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NUMBER OF EQUILIBRIUM STAGES IN BINARY DISTILLATION

NUMBER OF EQUILIBRIUM STAGES IN BINARY DISTILLATION. ANALYTICAL METHOD. LEWIS-SOREL METHOD. Over-all material balance:. V 1. q D. F = D + B (1). D, x D , h D. L 0 x 0 h 0. Component material balance:. F x F = D x D + B x B (2). F x F h F.

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NUMBER OF EQUILIBRIUM STAGES IN BINARY DISTILLATION

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  1. NUMBER OF EQUILIBRIUM STAGES IN BINARY DISTILLATION ANALYTICAL METHOD LEWIS-SOREL METHOD

  2. Over-all material balance: V1 qD F = D + B (1) D, xD, hD L0 x0 h0 Component material balance: F xF = D xD + B xB (2) F xF hF F xF = D xD + (F – D) xB (3) (4) qB B, xB, hB

  3. ASSUMPTIONS: • Equimolal overflow (5) and and (6) • The two components have equal and constant Hvap. • 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. V1 D L0 V2 L1 V3 L2 V4 L3 V5 L4 L5 ENRICHING SECTION D is calculated using eq. (4) V1 Providing that R is fixed, L0 is calculated using the equation defining reflux ratio: (5) Material balance around condenser: V1 = L0 + D (6) Since V1 is in equilibrium with L1, the composition (x1)of L1 is obtained from equilibrium data.

  5. V1 D L0 V2 L1 V3 L2 V4 L3 V5 L4 L5 Material balance around envelope A: A V2= L1+ D (7) Component balance: V2 y2 = L1 x1 + D xD (8) (9)

  6. The composition of V2, V3,. .., Vn is calculated by performing material balance in envelope A: A D xD Vn+1 yn+1= Lnxn+ D xD n (10) Vn+1 Ln n+1 (11) vF LF-1 F xF

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

  8. Equilibrium equation: Vn, yn Stage n Ln, xn

  9. vF LF-1 F xF FEED PLATE 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. yF+1 is calculated using eq. (14)

  10. STRIPPING SECTION The composition of Vm is calculated by performing material balance in envelope A: F xF (15) m m+1 (16) A (17) B xB (18)

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

  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 xn is calculated using equilibrium relationship, it is necessary to develop an equation correlating xn and yn. 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 L0 = R D = (2.5) (42.71) = 106.77  L1= L2 = . . . . = L0 = 106.77 V1 = L0 + D = 106.77 + 42.71 = 149.48  V1= V2 = . . . . = 149.48

  15. ENRICHING SECTION

  16. 5 V6 L5 F 6 A 7 FEED PLATE Feed is a saturated liquid: FL = F FV = 0 Material balance around feed plate (envelope A):

  17. Component balance around feed plate: y7 = 0.497

  18. STRIPPING SECTION Number of equilibrium stages = 11

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