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07/10/2014 Nur Istianah-KPP-Sentrifugasi-2014

1

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)


General principle


Klasifikasi centrifuge

Kapasitas

Labratory centrifuge

Preparative centrifuge

Kegunaan

Sedimenting

centrifuge

Filtering

centrifuge

Ultracentrifugation

Tubular bowl

Basket

Disk stack

Scroll decanter

Basket

Pusher

Baffle

Inverting bag

Cone screen


Tubular bowl

centrifuge

1000-15000rpm


Labratory centrifuge

Tubular bowl

centrifuge

500-2000 rpm


Better performance than turbular flow

Preparative centrifuge


Steve, 2007


Sedimenting centrifuge

07/10/2014 Nur Istianah-KPP-Sentrifugasi-2014 Steve, 2007


Filtering centrifuge


centrifugal filtration

1

centrifugal settling

2


Gas-solid cyclone

separator

3


horizontal axis scroll

decanter centrifuge

4


Peeler centrifuge

Pusher

centrifuge


Ultracentrifugation

1000-15000 rpm

Digunakan untuk

pemisahan atau

analisa campuran

makromolekul

(AUC). Ex: proteinRpm tinggi

menimbulkan

panas sehingga

memerlukan

cooling


Applications of centrifuges in food processing


Persamaan pada sentrifugasi


Persamaan pada centrifuge settling

20

• Settling: acceleration from gravity (Fg)

• Centrifuge: – acceleration from centrifugal force (Fc)

– circular motion and acceleration occurred from centrifugal force

ac = acceleration from centrifugal force (m/s2)

r = radial distance (m)

ω = angular velocity (rad/s)

2ra

c

Nur Istianah-KPP-Sentrifugasi-2014

Persamaan pada centrifuge settling

Centrifugal force (Fc)

21Nur Istianah-KPP-Sentrifugasi-2014

• 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 :

(1)

and

(2)

where N = rotational speed (rpm) ω= an angular velocity (rad s-1)

2mRFc

3060

2 NN


23

g force (gravities or g’s)

Nur Istianah-KPP-Sentrifugasi-2014

• The steady-state velocity of particles moving in a streamline flow under the action of an accelerating force

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.

18

)( 2

sls

t

Dgvfrom

18

)( 22

sls

t

Drv


Centrifugation time

• Time taken by the particle to move though the liquid layer is called residence time (tr).

dt

drVt


18

)( 22 rDv ss

t

18

)(22 ss rD

dt

dr


)(

ln18

18

)(ln

18

)(1

22

1

2

22

1

2

0

222

1

ss

r

r

ss

t

ss

r

r

D

r

r

t

tD

r

r

dtD

drr


Calculation of flow rate for continuous centrifuge

• flow rate (Q)

1

2

222

1

2

2

1

2

22

22

1

2

ln18

)()(

ln18

)(

)(

ln18

r

r

Dbrr

r

r

DVQ

D

r

r

V

t

VQ

ssss

ss

r


• 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


Example 1Find centrifugation time tr of a

particle d=1mm. In a centrifugeGiven

.25.0

.20.0

/1000

/1100

.101.8

995

3

3

4

mR

mR

mkg

mkg

sPa

RPMN

o

i

f

P

Ri

Ro


srad

N

/20.104

60

9952

60

2

Find ω


sec1025.3

1000110020.104001.0

)20.0/25.0ln(101.818

)/ln(18

3

22

4

22

r

r

fp

ior

t

t

d

rrt

Find time

tr of particle d=1mm. in centrifuge≥3.25x10-3sec


Example 2

Beer with a specific gravity of 1.042 and a viscosity of1.04x10-3 N s/m2 contains 1.5% solids which have adensity of 1160kg/m3. It is clarified at a rate of 240 l/hin a bowl centrifuge which has and operating volumeof 0.09 m3 and a speed of 10000 rev/min. The bowlhas a diameter of 5.5 cm and is fitted with a 4 cmoutlet. Calculate the effect on feed rate of an increasein bowl speed to 15000 rev/min and the minimumparticle size that can be removed at the higher speed.


• Solution

Initial flow rate

new flow rate

)/ln(18

)60/2( 22

1

1

io

fp

rr

DNVQ

)/ln(18

)60/2( 22

2

2

io

fp

rr

DNVQ


As all conditions except the bowl speed

remain the same,

Therefore,

Q2 = 0.15 l/s

2

2

2

2

1

2

2

1

2

)60/10000142.32(

)60/15000142.32(

)3600/240(

)60/2(

)60/2(

Q

N

N

Q

Q


To find the minimum particle size

mD

VN

rrQD

fp

io

8.61062.2

1020.1

09.0)10421160()60/15000142.32(

)]02.0/0275.0ln(1040.118[15.0

)()60/2(

)]/ln(18[

7

3

2

3

2

2

22



1

# A and B are dense and

light liquid,

rA, rB =outlet radius

rn= radius of neutral zone.

Fig 6.1. Separation of immiscible liquids

# t (s)=residence time

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

2

3

Separation of liquids

Example3

• A bowl centrifuge is used to break an oil-in-water emulsion. Determine the radius of theneutral zone in order to position the feed pipecorrectly. (Assume that the density of thecontinuous phase is 1000 kg/m3 and thedensity of the oil is 870 kg/m3. the outletradius from the centrifuge are 3 cm and 4.5cm).


• Solution

mr

r

r

n

n

n

098.0

130

783.0025.2

8701000

)03.0(870)045.0(1000 222


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