# Chapter 5 : Transport with a net convective flux¶

### Example 5.9 - Page No :166¶

In [3]:
# Variables
# given
v = 1.;  			 #[cm/sec] - volume velocity or bulk velocity
vol = 1.;  			 #[cm**3] - volume
na = 2.;  			 # moles of a
nb = 3.;  			 # moles of b
nc = 4.;  			 # moles of c
mma = 2.;  			 #molecular weight of a
mmb = 3.;  			 #molecular weight of b
mmc = 4.;  			 #molecular weight of c
ma = na*mma;  			 #[g] weight of a
mb = nb*mmb;  			 #[g] weight of b
mc = nc*mmc;  			 #[g] weight of c
NabyA = 2.+2;  			 #[mol/cm**2*s] - molar flux  =  diffusing flux +convected flux
NbbyA = -1.+3;  			 #[mol/cm**2*s] - molar flux  =  diffusing flux +convected flux
NcbyA = 0.+4;  			 #[mol/cm**2*s] - molar flux  =  diffusing flux +convected flux
NtbyA = NabyA+NbbyA+NcbyA;  			 #[mol/cm**2*s] - total molar flux

# Calculations
# on a mass basis,these corresponds to
nabyA = 4.+4;  			 #[g/cm**2*s]; - mass flux  =  diffusing flux +convected flux
nbbyA = -3.+9;  			 #[g/cm**2*s]; - mass flux  =  diffusing flux +convected flux
ncbyA = 0.+16;  			 #[g/cm**2*s]; - mass flux  =  diffusing flux +convected flux
ntbyA = nabyA+nbbyA+ncbyA;  			 #[g/cm**2*s] - total mass flux

# concentrations are expressed in molar basis
CA = na/vol;  			 #[mol/cm**3]
CB = nb/vol;  			 #[mol/cm**3]
CC = nc/vol;  			 #[mol/cm**3]
CT = CA+CB+CC;  			 #[mol/cm**3] - total concentration

# densities are on a mass basis
pa = ma/vol;  			 #[g/cm**3]
pb = mb/vol;  			 #[g/cm**3]
pc = mc/vol;  			 #[g/cm**3]
pt = pa+pb+pc;  			 #[g/cm**3]
Ua = NabyA/CA;  			 #[cm/sec];
Ub = NbbyA/CB;  			 #[cm/sec];
Uc = NcbyA/CC;  			 #[cm/sec];
# the same result will be obtained from dividing mass flux by density
Uz = (pa*Ua+pb*Ub+pc*Uc)/(pa+pb+pc);

# Results
print " Uz = %.3f cm/sec"%(Uz);
Uzstar = (NtbyA/CT);
print " Uz* = %.2f cm/sec"%(Uzstar);
print " For this Example  both Uz and Uz* are slightly greater than the volume \
velocity of 1cm/sec, because there is a net molar and \n mass diffusion in the positive direction."


 Uz = 1.034 cm/sec
Uz* = 1.11 cm/sec
For this Example  both Uz and Uz* are slightly greater than the volume  velocity of 1cm/sec, because there is a net molar and
mass diffusion in the positive direction.


### Example 5.10 - Page No :171¶

In [4]:
# Variables
# given (from Example  5.9)
na = 2.;  			 # moles of a
nb = 3.;  			 # moles of b
nc = 4.;  			 # moles of c
mma = 2.;  			 #molecular weight of a
mmb = 3.;  			 #molecular weight of b
mmc = 4.;  			 #molecular weight of c
ma = na*mma;  			 #[g] weight of a
mb = nb*mmb;  			 #[g] weight of b
mc = nc*mmc;  			 #[g] weight of c
NabyA = 2.+2;  			 #[mol/cm**2*s] - molar flux  =  diffumath.sing flux +convected flux
NbbyA = -1.+3;  			 #[mol/cm**2*s] - molar flux  =  diffusing flux +convected flux
NcbyA = 0.+4;  			 #[mol/cm**2*s] - molar flux  =  diffusing flux +convected flux
NtbyA = NabyA+NbbyA+NcbyA;  			 #[mol/cm**2*s] - total molar flux
vol= 1.
# Calculations
# on a mass basis,these corresponds to
nabyA = 4+4;  			 #[g/cm**2*s]; - mass flux  =  diffusing flux +convected flux
nbbyA = -3+9;  			 #[g/cm**2*s]; - mass flux  =  diffusing flux +convected flux
ncbyA = 0+16;  			 #[g/cm**2*s]; - mass flux  =  diffusing flux +convected flux

# concentrations are expressed in molar basis
CA = na/vol;  			 #[mol/cm**3]
CB = nb/vol;  			 #[mol/cm**3]
CC = nc/vol;  			 #[mol/cm**3]
CT = CA+CB+CC;  		 #[mol/cm**3] - total concentration

# densities are on a mass basis
pa = ma/vol;  			 #[g/cm**3]
pb = mb/vol;  			 #[g/cm**3]
pc = mc/vol;  			 #[g/cm**3]
Ua = NabyA/CA;  			 #[cm/sec];
Ub = NbbyA/CB;  			 #[cm/sec];
Uc = NcbyA/CC;  			 #[cm/sec];
U = (pa*Ua+pb*Ub+pc*Uc)/(pa+pb+pc);
Ustar = (NtbyA/CT);

# the fluxes relative to mass average velocities are found as follows
JabyA = CA*(Ua-U);  			 #[mol/cm**2*sec]
JbbyA = CB*(Ub-U);  			 #[mol/cm**2*sec]
JcbyA = CC*(Uc-U);  			 #[mol/cm**2*sec]

# Results
print " fluxes relative to mass average velocities are-";
print " Ja/A = %.4f mol/cm**2*sec"%(JabyA);
print " Jb/A = %.4f mol/cm**2*sec"%(JbbyA);
print " Jc/A = %.4f mol/cm**2*sec"%(JcbyA);
jabyA = pa*(Ua-U);  			 #[g/cm**2*sec]
jbbyA = pb*(Ub-U);  			 #[g/cm**2*sec]
jcbyA = pc*(Uc-U);  			 #[g/cm**2*sec]
print " ja/A = %.4f g/cm**2*sec"%(jabyA);
print " jb/A = %.4f g/cm**2*sec"%(jbbyA);
print " jc/A = %.4f g/cm**2*sec"%(jcbyA);

# the fluxes relative to molar average velocity are found as follows
JastarbyA = CA*(Ua-Ustar);  			 #[mol/cm**2*sec]
JbstarbyA = CB*(Ub-Ustar);  			 #[mol/cm**2*sec]
JcstarbyA = CC*(Uc-Ustar);  			 #[mol/cm**2*sec]
print " fluxes relative to molar average velocities are-";
print " Ja*/A = %.4f mol/cm**2*sec"%(JastarbyA);
print " Jb*/A = %.4f mol/cm**2*sec"%(JbstarbyA);
print " Jc*/A = %.4f mol/cm**2*sec"%(JcstarbyA);
jastarbyA = pa*(Ua-Ustar);  			 #[g/cm**2*sec]
jbstarbyA = pb*(Ub-Ustar);  			 #[g/cm**2*sec]
jcstarbyA = pc*(Uc-Ustar);  			 #[g/cm**2*sec]
print " ja*/A = %.4f g/cm**2*sec"%(jastarbyA);
print " jb*/A = %.4f g/cm**2*sec"%(jbstarbyA);
print " jc*/A = %.4f g/cm**2*sec"%(jcstarbyA);

 fluxes relative to mass average velocities are-
Ja/A = 1.9310 mol/cm**2*sec
Jb/A = -1.1034 mol/cm**2*sec
Jc/A = -0.1379 mol/cm**2*sec
ja/A = 3.8621 g/cm**2*sec
jb/A = -3.3103 g/cm**2*sec
jc/A = -0.5517 g/cm**2*sec
fluxes relative to molar average velocities are-
Ja*/A = 1.7778 mol/cm**2*sec
Jb*/A = -1.3333 mol/cm**2*sec
Jc*/A = -0.4444 mol/cm**2*sec
ja*/A = 3.5556 g/cm**2*sec
jb*/A = -4.0000 g/cm**2*sec
jc*/A = -1.7778 g/cm**2*sec


### Example 5.11 - Page No :176¶

In [4]:
import math
# Variables
# given
T = 0+273.15;  			 #[K] - temperature in Kelvins
pa2 = 1.5;  			 #[atm] - partial presuure of a at point2
pa1 = 0.5;  			 #[atm] - partial pressure of a at point 1
z2 = 20.;  			 #[cm] - position of point 2 from reference point
z1 = 0.;  			 #[cm] - position of point1 from reference point
p = 2.;  			 #[atm] - total pressure
d = 1.;  			 #[cm] - diameter
D = 0.275;  		 #[cm**2/sec] - diffusion coefficient
A = (math.pi*((d)**2))/4.;
R = 0.082057;  			 #[atm*m**3*kmol**-1*K**-1] - gas constant

# Calculations
# (a) using the formula Na/A = -(D/(R*T))*((pa2-pa1)/(z2-z1))
Na = (-(D/(R*T))*((pa2-pa1)/(z2-z1)))*(A)/(10.**6);
print " Na = %.2e kmol/sec \n The negative sign indicates diffusion from point 2 to point 1"%(Na);
pb2 = p-pa2;
pb1 = p-pa1;

# (b) using the formula Na/A = ((D*p)/(R*T*(z2-z1)))*ln(pb2/pb1)
Na = (((D*p)/(R*T*(z2-z1)))*math.log(pb2/pb1))*(A)/(10**6);

# Results
print " Na = %.2e kmol/sec"%(Na);
print " The induced velocity increases the net transport of A by the ratio of 10.6*10**-10 \
to 4.82*10**-10 or 2.2 times.This increse is equivalent to 120 percent"

 Na = -4.82e-10 kmol/sec
The negative sign indicates diffusion from point 2 to point 1
Na = -1.06e-09 kmol/sec
The induced velocity increases the net transport of A by the ratio of 10.6*10**-10 to 4.82*10**-10 or 2.2 times.This increse is equivalent to 120 percent


### Example 5.12 - Page No :178¶

In [10]:
# Variables
# given
T = 0+273.15;  			 #[K] - temperature in Kelvins
pa2 = 1.5;  			 #[atm] - partial presuure of a at point2
pa1 = 0.5;  			 #[atm] - partial pressure of a at point 1
z2 = 20.;  			 #[cm] - position of point 2 from reference point
z1 = 0.;  			 #[cm] - position of point1 from reference point
p = 2.;  			 #[atm] - total pressure
d = 1.;  			 #[cm] - diameter
D = 0.275;  			 #[cm**2/sec] - diffusion coefficient

# Calculations
A = (math.pi*((d)**2.))/4;
R = 0.082057;  			 #[atm*m**3*kmol**-1*K**-1] - gas consmath.tant
k = 0.75;

# umath.sing the formula (Na/A) = -(D/(R*T*(z2-z1)))*ln((1-(pa2/p)*(1-k))/(1-(pa1/p)*(1-k)))
NabyA = -(D/(R*T*(z2-z1)))*(2*0.7854)*math.log((1-(pa2/p)*(1-k))/(1-(pa1/p)*(1-k)))/(10**6);

# Results
print " Na/A = %.2e kmol/sec"%(NabyA);
print " Note that this answer is larger than the rate for equimolar counter diffusion \
but smaller tahn the rate for diffusion through a stagnant film. \nSometimes the\
rate for diffusin through a stagnant film can be considered as an upper bound\
if k ties between zero and one"

 Na/A = 1.38e-10 kmol/sec
Note that this answer is larger than the rate for equimolar counter diffusion but smaller tahn the rate for diffusion through a stagnant film.
Sometimes the rate for diffusin through a stagnant film can be considered as an upper bound if k ties between zero and one


### Example 5.13 - Page No :184¶

In [11]:
# Variables
# given
l = 4.;  			         #[m] - length of the tube
id_ = 1.6*10**-3;  			 #[m] - insid_e diameter
Nkn = 10.;  		    	 # - knudsen no.
Ma = 92.;  			         # - molecular weight of gas
mu = 6.5*10**-4;  			 #[kg/m*sec] - vismath.cosity
T = 300.;  	    		        #[K] - temperature
R = 8314.;    	    		    #[kPa*m**3*kmol**-1*K**-1] - gas consmath.tant
lambdaA = Nkn*id_;  			 #[m] mean free path

# Calculations
# for calculating pressure umath.sing the formula lamdaA = 32*(mu/p)*((R*T)/(2*pi*Ma))**(1/2)
p = 32*(mu/lambdaA)*((R*T)/(2*math.pi*Ma))**(1/2.);
patm = p/(1.01325*10**5);

# Results
print " p = %.2f kg/m*sec**2 = %.2f Pa = %.2e atm"%(p,p,patm);
print " The value of 10 for the knudsen number is on the border \
between Knudsen diffusion and transition flow";

 p = 85.39 kg/m*sec**2 = 85.39 Pa = 8.43e-04 atm
The value of 10 for the knudsen number is on the border  between Knudsen diffusion and transition flow