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This 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, L 0 is a saturated liquid Column pressure and reflux ratio are fixed,

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F LmLm L0L0 DxDDxD V m+1 qDqD m B qBqB p Overall mass balance: F = D + B

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ENVELOPE A V m+1 = L m + D V m+1 y m+1 = L m x m + D x D (1) (2) (3) This is an equation of a straight line on a plot of vapor composition versus liquid composition, where (L m /V m+1 ) is the slope and (Dx D /V m+1 ) is the intercept which passes through the point (x D, x D ) and (x m, y m+1 ), v1v1 v2v2 L1L1 L0L0 DxDDxD A F LmLm V m+1 v1v1 qDqD m

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Since 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,

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Since: V m = L m + D In term of R, equation (4) can be written as: (5)

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xxDxD

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ENVELOPE B (6) (7) (8) B qBqB p p+1

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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 (x B, x B ) and (x p, y p+1 ), This line can be drawn from point (x B, y B ) to point or with slope

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(10) (11) (12) and is calculated by material and enthalpy balance relationship around the feed plate, (13) The problem is, how to calculate and ? F LmLm VmVm

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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, x F = x i q = 0: saturated vapor feed, x F = y i q > 1: cold liquid feed, x F < x i q x i 0 < q < 1: two-phase feed, x F x i

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

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Component material balance of the feed: (15)

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Eq, (12) is the equation of the q line having a slope of q/(q – 1) and terminating at x F on the 45 line and at point (x i, y i ), Saturated liquid feed : q = 1: slope = Saturated vapor feed : q = 0: slope = 0 Cold liquid feed : q > 1: slope = + Superheated vapor feed: q < 1: slope = – Two-phase feed : 0 < q < 1: slope = –

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xFxF xDxD xBxB q = 1 q > 1 0 < q < 1 q = 0

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xFxF xDxD xBxB

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xFxF xDxD xBxB x 1, y 1 x 2, y 2 x 3, y 3 x 4, y 4 x 1, y 2 x 2, y 3 x 3, y 4

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MINIMUM REFLUX xFxF xDxD xBxB

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xFxF xDxD xBxB

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TOTAL REFLUX xFxF xDxD xBxB

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

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y x N = 11

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(b) Intercept = R min = 1,18

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(c) N = 8

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If 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,

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L m V m D, x D B, x B F, x F L0L0 S, x S Rectifying section Middle section Stripping section

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D, x D L0L0 S, x S V m+1 LmLm m MATERIAL BALANCE IN RECTIFYING SECTION Assuming constant molar overflow, then for the rectifying section the operating line is given by: (16)

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D, x D F, x F L0L0 S, x S Rectifying section Middle section MATERIAL BALANCE IN MIDDLE SECTION

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Overall: (19) (18) (17) Component: Since the side stream is normally removed as a liquid: For constant molal overflow: (20)

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VmVm LmLm S

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which is the mean molar composition of the overhead product and side streams, Since x S < x D 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

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MATERIAL BALANCE IN STRIPPING SECTION F, x F B, x B

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Overall: (23) (22) (21) Component: For constant molal overflow: (24) Equation (24) represents a line of slope, which passes through the point (x B, x B )

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F

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xBxB xDxD

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L m V m D, x D B, x B F 2, x F2 L0L0 F 1, x F1 Rectifying section Middle section Stripping section

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D, x D L0L0 F 1, x F1 V m+1 LmLm m MATERIAL BALANCE IN RECTIFYING SECTION Assuming constant molar overflow, then for the rectifying section the operating line is given by: (16)

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D, x D F 2, x F2 L0L0 F 1, x F1 Rectifying section Middle section MATERIAL BALANCE IN MIDDLE SECTION

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Overall: (27) (26) (25) Component: For constant molal overflow: (29) Equation (29) represents a line of slope

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Material balance around feed plate F 1 : F1F1 VmVm LmLm (30) (31)

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MATERIAL BALANCE IN STRIPPING SECTION F 2, x F2 B, x B

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Overall: (34) (33) (32) Component: For constant molal overflow: (35) Equation (24) represents a line of slope, which passes through the point (x B, x B )

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Material balance around feed plate F 2 : F2F2 (36) (37)

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EXAMPLE 3 Benzene (1) is to be separated from toluene (2) in a distillation column. Three feed streams are available: No. of stream123 Lb-mol/h Benzene mole fraction Fraction vaporized q A 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 R min, how many stages is required, and where should be the feed streams?

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SOLUTION D, x D B, x B F 2, z 2 L0L0 F 1, z 1 Rectifying section Middle section I Stripping section F 3, z 3 Middle section II

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Slope of feed lines: 1.Feed 1: 2.Feed 2: 3.Feed 3:

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

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Minimum Reflux: Column is operated at R = 1.25 R min = 1.2 (see the following graph)

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F1F1 F2F2 F3F3 Intercept = 0.5

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1.Rectifying section: 2.Middle section I:

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3.Middle section II: 4.Rectifying section:

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y x N =

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