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V1 qD Over-all material balance: D, xD, hD L0 x0 h0 F=D+B (1) Component material balance: F xF hF qB B, xB, hB F x F = D xD + B x B (2) F xF = D xD + (F – D) xB (3) F x F x B D xD xB (4) ASSUMPTIONS: Equimolal overflow V1 V2 Vn and VF1 VF2 . . . Vm (5) L 0 L1 . . . Ln and LF LF1 . . . Lm (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. ENRICHING SECTION D is calculated using eq. (4) V1 V1 V2 V3 V4 V5 L0 L1 L2 L3 L4 L5 Providing that R is fixed, L0 is calculated using the equation D defining reflux ratio: L0 R (5) D Material balance around condenser: V1 = L 0 + D (6) Since V1 is in equilibrium with L1, the composition (x1) of L1 is obtained from equilibrium data. Material balance around envelope A: A V2 = L 1 + D V1 V2 V3 V4 V5 L0 L1 L2 L3 L4 L5 D (7) Component balance: V2 y2 = L1 x1 + D xD (8) L1 D y2 x1 xD V2 V2 (9) A D xD Vn+1 F xF vF n Ln The composition of V2, V3,. .., Vn is calculated by performing material balance in envelope A: Vn+1 yn+1 = Ln xn + D xD (10) Ln D y n 1 xn xD Vn1 Vn1 (11) n+1 LF-1 L'F Since the molar liquid overflow is constant, Ln = L and Vn+1 = V: L D y n 1 x n x D V V (12) Eq. (12) relates the composition of the vapor rising to a plate to the composition of the liquid on the plate. Equilibrium equation: Vn, yn Stage n Ln, xn y 1 ,n K1 x 1 ,n y 2 ,n 1 y 1 ,n K2 x 2 ,n 1 x 1 ,n K2 1 x1,n 1 y1,n K2 1 x1,n 1 K1 x1,n x1,n K1 K2 1 K2 1 K2 x 1 ,n K1 K 2 FEED PLATE Material balance around feed plate: F VF1 LF1 VF LF vF (13) LF-1 F xF Component balance around feed plate: VF1 LF FxF VF1 yF1 LF1xF1 VFYF LFxF (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) STRIPPING SECTION The composition of Vm is calculated by performing material balance in envelope A: F xF Vm1 m Lm m+1 A B xB Vm1 Lm B (15) Vm1 ym1 Lm xm B xB (16) Lm B y m 1 xm xB Vm1 Vm1 (17) Lm B y m 1 x m x B Vm Vm (18) EXAMPLE 1 Pertinent data on the binary system heptane-ethyl benzene at 760 mm Hg are as follows. t, C 136.2 129.5 122.9 119.7 116.0 110.8 106.2 103.0 100.2 98.5 xH 0.000 0.080 0.185 0.251 0.335 0.487 0.651 0.788 0.914 1.000 yH 0.000 0.233 0.428 0.514 0.608 0.729 0.834 0.904 0.963 1.000 H -1.23 1.19 1.14 1.12 1.06 1.03 1.00 1.00 1.00 EB 1.00 1.00 1.02 1.03 1.05 1.09 1.15 1.22 1.27 -- 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. 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: x 0.771 y3 0.104y2 0.334y Assume 100 mole of feed is introduced: F x F x B 100 0.42 0.01 D 42.71 xD xB 0.97 0.01 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 ENRICHING SECTION Ln D 106.77 42.71 y n 1 x n x D x n 1 0.97 Vn Vn 149.48 149.48 yn1 0.7143 xn 0.277 3 xn 0.771 yn 2 0.104yn 0.334yn n yn xn 1 0.970 0.930 2 0.941 0.865 3 0.895 0.769 4 0.826 0.640 5 0.734 0.494 6 0.630 0.362 FEED PLATE Feed is a saturated liquid: FL = F FV = 0 LF L5 FL 106.77 100 206.77 5 V6 L5 F L6 L7 206.77 6 V7' A L'6 Material balance around feed plate (envelope A): F VF1 LF1 VF LF F V7 L5 V6 L6 100 V7 106.77 149.48 206.77 V7 149.48 7 Component balance around feed plate: F xF VF1 yF1 LF1 xF1 VF yF LF xF F xF V7 y 7 L5 x 5 V6 y 6 L6 x 6 V6 y 6 L 6 x 6 F x F L 5 x 5 y7 V7 y7 149.48 0.63 206.770.362 149.48 1000.42 106.770.494 y7 = 0.497 149.48 STRIPPING SECTION Lm B 206.77 57.29 y m1 x m x B xm 0.01 Vm Vm 149.48 149.48 ym1 1.3833 xm 0.0038 1.383261975 3 xm 0.771 ym m 7 8 9 10 11 2 0.104ym 0.334ym ym 0.497 0.321 0.165 0.073 0.030 xm 0.235 0.122 0.056 0.024 0.010 Number of equilibrium stages = 11 0.5729