Lecture 6 Single equilibrium stages (2) - CHERIC · Isothermal Flash (1) •Isothermal flash...
Transcript of Lecture 6 Single equilibrium stages (2) - CHERIC · Isothermal Flash (1) •Isothermal flash...
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Lecture Lecture 6. 6. ( )( )Single Equilibrium Stages (2)Single Equilibrium Stages (2)
[Ch. 4][Ch. 4][Ch. 4][Ch. 4]
• Multicomponent Flash, Bubble-Point, and Dew-Point Calculations
- Variables and equations in flash vaporization
- Isothermal flash
- Bubble and dew points
- Adiabatic flash
• Ternary Liquid-Liquid Systemsy q q y
- Carrier A and solvent C mutually insoluble
- Carrier A and solvent C partially soluble
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Flash VaporizationFlash Vaporization
• Flash : a single-equilibrium-stage distillation in which a feed is partially vaporized to give a vapor richer in the more-volatile components than the feed
• If the equipment is properly designed, the vapor and liquid leaving the flash drum are in equilibrium
Flash vaporization Partial condensation
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SingleSingle--Stage Equilibrium OperationStage Equilibrium Operation
• 3C + 10 variables
F V L T T T P P P Q
V, yi, hV, PV, TV
F, V, L, zi, yi, xi, TF, TV, TL, PF, PV, PL, Q
• 2C + 5 equationsPV, TV
Fzi QzihFTFP
Q
PF
L, xi, hL, PL, TL
• C + 5 degrees of freedom
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Common Sets of SpecificationsCommon Sets of Specifications
• C + 3 feed variables (F, TF, PF, zi) are known
• TV,PV Isothermal flash
2 additional variables can be specified
V, V
• V/F=0, PL Bubble-point T
V/F 1 P D i t T
V, yi, hV, PV, TV
• V/F=1, PV Dew-point T
• V/F=0, TL Bubble-point P
• V/F=1, TV Dew-point P
• Q=0 PV Adiabatic flash
FzihF
Q
Q 0, PV Adiabatic flash
• Q, PV Nonadiabatic flashTFPF
• V/F, PV Percent vaporization flash
L, xi, hL, PL, TL
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Isothermal Flash (1)Isothermal Flash (1)
• Isothermal flash calculation
- When the equilibrium temperature TV (or TL) and the equilibrium pressurethe equilibrium pressure PV (or PL) are specified
2C + 5 variables are 2C + 5 variables are determined from 2C + 5 equations
- Not straightforward because of nonlinear
tiequations
- Use the Rachford-Rice procedure when K-valuesprocedure when K values are independent of composition
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Isothermal Flash (2)Isothermal Flash (2)
(1 ){ } 0C
i iz Kf
• Solve iteratively by guessing values of between 0 and 1
1{ } 0
1 ( 1)i i
fK
• Solve iteratively by guessing values of between 0 and 1 until the function f{} = 0 )(xfN ’ h d
( )( 1) ( ) { }kk k f
)(xf
• Newton’s method
( ) ( )( )'{ }kf
2(1 )C z K( )2( )1
(1 )'{ }1 ( 1)
k i i
ki i
z KfK
x1x2x3x4
x( 1) ( ) ( )/ ( 0.0001) k k k
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Bubble and Dew Points (1)Bubble and Dew Points (1)
(1 ){ } 0C
i iz Kf
• At the bubble point = 0 and f{0} = 0
1{ } 0
1 ( 1)i i
fK
• At the bubble point, = 0 and f{0} = 0
{0} (1 ) 0i i i i if z K z z K 1i iz K i i i i
• At the due point, = 1 and f{1} = 0
(1 ){1} 0i i ii
i i ii i
z K zf zK K
1i
i i
zK
i i ii iK K i iK
• For a given feed composition, zi, the above equation can be used to find T for a specified P or to find P for a specified T
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Bubble and Dew Points (2)Bubble and Dew Points (2)
• How to determine K-values ?
(1) Plots of K-values for a specific T and P
(2) Equations for vapor-liquid equilibria(2) Equations for vapor-liquid equilibria
- Raoult’s law
Modified Raoult’s law
/sati iK P P
/satK P P- Modified Raoult s law
(3) Iterative calculations
/i i iK P P
{ } 1C
i ii
f P z K { } 1C
i
i i
zf PK
Method of false position
( 1) ( ){ } { }k kf P f P ( 1) ( )( 2) ( 1) ( 1)
( 1) ( )
{ } { }{ }/k k
k k kk k
f P f PP P f PP P
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Adiabatic (Q=0) FlashAdiabatic (Q=0) Flash
Start with guessed value of TV Start with guessed value of
Q
(wide-boiling mixtures) (close-boiling mixtures)
Q
Outer-loop
Guess Outer-loop
1
(1 ){ } 01 ( 1)
Ci i
Vi i
z Kf TK
Inner-loop
i
Inner-loop
{ } (1 ) 0 V V L Ff T h h h Satisfy ? { } (1 ) 0 V L Ff h h h Satisfy ?
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Ternary LiquidTernary Liquid--Liquid SystemsLiquid Systems
• Ternary mixtures that undergo phase splitting to form two separate liquid phases : using solubility differenceseparate liquid phases using solubility difference
• Extract : the exiting liquid phase that contains the solvent and the extracted solute
• Raffinate : the exiting liquid phase that contains the carrier, A, and the portion of the solute, B, that is not extracted
Solvent, S(component C)
Extract, E(components B C)
Solvent, S(component C)
Extract, E(components A B C)
Components A and C mutually insoluble Components A and C partially soluble
(component C) (components B, C) (component C) (components A, B, C)
Feed, F(components A, B)
Raffinate, R(components A, B)
Feed, F(components A, B)
Raffanite, R(components A, B, C)
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Components A and C mutually Components A and C mutually InsolubleInsolubleInsolubleInsoluble
• Solute material balance( ) ( ) ( )F E R( ) ( ) ( )F E RB A B B AX F X S X F
( ) ( )' / E RK X X ( ) ( )' E RB D BX K X/
BD B BK X X
XB: ratio of mass (or moles) of solute B, to mass (or moles) of the other component in F, R, or E
BB D BX K X
p , ,
K’DB: distribution coefficient defined in terms of mass or mole ratios
( )( )
FR B AX FX ( )
'B
R B AB
A D
XF K S
• Extraction factor E• Extraction factor, EB
' /BB D AE K S F E : (the extent to which the solute is extracted)
: fraction of B that is not extracted( ) ( )/ 1
1R F
B BB
X XE
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Components A and C Partially Components A and C Partially SolubleSolubleSolubleSoluble
• Equilateral triangular diagram
W t (A) th l l l (B)- Water (A), ethylene glycol (B), furfural (C)
- Above bubble-point pressure
Single phase region
Above bubble point pressure: no vapor phase
the two liquid phases have identical compositions (one phase)p
T li idTwo-liquid phase region
Miscibility limits for water-furfural
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Other LiquidOther Liquid--Liquid Equilibrium Liquid Equilibrium DiagramsDiagramsDiagramsDiagrams
Right triangular diagram Equilibrium solute diagram in mass fractions
Equilibrium solute diagram in mass ratios Janecke diagramq g g
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[Example] Water[Example] Water--GlycolGlycol--Furfural Furfural Equilibrium (1)Equilibrium (1)Equilibrium (1)Equilibrium (1)
A 45% by weight glycol (B)-55% water (A) solution is ( )contacted with twice its weight of pure furfural solvent (C)
at 25℃ and 101 kPa.
D t i th iti f th ilib i t t dDetermine the composition of the equilibrium extract and raffinate phases produced.
• Basis : 100 g of feed
Furfural, S 200 g 100% C
Extract, E
• Overall material balance
F + S = E + R
Feed, F 100 g55% A, 45% B
Raffinate, R
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[Example] Water[Example] Water--GlycolGlycol--Furfural Furfural Equilibrium (2)Equilibrium (2)Equilibrium (2)Equilibrium (2)
Locate the feed (F) and solvent (S) compositions(S) compositions
Define M, the mixing pointM = F + S = E + RM = F + S = E + R
Apply the inverse-lever-arm rule ( b l ) t fi d M i t(or mass balance) to find M point
( ) ( ) ( )
( ) ( )
( ) M F SC C C
S M
F S w Fw Sw( ) ( )
( ) ( )
S MC CM F
C C
w wF SMS w w MF
Find E and R along a tie line
A l th i lApply the inverse-lever-arm rule to find the amounts of E and R
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When Two Pairs of Components are When Two Pairs of Components are Partially SolublePartially SolublePartially SolublePartially Soluble
Miscibility boundaries t
Tie lines do not merge: th h iare separate
: two separate two-phase regions
: three-phase region, RST is formed
Miscibility boundaries and tie-line equilibria merge
As temperature is reduced, (a) (b) (c)g