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Sedimentation Characteristics - Novel Separation Processes - Lecture Notes, Study notes of Learning processes

some concept of Novel Separation Processes are Asymmetric Membrane, Centrifugal Separation Processes, Cloud Point Extraction, Colloidal Particles, Common Stationary Phase.Main points of this lecture are: Sedimentation Characteristics, Centrifuge, Centrifugal Force, Terminal Velocity, Density, Aqueous Solution, Physical Interpretation, Gravity Settler, Sedimentation Characteristics, Feed Rate

Typology: Study notes

2012/2013

Uploaded on 04/27/2013

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Chapter 7
1. For a centrifuge of radius 0.07m and rpm 10000, find out centrifugal force developed.
Ans. r=0.07 m2
2
60
2
0.001118
N
FcrN
Fg
=0.001118x0.07)108
=7826
Fc=7826Fg
2. For a centrifuge of radius 0.05m, and rpm of 12000, find out terminal velocity of a
particle of diameter 10 µm and density 1260 Kg/m3 in an aqueous solution.
Ans
rpm=12000
2 2 12000 1256 /
60 60
Nrad s


22
()
18
rd p
tp


r=0.05m
Dp=10x0-6 m;
p
=1260 kg/m3
210
1256 0.05 10 (1260 1000)
3
18 10
t


pf3

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Download Sedimentation Characteristics - Novel Separation Processes - Lecture Notes and more Study notes Learning processes in PDF only on Docsity!

Chapter 7

1. For a centrifuge of radius 0.07m and rpm 10000, find out centrifugal force developed.

Ans. r=0.07 m

2

N

F

c rN

Fg

  

=0.001118x0.07)

8

F c=7826F g

2. For a centrifuge of radius 0.05m, and rpm of 12000, find out terminal velocity of a

particle of diameter 10 μm and density 1260 Kg/m

3

in an aqueous solution.

Ans

rpm=

N

rad s

rd p

t p

r=0.05m

D p=10x

m;

p

 =1260 kg/m

3

t

= 0.114 m/s

3. What is physical interpretation of Σ value?

Ans: It is area (in m

2

) of a gravity settler that has the same sedimentation characteristics

as the centrifuge at the same feed rate.

4. We have to design a centrifuge that can handle 5 times flow rate in a prototype

centrifuge. If all other dimensions between two centrifuges are same, find the ratio of

the length of the centrifuge to be designed and prototype?

Ans

1

2

q b

q b

b

b

b 1: b 2 5 :

  1. A viscous solution contains with a density ρp =1500 Kg/m

3

is to be clarified by

centrifugation. Solution density is 1000Kg/m

3

and viscosity 20 CP. Centrifuge bowl is

with r 2 =0.02m and r1=0.008m and height, b=0.5m. Claculate critical diameter of

particle in exit stream, if N=1000 rpm and q (^) f =0.05 m

3

/hr.

Ans

N=1000 rpm

 N 

   rad/s

Bowl volume:

2 2 2 1

v   b r (  r )