Centrifugation calculation
Solution of Examples
Example 1. Separation of Cells Growing on Support
ρ s = 1.02 g / cm 2
Dextran
d = 150μm
bead
V= 50L
l = 1.5d
ρ = 1.00 g / cm 3
μ = 1.1cp(0.011g / cm sec)
(a) Settling time?
assumption : Beads quickly reach their maximum terminal velocity
t=

υg
(
length
) = time
lenght / time
Force balance
FD
FB
FG = F B + F D
Dextran
bead
FG − F B = F D
FG
4
FG = πR 3 ρ S g
3
4 3
4 3
πR ρS g = πR ρ g + 3πdμυ
3
3
4
FB = π R 3 ρ g
1 3
1 3
πR ρ S g = πR ρg + 3πdμυ
3
6
6
F D = 3π d μυ
Re < 1
1
π d 3 ( ρ S − ρ ) g = 3π d μυ
1
6
2
F D = κ ( ρυ 2 )[( π / 4 ) d 2 ] Re > 1
d
2
υ =
(ρS − ρ )g
18 μ
(0.015cm) 2 (1.02 − 1.00) g / cm 3 (980 cm / sec) 2
υg =
18(0.011g / cm sec)
υ g = 0 . 022 cm / sec
• check
Re =
dυρ
μ
(0.015cm)(0.022cm/ sec)1g / cm3
=
= 0.03< 1
(0.011g / cmsec)
V = πr 2
π
4
π

d  =V
t=
(  / 1 . 5 ) 2  = 50 × 10 3 cm 3
52 . 3 cm (min/ 60 sec)
t=
0 . 022 cm / sec
2
4
 = 52 . 3 cm
υg
= 39 . 6 min
(b) Estimate the time to reach this velocity
Force balance
Initial position
FG '−FB '−FD ' = FG − FB − FD
(υg= 0)
FG '−FB ' = FG − FB − FD
4 3
4 3
4 3
4 3
πR ρS g − πR ρ = πR ρS g − πR ρg + 3πdμυ
3
3
3
3
dvg 1 3
Step 1
1 3
πd ( ρS − ρ)
= πd ( ρS − ρ)g − 3πdμν
6
dt 6
Step 2
dvg
dt
Terminal velocity
(υg=constant)
(Initial condition : t=0, v=0)
=g−
3πdμν
πd 3 / 6( ρS − ρ )
1
dvg = dt
g −αυg
1
•α
− ln(g −αυg ) = t + c
α
=
18 μ
d /( ρ s − ρ )
1
c = − ln g
2
1
− ln(g −αvg ) = t + ln g
α
 gd2 (ρs − ρ) 
2
υg = 
1
−
exp(
−
18
μ
t
/
d
(ρs − ρ))

 18μ 
18μ
t >>1
2
d ((ρs − ρ))
[
d 2 ( ρs − ρ ) (0.015cm)2 (0.02g / cm3 )
t >>
=
18μ
18(0.01g / cmsec)
>> 2×10−5 sec
]
Example 2 Centrifugation of Yeast Cells.
(a) How long does it take to have
Spin =500r/min
3cm
ρ s = 1.05 g / cm 2
d = 8.0 μm
a complete separation ?
d2
υ =
(ρS − ρ )g
Water
18 μ
ρ = 1g / cm
μ = 1 cp ( 0 . 01 g / cm sec)
2
d
dr
(ρs − ρ)ω2r
= vω =
18μ
dt
2
10cm
1
d2
dr =
(ρs − ρ)dt
r
18μ
2
d
[lnr]10
(ρs − ρ)t
3 =
18μ
10cm d 2
ln
=
(ρs − ρ)ω2t
3cm 18μ
10cm
(8×10−4 cm)2
ln
=
⋅
3cm 18(0.01g / cmsec)
(0.05 g / cm 3 )(500 × 2π 2 / 60 sec)t
t = 2500sec
= 41.6 min
Example 3 Tubular Centrifugation of E.coli.
Bowl speed
16,000 r/min
r = 12.7cm
length
= 73.0cm
(a) Calculate the settling velocity υg
for the cells.
 2πR 2ω 2 
Q = vg 

g


Qg
vg =
2πR 2ω 2
Q = 200 liter/hr
z
r
200×103 cm3 / hr(980cm / sec2 )(hr / 3600sec)
vg =
2π (73cm)(12.7cm)2 ([2π (16,000)] / 60(sec/min)min)2
= 2.6 ×10−7 cm/ sec
Tubular Bowl Centrifuge Theory
Q = Av
dz
Q
=
dt π ( R02 − R12 )
d2
dr
2
=
(
ρ
−
ρ
)
ω
r
= vω
s
18μ
dt
 rϖ 2 
dr

= vg 
dt
 g 
 rω2  π (R02 − R12 ) Logarithmic mean radius

= vg 
R =
Q
 g 
 ω2  π (R02 − R12 )
1
R 02 − R 12
( R 0 + R 1 )( R 0 −
=
dr = vg  
dz
ln( R 0 / R 1 )
ln( R 0 / R 1 )
r
Q
g
 2πR 2ω 2 
 ω2  2 2
Q = vg 
vg π (R0 − R1 )

g
g


Q=  
ln(R0 / R1)
Q = v g []
dr dr / dt
=
dz dz/ dt
R
ln(
R1 )
− R1
R 0 / R1)
0
= 2R
2
(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.
d
Original cell
(μ, Qo)
1/2d
Disrupted cell
( 4μ, Qd)
2
2πR2ω2
]
Q = vg [
g
2πR2ω2
d2
(ρs − ρ)g(
)
Q=
18μ
g
Qd dd / μd
= 2
Qo do / μo
Qd (1/ 2d)2 / 4μd
=
Qo
d2 / μ
Qd 1
=
Qo 16
Example 4 Disc Centrifugation of Chlorella.
6000 r/min
vg =1.07×10-4
n= 80 (disc number)
Θ=40o
R 1 = 6cm
R 0 = 15.7cm
(a) estimate the volumetric capacity
for this centrifuge.
2πnω2 3 3

Q = vg 
(R0 − R1 ) cotΘ
 3g

2


 2π (6000) 
 2π (80)
−4
3
3
Q = 1.07×10 cm/ sec
× (15.7cm) − (6cm) cot(40)
2 

3(980cm/ sec ) 60(sec/min)min

[
= 3.1×104 cm3 / sec= 31liter/ sec
]
Disc Type Centrifuge Theory
 Q 
v0 = 
f ( y)

 n(2πr) 
dx
= v0 − vω sinΘ
dt
≅ v0
 Q 
f ( y)
=

 n(2πr) 
dy
= vω cosΘ
dt
 ω2r 
dy
 cosΘ
= vg 
dt
 g 
dy dy/ dt
=
dx dx/ dt
 2πrnvgϖ 2 )  2
=
r cosΘ
 Qgf( y) 
 2πrnvgϖ 2 ) 
2
=
(R0 − x sinΘ) cosΘ
 Qgf( y) 
2πrnω2 3 3

Q = vg 
(R0 − R1 ) cotΘ
 3g

Q = vg []
Example 5 Scale-up of Centrifugation in Yeast Separation.
Recombinant
protein
in suspension
Centrifuging
• Type of a centrifuge
= Bottle centrifuge
• Spin = 2000 r/min
• Time = 30 min
• R1 = 5 cm, R0 = 15 cm
Centrifuged
protein
(a) Somebody want to scale-up the size and type of centrifuge
for separation 10m3 of this suspension per day.
Solve)
Step 1)
Q = υ g [ ]
About υ g
dr
d2
= vω =
(ρs − ρ)ω2r
dt
18μ
ω 2r
dr
= vg
dt
g
1
ω2
dr = v g ( ) dt
r
g
1
ω2
R1 r d r = v g ( g ) dt
g ln(R0 / R1)
vg =
tω2
980cm/ sec2 ln(15/ 5)
vg =
[2π (2000) / 60sec]2 30min(60sec/min)
R0
= 1.53×10−5cm/ sec
• To use Gyro tester : accurate and reliable data
Figure. 1 Gyro tester
Operating condition: sample 10 ㎖
spin = 7500 min
interval time = 10 sec
980cm / sec2 ln(47 / 31)
vg =
[2π (7500) / 60 sec]2 40 sec
=1.65×10−5 cm/ sec
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
2πnϖ 2 3 3
=
(R0 − R1 ) cotΘ
3g
2πn182
 2π (8500) 
=
⋅
2 
3(980cm/ sec )  60sec 
[(4.7cm)3 − (2.1cm)3 ]cot51o
= 2.33×106 cm2
Maximum flow
Q Max = υ g []
= (1.65 × 10 −5 ) cm / sec( 2.33 × 10 6 ) cm 2
= 38 cm 3 / sec = 0 .038 liter / sec
(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)
Q
=( )
vg
10 × 10 6 cm 3 / hr 
1
m2 


=
−5
4
2 
1.65 × 10 cm / sec  3600 sec/ hr 10 cm 
= 17 ,000 m 2
twice
 = 34 ,000 m 2
Example 6 Centrifugal Filtration of Pogesterone.
Ρ0=0.016g/cm3
V=250cm3
Filtering
Product
•Af = 8.3 cm2
(slurry)
•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.
R0
Rc
R1
R0
 Ro 2
Ro 
μαρc Rc2
⋅ ( ) −1 − 2 ln 
t=
2
2
2
2ρω (Ro − R1 )  Rc
Rc 
= 51cm
i) Properties of the cake
Spin
530 r/min
t=
45 cm
μαρo V
( )2
2Δp A
μα(0.016g / cm3)
250 3
2 2
/
)
1920sec=
(
cm
cm
6
2
2(1atm)(1.01×10 g / cmsec atm) 8.3
μα = 2.67×108 sec−1
ii) mass balance
ρcπ (Ro2 − Rc2 ) = ρoV
1.09g / cm3π ((51cm)2 − Rc2 )45cm = (0.016g / cm3)1600×103cm3
Rc = 49.3cm
iii) filtration time
μαρcRc2
Ro 2
Ro
t=
⋅ [( ) −1− 2ln ]
2
2
2
Rc
2ρω (Ro − R1 ) Rc
2.67×108 sec−1(1.09g / cm3 )(49.3cm)2
51cm 2
=
⋅ [(
) −1 − 2 ln(51cm/ 49.3cm)]
530⋅ 2π 2
2(1g / cm)(
) [(51cm)2 − (45.5cm)2 ] 49.3cm
60sec
= 8.5min