Example 7.1
Consider the ternary system: Acetone (1) / Acetonitrile (2) / Nitromethane (3) for which:
ln P1S 14.5463 −
=
2940.46
2940.46
2972.64
ln P2S 14.5463 −
ln P3S 14.2043 −
=
=
;
;
t + 237.22
t + 209.00
t + 237.22
Pi S ( KPa ); t ( 0C ).
Calculate: (a) P, {y i } for a temperature = 80oC, x 1 = 0.3, x 2 = 0.3 (b) P, {x i }, for t = 70oC, y 1
= 0.5, y 2 = 0.3.
(a) For 80oC, P 1 S = 195.75, P 2 S = 97.84, P 3 S = 50.32 KPa. Thus:
=
P
x P 108.3KPa
∑=
S
i i
Next:
yi = xi Pi S / P
Thus:
=
y1 0.54,
=
y2 0.27,
=
y3 0.19
(b) For 70oC, P 1 S = 144.77, P 2 S = 70.37, P 3 S = 43.80 KPa
=
P 1/=
∑ yk / Pk S 81.4 KPa
Next:
xi = yi P / Pi S
Thus:
=
x1 0.28,
=
x2 0.34,
=
x3 0.38
Example 7.2
A liquid mixture containing equimolar amounts of benzene (1) /toluene (2) and ethylbenzene
(3) is flashed to conditions of T = 110oC, P = 90 kPa, determine the equilibrium mole
fractions {x i } and {y i } of the liquid and vapor phase formed and the molar fraction V of the
vapor formed. Assume that Raoult’s law applies. ln P sat ( Pa )= A −
B
t( K ) + C
0
A
B
C
Benzene
13.8594
2773.78
-53.08
Ethylbenzene
14.0045
3279.47
-59.95
Toluene
14.0098
3103.01
-53.36
At T = 383K, the saturation vapour pressures=
are: P1S 233.2;
=
99.1;
P2S =
P3S 47.1kPa
S
Thus:
=
K1 P=
=
K 2 1.1;
=
K 3 0.52
1 / P 2.6; similarly
For flash
n
Zi Ki
∑1 + V (K
i =1
Thus:
i
− 1)
=1
Z1K1
Z2 K2
Z3 K3
+
+
=
1 ..(1)
1 + V ( K1 − 1) 1 + V ( K 2 − 1) 1 + V ( K 3 − 1)
Z=
Z=
Z=
0.33
1
2
3
On substituting the values of K i and Z i by solving eqn. (1) on obtains: V = 0.834
Example 7.3
Methanol (1)-acetone (2) forms an azeotrope at 760 Torr with x1 = 0.2, T = 55.70C. Using van
Laar model predict the bubble pressure for a system with for x 1 = 0.1 at 55.70C.
log10 P1s =
8.0897 − [1582.271 / (t + 239.726)]; log10 P2s =
7.1171 − [1210.595 / (t + 229.664)]
Pi s (torr ); t ( 0C )
=
=
torr ; P2s 745.5torr
At
55.70C P1s 541.75
We assume that the vapour phase is ideal. Hence the VL equation is given by:
yi P = xiγ i Pi S
At the azeotropic condition: yi = xi
Hence,
γ i = P Pi S
/
Thus γ 1
=
1.4313,
=
ln γ 1
and γ 2 1.0318,
=
=
ln γ 2
0.3607
0.0137
The Van Laar parameters are estimated next using the azeotropic composition given by
=
x1 0.2
=
and x2 0.8 :
2
 x ln γ 2 
A12 =
ln γ 1 1 + 2
0.4786
 =
x1 ln γ 1 

2

x ln γ 1 
A21 =
ln γ 2 1 + 1
0.7878
 =
 x2 ln γ 2 
Thus ln γ 1
=
A12
A21
=
;ln γ 2
2
2

 A21 x2 
A12 x1 
1 +

1 +

A12 x1 
 A21 x2 

For x 1 = 0.1, γ 1 = 1.5219, γ 2 = 1.0032
s
torr, y1 γ=
∴P =
∑ xiγ i Pi s =
757.62=
0.1067
1 x1 P1 / P
Example 7.4
2
2
For a binary, the activity coefficients are ln γ 1 = Ax2 and ln γ 2 = Ax1 . Show that the system
(
s
s
forms an azeotrope when A > ln P2 / P1
)
=
γ 1 P=
P1s , γ 2 P P2s
Again
2
2
γ 1 Ax
=
γ 2 Ax
=
P P1s ;ln=
P P2s
ln=
2
1
∴ A ( x22 − x12 ) =
ln ( P2s P1s )
or A ( x2 − x1 ) = A (1 − 2 x1 ) = ln ( P2s P1s )
x1
or=
1 1

1 − ln ( P2s P1s ) 

2 A

For azeotropy 0 < x 1 <1
(
If x 1 = 0, then A = ln P2s P1s
)
If x 1 = 1, then A = − ln ( P2s P1s )
Thus for azeotropy to exist A > ln P2s P1s
Example 7.5
Estimate the vapour pressure of a substance “A’ using PR-EoS, at T = 428oK. For the
substance A: T C = 569.4 K, P C = 2.497 MPa, = 24.97 bar ω = 0.398.
T r = 0.7514
At this temperature for starting the iteration, assume Psat = 0.215 MPa
f ω = 0.37464 + 1.54226ω - 0.26992 ω2 = 0.94570
(
)
2
α PR = 1 + fω 1 − Tr  = 1.2677


o
3
0.45724 R 2TC2α PR 0.45724 x ( 8.314 Pa.m / mol K x 569.4 ) x1.2677
=
a =
PC
24.97 x 105
2
= 5.2024 Pa.m6/mol2
=
b
0.07780 RTC
=
PC
=
A aP
=
/ ( RT )
2
0.07780 x 8.314 x 569.4
= 1.4750 x 10-4 m3/mol
24.97 x105
5.2024 x 0.215 x 106
Assuming P = Psat:
(8.314 x 428.0 )
2
= 8.8398 x 10-2
1.4750 x 10-4 x 0.215 x 106
= 8.9151 x 10-3
B bP
/ RT
=
=
8.314 x 428
Z3+ α Z2 + β Z + γ γ = 0
α = -1+B = -0.9911
β = A-2B-3B2 = 7.0329 x 10-2
γ = -AB + B2+B3 = -7.0789 x 10-4
On solving Z 1 = 0.9151,
Thus ZV = 0.9151,
Z 2 = 1.2106 x 10-2,
Z 3 = 6.39 x 10-2
ZL = 1.2106 x 10-2.
=
Now by
PR-EoS: ln φ ( Z – 1) – ln
(Z − B) –
2
a
 Z + B(1 + 2) 
ln 

2 bRT  Z + B(1 − 2) 
∴ ∴ Putting Z = ZL
ln φ L φL = (1.2106 x 10-2) -1-ln [1.2106 – 8.9151 x 10-3]
–
2
 0.9151 + 8.9151 x 10-3 (1 + 2) 
5.2024
x ln 

-3
2 x 1.475 x 10-4 x 8.314 x 428
 0.9151 + 8.9151 x 10 (1 + 2) 
=
−0.0978 ⇒ φ L =
0.90 ⇒ f L =
φ LP =
0.19 MPa = –0.0978 ⇒ φL = 0.90 ⇒ fL = φLP =
0.19 MPa
Similarly putting Z = ZV gives ln φV = -0.0821 ⇒ ⇒ φV = 0.9212
or fV = 0.1981
→ → If fL > fV,
assumed value of P (=Psat) < actual Psat
→ If fL < fV, assumed value of P (=Psat) > actual P sat
P (revised) = P
fL
fV
For the present case P revised = 0.215 x
0.195
= 0.2116 MPa
0.1981
Use revised ‘P’ to recalculate A, B, {Z i }; thus:
A = aP/(RT)2 = 8.6999 x 10-2 ; B = bP/RT = 8.7742 x 10-3
∴ α = -6.8568 x 10-4; β = 6.922 x 10-2; γ = -6.8568 x 10-4
Resolving f(z) = 0 ⇒ Z 1 = 0.9166; Z 2 = 6.2907 x 10-4; Z 3 = 1.1711 x 10-2
Hence new ZV = 0.9166;
ZL = 1.1711 x 10-2
Using Z = ZV and Z = ZL respectively
New φ φV = 0.9225 ⇒ fV = (0.9225 x 0.2116) = 0.1952 MPa
Similarly φL = 0.9224 ⇒ fL ~ 0.1952 MPa
Thus
Psat ~ 0.2116 MPa
(at 428oK).
Example 7.6
A vapour mixture contains 20mol% methane (1), 30mol% ethane (2) , and rest propane (3), at
300C. Determine the dew composition.
Assume
read off the K factors from the charts.
Hence dew pressure = 2.15 MPa. The dew composition therefore corresponds to the values of
x i computed above.
Example 7.7
For the system of methane (1) and butane (2) compute the bubble pressure for a liquid phase
composition of x 1 = 0.2 at a temperature of 310K, using the PR-EOS.
P (guessed): 40.8 bar
y 1 (guessed) = 0.85; y 2 = 0.15;
First consider calculation of the species fugacity coefficients for the liquid phase as T, P and
x 1 (=0.2) are all know. For this one needs to solve for the cubic EOS with liquid phase
compositions.
T = 310K, P = 40.8bar
For each species the following estimates are made:
Table 1
Parameter
Methane (1)
Butane (2)
fω
3.9310 x 10-3
6.7229 x 10-3
a (Pa.m-6/mol2)
0.198207
1.811717
b (m3/mol)
0.000027
0.000072
With x 1 = 0.2, it follows that
am = x12 a1 + 2 x1 x2 (a1a2 )1/2 + x22 a2 =1.3592 Pa.m −6 / mo 2l
bm = x1b1 + x2b2 =0.000063 m3 / mol
=
Am
=
Bm
am P
= 0.8304
R 2T 2
bm P
= 0.10
RT
With the above values of A and B solve the cubic PR-EOS. The roots are as:
Z1 = 0.1471
Z 2 0.3764 + 0.5899i
=
Z 3 0.3764 − 0.5899i
=
The feasible root for the liquid phase is: ZL = 0.1471
Now using the generalized expression for species fugacity coefficients for PR-EOS:
 bi
bi
am
ai   Z + B (1 + 2) 
φi
ln=
( Z − 1) − ln( Z − B) +
 −2
 ln 

bm
am   Z + B (1 − 2) 
2 2bm RT  bm
Using: Z= ZL = 0.1471; B = B m
ln
φ1
=
am
b1 L
( Z − 1) − ln( Z L − Bm ) +
bm
2 2bm RT
 b1
a1   Z L + Bm (1 + 2) 
2
−

 ln 

am   Z L + Bm (1 − 2) 
 bm
whence : φˆ1L = 4.0271
Similarly: φˆ L = 0.0932
2
Next compute the fugacity coefficients for the vapour phase. The calculations are the same as
above except that x i is replaced with y i . (Note that the pure component properties remain the
same as in Table 1, since the T and R are the same, i.e, 310K and 40.8bar, respectively.
With y 1 = 0.85, it follows that
am = y12 a1 + 2 y1 y2 (a1a2 )1/2 + y22 a2 = 0.3323 Pa.m −6 / mo 2l bm = y1b1 + y2b2 =0.000033 m3 / mol
=
Am
am P
= 0.2030
R 2T 2
=
Bm
bm P
= 0.0528
RT
With the above values of A and B solve the cubic PR-EOS. The roots are as:
Z1 = 0.8537 Z 2 0.0467 + 0.0833i
=
Z 3 0.0467 − 0.0833i =
The feasible root for the liquid phase is: ZV = 0.8537
Now using the generalized expression for species fugacity coefficients for PR-EOS:
φi
ln=
 bi
bi
am
ai   Z + B (1 + 2) 
( Z − 1) − ln( Z − B) +
 −2
 ln 

bm
am   Z + B (1 − 2) 
2 2bm RT  bm
Using: Z= ZV = 0.8537; B = B m
ln
φ1
=
am
b1 V
( Z − 1) − ln( Z V − Bm ) +
bm
2 2bm RT
 b1
a1   Z V + Bm (1 + 2) 
2
−

 ln 

am   Z V + Bm (1 − 2) 
 bm
whence : φˆ1V = 0.9399
Similarly: φˆV = 0.5184
2
Therefore:
y1 φˆ1L 4.0271
= =
= 4.2846
K=
1
x1 φˆ1V 0.9399
0.8569
=
y1 K=
1 x1
y2 φˆ2L 0.0932
= =
= 0.1798
K=
2
x2 φˆV 0.5184
2
0.1438
=
y2 K=
2 x2
Thus : ∑ yi =1.0007
i
Therefore we may terminate the iteration at this point.
Bubble Pressure = 40.8bar
y 1 = 0.8569
Example 7.8
A concentrated binary solution containing mostly species 2 (but x 2 ≠ 1) is in equilibrium with
a vapor phase containing both species 1 and 2. The pressure of this two-phase system is 1 bar;
the temperature is 298.0K. Determine from the following data good estimates of x 1 and y 1 .
H 1 = 200 bar; P2sat = 0.10 bar.
= 200 bar,
= 0.1 bar,
= 1 bar
Assume that vapor phase is ideal at
= 1 bar. Assume Lewis - Randall rule applies to
concentrated species and Henry’s law to dilute species then:
And
Now
Or
Solving gives,
and