Question

A gas is separated from the air using an absorption column, which is randomly packed with 25mm metal Pall rings. Pure liquid solvent is used to reduce the mole fraction of gas A in air from 0.05 to 0.005 at a constant pressure and temperature. Total molar flow rates of entering gas and liquid to the column are 120 kmol/h and 180 kmol/h, respectively. The equilibrium relationship at operating conditions is y^∗ = 1.2 x and the gas side pressure drop in the column is selected as 200 Pa/m. By using the additional data given below calculate

a) diameter of the column,

b) percent of flooding gas velocity at operating conditions,

c) height of the column, when height of the one overall liquid transfer unit is 0.95 m.

MWG= 29 kg/kmol,

MWL= 18 kg/kmol,

ρG = 1.2 kg/m³ ,

ρL = 998 kg/m³ ,

µL= 1 cP

Answer 1

The column is packed with 25 mm metal pall rings

From handbook

For 25 mm metal pall rings

Packing factor (Fp) = 157 /m

G = 120 kmol/h

L = 180 kmol/h

 Kg/m³

Pg = 1.2

Kg/m³

PL = 998

Viscosity = 1 cp

The pressure drop of column = 200 Pa/m

1 Pa/m = 816.4 inches of water/ft

200 Pa/m = 0.2449 ~ 0.25 inches of water/ft

The relationship between pressure drop and flooding velocity can be given by following graph from handbook

0.60 0.40 Flooding line 0.20 Parometer of curves is pressure drop in inches of water/foot of pocked height 1.50 1.00 0.10 0.50 0.060 0.040 0.25 .62F+02 POPIT 0.020 0.10 0.010 0.006 0.05 0.004 0.002 1.0 4.0 60 10.0 2.0 0.4 0.6 0.2 0.001 0.01 0.02 0.04 006 0.1. 1/2 PI

X axis

0.5 LP9 G PL

= = 0.052013

07月 180 12 120_998

At X axis = 0.052013

∆P = 0.25

Y axis = 0.032.

Y axis =

G-FpH 0.2 POPL9

$\psi = 1$

0.032 GP(157)(0.001)0,2 (1.2)(998) (9.81)

G = 3.0875 Kg/m2s

G = (density of gas) (operating velocity)

operating Velocity (v) = (3 . 0875/1.2) =

2.5729 m/s

G = m/A

Molar flowrate of gas = 180 kmol/h

M. W of gas stream = 29 Kg/Kmol

Mass flowrate of gas = (120) (29) =3480 kg/h

3.0875= 3480/(A×3600)

A = 0.31309 m²

A = π(d²) /4

d = 0.6313 m

B)

At xaxis = 0.0052013 and flooding line we get

Y axis = 0.18

0.18 = GP(157)(0.001)0.2 (1.2) (998) (9.81)

G(max) = 7.322 Kg/m2s

Flooding velocity (vf) = G/(density of gas)

Vf = (7.322/1.2)

Vf = 6.1023 m/s

% flooding velocity = (v/vf) (100) = (2.5729/6.1023) (100) = 42.162%

C)

Given

H_toL = 0.95 m

y = 1.2(x)

y₁ = 0.05 , Y1 = (0.05/(1-0.05) )= 0.05263

y₂ = 0.005 , Y2 = (0.005/(1-0.005) =0.00502

Gs = G(1-y1) = (120) (1-0.05) =114 kmol/h

L= Ls = 180 kmol/h

Doing material balance

G (Y1 - Y2) = L (X1 – X2)

114(0.05263-0.00502) = 180(X1-0)

X1 = 0.030153

x1 = 0.02927

N_toL = (x₂ - x₁) /(x- x*) _lm

(1 – 2*)im (21 - 21*) - (22 - 22*) in (11-11) 12-12)

x1 will be in equilibrium with y1

x2 will be in equilibrium with y2

x1* = (0.05263/1.2) = 0.04385

x2* = (0.00502/1.2) = 0.004183

(1 – 1*)lm (0.02927 – 0.04385) - (0 -0.004183) in (0.02927–0.04385) (0-0.004183)

(x- x*) lm = -0.0083267

N_toL = (x2- x1) /(x- x*) _lm

N_toL = (0-(-0.02927)) /(-0.0083267)

N_toL = 3.51519

Z = H_toL × N_toL

Z = (0.95) (3.51519) = 3.3394 m

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