MULTICOMPONENT DISTILLATION

Operating pressure is defiend at 8.3 barDesired split

COMPONENT feed Mole fr Distillate BottomPropane 5 0.05 5 0.1Isobutane 15 0.15 15 0.3

split N butane 25 0.25 24 1split Isopentane 20 0.2 1 19

N pentane 35 0.35 0.1 35100 1 45.1 55.4

Observation 1Top of column has distillate composition at the dew pointBottom of the column has bottom composition at the bubble point.

Determine top of column temperatureDew point calculation Trial 1 Trial 2

y K value x trial K valueMoles Mole fr T= 100 T=60

Propane 5 0.1109 3.8 0.029175 2.3Isobutane 15 0.3326 2.05 0.162241 1.1N butane 24 0.5322 1.6 0.332594 0.85Isopentane 1 0.0222 0.85 0.026086 0.5N pentane 0.1 0.0022172949 0.65 0.003411 0.4

SUM 45.1 1 0.553507

Dew point of distillate is 60 degreesTop of the column temperature is 60 degree

Determine bottom of column temperatureBubble point calculation Trial 1 Trial 2

K value K valuex T=100 trial T = 120

Moles mole fraction y valuePropane 0.1 0.002 0 0 0Isobutane 0.3 0.005 0 0 0N butane 1 0.018 1.4 0.025271 2.15Isopentane 19 0.343 0.7 0.240072 1.1N pentane 35 0.632 0.5 0.315884 0.95

55.4 1 0.581227

Bubble point of the the bottom is 120 degreesBottom of the column is 120 degrees.

Summary : Top of the column operates at 60 degrees and bottom of the column operates at 120 degrees.The temperature difference is 60 degress across the column.Important assumption that we make is that the distillate condenser is a total condenser and does not sub cool

FIRST STAGE CALCULATIONS

Moles y1 Moles x0Propane 17.5 0.110865 12.5 0.110865Isobutane 52.5 0.332594 37.5 0.332594N butane 84 0.532151 60 0.532151Isopentane 3.5 0.022173 2.5 0.022173N pentane 0.1 0.002217 0.1 0.002217

V= 157.85 L= 112.75 0

65 degreey2 x1 K value

Propane 0.064671 0.046194 2.4Isobutane 0.293 0.277162 1.2N butane 0.594029 0.61878 0.86Isopentane 0.04593 0.055432 0.4N pentane 0.005913 0.007391 0.3

1.003542 CHECK SUM 1.004959

We have L0,V1,y1,x0,x1Unknowns are V2, L1 and y2Assume L0 = L1Assume V1 = V2Find y2 Vy2 + Lx0 = Vy1 + Lx1

y2 = (L/V)* (x1 - x0)+ y1

Energy balanceStream Lo L1 V1 V2Temp 60 65 65 70

EnthalpyPropane 20400 21000 34000 34800Isobutane 23400 24900 41000 41300N butane 25200 26000 43700 44200Isopentane 27500 28400 52000 52500

N pentane 30000 30700 54800 55000Propane H * L or H*V 2262 970.0665 3769 2251Isobutane 7783 6901.33 13636 12101N butane 13410 16088.28 23255 26256Isopentane 610 1574.279 1153 2411N pentane 67 226.9032 122 325

24131 25761 41935 43344

We had assumed equimolar counterflow which has to be corrected. We find values of V2 and L1

Energy balanceInput 2720750 + V2*43114Output L1*25591 + 6619480

V2*43114 - L1*25591 = 3898730

Mass BalanceV2. - L1 = 45.1

(L1+45)*43114 - L1*25591 = 389873017523 L1 = 1958600

L1 = 111.8V2 = 156.9

Iteration for correcting values of L1 and V2 instead of assumed value of L=L1=Lo and V=V1=V2

SECOND STAGE CALCULATIONS

V2y2

V3y3

y2 V2 x1 L1Propane 0.064671 10.15 0.046194 5.16Isobutane 0.293 45.96 0.277162 30.98N butane 0.594029 93.19 0.61878 69.16Isopentane 0.04593 7.21 0.055432 6.20N pentane 0.005913 0.9275571217 0.007391 0.826113

156.9 111.8

We assume1) V2 = V32) L1 = L23) Temp of stage 2 we take trial for 65 degree, 70 degree and 75 degree

From assumption 3, we can find x2 using y2 as in stage x2 and y2 are in equilibrium at 70 degree.(Dew point calculation)

Trial 1 Trial 2y2 K @ 70 C x2 K @ 65

Propane 0.064671 2.5 0.025868 2.4Isobutane 0.293 1.3 0.225384 1.2N butane 0.594029 0.9 0.660032 0.86Isopentane 0.04593 0.5 0.091859 0.4N pentane 0.005913 0.4 0.014782 0.3

1.017926

From trials, K has to be 70 degrees.

We find y3 based in assumption 1 and 2.

Required cut for the distillation

data requiredFeed compositionDistillate compositionBottom compositionK value (If not available, then vapour pressure)

Random low vaules are given instead of Zero so that in stage calculations, they don’t continue toremain as zero !!

Note 1Distillate is at 60 degrees.First stage temp is assumed higher at 65 degrees to reflect increasing temp tillthe bottom where temp is 120 degrees5 degree incresase per stage is assumed

0.0482020.302358

0.626060.044346

01.020966

Provided the condenser does not subcool and is a total condenser:FUNDATop of the column is at dew point of the distillate compositionBottom of the column is at the bubble point of the bottom composiiton

00

0.0388090.3772560.6001811.016245

Summary : Top of the column operates at 60 degrees and bottom of the column operates at 120 degrees.

Important assumption that we make is that the distillate condenser is a total condenser and does not sub cool

Reflux ratio ( L/D) 2.5V - L = DV - 112.75 = 45.1

V = 157.85L = 112.75

criteria x1 and y1 are in equilibriumx1 will be at the bubble point

L1x1

L2x2

we take trial for 65 degree, 70 degree and 75 degree

From assumption 3, we can find x2 using y2 as in stage x2 and y2 are in equilibrium at 70 degree.(Dew point calculation)

Trial 3x2 K @750.026946 2.75 0.0235170.244166 1.4 0.2092850.690731 1 0.5940290.114824 0.5 0.0918590.019709 0.4 0.0147821.096377 0.933472