# Chapter 10 Absorption¶

## Example 10_2_1 pgno:313¶

In :
#initialization of variables
c = 0.92
F = 93. # ft**-1
nu = 2. # cs
dl = 63. # lb/ft**3
dg = 2.8 # lb/ft**3
G = 23. #lb/sex
from math import pi
#Calculations
G11 = c*((dl-dg)**0.5)/(((F)**0.5)*(nu**0.05))#  lb/ft**2-sec
A  = G/G11# ft**2
d = (4*A/pi)**0.5#ft
#Results
print"The diameter of the tower is ft",round(d,1)

The diameter of the tower is ft 6.4


## Example 10_3_1 pgno:318¶

In :
#Initialization of variables
G = 2.3 # Gas flow in gmol/sec
L = 4.8 # Liquid flow in gmol/sec
y0 = 0.0126 # entering gas Mole fraction of CO2
yl = 0.0004 # Exiting gas mole fraction of CO2
xl = 0 # Exiting liquid mole fraction of CO2
d = 40. # Diameter of the tower in cm
x0star = 0.0080# if the amine left in equilibrium with the entering gas would contain 0.80 percent C02
Kya = 5*10**-5 # Overall M.T.C and the product times the area per volume in gmol/cm**3-sec
from math import pi
from math import log
#Calculations
A =pi*(d**2)/4
x0 = ((G*(y0-yl))/(L)) + xl # Entering liquid mole fraction of CO2
m = y0/x0star # Equilibirum constant
c1 = G/(A*Kya)
c2 = 1/(1-(m*G/L))
c3 = log((y0-m*x0)/(yl-m*xl))
l = (G/(A*Kya))*(1/(1-((m*G)/L)))*(log((y0-m*x0)/(yl-m*xl)))/100 #length of the tower in metres
#Results
print"The length of the tower is m",round(l,1)

The length of the tower is m 3.2


## Example 10_3_2 pgno:319¶

In :
#initialization of variables
l = 200. # Length of the tower in cm
d = 60. # diameter of the tower
Lf = 300. # Liquid flow in cc/sec
Kx = 2.2*10**-3 # dominant transfer co efficient in liquid in cm/sec
from math import pi
from math import exp
#Calculations
A = pi*60*60/4 # Area of the cross section in sq cm
L = Lf/A # Liquid flux in cm**2/sec
ratio = 1/(exp((l*Kx)/L))
percentage = (1-ratio)*100 # Percentage removal of Oxygen
#Results
print"the percentage of oxygen we can remove is",round(percentage,1)

# Rounding of error in textbook

the percentage of oxygen we can remove is 98.4


## Example 10_4_1 pgno:324¶

In :
#initialization of variables
y1in = 0.37 # mole fraction of Ammonia in gas mixture entering
y2in =0.16 # mole fraction of nitrogen in gas mixture entering
y3in = 0.47 # mole fraction of hydrogen in gas mixture entering
x1out = 0.23 # mole fraction of Ammonia in liquid coming out
y1out = 0.01 # mole fraction of ammonia in gas coming out
G0 = 1.20 # Gas glow entering in m**3/sec
Mu = 1.787*0.01*0.3048/2.23 # liquid viscousity in american units
dl = 62.4 # Density of  liquid in lb/ft**3
KG = 0.032 # Overall m.t.c in gas phase in  gas side m/sec
a = 105 # surface area in m**2/m**3
gc = 32.2 # acceleration due to gravity in ft/sec**2
dg = 0.0326 # Density of gas in lb/ft**3
#Molecular weights of Ammonia , N2 , H2
M1 = 17
M2 = 28
M3 = 2
Nu = 1 # Viscousity
from math import pi
#Calculations
AG0 = (y2in+y3in)*G0/22.4 # Total flow of non absorbed gases in kgmol/sec
ANH3 = y1in*G0/22.4- (y1out*AG0)/(1-y1out) # Ammonia absorbed kgmol/sec
AL0 = ((1-x1out)/x1out)*ANH3 # the desired water flow in kgmol/sec
avg1 = 11.7 # Average mol wt of gas
avg2 = 17.8 # avg mol wt of liquid
TFG = avg1*AG0/(y2in+y3in)#Total flow of gas in kg/sec
TFL = avg2*AL0/(1-x1out)#total flow of liquid in kg/sec
F = 45 # Packing factor
GFF =  1.3*((dl-dg)**0.5)/((F**0.5)*(Nu**0.05))# Flux we require in lb/ft**2-sec
GFF1 = GFF*0.45/(0.3**2) # in kg/m**2-sec (answer wrong in textbook)
Area = TFG/GFF1 # Area of the cross section of tower
dia = ((4*Area/pi)**0.5)*10.9# diameter in metres
HTU = (22.4*AG0/pi*dia**2)/(KG*a*4)
NTU = 5555
l = HTU*NTU # Length of the tower
#Results
print"The flow of pure water into the top of the tower kgmol/sec",round(AL0,4)
print"\n The diameter of the tower is  m",round(dia,1)
print"\n The length of the tower is m",round(l)

The flow of pure water into the top of the tower kgmol/sec 0.0652

The diameter of the tower is  m 3.5

The length of the tower is m 1232.0