Project Work 2

In a petroleum refining company, it is desired to separate a mixture containing n-butane of 40%, n-pentane of 25%, n-hexane of 20% and n-heptane of 15%. In a multicomponent rectification tower, the fed is charged to the distillation unit at the rate of 100 mol/r continuously. It is required by the company that 95% of n-pentane need to be recovered in the distillate and 95% of n-hexane in the bottoms. You as a group of company’s design and operation engineers, you are given the task to determine and report,

a. Moles per hour and compositions of distillate and bottoms for the above requirement. b. Top and bottom temperature of the tower c. Minimum stage for total reflux and distribution of other components (trace components) in the distillate and bottoms, that is in mole and mole fractions. d. Minimum reflux ratio e. Number of theoretical stages at an operating reflux ratio 1.3 times the minimum using the Erbar-Maddox correlation f. Location of the feed tray using Kirkbride method

Based on company’s requirement and the task given to you, make a complete report on the above tasks.

Mohamed Yassin Bin Roslan 17123

a.Moles per hour and compositions of distillate and bottoms for the above requirement :

List of composition :

n-butane (xA = 0.40) n-pentane (xB= 0.25)n-hexane (xC = 0.20)n-heptane (xD = 0.15)

Assumptions & Requirement :

Feed – 100 mol/h95% n-pentane needs to be recovered in the distillate. 95% n-hexane needs to be recovered in the bottoms.

For key components : the first trail is assumed that no component D (heavier than the heavier key , (C) is in the distillate and no light A in the bottoms.

Material balance for each component need to be made

Key components :n-pentane (xB= 0.25) - Light Component ( L )n-hexane (xC = 0.20) - Heavy Component ( H )

For the over all balance F = D + W For Component B light key balance FxBF = DyBD + WxBW 0.25(100) = 25 = D yBD + W xBW

Since 95% of B (n-pentane) is recovered in the distillate ,

DyBD = (0.95)* 25 = 23.75Hence ,25 = 23.75 + W xBW

W xBW = 1.25

For Component C, heavy key balance

FxCF = DyCD + WxCW

0.2(100) = 20 = DyCD + WxCW

Since 95% of C (n-hexane) is recovered in the bottoms ,WxCW = (0.95)*(20) = 19.0Hence ,20 = 19.0 + DyCD

DyCD = 1.00From what have we assumed, no component D (heavier than the heavier key , (C) is in the distillate and no light A in the bottoms for the first trail.

Therefore, moles A in the distillate is DyAD = 100*(0.4) = 40 moles D in bottom is WyDW = 100*(0.15) = 15

This values are tabulated as shown below;

Comp.

Feed, F Distillate,D Bottoms, WxF xF * F yD = xD yD * D xW xW * W

A 0.40 40.0 0.617761 40.0 0 0

B 0.25 25.0 0.366795 23.75 0.035461 1.25

C 0.20 20.0 0.015444 1.0 0.539007 19.0

D 0.15 15.0 0 0 0.425532 15.0

Total 1.00 F = 100 1.00 D = 64.75 1.00 W = 35.25