Absorption of ammonia from air by liquid water
Author's Data: Housam BINOUS
Department of Chemical Engineering
National Institute of Applied Sciences and Technology
Tunis, TUNISIA
Email: binoushousam@yahoo.com

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

V' = 57.8 (1 - 0.04)

V1 = V'/(1 - 0.005)

V2 = 57.8

V = (V1 + V2)/2

Equilibrium data

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}};

plt1 = 
      ListPlot[tbl, PlotJoined -> True, 
        PlotStyle -> RGBColor[1, 0, 0]]

eq=Interpolation[tbl,InterpolationOrder -> 1]

plt2 = 
      Plot[eq[x], {x, 0, 0.297}, PlotRange -> All, 
        PlotStyle -> RGBColor[0, 0, 1]]

Show[plt1, plt2]

Operating Line

Solve[y/(1 - y) ==  1.225 x/(1 - x) + 0.005025, 
      y]

y[x_] := ((201.` + 48799.` x)/(40201.` + 8799.` x))

plt3 = Plot[y[x], {x, 0, 0.3}]

Plot of equilibrium and opeating curves

Show[plt3, plt2, PlotRange -> {{0, 0.04}, {0, 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.

i = 0;

While[
      i < 11, {h =  0.0050 + i 0.0035, 
        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++}]

Table[{y[i], x[i], yi[i], xi[i]}, {i, 0, 10}] // TableForm

tbl2 = 
      Table[{y[i], 1/(y[i] - yi[i])}, {i, 0, 10}] // TableForm

tbl3 = Table[{y[i], 1/(y[i] - yi[i])}, {i, 0, 10}];

intg0=Interpolation[tbl3,InterpolationOrder -> 1]

Ny = Integrate[intg0[x], {x, 0.005, 0.04}]

tbl4 = 
      Table[Line[{{x[i], y[i]}, {xi[i], yi[i]}}], {i, 0, 10}]

Show[plt3, plt2, Graphics[{RGBColor[1, 0, 0], tbl4}], 
      PlotRange -> {{0, 0.04}, {0, 0.04}}]

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.

((1 - y[0]) - (1 - yi[0]))/
      Log[(1 - y[0])/(1 - yi[0])]

((1 - y[10]) - (1 - yi[10]))/
      Log[(1 - y[10])/(1 - yi[10])]

1 - y[0]

1 - y[10]

S = Pi Dt^2/4 /. Dt -> 0.747

Hy = 
      1/(0.995 + 0.96) (0.997 + 0.966) 56.8/S 1/0.0739 1/
          3600

Z, the column height, is equal to 2.26 m

Z = Hy Ny

Approximation for Ny using log mean of y-yi

lmdf = ((y[0] - yi[0]) - (y[10] - yi[10]))/
        Log[(y[0] - yi[0])/(y[10] - yi[10])]

(y[10] - y[0])/lmdf

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

i = 0;

While[
      i < 11, {h =  0.0050 + i 0.0035, 
        sol = FindRoot[{ystar ==  eq[x], y[x] ==  h}, {x, 
              0.001}, {ystar, 0.005}], y[i] = h, ystar[i] = sol[[2, 2]], 
        x[i] =  sol[[1, 2]], i++}]

Table[{y[i], x[i], ystar[i]}, {i, 0, 10}] // TableForm

tbl5b = 
        Table[Line[{{x[i], y[i]}, {x[i], ystar[i]}}], {i, 0, 
            10}];

Show[plt3, plt2, Graphics[{RGBColor[1, 0, 0], tbl5b}], 
      PlotRange -> {{0, 0.04}, {0, 0.04}}]

tbl5 = 
        Table[{y[i], 1/(y[i] - ystar[i])}, {i, 0, 10}];

intg1=Interpolation[tbl5,InterpolationOrder -> 1]

Noy = Integrate[intg1[x], {x, 0.005, 0.04}]

Computing the cross-sectional area

S = Pi Dt^2/4 /. Dt -> 0.747

Computing m the average slope of the equilibrium curve and Kya the overall mass transfer coefficient.

Fit[{{0, 0}, {0.0208, 0.0158}, {0.0258, 0.0197}, {0.0309, 
          0.0239}, {0.0405, 0.0328}, {0.0503, 0.0416}}, {1, x}, x]

m = 0.8273576398478655

sol = 
      Solve[1/X ==  1/0.0739 + 0.8273576398478655/0.169, X]

Kya = sol[[1, 1, 2]]

Computing the logmean of 1-y and 1-ystar

((1 - y[0]) - (1 - ystar[0]))/
      Log[(1 - y[0])/(1 - ystar[0])]

((1 - y[10]) - (1 - ystar[10]))/
      Log[(1 - y[10])/(1 - ystar[10])]

Height of a transfer unit

Hoy = 
      1/0.05426699461472708 56.8/
          S (0.997497955530854 + 0.9687879784288559)/(2 - y[0] - 
                y[10])/3600

Z, the column height, is equal to 2.31 m

Z = Hoy Noy