Question

An equimolar liquid mixture of benzene and toluene is separated into two product streams by distillation. A process flowchart and a somewhat oversimplified description of what happens in the process follow: Vapor-97% mole benzene A Reflux Overhead product Filmoi) 24 CONDENSER", mot Feed 100 mol 0.500 mol Bimal La DISTILLATION 0.500 mol/mol Bollup mol B[/mol) 1 Bottoms product REBOILER mol B/D/mol Inside the column a liquid stream flows downward and a vapor stream rises. At each point in the column some of the liquid vaporizes and some of the vapor condenses. The vapor leaving the top of the column, which contains 97 mole% benzene, is completely condensed and split into two equal fractions: one is taken off as the overhead product stream, and the other (the reflux) is recycled to the top of the column. The overhead product stream contains 93% of the benzene fed to the column. The liquid leaving the bottom of the column is fed to a partial reboiler in which 45% of it is vaporized. The vapor generated in the reboiler (the boilup) is recycled to become the nising vapor stream in the column, and the residual reboiler liquid is taken off as the bottom product stream. The compositions of the streams leaving the reboiler are governed by the relation Yb/(1 - YB) = 2.25 XB/(1-x₂) here y, and xg are the mole fractions of benzene in the vapor and liquid streams, respectively.

Answer 1

Solution - 0.49% - fromol F a. 5 molB/mol osmarrimo Here, given that. Tol(1-7B) 2.25 Relodive volatility. I. - / Cr-RB) This is the relative vcelatility. As per process engineer coolulalin. as 10o molter tee. calculation, F = D&B 100 = D&B F xe = xo.D & XB.B. .. Prox0.5= 0.97 DA Choo-D) X 0.03 Drso mollha! TB = somollars. where = teed rote in moll B = Bottom product (Toluenes in mollhor Do Distillate product (Bencene) in mollhr a = feed composition of benene x = Distillate composition of benene XB = Bottom composition of benzene go (1X) 1 56.97 2-25 (-0.977 Rm = 0.8 (1.94 - 0. 135) = 1.44 As per calculation, Ropt = 1.58 km = 1.58 1.44 - 2.166

vapor loud. V-CR+1) = (2.166 + 1) 50 = 158.3. moll for. lav-D - 158.3 - 50 - 108.3 mollhor. As per Benzene top product, can calculate, Benzene vapor velocity, at atmospheric pressure, : As per enperience of calculation, a benzene Toluene Boiling pt = 80.1 Boiling pt - 11100 Mol, wt = 7 8 19lmo) / Mol wt - 92.14 g/mol latent heat - 30.72 kJ/mol Lestent heut - 33.18 kJ/mol s vap of vap The mol volume is - PV, Tz - tho I amx22. 142x2.73 + 80.1) egtx 273 VPTi V = 28.638 malmol 3-MH/v2=2-73 kglm v=0.5494 mlsea. Japour load is 108.3 mol which to diame- terfcolumn is 9- Av de ou . 108.3 mollhom x 78. 1 kg 1 mol - Txoe x 0.5494 2.73 kg/m3 X 3600 see 0.86 m/sec= TA de ²x0.5494 = dig meter of idea 1.4126 m. Column To be calculation for Condenser capacity. S. 108.3 mollhom x 30.7215= 3326.976 kJ l bo mul - G. GUA & TlmX F as per thumb rule, Orca 50oX AX 80.1-32) xo.85

- 8).33 m2 Condenser capacity is .. A = 32 26.97 Ilhor x 1000 1500 g/m² hx (40.88) For Reboiler duty. Q = m Coat & M AT 50 XLS X (80.1-38) + SX25 here, feed temp is 38c. we cond consider, Specific heat capacity is st 2.5 kJ/kmoloc e benene boiling pt is 80.1°C - Tokene boiling pt is 111c. 92 25262.5+9125€ 14387.5 kJ/hr i. Total Reboiler heat duty. 9.792=9. : 3326.97+ 14387.50 17 7 14.47 kJ /hr. Required. Reboiler Capacity .. . = 177 14.47 x 16000 5) Ivo Jhr.mt ocx 34°C A = 651.25 m2) teboiler capacity. No. of stager for height of column ( 2o roll-NB) 04T Kong) log 3.019 0.352 except reboiler de minimum no. of stage- .. Nopt. = Nm x 2.5 = 20 theoretical. no. of plater. i total height is . . (20- 1X0.6 to 6to.r=12.6m.

2 Here we consider 0.6 m and 0.6m ar top section and bottom section. is plato sparing as well as
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