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SAMPLE PROBLEM #17: A total feed of 200 mol/h having an overall composition of 42 mol % heptane and 58 mol % ethyl benzene is to be fractionated at 101.3 kPa pressure to give a distillate containing 97 mol % heptane and a bottoms containing 1.1 mol % heptane. The feed enters the tower partially vaporized so that 40 mol % is liquid and 60 mol % vapor.
Cal alculate alcu cula late te
the th e fo foll following: llow owin ing g:
(a) Moles per hour of distillate and bottoms. (b) Minimum reflux ratio Rm. (c) Minimum steps and theoretical trays at total reflux. (d) Theoretical number of trays required for an operating reflux ratio of 2.5:1.
Equili Equilibrium lib briu ium m data ta are given below at 1 101. 01.32 .32 32 k kPa Pa abs pressu abs sure surre re fo for for r the m mole mo olle e fr frac fra acti action tion on h h--heptane -heptane heptane x xH H and y and yH: H: Temperature
xH
yH
K
rC
409.3
136.1
0.000
0.000
402.6
129.4
0.080
0.230
392.6
119.4
0.250
0.514
383.8
110.6
0.485
0.730
376.0
102.8
0.790
0.904
371.5
098.3
1.000
1.000
SOLUTION
total condenser L0
200 mol/h F (q = 0.4) xF = 0.42
Distilling column 101.3 kPa abs
R = 2.5
partial reboiler
D xD = 0.97
B xB = 0.011
D, B = ? Rmin = ? ntheo at (R = ), n min = ? ntheo at (R= 2.5) Solve for the y-intercept for the rectifying operating line y n
!
!
n
!
!
Solve for the slope of the feed line q-line slope
!
!
!
Plotting the McCabeMcCabe-Thiele -Thiele diagram
3
2
1
4 5
nF = 7
6 7
8 9 10 11
n = 11.3 xB
xF
xD
Doing the material balances
Total
balance F = 200 = D + B More volatile component balance xFF = xDD + xBB 0.42(200) = 0.97D + 0.011B Solving these two equations D = 85.3 mols/h B = 114.7 mols/h
Solving for R min
y-int for Rm =
B
F
y-int of rectifying line = !
R Rm
@
Rm =
Solving for n min 1 2 3
4
5
6
7
Nmin = 7.5 xB
xF
N = 7.5 = theoretical trays + reboiler = n + 1 @ n = 6.5 theoretical trays
xD
SUMMARY
nF = 7, n = 11.3 D = 85.3 mols/h and B 114.7 mols/h Rmin = 1.7 ntheo = 6.5