Q
Q
μ
μ
/
/
2
2
=
μ
μ
/
4
/
)
2
/
1
(
2
2
d
d
Q
Q d
o
d
=
16
1
=
o
d
Q
Q

11. Example 4 Disc Centrifugation of Chlorella.
6000 r/min
R1 = 6cm
R0 = 15.7cm
Θ=40o vg =1.07×10-4






Θ
−
= cot
)
(
3
2 3
1
3
0
2
R
R
g
n
v
Q g
ω
π
[ ]










−
×






×
= −
)
40
cot(
)
6
(
)
7
.
15
(
min
min)
(sec/
60
)
6000
(
2
)
sec
/
980
(
3
)
80
(
2
sec
/
10
07
.
1 3
3
2
2
4
cm
cm
cm
cm
Q
π
π
sec
/
31
sec
/
10
1
.
3 3
4
liter
cm =
×
=
(a) estimate the volumetric capacity
for this centrifuge.
n= 80 (disc number)

12. Disc Type Centrifuge Theory
Θ
−
= sin
0 ω
v
v
dt
dx
)
(
)
2
(
0 y
f
r
n
Q
v 





=

π
0
v
≅
)
(
)
2
(
y
f
r
n
Q






=

π
Θ
= cos
ω
v
dt
dy
Θ








= cos
2
g
r
v
dt
dy
g
ω
dt
dx
dt
dy
dx
dy
/
/
= Θ








= cos
)
(
)
2 2
2
r
y
Qgf
v
rn gϖ
π 
Θ
Θ
−








= cos
)
sin
(
)
(
)
2 2
0
2
x
R
y
Qgf
v
rn gϖ
π 






Θ
−
= cot
)
(
3
2 3
1
3
0
2
R
R
g
rn
v
Q g
ω
π
[ ]

= g
v
Q

13. Example 5 Scale-up of Centrifugation in Yeast Separation.
Recombinant
protein
in suspension
Centrifuged
protein
Centrifuging
• Type of a centrifuge
= Bottle centrifuge
• Spin = 2000 r/min
• Time = 30 min
• R1 = 5 cm, R0 = 15 cm
(a) Somebody want to scale-up the size and type of centrifuge
for separation 10m3 of this suspension per day.
ω
v
dt
dr
= r
d
s
2
2
)
(
18
ω
ρ
ρ
μ
−
=
g
r
v
dt
dr
g
2
ω
=
dt
g
v
dr
r
g )
(
1 2
ω
=
Solve) ]
[
= g
Q υ
Step 1) About g
υ

14. 2
1
0 )
/
ln(
ω
t
R
R
g
vg =
min)
sec/
60
min(
30
sec]
60
/
)
2000
(
2
[
)
5
/
15
ln(
sec
/
980
2
2
π
cm
vg =
sec
/
10
53
.
1 5
cm
−
×
=
Figure. 1 Gyro tester
• To use Gyro tester : accurate and reliable data
Operating condition: sample 10 ㎖
spin = 7500 min
interval time = 10 sec
dt
g
v
r
d
r
g
R
R
)
(
1 2
0
1
ω
=


15. sec
/
10
65
.
1 5
cm
−
×
=
sec
40
sec]
60
/
)
7500
(
2
[
)
31
/
47
ln(
sec
/
980
2
2
π
cm
vg =
Step 2) ]
[
About
• Disc bowl of the Gyro tester
Operating condition
n = 18 (disc number)
Θ = 51o (angle)
spin = 8500 r/min
R0=4.7cm, R1=2.1cm
Θ
−
=
 cot
)
(
3
2 3
1
3
0
2
R
R
g
nϖ
π
⋅






=
sec
60
)
8500
(
2
)
sec
/
980
(
3
18
2
2
2
π
π
cm
n
o
cm
cm 51
cot
]
)
1
.
2
(
)
7
.
4
[( 3
3
−
2
6
10
33
.
2 cm
×
=

16. Maximum flow
]
[
= g
Max
Q υ
2
6
5
)
10
33
.
2
sec(
/
)
10
65
.
1
( cm
cm ×
×
= −
sec
/
038
.
0
sec
/
38 3
liter
cm =
=
(b) At this point, we have a meeting with the client who change order
Order : Q = 1000ℓ/hr
removal of solids be continuous.(=>nozzle ejecting bowl type)
)
(
g
v
Q
=









×
×
= − 2
4
2
5
3
6
10
sec/
3600
1
sec
/
10
65
.
1
/
10
10
cm
m
hr
cm
hr
cm
2
000
,
17 m
= 2
000
,
34 m
=

twice

17. Example 6 Centrifugal Filtration of Pogesterone.
Ρ0=0.016g/cm3
V=250cm3
(slurry)
Product
Filtering
•Af = 8.3 cm2
•ΔP = 1 atm
•ρc = 1.09 (solid/cm3 cake)
• ρ= 1g/cm (slurry density)
• time = 32min(1920 sec)
(a) To use this experiment to estimate the time to filter 1600ℓ
of this slurry through a centrifugal filter.
45 cm
Spin
530 r/min
R0 Rc R1 R0 = 51cm
2
)
(
2 A
V
p
t o
Δ
=
μαρ
i) Properties of the cake






−
−
⋅
−
=
c
o
c
o
o
c
c
R
R
R
R
R
R
R
t ln
2
1
)
(
)
(
2
2
2
1
2
2
2
ρω
μαρ

18. 2
2
3
2
6
3
)
/
3
.
8
250
(
)
sec
/
10
01
.
1
)(
1
(
2
)
/
016
.
0
(
sec
1920 cm
cm
atm
cm
g
atm
cm
g
×
=
μα
1
8
sec
10
67
.
2 −
×
=
μα
V
R
R o
c
o
c ρ
π
ρ =
− 
)
( 2
2
3
3
3
2
2
3
10
1600
)
/
016
.
0
(
45
)
)
51
((
/
09
.
1 cm
cm
g
cm
R
cm
cm
g c ×
=
−
π
cm
Rc 3
.
49
=
ii) mass balance
iii) filtration time
]
ln
2
1
)
[(
)
(
2
2
2
1
2
2
2
c
o
c
o
o
c
c
R
R
R
R
R
R
R
t −
−
⋅
−
=
ρω
μαρ
)]
3
.
49
/
51
ln(
2
1
)
3
.
49
51
[(
]
)
5
.
45
(
)
51
[(
)
sec
60
2
530
)(
/
1
(
2
)
3
.
49
)(
/
09
.
1
(
sec
10
67
.
2 2
2
2
2
2
3
1
8
cm
cm
cm
cm
cm
cm
cm
g
cm
cm
g
−
−
⋅
−
⋅
×
=
−
π
min
5
.
8
=