Question

N-Heptane / Cycloheptane Separation.

A saturated liquid feed of 600 mol/hr containing 54 mole % n-heptane is to be fractionated at 101.325kPa (abs) in order to form a distillate containing 94 mole % n-heptane and a bottoms product of 10 mole % cycloheptane. Assume Raoult’s Law behavior applies (calculated results provided). The distillate reflux ratio to be used is 8:1.

a) Calculate the flowrates (mol/hr) of the distillate and bottoms.

b) Determine the minimum reflux ratio, RMIN and the minimum number of theoretical trays at total reflux.

c) Determine the theoretical number of trays and the feed tray number for the specified operation al design.

d) If each stage in the column is 75% efficient how many actual stages will be required?

e) If the column is 75% efficient overall how many stages will be required?

x	y
1.000	1.000
0.857	0.915
0.724	0.825
0.599	0.728
0.483	0.625
0.374	0.515
0.273	0.399
0.177	0.275
0.087	0.144
0.003	0.004

Answer 1

Here there is a mistake for finding out the minimum reflux ratio of part (b) please reconsider the part i have uploaded here

for part(b) the minimum reflux ratio is 1.68571

since by HomeworkLib guidelines we are to solve first four subparts of a question unless specified. Thank you!

solution mi-heptane + 1, cycloheptane 72 b) ecall mole balance F- 0+B) Component mole balance Wood FXe=D2g,+ BAB, Iris Rex, nos are the liquid mole fraction of component FD and B are the molar flow rates dritto F=600 mollha. D+ B - 600 0-94D+0.9 B = 600 X 0.54 0.94D+0.28 -324... D + B = 600... . D = 314.2857 mol ha B - 285.7142 mol/ha love best the use and The Rectifying section operating line is on + 20 y Rt. RL Latest h e 2- KF . The feed line is given as y = var It Rimin the equilibrium line q=1 ay oo Rimin = 0.35 Romin + 1 Rminth e t ss kime 18 -

Rmin = 0.538. At total Refluxe the operating lines. all together lied on the diagonal line Hence The mini mum number of Trays are c) of the Reflux Ratio is 8. a Rectifying d) Efficiency of tray = 0.75 Theoretical Number of stages - 10 Actual number of Stages - 0.45 = 13-333 : cdetual number of stages $14. Tits we are to solve first four subparts of a question unles specified by Chegg guidelines. Thank you! +0.94 - section y = 871 871 y = $x +0.10444 It should intersect with the q line- and pass through the points (0.94,0-94) and (0,0. 10w4u. The stripping section should abo pass through (0.1,0.1) and intersect pass the onestion point of Rectifying section the interse operating line & q-line. By Mc- lake Thiele method The theoretical number of trays are 10 And the feed tray is at 5th tray as q-line intersects that p part each downward Triangle reprenuto a tray

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