adv. thermo assignment
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Modelling of diethyl ether and ethene ideal behaviour. Modelling is done using Raoult's and Dalton's law and aTRANSCRIPT
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Assignment Title: Vapour – Liquid Equilibrium
Subject Number: CHEN9007
Subject Name: Advanced Thermodynamics and reactor engineering
Student Name: Rocky Tran
Lecturer/Tutor: Anthony Stickland
Due Date: 16th October,2015
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Executive Summary
The vapour – liquid equilibrium behaviour of diethyl ether and ethylene is modelled and the mixture is
determined whether it is ideal or not. The mixture is simulated using the Antione equation and Aspen
under isothermal and isobaric conditions. A mixture of diethyl ether and ethylene is easily separable
as the vapour pressure of ethane is much higher than diethyl ether. An inlet feed with an ethylene
liquid fraction of 0.5 requires 2 theoretical stages under both isothermal and isobaric conditions
assuming Raoult’s and Dalton’s laws apply. Modelling on Aspen using an NRTL fluid package
indicates that the mixture is ideal with activity coefficients equal to 1.
Vapour Pressure of Diethyl Ether and Ethylene
Vapour pressures are calculated using the Antione equation which is shown in equation (1). In the
Antione equation P represents pressure which is in units kPa, T represents temperature in Celsius
and Antione coefficients are represented by A,B and C.
log10 𝑃 = 𝐴 −𝐵
𝐶 + 𝑇, (1)
All Antione coefficients for Ethylene and Diethyl ether and shown in Table (1) below. The
temperature ranges where these values are applicable are also shown in the table. All Antione
coefficients are obtained from Yaws et Al. (2005).
Component A B C Tmin Tmax
Diethyl ether 6.966 649.806 262.73 -169.14oC 9.21oC
Ethylene 7.046 1112.55 232.657 -116.3oC 193.55oC
The vapour pressures of the two components are plotted between -116oC – 9.2oC and shown in
Figure (1). From this data, this gives a Tavg value of -53.4oC (219.6K) and a geometric mean pressure
of 29.92 kPa.
Figure 1 Vapour pressures of Ethylene and diethyl ether using Antione equation between temperature range of -116 - 9.2C (157K – 282.2K)
0.0001
0.001
0.01
0.1
1
10
100
1000
10000
150 170 190 210 230 250 270 290
Vap
ou
r P
ress
ure
(kP
a)
Temperature (K)
Vapour pressure of Ethylene and Diethyl ether at different temperatures
Ethylene
DiethylEther
Figure 3 Liquid and Vapour compositions in the liquid and vapour phases assuming Raoult’s and Dalton’s Law applies at a constant temperature of 219.6K
Vapour – Liquid Equilibrium under isobaric and isothermal conditions
The mixture of diethyl ether and ethylene is assumed to be to ideal therefore Raoult’s and Dalton’s
laws are applicable. From Raoult’s Law the pressure in the liquid phase can be calculated according
to equation (2) while the pressure in the vapour phase can be calculated according to equation (3).
Component A is diethyl ether and component B is ethylene, Pvap represent the vapour pressures of
either component A or B and xb represents the liquid fraction of ethylene.
𝑃 = 𝑥𝑏𝑃𝑏𝑣𝑎𝑝
+ (1 − 𝑥𝑏)𝑃𝑎𝑣𝑎𝑝
, (2)
𝑃 =1
𝑦𝑏
𝑃𝑏𝑠𝑎𝑡 +
1 − 𝑦𝑏
𝑃𝐴𝑠𝑎𝑡
, (3)
In Figure (2), the mixture is modelled under isothermal conditions at the average temperature of
219.6K. The corresponding liquid and vapour fractions are also shown in Figure (3) under isothermal
conditions calculated using equation (4). The vapour fraction is represented by yi while the liquid
fraction is represented by xi .
𝑦𝑖 =70.61𝑥𝑖
969.69𝑥𝑖 + 0.9222, (4)
Figure 2 The liquid and vapour fractions at different pressures at an isothermal temperature of 219.6K assuming Raoult’s and Dalton’s laws apply
0
200
400
600
800
1000
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Pre
ssu
re (
kPa)
Mole Fraction
Pressure vs composition at constant temperature
Liquid
Vapour
140
160
180
200
220
240
260
280
300
0 0.2 0.4 0.6 0.8 1
Tem
per
atu
re (
K)
Mole Fraction
Temperature vs composition at constant pressure
Similarly, the mixture is modelled under isobaric conditions under the geometric mean pressure of
29.92 kPa. The liquid fractions can be calculated according to equation (5) while the vapour fractions
can be calculated by equation (6). Figure (4) shows the variations of liquid and vapour fractions
under different temperatures. Figure (5) shows the vapour – liquid composition diagram.
𝑥𝑖 =𝑃 − 𝑃𝐴
𝑣𝑎𝑝
𝑃𝐵 − 𝑃𝐴𝑣𝑎𝑝 , (5)
𝑦𝑖 =𝑥𝑖𝑃𝑣𝑎𝑝
𝑃, (6)
Figure 4 The liquid and vapour fractions at different temperatures at an isobaric pressure of 29.92 kPa assuming Raoult’s and Dalton’s laws apply
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
Vap
ou
r Fr
acti
on
of
Eth
ene
Liquid Fraction of Ethene
Liquid/Vapour Compositions at constant pressure
Vapour
Liquid
Figure 5 Liquid and Vapour compositions in the liquid and vapour phases assuming Raoult’s and Dalton’s Law applies at a constant pressure of 29.92 kPa
Vapour – Liquid Separation
A mixture with an ethylene liquid fraction of 0.5 is assumed to fed into the distillation column and
the theoretical number of plates are determined. The desired vapour fraction of ethylene is 0.99 in
the distillate while the liquid fraction of ethylene should be less than 0.01 in the bottoms product.
Figure (6) shows the distillation process under isothermal conditions. For each stage the
corresponding pressures are calculated and shown in Figure (7). The theoretical number of stages
needed is 2.
Figure 6 Distillation of mixture with ethylene liquid fraction of 0.5 under isothermal conditions assuming Raoult’s and Dalton’s Law applies.
FIgure 7 Pressures at each stage of the distillation with an inlet ethylene liquid fraction of 0.5. An ideal mixture is assumed so Raoult’s and Dalton’s laws apply.
The distillation process is also simulated under an isobaric pressure of 29.92 kPa. From Figure (8) the
number of theoretical stages needed is 2. Figure (9) shows the temperature of each stage of the
distillation process.
Figure 8 Distillation of mixture with ethylene liquid fraction of 0.5 under isobaric conditions assuming Raoult’s and Dalton’s Law applies.
Figure 9 Temperatures at each stage of the distillation with an inlet ethylene liquid fraction of 0.5. An ideal mixture is assumed so Raoult’s and Dalton’s laws apply.
Simulation of Ethylene and Diethyl Ether Mixture
The calculations performed using the Antione equation were based on an ideal mixture. However, a
non – ideal solution have an effect on the number of stages needed in distillation as it may
increase/decrease the number of stages. The ethylene and diethyl ether mixture is modelled on
Aspen and the T-x-y diagram is shown in Figure (10). The mixture is modelled using an NRTL fluid
package. From Figure (10) we observe a similar diagram as Figure (9) which may highlight ideal
behaviour. In Table (2) the values of the activity coefficients are given by gamma. Therefore, an
assumption of an ideal solution is valid.
Figure 10 Aspen simulation of an ethylene and diethyl ether mixture assuming constant pressure of 29.92 kPa using an NRTL fluid package.
Table 2 Values of activity coefficients for Aspen simulation using a NRTL fluid package.
Conclusion
The mixture of ethylene and diethyl ether were successfully modelled. Initially the mixture was
modelled as an ideal mixture using Raoult’s and Dalton’s laws which were used to calculate vapour
and liquid fractions. The vapour pressure of ethylene is extremely high therefore the mole fraction
of ethylene in the vapour phase is usually 1. The activity coefficients of 1 demonstrated in Aspen
indicate that the mixture can be approximated by an ideal mixture. Two theoretical stages are
needed to separate an inlet feed mixture with an ethylene liquid fraction of 0.5.
References
Yaws, G.L, Narasimhan, P.K & Gabula, C. 2005. Yaws’ Handbook of Antione Coefficients for Vapor
Pressure. Knovel, Oxford, UK.
Appendices
Derivation of 3
𝑥𝐴 + 𝑥𝐵 = 1
𝑦𝐴𝑃
𝑃𝐴𝑣𝑎𝑝 +
𝑦𝐵𝑃
𝑃𝐵𝑣𝑎𝑝 = 1
𝑃 =1
𝑦𝐴
𝑃𝐴𝑣𝑎𝑝 +
𝑦𝐵
𝑃𝐵𝑣𝑎𝑝
Derivation of 4
𝑃𝐵𝑣𝑎𝑝
= 970.61𝑘𝑃𝑎, 𝑃𝐴𝑣𝑎𝑝
= 0.922𝑘𝑃𝑎
𝑃 = 𝑥𝐵(970.61) + (1 − 𝑥𝐵)(0.922)
𝑃 = 969.69𝑥𝐵 + 0.9222
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑖𝑛𝑡𝑜 𝑦𝐵 =𝑥𝐵𝑃𝐵
𝑣𝑎𝑝
𝑃
𝑦𝑏 =970.61𝑥𝐵
969.69𝑥𝐵 + 0.9222
Derivation of 5
𝐏 = 𝐏𝐁𝐯𝐚𝐩
𝐱𝐁 + 𝐏𝐀𝐯𝐚𝐩
𝐱𝐀
𝐏 = 𝑷𝑩𝒗𝒂𝒑
𝒙𝑩 + 𝑷𝑨𝒗𝒂𝒑
(𝟏 − 𝒙𝑩)
𝒙𝑩 =(𝑷 − 𝑷𝑨
𝒗𝒂𝒑)
𝑷𝑩 − 𝑷𝑨𝒗𝒂𝒑