2.  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.

3. The following data are needed: 1.Heat capacity as a function of temperature, composition and pressure. 2.Heat of mixing and dilution as a function of temperature and composition. 3.Latent heats of vaporization as a function of composition and pressure or temperature. 4.Bubble-point temperature as a function of composition and pressure.

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

5. Enthalpy-composition diagram

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

7. EXAMPLE 2 Devise an enthalpy-concentration diagram for the heptane-ethyl benzene system at 760 mm Hg, using the pure liquid at 0  C as the reference state and assuming zero heat of mixing. SOLUTION heptaneethyl benzene TB (  C) 136.298.5 C P (cal/mole K)51.943.4 (cal/mole) 75758600

8. t,  C xHxH yHyH HH  EB 136.20.000 --1.00 129.50.0800.2331.231.00 122.90.1850.4281.191.02 119.70.2510.5141.141.03 116.00.3350.6081.121.05 110.80.4870.7291.061.09 106.20.6510.8341.031.15 103.00.7880.9041.001.22 100.20.9140.9631.001.27 98.51.000 1.00--

9. t = 136.2  C x H = 0.0x EB = 1.0 = 0 + 1.0 (43.4) (136.2 – 0) + 0 = 5,911 cal/mole mix = x H H + x EB EB = 0 + 1.0 (8,600) = 8,600 H = h + mix = 5,920 + 8,600 = 14,511 cal/mole

10. t = 129.5  C x H = 0.08x EB = 0.92 = 0.08 (51.9) (129.5) + 0.92 (43.4) (136.2) = 5,708 cal/mole mix = x H H + x EB EB = 0.08 (7,575) + 0.92 (8,600) = 8,518 H = h + mix = 5,708 + 8,518 = 14,226 cal/mole The computations are continued until the last point where x H = 1.0 and x EB = 0.0

11. t,  C xHxH h (cal/mole) H (cal/mole) 136.20.0005,91114,511 129.50.0805,70814,226 122.90.1855,52713,937 119.70.2515,45013,793 116.00.3355,36513,621 110.80.4875,26713,368 106.20.6515,19713,129 103.00.7885,16012,952 100.20.9145,12712,790 98.51.0005,11212,687

12. Vapor 2 Phase Liquid Saturated liquid Saturated vapor

13. qF V L Over-all material balance: F = V + L F x F = V y + L x F h F = 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 added Component material balance Enthalpy balance:

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

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

16.    h hFhF H L F V x xFxF y Enthalpy-concentration lines – adiabatic, q = 0 According 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.

17. L F V H hFhF h x xFxF y A B LEVER-ARM RULE PRINCIPLE Consider triangle LBV Similarly:

18. D, x D, H L D B, x B, H L B FxFHFFxFHF qDqD qBqB V1V1 L0x0HL0L0x0HL0 Over-all material balance: F = D + B (16) Component material balance: F x F = D x D + B x B (17) F x F = D x D + (F – D) x B (18) (19) Enthalpy balance: (20)

19. V 1 = L 0 + D (21) V 1 y 1 = L 0 x 0 + D x 0 (22) Material balance around condenser: Component material balance: Enthalpy balance: A V1V1 L1L1 L0L0 DxDDxD qDqD q D + V 1 H 1 = L 0 h 0 + D h D (23)

20. Combining eqs. (21) and (24): Internal reflux is shown as: (25) (26) Designating: V 1 H 1 = L 0 h 0 + D (h D – Q D ) (24)

21. 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) A V m+1 LmLm L0L0 DxDDxD qDqD m

22.  L 0, D h 0, h D V1V1 H1H1  (h D – Q D ), x D h D – Q D – H 1 H 1 – h D y 1, x 0, x D H or h 

23. V 1 = L 0 + D (28) y 1 V 1 = x 0 L 0 + y D D(29) Material balance: Component material balance: Enthalpy balance: q D + V 1 H 1 = L 0 h 0 + D H D (30) Designating: V 1 H 1 = L 0 h 0 + D (h D – Q D )(31) A v1v1 v2v2 v3v3 vFvF L1L1 L2L2 L F-1 L0L0 D, y D FxFFxF qDqD

24. 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)

25. h 0, x 0 V1V1 H D, y D  (H D – Q D ), y D H D – Q D – H 1 H 1 – h 0 y 1, x 0, y D H or h D   

26. V1V1 V3V3 V n+1 L1L1 L2L2 LnLn L0L0 DxDDxD qDqD The material balance equation maybe rearranged in the from of difference: L 0 – V 1 = L 1 – V 2 = L 2 – V 3 =.... = L m – V m+1 = – D =   (35) L 0 – V 1 = – D =  n V2V2

27. Combining eqs. (35) and (36): x D = x  (37) For the component material balance: L 0 x 0 – V 1 y 1 = L 1 x 1 – V 2 y 2 = L 2 x 2 – V 3 y 3 =.... = L m x m – V m+1 y m+1 = – D x D =  x  (36) L 0 x 0 – V 1 y 1 = – D x D =  x 

28. For the enthalpy balance: L 0 h 0 – V 1 H 1 = L 1 h 1 –V 2 H 2 = L 2 h 2 –V 3 H 3 =.... = L m h m – V m+1 H m+1 = – D (h D – Q D ) =  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,  ([x D, (h D – Q D )] 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  = h D – Q D (39)

29. PLATE-TO-PLATE GRAPHICAL PROCEDURE FOR DETERMINING THE NUMBER OF EQUILIBRIUM STAGES: 1.Use R, x D, H D or h D to establish the location of point  with x  = x D and h  = h D – Q D or h  = H D – Q D 2.Use Equilibrium data alone to establish the point L 1 at (x 1, h 1 ). Since L 1 is assumed to be a saturated liquid, x 1 must lie on the saturated-liquid line. 3.Draw the operating line between L 1 and . This line intersects the saturated-vapor line at V 2 (y 2, H 2 ). 4.Repeat steps 2 and 3 until the feed plate is reached.

30.  x or y H or h V1V1 V2V2 V3V3 V4V4 D L1L1 L2L2 L3L3  (x , h  )

31. BxBBxB qBqB m N

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

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

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

35. 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, [x B, (h B – Q B )]. 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.

36. F = D + B (45) Combining eq. (45) with eqs. (35) and (40) gives: (46) Equation (46) implies that lies on the extension of the straight line passing through F and . Q B is usually not known. It can be derived from over-all material balance:

37. PLATE-TO-PLATE GRAPHICAL PROCEDURE FOR DETERMINING THE NUMBER OF EQUILIBRIUM STAGES: 1.Draw a straight line passing through F and . 2.Draw a vertical straight line at x B all the way down until it intersects the extension of line F  in 3.Assuming the reboiler to be an equilibrium stage, the vapor V M+1 is in equilibrium with the bottom stream. 4.Use equilibrium data alone to establish the value of y m+1 on the saturated-vapor line. 5.Draw the operating line between L m (x m, h m ) and V M+1. This line intersects the saturated-liquid line at 6.Repeat steps 4 and 5 until the feed plate is reached.

38.  x or y H or h hBhB LMLM L M-1

39. D B F qDqD qBqB V1V1 L0L0  The 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.

40. H or h F 1 2 3 4 5 67 8 9  xFxF xDxD xBxB 

41. V1V1 y1y1 x1x1 L1L1

42. EXAMPLE 3 Using the enthalpy-concentration diagram from Example 2, determine the following for the conditions in Example 1, assuming a saturated liquid feed. a.The number of theoretical stages for an operating reflux ratio of R = L 0 /D = 2.5 b.Minimum reflux ratio L 0 /D. c.Minimum equilibrium stages at total reflux. d.Condenser duty feeding 10,000 lb of feed/hr, Btu/hr. e.Reboiler duty, Btu/hr. SOLUTION (a)From the graph: h D = h 0 = 5,117 cal/mole H 1 = 12,723 cal/mole Q D = – 26,621 cal/mole

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

44.  F

45.  = 21,700 cal/mole F = 1.18 (b)

46. F 1 23 45 6 7 N = 7 (c)

47. (d)h D – Q D = h  = 31,738 cal/mole h D = 5,117 cal/mole Q D = – 26,621 cal/mole = – 1,981,843 Btu/hr (e) h B – Q B = – 14,350 cal/mole h B = 5,886 cal/mole Q B = 14,350 + 5,886 = 20,236 cal/mole = 2,631,751 cal/mole