Chapter 3
Example 3.2-21. ---------------------------------------------------------------------------------In an experimental study of the absorption of ammonia by water in a wetted-wall column, the
value of KG was found to be 2.75×10-6 kmol/m2⋅s⋅kPa. At one point in the column, the
composition of the gas and liquid phases were 8.0 and 0.115 mole % NH3, respectively. The
temperature was 300 K and the total pressure was 1 atm. Eighty-five percent of the total
resistance to mass transfer was found to be in the gas phase. At 300 K, ammonia-water
solution follow Henry’s law up to 5 mole % ammonia in the liquid, with m = 1.64 when the
total pressure is 1 atm. Calculate the individual film coefficients and the interfacial
concentrations.
Solution ----------------------------------------------------------------------------------------NA = KG(pA − pA*) = KGP(yA − yA*) = Ky(yA − yA*)
Ky = KGP = (2.75×10-6)(101.3) = 2.786×10-4 kmol/m2⋅s
Since
m
1
1
=
+
for gas phase resistance that accounts for 85% of the total
kx
Ky
ky
resistance:
1/ k y
1/ K y
= 0.85 ⇒ ky = Ky/0.85 = 2.786×10-4/0.85 = 3.28×10-4 kmol/m2⋅s
m / kx
= 0.15 ⇒ kx = mKy/0.15 = 1.64×2.786×10-4/0.15 = 3.05×10-3 kmol/m2⋅s
1/ K y
yA* = mxA = 1.64×1.15×10-3 = 1.886×10-3
NA = Ky(yA − yA*) = 2.786×10-4(0.08 − 1.886×10-3) = 2.18×10-5 kmol/m2⋅s
2.18 × 10 −5
NA
NA = ky(yA − yAi) ⇒ yAi = yA −
= 0.080 −
= 0.01362
3.28 × 10 −4
ky
yAi = m xAi ⇒ xAi =
1
yAi
0.01362
=
= 8.305×10-3
m
1.64
Benitez, J. Principle and Modern Applications of Mass Transfer Operations, Wiley, 2009, p. 169
3-19
Example 3.2-32. ---------------------------------------------------------------------------------A wetted-wall absorption tower is fed with water as the wall liquid and an ammonia-air
mixture as the central-core gas. At a particular point in the tower, the ammonia concentration
in the bulk gas is 0.6 mole fraction, that in the bulk liquid is 0.12 mole fraction. The
temperature is 300 K and the pressure is 1 atm. Ignoring the vaporization of water, calculate
the local ammonia mass-transfer flux. Data: kx = 3.5/(1 − xA)lm mol/m2⋅s, and ky = 2.0/(1 −
yA)lm mol/m2⋅s; equilibrium relation: yAi = 10.5xAi[0.156 + 0.622xAi(5.765xAi − 1)].
Solution ----------------------------------------------------------------------------------------The molar flux of ammonia (A) is given by
NA = ky
x Ai − xA
y A − y Ai
= kx
(1 − xA ) − (1 − x Ai )
(1 − y Ai ) − (1 − y A )
 1 − xA 
 1 − y Ai 
ln 
ln 


 1 − yA 
 1 − xAi 
 1 − y Ai 
 1 − xA
ky ln 
 = kx ln 
 1 − yA 
 1 − x Ai
 1 − xA 
yAi = 1 − (1 − yAi) 

 1 − x Ai 

 1 − xA 
1 − y Ai
=
⇒

1 − yA

 1 − x Ai 
kx / ky
kx / ky
1.75
 0.88 
= 1 − 0.4 

 1 − x Ai 
(E-1)
yAi and xAi are the solutions of Eq. (E-1) and the equilibrium relation:
yAi = 10.5xAi[0.156 + 0.622xAi(5.765xAi − 1)]
(E-2)
The following Matlab codes solve Eqs. (E-1) and (E-2) and plot out these equations. The
intersection of the two curves from these equations also provides the solution.
v=fminsearch('f3d2d3',[0.5 0.5]);
xi=v(1);yi=v(2);
fprintf('xAi = %8.3f, yAi = %8.3f\n',xi,yi)
x=0:0.01:0.28;
y1=1-0.4*(0.88./(1-x)).^1.75;
y2=10.5*x.*(0.156 + 0.622*x.*(5.765*x-1));
plot(x,y1,x,y2)
xlabel('x_A');ylabel('y_A')
grid on
function ff=f3d2d3(v)
xi=v(1);yi=v(2);
f1=yi-1+0.4*(0.88/(1-xi))^1.75;
f2=yi-10.5*xi*(0.156 + 0.622*xi*(5.765*xi-1));
2
Benitez, J. Principle and Modern Applications of Mass Transfer Operations, Wiley, 2009, p. 171
3-20
ff=f1*f1+f2*f2;
>> e3d2d3
xAi = 0.231, yAi =
0.494
The interfacial concentrations are: xAi = 0.231 and yAi = 0.494. The molar flux of A is then
 1 − y Ai 
 1 − 0.494 
2
NA = ky ln 
 = 2 ln 
 = 4.7 mol/m ⋅s
1
−
y
1
−
0.600



A 
Example 3.2-43. ---------------------------------------------------------------------------------A mixture of methanol (substance 1, the more volatile) and water (substance 2) is being
distilled in a packed tower. At a point along the tower where the temperature is 360 K, the
methanol content of the bulk of the gas phase is 36 mol%; that of the bulk of the liquid phase
is 20 mol%. Assume equimolar counter diffusion with mass transfer coefficients ky = 0.0017
kmol/m2⋅s and ky = 0.0017 kmol/m2⋅s, estimate the local flux of methanol from the liquid to
the gas phase. Solve the problem by plotting equilibrium curve at 360 K. The activity
coefficients for this system are given by:
 ∆12
∆ 21 
V2
 −a 
−
ln γ1 = − ln(x1 +x2∆12) + x2 
exp  12 
 ; ∆12 =
V1
 RT 
 x1 + x2 ∆12 x2 + x1∆ 21 
3
Benitez, J. Principle and Modern Applications of Mass Transfer Operations, Wiley, 2009, p. 174
3-21
 ∆12
∆ 21 
V1
 −a 
−
ln γ2 = − ln(x2 +x1∆21) + x1 
exp  21 
 ; ∆21 =
V2
 RT 
 x1 + x2 ∆12 x2 + x1∆ 21 
Recommended values of the parameter for this system are: V1 = 40.73 cm3/mol, V2 = 18.07
cm3/mol, a12 = 107.38 cal/mol, and a21 = 469.55 cal/mol. The vapor pressure for methanol
and water can be determined from the following equations:
ln P1vap = 16.5938 −
3644.3
T − 33
ln P2vap = 16.2620 −
3800.0
T − 47
Solution ----------------------------------------------------------------------------------------The equilibrium curve at 360 K is obtained from the following steps:
1)
2)
3)
4)
Choose a value of mole fraction in the liquid phase, x1, between 0.1 and 0.25;
Evaluate the activity coefficients;
Evaluate the vapor pressures
Evaluate the partial pressures:
p1 = γ1x1P1vap, and p2 = γ2x2P2vap
5) The mole fraction in the vapor phase, y1, is then p1/( p1 + p2)
The local flux is obtained from the following expressions:
N1 = kx(x − xi) = ky(yi − y)
yi = y +
0.0149
kx
(x − xi) = 0.36 +
(0.2 − xi)
0.0017
ky
(E-1)
The above equation is a straight line that can be plotted on the graph with the equilibrium
curve. The intersection of flux equation (E-1) and the equilibrium curve provides the
interfacial compositions xi and yi.
The Matlab codes, listed in Table E-1, plots the equilibrium curve and flux equation (E-1).
From the closed up view of the graph (Figure E-2), the interfacial compositions are
xi = 0.17884
and yi = 0.54545
The local flux is then
N1 = kx(x − xi) = 0.0149(0.2 − 0.17884) = 3.15×10-4 kmol/m2⋅s
3-22
--------- Table E-1 Matlab codes to plot equilibrium curve and flux equation ---------
R=1.987;T=360;v1=40.73;v2=18.07;a12=107.38;a21=469.55;
RT=R*T;
d12=v2*exp(-a12/(RT))/v1;d21=v1*exp(-a21/(RT))/v2;
x1=0.1:0.01:0.25;x2=1-x1;
con=d12./(x1+x2*d12)-d21./(x2+x1*d21);
gam1=exp(-log(x1+x2*d12)+x2.*con);
gam2=exp(-log(x2+x1*d21)-x1.*con);
pv1=exp(16.5938-3644.3/(T-33));
pv2=exp(16.2620-3800.0/(T-47));
p1=pv1*gam1.*x1;p2=pv2*gam2.*x2;
y1=p1./(p1+p2);
plot(x1,y1)
x=0.15;
y=0.36+(0.2-x)*.0149/.0017;
plot(x1,y1,[x 0.2],[y 0.36],'--')
xlabel('x_A');ylabel('y_A')
grid on
legend('Equilibrium','Flux equation')
---------------------------------------------------------------------------
Figure E-1 Graphical solution
3-23
Figure E-2 Graphical solution (closed-up view)
The interfacial compositions xi and yi can be obtained directly from the following Matlab
codes without graphing:
function ff=f3d2d4(v)
x1=v(1);y1=v(2);x2=1-x1;
R=1.987;T=360;v1=40.73;v2=18.07;a12=107.38;a21=469.55; RT=R*T;
d12=v2*exp(-a12/(RT))/v1;d21=v1*exp(-a21/(RT))/v2;
con=d12/(x1+x2*d12)-d21/(x2+x1*d21);
gam1=exp(-log(x1+x2*d12)+x2*con);
gam2=exp(-log(x2+x1*d21)-x1*con);
pv1=exp(16.5938-3644.3/(T-33));
pv2=exp(16.2620-3800.0/(T-47));
p1=pv1*gam1*x1;p2=pv2*gam2*x2;
f1=y1-p1/(p1+p2);
f2=y1-0.36-0.0149*(0.2-x1)/0.0017;
ff=f1*f1+f2*f2;
>> fminsearch('f3d2d4',[0.5 0.5])
ans =
0.1788
0.5455
3-24