3 Mccabe Thiele Method

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NUMBER OF EQUILIBRIUM STAGESIN BINARY DISTILLATIONGRAPHICAL METHODMcCABE-THIELE METHODThis method is based on the Lewis modification of the Sorel method, It assumes equimolal overflow in the rectifying section, in the stripping section, and equimolal latent heats,L0 is a saturated liquidColumn pressure and reflux ratio are fixed,FLmL0DxDVm+1qDmBqBpOverall mass balance:F = D + B

ENVELOPE AVm+1 = Lm + DVm+1 ym+1 = Lm xm + D xD

(1)(2)(3)This is an equation of a straight line on a plot of vapor composition versus liquid composition, where (Lm/Vm+1) is the slope and (DxD/Vm+1) is the intercept which passes through the point (xD, xD) and (xm, ym+1),v1v2L1L0DxDAFLmVm+1v1qDm4Since all L values are equal and all V values are equal (due to constant molal overflow assumption:

(4)Equation (4) is the operating line or material balance line for the rectifying section,Since:

Vm = Lm + D

In term of R, equation (4) can be written as:

(5)xxD

ENVELOPE B

(6)(7)(8)BqB

pp+1

Since all L values are equal and all V values are equal (due to constant molal overflow assumption:

(9)Equation (9) is the operating line or material balance line for the stripping section,This is an equation of a straight line with slope and intercept passing through (xB, xB) and (xp, yp+1),This line can be drawn from point (xB, yB) to point or with slope

(10)(11)(12) and is calculated by material and enthalpy balancerelationship around the feed plate,

(13)The problem is, how to calculate and ?FLmVm

q is the number of moles of saturated liquid formed on the feed plate by the introduction of 1 mole of feed:q = 1: saturated liquid feed, xF = xiq = 0: saturated vapor feed, xF = yiq > 1: cold liquid feed, xF < xiq < 1: superheated vapor, xF > xi0 < q < 1: two-phase feed, xF xi

Substituting eqs, (10) and (13) to eq, (9) yields: (14)This equation gives the slope of the operating line in the stripping section asThere is an easier way to draw the operating line in the stripping section, i,e, by using the q-line, which started from point (xF, yF = xF),

Component material balance of the feed:

(15)Eq, (12) is the equation of the q line having a slope of q/(q 1) and terminating at xF on the 45 line and at point (xi, yi),Saturated liquid feed : q = 1: slope = Saturated vapor feed : q = 0: slope = 0Cold liquid feed : q > 1: slope = +Superheated vapor feed: q < 1: slope = Two-phase feed : 0 < q < 1: slope = xF xDxB q = 1 q > 10 < q < 1q = 0xF xDxB

xF xDxB x1, y1x2, y2x3, y3x4, y4x1, y2x2, y3x3, y4MINIMUM REFLUXxF xDxB

MINIMUM REFLUXxF xDxB

19TOTAL REFLUXxF xDxB EXAMPLE 2Using the data of EXAMPLE 1, determine:The number of equilibrium stages needed for saturated-liquid feed and bubble-point reflux with R = 2,5 using McCabe-Thiele graphical methodRminMinimum number of equilibrium stages at total reflux.SOLUTION(a)The slope of the operating line in the rectifying section:

yxN = 11

(b)Intercept =

Rmin = 1,18 (c)

N = 8SIDE PRODUCTIf a product of intermediate composition is required, a vapor or a liquid side stream can be withdrawn,This kind of column configuration is typical of the petrochemical plants, where the most common running unit operation is the fractional distillation,This consists in splitting a mixture of various components, the crude oil, into its components, Because of their different boiling temperatures, the components (or so-called fractions) of the crude oil are separated at different level (i,e, plate) of the column, where different boiling temperatures are present,The fractions are then withdrawn from the plate where they form, therefore the column presents numerous side streams,

Lm VmD, xD B, xBF, xFL0 S, xSRectifying sectionMiddle sectionStripping section

D, xD L0 S, xSVm+1LmmMATERIAL BALANCE IN RECTIFYING SECTIONAssuming constant molar overflow, then for the rectifying section the operating line is given by:

(16)D, xD F, xFL0 S, xSRectifying sectionMiddle section

MATERIAL BALANCE IN MIDDLE SECTIONOverall:

(19)(18)(17)Component:Since the side stream is normally removed as a liquid:

For constant molal overflow:

(20)VmLm

S

which is the mean molar composition of the overhead product and side streams,

Since xS < xD and , this additional operating line cuts the line y = x at a lower value than the operating line though it has a smaller slope,

Equation (20) represents a line of slope , which passes through the point

MATERIAL BALANCE IN STRIPPING SECTIONF, xFB, xB

Overall:

(23)(22)(21)Component:For constant molal overflow:(24)

Equation (24) represents a line of slope , which passes through the point (xB, xB)

F

xBxDMULTIPLE-FEEDLm VmD, xD B, xBF2, xF2L0 F1, xF1Rectifying sectionMiddle sectionStripping section

D, xD L0 F1, xF1Vm+1LmmMATERIAL BALANCE IN RECTIFYING SECTIONAssuming constant molar overflow, then for the rectifying section the operating line is given by:

(16)D, xD F2, xF2L0 F1, xF1Rectifying sectionMiddle section

MATERIAL BALANCE IN MIDDLE SECTIONOverall:

(27)(26)(25)Component:For constant molal overflow:

(29)Equation (29) represents a line of slope

Material balance around feed plate F1 :

F1VmLm

(30)(31)

MATERIAL BALANCE IN STRIPPING SECTIONF2, xF2B, xB

Overall:

(34)(33)(32)Component:For constant molal overflow:(35)

Equation (24) represents a line of slope , which passes through the point (xB, xB)

Material balance around feed plate F2 :

F2

(36)(37)

EXAMPLE 3Benzene (1) is to be separated from toluene (2) in a distillation column. Three feed streams are available:No. of stream123Lb-mol/h200500300Benzene mole fraction0.70.50.2Fraction vaporized0.250.00.5q0.751.00.5A high-purity product containing 98-mol% benzene is required. Toluene is to be recovered at 95 mol % purity. A total overhead condenser is to be used. The column is designed to operate at R = 1.25 Rmin, how many stages is required, and where should be the feed streams?T (K)x1y13850.0000.0003800.1000.2103750.2300.4103700.3650.5753650.5250.7253600.7000.8503550.9100.9653531.0001.000SOLUTIOND, xD B, xBF2, z2L0 F1, z1Rectifying sectionMiddle section IStripping sectionF3, z3Middle section IISlope of feed lines:Feed 1:

Feed 2:

Feed 3:

OVERALL MATERIAL BALANCE

COMPONENT MATERIAL BALANCE

(a)(b)Simultaneous solution of eqs. (a) and (b) results in: D = 430.1andB = 569.9

Minimum Reflux:

Column is operated at R = 1.25 Rmin = 1.2(see the following graph)F1F2F3Intercept = 0.5Rectifying section:

Middle section I:

Middle section II:

Rectifying section:

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xy

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Sheet100000000000000000000000000000000000000000000000000000000

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Sheet3

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MBD00096B53.unknown

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Chart500000.080.2330.080.080.1850.4280.1850.1850.2510.5140.2510.2510.3350.6080.3350.3350.4870.7290.4870.4870.6510.8340.6510.6510.7880.9040.7880.7880.9140.9630.9140.91411110.980.9800.980.980.980.980.980.010.010.0100.010.010.010.01

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Sheet3

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xy

MBD000DE40B.unknown

MBD00113F0F.unknown

MBD0011FC61.unknown

MBD00111B34.unknown

MBD00045BF0.unknown

MBD00096B53.unknown

MBD00028449.unknown