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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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Student signature …………………………………………………………………… Date …16/10/15…………………………………

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587276

Group Code (if applicable):

Vapour Liquid Equilibrium of

Diethyl Ether and Ethylene

Rocky Tran

587276

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