Lecture 6.
Single Equilibrium Stages (2)
( )
[Ch. 4]
• Multicomponent Flash, Bubble-Point, and
Dew-Point Calculations
-
Variables and equations in flash vaporization
Isothermal flash
Bubble and dew points
Adiabatic flash
• Ternaryy Liquid-Liquid
q
q
Systems
y
- Carrier A and solvent C mutually insoluble
- Carrier A and solvent C partially soluble
Flash 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
SingleSingle
-Stage Equilibrium Operation
• 3C + 10 variables
F V
F,
V, L,
L zi, yi, xi, TF, TV, TL, PF, PV, PL, Q
V, yi, hV,
PV, TV
F
zi
hF
TF
PF
• 2C + 5 equations
Q
L, xi, hL,
PL, TL
• C + 5 degrees of freedom
Common Sets of Specifications
• C + 3 feed variables (F, TF, PF, zi) are known
 2 additional variables can be specified
V, yi, hV,
PV, TV
F
zi
hF
TF
PF
Q
L, xi, hL,
PL, TL
• TV,,PV
Isothermal flash
• V/F=0, PL
Bubble-point T
• V/F=1,
V/F 1 PV
D
Dew-point
i tT
• V/F=0, TL
Bubble-point P
• V/F=1, TV
Dew-point P
• Q
Q=0
0, PV
Adiabatic flash
• Q, PV
Nonadiabatic flash
• V/F, PV
Percent vaporization
flash
Isothermal Flash (1)
• Isothermal flash calculation
- When the equilibrium
temperature TV (or TL) and
the equilibrium pressure
PV (or PL) are specified
 2C + 5 variables are
determined from 2C + 5
equations
- Not straightforward
because of nonlinear
equations
ti
- Use the Rachford-Rice
procedure when K-values
K values
are independent of
composition
Isothermal Flash (2)
zi (1  Ki )
f {}  
0
i 1 1   ( K i  1)
C
• Solve iteratively by guessing values of  between 0 and 1
until the function f{} = 0
f (x)
• Newton’s
N
’ method
h d

( k 1))

f (x)
f { }

f '{ ( k ) }
(k )
(k )
C
f '{ }  
(k )
i 1
zi (1  K i ) 2
1   ( K i  1) 
(k )
2
 ( k 1)   ( k ) /  ( k )   ( 0.0001)
x
x4 x3 x2 x1
Bubble and Dew Points (1)
zi (1  Ki )
f {}  
0
i 1 1   ( K i  1)
C
• At the bubble point
point,  = 0 and f{0} = 0
f {0}   zi (1  K i )   zi   zi K i  0
i
i

i
z K
i
i
1
i
• At the due point,  = 1 and f{1} = 0
zi (1  K i )
zi
    zi  0
f {1}  
Ki
i
i Ki
i

zi
i K  1
i
• 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
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
- Raoult’s law K i  Pi sat / P
K i   i Pi sat / P
- Modified Raoult
Raoult’s
s law
(3) Iterative calculations
C
f {P}   zi K i  1
i
C
f {P}  
i
zi
1
Ki
Method of false position
P ( k  2)
( k 1)
(k )

f
{
P
}

f
{
P
}
( k 1)
( k 1)
P
 f {P }/ 

( k 1)
(k )
P
P



Adiabatic (Q=0) Flash
Start with guessed value of TV
Start with guessed value of 
(wide-boiling mixtures)
(close-boiling mixtures)
Q
Q
Outer-loop
Outer-loop
Guess 
zi (1  Ki )
0
i 1 1   ( K i  1)
C
f {TV }  
Inner-loop
f {TV }  hV  (1   )hL  hF  0
Satisfy ?
Inner-loop
f {}  hV  (1   )hL  hF  0
Satisfy ?
Ternary LiquidLiquid-Liquid Systems
• Ternary mixtures that undergo phase splitting to form two
separate 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
Components A and C mutually insoluble
Solvent, S
(component C)
Feed, F
(components A, B)
Extract, E
B C)
(components B,
Raffinate, R
(components A, B)
Components A and C partially soluble
Solvent, S
(component C)
Extract, E
A B,
B C)
(components A,
Feed, F
(components A, B)
Raffanite, R
(components A, B, C)
Components A and C mutually
Insoluble
• Solute material balance
X B( F ) FA  X B( E ) S  X B( R ) FA
K 'DB  X B( E ) / X B( R )
 X B( E )  K 'DB X B( R )
XB: ratio of mass (or moles) of solute B, to mass (or moles) of the other
component
p
in F,, R,, or E
K’DB: distribution coefficient defined in terms of mass or mole ratios
X B( R )
X B( F ) FA

FA  K 'DB S
• Extraction factor
factor, EB
E B  K 'DB S / FA
X
( R)
B
/X
(F )
B
1

1  EB
E  : (the extent to which the solute is extracted) 
: fraction of B that is not extracted
Components A and C Partially
Soluble
• Equilateral triangular diagram
-W
Water
t (A),
(A) ethylene
th l
glycol
l
l (B)
(B),
furfural (C)
bubble point pressure
- Above bubble-point
: no vapor phase
Single phase
region
the two liquid phases have
identical compositions
(one phase)
p
Two-liquid
T
li id
phase region
Miscibility limits for water-furfural
Other Liquid
Liquid-Liquid Equilibrium
Diagrams
Right triangular diagram
Equilibrium solute diagram in mass fractions
Equilibrium
q
solute diagram
g
in mass ratios
Janecke diagram
g
[Example] Water
Water-GlycolGlycol-Furfural
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
Determine
i th
the composition
iti off th
the equilibrium
ilib i
extract
t t and
d
raffinate phases produced.
Furfural, S 200 g
100% C
Extract, E
• Basis : 100 g of feed
• Overall material balance
F+S=E+R
Feed, F 100 g
55% A, 45% B
Raffinate, R
[Example] Water
Water-GlycolGlycol-Furfural
Equilibrium (2)
Locate the feed (F) and solvent
(S) compositions
Define M, the mixing point
M=F+S=E+R
Apply the inverse-lever-arm rule
( mass b
(or
balance)
l
) tto fi
find
d M point
i t
( F  S ) wC( M )  FwC( F )  SwC( S )
F wC( S )  wC( M ) SM
 (M )

(F )
S wC  wC
MF
Find E and R along a tie line
Apply
A
l the
th iinverse-leverl
arm rule to find the
amounts of E and R
When Two Pairs of Components are
Partially Soluble
Miscibility boundaries
are separate
t
: two separate twophase regions
Miscibility boundaries
and tie-line equilibria
merge
g
Tie lines do not merge
: three-phase
th
h
region,
i
RST is formed
As temperature is
reduced,
(a)  (b)  (c)