Hugo Cloudt Advanced Separation Process Design (BPE-36806)
Summary lectures & reader Advanced Separation Process Design
(BPE-36806)
Lectures Evaporation & distillation
Lecture 1 – One single evaporator
Unit operations in a separation process can be classified into 2 types:
1. Phase separations: unit operations in a separation process in which two or more phases are
separated from each other. -> 2 types:
a. Phase separation by applying a barrier (e.g. filtration).
b. Phase separation by applying a force field (e.g. sedimentation, centrifugation).
2. Molecule separations: unit operations in a separation process in which two or more different types
of molecules are separated from each other. -> 4 types:
a. Phase addition (e.g. extraction): molecule separation which occurs due to the addition of a phase.
b. Phase generation (e.g. evaporation): molecule separation which occurs due to the generation of a
phase.
c. Molecule separation by applying a barrier (e.g. ultrafiltration).
d. Molecule separation by applying a force field (e.g. electrophoresis).
Evaporation: unit operation in which compounds are separated when heat is supplied because they
have different boiling points and therefore different volatilities.
- Molecule separation which is a phase generation, because due to the fact that heat is supplied
molecules of the liquid phase start to form a gas phase.
- Opposite of evaporation = condensation: unit operation in which compounds are separated when
heat is withdrawn because they have different boiling points and therefore different volatilities.
Main drawbacks of evaporation:
1. The high energy costs due to the necessary extensive heat supply.
2. The exposure of the molecules in the mixture to be separated to high temperatures.
Evaporation can be used for different purposes in a separation process:
1. To concentrate a product.
2. To separate a solvent from the product.
3. To fractionate a mixture of components into the individual components.
4. To dry a wet solid product.
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Ideally, a single evaporator in a separation process separates the components to be separated by
evaporation completely.
E.g. separation of ethanol from water by evaporation in a single evaporator in the ideal situation:
x = molar fraction of ethanol in the liquid phase.
y = molar fraction of ethanol in the gas phase.
1 - x = molar fraction of water in the liquid phase.
1 - y = molar fraction of water in the gas phase.
“x” is thus always used to denote the molar fraction of a component
in the liquid phase and “y” is thus always used to denote the molar
fraction of a component in the gas phase.
However, in reality a single evaporator in a separation process will never be able to separate the
components to be separated by evaporation completely, while it is imposed by thermodynamics that
always some of the less volatile component (“heavy” component) will evaporate and thus go in the
gas phase and always some of the more volatile component (“light” component) will not evaporate
and thus stay in the liquid phase.
E.g. separation of ethanol from water by evaporation in a single evaporator in reality:
Phase diagram: graph of pressure (p) as function of the temperature (T) for a pure component which
shows in which phase that pure component is at a certain combination of pressure and temperature.
The lines in the phase diagram are
equilibrium curves, because they
describe the thermodynamic
equilibrium between the two
adjacent phases/states of the
component for which the phase
diagram is made.
Special points in a phase diagram:
1. Triple point: the certain combination of pressure and temperature at which the pure component
for which the phase diagram is occurs in all three possible phases (solid, liquid and gas) at the same
time.
2. Critical point: the certain combination of pressure (critical pressure (pcrit)) and temperature (critical
temperature (Tcrit)) at which the distinction between gas and liquid becomes meaningless while at
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that certain combination of pressure and temperature (or at a combination of p > pcrit and T > Tcrit)
the phase of the mixture cannot be recognised as either gas or liquid but the phase is a sort of
combination of the gas phase and the liquid phase.
Mathematical model of one x = molar fraction of the more volatile component (“light component”) in the
single evaporator: liquid phase.
1. Schematic drawing: y = molar fraction of the more volatile component (“light component”) in the
gas phase.
1 - x = molar fraction of the less volatile component (“heavy component”) in the
liquid phase.
1 - y = molar fraction of the less volatile component (“heavy component”) in the
gas phase.
NQ = the energy which is supplied as heat (J s-1 = W)
2. Equations:
A. Mass balances:
- Mass balance over the more volatile component (“light” component (1)):
- Mass balance over the less volatile component (“heavy” component (2)):
For enthalpy (h) the liquid phase is in
B. Enthalpy balance: energy balance which takes only the ingoing
thermodynamics usually used as the
and outgoing enthalpy (h) into account.
reference state. Therefore the enthalpy
of vaporisation/evaporation heat/latent
With the following equations for the enthalpies:
heat (ΔhVAP) has to be taken into account
for calculating the enthalpy of a gas (hG)
and not for calculating the enthalpy of a
liquid (hL).
T0 = reference temperature (K) -> the standard T0 used in ΔhVAP = the heat which has to be supplied
thermodynamics is 298 K (25 °C), but you can in principle choose to get a certain component which is in
which temperature you want to use as T0, however the enthalpy of the liquid phase and has a certain
vaporisation/evaporation heat/latent heat (ΔhVAP) is usually given at temperature and pressure completely in
a T0 of 298 K (25 °C) the gas phase where it has the same
and therefore you usually have to use this T0. temperature and pressure.
C. Equations for the thermodynamic equilibria:
- Equation for the thermodynamic equilibrium between the liquid phase and the gas phase of the
more volatile component (“light” component (1)):
p = overall/total pressure. py = partial pressure
- Equation for the thermodynamic equilibrium between the liquid phase and the gas phase of the less
volatile component (“heavy” component (2)):
γL,i = activity coefficient of component i in the liquid phase. -> can be given by various models (e.g.
Margules model).
pSAT = saturation pressure/vapour pressure: pressure exerted by a gaseous component when there is
thermodynamic equilibrium for this component between being in the gas phase and in the liquid
phase (in this thermodynamic equilibrium the gas phase is saturated with the gaseous component).
- The higher pSAT of a compound, the more volatile the compound is.
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- Function of the temperature (T), which can be given by various models:
a. Clausius-Clapeyron’s model: model for the saturation pressure (pSAT) which is derived from
thermodynamic theory. -> not so convenient to use.
b. Antoine’s model: model for the saturation pressure (pSAT) which is an empirical model while it is
derived from experimental observations. -> more convenient to use and therefore more often used.
A, B and C = constants which are empirically derived for a certain compound.
The equations for the thermodynamic equilibria in a single evaporator can also be visualised in a
phase diagram specifically for the liquid-gas phase transition (evaporation) of a mixture of two
components in an evaporator:
2. Phase diagram specifically for the liquid-gas phase transition (evaporation) of a mixture of two
components in an evaporator with x or y as function of T:
E.g. for a mixture of water (boiling point = 100 °C) and methanol (boiling point = 65 °C):
In this graph for the mixture of water and methanol two special temperatures can be clearly
distinguished:
a. Bubble point/bubble temperature (Tbubble): the temperature at which in theory the first gas bubble
is formed and at which evaporation thus starts. -> Tbubble of the mixture of water and methanol = 80
°C
b. Dew point/dew temperature (Tdew): the temperature at which there is in theory only one
remaining droplet of liquid and at which evaporation thus ends because it is completed. -> Tdew of the
mixture of water and methanol = 95 °C
2. Phase diagram specifically for the liquid-gas phase transition (evaporation) of a mixture of two
components in an evaporator with T as function of x or y:
G
Dark blue line = bubble point line: line for Tbubble as function
of x.
G-L Light blue line = dew point line: line for Tdew as function of y.
L
Pure component -> boiling point: the temperature at which the pure component starts to boil and in
which the transition from the liquid phase into the gas phase starts occurring.
Mixture of components -> boiling range: the temperature range in which the mixture of components
boils and in which the transition from the liquid phase into the gas phase for the components of the
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mixture starts occurring.
> Boiling range = from bubble point (Tbubble) to dew point (Tdew)
> Start of the boiling range (the bubble point (T bubble)) is always a temperature which is higher than
the boiling point of the pure component with the lowest boiling point (in the case of the mixture of
water and methanol the start of the boiling range (the bubble point (T bubble)) is 80 °C, while the
boiling point of the pure component with the lowest boiling point (methanol) is 65 °C!).
For each mixture of two components there is a unique phase diagram specifically for the liquid-gas
phase transition (evaporation) occurring in an evaporator.
Solving the mathematical model of one single evaporator:
1. Find out if solving the mathematical model of the single evaporator is possible by comparing the
number of independent equations to the number of unknown variables:
a. Number of independent equations < number of unknown variables -> solving of the mathematical
model is NOT possible, you cannot determine all unknown variables. -> start looking for additional
equations or solve by trial & error.
b. Number of independent equations ≥ number of unknown variables -> solving of the mathematical
model is possible, you can determine all unknown variables. -> proceed as indicated below!
2. Find x from the equations for the thermodynamic equilibria in one of the following ways:
a. Manually in one of the following ways:
> By substitution
> By using the echelon method
> By first summing up both equations and subsequently solving the resulting equation:
b. With Mathcad by using “Given Find”.
3. Use the found x to find y from one of the equations for the thermodynamic equilibria.
4. Solve the mass balances to find FL and FG in one of the following ways:
a. By substitution
b. By using the echelon method
c. With Mathcad by using “Given Find”.
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Found FL:
If desired FL can be used to calculate YL (the efficiency of the separation by evaporation for the liquid
phase):
Found FG:
If desired FG can be used to calculate YG (the efficiency of the separation by evaporation for the gas
phase):
5. Find hIN, hL and hG from the equations for them:
6. Find NQ from the enthalpy balance:
Performance indicators of a unit operation in a separation process:
1. Efficiency (Y):
The given formulas in symbols
are examples of formulas for
calculating respectively the
2. Concentration factor (CF) efficiency (Y), concentration
factor (CF) and the selectivity
(Φ), in the case the unit
operation is an evaporator:
3. Selectivity (φ)
A good unit operation in a separation process has a:
1. High efficiency -> much product is recovered and not much product is lost (economically good)
2. High concentration factor -> much higher concentration of product in the outgoing stream than in
the ingoing stream of the separator (facilitates further processing)
3. High selectivity -> much higher product-solvent ratio in the outgoing stream than in the ingoing
stream of the separator
A single separator seldom scores well on all goals (high efficiency, high concentration factor, high
selectivity)!
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Efficiency whole process = efficiencies single steps of process multiplied
Concentration factor whole process = concentration factors single steps of process multiplied
Lecture 2 – Cascade of evaporators
When the boiling temperatures of the two components in a mixture are close to each other a single
evaporator or condenser is unable to achieve both a high efficiency and a high concentration factor,
to overcome this problem and thus be able to achieve both a high efficiency and a high concentration
factor by evaporation or condensation single evaporators or single condensers are coupled to each
other to form a cascade.
Single evaporation stages (evaporators) or condensation stages (condensers) can be coupled to each
other to form 2 different types of cascades:
1. Cross-current cascade: multiple individual evaporation stages (evaporators) or condensation
stages (condensers) which are coupled to each other by either the liquid flow or the gas flow. -> 2
types:
a. Cross-current evaporation cascade: cascade in which the individual stages are coupled to each
other by the liquid flow and in which in each individual stage therefore evaporation takes place (the
individual stages are therefore evaporators).
T1 < T2 < T 3
x = molar fraction of the “light” component in the
liquid phase in the evaporators.
y = molar fraction of the “light” component in the gas
phases created by the evaporators.
b. Cross-current condensation cascade: cascade in which the individual stages are coupled to each
other by the gas flow and in which in each individual stage therefore condensation takes place (the
individual stages are therefore condensers).
T1 > T2 > T3
x = molar fraction of the “light” component in the
liquid phases created by the condensers.
y = molar fraction of the “light” component in the gas
phase in the condensers.
2. Counter-current cascade: multiple individual evaporation stages (evaporators) or condensation
stages (condensers) which are coupled to each other because both the liquid flow and the gas flow
move from one stage to the other & in which the liquid flow and the gas flow move in the opposite
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direction of each other through the coupled evaporators. -> 2 types:
a. Counter-current evaporation cascade: cascade in which in the individual stages evaporation takes
place and in which the individual stages are thus evaporators.
T 1 < T2 < T 3
b. Counter-current condensation cascade: cascade in which in the individual stages condensation
takes place and in which the individual stages are thus condensers.
T1 > T 2 > T 3
Cross-current cascades and counter-current cascades are both able to achieve a higher concentration
factor than individual evaporators or condensers, however in general counter-current cascades are
also able to achieve a high efficiency while cross-current cascades have a low efficiency while they
cannot achieve a high efficiency (due to the fact that in a cross-current evaporation cascade a
relevant amount of the “heavy” component is lost in the created gas phases and that in a cross-
current condensation cascade a relevant amount of the “light” component is lost in the created
liquid phases), therefore the counter-current cascade is usually used in separation processes.
A certain individual stage i in a counter-current cascade is, just like each other individual stage in that
counter-current cascade described by the following equations:
1. Equations for the thermodynamic equilibria in the individual stage i:
- Equation for the thermodynamic equilibrium of the “light” component (1):
𝑝𝑦𝑖 = 𝑝𝑆𝐴𝑇,1 (𝑇𝑖 )𝛾1 (𝑥𝑖 )𝑥𝑖
- Equation for the thermodynamic equilibrium of the “heavy” component (2):
𝑝(1 − 𝑦𝑖 ) = 𝑝𝑆𝐴𝑇,2 (𝑇𝑖 )𝛾2 (𝑥𝑖 )(1 − 𝑥𝑖 )
2. Mass balances over the individual stage i:
- Mass balance over the “light” component in the individual stage i:
- Mass balance over the “heavy” component in the individual stage i:
𝐹𝐺,𝑖+1 (1 − 𝑦𝑖+1 ) + 𝐹𝐿,𝑖−1 (1 − 𝑥𝑖−1 ) = 𝐹𝐺,𝑖 (1 − 𝑦𝑖 ) + 𝐹𝐿,𝑖 (1 − 𝑥𝑖 )
- Overall mass balance over the individual stage i:
3. Enthalpy balance over the individual stage i:
No NQ is included into this enthalpy balance, while in a counter-current cascade the energy required
to establish the evaporation (+NQ) or the energy withdrawn to establish the condensation (-NQ) is in
theory completely delivered by respectively the hot gas flow or the cold liquid flow.
All equations describing the whole counter-current cascade (so all the equations which result from
each individual stage being described by the above mentioned set of equations) can be solved stage-
by-stage when:
1. The properties of the “light” component (1) and the “heavy” component (2) are quite similar:
When this is the case the expressions are valid and can thus safely be assumed:
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2. The gas flow rate (FG) and the liquid flow rate (FL) are assumed to be constant over the whole
counter-current cascade.
Solving the equations describing the whole counter-current cascade can be done for 2 different types
of calculations:
1. Design calculation: calculation of the number of stages (N) the counter-current cascade needs to
have to reach a certain desired performance (e.g. efficiency, concentration factor).
2. Performance calculation: calculation of the performance (e.g. efficiency, concentration factor) the
counter-current cascade has when it has a certain number of stages (N).
Design calculation
A. The T vs (x,y) diagram describing the thermodynamic equilibrium between the “light” component
in the gas phase and the “light” component in the liquid phase is converted to a y vs x diagram
describing this thermodynamic equilibrium (the equilibrium curve):
B. Solving the equations of each individual stage is started at the last stage of the counter-current
cascade (stage N) and is done by following the algorithm below:
a. Finding TN (temperature of stage N) by trial & error from the equation resulting from summing up
the equations describing the thermodynamic equilibria in stage N.
b. Finding yN from one of the equations describing the thermodynamic equilibria in stage N.
c. Finding xN-1 (the molar fraction of the “light” component in the liquid entering stage N and leaving
stage N-1) from one of the mass balances over stage N.
d. Use the found xN-1 to solve the equations of the next stage in the counter-current cascade (N-1
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stage) by taking the same steps as described above for stage N, so:
> Find TN-1 by trial & error from the equation resulting from summing up the thermodynamic
equilibria in stage N-1.
> Find yN-1 from one of the equations describing the thermodynamic equilibria in the stage.
> Find xN-2 from one of the mass balances over the stage.
e. Solve the equations of each stage which is next in line thus by:
> Find T of the stage by trial & error from the equation resulting from summing up the
thermodynamic equilibria in the stage.
> Find y in the gas flow leaving the stage from one of the equations describing the thermodynamic
equilibria in the stage.
> Find x in the liquid flow entering the stage from one of the mass balances over the stage.
f. Keep repeating the steps above until:
> The found x is lower then or close to x0.
> The drawn line in the y vs x diagram (the operating line) is drawn until approximately x0 and
therefore ends in approximately x0.
g. Count the number of stages from:
> The calculation by counting the number of times the steps for solving the equations of a certain
stage is repeated.
> The y vs x diagram by counting the number of times a step from the operating line to the
equilibrium curve to the operating line (step from the operating line to the equilibrium curve to the
operating line = 1 stage) is taken.
The smaller the gap between the operating line and the equilibrium curve, the larger the number of
stages required to achieve a certain extent of separation. -> the gap between the operating line and
the equilibrium curve depends on the value of FG:
The lower FG, the smaller the gap between the operating line and the equilibrium curve, the larger
