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
Nur Istianah-KPP-Sentrifugasi-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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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) 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 2N 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
8.110 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
2N 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 8.110 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 (2N1 / 60) 2 D 2 p f 18 ln( ro / ri )
new flow rate
Q2
V (2N 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 (2N 2 / 60) 2 Q1 (2N1 / 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 (2N 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 6.8m 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.
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 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
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Nur Istianah-KPP-Sentrifugasi-2014