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SENTRIFUGASI
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Nur Istianah-KPP-Sentrifugasi-2014
Sentrifugasi
Proses pemisahan solid dari liquid dengan prinsip grafitasi.
Densitas solid harus lebih besar dari densitas liquid
Peran gaya sentrifugal:
1. Mendorong partikel kecil agar mengendap
2. Menahan brownian motion
3. Mencegah arah free convection fluida
4. Mengurangi penumpukan “cake” pada screen (untuk
centrifugal filtration)
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General principle
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Klasifikasi centrifuge
Labratory centrifuge
Kapasitas
Preparative centrifuge
Sedimenting
centrifuge
Kegunaan
Filtering
centrifuge
Ultracentrifugation
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Tubular bowl
Basket
Disk stack
Scroll decanter
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
Basket
Pusher
Baffle
Inverting bag
Cone screen
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Klasifikasi centrifuge
Labratory centrifuge
Tubular bowl
centrifuge
1000-15000rpm
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Preparative centrifuge
Tubular bowl
centrifuge
500-2000 rpm
Better performance
than turbular flow
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Klasifikasi centrifuge
Sedimenting centrifuge
Steve, 2007
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Klasifikasi centrifuge
Filtering centrifuge
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centrifugal filtration
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centrifugal settling
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3
Gas-solid cyclone
separator
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4
horizontal axis scroll
decanter centrifuge
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Pusher
centrifuge
Peeler centrifuge
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Ultracentrifugation
1000-15000 rpm
Digunakan untuk
pemisahan
atau
analisa campuran
makromolekul
(AUC). Ex: protein
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Rpm tinggi
menimbulkan
panas sehingga
memerlukan
cooling
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Applications of centrifuges in food processing
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Persamaan pada sentrifugasi
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Persamaan pada centrifuge settling
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Persamaan pada centrifuge settling
• Settling: acceleration from gravity (Fg)
• Centrifuge:
– acceleration from centrifugal force (Fc)
– circular motion and acceleration occurred from
centrifugal force
ac
 r
2
ac = acceleration from centrifugal force (m/s2)
r = radial distance (m)
ω = angular velocity (rad/s)
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Centrifugal force (Fc)
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• The centrifugal force, Fc acting on an object of mass
m, rotating in a circular path of radius R, at an
angular velocity of ω is :
Fc  mR
2
(1)
and
2N N


60
30
(2)
where N = rotational speed (rpm) ω= an angular
velocity (rad s-1)
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g force (gravities or g’s)
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• The steady-state velocity of particles moving in a
streamline flow under the action of an
accelerating force
g (  s   l ) Ds2
from vt 
18
r 2 (  s   l ) Ds2
vt 
18
Where vt=terminal velocity of particle; ρs and ρl =
density of solid and liquid ; r = distance of the
particle from center of rotation;µ = viscosity of
liquid.
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Centrifugation time
• Time taken by the particle to move though the
liquid layer is called residence time (tr).
dr
Vt 
dt
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 ( s   )D r
vt 
18
2
2
s
dr D r (  s   )

dt
18
2
s
2
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D  ( s   )
1
dt
r r dr 

18

0
1
r2
2
s
2
t
r2 D  (  s   )
ln 
 tr
r1
18
2
s
2
r2
18 ln
r1
tr  2 2
Ds  (  s   )
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Calculation of flow rate for continuous centrifuge
• flow rate (Q)
V
Q 
tr
V
r2
18 ln
r1
Ds2 2 (  s   )
V  Ds2 2 (  s   )  (r22  r12 )b  Ds2 2 (  s   )
Q

r2
r2
18 ln
18 ln
r1
r1
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• r1 = inside radius (m)
• r2 = outside radius (m)
• b = height of
centrifuge(m)
• µ = viscosity (Pa.s)
• ω = an angular velocity
(rad s-1)
• ρs = density of solid
(kg/m3)
• ρ = density of liquid
(kg/m3)
• Ds= diameter of
particle(m)
• V(m3)=operating
volume of the
centrifuge
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Example 1
Find centrifugation time tr of a
particle d=1mm. In a centrifuge
Given
N  995 RPM
  8.110  4 Pa.s
 P  1100kg / m 3
Ri
Ro
 f  1000kg / m 3
Ri  0.20m.
Ro  0.25m.
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Find ω
2N

60
2    995

60
  104.20rad / s
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Find time
18 ln( ro / ri )
tr  2 2
d   p   f 
4
18  8.110  ln(0.25 / 0.20)
tr 
2
2
0.001  104.20  1100  1000
3
t r  3.25 10 sec
tr of particle d=1mm. in centrifuge≥3.25x10-3sec
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Example 2
Beer with a specific gravity of 1.042 and a viscosity of
1.04x10-3 N s/m2 contains 1.5% solids which have a
density of 1160kg/m3. It is clarified at a rate of 240 l/h
in a bowl centrifuge which has and operating volume
of 0.09 m3 and a speed of 10000 rev/min. The bowl
has a diameter of 5.5 cm and is fitted with a 4 cm
outlet. Calculate the effect on feed rate of an increase
in bowl speed to 15000 rev/min and the minimum
particle size that can be removed at the higher speed.
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• Solution
Initial flow rate
Q1 
V  (2N1 / 60) 2 D 2  p   f 
18  ln( ro / ri )
new flow rate
Q2 
V  (2N 2 / 60) 2 D 2  p   f 
18  ln( ro / ri )
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As all conditions except the bowl speed
remain the same,
Q2 (2N 2 / 60) 2

Q1 (2N1 / 60) 2
Q2
(2  3.142 15000 / 60) 2

(240 / 3600) (2  3.142 10000 / 60) 2
Therefore,
Q2 = 0.15 l/s
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To find the minimum particle size
Q2 [18 ln( ro / ri )]
D 
(2N 2 / 60) 2 (  p   f )V
2
0.15[18 1.40 10 3  ln(0.0275 / 0.02)]

(2  3.142 15000 / 60) 2 (1160  1042)0.09
3
1.20 10
D 
 6.8m
7
2.62 10
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Separation of liquids
1
# A and B are dense and
light liquid,
rA, rB =outlet radius
rn= radius of neutral zone.
2
Ω = angular velocity,
Q = volumetric flowrate,
V = operating volume of the centrifuge,
D = diameter of the particle,
r2 = radius of light phase outlet,
r1 = radius of dense phase outlet,
N =speed of rotation
3
Fig 6.1. Separation of immiscible liquids
# t (s)=residence time
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Example3
• A bowl centrifuge is used to break an oil-inwater emulsion. Determine the radius of the
neutral zone in order to position the feed pipe
correctly. (Assume that the density of the
continuous phase is 1000 kg/m3 and the
density of the oil is 870 kg/m3. the outlet
radius from the centrifuge are 3 cm and 4.5
cm).
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• Solution
1000(0.045)  870(0.03)
rn 
1000  870
2.025  0.783
rn 
130
rn  0.098m
2
2
2
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THANKS FOR YOUR ATTENTION
The best person is one give something useful always
07/10/2014
Nur Istianah-KPP-Sentrifugasi-2014