Page 275 - Process Equipment and Plant Design Principles and Practices by Subhabrata Ray Gargi Das

276 Chapter 10 Absorption and stripping

We decide L op ¼ 1.5 L min ¼ 9400.65 kmol ¼ 169,211.7 kg, as usually the economic optimum is
in the limit: 1.5 L min L op 3 L min
At this operating point, x 1 ¼ (17.86 e 0.893) / 9400.65 ¼ 0.001802.
x 2 , y 2 and y 1 have the original values, as shown in Fig. P10.2B. The flow rates and the concen-
trations are also shown in the figure along with a control volume cutting across an arbitrary elevation in
the bed where the entering gas and the exiting liquid have compositions of y op and x op mole fraction
respectively.

kmol mf kg
0.893 0.00553(y ) 57.15
SO 2 2
Air 160.71 0.99447 4660.59
161.603 1.0 4717.74
kmol mf
SO 2 – 0(x )
2
H O 9400.65 1.0
2
30°C
9400.65 1.0
1atm
L
1.5x Control volume
G
min
kmol mf kg
16.967 0.00182(x ) 1085.89
SO 2 1
y op x op H 2 O 9400.65 0.99818 169211.7
9417.617 1.0 170297.59
kmol mf kg
SO 2 17.86 0.00553(y 1 ) 1143.04
Air 160.71 0.99447 4660.59
178.57 1.0 5803.63
FIGURE P10.2B
Material balance for operation with 1.5 (L/G) min .

Operating line equation based on the solute balance across the control volume shown in
Fig. P10.2B is e
!

y op y 2 x op x 2
160:71 ¼ 9400:65
1 y op 1 y 2 1 x op 1 x 2
or,
!

x op y op
0:005561 ¼ p; say
¼ 0:0171
1 x op 1 y op
or,
p

x op ¼
1 þ p