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Prin
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Prin
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VST rz10VST rz10
1/1081/108
Principles of process technologyPrinciples of process technologyPart 2: Mass transfer and separation Part 2: Mass transfer and separation
processesprocesses
MPCE Course MPCE Course ””Principles of Process technologyPrinciples of Process technology”” –– 414102 414102 -- Helle/Zevenhoven/SaxHelle/Zevenhoven/Saxéénn
Ron ZevenhovenRon ZevenhovenÅÅbo Akademi Universitybo Akademi University
Thermal and Flow Engineering Laboratory Thermal and Flow Engineering Laboratory / V/ Väärmerme-- och stroch ströömningsteknikmningstekniktel. 3223 ; [email protected]. 3223 ; [email protected]
http://users.abo.fi/rzevenho/kursRZ.htmlhttp://users.abo.fi/rzevenho/kursRZ.html
Prin
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Prin
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VST rz10VST rz10
2/1082/108
Mass transfer process examplesMass transfer process examples•• Drying of a solid using heat or dry gasDrying of a solid using heat or dry gas•• Adsorption of gas or liquid using a solidAdsorption of gas or liquid using a solid•• Distillation using heat to create a vapour phaseDistillation using heat to create a vapour phase•• Gas absorption using a liquid absorbentGas absorption using a liquid absorbent•• Extraction of liquid or solid using a solvent that Extraction of liquid or solid using a solvent that
creates immiscible phasescreates immiscible phases•• Crystallisation using cooling Crystallisation using cooling
to create a solid phaseto create a solid phaseNoteNote the the ””support phasesupport phase””: : solvent, sorbent, heat, ....solvent, sorbent, heat, ....
Pic
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: http
://w
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Overview of lectures Overview of lectures 3x2 h3x2 hMass transferMass transfer
5.5. Diffusion, drift flow Diffusion, drift flow (see part 1)(see part 1)6.6. Phase equilibriumPhase equilibrium7.7. Convective mass transfer Convective mass transfer
Heat / mass transfer analogy Heat / mass transfer analogy 8.8. NonNon--stationary heat / mass transferstationary heat / mass transfer
Separation processesSeparation processes9.9. Separation processes, equilibrium stagesSeparation processes, equilibrium stages10.10. Absorption / desorption (stripping)Absorption / desorption (stripping)11.11. Continuous distillation Continuous distillation (McCabe(McCabe--Thiele)Thiele)
12.12. Packed columnsPacked columns
Note: in these Note: in these lectures, only lectures, only 1 1 -- dimensional, dimensional, mostly stationary mostly stationary cases are cases are considered.considered.
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6. Phase equilibrium6. Phase equilibrium(gas(gas--gas, gasgas, gas--liquid, liquidliquid, liquid--liquid)liquid)
HenryHenry’’s Law, Raoults Law, Raoult’’s Laws Law
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Phase equilibrium Phase equilibrium /1/1
Gas / solidequilibrium
Liquid water in equilibrium with ice and water vapour
Gas / liquidequilibrium
Liquid / solidequilibrium
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Mass transfer and equilibriumMass transfer and equilibriumDrying of wet gas in an glycol absorberDrying of wet gas in an glycol absorber
cH2O
wet gas dry gas
time
cH2Oin liq
cH2O,eq
Equilibrium determined bythermodynamics
Rate determined bytransport processes and
equipment design
glycol
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Humid air above a lake.Humid air above a lake.
x = fraction in liquid, y = fraction in gas, x = fraction in liquid, y = fraction in gas, z = position coordinate, T = temperaturez = position coordinate, T = temperature
z
x, y, T
xH2O = 1
yH2O << 1 TBoundary layer
gas side
Boundary layerliquid side
Equilibriumat surface
Coolingdue to watervaporisation
and: someand: someair dissolvesair dissolvesin the lake !in the lake !
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GasGas--liquid phase equilibrium: liquid phase equilibrium: RaoultRaoult’’s laws law
•• Assume a liquid mixture of components A, B, Assume a liquid mixture of components A, B, C, .... at a temperature T. C, .... at a temperature T.
•• ””AA”” occupies a occupies a largelarge fraction, say fraction, say xxAA > 5 %> 5 %•• At temperature T, saturation pressure of pure At temperature T, saturation pressure of pure
substance A substance A (i.e. A vapour above liquid A)(i.e. A vapour above liquid A) is pis pAA00
•• For the mixture, the vapour pressure of A For the mixture, the vapour pressure of A equals equals ppAA = y= yAA.p.ptottot = x= xAA.p.pAA
00
•• For a twoFor a two--component mixture component mixture of A and B: of A and B:
•• ppBB=x=xBB.p.pBB00 = (1= (1--xxAA).p).pBB
00 , , using xusing xAA+x+xBB =1=1
Pic
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: http
://w
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/raou
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GasGas--liquid phase equilibrium: liquid phase equilibrium: HenryHenry’’s laws law
•• Assume a liquid mixture of components A, B, Assume a liquid mixture of components A, B, C, .... at a temperature T. C, .... at a temperature T.
•• ””AA”” occupies a occupies a smallsmall fraction, say fraction, say xxAA < 5 %< 5 %•• For the mixture, the vapour pressure of A For the mixture, the vapour pressure of A
equals equals ppAA = y= yAA.p.ptottot = H= HcAcA.x.xAAwith with Henry constantHenry constant HHcc(unit: Pa, bar, ....) (unit: Pa, bar, ....)
•• yyA A /x/xAA = H= HcAcA/p/ptottot= = ββ, y, yAA = = ββ.x.xAAdistribution coefficient distribution coefficient ββ (mol/mol)(mol/mol)
•• HHcc is a function of temperature, but is a function of temperature, but independent of pressure at pindependent of pressure at ptottot < 5 bar.< 5 bar. P
ictu
re: h
ttp://
ww
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/aer
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ExampleExample: : WaterWater--ammonia vapour /1ammonia vapour /1•• A mixture of ammonia and waterA mixture of ammonia and water•• 4040°°C, total pressure unknownC, total pressure unknown•• Liquid composition:Liquid composition:
70 %70 %--mol NHmol NH3 3 + 30 %+ 30 %--mol Hmol H22OO•• What is the composition of the What is the composition of the
gas, ygas, yNH3NH3, y, yH2OH2O ??•• At equilibrium, no driving forces At equilibrium, no driving forces
or temperature gradients, or temperature gradients, xxNH3NH3 + x+ xH2OH2O = 1, y= 1, yNH3NH3 + y+ yH2OH2O = 1= 1butbut xxNH3NH3 ≠≠ yyNH3NH3, x, xH2OH2O ≠≠ yyH2OH2O !!!!!!
xH2O = 0.3 mol/molxNH3 = 0.7 mol/mol
T = 40°CPtot = ?
GASLIQUID
Relative volatility of NH3 with respect to water: α = (yNH3 / xNH3) / (yH2O / xH2O) = KNH3 / KH2O at equilibrium
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ExampleExample: : WaterWater--ammonia vapour /2ammonia vapour /2•• Gas above an NHGas above an NH33/H/H22O liquid mixture with O liquid mixture with
xxNH3NH3 = 0.7 and x= 0.7 and xH2OH2O = 0.3, T = 40= 0.3, T = 40°°C.C.•• Questions: pressure & equilibrium composition of gas ?Questions: pressure & equilibrium composition of gas ?•• From From tabelised datatabelised data for saturation pressures, at 40for saturation pressures, at 40°°C :C :
pp°°H2OH2O = 7.348 kPa; p= 7.348 kPa; p°°NH3NH3 = 1554.33 kPa= 1554.33 kPa•• xx--values for liquid >> 5 % : use values for liquid >> 5 % : use RaoultRaoult’’s Laws Law ::
ppH2OH2O = x= xH2OH2O·· pp°°H2OH2O = 0.3 = 0.3 ·· 7.348 kPa = 2.22 kPa7.348 kPa = 2.22 kPappNH3NH3 = x= xNH3NH3·· pp°°NH3NH3 = 0.7 = 0.7 ·· 1554.33 kPa = 1088.3 kPa1554.33 kPa = 1088.3 kPa
•• PPtotaltotal = p= pH2OH2O + p+ pNH3NH3 = 1090.25 kPa = 10.9025 bar= 1090.25 kPa = 10.9025 bar•• yyH2OH2O = 2.22 kPa / 1090.25 kPa = 0.002 = 0.2 %= 2.22 kPa / 1090.25 kPa = 0.002 = 0.2 %--vv
yyNH3NH3 = 1088.3 kPa / 1090.25 kPa = 0.998 = 99.8 %= 1088.3 kPa / 1090.25 kPa = 0.998 = 99.8 %--vvSource: ÇB98
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HenryHenry’’s law constantss law constants
Source: Atkins, 1983
Source: Çengel & Boles, 1998
Henry coefficients in waterMPa 290 K 300 K 310 K 320 K 330 K 340 KH2S 44 56 70 83 98 114CO2 128 171 217 272 322 --- O2 3800 4500 5200 5700 6100 6500H2 6700 7200 7500 7600 7700 7600CO 5100 6000 6700 7400 8000 8400Air 6200 7400 8400 9200 9900 10400N2 7600 8900 10100 11000 11800 12400
Henry coefficients at 25°C
mm Hg in water in benzene MPa in water in benzene
H2 5.34×107 2.75×106 H2 7119.4 366.6N2 6.51×107 1.79×106 N2 8679.3 238.6
O2 3.30×107 O2 4399.6
CO2 1.25×106 8.57×104 CO2 166.7 11.4CH4 3.14×105 4.27×105 CH4 41.9 56.9
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Air above a lakeAir above a lake
x = fraction in liquid, y = fraction in gas, x = fraction in liquid, y = fraction in gas, z = position coordinate, T = temperaturez = position coordinate, T = temperature
z
x, y, T
xH2O ≈ 1
yH2O << 1 TEquilibriumat surface
Concentrationjump
and: someair dissolvesin the lake !
OH
OHOH x
yK2
22 = :mequilibriu at yO2 ≈ 0.21
yN2 ≈ 0.79
xO2 << 0.01xN2 << 0.01
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Water saturation tables Water saturation tables
Source: Çengel & Boles, 1998
See
also
cou
rse
PTG
(20
06 #
4)
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ExampleExample: : Water in air above a lakeWater in air above a lake
•• Assume a lake with T=17Assume a lake with T=17°°C, pC, ptottot = 92 kPa at = 92 kPa at water surface level.water surface level.
•• At the surface, the water will be saturated with At the surface, the water will be saturated with water, which means pwater, which means pH2OH2O = p= p°°H2OH2O = 1920 Pa.= 1920 Pa.
•• Fraction of water in the air at the water surface Fraction of water in the air at the water surface at equilbrium with the air above is then at equilbrium with the air above is then yyH2OH2O = p= pH2OH2O/p/ptottot = 1.92 kPa / 92 kPa = 2.09 %= 1.92 kPa / 92 kPa = 2.09 %note:note: % = %% = %--v (volume %)v (volume %)
Source: ÇB98
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ExampleExample: : Air dissolved in lake waterAir dissolved in lake water
•• Assume the same lake, T=17Assume the same lake, T=17°°C, pC, ptottot = 92 kPa at water = 92 kPa at water surface level. At the surface psurface level. At the surface pH2OH2O = p= p°°H2OH2O = 1920 Pa.= 1920 Pa.
•• A small amount (<< 5 %A small amount (<< 5 %--vol) of air will be dissolved in vol) of air will be dissolved in the water: use the water: use HenryHenry’’s Laws Law to calculate the equilibrium:to calculate the equilibrium:
•• At T = 290 K, HAt T = 290 K, Hc AIR,waterc AIR,water = 6200 MPa= 6200 MPa•• ppAIRAIR = p= ptottot –– ppH2OH2O = 90.008 kPa= 90.008 kPa•• xxAIR, water sideAIR, water side = p= pAIR,air sideAIR,air side / H/ Hc AIR, waterc AIR, water
= 0.090008 MPa / 6200 MPa = 1.45= 0.090008 MPa / 6200 MPa = 1.45××1010--5 5 = 0.00145 %= 0.00145 %--vv•• This means 1.45 moles air (molar mass ~29 kg/kmol) in This means 1.45 moles air (molar mass ~29 kg/kmol) in
100000 moles water (molar mass 18 kg/kmol), which 100000 moles water (molar mass 18 kg/kmol), which means 23.4 mg air / kg watermeans 23.4 mg air / kg water Source: ÇB98
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TwoTwo--component phase diagram component phase diagram (G/L)(G/L)
P, T, x – diagram for a binary gas-liquid system
xxAA = 1= 1--xxBByyAA = 1= 1--yyBB
Picture: T68
critical pointspure A, B
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Constant pressure; the Constant pressure; the xx--y diagramy diagramBinary vapourBinary vapour--liquid equilibriumliquid equilibrium
Temperature
Picture: T68
y
x
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Relative volatility, Relative volatility, αα
•• Raoult Raoult (for not(for not--small xsmall xii)): : yyAA=x=xAAppAA°°, y, yBB=x=xBBppBB°°
•• xxBB = 1= 1--xxAA, y, yBB = 1= 1--yyAA•• yyBB/y/yAA= = αα ··xxBB/x/xAAαα = relative volatility= relative volatilityαα = = ppBB°°/p/pAA°° = = αα (T)(T)
•• result:result:(1(1--yyAA)/y)/yAA = = αα ··((11--xxAA)/x)/xAA
yyAA = = αα··xxAA / (1+ (/ (1+ (αα--1)1)··xxAA))
α =
Picture: WK92
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7.7. Convective mass transferConvective mass transferHeat / mass transfer analogy Heat / mass transfer analogy
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Variables & transfer coefficientsVariables & transfer coefficients
Source: BMH99
For mass transfer of species A in B, Sh = ShAB is often written as NuAB
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Gradients, driving forces Gradients, driving forces /1/1
T1
T2
Heat transfercoefficient h2Heat transfer
coefficient h1
dHeat
conductivityλ
T’1
T’2
21
222111
21
21
1++1=1→
−=−=
−=
−===
hλd
hh
)TT(h)TT(h
)TT(dλ
)TT.(hTΔ.hΦ
total
''
''
total
total"h
Heat flux Φh” (W/m2), local and overall heat transfer coeffients h
Note: temperature is continuous
heat
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Gradients, driving forcesGradients, driving forces /2/2
A temperature A temperature profile, heat profile, heat
transfer 1transfer 1→→ 22
T2
Ti
T1
µ2
µi
µ1c1
c2
c1,i
c2,i
c1
c1,i
c2,i
c2
A chemical A chemical potential profile, potential profile,
mass transfer 1mass transfer 1→→ 22
Concentration profiles, Concentration profiles, mass transfer 1 mass transfer 1 →→ 22
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Example:Example: mass transfer coefficient mass transfer coefficient /1/1
Picture: T06
( ) ( )
( ) ( )
k. tcoefficien transfermass and Ð tcoefficien diffusion with
-ÐÐ
as described be can boundary a at s)(mol/m"m transferMass
h. tcoefficien transfer heat and tyconductivi heat with
-
as described be can boundary a at )(W/mq" transfer Heat
2A
2
∞0=
∞0=
∞0=
∞0=
−=⇒−⋅===
−=⇒−⋅===
AAsy
AAAs
y
AA"A
sy
sy
ccdy
dckcckdy
dc-A
m m
λ
TTdydTλhTTh
dydT-λ
AQ " Q
y
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Example:Example: mass transfer coefficient mass transfer coefficient /1/1
•• At and near the surface of At and near the surface of a pan of water, a pan of water, measurements of the measurements of the partial pressure of water partial pressure of water are made. The results are are made. The results are given in the Figure.given in the Figure.
•• CalculateCalculate the mass the mass transfer coefficient k (m/s).transfer coefficient k (m/s).
•• The system is isothermal.The system is isothermal.•• Diffusion coefficient Diffusion coefficient ĐĐ = =
ĐĐwater vapour in airwater vapour in air = at 298 K = = at 298 K = 2.602.60××1010--55 mm22/s/s
Source: IDBL06
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Example:Example: mass transfer coefficient mass transfer coefficient /2/2
•• At the surface, the water partial pressure = saturated vapour At the surface, the water partial pressure = saturated vapour pressure = 0.10 atm pressure = 0.10 atm →→ T = 319 K (water/steam tables). T = 319 K (water/steam tables).
•• The mass transfer coefficient k is calculated usingThe mass transfer coefficient k is calculated using
•• ĐĐ (319K) = (319K) = ĐĐ (298K) (298K) ×× (319/298)(319/298)1.51.5 = 2.88= 2.88××1010--55 mm22/s/s•• k = k = --2.882.88××1010--55 mm22/s /s ··(0 (0 –– 0.1) atm / (0.003 0.1) atm / (0.003 –– 0) m / (0.1 0) m / (0.1 –– 0.02) atm 0.02) atm
= 0.012 m/s = 0.012 m/s
( ) ( ) /RTpnA/Vc with -Ð-Ð AA ==−=−= ∞0=
∞0=
AAsy
As
y
ppdy
dpccdydck
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Mass Mass ↔↔ heat transfer analogyheat transfer analogyImportant differences heat transfer / mass transferImportant differences heat transfer / mass transfer–– The interface between media or phases is usually The interface between media or phases is usually
mobile during mass transfer.mobile during mass transfer.–– In liquids and solids, the diffusion coefficients In liquids and solids, the diffusion coefficients ĐĐ (m(m22/s) /s)
are (much) smaller the heat diffusivityare (much) smaller the heat diffusivity aa (m(m22/s) and /s) and kinematic viscosity kinematic viscosity νν (m(m22/s). /s).
–– Diffusion can induce a drift flow.Diffusion can induce a drift flow.–– Instead of concentration (c), chemical potential (Instead of concentration (c), chemical potential (µµ) )
should actually be used. Using concentration gives should actually be used. Using concentration gives •• diffusion coefficients dependent on concentrationsdiffusion coefficients dependent on concentrations•• discontinuous concentration profiles across phase boundariesdiscontinuous concentration profiles across phase boundaries
– There seems to be no mass analogy of heat radiation transfer.....
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Mass transfer and boundary layersMass transfer and boundary layers
•• For mass transfer For mass transfer between two phases A between two phases A and B, concentration and B, concentration gradients usually only gradients usually only exist near the physical exist near the physical boundary that boundary that separates A and Bseparates A and B
•• Thus, the driving forces Thus, the driving forces are active only in are active only in boundary layers at the boundary layers at the separating surfaceseparating surface
Equilibrium betweencA,interfaceA
and cA,interfaceB
AΦΦ˝̋mol Amol A →→
boundary layermedium A
phaseboundary
boundary layermedium B
cA,bulkA
cA,bulkB
cA,interfaceA
cA,interfaceB
B
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Mass transfer coefficient Mass transfer coefficient /1/1
•• Mass flow species A: Mass flow species A: ṅṅAA = = ΦΦAA mol/smol/s
•• Mass transfer rate per area: Mass transfer rate per area: ṄṄAA = = ṅṅAA/a =/a = ΦΦ””AA mol/(mmol/(m22··s)s)
•• Mass transfer coefficients, k, Mass transfer coefficients, k, (unit: m/s) for both sides of (unit: m/s) for both sides of the interface:the interface:ṄṄAA = k= kxx ··(c(c1,i1,i--cc11) = k) = kyy··(c(c22--cc2,i2,i))
1 (L) 2 (G)
yiC2.i
yC2
xiC1.i
L,G example
xC1
interface a
ṄṄAA
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Mass transfer coefficient Mass transfer coefficient /2/2
•• But for a total (overall) mass But for a total (overall) mass transfer coefficient ktransfer coefficient ktot tot ::oo It is very difficult to determine It is very difficult to determine
concentrations at interfacesconcentrations at interfacesoo ṄṄAA = k= ktottot(x(x--y) or y) or ṄṄAA = k= ktottot(c(c22--cc11) )
makes no sensemakes no sense
•• More important instead: how More important instead: how are the concentrations are the concentrations related to saturation in their related to saturation in their phase: x phase: x ↔↔ xxsatsat, y , y ↔↔ yysatsat
1 (L) 2 (G)
yiC2.i
yC2
xiC1.i
L,G example
xC1
interface a
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Mass transfer coefficient Mass transfer coefficient /3/3
•• Deviation from equilibrium Deviation from equilibrium gives transport gives transport →→ use use information for information for concentrations at equilibrium: concentrations at equilibrium: cc1,i1,i = m= m··cc2,i2,i with with distribution distribution coefficient mcoefficient m (from Raoult(from Raoult’’s or s or HenryHenry’’s law, for example)s law, for example)
•• Eliminate interface values in Eliminate interface values in ṄṄAA = k= kxx ··(c(c1,i 1,i -- cc11) = k) = kyy··(c(c2 2 -- cc2,i2,i) ) →→ CC2,i2,i = (k= (kyycc2 2 + k+ kxxcc11)/(mk)/(mkx x + k+ kyy), or), or→→ CC1,i1,i = (k= (kyycc2 2 + k+ kxxcc11)/(k)/(kx x + k+ kyy/m)/m)
1 (L) 2 (G)
yiC2.i
yC2
xiC1.i
L,G example
xC1
interface a
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Mass transfer coefficient Mass transfer coefficient /4/4
•• The result is:The result is:
•• Replace cReplace c11/m = c/m = c22*, which is *, which is the concentration in phase 2 the concentration in phase 2 if in equilibrium with phase 1, if in equilibrium with phase 1, or similarly m.cor similarly m.c22 = c= c11**··::ṄṄAA = K= Kyy(c(c2 2 -- cc22*) = K*) = Kxx(c(c11* * -- cc11))
( ) ( )1212
1−
12
12
1−
−⋅=−⋅⎟⎟⎠
⎞⎜⎜⎝
⎛ 1+=
⎟⎠⎞
⎜⎝⎛ −=⎟
⎠⎞
⎜⎝⎛ −⎟
⎟⎠
⎞⎜⎜⎝
⎛
⋅1+1=
ccmKccmkk
mN
mccK
mcc
kmkN
xxy
A
yxy
A
1 (L) 2 (G)
yiC2.i
yC2
xiC1.i
L,G example
xC1
interface a
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The film model The film model /1/1•• SteadySteady--state diffusion, no state diffusion, no
accumulationaccumulation
•• For a boundary layer For a boundary layer thickness thickness δδcc, , with c=cwith c=cA0A0 @ @ x=0 and c=cx=0 and c=cA1A1 @ x=@ x=δδcc the the concentration profile is concentration profile is
c1A0A
1AA y1cccc
δ−=
−−
dyy
y
dydc
dydc
+
−
−
Ð
Ð
y y +dy0=−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−≈
⎟⎟⎠
⎞⎜⎜⎝
⎛−−⎟
⎟⎠
⎞⎜⎜⎝
⎛−
2
2
+
dycd
dydydcd
dy
dydc
dydc
ydyy
ÐÐ
ÐÐ
diffusion
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The film model The film model /2/2
•• A mass transfer coefficient can A mass transfer coefficient can be linked to the film model: be linked to the film model: ṄṄAA= = ΦΦ””A,molA,mol = k= k··(c(cA0A0--ccA1A1) = ) = -- ĐĐAA··dcdcAA/dy = /dy = ĐĐAA··(c(cA0A0--ccA1A1)/)/δδcc
which gives k = which gives k = ĐĐAA//δδcc
•• Thus, the boundary layer Thus, the boundary layer thickness can be estimated if thickness can be estimated if k and k and ĐĐAA are known.are known.
•• The mass transfer limitations The mass transfer limitations are concentrated in a wellare concentrated in a well--defined region.defined region.
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The film model The film model /3/3
Phase boundary concentration gradientsPhase boundary concentration gradients(a)(a) Film theoryFilm theory (b)(b) Closer to realityCloser to reality
Source: SH06Source: SH06
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Mass transfer, forced convection Mass transfer, forced convection /1/1
•• BUT: Often boundary layer thickness BUT: Often boundary layer thickness δδ (and (and sometimes even its position) is unknown sometimes even its position) is unknown
•• A correlation for Sherwood number (Sh) based on A correlation for Sherwood number (Sh) based on dimensional analysis can be used, relating Sh, dimensional analysis can be used, relating Sh, δδ, , diffusion coefficient D, mass transfer coefficient k and diffusion coefficient D, mass transfer coefficient k and size scale d as Sh = ksize scale d as Sh = k··dd//Đ Đ = d/= d/δδ
•• Sh is the mass transfer analogue of Nusselt number Sh is the mass transfer analogue of Nusselt number for heat transfer Nu = hfor heat transfer Nu = h··d/d/λλ with heat transfer coefficient hwith heat transfer coefficient h
•• Important situations for mass transfer with convection:Important situations for mass transfer with convection:–– Internal forced flow: inside a tube (laminar or turbulent)Internal forced flow: inside a tube (laminar or turbulent)–– External forced flow: around an obstacle (laminar or turbulent)External forced flow: around an obstacle (laminar or turbulent)–– Natural convectionNatural convection
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Sherwood number Sherwood number •• Diffusion from a spherical particle with radius R in a Diffusion from a spherical particle with radius R in a
stagnant medium (i.e. no flow), with concentration c=cstagnant medium (i.e. no flow), with concentration c=c00at the surface of the sphere and c=cat the surface of the sphere and c=c∞∞ far away from far away from the sphere: the sphere: ΦΦAA = k= kAA.a.(c.a.(c00--cc∞∞) with surface a=4) with surface a=4ππRR²², , mass transfer coefficient kmass transfer coefficient kAA (m/s)(m/s)
•• The diffusion equation The diffusion equation ΦΦAA = = --ĐĐAA··4 4 ππrr²².dc/dr = constant .dc/dr = constant gives the concentration profile around the sphere:gives the concentration profile around the sphere:(c(c--cc∞∞)/(c)/(c00--cc∞∞) = R/r ) = R/r →→ dc/dr = dc/dr = --(c(c00--cc∞∞))··R/rR/r²²
•• For r=R: For r=R: ΦΦAA = k= kAA.a.(c.a.(c00--cc∞∞) = ) = ĐĐAA.a.(c.a.(c00--cc∞∞)/R)/R•• kkAA = = ĐĐAA/R = 2/R = 2·· ĐĐAA/d, with d=2R./d, with d=2R.•• Ratio Ratio kkAA··d/ d/ ĐĐAA is referred to as is referred to as Sherwood numberSherwood number, ,
Sh, (the analogue of Nusselt number Nu for heat Sh, (the analogue of Nusselt number Nu for heat transfer) with Sh =2 for a sphere in a stagnant medium transfer) with Sh =2 for a sphere in a stagnant medium
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Example:Example: a naphtalene spherea naphtalene sphere
•• A DA D0 0 = 1 cm mothball= 1 cm mothball, , is suspended in still air is suspended in still air and vaporises. The vapour pressure of and vaporises. The vapour pressure of naphtalene is ~ 0.05 mm Hg, the density is naphtalene is ~ 0.05 mm Hg, the density is ρρ = = 1150 kg/m1150 kg/m33, the diffusion coefficient of , the diffusion coefficient of napthalene in air at ambient conditions is napthalene in air at ambient conditions is ĐĐ = = 77××1010--66 mm²²/s./s.
•• Question:Question: How long does it take for the sphere How long does it take for the sphere to be reduced to half its original diameter?to be reduced to half its original diameter?
•• Note: 1 atm = 760 mm Hg = 101325 Pa; molar Note: 1 atm = 760 mm Hg = 101325 Pa; molar mass naphtalene M = 0.128 kg/mol.mass naphtalene M = 0.128 kg/mol.
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Example:Example: a naphtalene spherea naphtalene sphereAnswer:Answer:
•• Mass transfer for a sphere in a stagnant medium: Mass transfer for a sphere in a stagnant medium: Sh = 2 = kSh = 2 = k··DD//ĐĐ
•• Concentration c = p/RT (mol/mConcentration c = p/RT (mol/m33); c = c* = p*/RT at the ); c = c* = p*/RT at the spheresphere’’s surface, c = 0 far from the sphere.s surface, c = 0 far from the sphere.
•• Molar flow rate: Molar flow rate: ΦΦmolmol= k= k··aa··c*= 2c*= 2··((ĐĐ/D)/D)··ππDD²²··p*/RT p*/RT ((mol/s), with a=mol/s), with a=ππDD²² (m(m22), and ), and ΦΦmm = M= M··ΦΦmolmol (kg/s) (kg/s) gives gives ΦΦmm=2=2ππ··ĐĐ··DD··MM··p*/RT = p*/RT = --dm/dt = dm/dt = --d(d(ρ·ρ·V)/dt = V)/dt = --ρ·ρ·ddV/dt V/dt
•• With V=(With V=(ππ/6)/6)··DD³³, dV/dt = , dV/dt = ½½DD²²··ππ··dD/dt gives for the dD/dt gives for the diameter D as function of time, D=Ddiameter D as function of time, D=D00 @ t=0:@ t=0:t = t = ρρ··(D(D00²²--DD²²))··RT/(8RT/(8··MM··ĐĐ··p*) p*) →→ t = 51 days for D=t = 51 days for D=½½DD00
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Mass Mass ↔↔ heat transfer analogyheat transfer analogy
Heat transferHeat transfer•• Nu = f(Re, Pr, L/D, Gr,..)Nu = f(Re, Pr, L/D, Gr,..)•• Convection around a sphere:Convection around a sphere:
Nu = 2 + 0.6ReNu = 2 + 0.6Re½½PrPr⅓⅓
•• Transfer from a wall and a Transfer from a wall and a turbulent flow: 2000 < Re < 10turbulent flow: 2000 < Re < 1055
and Pr > 0.7and Pr > 0.7Nu = 0.027ReNu = 0.027Re0.80.8PrPr0.330.33((ηη//ηηwallwall))1/71/7
•• General: Nu = CReGeneral: Nu = CRemmPrPrnn, where , where m = 0.33 .. 0.8, n m = 0.33 .. 0.8, n ≈≈ 0.330.33
Mass transferMass transfer•• Sh = f(Re, Sc, L/D, Gr,..)Sh = f(Re, Sc, L/D, Gr,..)•• Convection around a sphere:Convection around a sphere:
Sh = 2 + 0.6ReSh = 2 + 0.6Re½½ScSc⅓⅓
•• Transfer from a wall and a Transfer from a wall and a turbulent flow: 2000 < Re < 10turbulent flow: 2000 < Re < 1055
and Sc > 0.7and Sc > 0.7Sh = 0.027ReSh = 0.027Re0.80.8ScSc0.330.33
•• General: Sh = CReGeneral: Sh = CRemmScScnn, , where m = 0.33 .. 0.8, n where m = 0.33 .. 0.8, n ≈≈ 0.330.33
•• ChiltonChilton--Colburn analogiesColburn analogies, heat and mass transfer values j, heat and mass transfer values jHH, j, jDD::jjHH = NuRe= NuRe--11PrPr--⅓⅓ jjDD = ShRe= ShRe--11ScSc--⅓⅓
jH = jD = CRem-1 = ½ƒ ƒ = Fanning friction factor for pipe flow
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Example:Example: WettedWetted--wall columnwall column•• Sh = 0.023Sh = 0.023··ReRe0.810.81··ScSc0.44 0.44
(2000 < Re 35000, 0.6 < Sc < 2.5)(2000 < Re 35000, 0.6 < Sc < 2.5)•• Sh = 0.0096Sh = 0.0096··ReRe0.9130.913··ScSc0.3460.346
(10000 < Re 100000, 432 < Sc < 97600)(10000 < Re 100000, 432 < Sc < 97600)
•• Example:Example:Calculate the mass transfer (gas) film Calculate the mass transfer (gas) film ththííckness ckness δδ for 1) water and 2) ethanol for 1) water and 2) ethanol evaporating into air in a 2evaporating into air in a 2”” inner diameter inner diameter wettedwetted--wall column. Re = 20000, Scwall column. Re = 20000, ScH2OH2O--airair= 0.573, Sc= 0.573, Scethanolethanol--airair = 1.14= 1.14Answer Answer (using the first expression for Sh)(using the first expression for Sh): : 1) Sh = 31.3 = k1) Sh = 31.3 = k··DDtubetube/ / ĐĐH2O H2O –– air air = D= Dtubetube//δδgives gives δδ = 0.064= 0.064”” = 1.6 mm= 1.6 mm2) Sh = 1.14 gives 2) Sh = 1.14 gives δδ = 0.047= 0.047”” = 1.2 mm= 1.2 mm
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8.8. NonNon--stationary heat / mass stationary heat / mass transfertransfer
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FourierFourier’’s Law s Law /3/3
•• For a general case with a 3For a general case with a 3--dimensional temperature dimensional temperature gradient T = (gradient T = (∂∂T/T/∂∂x,x,∂∂T/T/∂∂y,y,∂∂T/T/∂∂z), Fourierz), Fourier’’s Law gives s Law gives (for constant (for constant λλ)) for the heat flux for the heat flux QQ”” = = -- λλ TT
•• The the temperature field inside the The the temperature field inside the conducting medium can be written as conducting medium can be written as T = T(t, T = T(t, xx) with time t and 3) with time t and 3--dimensional dimensional location vector location vector xx
•• For stationary heat transfer For stationary heat transfer ∂∂T/T/∂∂t = 0 at t = 0 at each position each position xx
•• The heat transfer vector is perpendicular The heat transfer vector is perpendicular to the isothermal surfacesto the isothermal surfaces
•• Note that material property Note that material property λλ is a is a function of temperature: function of temperature: more accurately more accurately QQ”” = = -- λλ(T)T(T)T
∆∆∆∆.
Figure: JK05
∆∆ Q is a vector with direction - T
∆∆..
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Transient heat conduction 1Transient heat conduction 1--D D /1/1
•• For For 11--dimensional transient heat dimensional transient heat conductionconduction in a balance volume in a balance volume dV with mass dm =dV with mass dm =ρρ··dV =dV =ρρ··AA··dx :dx :
2
2
2
2
xTa
tT
xTA
xxTA
tTAc
xTA
xQ
tTcA
dxxQ
tTcdmQQ outin
∂∂⋅=
∂∂
⇒∂∂⋅⋅λ+=
∂
⎟⎠⎞
⎜⎝⎛
∂∂⋅⋅λ∂
−=∂∂⋅⋅⋅ρ⇒
∂∂⋅⋅λ=
∂∂−=
∂∂⋅⋅⋅ρ⇒⋅=
∂∂−=
∂∂⋅⋅=−
-
-Q Laws Fourier' with
Aρdm/dx with
x
w
L
dx
Q.
A = L·w
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Transient heat conduction 1Transient heat conduction 1--D D /2/2
•• Case 1Case 1: Assume a material with flat boundary at x=0, : Assume a material with flat boundary at x=0, infinite length in xinfinite length in x--direction, with T=Tdirection, with T=T00 at all x at all x
•• At time tAt time t≥≥0 the temperature at x=0 is increased to 0 the temperature at x=0 is increased to T=TT=T11 and heat starts to enter (diffuse into) the and heat starts to enter (diffuse into) the material. At xmaterial. At x→∞→∞, T stays at T, T stays at T00..
conditions initial and boundary←
∂∂=
∂∂
2
2
xTa
tT
Picture: BMH99
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Transient heat conduction 1Transient heat conduction 1--D D /3/3
•• With dimensionless variables With dimensionless variables θθ = (T= (T--TT00)/(T)/(T11--TT00) )
and and ξξ = x / (4at)= x / (4at)½½
this gives the following solution:this gives the following solution:
)(yerfde
deTTTT
y
atx
=ξπ
ξπ
=−−=θ−
∫
∫
ξ−
ξ−
0
4
001
1
2
2
2
21
with
erf(x) ≈ 1 - exp(- 1.128x - 0.655x2 - 0.063x3)
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Transient heat conduction 1Transient heat conduction 1--D D /4/4
•• At x = 0 the slope of the At x = 0 the slope of the penetration profile lines penetration profile lines equals equals
∂∂T/T/∂∂x = x = --(T(T11--TT00)/()/(ππat)at)½½
where x = (where x = (ππat)at)½½
is referred to as is referred to as penetration depth.penetration depth.
•• Fourier number Fo Fourier number Fo is (for heat transfer) defined asis (for heat transfer) defined asFo = at/dFo = at/d2 2 = t /(d= t /(d22/a)) for a medium with thickness d /a)) for a medium with thickness d
•• Fo gives the ratio between time t and the penetration time dFo gives the ratio between time t and the penetration time d22/a /a
•• The penetration depth concept is valid for Fo < 0.1The penetration depth concept is valid for Fo < 0.1Picture: BMH99
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9. Separation processes 9. Separation processes (gas(gas--gas, gasgas, gas--liquid, liquidliquid, liquid--liquid):liquid):
equilibrium stages equilibrium stages
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Separation of mixtures: stages Separation of mixtures: stages /1/1
For example: For example: separating phenol from water (L) separating phenol from water (L) by adding benzene (V) in a by adding benzene (V) in a separation funnel.separation funnel.x = phenol conc. in L, y = phenol conc. in Vx = phenol conc. in L, y = phenol conc. in V
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Separation of mixtures: stages Separation of mixtures: stages /2/2
•• The ratio phenol transferred / phenol notThe ratio phenol transferred / phenol not--transferredtransferred= Vy/Lx = KV/L; can be referred to as S = KV/L = Vy/Lx = KV/L; can be referred to as S = KV/L separation factorseparation factor, for mass w : Vy/Lx = Sw / w = S , for mass w : Vy/Lx = Sw / w = S
•• The fraction f The fraction f notnot transferred from L to V, or transferred from L to V, or not not ””removedremoved”” from L: S = (1from L: S = (1--f) / f f) / f →→ f = 1/(1+S)f = 1/(1+S)
•• S can be changed by varying V (or L), or changing K S can be changed by varying V (or L), or changing K
At ambient conditions (1 bar, 20At ambient conditions (1 bar, 20°°C)C)the distribution coefficient for the distribution coefficient for phenol across the phases equalsphenol across the phases equalsK = y/x = 2 K = y/x = 2 →→ if L=V then S = 2 if L=V then S = 2
Phenol Phenol notnot removed = 1/3 = 33.33% removed = 1/3 = 33.33%
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Separation of mixtures: stages Separation of mixtures: stages /3/3
Adding the benzene Adding the benzene support phase to the support phase to the water/phenol in two water/phenol in two stepssteps
Phenol not removed Phenol not removed = w/4w = 25% = w/4w = 25%
S = 1
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Separation of mixtures: stages Separation of mixtures: stages /4/4
Mixing the water/phenol Mixing the water/phenol and the benzene support and the benzene support phase in two stepsphase in two steps
Phenol not removed Phenol not removed = 3.33w / 18w = 19% = 3.33w / 18w = 19%
S = 2
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Separation Separation and and
equilibriumequilibriumstagesstagesExampleExample
FourFour--stage stage processing for Cu processing for Cu recovery from Curecovery from Cu--
containing orecontaining ore(liquid / liquid)(liquid / liquid)
release from ore
extraction
recovery
electrolysis
Diluted H2SO4
Organic solvent
Concentrated H2SO4
Cu
Cu - ore
Numbers are concentrations kg Cu/m3
2,0
2,2
0,1
0,2
50 40
Source: WK92
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Phase equilibrium stagesPhase equilibrium stages /1/1
The extraction and the recovery process are The extraction and the recovery process are combinations of mixing tanks and settling tankscombinations of mixing tanks and settling tanks
extractionfeed refine
solventloaded solvent
Picture: WK92
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Phase equilibrium stagesPhase equilibrium stages /2/2
•• Cu is transferred from feed stream L to solvent (Cu is transferred from feed stream L to solvent (””support phasesupport phase””) V; ) V; the extraction unit can be described as a the extraction unit can be described as a series of equilibrium stagesseries of equilibrium stages..•• Equilibrium constant for Cu in feed stream and solvent stream iEquilibrium constant for Cu in feed stream and solvent stream is s
K = yK = yCuCu/x/xCuCu. Thus, for an equilibrium stage n: y. Thus, for an equilibrium stage n: ynn = Kx= Kxnn
•• Streams L and V are often roughly constantStreams L and V are often roughly constantLL11≈≈ LL22 ≈≈ .. .. ≈≈ LLnn = L; V= L; V11 ≈≈ VV22 ≈≈ ......≈≈ VVnn = V = V →→ separation factor S = Kseparation factor S = K··V/LV/L
extractionfeed refine
solventloaded solvent
equilibriumstage n
Lnxn
equilibriumstage n-1
equilibriumstage n+1
Ln+1xn+1
Vnyn Vn-1yn-1
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Example: phase equilibrium stageExample: phase equilibrium stage
•• For example: phase equilibrium constant K=5; For example: phase equilibrium constant K=5; L = 1000 kg/s, V = 800 kg/s L = 1000 kg/s, V = 800 kg/s →→ S = KV/L= 4.S = KV/L= 4.Cu in feed xCu in feed x00 = 0.2 %= 0.2 %--wt = 0.002 kg/kgwt = 0.002 kg/kg
•• For clean solvent, yFor clean solvent, y22=0, then Vy=0, then Vy11 / Lx/ Lx11 = S= S•• →→ Mass balance gives LxMass balance gives Lx00 = (S+1)w= (S+1)w--rr
LxLx00+Vy+Vy22 = Lx= Lx00 = Lx= Lx11+Vy+Vy11 = Lx= Lx11(1+KV/L) = Lx(1+KV/L) = Lx11(1+S)(1+S)→→ LxLx00 = 2 kg Cu/s w = 2 kg Cu/s w →→ LxLx11 = Lx= Lx00 / (1+S) = 0.4 kg Cu/s/ (1+S) = 0.4 kg Cu/s
Picture: WK92
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Phase equilibrium stages Phase equilibrium stages /3/3
Calculation procedure Calculation procedure
•• Number of equilibrium stages is determined Number of equilibrium stages is determined by the result w in the refine stream, and K, V by the result w in the refine stream, and K, V and L (i.e. the separation factor S)and L (i.e. the separation factor S)
•• NoteNote:: S= KV/L if V is support phase S= KV/L if V is support phase S = L/KV if L is support phaseS = L/KV if L is support phase
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10. Separation processes10. Separation processes::Absorption/stripping processesAbsorption/stripping processes
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Absorption & strippingAbsorption & stripping•• AbsorptionAbsorption (or: gas absorption, gas scrubbing, gas (or: gas absorption, gas scrubbing, gas
washing): contacting a gas mixture with a liquid (the washing): contacting a gas mixture with a liquid (the ””absorbentabsorbent”” or or ””solventsolvent””) in order to selectively ) in order to selectively dissolve one (or more) components from the gas in dissolve one (or more) components from the gas in the liquidthe liquid
•• StrippingStripping (or: desorption) is the opposite process: a (or: desorption) is the opposite process: a liquid mixture is contacted with a gas in order to liquid mixture is contacted with a gas in order to selectively remove one (or more) components from selectively remove one (or more) components from the liquid to the gasthe liquid to the gas
•• Often absorbers and strippers are coupled, as to allow Often absorbers and strippers are coupled, as to allow for for recoveryrecovery (and re(and re--use) of solvent.use) of solvent.
•• The The type of equipmenttype of equipment used depends on the used depends on the relative relative amountsamounts of liquid and gas mass streamsof liquid and gas mass streams
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COCO22 strippingstripping
Picture: SA05Picture: SA05
Natural gas + CO2
Liquid solventfor example
alkanol amine
Natural gas
Liquid solvent+ CO2
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Process example and equipmentProcess example and equipment
↑ Typical absorption process for acetone
→ Industrial equipment: (a) tray tower; (b) packed column; (c) spray tower; (d) bubble colum Pictures: SH06
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A performated tray columnA performated tray column
Especially Especially suitablesuitable if total gas and liquid mass streams if total gas and liquid mass streams are roughly the same and are more or less constantare roughly the same and are more or less constant
tray
gas
gas
liquid
liquid
liquid
gasPicture: WK92
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VapourVapour--liquid stream relationshipsliquid stream relationships
(a)(a) Mass balanceMass balance →→ working line in x,y plotworking line in x,y plot(b)(b) EquilibriumEquilibrium →→ equilibrium line or curve in x,y plot equilibrium line or curve in x,y plot
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Absorber/stripper x,y diagramAbsorber/stripper x,y diagramContinuous steadyContinuous steady--state process in a state process in a countercurrent countercurrent cascadecascade with with equilibrium stages: equilibrium stages:
absorber absorber
stripperstripper
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65/10865/108Graphical determination of the Graphical determination of the number of equilibrium stagesnumber of equilibrium stages
(a)(a) AbsorberAbsorberx1 = liquid at equilibrium
with gas y = y1
y2 = gas that exchanges mass with liquid x = x1
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66/10866/108Graphical determination of the Graphical determination of the number of equilibrium stagesnumber of equilibrium stages
(b) Stripper(b) Stripper
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Absorber: Absorber: wet gas drying /1wet gas drying /1
•• Drying of wet methane Drying of wet methane with diwith di--ethylene glycol ethylene glycol (DEG)(DEG)
•• Regeneration of wet Regeneration of wet DEG with dry NDEG with dry N22
•• DEG boiling point at 1 DEG boiling point at 1 bar = 245bar = 245°°C;C;
•• DEG molar mass DEG molar mass = 106 kg/kmol= 106 kg/kmol
Dry DEG
Wet DEG
ABSORPTION
DESORPTION
Wet CH4
Dry CH4 Dry N2
Wet N2Picture: WK92
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Absorber:Absorber:wet gas drying /3wet gas drying /3
example:example:HH22O in DEG at 72O in DEG at 72°°C, 40 barC, 40 bar
low concentration low concentration HHc c forfor HH22OO in DEG = 0.02 MPain DEG = 0.02 MPa
gives equilibrium constantgives equilibrium constantK = yK = yH2OH2O/x/xH2OH2O = H= HcH2OcH2O / p/ ptotaltotal= 0.02 MPa/4 MPa = 0.005= 0.02 MPa/4 MPa = 0.005
Picture: WK92
Hen
ry c
oeffi
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sev
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gas
es in
wat
eran
d fo
r wat
er in
DEG
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Absorber:Absorber:wet gas drying /4wet gas drying /4
process data:process data:HH22O in wet gas = yO in wet gas = yN+1N+1
= 0.001 mol/mol = 0.001 mol/mol HH22O in dry gas = yO in dry gas = y11= 2= 2××1010--4 4 mol/molmol/mol
HH22O in O in ““drydry”” DEG = xDEG = x00= = 22××1010--22 mol/mol, mol/mol,
and L/V = 0.01 (kmol/s)/(kmol/s)and L/V = 0.01 (kmol/s)/(kmol/s)Total mass balance givesTotal mass balance gives
(x(xNN –– xx00))··L = (yL = (yN+1N+1-- yy11))··V V →→ xxNN = 0.1= 0.1Separation factor S = L/KV = 2 Separation factor S = L/KV = 2 Picture: after SH06
For the first stage, equilibrium between exit gas and exit vapour implies K = y1/x1 →x1 = y1/K = 2×10-4 / 0.005 gives x1 = 0.04
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Absorber: Absorber: wet gas drying /5wet gas drying /5
•• Working line VWorking line V··(y(y--yy11) = L) = L··(x(x--xx00) and equilibrium line) and equilibrium lineallow for calculation of theoretical equilibrium stagesallow for calculation of theoretical equilibrium stages
•• Working line requires V/L ratio + one point (xWorking line requires V/L ratio + one point (xii, y, yi+1i+1) )
equilibriumline y = Kx
working liney = y1 + L/V(x-x0)
x1
y1
x0 = 0.02 xn = 0.10 y1 = 2×10-4 yn+1 = 0.001
Note: Kremser equation: f = 0.2, S = 2 → N = 1.6 (?!)
Picture: WK92
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•• Reducing the amount of solvent (liquid) by 50% gives Reducing the amount of solvent (liquid) by 50% gives L/V = 5L/V = 5××1010--33 →→ Separation factor S Separation factor S →→ S = 1S = 1
Equilibriumline
Operating line
Absorber: Absorber: wet gas drying /6wet gas drying /6
Picture: WK92
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Absorber: Absorber: wet gas drying /7wet gas drying /7
•• Reducing the amount of solvent (liquid) further to L/V Reducing the amount of solvent (liquid) further to L/V = 3= 3××1010--33 →→ Separation factor S Separation factor S →→ S = 0.6S = 0.6
•• The separation process can not be accomplishedThe separation process can not be accomplished
Equilibriumline
Operating line
Picture: WK92
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Absorber: Absorber: wet gas drying /8wet gas drying /8
Equilibriumline
Operating line
This givesSmin = 0.888for this caseand N → ∞
Picture: WK92
•• The The minimum amount of liquid, minimum amount of liquid, oror (L/V)(L/V)minmin,, can be can be found by crossing the operating line with the found by crossing the operating line with the equilibrium line at outgoing gas specification yequilibrium line at outgoing gas specification yN+1N+1..
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Stripper: Stripper: wet gas drying /9wet gas drying /9
Dry DEG
Wet DEG
ABSORPTION
DESORPTION
Wet CH4
Dry CH4 Dry N2
Wet N2
Dry DEG
Wet DEG
ABSORPTION
DESORPTION
Wet CH4
Dry CH4 Dry N2
Wet N2
x0
xN
y´0
yN+1
y1
•• For the desorption, the wet DEG For the desorption, the wet DEG is stripped with dry nitrogen at is stripped with dry nitrogen at 120120°°C, 0.1 MPa.C, 0.1 MPa.
•• For HFor H22O vapour in DEG under O vapour in DEG under these conditions, Hthese conditions, Hc c HH22OO in DEG = in DEG = 0.2 MPa 0.2 MPa (see diagram p. 15/24)(see diagram p. 15/24)
•• This gives equilibrium constant K This gives equilibrium constant K = y= yH2OH2O/x/xH2OH2O = H= HcH2OcH2O / p/ ptotaltotal = = 0.2 MPa/0.1 MPa = 20.2 MPa/0.1 MPa = 2
•• How many stages for yHow many stages for y’’00 = 0 in the = 0 in the dry Ndry N22, and V, and V´́/L = 2/L = 2½½××((VV´́/L)/L)minmin ?? Picture:
WK92
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Stripper: Stripper: wet gas drying /10wet gas drying /10
•• L/VL/V´́maximummaximum = V= V´́/L/Lminimumminimum = 0.4 = 0.4 →→ ×× 22½½ gives gives VV´́/L = 1/L = 1•• Draw up operating line and count stages: N = 2Draw up operating line and count stages: N = 2
Equilibriumline
Operating line
Note: operating line now underequilibrium line
Picture: WK92
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11. Separation processes11. Separation processes::continuous distillationcontinuous distillation
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Continuous Continuous distillationdistillation
Pictures: T68
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Distillation, principleDistillation, principle•• Liquid with composition Liquid with composition
aa11 boils at temperature boils at temperature TT22, giving vapour with , giving vapour with composition acomposition a’’22 which which is is enrichedenriched in in component Acomponent A
•• Taking out and cooling Taking out and cooling vapour avapour a’’22 gives liquid gives liquid aa44 at temperature Tat temperature T44
•• In equilibrium with In equilibrium with liquid aliquid a44 is vapour ais vapour a’’44, , again again further enrichedfurther enrichedin component Ain component A Picture: A83
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Continuous distillation Continuous distillation (binary) /1(binary) /1
•• Separating a mixture Separating a mixture of 2 components A of 2 components A and B, with A more and B, with A more volatilevolatile
•• Liquid for Liquid for absorption sectionabsorption sectionproduced by produced by condensing some top condensing some top productproduct
•• Gas for Gas for stripping stripping sectionsection produced by produced by boiling some bottom boiling some bottom productproduct
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
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Continuous distillationContinuous distillation /2/2
•• Roughly Roughly equimolar equimolar exchangeexchange: 1 mol A : 1 mol A liquid liquid →→ gas gas ««--»»1mol 1mol B gas B gas →→ liquidliquid
•• As a result: As a result: Gas and liquid Gas and liquid streams ~ constant in streams ~ constant in each section: L and V each section: L and V in top section, Lin top section, L’’ and and VV’’ in bottom sectionin bottom section
•• FeedFeed enters there enters there where it is similar to where it is similar to the mixture inside the mixture inside
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
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Continuous distillationContinuous distillation /3/3•• Some general considerations:Some general considerations:
–– The equimolar exchange A The equimolar exchange A ↔↔ B requires balanced energy B requires balanced energy exchange as well: vaporisation heats of A and B should be exchange as well: vaporisation heats of A and B should be roughly the sameroughly the same
–– Pressure Pressure and theand the temperature rangestemperature ranges must be chosen:must be chosen:•• Below the critical pressures (and temperatures) of the Below the critical pressures (and temperatures) of the
componentscomponents•• The components must be thermally stable (for hydrocarbon The components must be thermally stable (for hydrocarbon
fractions of crude oil: below 300 ~ 350fractions of crude oil: below 300 ~ 350°°C)C)•• The temperature at the top of the column should preferably be The temperature at the top of the column should preferably be
> 40> 40°°C, allowing for heat transfer with surrounding air or waterC, allowing for heat transfer with surrounding air or water–– Equilibrium data is needed, and data on the feed (see later!) Equilibrium data is needed, and data on the feed (see later!)
and the required products specificationsand the required products specifications
→→ The necessary gas and liquid streams, the number The necessary gas and liquid streams, the number of stages and heat consumption can calculated.of stages and heat consumption can calculated.
!
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Example:Example:CC33 –– iCiC44
distillation distillation /1/1Give the Give the ranges for ranges for temperature and temperature and pressurepressure for a propane for a propane --isobutane distillation. isobutane distillation. Data:Data:1) Equilibrium data 1) Equilibrium data →→2) Components stable up2) Components stable up
to 300to 300°°CC3) Critical pressure 42 bar3) Critical pressure 42 bar
for Cfor C33, 37 bar for iC, 37 bar for iC44..
vapour pressures C1 ... C10Raoults law: K = y/x = p°/ptotal
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Example:Example:CC33 –– iCiC44
distillation distillation /2/2aa
Answer:Answer:
pressures < criticalpressures < criticalpressures allowable pressures allowable
for Cfor C33 –– iCiC44 : p < 37 bar: p < 37 bar
37 bar boiling points 37 bar boiling points 85 85 -- 140140°°C: stability OKC: stability OK
Tboil iC4at 37 bar
Tboil C3at 37 bar
Raoults law: K = y/x = p°/ptotal
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Example:Example:CC33 –– iCiC44
distillation distillation /2/2bb
Answer:Answer:
Chose Chose 4040°°C as the C as the temperatur at the temperatur at the condensorcondensor..
At TAt Tminmin = 40= 40°°C, the C, the vapour pressure of the vapour pressure of the more volatile more volatile componentcomponent which is Cwhich is C33
= 14 bar = 14 bar →→ p p ≥≥ 14 bar 14 bar
Then TThen Tboilboil ~ 81~ 81°°C for iCC for iC44
Tboil C3at 14 bar
p° C3at 40°C
Tboil iC4at 14 bar
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Distillation: the x,y diagram Distillation: the x,y diagram /1/1
Mass balance Mass balance top sectiontop section for tray i: for tray i: ((x, y for volatile component)x, y for volatile component)
with distillate product D (kmol/s)with distillate product D (kmol/s)With LWith Li+1 i+1 ~ L~ Lii = L, V= L, Vi+1i+1 ~ V~ Vii = V := V :
LLii = R= RDD··D and VD and Vii = D + L= D + Lii, gives, gives
Dii xVDx
VLy += 1+
i+1i
V(kmol/s)
L(kmol/s)
RefluxRD·D
(kmol/s)
DD
iD
Di x
Rx
RRy
+11+
+1= 1+
Dii xVDx
VLy += 1+
Diiii DxxLVy += 1+1+
Picture: after WK92
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Distillation: the x,y diagram Distillation: the x,y diagram /2/2Mass balance Mass balance bottom sectionbottom section for for tray j: (tray j: (x, y for volatile component)x, y for volatile component)
with distillate product D (kmol/s)with distillate product D (kmol/s)With LWith L’’j+1 j+1 ~ L~ L’’jj = L= L’’, V, V’’j+1j+1 ~ V~ V’’jj = V= V’’ ::
LL’’jj = V= V’’jj + B and V+ B and V’’jj = R= RBB··B, givesB, gives
Dii xVDx
VLy += 1+
j+1j
V’(kmol/s)
L’(kmol/s)
RefluxRB·B
(kmol/s)
BB
jB
Bj x
Rx
RRy 1−+1= 1+
Bjj x'VBx
'V'Ly −= 1+
1+1+=+ jjBjj xLBxVy
Picture: after WK92
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Operation line top sectionOperation line top section
Source: T68, see also SH06Source: T68, see also SH06
condenser
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3
n-hexane - n-octane 1 atm
0.0
0.2
0.4
0.6
0.8
1.0
0.0 0.2 0.4 0.6 0.8 1.0x n-hexane (mol/mol)
y n-
hexa
ne (m
ol/m
ol)
Locating flows in x,y diagramsLocating flows in x,y diagrams
1
2
Heat
xD
x1
x2
y1
y2
y3
xd,yd
x1,y2
x1,y1x2,y2
x2,y3
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Operation line bottom sectionOperation line bottom section
Source: T68, see also SH06Source: T68, see also SH06
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McCabe McCabe --Thiele diagram Thiele diagram /1/1
•• Working line top Working line top sectionsection with slope with slope L/V = RL/V = RDD/(R/(RDD+1) < 1+1) < 1
•• Working line Working line bottom sectionbottom section with with slope Lslope L’’/V/V’’ = = (R(RBB+1)/R+1)/RBB >1>1
•• Product xProduct xBB=y=yBB, x, xDD=y=yDDon x=y diagonal lineon x=y diagonal line
•• Next:Next: the feedthe feed (x(xFF,y,yFF))
Picture: WK92
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McCabe McCabe --Thiele diagram Thiele diagram /2/2•• A third line in the x,y A third line in the x,y
diagram is the sodiagram is the so--called called qq--line through feed line through feed point (xpoint (xFF,y,yFF) and ) and crossing point (xcrossing point (xSS,y,ySS))of operating lines for of operating lines for top and bottom sectiontop and bottom section
•• For point xFor point xss, y, yss ::
xS xF
yS
yF
slope-q/(1-q)
where where ΔΔV = VV = V--VV’’ is vapor entering with the feed, and is vapor entering with the feed, and ΔΔL = LL = L’’--L is liquid entering with the feedL is liquid entering with the feed
FSS xq
xqqy
1−1−
1−=
BDss
Bss
Dss
xBxDLΔxVΔyxBx'L'VyxDxLVy
⋅+⋅+⋅−=⋅⋅−⋅=⋅⋅+⋅=⋅
Picture: after WK92
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McCabe McCabe --Thiele diagram Thiele diagram /3/3
•• q = fraction of feed that q = fraction of feed that gives liquid on the gives liquid on the feeding trayfeeding tray: L: L’’= L+q= L+q··F,F,
•• FF··q = q = ΔΔL, FL, F··(1(1--q) = q) = ΔΔVV
•• q is related to the q is related to the energyenergyneeded to convert the feed needed to convert the feed completely into vapour. With completely into vapour. With enthalpies H for saturated enthalpies H for saturated liquid and saturated gas it is liquid and saturated gas it is found thatfound that
q >1q = 1
0 < q < 1q = 0q < 0
sat,Lsat,G
Fsat,G
HHHH
q−
−=
Picture: after T68
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Distillation: feed temperatureDistillation: feed temperature
q = fraction of feed that gives liquid on the feeding trayq = fraction of feed that gives liquid on the feeding trayq = 0 ~ saturated gas; q = 1 ~ saturated liquidq = 0 ~ saturated gas; q = 1 ~ saturated liquid
Picture: WK92
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Having determined Having determined •• the equilibrium curve;the equilibrium curve;•• the operation lines for top the operation lines for top and bottom section;and bottom section;•• the qthe q--line, line, the the number of equilibrium number of equilibrium stagesstages can be counted.can be counted.
Note: the qNote: the q--line is fixed by (xline is fixed by (xFF,y,yFF) ) and q. In practice the operaton and q. In practice the operaton line for the bottom section is line for the bottom section is usually determined from the usually determined from the operation line for the top section operation line for the top section and the qand the q--line !line !
McCabe McCabe --Thiele diagramThiele diagram /4/4
Example: 4 stages bottom section +5.4 stages top section, total 10 stages
Picture: WK92
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Location of the feed trayLocation of the feed tray
(a)(a) feed on 7feed on 7thth tray from top; tray from top; (b)(b) feed on 4feed on 4thth traytrayfrom top; from top; (c)(c) feed on 5feed on 5thth tray (optimum)tray (optimum)
above feed tray:above feed tray:count with operationcount with operationline for top section;line for top section;
below feed tray:below feed tray:continue with continue with operation line operation line for bottom sectionfor bottom section
Picture: after T68
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McCabe McCabe --Thiele diagramThiele diagram /5/5
•• The point (xThe point (xSS,y,ySS) may be on ) may be on the equilibrium line, if L/V = the equilibrium line, if L/V = (L/V)(L/V)minmin, or L/V = R, or L/V = Rminmin/(R/(Rminmin+1)+1)reflux ratio too small reflux ratio too small (not (not enough liquid for absorption)enough liquid for absorption)
•• This results in a problematic This results in a problematic ””pinchpinch”” point (xpoint (xSS,y,ySS) = (x*,y*) ) = (x*,y*) where N = where N = ∞∞ equilibrium equilibrium stages are needed: stages are needed: RRminmin/(R/(Rminmin+1) = (x+1) = (xDD--y*)/(xy*)/(xDD--x*),x*),or: Ror: Rminmin = (x= (xDD--y*)/(y*y*)/(y*--x*)x*)
•• The The optimal reflux ratiooptimal reflux ratio R R depends on costs, depends on costs, typically R = 1.1~1.3 Rtypically R = 1.1~1.3 Rminmin
(x*,y*)
Picture: after WK92
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Example:Example: CC33 –– iCiC44 distillation distillation //44
•• For the CFor the C33 ––iCiC44 distillation distillation (see above)(see above), at p = 14 bar, , at p = 14 bar, with F = 500 mol/s, xwith F = 500 mol/s, xFF = 0.40, x= 0.40, xDD = 0.95, x= 0.95, xBB = 0.1 and = 0.1 and q = 2/3:q = 2/3:–– Produce an (x,y) equilibrium diagram using the Produce an (x,y) equilibrium diagram using the
nomogram given above. nomogram given above. –– Determine the minimal (top section) reflux RDetermine the minimal (top section) reflux R–– Construct the two operation lines in the (McCabe Construct the two operation lines in the (McCabe
––Thiele) diagram for R = 1.2 Thiele) diagram for R = 1.2 ×× RRminmin
–– Determine the number of equilibrium stages in Determine the number of equilibrium stages in bottom (includes feed) and top section.bottom (includes feed) and top section.
–– Calculate the product streams (in mol/s or kg/s)Calculate the product streams (in mol/s or kg/s)
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Answer part 1:Answer part 1:•• For the (x,y) equilibrium plot; For the (x,y) equilibrium plot;
temperature varies from 40temperature varies from 40°°C C at the condensor to 81at the condensor to 81°°C at the C at the reboiler. p = 14 barreboiler. p = 14 bar
•• For temperatures in this range, For temperatures in this range, find Kfind KC3C3 and Kand KiC4iC4, and calculate , and calculate x and y. x and y. For example: At 1.4 MPa for For example: At 1.4 MPa for 6060°°C KC KC3C3 = 1.45 and K= 1.45 and KiC4iC4 = = 0.70. Thus 0.70. Thus 1)1) y/x = 1.45 and y/x = 1.45 and 2)2) (1(1--y)/(1y)/(1--x) = 0.7. x) = 0.7. Fill in y = 1.45x in Fill in y = 1.45x in 2)2) givesgives(1(1--1.45x)=0.7(11.45x)=0.7(1--x). This gives x). This gives result x=0.40, y=0.58.result x=0.40, y=0.58. et ceteraet cetera
Example:Example: CC33 –– iCiC44 distillation distillation /5/5
C3 –iC4 liquid – vaporequilibrium
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Example:Example: CC33 –– iCiC44 distillation distillation /5a/5a
•• Method 1: for several T:Method 1: for several T:x = (Kx = (KiC4iC4 –– 1)/(K1)/(KiC4iC4 –– KK33) and) andy = Ky = KC3C3··x gives x gives
(40(40°°C,C,1,1), (451,1), (45°°C,0.8,0.9)C,0.8,0.9)(50(50°°C,0.67,0.82), (55C,0.67,0.82), (55°°C,0.54,0.71),C,0.54,0.71),(60(60°°C,0.40,0.58), (65C,0.40,0.58), (65°°C,0.31,0.48),C,0.31,0.48),(70(70°°C,0.20,0.34), (75C,0.20,0.34), (75°°C,0.11,0.20),C,0.11,0.20),(80(80°°C,0.05,0.10)C,0.05,0.10)
•• Method 2: average Method 2: average temperature ~ 60temperature ~ 60°°C,C,αα == KKc3c3 / K/ KiC4iC4 = 2.07= 2.07plot y = plot y = αα··x /(1 + (x /(1 + (αα--1)1)··x)x)
(at 40(at 40°°C: C: αα= 2.43, at 8= 2.43, at 800°°C: C: αα=1.95)=1.95)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1x
y
relative volatility 1.9 ... 2.4
relative volatility 2.07
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Example:Example: CC33 –– iCiC44 distillation distillation /6/6
Answer part 2:Answer part 2:•• See McCabeSee McCabe--Thiele Thiele
diagram:diagram:RRmin min = 2.45, R = 2.93= 2.45, R = 2.93
•• Top: 7.7 (8) stages, Top: 7.7 (8) stages, bottom 7.1 (8) stagesbottom 7.1 (8) stages
•• Mass balances Mass balances F = D + B, andF = D + B, andFF··xxFF = D= D··xxDD + B+ B··xxBB
gives via 1gives via 1--D/F = B/F D/F = B/F and D/F = (xand D/F = (xFF--xxBB)/(x)/(xDD--xxBB))
D = 0.353D = 0.353··F = 176 mol/s F = 176 mol/s B = 0.647B = 0.647··F = 324 mol/sF = 324 mol/s
C3 – iC4 separation at 14 bar
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Restrictions McCabeRestrictions McCabe--Thiele methodThiele method•• Molar heats of vaporisation should not differ more Molar heats of vaporisation should not differ more
than ~10%than ~10%•• Heats of solution are negligibleHeats of solution are negligible•• Relative volatilities should be 1.3 < Relative volatilities should be 1.3 < αα < 5< 5•• Reflux ratioReflux ratio’’s should be R > 1.1s should be R > 1.1··RRminmin•• Number of trays N < 25 preferably.Number of trays N < 25 preferably.
•• Otherwise: operating lines are presumably not Otherwise: operating lines are presumably not straight: use a more exact method. straight: use a more exact method. for example, Ponchon for example, Ponchon -- Savarit, based on the enthalpySavarit, based on the enthalpy--composition, or H,x chartcomposition, or H,x chart-- see next slidesee next slide
Source: Coulson & Richardson vol. 2 (1983)Source: Coulson & Richardson vol. 2 (1983)
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PonchonPonchon--Savarit method Savarit method (intro)(intro)
For more detail: For more detail: advanced coursesadvanced courses, or literature., or literature.
h,x diagramsh,x diagrams
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12. Separation processes12. Separation processes::packed tower columns packed tower columns
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Pictures:T68
Packed tower columns Packed tower columns /1/1
””MellapakMellapak””http://www.sulzerchemtech.comhttp://www.sulzerchemtech.com
http://www.raschig.de/http://www.raschig.de/
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Packed tower columns Packed tower columns /2/2
•• Mass transfer from gas to liquid or Mass transfer from gas to liquid or vice versa vice versa where where the liquid forms a (thin) film on the surface of packing the liquid forms a (thin) film on the surface of packing material elements, creating a large contact surface material elements, creating a large contact surface ””aa””(m(m22 / m/ m33 apparatus)apparatus)
•• For relatively small For relatively small amounts of material amounts of material transferred (say, < 2% of transferred (say, < 2% of the streams) the process the streams) the process may be considered may be considered isotherm isotherm (vaporisation and (vaporisation and condensation have a heat condensation have a heat effect!)effect!) and streams V and streams V and L may be and L may be considered constant.considered constant. Picture: W92
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Packed tower columnsPacked tower columns
•• In a packed column no discrete, identifiable In a packed column no discrete, identifiable stages are seenstages are seen
•• Similar to a plate column the goal is, however, Similar to a plate column the goal is, however, not that equilbrium is reached at each stagenot that equilbrium is reached at each stage
•• More importantly, mass transfer is accomplished More importantly, mass transfer is accomplished that brings the phases closer to equilibriumthat brings the phases closer to equilibrium
Pictures:↑ BSL60← K71
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Mass balance for a packed columnMass balance for a packed column•• Mass balance for a height section dMass balance for a height section dℓℓ ::
ṅṅ = V= V··dy = kdy = kGG··(c(cGG--ccGiGi))··aa··AA··ddℓℓ= k= kLL··(c(cLiLi--ccLL))··aa··AA··ddℓℓ = L= L··dx dx
with mass transfer coefficients kwith mass transfer coefficients kGG and kand kLL
for gas and liquid side, crossfor gas and liquid side, cross--section area section area A and packing surface per volume a.A and packing surface per volume a.
•• With With overalloverall mass transfer coefficients kmass transfer coefficients koGoG
= K= KGG and kand koLoL = K= KLL, (eliminating the phase , (eliminating the phase interface concentrations), for the gas side:interface concentrations), for the gas side:
VV··dy = Kdy = KGG··(y*(y*--y)y)··(p/R(p/R··T)T)··aa··AA··ddℓℓ
yinyin∫∫ yout yout dy/(y*dy/(y*--y) = y) = 00∫∫ZZ (1/V)(1/V)··KKGG··(p/R(p/R··T)T)··aa··AA··ddℓℓ
and similar for the liquid phaseand similar for the liquid phase…………
Z
Picture: SH06
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SourcesSources•• A83:A83: P.W. Atkins P.W. Atkins ””PhysicalPhysical chemistrychemistry”” 2nd Ed. Oxford Univ. Press (1983)2nd Ed. Oxford Univ. Press (1983)•• BMH99:BMH99: BeekBeek, W.J., , W.J., MuttzallMuttzall, K.M.K., van , K.M.K., van HeuvenHeuven, J.W. , J.W. ””Transport Transport
phenomenaphenomena”” Wiley, 2nd ed. (1999) Wiley, 2nd ed. (1999) •• ÇÇB98:B98: Y. A. Y. A. ÇÇengelengel, M.A Boles , M.A Boles ““ThermodynamicsThermodynamics”” 3rd Ed. McGraw3rd Ed. McGraw--Hill Hill
(1998)(1998)•• CR83CR83: : CoulsonCoulson, J.M., Richardson, J.F., , J.M., Richardson, J.F., BackhurstBackhurst, J.R., , J.R., HarkerHarker, J.H. , J.H.
““Chemical Engineering, Vol. 2 : Unit OperationsChemical Engineering, Vol. 2 : Unit Operations”” PergamonPergamon Press, Press, Oxford (1983)Oxford (1983)
•• SH06:SH06: J.D. J.D. SeaderSeader, E.J. Henley , E.J. Henley ””Separation process Separation process principlesprinciples””, Wiley, , Wiley, 2nd ed. (2006)2nd ed. (2006)
•• SSJ84:SSJ84: J.M. Smith, E. J.M. Smith, E. StammersStammers, L.P.B.M , L.P.B.M JanssenJanssen ””FysischeFysischeTransportverschijnselenTransportverschijnselen II”” TU TU DelftDelft, D.U.M. (1984), D.U.M. (1984)
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