2. D, x D, h D B, x B, h B FxFhFFxFhF qDqD qBqB V1V1 L0x0h0L0x0h0 Over-all material balance: F = D + B (1) Component material balance: F x F = D x D + B x B (2) F x F = D x D + (F – D) x B (3) (4)

3. ASSUMPTIONS:  Equimolal overflow and (5) (6)  The two components have equal and constant  H vap.  The sensible enthalpy changes of the vapor and liquid are negligible compared to the latent heats.  The binary mixture behaves as an ideal solution.  The stages are adiabatic except at designated locations (condenser and reboiler).  The pressure is constant throughout the column.

4. ENRICHING SECTION V1V1 V2V2 V3V3 V4V4 V5V5 L1L1 L2L2 L3L3 L4L4 L0L0 D L5L5 D is calculated using eq. (4) Providing that R is fixed, L 0 is calculated using the equation defining reflux ratio: (5) Material balance around condenser: V 1 = L 0 + D(6) Since V 1 is in equilibrium with L 1, the composition (x 1 ) of L 1 is obtained from equilibrium data. V1V1

5. Material balance around envelope A: V 2 = L 1 + D (7) V1V1 V2V2 V3V3 V4V4 V5V5 L1L1 L2L2 L3L3 L4L4 L0L0 D L5L5 A Component balance: V 2 y 2 = L 1 x 1 + D x D (9) (8)

6. The composition of V 2, V 3,..., V n is calculated by performing material balance in envelope A: V n+1 y n+1 = L n x n + D x D (10) vFvF L F-1 DxDDxD A FxFFxF n LnLn V n+1 n+1 (11)

7. Eq. (12) relates the composition of the vapor rising to a plate to the composition of the liquid on the plate. Since the molar liquid overflow is constant, L n = L and V n+1 = V: (12)

8. Equilibrium equation: Stage n V n, y n L n, x n

9. FEED PLATE vFvF L F-1 FxFFxF Material balance around feed plate: (13) Component balance around feed plate: (14) VF+1 and LF are calculated using eqs. (13) along with the information about thermal condition of the feed. y F+1 is calculated using eq. (14)

10. STRIPPING SECTION (15) The composition of V m is calculated by performing material balance in envelope A: BxBBxB A FxFFxF m+1 m (17) (16) (18)

11. EXAMPLE 1 Pertinent data on the binary system heptane-ethyl benzene at 760 mm Hg are as follows. 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--

12. A feed mixture composed of 42 mole % heptane, 58 mole % ethyl benzene is to be fractionated at 760 mm Hg to produce distillate containing 97 mole % heptane and a residue containing 99 mole % ethyl benzene. Using (L/D) = 2.5, determine the number of equilibrium stages needed for a saturated liquid feed and bubble-point reflux.

13. SOLUTION Since x n is calculated using equilibrium relationship, it is necessary to develop an equation correlating x n and y n. Based on the available data, an equation is established: Assume 100 mole of feed is introduced:

14. B = F – D = 100 – 42.71 = 57.29 L 0 = R D = (2.5) (42.71) = 106.77  L 1 = L 2 =.... = L 0 = 106.77 V 1 = L 0 + D = 106.77 + 42.71 = 149.48  V 1 = V 2 =.... = 149.48

15. ENRICHING SECTION nynyn xnxn 10.9700.930 20.9410.865 30.8950.769 40.8260.640 50.7340.494 60.6300.362

16. FEED PLATE Feed is a saturated liquid: F L = F F V = 0 Material balance around feed plate (envelope A): 6 5 7 F V6V6 L5L5 A

17. Component balance around feed plate: y 7 = 0.497

18. STRIPPING SECTION mymym xmxm 70.4970.235 80.3210.122 90.1650.056 100.0730.024 110.0300.010 1.3832619750.5729 Number of equilibrium stages = 11