For a non-ideal system, where the molar latent heat is no longer constant and where there is a substantial heat of mixing, the calculations become much more tedious.For binary mixtures of this kind a graphical model has been developed by RUHEMANN, PONCHON, and SAVARIT, based on the use of an enthalpy-composition chart.It is necessary to construct an enthalpy-composition diagram for particular binary system over a temperature range covering the two-phase vapor-liquid region at the pressure of the distillation.

The following data are needed:Heat capacity as a function of temperature, composition and pressure.Heat of mixing and dilution as a function of temperature and composition.Latent heats of vaporization as a function of composition and pressure or temperature.Bubble-point temperature as a function of composition and pressure.

(1)In regular / ideal mixtures: (2)For gaseous / vapor mixtures at normal T and P:(3)Enthalpy of liquid:

Enthalpy-composition diagram

The equations used to calculate enthalpy of liquid and vapor are:(3)(4)(5)(6)

EXAMPLE 2Devise an enthalpy-concentration diagram for the heptane-ethyl benzene system at 760 mm Hg, using the pure liquid at 0C as the reference state and assuming zero heat of mixing.SOLUTION

heptaneethyl benzene TB (C)136.298.5CP (cal/mole K)51.943.4 (cal/mole)75758600

t, CxHyHHEB136.20.0000.000--1.00129.50.0800.2331.231.00122.90.1850.4281.191.02119.70.2510.5141.141.03116.00.3350.6081.121.05110.80.4870.7291.061.09106.20.6510.8341.031.15103.00.7880.9041.001.22100.20.9140.9631.001.2798.51.0001.0001.00--

t = 136.2CxH = 0.0xEB = 1.0= 0 + 1.0 (43.4) (136.2 0) + 0 = 5,911 cal/molemix = xH H + xEB EB = 0 + 1.0 (8,600) = 8,600H = h + mix = 5,920 + 8,600 = 14,511 cal/mole

t = 129.5CxH = 0.08xEB = 0.92= 0.08 (51.9) (129.5) + 0.92 (43.4) (136.2) = 5,708 cal/molemix = xH H + xEB EB = 0.08 (7,575) + 0.92 (8,600) = 8,518H = h + mix = 5,708 + 8,518 = 14,226 cal/moleThe computations are continued until the last point where xH = 1.0 and xEB = 0.0

t, CxHh(cal/mole)H(cal/mole)136.20.0005,91114,511129.50.0805,70814,226122.90.1855,52713,937119.70.2515,45013,793116.00.3355,36513,621110.80.4875,26713,368106.20.6515,19713,129103.00.7885,16012,952100.20.9145,12712,79098.51.0005,11212,687

Vapor2 PhaseLiquidSaturated liquidSaturated vapor

Chart4

5911.0814511.08

5708.3614226.36

5527.1202513937.49525

5450.3599513793.08495

5364.7113621.335

5267.376613368.2016

5196.737713129.4627

5160.09412952.394

5127.133812790.2838

5112.1512687.15

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42.7083333333

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106.7708333333149.4791666667000

0.080.080.233

0.1850.1850.428

0.2510.2510.514

0.3350.3350.608

0.4870.4870.729

0.6510.6510.834

0.7880.7880.904

0.9140.9140.963

111

00.031

0.010.031

0.010

0.010.031

Sheet1

x

y

Sheet2

0.5342845483

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0.0305852728

Sheet3

8518hLHV

606136.205,91114,511

7912129.50.085,70814,226

122.90.1855,52713,937

119.70.2515,45013,793

1160.3355,36513,621

110.80.4875,26713,368

106.20.6515,19713,129

1030.7885,16012,952

100.20.9145,12712,790

98.515,11212,687

Sheet3

x

H

MBD00045BF0.unknown

MBD00096B53.unknown

MBD00028449.unknown

Over-all material balance:F = V + LF xF = V y + L xF hF = V H + L h(7)(8)(9)The enthalpy-concentration diagram may be used to evaluate graphically the enthalpy and composition of streams added or separated.Steady-state flow system with phase separation and heat addedComponent material balanceEnthalpy balance:

For adiabatic process, q = 0:V (H hF) = L (hF h)V (y xF) = L (xF x)(10)(12)(11)(13)Substituting eq. (7) to (9) gives:Substituting eq. (7) to (8) gives:

(14)Substituting eq. (12) to (13) gives:Eq. (14) can be rearranged:(15)

Enthalpy-concentration lines adiabatic, q = 0According to eq. (15), the slopes of both lines are the same.Since both lines go through the same point (F), the lines lie on the same straight line.

LEVER-ARM RULE PRINCIPLEConsider triangle LBVSimilarly:

Over-all material balance:F = D + B (16)Component material balance:F xF = D xD + B xB (17)F xF = D xD + (F D) xB (18)(19)Enthalpy balance:(20)

V1 = L0 + D(21)V1 y1 = L0 x0 + D x0 (22)Material balance around condenser:Component material balance:Enthalpy balance:AV1L1L0DxDqDqD + V1 H1 = L0 h0 + D hD(23)

Combining eqs. (21) and (24): Internal reflux is shown as:(25)(26)Designating:V1 H1 = L0 h0 + D (hD QD)(24)

Internal reflux between each plate, until a point in the column is reached where a stream is added or removed, can be shown as:(27)

L0, Dh0, hDV1 H1(hD QD), xDhD QD H1H1 hD y1, x0, xDH or h

V1 = L0 + D(28)y1 V1 = x0 L0 + yD D(29)Material balance:Component material balance:Enthalpy balance:qD + V1 H1 = L0 h0 + D HD(30)Designating:V1 H1 = L0 h0 + D (hD QD)(31)

Combining eqs. (28) and (30): Internal reflux is shown as:(32)(33)Internal reflux between each plate, until a point in the column is reached where a stream is added or removed, can be shown as:(34)

h0, x0V1 HD, yD(HD QD), yDHD QD H1H1 h0 y1, x0, yDH or hD

The material balance equation maybe rearranged in the from of difference:L0 V1 = L1 V2 = L2 V3 = . . . . = Lm Vm+1= D = (35)L0 V1 = D = nV2

Combining eqs. (35) and (36):xD = x(37)For the component material balance:L0 x0 V1 y1 = L1 x1 V2 y2= L2 x2 V3 y3 = . . . . = Lm xm Vm+1 ym+1= D xD = x(36)L0 x0 V1 y1 = D xD = x

For the enthalpy balance:L0 h0 V1 H1 = L1 h1 V2 H2 = L2 h2 V3 H3 = . . . . = Lm hm Vm+1 Hm+1 = D (hD QD) = h(38)These 3 independent equations [eqs. (35), (36), and (37)] can be written for rectifying section of the column between each plate.On the enthalpy scale and on the composition scale, the differences in enthalpy and in composition always pass through the same point, ([xD, (hD QD)]This is designated as point , the difference point, and all lines corresponding to the combined material and enthalpy balance equations (operating line equations) for the rectifying section of the column pass through this intersection.Combining eqs. (23) and (35):h = hD QD(39)

PLATE-TO-PLATE GRAPHICAL PROCEDURE FOR DETERMINING THE NUMBER OF EQUILIBRIUM STAGES:Use R, xD, HD or hD to establish the location of point with x = xD and h = hD QD or h = HD QD Use Equilibrium data alone to establish the point L1 at (x1, h1). Since L1 is assumed to be a saturated liquid, x1 must lie on the saturated-liquid line.Draw the operating line between L1 and . This line intersects the saturated-vapor line at V2 (y2, H2).Repeat steps 2 and 3 until the feed plate is reached.

mN

The material balance equation maybe rearranged in the from of difference:(40)

For the component material balance:(41)Combining eqs. (40) and (42):(42)

For the enthalpy balance:(43)Combining eqs. (40) and (43):(44)

These 3 independent equations [eqs. (40), (41), and (43)] can be written for stripping section of the column between each plate.On the enthalpy scale and on the composition scale, the differences in enthalpy and in composition always pass through the same point, [xB, (hB QB)].This is designated as point , the difference point, and all lines corresponding to the combined material and enthalpy balance equations (operating line equations) for the stripping section of the column pass through this intersection.

F = D + B(45)Combining eq. (45) with eqs. (35) and (40) gives:(46)QB is usually not known. It can be derived from over-all material balance:

PLATE-TO-PLATE GRAPHICAL PROCEDURE FOR DETERMINING THE NUMBER OF EQUILIBRIUM STAGES:Draw a straight line passing through F and . Draw a vertical straight line at xB all the way down until it intersects the extension of line F in Assuming the reboiler to be an equilibrium stage, the vapor VM+1 is in equilibrium with the bottom stream. Use equilibrium data alone to establish the value of ym+1 on the saturated-vapor line. Draw the operating line between Lm(xm, hm) and VM+1. This line intersects the saturated-liquid line at Repeat steps 4 and 5 until the feed plate is reached.

LMLM-1

DBFqDqBV1L0The construction may start from either side of the diagram, indicating either the condition at the top or the bottom of the column.Proceed as explained in previous slides.In either case, when an equilibrium tie line crosses the line connecting the difference points through the feed condition, the other difference point is used to complete the construction.

EXAMPLE 3Using the enthalpy-concentration diagram from Example 2, determine the following for the conditions in Example 1, assuming a saturated liquid feed.The number of theoretical stages for an operating reflux ratio of R = L0/D = 2.5Minimum reflux ratio L0/D.Minimum equilibrium stages at total reflux.Condenser duty feeding 10,000 lb of feed/hr, Btu/hr.Reboiler duty, Btu/hr.SOLUTION(a)From the graph: hD = h0 = 5,117 cal/mole H1 = 12,723 cal/moleQD = 26,621 cal/mole

h = hD QD = 5,117 ( 26,621) = 31,738 cal/mole The coordinate of point is:x = xD = 0.97Draw a straight line passing through and F. Extend the line until it intersects a vertical line passing through xB, atDraw operating lines and equilibrium lines in the whole column using the method explained in the previous slides.Number of stages = 11

= 21,700 cal/mole= 1.18(b)

N = 7(c)

Chart5

00

0.080.233

0.1850.428

0.2510.514

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0.4870.729

0.6510.834

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0.9140.963

11

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0.2510.2510.514

0.3350.3350.608

0.4870.4870.729

0.6510.6510.834

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111

00.031

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0.010

0.010.031

Sheet1

0000

0000

0000

0000

0000

0000

0000

0000

0000

0000

0000

0000

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Sheet2

0.5342845483

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Sheet3

8518hLHV

606136.205,91114,511

7912129.50.085,70814,226

122.90.1855,52713,937

119.70.2515,45013,793

1160.3355,36513,621

110.80.4875,26713,368

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000

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111

Sheet3

H

x

y

MBD00045BF0.unknown

MBD00096B53.unknown

MBD00028449.unknown

(d)hD QD = h = 31,738 cal/molehD = 5,117 cal/moleQD = 26,621 cal/mole= 1,981,843 Btu/hr (e)hB QB = 14,350 cal/molehB = 5,886 cal/moleQB = 14,350 + 5,886 = 20,236 cal/mole= 2,631,751 cal/mole