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Module :
Centrifugal Separation processes and
their calculations
Dr. Sirshendu De
Professor, Department of Chemical Engineering
Indian Institute of Technology, Kharagpur
Keywords:
Separation processes, membranes, electric field assisted separation, liquid membrane, cloud point extraction, electrophoretic separation, supercritical fluid extraction
Centrifugal Separation Processes
In centrifugal separation processes, centrifugal force is used to induce separation. Fig. 7.1 demonstrates stages during centrifugal separation.
Slurry Liquid-liquid feed
Heavyliquid Light liquidfraction
Slurry
Soli Liquid
Fig. 7.1: Centrifugal separation: (a) the solid particles experience a force directed towards the wall; (b) In case of solid-liquid slurry, solid particles are pushed towards the wall and get collected; (c) In case of liquid-liquid system, heavier liquid having higher density are pushed towards the wall and can be collected.
Principle:
Centrifugal force acting on particle of mass m at a radial location r is F c = m ω^2 r (7.1)
The angular velocity is related to the linear velocity as, ω^ =^ vr (7.2)
If rotating speed is N rev/min, then ω = 260 π^ N and the centrifugal force in Newton is
calculated as,
(^4 2 2) 0.01097 2 c 3600 F = m^ π N r = mrN (7.3)
Fig. 7.2: Schematic of geometry in centrifuge The feed enters from the bottom at uniform velocity. The length of bowl is b. A particle is removed if it has sufficient residence time to reach the wall. At the end of residence time, particle is at a distance r B from the axis of rotation. If rB is less than r2, the particle leaves the bowl with the fluid. If rB = r 2 , it is deposited on the wall and effectively removed from the liquid. In Sokes law regime (Re<0.1), terminal velocity is expressed in gravitational field as,
( )
2 18
vt gD^ p ρ p ρ
The terminal velocity in the centrifugal force field becomes,
(
2 2 18 vt^ ω^ rD^ p ρ (^) p ρ)
Feed
r 2
rB
r
Liquid Surface
b
Particle
Trajectory
The terminal velocity in this field can be expressed as vt = drdt. Therefore, by integrating
the above equation after equating vt = drdt ,
( )
2 (^02 )
tT (^) 1 8 r p p r
d t d r D r
μ ∫ =^ ω ρ − ρ ∫ (7.8)
( ) 2 2 2 1
T 1 8^ ln
p p
t r
D r
μ ω ρ ρ
= ⎛^ ⎞
− ⎜⎝^ ⎟⎠ (7.9)
If the volumetric flow rate at the steady state of the liquid is q and V is the volume of the liquid column in the centrifuge, then residence time can be expressed as,
tT = (^) Vq = π b r ( 22 − r 1^2 ) (7.10)
Combining Eqs. (7.9) and (7.10), the volumetric flow rate is obtained in terms of known variables.
2
1 8 ln 1
q p^ Dp b
r
r
= −^ ⎡
r − r )⎤⎦ (7.11)
Particles with diameter smaller than Dp calculated from the above equation, will not reach the wall of the bowl and will go out with the exit liquid. Larger particles reach the wall and are removed from the liquid. A cut point/ critical diameter Dpc is defined as the diameter of the particle that reaches half the distance between r 1 and r 2 within residence time, tT.
at t=0, r =^ r^1^^ + 2 r^2 (7.12)
at t=tT, , r=r 2 (7.13)
Where, vt ug is the terminal velocity under gravity (^2) ( 22 12 ) 2 1 2 2 ln^2
b r r g r r r
Σ = ω π^ −
(m^2 ) (7.16)
∑ is a physical characteristic of centrifuge, not the fluid-particle system.
Physical interpretation of ∑
∑ has the unit of area. It is area in m^2 of a gravitational settler that will leave same sedimentation characteristics as the centrifuge at the same feed rate. For scaling up, from lab to pilot scale,
1 2 1 t t 1 2 v = v ⇒ q^ = q Σ Σ
This is dependable if centrifugal forces between the two are within a factor of 2 from each other. If different centrifuges are used efficiency factors have to be used as,
1 1 1 2 2
q q E Σ = E Σ
These efficiency factors are determined experimentally.
Separation of liquids
Separation if liquids are relevant in various applications like, emulsions in food product, fruit juice processing and relevant industries. This is also common in dairy industries for separation of skim milk and cream, etc. Tubular bowl centrifuge is quite common equipment in this regard. Fig. 7.3 shows the one half of this kind of centrifuge with various layers of fluids with geometrical details.
Feed
Heavy liquid, ρH
Light liquid, ρL
Lighter Product
Heavier Product
r 4
r 3
r 2
r 1
Liquid-liquid interface
Fig. 7.3: Schematic of tubular centrifuge with geometrical details showing various liquid layers
In the above figure, following are the details of geometry. r 1 is the location of lighter liquid; r 2 is the location of liquid liquid interface; r 3 is the location of weir; r 4 is the location of heavy liquid downstream/product. Location of interface can be calculated by a balance of pressure in two layers. Force on the fluid located at a radial location r,
F c = mr ω^2 (7.19)
Differential force on a differential thickness dr is calculated as
dFc = dm r. ω^2 (7.20)
Differential mass can be expressed as, dm = (^) ( 2 π brdr (^) )ρ (7.21)
Where, b≈ height of bowl. The pressure force can be expressed as,
Solution:
2 2 (^23000 ) 2410 60 60
ω = π^ N = π = rad/s
Bowl volume V, V = π b r ( 2^2 − r 12 ) = π (^) ( 0.197 (^) )⎡⎣0.02225^2 −0.00716^2 ⎤⎦ = 2.747 *10−^4 m^3
μ = 10 −^3 *100 = 0.1pa.s
q c (^) 3600 = = − m^3 /s Dpc= 0.746 μm Calculate vt, check NRe<1.
- A viscous solution contains particles with a density is to be clarified by
centrifugation. The solution density is and its viscosity is 80 cp. The centrifuge has a bowl with and
ρ p = 1200 kg / m^3
ρ = 850 kg / m^3
r 2 (^) = 0.02 m r 1 (^) = 0.01 m and height b=0.25 m. Calculate the critical particle diameter of the largest particles in the exit stream if N=15000 rpm and flow rate q=0.002 m^3 /hr?
Solution:
ω = π^^ N^ = π× = rad s
The bowl volume, V = π b r ( 2^2 − r 12 ) = π× 0.25 0.02 ( 2 −0.01^2 ) = 2.355 × 10 −^4 m^3 0.002 (^3) / 5.56 10 7 / q flowrate (^) 3600 m s m s = = = × −^3
7 2 (^ )^24 3
18 80 10 ln 2 0. 0.01 0.
− Dp c − −
× = − × ×
× × ⎛⎜^ × ⎞⎟
= 84.16 × 103 Dp^2 c
Dp c = 2.57 μ m
- In the above problem, we would like to scale up the centrifuge. We have to design a centrifuge such that it can handle 1.5 times q in Question (3). r 1 and r 2 remains same, find the length of the centrifuge. Both the centrifuges have same rotational speed?
Solution:
1 2 1 2
q = q
1 1 2 2
q
q
=^ ∑
2 1 2
2 ln^2
b r r
g r
r r
∑ = ω π^ −
1 1
q b
q^ = b
b 2 = 1.5 b 1 = 1.5 × 0.25 m
b 2 (^) =0.375 m
References:
- C. J. Geankoplis, Transport Processes and Unit Operations, Prentice Hall of India, New Delhi, 1997.
