8/11/2019 Ms Exercise10

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Prof M. MazzottiThermische Verfahrenstechnik I binary distillation in packed column

Binary distillation in packed column

A process engineer visiting a company of a competitor was told that a packed column (innerdiameter 0.5m) is used to separate a binary liquid mixture. During his visit he also saw that thecolumn was running with a reflux ratio of R = 1.5. The manufacturer of the packing material X

provides the following HTU-diagram:

Figure 1: HTU-diagram of packing material X

From scientific publications the equilibrium curve of the binary liquid mixture is available:

y =x2 + 2x (1)

also the vapor density of the mixture is known: v = 4.1kg/m3

In a flyer, provided by the company, the following information are printed:

concentration in the product stream: x1,D = 0.82

concentration in the bottom stream: x1,B = 0.03

product stream: F = 77 kg/h

feed concentration: xF= 0.4

With these information, the process engineer is able to calculate the height of the packing materialin the column of the competitor. Which height did he get?

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8/11/2019 Ms Exercise10

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Prof M. MazzottiThermische Verfahrenstechnik I binary distillation in packed column

Guideline for the solution

N T Uov,G = N T Urectov,G+N T U

stripov,G =

y=xD

y=xB

dy

yy (2)

Equilibrium:

y =x2 + 2x (3)

Operating line:

y= px+q (4)

Rectification part:

p= R

R+ 1 (5)

q= xp1 +R = x1,D

1 + 1.5 (6)

Using equation 5 and 6 in 4:

y= 0.6x+ 0.328; dy/dx= 0.6 (7)

N T Urectov,G =

y=xD

y=xF

dx0.6yy =0.6

y=xD

y=xF

dx

x2 1.4x+ 0.328 (8)

Substitution:

X=x2

1.4x+ 0.328 (9)

= 4cb2 = 40.3281.42 =0.648 (10)

N T Urectov,G =0.6y=xD

y=xF

dx

X =0.6 1

sqrtln2x+bsqrt2x+b+sqrt= 1.89 (11)

Stripping part:

p= yFx1,BxFx1,B

(12)

q= yFpxF =0.014 (13)

yF =RxF+x1,D

1 +R = 0.568 (14)

This leads to

y = 1.454x0.014; dy/dx= 1.454 (15)

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8/11/2019 Ms Exercise10

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Prof M. MazzottiThermische Verfahrenstechnik I binary distillation in packed column

N T Ustripov.G =1, 454y=xF

y=xB

dx

x2 0.546x0.014 (16)

= 4cb2 = 4(0.014)(0.546)2 =0.354116 (17)

= 0.595 (18)

N T Ustripov.G = 7.88 (19)

Using equation 2:

N T Uov.G = 1.89 + 7.88 = 9.77 (20)

with

wD =

DAD = (1 +R)pAD = 2.5

77436000.52 4.1= 0.06642m/s (21)

F= 0.06642

4.1 = 0.1345 (22)

Using diagram in figure 1 leads to H T Uov,G=0.528 m

H= H T Uov,GN T Uov,G = 5.16m (23)

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