Thermodynamics of separation processes 3.1 ... - Åbo Akademi

Process EngineeringThermodynamics
course # 424304.0 v. 2013
Thermodynamics of separation
processes
Ron Zevenhoven
Åbo Akademi University
Thermal and Flow Engineering Laboratory / Värme- och strömningsteknik
tel. 3223 ; ron.zevenhoven@abo.fi
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 1/82
3.1 Separation and entropy
ÅA 424304
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Separation process and entropy /1
Separation processes involve de-mixing of species or phases,
in thermodynamic sense resulting in a decrease of the total
entropy of these
∑ Entropy of components or phases < Entropy of mix
One way of acccomplishing a separation is to create more
phases; Gibbs’ phase rule gives for the degrees of freedom,
f, for a system of p phases and n components: f = n + 2 – p.
f↑ gives ∆S↑; f↓ gives ∆S↓.
A new phase can be created by 1) adding a new component
(absorption/desorption, extraction), or 2) by adding or removing
energy (e.g. heat) (distillation, crystallisation)
Creating an ”un-equilibrium” gives rise to a separation
(or a chemical reaction) if ∆G = ∆H-T·∆S < 0, or ∆S >∆H/T
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Separation process and entropy /2
m mix
s mix
Q out
T out
Q in
T in
m A
s A
m B
s B
Steady - state mass balance, 1st Law, 2nd Law :
mmix
m m A B
m h
mix mix Qin
m h
m h
Q
A A B B out
Qout
m s m s
A A B B
Tout
Qin
m s mix mix
Tin
Q Q
out in m s mix mix
T T
m s m s
A A B B
out
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in
m mix
s mix
Qout Tout Distillation Crystallisation
m liquid
s liquid
m crystals
s crystals
4/82

Separation process and entropy /3
Clearly, some entropy must be produced to
1) compensate for ∆S < 0 resulting from the separation, and
2) create some driving force ∆G < 0, ∆S > ∆H/T.
Proper engineering requires minimisation of this entropy
production ∆S and the exergy losses T°·∆S, and balancing
this against economical considerations
It was found that a uniform production of entropy is
preferable: the equipartitioning principle (TK87, B97,
SAKS04) – see also macroscopic organisation in natural processes!
For example, for a heat exchanger
this implies that the driving force
∆(1/T) = constant.
It can be extended to considering
stable, non-equilibrium states S > Smin, with stability criterion dṠgen/dt < 0.
Stable, unstable,
neutral equilibrium
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Separation process and entropy /4
Example taken from
TK87
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Pic: http://www.diracdelta.co.uk/science/source/e/q/equilibrium/source.html

Separation process and entropy /5
µ = chemical potential
Examples taken from
TK87
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3.2 Binary Distillation I
(the McCabe-Thiele procedure)
ÅA 424304
Assumed pre-knowledge: Continuous distillation
see course 424302 Massöverföring & separationsteknik, #12
http://users.abo.fi/rzevenho/MOFST12-OH12.pdf
or re-wrap course 424102 Principles of process engineering
(which summarises courses 424101, 424302 and 424300)
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From MÖF-ST
ÅA 424302
From MÖF-ST
ÅA 424302
Continuous distillation (binary) /1
Separating a mixture of
2 components A and
B, with A more volatile
Liquid for absorption
section produced by
condensing some top
product
Gas for stripping
section produced by
boiling some bottom
product
Picture: WK92
Top product:
more volatile
component
Top section:
rectifying section
absorbing the less
volatile component
Bottom section:
stripping section
desorbing the more
volatile component
Bottom product:
less volatile
component
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Mass balance and equilibrium:
absorption
y in
y out
slope L/V
x in
x out
y i = Kx i
mass balance: V·y i+L·x i= V·y i+1 + L·x i-1
→ working line y i+1 – y i = (L/V)·(x i-1 –x i)
V
y out
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y i
y i+1
V
i
x i-1
x i
L
V
y in
L
x in
L
x out

From MÖF-ST
ÅA 424302
From MÖF-ST
ÅA 424302
Mass balance and equilibrium:
desorption / stripping
y out
y in
slope L/V
x out
y i = Kx i
x in
mass balance: V·y i+L·x i= V·y i+1 + L·x i-1
→ working line y i+1 – y i = (L/V)·(x i-1 –x i)
V
y out
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y i
y i+1
Continuous distillation (binary) /2
Roughly equimolar
exchange:
1 mol A liquid → gas
«-» 1mol B gas → liquid
As a result: Gas and
liquid streams ~ constant
in each section:
L and V in top section,
L´ and V´ in bottom
section
Feed enters there where
it is similar to the mixture
inside
Picture: WK92
V
i
x i-1
x i
L
V
y in
L
x in
L
x out
Top product:
more volatile
component
Top section:
rectifying section
absorbing the less
volatile component
Bottom section:
stripping section
desorbing the more
volatile component
Bottom product:
less volatile
component
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From MÖF-ST
ÅA 424302
From MÖF-ST
ÅA 424302
Continuous distillation /3
Some general considerations and assumptions:
– The equimolar exchange A ↔ B requires balanced energy exchange
as well: vaporisation heats of A and B should be roughly the same
– Pressure and the temperature ranges must be chosen:
• Below the critical pressures and temperatures of the components
• The components must be thermally stable (for hydrocarbon fractions
!
of crude oil: below 300 ~ 350°C)
• The temperature at the top of the column should preferably be > 40°C,
allowing for heat transfer with surrounding air or water
– Equilibrium data is needed, + data for feed and products
specifications
→ The necessary gas and liquid streams, the number of stages and
heat consumption can be calculated
A simple procedure based on local equilibrium and mass balances
was suggested by McCabe and Thiele
Equilbrium
data
Example:
C 3 –iC 4 /1
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pressures < critical
pressures allowable
for C 3 –iC 4 : p < 37 bar
37 bar boiling points
85 - 140°C: stability OK
Raoults law:
y·p total = x·p°
K = y/x = p°/p total
T boil iC 4
at 37 bar
T boil C 3
at 37 bar
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From MÖF-ST
ÅA 424302
Equilbrium
data
Example:
C 3 –iC 4 /2
Chose 40°C as the
temperatur at the
condensor.
At T min = 40°C, the
vapour pressure of
the more volatile
component which is C 3
= 14 bar → p ≥ 14 bar
Then T boil ~ 81°C for iC 4
p° C 3
at 40°C
T boil iC 4
at 14 bar
T boil C 3
at 14 bar
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Binary distillation, McCabe-Thiele /1
The McCabe-Thiele x,y diagram
procedure for continuous binary
distillation is limited to cases where the
vapour and liquid mass flows are
≈ constant in both the top and bottom
section of the column.
This requires that the vaporisation
/ condensation heats for the two
components are ≈ the same, and that
mixing heat effects are negligible.
For example EtOH/H 2O: ∆ vaph(EtOH) ≈
∆ vaph(H 2O) ≈ 40 kJ/mol
elementary courses (e.g. ÅA424302, Z12)
Pic: WK92
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From MÖF-ST
ÅA 424302
Binary distillation, McCabe-Thiele /2
q = fraction of feed that
gives liquid on the feeding
tray: L´= L + q· F,
F· q = ∆L and F· (1-q) = ∆V
q is related to the energy
needed to convert the feed
completely into vapour. With
enthalpies H for saturated
liquid and saturated gas it is
found that
q
H
H
G,
sat
G,
sat
H
H
L,
sat
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F
q >1
q = 1
0 < q < 1
q = 0
q < 0
Picture:
after T68
Binary distillation, McCabe-Thiele /3
The McCabe-Thiele x,y
diagram procedure is thus
based on mass balances
(giving straight operating
lines) and chemical
equilibrium data
If this method cannot be
used, a graphical method
based on enthalpy –
composition (h,x or h,ξ)
diagrams as given by
Ponchon – de Savarit can
be used, based on local
equilibrium, mass balances
and energy balances
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17/82
elementary courses (e.g. ÅA424302, Z12
Pic: http://commons.wikimedia.org/wiki/File:McCabe-Thiele_diagram.png

From MÖF-ST
ÅA 424302
Restrictions McCabe-Thiele method
Molar heats of vaporisation should
not differ more than ~10%
Heats of dissolution or mixing are
negligible
Relative volatilities should be
1.3 < α < 5
Reflux ratio’s (L/V) should
be R > 1.1· R min
Number of trays N

Mixing heat for binary mixtures
ξ kg B
1-ξ kg A
1 kg mix
For this mixing process:
h = ξ·ha + (1-ξ)·hb + qmix with mixing heat qmix to keep temperature
constant
Mixing heat data can be
used to produce h,x or h,ξ
diagrams
ξ = mass fraction for the
most volatile compound
x = molar fraction for the
most volatile compound
With molar masses M a
and M b :
x
ξ Mb
ξ M
ξ
x Ma
x M
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Mixing heat, h,ξ diagram /1
NH 3-H 2O
↑ Mixing heat data for NH3-H2O and for EtOH-H2O (liq)
Schematic h,x or h,ξ diagram → for
three temperatures (t1, t2, t3) EtOH-
H 2O
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a
b
;
Pics: B01

Mixing heat, h,ξ diagram /2
Determination of heat of mixing, q mix, versus composition, ξ:
a boiling liquid mixture (mass m, composition ξ f, enthalpy h f) is fed
into a calorimeter; a small amount of heat dq gives a small
amount of vapour, dm (composition ξ d, enthalpy h d)
Mass balance, heat balance:
gives (eliminate m) q mix= dq/dm =
Pic: B01
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Mixing heat, h,ξ diagram /3
liquid
gas
q mix
Construction of condensation line from
boiling line, heat of vaporisation and heat of
mixing
For point P, which demixes into liquid F
and gas D, the relative amounts are
G/L = FP / PD (lever rule)
Schematic h,ξ diagram:
- Condensation line c
- Boiling line b
- Isotherms i
Pics: B01
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c
i
i
b

s,ξ diagram
For 2nd Law analysis, exergy analysis or for tracking down
irreversibilities in mixture applications (for example
mixtures of refrigerants) s,ξ or s,x diagrams may be used.
Note that during adiabatic mixing hmix = ha+ hb, but
smix = sa + sb+ ∆smix s,x and
∆s mix for
two ideal
gases
x
Pics: B01
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3.4 Flash vaporisation
x
ÅA 424304
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Flash vaporisation, h,x diagram /1
Consider a flash vaporisation process, in which a binary
liquid at temperature T0, pressure p0 is brought to a lower
pressure p1 using a throttling valve. Temperature drops to
T1. Part of the liquid vaporises, giving a vapour that is more
rich in the more volatile component.
Mass balance: F = G + L
Mass balance volatile
component:
xF·F = y·G + x· L
Heat balance:
hF· F = hG· G + hL· L, with
heat of vaporisation
∆vaph = hG-hL at T0. Pic: Z87
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Flash vaporisation, h,x diagram /2
Initially, at p0 the feed
point A with x = xF is
in the liquid region of
the h,x diagram
After the pressure
reduction point A lies in
the 2-phase region,
giving liquid (L) C + gas
(G) B
Amount ratio G/L
equals = CA /AB, or
G
L
x x F
y x
F
h
h
F
G
hL
h
F
G
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L
Pic: Z87
28/82

Flash vaporisation, example /1
220°F = 104°C
170°F = 77°C
K71- Fig 2.11
K71- Fig 2.12
1 BTU = 1055,06 J 100 BTU/lb = 2,326 kJ/kg
Pics: K71
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Flash vaporisation, example /2
v
w
K71- Fig 2.11
K71- Fig 2.12
v w
Lever rule – see for example:
http://www.doitpoms.ac.uk/tlplib/phase-diagrams/lever.php
Pics: K71
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3.5 Binary Distillation II
(the Ponchon-Savarit method)
ÅA 424304
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Ponchon-Savarit method /1
Continuous distillation
Feed F(kg/s), xf (kg/kg)
Top product D, xd Bottom product W, xw Upward vapour flows V
(kg/s), downward liquid
flows L (kg/s)
Liquid mass fractions x,
vapour mass fractions y
Reboiler heat input QB (kW), condenser heat
output QC (kW)
Pic: CR83
Top section I, index n (downwards)
Bottom section II, index m (downwards)
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Ponchon-Savarit method /2
HL and HV are enthalpies of
liquid and vapour streams
(kJ/kg)
Mass balances top section
stages (total, and for the most
volatile species of the two)
*) Pic: CR83
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Ponchon-Savarit method /3
Heat balances (top section)
H L d = H of liquid D (x = x d)
Constant H´ d = H L d + Q C/D:
**)
Pic: CR83
”leaving” system via top:
D·HL d + QC Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 34/82

From the mass balances *)
From mass balance *) and heat
balances **)
Ponchon-Savarit method /4
x
Pic: CR83
Functions yn(xn-1) are all lines
that go through point (pole) N =
(xd, H´ d ) in the (x,H) plot
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Ponchon-Savarit method /5
The lines y n(x n-1) can be put
in the h,x or h,ξ diagram
(see next slide)
Similarly, mass and heat
balances for the bottom
section:
Pic: CR83
Top section I, index n (downwards)
Bottom section II, index m (downwards)
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”leaving”
system via
bottom:
W·H L w -Q B
Ponchon-Savarit method /6
HL w = H of liquid W
(x = xw) Constant
H´ w = HL w -QB/W This then gives
Note: this implies q = 1
Functions y m(x m-1) , all lines
go through point (pole)
M = (x w, H´ w ) in (x,H) plot
Pic: CR83
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Ponchon-Savarit method /7
A stream N can
be visualised with
composition = x d ,
enthalpy = H´ d ,
mass = V n –L n-1
Similarly, M can be
visualised as a
stream with
composition = x w ,
enthalpy = H´ w ,
mass = L m-1 –V m
Then also: F = M + N, and F·x F = M·x w + N·x d
Pic: CR83
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Ponchon-Savarit method /8
The necessary
number of
theoretical
stages can be
determined
graphically as
shown
Feed point F is shown
on the boiling line
(here, q =1)
V1 is vapour from first
plate, L1 is liquid from
first plate to second
plate, et c.
Here, 7 theoretical stages needed.
Pic: CR83
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Ponchon-Savarit method /9
The heat removed
in the condenser
per unit mass D is
q c = Q c/D.
Lowering N N´
means that more
stages are needed
At N Nm the
number of stages
∞ Here, 7 theoretical stages needed.
Pic: CR83
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Ponchon-Savarit method /10
Using q c = Q c/D and
the minimum heat
requirements can be found:
Increasing the reflux ratio
requires more heat to be
removed, point N
upwards, reducing the
number of stages
Pic: CR83
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Ponchon-Savarit method Example /1
1 kg/s
x f = 0.3
x d = 0.995
R =
1.08·R min
x w = 0.100
Mass balances
Source: CR83
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Source:
CR83
Ponchon-
Savarit
method
Example /2
h, ξ diagram
NH 3-H 2O
1.013 MPa =
10 atm
Source: CR83
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Ponchon-Savarit method Example /3
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3.6 Ideal distillation column
analysis
ÅA 424304
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Analysis of ideal column /1
Exergy analysis shows that
the minimum work for
separation of 1 mol of
mixture equals
wmin = ∆ex = T°·∆s
For distillation with
reboiler heat Qin and
condenser heat Qout, the
exergy loss –∆Ex can be
approximated as
o
1
Ex
Qin
T
Ttop
T
(if H F ≈ H B + H D !)
bottom
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1
Thus, the separation is
accomplished by
degrading Q in at T in
to Q out (= Q in) at T out < T in
Pic: SAKS04
46/82

Analysis of ideal column /2
The Clausius Clapeyron expression
for phase transitions
based on the slope of the
two-phase co-existence
curve in a p-T diagram,
dp/dT = ∆S/∆V ~ ∆vapH/(T∆V) gives for ideal gases:
p
vapH
dT p
ln
2 R T p
sat,
2
R
dp sat,
1 vap
sat
H 1
T
Also, for an ideal mixture the relative
volatility for two similar* components is
* Chemically similar, boiling points not
too far apart, α ~ O(1)
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1
1
T2
12
Analysis of ideal column /3
This then gives, for the distillation
temperature range (T top, T bottom):
ln
12
R
vap
H
1 1
Tbottom
T
top
p
p
sat
1
sat
2
Pic: SAKS04
which if α ~ 1 (and using
ln(1+z) ≈ z if z «1) simplifies to: Pic: SAKS04
12
1
R
vap
H
1 1
Tbottom
T
top
47/82
This requires that
- heat Q in is supplied using
an Carnot heat pump
-heat Q out is converted into
work in a Carnot process,
- minimum reflux is used
48/82

Analysis of ideal column /4
Assume a propane C3, propene C =
3
distillation with xF = ½, q = 1,
B/F = D/F = ½ (mol/s)/(mol/s)
Note: for this equimolar mix,
∆smix is at its maximum (if ideal)
Then per mole of feed, L/F = r· D/F
= ½·r, (r = reflux ratio* = L/D) above the
feed, L/F = (½·r + 1) below the feed, and
V/F = ½·(r + 1) in the whole column.
Then the heat input at the reboiler is equal
to
Qin/F = V/F·∆vapH = ½· (r+1)·∆vapH ≈ Qout/F 7 feb 13 Åbo Akademi Univ - Thermal and Flow Engineering
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Analysis of ideal column /5
The minimum work required
to separate the mixture can be
found by considering a heat
pump between T out and T in
– see Figure – and exergy
analysis gives (per mole feed):
w
R T
ln2
for an equimolar mixture
This can be used to define the
minimum reflux ratio rmin: W
sep
sep
F
min
in
Q
F
o
R
T
x
o
o
R
T
x
½
ln x
2ln2
ln
1
T
r1Hr 1
min
i
i
i
i
ln x
i
vap
i
W
F
min
min
min
in
Q
F
T
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Piispankatu 8, 20500 Turku
12
o
top
T
1
bottom
(α < 4 !)
Pic: SAKS04
* Using ” r ” to
distinguish it
from gas
constant R.
49/82
Pic:
SAKS04
100%
efficient!
(reversible)
50/82

3.7 Real distillation column
analysis I: efficiency
ÅA 424304
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Analysis of real column /1
What will be the efficiency of
a real column for this case?
Heat transfer requires
temperature differences!
Qin is supplied at 377K,
Qout goes out at 298K
The minimum heat
input for the separation is
Q min in / F = ½·(r min+1)·∆ vapH with
r min = 9.64 (α 12 ≈ 1.14), per mole feed
Deviations from ideal thermodynamics
r real = 15.9 (α 12 ≈ 1.09)
Q real in / F = ½ ·(r real+1) · ∆ vapH
r real/r min = 1.65 Pic: SAKS04
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Analysis of real column /2
Bottom section ∆T = 46 K, top section ∆T = 22 K
The total thermodyn. eff. of the column is calculated as:
which gives η overall = 0.093 = 9.3%
The main sources of the
inefficiency are the driving forces
Δ(1/T) in reboiler and condenser
The ratio Q min
in /Qin
real = 0.63 reflects
exergy losses inside the column
Pic: SAKS04
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Analysis of real column /3
The amount of exergy (work) lost in the column + the
exergy used for the separation = the difference between the
exergies of the reboiler heat and the condenser heat
For the minimum work (exergy) needed it was shown that
so that the work (exergy) lost
in the column is equal to
see Table on next slide
Pic: SAKS04
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Analysis of real column /4
Efficiency of the distillation column: summary
Pic: SAKS04
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 55/82
Analysis of real column /5
Efficiency of the distillation column: summary
Reboiler
Column
Condenser
Previous table, rated by normalising
with
real
in
7 feb 13 Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
Q
T
1
T
0
R
Pic: SAKS04
56/82

Binary distillation
efficiency: overview
Source: OFRGJ03
Maximum efficiency:
where x = mole fraction of
the more volatile component
Reboiler heat duty:
where r = reflux ratio, D
= distillate product flow
For Q REB ≈ Q COND:
Reboiler duty exergy:
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 57/82
3.8 Real distillation column
analysis II: exergy analysis
ÅA 424304
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 58/82

Exergy analysis /1
An analysis can also be made based on exergy flows:
Note that still no enthalpy
of mixing is considered,
and ideal mixing
entropy -R·∑x·lnx.
Flowsheeting
programs (should)
give values for H’s, S’s,
and Q’s, making it easy to
calculate exergy streams
No chemistry physical exergy is sufficient !
Pic: SAKS04
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Exergy analysis /2
Exergy balance for the equimolar C 3/C 3 = distillation
* reversible case:
* real case
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 60/82

Exergy analysis /3
Exergy streams for ideal and real case: summary
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 61/82
Exergy analysis /4
The main difference between ideal and real
performances are Ex(Q1) and Ex(Q2); note that Ex(Q2) = 0 in the real case.
The difference between the two cases as given on the
previous two slides is
+
or, normalised per unit input
exergy:
(gives again value 0.907 efficiency 0.093)
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 62/82
= 0
Pic: SAKS04

Exergy analysis /5
The analysis showed that the main losses were at the
reboiler and the condenser, due to heat transfer over
large temperature differences ∆T
Instead of trying to
reduce these ∆T’s,
waste heat from
another process unit
may be available
Heat integration,
which at the same
time saves steam
(for example)
Pic: SAKS04
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Example: H 2O/EtOH distillation /1
A 5%/95% (w/w) EtOH/H2O mix is to be distilled into a 90% EtOH
top product and pure water bottom product
Feed (18 kg/s) and bottom product (17 kg/s) exchange heat: feed is
heated 20 70°C; bottom product cools 100 48.3°C; the heat
exchanged is 3690 kJ/kg top product (1 kg/s)
The reflux cooling heat qr = 2640 kJ/kg top product; the heat
reboiler qb = 5620 kJ/kg top product
Tempertures q r = 79°C; q b = 104°C
m·c p (top) = 1·2.6 kJ/K, m·∆ vapH (top) = 980
kJ
m·c p (bottom) = 17·4.2 kJ/K
Source: B01, H84
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 64/82

Example: H 2O/EtOH distillation /2
The minimum work needed for making a 90% EtOH mix from a
5% EtOH mix: see the mass balance in moles.
The mole fractions for water are 0.9798 for the feed, 0.2212 for
the distillate and 1.00 for the bottom product; this then gives with
the molar amounts as indicated:
See section 1.6 of
Exergy analysis
for the minimum work per kg top product.
Source: B01, H84
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 65/82
Example: H 2O/EtOH distillation /3
Assuming surroundings temperature T° = 293K an exergy analysis
can be made (exergy flows per kg top product):
Reflux cooler
Boiling top product
Top product heat-up
Bottom product
Bottom product after heat exchange
Source: B01, H84
.
.
.
980 kJ/kg for
10% H 2O + 90%
EtOH
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 66/82

Example: H 2O/EtOH distillation /4
Assume that the bottom heat q b is delivered by an ideal (Carnot)
heat pump: 5620· (1-293/377) = 1252 kJ/kg top product exergy
(Note the difference 5620 1252!)
This goes to the cold bottom product (7%), top product (14%),
reflux cooler (35%) and separation work (4%), total 60%.
Thus, 1252 - 896 = 356 kJ is lost as irreversibilities.
For the heat exchanger, the losses per kg top product are
which means that 70 kJ/kg top product is lost in the column itself;
which is ~ 40% of the minimum separation work
(186 kJ/kg top product.) Source: B01, H84
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 67/82
3.9 Distillation + heat pump and
other improvements
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One option: feed stream splitting
Also known as hybrid distillation
A membrane is used to split the feed into two
streams, fed into the column at different locations
Part of the separation is done (less irreversible) by the
membrane lower reflux ratio and/or less stages
Pics: SAKS04
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A few other options
Some GENERAL considerations for improving
distillation column efficiency:
Process control
Feed pre-heat
Reduce pressure
drop by optimising
the trays (or packing)
Separate to lower
purity and combine
Pic, Source: SAKS04
with another separation process
Use intermediate reboilers and condensers, bringing
operating lines closer to the equilibrium line
(but this may require more stages or trays)
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Double columns
Pic, Source: SAKS0
Paraffin /olefin separations are very energy intensive.
One example is propane / propylene, or C3 (atm. b.p. - 42.1°C)
/ C =
3 (atm. b.p. - 47.7°C). For a single-column tray column
distillation process
150-200 trays would be
needed, giving a very
high (> 100 m)
column, with reflux
condensing using air or water
A double-column process
is preferred, with smaller (shorter)
columns, using hot and cold
water for condensing (column 2) and reboiling (column 1).
Note that reboiler for column 2 = condenser for column 1.
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 71/82
Distillation column + heat pump /1
Improved efficiency can be obtained by using a heat
pump: the condenser is removed and the reflux is
obtained after heat exchange in the reboiler and with
the feed, and throttling
Drawback: complexity costs...
Source: B01, H84
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 72/82

Distillation column + heat pump /2
Returning to the EtOH/water distillation (see the end of the
previous section): the top product is compressed so that its
temperature rises from 79 to 104°C (which implies p2 = 2 bar)
It can then give off heat to the water in the bottom of the column
at 100°C. See points 1,2,3,4 in the h,ξ diagram below.
The h, ξ diagram gives masses δ:φ:1 = (*-4):(*-1):(1-4) = 3.7:2.7:1
a.k.a.
Mechanical
Vapour
Recompression
(MVR)
Source: B01, H84
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 73/82
*
pole
point
Distillation column + heat pump /3
For the compressor work wc = δ·(h2-h1), and for the bottom heat
exchanger ∆qb = δ·(h3-h2) which is then found to be 3580 kJ per
kg top product. This is 64% of the conventional process value as
in the previous section.
Note compressor losses; and that stream 3 may be used to
preheat the top product before the compressor.
Source: B01, H84
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 74/82
*
pole
point

Two-feed distillation
Two-feed, or two-enthalpy feed can
Allow for lowering the number of stages, N (same L/D, x D, x B)
Improve the separation, better XD, XB (same L/D, N)
Allow for lowering reflux ratio r = L/D (same N, xD, xB) 50% energy savings possible, but modifications are needed
Source: D04, WP93
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Adiabatic versus diabatic distillation
In conventional distillation, heat is only
added/removed at reboiler/condensor;
no heat is added/removed on the trays:
adiabatic operation
In diabatic operation, heat can
be added/removed by heat
exchange, in principle on each tray
Heat is removed from each tray above
feed, added to each tray below feed.
Reboiler much smaller or not needed.
More complicated (‹-› process control),
but more efficient: lower entropy
production (decreases with increasing
heat exchange area)
Source: KBJG10, J10
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 76/82

HIDiC Heat integrated distillation column
Rectification (”top”) section is operated at a higher
pressure (= pressure of conventional column) than the
stripping (”bottom”) section
Heat is transferred from rectification to stripping section
Comparison for C 3 / C 3 = splitter
Source:
J10, OFRGJ03
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 77/82
Final remarks
Distillation accounts for ~3% of world energy consumption
Heat integration can give significant energy efficiency
improvements (but increased costs, more process control)
Fully diabatic operation of a distillation column won’t be
practically, not all temperature levels will be available
Most important are the heat exchanger at the lowest
and highest trays, these can be integrated using for
example a heat pump.
Maybe integration with other columns is possible
Further improvement can be obtained by
varying column cross sectional area:
in a diabatic column rectifier section the
vapour flow decreases upwards, in the
stripper section it is the opposite
Source: KBJG10, J10
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 78/82

Sources
B01: Bart, G.C.J. Advanced thermodynamics (in Dutch) course compendium Delft Univ. of
Technol., Delft (2001)
B97: Bejan, A. Advanced engineering thermodynamics John Wiley & Sons (1997) Chapter 3
CRBH83 J.M. Coulson, J.F. Richardson , J.R. Blackhurst, J.H. Harker ”Chemical engineering”,
vol. 2, 3rd ed. Pergamon Press (1983) Chapter 11
D04: Demirel, D. ”Thermodynamic analysis of separation systems” Separation Sci. &
Technol. 39 (16) 3897-3942 (2004)
H84: Hoogendoorn, C.J. Advanced thermodynamics (in Dutch) course compendium Delft
Univ. of Technol., Delft (1984)
HS92: Humphrey, J.L, Siebert, A.F. ”Separation technologies: an opportunity for energy
savings” Chem Eng Prog 88 (March 1992) 32-41 (1992)
J10: Jana, A.K., ”Heat integrated distillation operation” Appl. Energy 87, 1477-1494 (2010)
K71 King, C.J. ”Separation processes” McGraw-Hill (1971)
KBJG10: Kjelstrup, S., Bedeaux, D., Johanessen, E., Gross, J. ”Non-equilibrium
thermodynamics for engineers”, World Scientific (2010)
OFRGJ03: Olujic, Z, Fakhri, F., de Graauw, J., Jansens, P.J. ”Internal heat integration – the
key to an energy conserving distillation column” J. Chem Technol Biotechnol 78, 241-248
(2004)
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 79/82
Sources cont’d
SAKS04: de Swaan Arons, J., van der Kooi, H., Sankaranarayanan, K. ”Efficiency and
sustainability in the efficiency and chemical industries.” Marcel Dekker, New York (NY)
2004
SH06 J.D. Seader, E.J Henley ”Separation process principles” John Wiley, 2nd edition (2006)
chapter 17
TK87: Tondeur, D., Kvaalen, E. ”Equipartition of entropy production. An optimality
criterion for transfer and separation processes” Ind. Eng. Chem. Res. 87 (26) 50-56 (1987)
T68 R.E. Treybal ”Mass transfer operations” McGraw-Hill 2nd edition (1968)
WK92 J.A. Wesselingh, H.H. Kleizen ”Separation processes” (in Dutch: Scheidingsprocessen)
Delft University Press (1992)
WP93: Wankat, P.C., Kessler, D.P. ”Two-feed distillation: same composition feeds with
different enthalpies” Ind. Eng. Chem. Res. 32, 3061-3067 (1993)
Z87 F. Zuiderweg ”Physical separation methods” (in Dutch: Fysische Scheidingsmethoden) TU
Delft (1987) (vol. 1)
Z12: Zevenhoven, R. ”Massöverföring & separationsteknik” / ”Mass transfer & separation
technology” course material 424302 (in English !) Åbo Akademi Univ. (2012) Chapter 12
ÖS97 G. Öhman, H. Saxén ”Transportprocesser” (in Swedish) Åbo Akademi Värmeteknik
(1997) 3rd ed., Chapter 3
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7 feb 13 80/82

100 PSIA =
6.8 atm
1 BTU/lb
mole =
2.326 J/mol
HL, HG kJ/mol
80
70
60
50
40
30
20
10
Appendix:
h,ξ & x,y
diagram
NH 3 –H 2O
6.8 atm
ÅA 424304
Source:
http://www.chbmeng.ohio-state.edu/~koelling/523/Appendix_d17.pdf
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7.2.2013 81
Appendix: h,ξ diagram n-C 6 / n-C 8
n-C 6 - n-C 8 101.3 kPa
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
x,y, -
1
HL, HG kJ/mol
120
110
100
90
80
70
60
50
n-C 6 - n-C 8 101.3 kPa
40
30
20
10
0
-10
-20
-30
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x,y, -
ÅA 424304
T °C x - HL kJ/mol y - HG kJ/mol
68.7 1 12.98 1 41.87
69.35 0.9772 13.22 0.9963 42.03
70 0.9546 13.45 0.9923 42.18
75 0.8017 15.19 0.9609 43.39
80 0.6684 16.94 0.9237 44.7
85 0.5566 18.65 0.8788 46.14
90 0.4566 20.36 0.8235 47.74
95 0.3705 22.05 0.7572 49.53
100 0.2917 23.75 0.6756 51.57
105 0.2221 25.44 0.5842 53.8
110 0.1591 27.13 0.4754 56.33
115 0.1035 28.8 0.3525 59.14
120 0.0538 30.46 0.2088 62.33
122.9 0.0262 31.43 0.1096 64.46
125.8 0 32.39 0 66.78
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 7.2.2013 82