2014 Nur Istianah-KPP-Sentrifugasi-2014

SENTRIFUGASI

07/10/2014

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 07/10/2014


   

Tubular bowl Basket Disk stack Scroll decanter

    

Basket Pusher Baffle Inverting bag Cone screen 3

Klasifikasi centrifuge Labratory centrifuge

Tubular bowl centrifuge 1000-15000rpm 07/10/2014


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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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Steve, 2007

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centrifugal filtration

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centrifugal settling 07/10/2014


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Gas-solid cyclone separator

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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) Nur Istianah-KPP-Sentrifugasi-2014

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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) Nur Istianah-KPP-Sentrifugasi-2014

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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. Nur Istianah-KPP-Sentrifugasi-2014

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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   ) Nur Istianah-KPP-Sentrifugasi-2014

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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 Nur Istianah-KPP-Sentrifugasi-2014

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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 Nur Istianah-KPP-Sentrifugasi-2014

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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. Nur Istianah-KPP-Sentrifugasi-2014

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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 Nur Istianah-KPP-Sentrifugasi-2014

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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 Nur Istianah-KPP-Sentrifugasi-2014

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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 Nur Istianah-KPP-Sentrifugasi-2014

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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.

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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 07/10/2014


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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


2014 Nur Istianah-KPP-Sentrifugasi-2014

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