Question

A mixture of hydrocarbons (10mol% methane, 20 mol% ethane, 30mol% propane, 15mol% isobutane, 20 mol% n-butane, 5 mol% n-pentane) is flashed into a separator at 80 F and 150 psia. What fraction leaves the separator as liquid and what is its composition?

Answer 1

Vapor-Liquid Equilibrium Calculations Using K Values

The vapor-liquid equilibrium constant or distribution coefficient for component A is defined as

                                                                                                                     

where yA = mole fraction of A in the vapor phase and xA = mole fraction of A in the liquid phase

For light hydrocarbon systems the K values have been determined semi-emperically and can be evaluated from the equations given below.

In general, K is a function of temperature, pressure, and composition.

Equilibrium K values for light hydrocarbon systems

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(1) ln K = -A/T² + B - C ln(P) + D/P², where P is in psia, T is in ^oR (THIS IS IMPORTANT)

compound                   A                  B                     C                     D              

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Methane                292860            8.2445             .8951               59.8465          

Ethane                   687248.2         7.90694           .866                 49.02654              

Propane                 970688.6         7.15059           .76984             6.90224          

i-Butane                1166846          7.72668           .92213             0                     

n-Butane               1280557          7.94986           .96455             0                     

n-Pentane              1524891          7.33129           .89143             0                     

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80^oF is 539.67^oR

Substituting and calculating we get

K methane is 15.74

K for ethane is 3.35

K for propane is 0.96

K for isobutane is 0.41

K for n-butane is 0.28

K for n-pentane is 0.093

From these K values we can calculate the mole fraction of these components in the liquid

KA = yA/xA

where yA = mole fraction of A in the vapor phase and xA = mole fraction of A in the liquid phase

So for Methane it is

15.746 = yA/xA

xA = yA/15.746

if yA = 1 xA = 1/15.746 This is the mole fraction of methane now since methane is 10mol% in the liquid it will be 0.0063mol%

Similarly for ethane it is 0.0596mol% in liquid

Propane it is 0.312mol%

i-butane it is 0.369mol%

n-butane is 0.719 mol%

n-pentane is 0.53 mol%

The mole % sum in liquid is 2.002

Next we calculate the total sum of the material from the mole % mentioned in the problem using the molecular weight of the respective compounds so it is 1.6 for 10mol% methane (MW 16), 6 for 20mol% ethane (MW 30), 13.2 for 30 mol% propane (MW 44) 8.7 for 15mol% isobutane (MW 58), 11.6 for 20 mol% n-butane(MW 58), 3.6 for 5 mol% n-pentane (MW 72). The sum of this is 44.7.

So the fraction in liquid is 0.0447

Its composition can be obtained by dividing the individual mole % in the liquid shown above with the total sum and converting into % for example for methane it is (0.006351/2.002)*100 = 0.317%

or for pentane it is (0.535/2.002)*100 = 26.75%

So the final composition is methane 0.317mol%, Ethane 2.98mol%, propane 15.58mol%, i-butane 18.43mol%, n-butane 35.93 mol% and n-pentane 26.74 mol%.

I did all calculations using excel ans the calculations are complicated, so I have not shown all intermediate calculations but only the final values.