VLE

CPE553 CHEMICAL ENGINEERING THERMODYNAMICS 3/22/2013

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VAPOR LIQUID EQUILIBRIUM

VLE FROM K-VALUE CORRELATION

FLASH CALCULATION

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K-value or equilibrium ratio Ki is a measure of the tendency of a given

chemical species to partition itself between liquid and vapor phases.

It serve as a measure of the “lightness” of a constituent species, i.e., of its

tendency to favor the vapor phase.

When Ki is greater than unity, species i exhibits a higher concentration in

the vapor phase; when less, a higher concentration in the liquid phase, and

is considered a “heavy” constituent.

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ii

i

yK

x(10.10)

The use of K-values allow elimination of one set of mole fractions {yi} or

{xi} in favor of the other.

Reference to Eq. (10.1) shows that the K-value for Raoult’s law is

and reference to Eq. (10.5) shows that for modified Raoult’s law it is

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sat

ii

PK

P(10.11)

sat

i ii

PK

P(10.12)


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According to Eq. (10.10), yi = Kixi. Summation with iyi = 1 yields

For bubblepoint calculations, where the xi are known, the problem is to

find the set of K-values that satisfies Eq. (10.13).

Alternatively, Eq. (10.10) can be written, xi = yi/Ki. Summation with ixi = 1

yields

For dewpoint calculations, where the yi are known, the problem is to find

the set of K-values that satisfies Eq. (10.14).

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1i i

i

K x(10.13)

1i

i i

y

K (10.14)

Equations (10.11) and (10.12) together with Eq. (10.10) represent

alternative forms of Raoult’s law and modified Raoult’s law.

Raoult’s law expresses K-values as functions of just T and P, independent of

the composition of the liquid and vapor phases. This allows K-values to be

calculated and correlated as functions of T and P.

For mixtures of light hydrocarbons and other simple molecules, in which

the molecular force fields are relatively uncomplicated, correlation of this

kind have approximate validity.

Figures 10.13 and 10.14, show monographs for the K-values of light

hydrocarbons as functions of T and P, prepared by Dadyburjor. They do

allow for an average effect of composition, but the essential basis is

Raoult’s law.

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For a mixture of 10 mole % methane, 20 mole % ethane, and 70 mole %

propane at 10oC (283.15K), determine the

(a) dewpoint pressure

(b) bubblepoint pressure

The K-values are given by Fig. 10.13.

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P = 6.9 bar P = 10.34 bar P = 8.7 bar

Species yi Ki yi/Ki Ki yi/Ki Ki yi/Ki

Methane 0.1 20 0.005 13.2 0.008 16 0.006

Ethane 0.2 3.25 0.062 2.25 0.089 2.65 0.075

Propane 0.7 0.92 0.761 0.65 1.077 0.762 0.919

(yi/Ki) = 0.828 (yi/Ki) = 1.174 (yi/Ki) = 1.000

Solution:

(a) When the system is at its dewpoint, only an insignificant amount of liquid

is present, and the given mole fractions are values of yi. For the given

temperature, the K-values depend on the choice of P, and by trial we find

the value for which Eq. (10.14) is satisfied. Results for several values of P

are given as follows:

The results given in the last two columns show that Eq. (10.14) is satisfied

when P = 8.7 bar. This is the dewpoint pressure, and the composition of

the dew is given by the values of xi = yi/Ki listed in the last column of the table.

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P = 26.2 bar P = 27.6 bar P = 26.54 bar

Species xi Ki Kixi Ki Kixi Ki Kixi

Methane 0.1 5.6 0.56 5.25 0.525 5.49 0.549

Ethane 0.2 1.11 0.222 1.07 0.214 1.1 0.22

Propane 0.7 0.335 0.235 0.32 0.224 0.33 0.231

Kixi = 1.017 Kixi = 0.963 Kixi = 1.000

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(b) When the system is at its bubblepoint, the given mole fractions becomes

values of xi. In this case we find by trial the value of P for which the K-values

satisfy Eq.(10.13). Results for several values of P are given in the following

table.

Equation (10.13) is satisfied when P = 26.54 bar. This is the bubblepoint

pressure. The composition of the bubble of vapor is given by yi = Kixi as

shown in the last column.

An important application of VLE is the flash calculation.

The name ‘flash’ originates from the fact that a liquid at a pressure equal to or greater than its bubblepoint pressure “flashes” or partially evaporates when the pressure is reduced, producing a two phase system of vapor and liquid in equilibrium.

P, T-flash calculation refers to any calculation of the quantities and compositions of the vapor and liquid phases making up a two-phase system in equilibrium at known T, P, and overall composition.

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Consider a system containing one mole of nonreacting chemical species with an overall composition represented by the set of mole fractions {zi}. Let L be the moles of liquid, with mole fractions {xi}, and let V be the moles of vapor, with mole fraction {yi}. The material balance equations are:

Combining these equations to eliminate L gives:

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1L V

1, 2, ..., i i iz x L y V i N

1 1, 2, ..., i i iz x V y V i N (10.15)

Substituting xi = yi/Ki, and solving for yi yields:

Because iyi = 1, eq. (10.16) is summed over all species:

The initial step in solving a P, T-flash problem is to find the value of V which satisfies this equation.

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1 1

i ii

i

z Ky

V K

1

1 1

i i

i i

z K

V K (10.17)

(10.16)

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The system acetone(1)/acetonitrile(2)/nitromethane(3) at 80oC (353.15K)

and 110 kPa has the overall composition, z1 = 0.45, z2 = 0.35, z3 = 0.20.

Assuming that Raoult’s law is appropriate to this system, determine L, V, {xi},

and {yi}. The vapor pressures of the pure species at 80oC (353.15K) are:

P1sat = 195.75 kPa P2

sat = 97.84 kPa P3sat = 50.32 kPa

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Solution:

Determine whether the system is in two phase region (Pdew<P<Pbubl), then

only flash calculation can be made.

BUBL P calculation

{zi} = {xi}

By eq. (10.2),

Pbubl = x1P1sat + x2P2

sat + x3P3sat

= (0.45)(195.75) + (0.35)(97.84) +(0.20)(50.32)

= 132.40 kPa

DEW P calculation

{zi} = {yi}

By eq. (10.3),

Because 101.52 kPa <110 kPa < 132.4 kPa, so the system is in two phase

region, and flash calculation can be made.

1 1 2 2 3 3

1101.52kPasat sat satdewP

y P y P y P


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By eq. (10.11), Ki = Pisat/P;

K1 = 1.7795 K2 = 0.8895 K3 = 0.4575

Substitute known values into eq. (10.17),

Solution for V by calculator yields V = 0.7367 mol

Thus, L = 1 – V = 0.2633 mol

0.45 1.7795 0.35 0.8895 0.20 0.45751

1 0.7795 1 0.1105 1 0.5425V V V

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1 1

i i

i i

z K

V K

By eq. (10.16) and eq. (10.10),

1 1

i ii

i

z Ky

V K i

i

i

yx

K

y1 = 0.5087 y2 = 0.3389 y3 = 0.1524

x1 = 0.2859 x2 = 0.3810 x3 = 0.3331

For the system described in Ex. 10.4, what fraction of the system is vapor

when the pressure is 13.8 bar and what are the compositions of the

equilibrium vapor and liquid phases?

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Solution:

The given pressure lies between the dewpoint and bubblepoint pressures

established for this system in Ex. 10.4. The system therefore consists of two

phases. With K-values from Fig. 10.13, the procedure is to find that value of

V for which Eq. (10.17) is satisfied.

By calculator, V = 0.236 mol.

Species zi Ki

Methane 0.1 9.0

Ethane 0.2 1.7

Propane 0.7 0.5

0.1 9.0 0.2 1.7 0.7 0.51

1 8.0 1 0.7 1 0.5V V V

1

1 1

i i

i i

z K

V K (10.17)

The phase compositions are:

y1 = 0.312 y2 = 0.292 y3 = 0.397

x1 = 0.035 x2 = 0.172 x3 = 0.794

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By eq. (10.16) and eq. (10.10),

1 1

i ii

i

z Ky

V K i

i

i

yx

K


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Smith, J.M., Van Ness, H.C., and Abbott, M.M. 2005. Introduction to

Chemical Engineering Thermodynamics. Seventh Edition. Mc

Graw-Hill.

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PREPARED BY:

MDM. NORASMAH MOHAMMED MANSHOR

FACULTY OF CHEMICAL ENGINEERING,

UiTM SHAH ALAM.

[email protected]

03-55436333/019-2368303

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