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Absorption of Ammonia in a Countercurrent Stage Tower
based upon Problem 10.6-4 , p. 692, in Geankopolis (2003)
Problem Statement
The gas stream from a chemical reactor contains 25 mol% ammonia and the rest
inert gases. The total gas flow rate is 181.4 kmol/hr
The exit gas can have a maximum of only 2.0 mol% ammonia. Using a
countercurrent absorption tower, scrub the feed gas with water containing 0.5 mol%
ammonia at 1.5 the minimum absorbent rate. Operate the tower isothermally and
isobarically at 303 K and 101.3 kPa. Under these conditions, vapor-liquid
equilibrium follows Henry's law with an equilibrium constant yeq/x=1.153. The tower
is packed with 25 mm ceramic Raschig rings, which have a volumetric mass transfer
coefficient Kya=12 lbmol/(ft3hr y)
Determine the following. Answers in parentheses.
1. The minimum absorbent flow rate (157 kmol/hr)
2. The number of equilibrium stages required at an absorbent flow rate 1.5 times
the minimum, using graphical step-off. (5)
3. The ammonia recovery (93.9%).
4. The number of equilibrium stages using an algebraic method. (5.184)
5. The tower diameter at 70% flood velocity (3.84 ft)
6. Tower packed height using the HETP method (7.65 ft)
7. Tower packed height using the HTUOG/NTUOG method (36.8 ft)
8. Total height of the tower using the packed height from number 6 (21.1 ft).
9. The total pressure drop in the tower at operating conditions (0.271 psi).




by T. Wiesner
3/6/2018 11:57 AM
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Given
inlet ammonia mole fraction in y1 := 25% = 0.25
gas
kmol
total vapor feed V := 181.4
hr
desired exit gas composition y2 := 2% = 0.02

operating temperature T := 303K
operating pressure P := 101.3kPa
Henry's Law correlation (cf. Sander, R. (2015). "Compilation of Henry's law constants (version 4.0)
for water as solvent." Atmospheric Chemistry and Physics 15(8): 4399-4981.)

csat_L d 1 1  1
correlation Hcp ( T ) = = Href exp  ln ( H cp)  -  =
pg 1  T Tref  HA ( T )
d
 T 
mol
Henry's Law constant Href := 0.59
for ammonia at 3
m Pa
Tref := 298.15K
d
temperature dependence of B= ln ( H cp) B := 4200K
Henry's coefficient 1
d
T

correlation Hcp ( T ) := Href exp B
  1 - 1  Hcp ( T ) = 47.714
mol

  T Tref  L atm

conversion to
mole
55
ye ye P pg pg pg L
Hcc = = = = = =
x xP moles_A moles_A P Hcp P
P csat_L
moles_liq L mole
P 55
moles_water L
55
L
mole
55
L
Hcc ( T , P) :=
Hcp ( T ) P

equilibrium K-value meq := Hcc ( T , P) = 1.153

by T. Wiesner
3/6/2018 11:57 AM
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Find
minimum absorbent flow rate Ls_min
number of equilibrium stages at an absorbent Ls = 1.5Ls_min
flow rate 1.5 times the minimum.
ammonia recovery rA
tower diameter at 70% flooding velocity DT
tower packed height Z
tower total height H
tower pressure drop ΔP tot
Analysis
Specify the terminal concentrations on the tower.

exit gas composition. y2 = 0.02
y2
exit gas composition in mole ratios Y2 := Y2 = 0.020408
1 - y2
y1
inlet gas composition in mole ratios Y1 := Y1 = 0.333333
1 - y1
kmol
flow of solute-free gas Vs := ( 1 - y1 ) V Vs = 136.05
hr
x2
solute concentration in inlet liquid x2 := 0.5% X2 := = 0.005
1 - x2

Plot the equilibrium line and the operating line for minimum absorbent on solute-free coordinates.
Yeq
1Yeq solve , Yeq X meq
equilibrium relationship meq 
X simplify X - X m eq  1
1 X
meq X Y1
solve in terms Yeq ( X ) := X1_max := = 0.277
of mole ratios 1  ( 1 - meq) X meq( 1  Y1 ) - Y1

Vs( Y1 - Y2 ) kmol
minimum absorbent Ls_min := Ls_min = 156.617
flow rate X1_max - X2 hr

slope of operating line at Ls_min
= 1.151
minimum absorbent Vs

by T. Wiesner
3/6/2018 11:57 AM
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Ls_min
operating line for Yop_min ( X ) := Y1 -
Vs
( X1_max - X )
minimum absorbent

multiplier of minimum α := 1.5
liquid rate
kmol
actual absorbent flow Ls := α Ls_min Ls = 234.926
hr
Ls Ls
slope of actual
Vs
= 1.727 operating line Yop ( X ) := Y2 
...