CHEN90012, Design and Construction of Equipment, Tutorial 9

Please attempt to solve the tutorial question during your assigned tutorial time. The Tutor

will work through the solutions during the second half of the Tutorial session. No marks are

given for the Tutorials.

A sieve plate distillation column must operate under the following conditions in the section

below the feed.

Liquid flow rate 38,000 kg/hr

Vapour flow rate 31,500 kg/hr

Liquid density 880 kg/m3

Vapour density 3.4 kg/m3

Liquid surface tension 20 mN/m

Assume that A = 0.1 A and also assume A =A . Operate at 75% of flooding. The plate

h p, p a

configuration is single pass cross flow. The stainless steel plates should be 3 mm thick. Use

the standard rules of thumb when other assumptions need to be made.

Determine the following aspects of the plate distillation column:

a) Column diameter.

b) Pressure drop per plate in Pa.

c) What is the maximum hole diameter that can be used (when operating at 75% turndown)

which will not weep?

d) If you change your hole size from the standard one chosen when you calculated a) and b)

above, to the result found in c) what else will change?

You may find the following figures and equations useful.

___________________________________________________________________________

PD 0.2

K  

t m  2fP 1b  b 

K  20

1a

PD

t   

m 2feM  P u  K L V

f 1 

V

gH D

t  L

m 2fJ 103 Fraction of Flooding  K 4operating

K

4flooding

PRM

t 

 

2f0.2P 0.1

 

13.1(G)2F  L 

w p  

1 R1/2 K

4



 (  )

L

M  3   V L V

4  r  

  

K 0.9 25.4d

u  2 h

R = (T

1

– T 2)/( t

2

– t 1) h 0.5

V

S = (t – t )/T –t )

2 1 1 1 h = h +h + h + h

t d w ow r

(T t )(T t )

T  1 2 2 1 h b = h t +h w + h ow + h dc

LM

T t 

ln 1 2  2

  u  

T t

 2 1  h

d

51  Ch  V

 

o L

h d

Nu  i e

k

12.5103

f h 

r 

L

0.14

  

Nu  j

h

RePr0.33 





 L

2

 W  h

dc

166 

 A





Note L’

L M must be in

C 

Pr  p

2

kg/s

k  L  3

f h 750 

ow  l 

 

  L   m  v2 L w

P N 8j    2.5 L

p  f d    2 L(x 1-x 2) = G(y 1-y 2)

 i W 

     gd3

  D L  0.14 v2 Ga  g s g

P8j  s    L 2





f d el

B



W

 



2

 1  U d 1.75U d 2

P(1 )gh Ga 150 mf mf g   mf g 

mf s mf 3  3   

mf mf

d lnd o  ax2 +bx +c =0

1 1 d  1 d  o  d   1 1

  o   o  i  

U h  d   h  d   2k h h b b2 4ac

i i fi i o fo x 

2a

L 

F  V

LV V 

L