CHEN90012, Design and Construction of Equipment, Tutorial 10

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.

Water is to be used to remove SO from a stream of air. The water and the contaminated air

2

are contacted in a counter current packed absorption column. The absorber must remove at

least 98 mole% of the total SO from the air stream. Fresh water containing no SO enters

2 2

the column at the top. To maintain optimum operating conditions, the water leaving the

column should contain no more than 0.5 mole% SO .

2

The following data are available:

Water feed rate = 20,000 kg/hr

Composition of gas feed = 95 mole% air and 5 mole% SO

2

 (20oC) = 1.18 kg/m3

air

 (20oC) = 1000 kg/m3

H2O

 (20oC) = 1.0 x 10-3 kg/(m.s)

H2O

Operating pressure = 1 atm

Operating temperature = 20oC

Packing = 1 inch (25 mm) ceramic Intalox saddles

Pressure drop in the bed = 42 mm water/m packing (irrigated)

Molecular weight (air) = 29 kg/kmole

Molecular weight (H O) = 18 kg/kmole

2

Molecular weight (SO ) = 64 kg/kmole

2

Height of packing in tower = 4.5 m

a) Determine the mole fraction of SO in the exit gas, y

2 2

b) Determine the maximum gas mass flow rate, G’ (kg/s).

c) Determine the percentage of flooding.

d) Determine the superficial gas velocity, G* .

w

e) Determine the diameter of the packed column.

f) Determine the pressure drop across the irrigated packed bed (Pa).

Where G* = gas mass flow rate per unit column cross-sectional area, kg/m2s

w

L* = liquid mass flow rate per unit column

w

 = liquid viscosity, kg/ms

L

 = gas density

V

   = liquid density

L

L’ = liquid mass flow rate, kg/s

G’ = gas mass flow rate, kg/s

L = liquid molar flow rate, kmole/s

G = gas molar flow rate, kmole/s

x = mole fraction of ethanol in liquid phase

i

y = mole fraction of ethanol in gas phase

i

i=1 = bottom of tower

i=2 = top of tower

For dilute systems, containing only small quantities of SO  may be approximated by 

2, V air

and  may be approximated by  and the average molecular weight of the gas can be

L H2O

assumed to be equal to air and the average molecular weight of the liquid can be assumed to

be equal to water.

You may find the following figures and equations useful.

LL’’ == 2200,,000000 kkgg//hhrr

xx == 00..00 yy == ????

22 22

22

9988%% rreemmoovvaall

11

xx == 00..000055 yy == 00..0055

11 11

GG’’ == ??

Note: Surface area (a) = S

B

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