Absorption of ammonia from air by liquid water
Authors: Housam Binous and Ahmed Bellagi

We computes the height of a packed tower used for the absorption of ammonia from air by liquid water at 20 °C and 1 atm . Two different methods are used to compute Hy Ny and Hoy Noy . The methods are based respectively on local and overall mass transfer coefficients . The problem is taken from Tranport Processes and Unit Operation by Geankoplis (3 rd Edition, Prentice Hall, 1993) .
Problem 10-6-6 from Geankoplis, Transport Processses and Unit Operations (Prentice Hall, 3rd Edition, 1993)
​
Inlet gas: air with 4 mol % NH3 and flow rate equal to 57.8 kg mol/hr
Oulet gas: air with 0.5 mol %
Flow rate of inlet pure liquid water is 68 kg mol/h
T=293 K and P=1 atm
Column diameter=0.747 m
Mass transfer coefficients: kya=0.0739 and kxa=0.169 Kg mol/(s m3)
​
Average gas flow rate
In[]:=
V'=57.8(1-0.04)
Out[]=
55.488
In[]:=
V1=V'/(1-0.005)
Out[]=
55.7668
In[]:=
V2=57.8
Out[]=
57.8
In[]:=
V=(V1+V2)/2
Out[]=
56.7834
Equilibrium data
In[]:=
tbl={{0,0},{0.0208,0.0158},{0.0258,0.0197},{0.0309,0.0239},{0.0405,0.0328},{0.0503,0.0416},{0.0737,0.0657},{0.0960,0.0915},{0.137,0.150},{0.175,0.218},{0.210,0.298},{0.241,0.392},{0.297,0.618}};
In[]:=
plt1=ListPlot[tbl,JoinedTrue,PlotStyleRGBColor[1,0,0]]
Out[]=
0.05
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0.1
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0.6
In[]:=
ListPlot[{{0,0},{0.0208`,0.0158`},{0.0258`,0.0197`},{0.0309`,0.0239`},{0.0405`,0.0328`},{0.0503`,0.0416`},{0.0737`,0.0657`},{0.096`,0.0915`},{0.137`,0.15`},{0.175`,0.218`},{0.21`,0.298`},{0.241`,0.392`},{0.297`,0.618`}},JoinedTrue,PlotStyle
]
Out[]=
0.05
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0.30
0.1
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0.6
In[]:=
eq=Interpolation[tbl,InterpolationOrder1]
Out[]=
InterpolatingFunction
Domain: {{0.,0.297}}
Output: scalar

In[]:=
plt2=Plot[eq[x],{x,0,0.297},PlotRangeAll,PlotStyleRGBColor[0,0,1]]
Out[]=
0.05
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In[]:=
Show[plt1,plt2]
Out[]=
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In[]:=
ShowListPlot[{{0,0},{0.0208`,0.0158`},{0.0258`,0.0197`},{0.0309`,0.0239`},{0.0405`,0.0328`},{0.0503`,0.0416`},{0.0737`,0.0657`},{0.096`,0.0915`},{0.137`,0.15`},{0.175`,0.218`},{0.21`,0.298`},{0.241`,0.392`},{0.297`,0.618`}},JoinedTrue,PlotStyle
],
0.00
0.0
0.3

Out[]=
0.05
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Operating Line
In[]:=
Solve[y/(1-y)1.225x/(1-x)+0.005025,y]
Solve
:Solve was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result.
Out[]=
y
201.+48799.x
40201.+8799.x

In[]:=
y[x_]:=
201.`+48799.`x
40201.`+8799.`x
In[]:=
plt3=Plot[y[x],{x,0,0.3}]
Out[]=
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0.35
Plot of equilibrium and opeating curves
In[]:=
Show[plt3,plt2,PlotRange{{0,0.04},{0,0.04}}]
Out[]=
0.01
0.02
0.03
0.04
0.01
0.02
0.03
0.04
Computing the height of packing using Z=Hy Ny and local mass transfer coefficient kya
Computing Ny : the number of transfer units. Thus, we need to compute the interfacial concentrations using kya and kxa.
In[]:=
i=0;
In[]:=
While[i<11,{h=0.0050+i0.0035,sol=FindRoot[{(y[x]-yi)/(x-xi)-2.287((1-y[x])-(1-yi))/Log[(1-y[x])/(1-yi)]Log[(1-x)/(1-xi)]/((1-x)-(1-xi)),yieq[xi],y[x]h},{x,0.001},{xi,0.005},{yi,0.002}],y[i]=h,yi[i]=sol[[3,2]],xi[i]=sol[[2,2]],x[i]=sol[[1,2]],i++}]
Computing Hy : the height of a transfer unit. We need (1-y)iM
log nean of 1-y and 1-yi. We also have to compute the average
of 1-y. S is the cross-sectional area. Dt is the tower diameter.
Z, the column height, is equal to 2.26 m
Approximation for Ny using log mean of y-yi
Computing the height of packing using Z=Hoy Noy and overall mass transfer coefficient Kya
Computing the number of transfer units Noy using y-ystar
Computing the cross-sectional area
Computing m the average slope of the equilibrium curve and Kya the overall mass transfer coefficient.
Computing the logmean of 1-y and 1-ystar
Height of a transfer unit
Z, the column height, is equal to 2.31 m