Exercise 4.
Subject: Vapor-liquid equilibrium for n-heptane (A) - toluene (B) system at 1 atm
Given: Vapor pressure data for n-heptane and toluene, and experimental T-y-x data.
Assumptions: Raoult's and Dalton's laws
Find: (a) x-y plot based on n-heptane, the most volatile component.
(b) T-x bubble-point plot.
(c) A,B and K-values plotted against temperature.
(d) x-y plot based on the average A,B.
(e) Comparison of x-y and T-x-y plots with experimental data.
Analysis: (a) To calculate y-x and T-x-y curves from vapor pressure data, using Raoult's and
Dalton's laws. Eq. (2-47) applies, as well as the sum of the mole fractions in the phases in
equilibrium. Thus,
KA
s
yA PA T lq , KB
s
yB PB T lq (1, 2)
xA P xB P
yA yB 1 , xA xB 1 (3, 4)
Equations (1) to (4) can be reduced to the following equations for the mole
fractions of n-heptane (A) in terms of the K-values:
1 KB
xA , yA KA xA (5, 6)
KA KB
If the given vapor pressure data in Exercise 4.8 for toluene, and this exercise for
n-heptane are fitted to Antoine equations, we obtain:
FG 2855.27 I
J
H
PAs exp 15.7831
T 213.64 K
(7)
F
expG 17.2741
3896.3 I
J
PBs
H T 255.67 K
(8)
Where vapor pressure is in torr and temperature is in oC. Solving, Eqs. (1) to (8),
Exercise 4.9 (continued)
Analysis: (a) (continued)
T, oC Ps of A, torr Ps of B, torr KA KB xA yA A,B
98.4 760.0 528.7 1.0000 0.6956 1.000 1.000 1.438
99.0 773.0 538.3 1.0172 0.7083 0.944 0.961 1.436
100.0 795.9 555.2 1.0472 0.7305 0.851 0.891 1.434
101.0 819.2 572.5 1.0780 0.7533 0.760 0.819 1.431
102.0 843.1 590.2 1.1094 0.7766 0.671 0.745 1.428
103.0 867.6 608.4 1.1415 0.8006 0.585 0.668 1.426
104.0 892.6 627.1 1.1744 0.8251 0.501 0.588 . 1.423
105.0 918.1 646.2 1.2080 0.8503 0.418 0.506 1.421
106.0 944.2 665.8 1.2424 0.8761 0.338 0.420 1.418
107.0 970.9 685.9 1.2775 0.9025 0.260 0.332 1.415
108.0 998.1 706.5 1.3133 0.9296 0.184 0.241 1.413
109.0 1026.0 727.5 1.3500 0.9573 0.109 0.147 1.410
110.0 1054.4 749.1 1.3874 0.9856 0.036 0.050 1.408
110.5 1068.9 760.1 1.4064 1.0001 0.000 0.000 1.406
From this table, an x-y plot is given below.
(b) From the above table, a T-x-y plot is given below. The x-curve is the bubble-point curve,
while the y-curve is the dew-point curve.
(c) A graph of relative volatility and K-values as a function of temperature is given on the next
page.
(d) From the above table, the arithmetic average relative volatility, using the extreme values
is:(A,B)avg = (1.438 + 1.406)/2 = 1.422
Exercise 4.9 (continued)
Analysis: (a) (continued)
Exercise 4.9 (continued)
Analysis: (c) and (d) (continued) Relative Volatility and K-Values
For a constant relative volatility, a rearrangement of Eq. (4-10) applies. For A,B = 1.422,
A,B xA 1422
. xA
yA
c h
1 xA A,B 1 1 0.422 xA
Solving this equation for values of xA = 0 to 1.0 gives the following:
xA yA
0 0.0000
0.1 0.1364
0.2 0.2623
0.3 0.3787
0.4 0.4867
0.5 0.5871
0.6 0.6808
0.7 0.7684
0.8 0.8505
0.9 0.9275
1 1.0000
Exercise 4.9 (continued)
Analysis: (c) and (d) (continued)
y-x Plot for an average relative volatility
0.9
Mole fraction n-heptane in vapor
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Mole fraction n-heptane in liquid
(e) Raoult’s law calculations compared to experimental are as follows:
Raoult’s law Experimental
o
T, C xA yA xA yA
110.75 -0.018 -0.026 0.025 0.048
106.80 0.276 0.350 0.129 0.205
104.50 0.459 0.547 0.250 0.349
102.95 0.589 0.672 0.354 0.454
101.35 0.729 0.793 0.497 0.577
99.73 0.876 0.910 0.692 0.742
98.90 0.954 0.967 0.843 0.864
98.50 0.992 0.995 0.940 0.948
98.35 1.007 1.005 0.994 0.993
The Raoult’s law values are in very poor agreement with the experimental values.
Exercise 4.9 (continued)
Analysis: (e) (continued)
Exercise 4.9 (continued)
Analysis: (e) (continued)
Comparison with Experimental Data
Exercise 4.12
Subject: Vaporizer and condenser heat duties for benzene (A) -toluene (B) mixtures, using an
enthalpy-concentration diagram.
Given: P = 1 atm. Vapor pressure data. Saturated liquid and vapor enthalpy data.
Assumptions: Raoult's law.
Find: (a) Construct an h-x-y plot.
(b) Heat duty for 50 mol% vaporization of a 30 mol% A mixture, starting from liquid
saturation temperature. Heat duty to condense the vapor and subcool it 10 oC.
Analysis: (a) First, compute the vapor and liquid equilibrium compositions at 1 atm and
temperatures from 60 to 100oC using Raoult's law with the vapor pressure data. Eq.
(2-47) applies, as well as the sum of the mole fractions in the phases in equilibrium. Thus,
KA
s
yA PA T lq , KB
s
yB PB T lq (1, 2)
xA P xB P
yA yB 1 , xA xB 1 (3, 4)
Equations (1) to (4) can be reduced to the following equations,
1 KB
xA , yA KA xA (5, 6)
KA KB
Vapor pressure data in Exercises 4.6 for benzene, and 4.8 for toluene give Antoine equations,
FG 2602.34 IJ FG 3896.3 IJ
H
PAs exp 155645
.
T 211271
.
,
K H
PBs exp 17.2741
T 255.67 K (7, 8)
Where vapor pressure is in torr and temperature is in oC. Solving, Eqs. (1) to (8),
T, oC Ps of A, torr Ps of B, torr KA KB xA yA
80.1 759.9 290.0 0.9998 0.3816 1.000 1.000
85.0 880.8 342.7 1.1590 0.4510 0.775 0.899
90.0 1019.1 404.4 1.3409 0.5321 0.579 0.776
95.0 1173.4 474.9 1.5439 0.6249 0.408 0.630
100.0 1345.0 555.2 1.7697 0.7305 0.259 0.459
105.0 1535.0 646.2 2.0198 0.8503 0.128 0.259
110.5 1766.8 760.1 2.3248 1.0001 0.000 0.000
This covers the temperature range of co-existence of vapor and liquid.
Exercise 4.12 (continued)
Analysis: (a) (continued)
Molecular weights are MA = 78 and MB = 92
For a given temperature, compute saturated liquid-phase mixture enthalpies in kJ/kg of
mixture from,
xA M A hLA (1 xA ) M BhLB
hL (9)
xA M A (1 xA ) M B
yA M A hVA (1 yA ) M BhVB
Similarly for the vapor, hV (10)
yA M A (1 yA ) M B
Will have to interpolate and extrapolate given saturated enthalpy data. Liquid enthalpy data are
linear with temperature, therefore, it is found that:
hLA 185
. T 32 , hLB 185
. T 34 (11, 12)
Vapor enthalpy data are not quite linear, but fit the following quadratic equations:
hVA 427 0.85T 0.0025T 2 , hVB 411 0.85T 0.0025T 2 (13, 14)
T, oC xA yA (hL)A , (hL)B , (hV)A , (hV)B , hL , hV ,
kJ/kg kJ/kg kJ/kg kJ/kg kJ/kg kJ/kg
80.1 1.000 1.000 116.2 114.2 511.1 495.1 116.2 511.1
85.0 0.775 0.899 125.3 123.3 517.3 501.3 124.7 515.4
90.0 0.579 0.776 134.5 132.5 523.8 507.8 133.6 519.7
95.0 0.408 0.630 143.8 141.8 530.3 514.3 142.5 523.8
100.0 0.259 0.459 153.0 151.0 537.0 521.0 151.5 527.7
105.0 0.128 0.259 162.3 160.3 543.8 527.8 160.5 531.5
110.5 0.000 0.000 172.4 170.4 551.5 535.5 170.4 535.5
Plots of h in kJ/kg mixture as a function of saturated vapor and liquid mole fractions, and y-x are
given on the next page.
(b) Take a basis of 1 kmol of 30 mol% A - 70 mol% B feed mixture. Then,
kg A = (0.30)(78) = 23.4 kg and kg B = (0.70)(92) = 64.4 kg or 87.8 kg total feed.
Use y-x diagram to obtain compositions of vapor and liquid for 50 mol% vaporized.
From the equation above Eq. (4-6), the slope of the q-line is [(V/F)-1]/(V/F) = (0.5-1.0)/0.5 = -1.
The construction is shown on the y-x diagram, where the intersection with the equilibrium curve
gives xA = 0.22 and yA = 0.38. The mass of liquid = (0.22)(0.5)(78) + (0.78)(0.5)(92) = 44.5 kg.
The mass of vapor = 87.8 - 44.5 = 43.3 kg. On the h-x-y diagram, Point A is the saturated liquid
feed with hL = 150 kJ/kg of feed. Point C is the liquid remaining after 50 mol% vaporization,
with hL, = 158 kJ/kg. Since 44.5/87.8 or 0.507 of the feed is left as liquid, this is equivalent to
(0.507)(158) = 80 kJ/kg feed. Point D is the vapor, with hV = 540 kJ/kg vaporized. Since 0.493
Exercise 4.12 (continued)
Analysis: (b) (continued)
of the feed is vaporized, this is equivalent to (0.493)(540) = 266 kJ/kg feed. Therefore, the
energy required for partial vaporization = 266 + 80 - 150 = 196 kJ/kg of feed.
Point B is the combined vapor and liquid phases after partial vaporization. Point E is
condensed vapor as saturated liquid, with an enthalpy of 145 kJ/kg. This is equivalent to
(0.493)(145) = 71 kJ/kg of feed. Therefore, the condenser duty = 266 - 71 = 195 kJ/kg feed.
Point F is 10oC subcooled condensate, where the enthalpy change from saturation, based
on a liquid specific heat of 1.85 kJ/kg-oC, is 1.85(10)(0.493) = 9 kJ/kg feed. Therefore, the
condenser duty is now 195 + 9 = 204 kJ/kg feed.
Exercise 4.12 (continued)
Analysis: (a) Enthalpy – Composition Diagram
Exercise 4.18
Subject: Derivation of equilibrium flash equations for a binary mixture (1, 2).
Given: Equilibrium equations for a flash in Section 4.3.2.
Find: Derive the equation for = V/F.
Analysis: First derive the equation for = V/F. From Eq. (4-26).
b g b gb
z1 1 K1
1 z1 1 K2 g
0
b g b
1 K1 1 1 K2 1 g (1)
Solving Eq. (1) for , and simplifying,
b g b gb g
z1 1 K1 1 z1 1 K2
b gb g
z1 K1 K2 / 1 K2 1
b gb g b gb gb
z1 1 K1 K2 1 1 z1 1 K2 K1 1 g K1 1
(3)
From Eqs. (4-22) to (4-24) and some simplification gives equations for the mole fractions.
Exercise 4.19
Subject: Flash vaporization of a benzene (A) - toluene (B) mixture for A-B = 2.3.
Given: Feed is 40 mol% A and 60 mol% B.
Find: Percent of A in the equilibrium vapor if 90% of the toluene leaves in the liquid by
graphical means.
Analysis: For constant relative volatility, a rearrangement of Eq. (4-10) applies,
A,B xA
yA
c
1+ xA A,B 1h
Solving this equation for yA as a function of xA ,
xA yA
0.1 0.2035
0.2 0.3651
0.3 0.4964
0.4 0.6053
0.5 0.6970
0.6 0.7753
0.7 0.8429
0.8 0.9020
0.9 0.9539
A plot of the calculated equilibrium curve is given below. To use this plot for a graphical
solution of the equilibrium, draw a q-line, using the following equation for an assumed value of
= V/F and check the resulting % recovery of toluene in the liquid. Vary until the %
recovery = 90%. Then compute, for the corresponding , the % recovery of benzene in the
vapor.
FG 1IJ x FG 1 IJ z FG 1IJ x FG 1 IJ 0.40
yA
H K H K H K H K
A A A (1)
Exercise 4.19 (continued)
Analysis: (continued)
Basis: F = 100 moles, 60 moles toluene (B). Want 0.9(60) = 54 moles B in liquid. Therefore,
60 - 54 = 6 moles B in vapor. Therefore, want (nB)V = yBV = (1 - yA)100Then compute %
recovery of benzene in vapor = (nA)V/40 x 100% = yAV/40 x 100% = 2.5 yAx 100%. The
following are typical values for the trial and error procedure, with the final result at the bottom.
Assumed yA xA (nB)V , % recovery
moles of A in vapor
0.3 0.54 0.35 13.8 40.5
0.2 0.56 0.36 8.8 28.0
0.15 0.575 0.37 6.4 21.6
0.142 0.58 0.375 6.0 20.6
Exercise 4.25
Subject: Partial condensation of a gas mixture at 120oF and 300 psia.
Given: Gas at 392oF and 315 psia, with a composition in kmol/h of 72.53 N2, 7.98 H2, 0.13
benzene, and 150 cyclohexane. The gas is cooled and partial condensed to 120 oF and 300 psia,
followed by phase separation.
Find: Equilibrium vapor and liquid flow rates and compositions.
Analysis: The flash calculations are made conveniently with a process simulator, using an
appropriate K-value correlation. The following results were obtained with CHEMCAD, using
the Chao-Seader, Grayson-Streed (CSGS) method for K-values. Results using the SRK EOS
should be similar.
Component CSGS Ki fi , kmol/h i , kmol/h li , kmol/h
Hydrogen 79.7 72.53 70.82 1.71
Nitrogen 7.54 7.98 6.36 1.62
Benzene 0.024 0.13 0.0016 0.1284
Cyclohexane 0.022 150.00 1.67 148.33
Exercise 4.26
Subject: Rapid determination of phase condition without making a flash calculation.
Given: A hydrocarbon mixture at 200oF and 200 psia, with a composition in lbmol/h of 125
C3, 200 nC4, and 175 nC5, and K-values at these conditions.
Find: Phase(s) present without making a flash condition.
C
Analysis: From Eq. (4-30), have a subcooled liquid if z K
i 1
i i 1.
C
zi
From Eq. (4-31), have a superheated vapor if Ki 1
1.
i
Component fi zi Ki ziKi zi/Ki
C3 125 0.25 2.056 0.514 0.122
nC4 200 0.40 0.925 0.370 0.432
nC5 175 0.35 0.520 0.182 0.673
Total: 500 1.00 1.066 > 1 1.227 > 1
Therefore stream is partially vaporized. Both vapor and liquid phases present.
