Thermodynamics of separation processes - Åbo Akademiweb.abo.fi/~rzevenho/PET15-3-SEP.pdf ·...

28 jan 15Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 1/82

Thermodynamics of separation processes

Ron ZevenhovenÅbo Akademi University

Thermal and Flow Engineering Laboratory / Värme- och strömningstekniktel. 3223 ; [email protected]

Process EngineeringThermodynamicscourse # 424304.0 v. 2015

ÅA 424304


3.1 Separation and entropy


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 ΔS↑; f↓ gives ΔS↓.

A new phase can be created by 1) adding a new component(absorption/desorption, extraction), or 2) by adding or removingenergy (e.g. heat) (distillation, crystallisation)

Creating an ”un-equilibrium” gives rise to a separation (or a chemical reaction) if ΔG = ΔH-T·ΔS < 0, or ΔS >ΔH/T

28 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku

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mmix

smix

Qin

Tin

Qout

Tout

mA

sA

mB

sB

BBAAmixmix

in

in

out

out

mixmix

in

in

BBAA

out

out

outBBAAinmixmix

BAmix

smsmsmTQ

TQ

smTQ

smsmTQ

QhmhmQhm

mmm

:Law 2nd Law, 1st balance,mass state-Steady

Qout

Tout

mmix

smix

mliquid

sliquid

mcrystals

scrystals

Distillation Crystallisation



Clearly, some entropy must be produced to 1) compensate for ΔS < 0 resulting from the separation, and 2) create some driving force ΔG < 0, ΔS > ΔH/T.

Proper engineering requires minimisation of this entropyproduction ΔS and the exergy losses T°·ΔS, and balancingthis 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 exchangerthis implies that the driving forceΔ(1/T) = constant.

It can be extended to consideringstable, non-equilibrium states S > Smin, with stability criterion dṠgen/dt < 0.

Stable, unstable,neutral equilibrium

Pic

: ht

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Example taken from TK87



Examples taken from TK87


µ = chemical potential

ÅA 424304


3.2 Binary Distillation I(the McCabe-Thiele procedure)

Assumed pre-knowledge: Continuous distillationsee course 424302 Massöverföring & separationsteknik, #12http://users.abo.fi/rzevenho/MOFST12-OH12.pdfor re-wrap course 424102 Principles of process engineering(which summarises courses 424101, 424302 and 424300)


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

Top section:rectifying sectionabsorbing the less volatile component

Bottom section:stripping sectiondesorbing the more volatile component

Bottom product:less volatile component

Top product: more volatile component

Picture: WK92

Fro

m M

ÖF

-ST

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Mass balance and equilibrium: absorption

yi = Kxi

Lxin

Lxout

Vyout

Vyinxoutxin

yout

yin

i

yi+1

yi

xi

xi-1

LV

mass balance: V·yi+L·xi= V∙yi+1 + L∙xi-1

→ working line yi+1 – yi = (L/V)·(xi-1 – xi)

slope L/V

Fro

m M

ÖF

-ST

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Mass balance and equilibrium: desorption / stripping

yi = Kxi

Lxin

Lxout

Vyout

Vyinxinxout

yin

yout

i

yi+1

yi

xi

xi-1

LV

mass balance: V·yi+L·xi= V∙yi+1 + L∙xi-1

→ working line yi+1 – yi = (L/V)·(xi-1 – xi)

slope L/V

Fro

m M

ÖF

-ST

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Continuous distillation (binary) /2

Roughly equimolarexchange: 1 mol A liquid → gas «-» 1mol B gas → liquid

As a result: Gas and liquid streams ~ constantin each section:

L and V in top section, L´ and V´ in bottom section

Feed enters there whereit is similar to the mixtureinside

Top section:rectifying sectionabsorbing the less volatile component

Bottom section:stripping sectiondesorbing the more volatile component

Bottom product:less volatile component

Top product: more volatile component

Picture: WK92

Fro

m M

ÖF

-ST

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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 productsspecifications

→ 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 balanceswas suggested by McCabe and Thiele

!

Fro

m M

ÖF

-ST

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EquilbriumdataExample:C3 – iC4 /1

pressures < critical pressures allowablefor C3 – iC4 : p < 37 bar

37 bar boiling points 85 - 140°C: stability OK

Tboil iC4

at 37 bar

Tboil C3

at 37 bar

Raoults law: y·ptotal = x·p°

K = y/x = p°/ptotal

Fro

m M

ÖF

-ST

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Chose 40°C as the temperatur at the condensor.

At Tmin = 40°C, the vapour pressure of the more volatile component which is C3

= 14 bar → p ≥ 14 bar

Then Tboil ~ 81°C for iC4

Tboil C3

at 14 bar

p° C3

at 40°C

Tboil iC4

at 14 bar

EquilbriumdataExample:C3 – iC4 /2

Fro

m M

ÖF

-ST

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Binary distillation, McCabe-Thiele /1

Pic: WK92

The McCabe-Thiele x,y diagram procedure for continuous binarydistillation 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 twocomponents are ≈ the same, and that mixing heat effects are negligible.

For example EtOH/H2O: Δvaph(EtOH) ≈ Δvaph(H2O) ≈ 40 kJ/mol elementary courses (e.g. ÅA424302, Z12)


17/82


q = fraction of feed that gives liquid on the feedingtray: L´= L + q· F,

F· q = ΔL and F· (1-q) = ΔV

q is related to the energyneeded to convert the feedcompletely into vapour. With enthalpies H for saturatedliquid and saturated gas it is found that

q >1q = 1

0 < q < 1q = 0q < 0

sat,Lsat,G

Fsat,G

HH

HHq

Picture: after T68

Fro

m M

ÖF

-ST

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The McCabe-Thiele x,ydiagram procedure is thusbased on mass balances(giving straight operating lines) and chemicalequilibrium data

If this method cannot be used, a graphical methodbased on enthalpy –composition (h,x or h,ξ) diagrams as given by Ponchon – de Savarit canbe used, based on localequilibrium, mass balancesand energy balances

Pic

: http

://co

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iki/F

ile:M

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agra

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elementary courses(e.g. ÅA424302, Z12)


Restrictions McCabe-Thiele method

Molar heats of vaporisation shouldnot differ more than ~10%

Heats of dissolution or mixing are negligible

Relative volatilities should be 1.3 < α < 5

Reflux ratio’s (L/V) shouldbe R > 1.1· Rmin

Number of trays N << 50 preferably

Otherwise: operating lines are presumably not straight use a more exact method that includes an energy balanceSource: CR83

Fro

m M

ÖF

-ST

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: http

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ÅA 424304


3.3 h,x / h,ξ, and s,x / s,ξ diagrams


Mixing heat for binary mixtures

For this mixing process: h = ξ·ha + (1-ξ)·hb + qmix

with mixing heat qmixto keep temperatureconstant

Mixing heat data can be used to produce h,x or h,ξdiagrams

ξ kg B

1-ξ kg A

1 kg mixξ = mass fraction for the most volatile compoundx = molar fraction for the most volatile compound

With molar masses Ma

and Mb :

b

a

a

b

MM

xx

ξ

;

MM

ξξ

x


↑ 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-H2O

NH3-H2O

Pics: B01

Mixing heat, h,ξ diagram /1



Determination of heat of mixing, qmix, versus composition, ξ: a boiling liquid mixture (mass m, composition ξf, enthalpy hf) is fedinto a calorimeter; a small amount of heat dq gives a small amount of vapour, dm (composition ξd, enthalpy hd)

Mass balance, heat balance:

gives (eliminate m) qmix= dq/dm =

Pic: B01



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) Pics: B01

Schematic h,ξ diagram:- Condensation line c- Boiling line b- Isotherms i

qmix

liquidgas

c

b

i

i


s,ξ diagram

For 2nd Law analysis, exergy analysis or for tracking downirreversibilities in mixture applications (for examplemixtures of refrigerants) s,ξ or s,x diagrams may be used.

Note that during adiabatic mixing hmix = ha+ hb, butsmix = sa + sb+ Δsmix

x

s,x andΔsmix for two ideal gases

xPics: B01

ÅA 424304


3.4 Flash vaporisation


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 Δvaph = hG - hL at T0.

Pic: Z87


28/82

Flash vaporisation, h,x diagram /2

Initially, at p0 the feedpoint A with x = xF is in the liquid region of the h,x diagram

After the pressure reduction to p1, point A lies in the 2-phase region, giving liquid (L) C + gas (G) B

Amount ratio G/L equals = CA /AB, or

Pic: Z87FG

LF

F

F

hhhh

xyxx

LG

L

G


Flash vaporisation, example /1

K71- Fig 2.11

K71- Fig 2.12

Pics: K711 BTU = 1055,06 J 100 BTU/lb = 2,326 kJ/kg

220°F = 104°C170°F = 77°C


K71- Fig 2.11

K71- Fig 2.12

Flash vaporisation, example /2

Pics: K71

v

w

v w

Lever rule – see for example:http://www.doitpoms.ac.uk/tlplib/phase-diagrams/lever.php

ÅA 424304


3.5 Binary Distillation II (the Ponchon-Savarit method)


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 liquidflows L (kg/s)

Liquid mass fractions x, vapour mass fractions y

Reboiler heat input QB(kW), condenser heat output QC (kW)

Top section I, index n (downwards)Bottom section II, index m (downwards)

Pic: CR83



HL and HV are enthalpies of liquid and vapour streams(kJ/kg)

Mass balances top sectionstages (total, and for the mostvolatile species of the two)

*) Pic: CR83



Heat balances (top section)

HLd = H of liquid D (x = xd)

Constant H´d = HLd + QC/D:

**)Pic: CR83

”leaving” system via top:D·HL

d + QC


From the mass balances *)

From mass balance *) and heat balances **)


Functions yn(xn-1) are all linesthat go through point (pole) N = (xd, H´d ) in the (x,H) plot

Pic: CR83x



The lines yn(xn-1) can be putin the h,x or h,ξ diagram (seenext slide)

Similarly, mass and heat balances for the bottom section:

Top section I, index n (downwards)Bottom section II, index m (downwards)

Pic: CR83



Functions ym(xm-1) , all linesgo through point (pole) M = (xw, H´w ) in (x,H) plot

HLw = H of liquid W

(x = xw) Constant

H´w = HLw - QB/W

This then gives

Pic: CR83

Note: this implies q = 1

”leaving” system via

bottom:W·HL

w - QB



A stream N canbe visualised with composition = xd , enthalpy = H´d , mass = Vn – Ln-1

Similarly, M can be visualised as a stream with composition = xw , enthalpy = H´w , mass = Lm-1 – Vm

Pic: CR83 Then also: F = M + N, and F∙xF = M∙xw + N∙xd



The necessarynumber of theoreticalstages can be determinedgraphically as shown

Feed point F is shownon the boiling line(here, q =1)

V1 is vapour from first plate, L1 is liquid from first plate to second plate, et c.

Pic: CR83Here, 7 theoretical stages needed.



The heat removedin the condenserper unit mass D is qc = Qc/D.

Lowering N N´means that morestages are needed

At N Nm the number of stages ∞ Pic: CR83Here, 7 theoretical stages needed.


Using qc = Qc/D and

the minimum heat requirements can be found:

Pic: CR83

Increasing the reflux ratiorequires more heat to be removed, point N upwards, reducing the number of stages



Ponchon-Savarit method Example /1

1 kg/sxf = 0.3

xd = 0.995

xw = 0.100

R = 1.08∙Rmin

Mass balances

Source: CR83


Ponchon-Savarit methodExample /2

h, ξ diagramNH3-H2O1.013 MPa = 10 atm

Source: CR83


Ponchon-Savarit method Example /3

Source: CR83

ÅA 424304


3.6 Ideal distillation column analysis


46/82

Analysis of ideal column /1

Exergy analysis shows that the minimum work for separation of 1 mol of mixture equalswmin = Δex = T°·Δs

For distillation with reboiler heat Qin and condenser heat Qout, the exergy loss –ΔEx can be approximated as

(if HF ≈ HB + HD !)Pic: SAKS04

bottomtop

oin TT

TQEx11

Thus, the separation is accomplished by degrading Qin at Tin

to Qout (= Qin) at Tout < Tin


47/82


The Clausius Clapeyron expression for phase transitionsbased on the slope of the two-phase co-existence(boiling) curve in a p - T diagram,

dp/dT = ΔS/ΔV ~ ΔvapH/(TΔV)

gives for ideal gases:

Also, for an ideal mixture the relative volatility for two similar* components is

sat

sat

p

p

2

112

Pic: SAKS04

212

12

11

TTR

H

p

p

T

dT

R

H

p

dp vap

sat

satvap

sat ,

,ln

* Chemically similar, boiling points not too far apart, α ~ O(1)


48/82


This then gives, for the distillationtemperature range (Ttop, Tbottom):

which if α ~ 1, and usingln(1+z) ≈ z if z «1, it simplifies to: Pic: SAKS04

topbottom

vap

TTR

H 1112ln

topbottom

vap

TTR

H 11112

This requires that- heat Qin is supplied usinga Carnot heat pump -heat Qout is converted intowork in a Carnot process, - minimum reflux is used


49/82


Assume a propane C3, propene C3=

distillation with xF = ½, q = 1, B/F = D/F = ½ (mol/s)/(mol/s)

Note: for this equimolar mix, Δsmix 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 equalto Qin/F = V/F·ΔvapH = ½· (r+1)·ΔvapH ≈ Qout/F

Pic: SAKS04

* Using ” r ” to distinguish it from gas constant R.


50/82


The minimum work requiredto separate the mixture can be found by considering a heat pump between Tout and Tin– see Figure – and exergyanalysis gives (per mole feed):

This can be used to define the minimum reflux ratio rmin:

1ln

2ln2 1½

11ln

12minmin

min

minmin

rHr

F

Q

TTT

F

Q

F

WxxTR

F

W

vapin

bottomtop

oin

iii

osep

mixture equimolar an for 2ln

ln

o

iii

osep

TR

xxTRw

100% efficient!(reversible)

Pic: SAKS04

(α < 4 !)

ÅA 424304


3.7 Real distillation column analysis I: efficiency


Analysis of real column /1

What will be the efficiency of a real column for this case?

Heat transfer requirestemperature differences!

Qin is supplied at 377K, Qout goes out at 298K

The minimum heat input for the separation is

Qminin / F = ½·(rmin+1)·ΔvapH with

rmin = 9.64 (α12 ≈ 1.14), per mole feed

Deviations from ideal thermo-dynamics rreal = 15.9 (α12 ≈ 1.09)

Qrealin / F = ½ ·(rreal+1) · ΔvapH

rreal/rmin = 1.65 Pic: SAKS04



Bottom section ΔT = 46 K, top section ΔT = 22 K

The total thermodyn. eff. of the column is calculated as:

which gives ηoverall = 0.093 = 9.3%

The main sources of the inefficiency are the driving forcesΔ(1/T) in reboiler and condenser

The ratio Qinmin/Qin

real = 0.63 reflectsexergy losses inside the column

Pic: SAKS04

1

1

real

min

r

r



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) lostin the column is equal to

see Table on next slide

Pic: SAKS04



Efficiency of the distillation column: summary

Pic: SAKS04


56/82


R

realin T

TQ

0

1 withgnormalisin by rated table, Previous

Efficiency of the distillation column: summary

Pic: SAKS04

Reboiler

Column

Condenser

Binary distillationefficiency: overview

Source: OFRGJ03


Maximum efficiency:

where x = mole fraction ofthe more volatile component

Reboiler heat duty:

where r = reflux ratio, D = distillate product flow

For QREB ≈ QCOND:

Reboiler duty exergy:

ÅA 424304


3.8 Real distillation column analysis II: exergy analysis


Exergy analysis /1

An analysis can also be made based on exergy flows:

Note that still no enthalpyof mixing is considered, and ideal mixingentropy -R·∑x·lnx.

Flowsheetingprograms (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


Exergy analysis /2

Exergy balance for the equimolar C3/C3= distillation

* reversible case:

* real case


Exergy analysis /3

Exergy streams for ideal and real case: summary

= 0


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)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 ΔT

Instead of trying to reduce these ΔT’s, waste heat from another process unitmay be available Heat integration, which at the same time saves steam(for example)

Pic: SAKS04


Example: H2O/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 qr = 79°C; qb = 104°Cm∙cp (top) = 1·2.6 kJ/K, m∙ΔvapH (top) = 980 kJm∙cp (bottom) = 17·4.2 kJ/K

Source: B01, H84


Example: H2O/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:

for the minimum work per kg top product.

Source: B01, H84See section 1.6 ofExergy analysis



Assuming surroundings temperature T° = 293K an exergy analysiscan 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% H2O + 90% EtOH

.

.

.



Assume that the bottom heat qb 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

ÅA 424304


3.9 Distillation + heat pump and other improvements


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


A few other options

Some GENERAL considerations for improvingdistillation column efficiency:

Process control Feed pre-heat Reduce pressure

drop by optimisingthe trays (or packing)

Separate to lowerpurity and combinewith another separation process

Use intermediate reboilers and condensers, bringingoperating lines closer to the equilibrium line(but this may require more stages or trays)

Pic, Source: SAKS04


Double columns Paraffin /olefin separations are very energy intensive. One example is propane / propylene, or C3 (atm. b.p. - 42.1°C)

/ C3= (atm. b.p. - 47.7°C). For a single-column tray column

distillation process 150-200 trays would be needed, giving a veryhigh (> 100 m) column, with refluxcondensing using air or water

A double-column process is preferred, with smaller (shorter)

columns, using hot and coldwater for condensing (column 2) and reboiling (column 1). Note that reboiler for column 2 = condenser for column 1.

Pic, Source: SAKS0


Distillation column + heat pump /1

Source: B01, H84

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: complexitycosts...


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 itstemperature 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 columnat 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

Source: B01, H84

* pole point

a.k.a.MechanicalVapourRecompression(MVR)


Distillation column + heat pump /3 For the compressor work wc = δ·(h2-h1), and for the bottom heat

exchanger Δqb = 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

* pole point

Two-feed distillation

Two-feed, or two-enthalpy feed can Allow for lowering the number of stages, N (same L/D, xD, xB)

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

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 canbe added/removed by heat exchange, in principle on each tray

Heat is removed from each tray abovefeed, added to each tray below feed. Reboiler much smaller or not needed.

More complicated (‹-› process control), but more efficient: lower entropyproduction (decreases with increasingheat exchange area)


Source: KBJG10, J10

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 sectionComparison for C3 / C3

= splitter

Source: J10, OFRGJ03


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


Sources 3 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)


Sources 3 /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 optimalitycriterion 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

ÅA 424304

Appendix: h,ξ & x,ydiagram NH3 – H2O6.8 atm

28.1.2015Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 81

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100 PSIA = 6.8 atm

1 BTU/lbmole = 2.326 J/mol

ÅA 424304

Appendix: h,ξ diagram n-C6 / n-C8

28.1.2015Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 82

n-C6 - n-C8 101.3 kPa

0

10

20

30

40

50

60

70

80

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x,y, -

HL, H

G

kJ

/mo

l

n-C6 - n-C8 101.3 kPa

-30

-20

-10

0

10

20

30

40

50

60

70

80

90

100

110

120

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x,y, -

HL, H

G

kJ

/mo

l

T °C x - HL kJ/mol y - HG kJ/mol68.7 1 12.98 1 41.87

69.35 0.9772 13.22 0.9963 42.0370 0.9546 13.45 0.9923 42.1875 0.8017 15.19 0.9609 43.3980 0.6684 16.94 0.9237 44.785 0.5566 18.65 0.8788 46.1490 0.4566 20.36 0.8235 47.7495 0.3705 22.05 0.7572 49.53

100 0.2917 23.75 0.6756 51.57105 0.2221 25.44 0.5842 53.8110 0.1591 27.13 0.4754 56.33115 0.1035 28.8 0.3525 59.14120 0.0538 30.46 0.2088 62.33

122.9 0.0262 31.43 0.1096 64.46125.8 0 32.39 0 66.78

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