Introduction to Mass Transfer Operations (5 of 5)

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▪ Flotation
▪ Flocculation
▪ Filtration
▪ Decanting
▪ Centrifugation
▪ Cyclone Separation
▪ Magnetic Separation
▪ Settling & Sedimentation
▪ Mechanical Size Reduction
▪ Sieve System

▪ This section does not deal with separations where one or more chemicals are
removed from a feed mixture
▪ Instead
▪ it describes mechanical devices used to separate one bulk phase from another
▪ Bulk phases involving mostly solids and fluids
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▪ The main issue in this type of separations are:
▪ spectrum of particle sizes
▪ This is important to identify, as they will be the main focus of:
▪ the devices that could be used to separate
▪ This requires the understanding the force balances
▪ Several models & settling equations are based on these:
▪ Newton’s law
▪ Stokes’ law
▪ Brownian motion
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▪ Most applications of mechanical-physical separation involve particles or droplets:
▪ Removing airborne liquids, solid particles, microorganisms, and vapors from air streams
▪ Cleaning gas streams & to prevent contamination
▪ Separating entrained liquids from vapor streams as in a flash distillation chamber, or partial
condenser.
▪ Condensing vapors from air streams when downstream conditions favor an undesirable
condensation.

▪ Continued…:
▪ Eliminating pollutant particles, mists, and fogs from
gases that are vented to the atmosphere from
manufacturing plants.
▪ Removing droplets of one liquid suspended in another
as in hydrocarbon-water decanters.
▪ Recovering, as a cake, solid particles suspended in
liquids, by means of plate-and-frame, drum, leaf, and
other filters; and determining cake wash cycles.
▪ Operating filters at constant pressure and variable
rates, using pump characteristic curves.
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▪ Flotation
▪ Flocculation
▪ Decanting
▪ Filtration
▪ Centrifugation
▪ Sieve System

▪ Flotation
▪ Flocculation
▪ Decanting
▪ Filtration
▪ Centrifugation
▪ Sieve System
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▪ Common Equipment:
▪ Settlers
▪ Decanters
▪ Coalescers
▪ Vanes
▪ Centrifuges
▪ Demisters & Mesh pads
▪ Knock-out drums
▪ Electrostatic precipitators
▪ Cyclones
▪ Impingement separators
▪ Bag filters, and drum, plate-and-frame, and vacuum filters
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▪ Common Equipment:
▪ Settlers
▪ Decanters
▪ Coalescers
▪ Vanes
▪ Centrifuges
▪ Demisters & Mesh pads
▪ Knock-out drums
▪ Electrostatic precipitators
▪ Cyclones
▪ Impingement separators
▪ Bag filters, and drum, plate-and-frame, and vacuum filters

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▪ Component Recovery (Key Component, etc)
▪ Split Factor, Split Ratio, Split Fraction
▪ Purity & Concentration
▪ Separation Factor & Sequences
▪ Method Selection
▪ Exercise

▪ There are many variables to track and keep pace when doing a separation process.
▪ Identifying the best split fractions, ratios, purity, concentrations and methodologies
is key for a feasible Process Separation
▪ Some examples:
▪ Key Components (light key, heavy key)
▪ Split Fraction
▪ Split Ratio
▪ Separation Factor
▪ Recovery
▪ Product Purity
▪ Separation trains
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▪ In this section, we will cover the most common variables and concepts.
▪ An introduction to Separation Trains is also considered.

▪ If no chemical reaction occurs and the process operates in a continuous, steady-
state fashion, then:
▪ for each component i, in a mixture of C components the molar (or mass) flow rate in the
feed, ni
(F), equals the sum of the product molar (or mass) flow rates, ni
(p), for that
component in the N product phases, p:
( ) ( ) (1) (2) ( 1)
1
...
N
F p N
i i i i i
p
n n n n n −
=
= = + + +
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▪ Thus, in order to solve this equation ( )
▪ For values of ni
(p) from specified values of ni
(F):
▪ an additional N- 1 independent expressions involving ni
(p) are required.
▪ This gives a total of NC equations in NC unknowns.
▪ If a single-phase feed containing C components is separated into N products:
▪ C(N-1) additional expressions are needed.
▪ If more than one stream is fed to the separation process:
▪ ni
(F) is the summation for all feeds.
( ) ( ) (1) (2) ( 1)
1
...
N
F p N
i i i i i
p
n n n n n −
=
= = + + +

▪ Chemical plants are designed and operated to meet specifications given as:
▪ component recoveries
▪ product purities.

▪ The feed is the bottoms product from a reboiled absorber
used to deethanize
▪ i.e., remove ethane and lighter components from a mixture
of petroleum refinery gases and liquids.

▪ The separation process of choice:
▪ A sequence of three multistage distillation columns
▪ Where feed components are rank-listed by decreasing
volatility:
▪ Hydrocarbons heavier
▪ i.e., of greater molecular weight than n-pentane
▪ In nC4 – iC4 range
▪ Finally, C3
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▪ The three distillation columns are to be separated into
four products:
▪ C5-rich bottoms
▪ C3-rich distillate
▪ iC4-rich distillate
▪ nC4-rich bottoms

▪ For each column:
▪ Feed components are partitioned between the overhead
and the bottoms according to a split fraction or split ratio
that depends on:
▪ (1) the component thermodynamic and transport properties
▪ (2) the number of stages
▪ (3) the vapor and liquid flows through the column.

▪ The split fraction, SF, for component “I” in separator “k”
is the fraction found in the first product:
▪ Which reads:
▪ Sfi,k = Split Fraction of species “I” in separator “k” is given
by:
▪ The moles of species “i” in stream 1
▪ The moles of species “i” in feed stream
(1)
,
, (F)
,
i k
i k
i k
n
SF
n
=
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▪ A split ratio, SR, between two products can be defined as
well.
▪ Given:
▪ Where:
▪ n(2) refers to a component flow rate in the second product.
(1)
,
, (2)
,
i k
i k
i k
n
SR
n
=

▪ Mathematically, we can prove that:
▪ Therefore:
(1)
,
, (2)
,
i k
i k
i k
n
SR
n
=
(1)
,
, (F)
,
i k
i k
i k
n
SF
n
=
(1)
, ,
, (2)
, ,(1 )
i k i k
i k
i k i k
n SF
SR
n SF
= =
−

▪ Given the following data on streams:
▪ Perform Split Fraction and Split Ratios of the products in each
column
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▪ Given the following data on streams:
▪ Perform Split Fraction and Split Ratios of the products in each
column
Component 1 2 3 4 5 6 7
C2 0.60 0.00 0.60 0.60 0.00 0.00 0.00
C3 57.00 0.00 57.00 54.80 2.20 2.20 0.00
iC4 171.80 0.10 171.70 0.60 171.10 162.50 8.60
nC4 227.30 0.70 226.60 0.00 226.60 10.80 215.80
iC5 40.00 11.90 28.10 0.00 28.10 0.00 28.10
nC5 33.60 16.10 17.50 0.00 17.50 0.00 17.50
C6+ 205.30 205.30 0.00 0.00 0.00 0.00 0.00
Total 735.60 234.10 501.50 56.00 445.50 175.50 270.00
Stream
Units → lbmol/h

▪ Calculating Split Fraction
Component 1 2 3 4 5 6 7
C2 0.60 0.00 0.60 0.60 0.00 0.00 0.00
C3 57.00 0.00 57.00 54.80 2.20 2.20 0.00
iC4 171.80 0.10 171.70 0.60 171.10 162.50 8.60
nC4 227.30 0.70 226.60 0.00 226.60 10.80 215.80
iC5 40.00 11.90 28.10 0.00 28.10 0.00 28.10
nC5 33.60 16.10 17.50 0.00 17.50 0.00 17.50
C6+ 205.30 205.30 0.00 0.00 0.00 0.00 0.00
Total 735.60 234.10 501.50 56.00 445.50 175.50 270.00
Stream
SF-C1
(1)
,
, (F)
,
i k
i k
i k
n
SF
n
=
Component SF-C1 SF-C2 SF-C3
C2 1.0000 1.0000 #DIV/0!
C3 1.0000 0.9614 1.0000
iC4 0.9994 0.0035 0.9497
nC4 0.9969 0.0000 0.0477
iC5 0.7025 0.0000 0.0000
nC5 0.5208 0.0000 0.0000
C6+ 0.0000 #DIV/0! #DIV/0!
Total 5.2197 #DIV/0! #DIV/0!
Column
SF-C2
(1)
,
, (F)
,
i k
i k
i k
n
SF
n
=
SF-C3
(1)
,
, (F)
,
i k
i k
i k
n
SF
n
=
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▪ Calculating Split Ratio
(1)
, ,
, (2)
, ,(1 )
i k i k
i k
i k i k
n SF
SR
n SF
= =
−
Component 1 2 3 4 5 6 7
C2 0.60 0.00 0.60 0.60 0.00 0.00 0.00
C3 57.00 0.00 57.00 54.80 2.20 2.20 0.00
iC4 171.80 0.10 171.70 0.60 171.10 162.50 8.60
nC4 227.30 0.70 226.60 0.00 226.60 10.80 215.80
iC5 40.00 11.90 28.10 0.00 28.10 0.00 28.10
nC5 33.60 16.10 17.50 0.00 17.50 0.00 17.50
C6+ 205.30 205.30 0.00 0.00 0.00 0.00 0.00
Total 735.60 234.10 501.50 56.00 445.50 175.50 270.00
Stream
Units → lbmol/h
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(1)
, ,
, (2)
, ,(1 )
i k i k
i k
i k i k
n SF
SR
n SF
= =
−
C2 1.0000 1.0000 #DIV/0!
C3 1.0000 0.9614 1.0000
iC4 0.9994 0.0035 0.9497
nC4 0.9969 0.0000 0.0477
iC5 0.7025 0.0000 0.0000
nC5 0.5208 0.0000 0.0000
C6+ 0.0000 #DIV/0! #DIV/0!
Total 5.2197 #DIV/0! #DIV/0!
Column
C2 #DIV/0! #DIV/0! #DIV/0!
C3 #DIV/0! 24.9091 #DIV/0!
iC4 1717.000 0.0035 18.8953
nC4 323.7143 0.0000 0.0500
iC5 2.3613 0.0000 0.0000
nC5 1.0870 0.0000 0.0000
C6+ 0.0000 #DIV/0! #DIV/0!
Total #DIV/0! #DIV/0! #DIV/0!
Column

(1)
, ,
, (2)
, ,(1 )
i k i k
i k
i k i k
n SF
SR
n SF
= =
−
C2 1.0000 1.0000 #DIV/0!
C3 1.0000 0.9614 1.0000
iC4 0.9994 0.0035 0.9497
nC4 0.9969 0.0000 0.0477
iC5 0.7025 0.0000 0.0000
nC5 0.5208 0.0000 0.0000
C6+ 0.0000 #DIV/0! #DIV/0!
Total 5.2197 #DIV/0! #DIV/0!
Column
C2 #DIV/0! #DIV/0! #DIV/0!
C3 #DIV/0! 24.9091 #DIV/0!
iC4 1717.000 0.0035 18.8953
nC4 323.7143 0.0000 0.0500
iC5 2.3613 0.0000 0.0000
nC5 1.0870 0.0000 0.0000
C6+ 0.0000 #DIV/0! #DIV/0!
Column
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▪ Watch how other instructors use this example!
▪ https://youtu.be/1LfyZE1YWsM?t=3920
▪ 1:05:20

▪ Typically, there is at least 1 key component which we base the separation.
▪ More likely, there are TWO main key components:
▪ Light key → will be separated on tops
▪ Heavy key → will be separated on bottoms
▪ An example of a distillation of:
▪ Water & Ethanol:
▪ Heavy key → will not vaporize readily (water)
▪ Light key → will vaporize readily (ethanol)

▪ IN many processes, we want to purify or recover a Key Component.
▪ The concept of Product Recovery is related to this:
▪ Total amount of product going “in”
▪ Total amount of product “separated” or “recovered”
" " cov
% cov 100%
" "
moles i re ered
re ery x
total moles i
=
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▪ In many processes, we might not be able to recover 100% of the species “i” in the
feed.
▪ These are said to be “losses”
▪ Product Loss is very common and is also an issue to consider when designing a
process.
" " Pr
% 100%
" " in
moles i not in oduct stream
Loss x
total moles i Feed
=

▪ The concept of product purity is related to the % of composition of our product of interest in the
final stream or product stream.
▪ Typically, the higher purity, the better
▪ In engineering “high” purities are related to:
▪ 95%
▪ 99%
▪ 99.9% …. 99.99999%
▪ Impurities are all the other materials affecting the product purity.
▪ 5%
▪ 1%
▪ 0.1% …. 0.0000001% or even ppm
" " cov
% " " 100%
moles i re ered
Purityof i x
total molesin stream
=

▪ Sometimes, product recovery must be sacrificed in order to achieve a higher purity
or vice versa:
▪ A product’s purity must be sacrificed (decreased) in order to make the process
feasible.
▪ Trade-offs are very common in process design.
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▪ Separation Sequencing
▪ Heuristics & Thumb Rules
▪ Separation Factor
▪ Ex. Separation Factor
▪ Separation Feasibility
▪ Ex. Feasibility of a Separation Method

▪ The three-column recovery process (right):
▪ Is only one of five (1/5) alternative sequences of distillation
operations
▪ Since each column has a single feed and produces an overhead
product and a bottoms product:
▪ Multiple sequencing might take place.

▪ Example:
▪ Consider a hydrocarbon feed that consists:
▪ propane (C3)
▪ isobutane (iC4)
▪ n-butane (nC4)
▪ isopentane (iC5)
▪ n-pentane (nC5).
▪ A sequence of distillation columns is to be used to separate the
feed into three nearly pure products of:
▪ Individual Product Lines: C3, iC4, and nC4
▪ A final (multicomponent) product of iC5/nC5.
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▪ All 5 possible sequences:
▪ C3 – nC4 – iC4 – nC5 – iC5

▪ A diagram on all possible outcomes:
▪ 4 components
▪ Two Product Separation
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▪ It is impossible to “set” a set of rules for processes
▪ Engineers use heuristics which typically help in quick/informal design of Processes
▪ They are based on straight forward logic, which must be then analyzed thoroughly.
▪ These must be used for initial screening and must be useful and easy to apply, as
well as not requiring column design or cost estimation.
▪ As you can imagine:
▪ These heuristics sometimes conflict with each other
▪ Thus a clear choice is not always possible
▪ This is great for ENGINEERS!

1. Remove unstable, corrosive, or chemically reactive components early in the
sequence.
2. Remove final products one by one as overhead distillates.
3. Remove, early in the sequence, those components of greatest molar percentage
in the feed.
4. Make the most difficult separations in the absence of the other components.
5. Leave for later in the sequence those separations that produce final products of
the highest purities.
6. Select the sequence that favors near-equimolar amounts of overhead and bottoms
in each column.
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▪ Other “Thumbrules” to consider:
▪ Remove most plentiful impurities 1st
▪ Remove the easiest to remove impurities 1st
▪ Make the most difficult & expensive separation later
▪ Select processes that make use of greatest differences in the properties of the product
▪ Select the sequence processes exploiting different separations

▪ Some separation operations are incapable of making a sharp split between key
components
▪ This will affect the desired recovery of only a single component.
▪ The split ratio (SR), split fraction (SF), recovery, or purity that can be achieved for
the single key component depends on a number of factors
▪ For the simplest case of a single separation stage, these factors include:
▪ (1) the relative molar amounts of the two phases leaving the separator
▪ (2) thermodynamic, mass transport, and other component properties.

▪ For multistage separators, additional factors are the number of stages and their
configurations.
▪ If the feed enters near the middle of the column as in distillation it has both
enriching and stripping sections, and it is often possible to achieve a sharp
separation between two key components.
▪ The enriching section purifies the light key and the stripping section purifies the
heavy key.

▪ For these, a measure of the relative degree of separation between two key
components, i and j, is the separation factor, SP.
▪ SP is defined in terms of the component splits as measured by the compositions of
the two products, (1) and (2):
▪ C: Measurement of composition
(1) (2)
,j (1) (2)
/
/
i i
i
j j
C C
SP
C C
=
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▪ SP is readily converted to the following forms in terms of split fractions or split
ratios
▪ Achievable values of SP depend on the number of stages and the properties of
components i and j.
▪ In general, components i and j and products 1 and 2 are selected so that SPi,j > 1.0.
,j
i
i
j
SR
SP
SR
=
,j
/
(1 ) / (1 )
i j
i
i j
SF SF
SP
SF SF
=
− −

▪ Then, a large value corresponds to a
relatively high degree of separation or
separation factor, and a small value close to
1.0 corresponds to a low degree of separation
factor.
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▪ For example:
▪ if SP = 10,000 and SRi = 1/SRj
▪ then, from SRi = 100 and SRj = 0.01
▪ Results to a corresponding to a sharp separation.
▪ However:
▪ if SP = 9 and SRi = 1/SRj, then SRj = 3 and SRj = 1/3
▪ This Results ina corresponding to a non-sharp
separation.

▪ Calculate the Separation Factor of the most relevant key-components in the
Hydrocarbon Process we have been working on.
Component 1 2 3 4 5 6 7
C2 0.60 0.00 0.60 0.60 0.00 0.00 0.00
C3 57.00 0.00 57.00 54.80 2.20 2.20 0.00
iC4 171.80 0.10 171.70 0.60 171.10 162.50 8.60
nC4 227.30 0.70 226.60 0.00 226.60 10.80 215.80
iC5 40.00 11.90 28.10 0.00 28.10 0.00 28.10
nC5 33.60 16.10 17.50 0.00 17.50 0.00 17.50
C6+ 205.30 205.30 0.00 0.00 0.00 0.00 0.00
Total 735.60 234.10 501.50 56.00 445.50 175.50 270.00
Stream

▪ From the SR Data con Columns…
Column
Key/Comp SepFac-C1 SepFac-C2 SepFac-C3
nC4/iC5 137.0890
C3/iC4 7103.2424
iC4/nC4 377.5571
Column
Component SR-C1 SR-C2 SR-C3
C2 #DIV/0! #DIV/0! #DIV/0!
C3 #DIV/0! 24.9091 #DIV/0!
iC4 1717.000 0.0035 18.8953
nC4 323.7143 0.0000 0.0500
iC5 2.3613 0.0000 0.0000
nC5 1.0870 0.0000 0.0000
C6+ 0.0000 #DIV/0! #DIV/0!
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▪ Analysis:
▪ The SP in Column C1 is small because:
▪ the split for the heavy key, iC5H12, is not sharp.
▪ The largest SP occurs in Column C2, where:
▪ the separation is relatively easy because of the large volatility
difference.
▪ Much more difficult is the butane-isomer split in Column C3:
▪ where only a moderately sharp split is achieved.
Column
Key/Comp SepFac-C1 SepFac-C2 SepFac-C3
nC4/iC5 137.0890
C3/iC4 7103.2424
iC4/nC4 377.5571
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▪ Only an introduction to the separation-selection
process is given here.
▪ These deal with:
▪ feed and product conditions
▪ property differences
▪ characteristics of the candidate separation operations
▪ economics.

▪ The most important feed conditions are:
▪ Composition
▪ Flow rate
▪ All the other conditions
▪ temperature, pressure, and phase
▪ can be altered to fit a particular operation
▪ However:
▪ Feed vaporization
▪ Condensation of a vapor feed
▪ Compression of a vapor feed
▪ These add significant energy costs to chemical processes.
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▪ Separations using barriers or solid agent:
▪ Perform best on dilute feeds.
▪ The cost of recovering and purifying a chemical depends
strongly on:
▪ Initial Concentration of the feed.
▪ Typically:
▪ the more dilute the feed → the higher the product price.

▪ When a very pure product is required:
▪ large differences in volatility or solubility or significant numbers of stages are needed for
chemicals in commerce.
▪ For biochemicals, especially proteins, very expensive separation methods may be
required.

▪ Operations based on barriers are more expensive than operations based on the use
of a solid agent or the creation or addition of a phase.
▪ All separation equipment is limited to a maximum size.
▪ For capacities requiring a larger size:
▪ parallel units must be provided.
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▪ Except for size constraints or fabrication problems:
▪ capacity of a single unit can be doubled
▪ → for an additional investment cost of about 60%.
▪ If two parallel units are installed:
▪ the additional investment is 100%.
▪ For new processes:
▪ it is never certain that product specifications will be met.

▪ Providing multiple stages and whether parallel units may be required is an issue to
account depending on the processes:
▪ Distillation & absorption → Easy to add stages
▪ Crystallization & Drying → Not so easy
▪ Parallel Unit requirements:
▪ Distillation & Absorption → No need, use stages
▪ Crystallization & Drying → Required, no stages available
▪ Maximum equipment size is determined by:
▪ height limitations
▪ shipping constraints
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▪ The process having:
▪ lowest operating
▪ Lowest maintenance
▪ Lowest capital costs
▪ Provided it is:
▪ Controllable
▪ Safe
▪ Nonpolluting
▪ meet specifications.

▪ For each of the following binary mixtures, a separation operation is suggested.
▪ Explain why the operation will or will not be successful.
▪ (a) Separation of air into oxygen-rich and nitrogen-rich products by distillation.
▪ (b) Separation of m-xylene from p-xylene by distillation.
▪ (c) Separation of benzene and cyclohexane by distillation.
▪ (d) Separation of isopropyl alcohol and water by distillation.
▪ (e) Separation of penicillin from water in a fermentation broth by evaporation of the water.

▪ Solution of (a) Separation of air into oxygen-rich and nitrogen-rich products by
distillation.
• The normal boiling points of O2 (183C) and N2 (195.8C) are
sufficiently different that they can be separated by
distillation, but elevated pressure and cryogenic
temperatures are required.
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▪ Solution of (a) Separation of air into oxygen-rich and nitrogen-rich products by
distillation.
• At moderate to low production rates, they are usually
separated at lower cost by either adsorption or gas permeation
through a membrane.
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▪ Solution of (b) Separation of m-xylene from p-xylene by distillation.
▪ The close normal boiling points of m-xylene (139.3C) and pxylene (138.5C) make
separation by distillation impractical.
▪ However, their widely different melting points of 47.4C for m-xylene and 13.2C for
p-xylene make crystallization the separation method of choice.

▪ Solution of (c) Separation of benzene and cyclohexane by distillation.
▪ The normal boiling points of benzene (80.1C) and cyclohexane (80.7C) preclude a
practical separation by distillation.

▪ Solution of (c) Separation of benzene and cyclohexane by distillation.
• Their melting points are also close, at 5.5C for benzene and
6.5C for cyclohexane, making crystallization also impractical.
• The method of choice is to use distillation in the presence of
phenol (normal boiling point of 181.4C), which reduces the
volatility of benzene, allowing nearly pure cyclohexane to be
obtained.
• The other product, a mixture of benzene and phenol, is
readily separated in a subsequent distillation operation.
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▪ Solution of (d) Separation of isopropyl alcohol and water by distillation.
▪ The normal boiling points of:
▪ isopropyl alcohol (82.3C)
▪ water (100.0C)
▪ seem to indicate that they could be separated by distillation.
▪ However:
▪ they cannot be separated in this manner because they form a minimum-boiling azeotrope
▪ It forms at 80.4C and 1 atm of 31.7 mol% water and 68.3 mol% isopropanol.

▪ Solution of (d) Separation of isopropyl alcohol and water by distillation.
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▪ A feasible separation method is to distill the mixture in the presence of:
▪ benzene, using a two-operation process.
▪ The first step produces almost pure isopropyl alcohol and a heterogeneous
azeotrope of the three components.
▪ The azeotrope is separated into two phases, with the benzene-rich phase recycled
to the first step and the water-rich phase sent to a second step.
▪ Here, almost pure water is produced by distillation, with the other product recycled
to the first step.

▪ Solution of (e) Separation of penicillin from water in a fermentation broth by
evaporation of the water.
▪ Penicillin has a melting point of 97C:
▪ but decomposes before reaching the normal boiling point.
▪ Thus, it would seem that it could be isolated from water by evaporation of the water.

▪ However:
▪ penicillin and most other antibiotics are heat-sensitive, so a near-ambient temperature must be
maintained.
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▪ Thus:
▪ water evaporation would have to take place at impractical, high-vacuum conditions.
▪ A practical separation method is liquid– liquid extraction of the penicillin with n-butyl
acetate or n-amyl acetate.

1. Introduction
2. Overview of Separation Processes
3. Separation by Phase Creation & Phase Addition
4. Separation by Barrier
5. Solid Agent Addition
6. Separation by External Force/Gradient
7. Separations involving Solids
8. Mechanical-Physical Separations
9. Concepts in Separation Technology
10. Conclusion

1. History & Status Quo
2. ESA & MSA
3. Classification of MTO
1. Method
2. Phase
4. Equilibrium vs. Rate Based Processes

▪ Separation by Phase Addition/Creation
▪ Distillation
▪ Flash, Fractional, Batch Distillations
▪ Extractive, Salt, Reactive Distillation
▪ Gas Absorption
▪ Reboiled, Refluxed
▪ Stripping (Reboiled, Refluxed, etc.)
▪ Liquid-Liquid Extraction
▪ One or Two Solvents
▪ Drying, Humidification, Evaporation, Crystallization, Desublimation, Leaching*
*Technically Phase Addition/Creation →
“Separations involving Solids”

1. Membrane Theory
2. Industrial Processes:
1. Osmosis
2. Reverse osmosis
3. Dialysis/Electrodialysis
4. Gas permeation
5. Pervaporation
6. Microfiltration
7. Ultrafiltration/HyperFiltration
8. Liquid membranes

▪ 5.1 Adsorption Columns
▪ 5.2 Ion Exchange
▪ 5.3 Chromatography

▪ 6.0 Centrifugation (Studied in “Mechanical – Physical Separations”)
▪ 6.1 Thermal Diffusion
▪ 6.2 Electrolysis
▪ 6.3 Electro-dialysis
▪ 6.4 Electrophoresis

▪ 7.1 Leaching
▪ 7.2 Washing
▪ 7.3 Drying
▪ 7.4 Evaporation
▪ 7.5 Crystallization

▪ More Engineering Courses
▪ Process Design & Simulation
▪ Aspen Plus
▪ Aspen HYSYS
▪ Unit Operations
▪ Gas Absorption & Stripping
▪ Flash Distillation
▪ Binary Distillation
▪ Oil & Gas
▪ Petrochemical Industries
▪ Petroleum Refining

Introduction to Mass Transfer Operations (5 of 5)

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