Chapter 2 : Diffusion¶
Example 2.1.a¶
In [30]:
# variable declaration
rA=.3798;
rB=.3941;
# Calculation
rAB=(rA+rB)/2; #molecular seperation at collision
ebyk_A=71.4;
ebyk_B=195.2;
ebyk_AB=(ebyk_A/ebyk_B)**.5; #energy of molecular attraction
pt=1.013*10**5; #absolute total pressure in pascal
T=298; #absolute temperature in kelvin
s=T/ebyk_AB; #collision function
#from chart f(T/ebyk_AB) = 0.5 let it be = x
x=.5; #collision function
MA=28; #molecular weight of nitrogen
MB=44; #molecular weight of carbondioxide
Mnew=((1./MA)+(1./MB))**.5;
Dab= 10**-4*(1.084-.249*(Mnew))*T**1.5*((Mnew))/(pt*x*rAB**2);
# Result
print " the diffisivity of nitrogen-carbondioxide is :%f *10**-5 m**2/s"%(Dab/10**-5)
#end
Example 2.1.b¶
In [31]:
#"part(ii)"
rA=.3339;
rB=.3711;
# Calculation
rAB=(rA+rB)/2; #molecular seperation at collision
ebyk_A=344.7;
ebyk_B=78.6;
ebyk_AB=(ebyk_A/ebyk_B)**.5; #energy of molecular attraction
pt=200.*10**3; #absolute total pressure in pascal
T=298.; #absolute temperature in kelvin
s=T/ebyk_AB; #collision function
#from chart f(T/ebyk_AB) = 0.62 let it be = x
x=0.62; #collision function
MA=36.5; #molecular weight of hydrogen chloride
MB=29; #molecular weight of air
Mnew=((1./MA)+(1./MB))**.5;
Dab=10.**-4*(1.084-.249*(Mnew))*T**1.5*((Mnew))/(pt*x*rAB**2);
# Result
print "\n the diffisivity of hydrogen chloride-air is :%f *10**-6 m**2/s"%(Dab/10**-6)
#end
Example 2.2¶
In [32]:
# variable declaration
u=1.145*10**-3; #viscosity of water1.145cp
v_a=5*.0148+12*.0037+1*.0074; #by kopp's law
t=288; #temperature of water in kelvin
MB=18; #molecular weight of water
phi=2.26; #association parameter for solvent-water
# Calculation
D_ab=(117.3*10**-18)*((phi*MB)**.5)*(t)/(u*(v_a)**.6);
# Result
print "\n the diffusivity of isoamyl alcohol is :%f *10**-9 m**2/s"%(D_ab/10**-9)
#end
Example 2.3¶
In [33]:
pa1=(33./760)*1.013*10**5; #vapour pressure of ccl4 at 273 in pascal
pa2=0;
d=1.59;
import math #density of liquid ccl4 in g/cm**3
#considering o2 to be non diffusing and with
T=273.; #temperature in kelvin
pt=(755./780)*1.013*10**5; #total pressure in pascal
z=.171; #thickness of film
a=.82*10**-4; #cross-sectional area of cell in m**2
v=.0208; #volume of ccl4 evaporated
t=10.; #time of evaporation
MB=154.;
# Calculation
#molecular wght of ccl4
rate=v*d/(MB*t); #.0208cc of ccl4 is evaporating in 10hrs
Na=rate*10.**-3/(3600.*a); #flux in kmol/m**2*S
D_ab=Na*z*8314*273./(pt*(math.log((pt-pa2)/(pt-pa1)))); #molecular diffusivity in m**2/s
# Result
print "the diffusivity of ccl4 through oxygen:%.2f *10**-6 m**2/s"%(D_ab/10.**-6)
#end
Example 2.4¶
In [34]:
z=.0305*10**-3; #wall thickness sorrounding the crystal
x1=0.0229;
w1=160.; #molecular weight of copper sulphate
w2=18.; #molecular weight of water
Dab=7.29*10**-10; #diffusivity of copper sulphatein m**2/s
#av=d/m
# Calculation
Mavg=x1*w1+(1-x1)*w2; #average molecular wght of solution
d1=1193; #density of copper sulphate solution
av1=d1/Mavg; #value of (d/m) of copper solution
#for pure water
d2=1000.; #density of water
m2=18.; #molecular wght of water
av2=d2/m2; #value of (d/m) of water
allavg=(av1+av2)/2.; #average value of d/m
xa2=0;
import math
# Result
Na=Dab*(allavg)*math.log((1-xa2)/(1-x1))/z; #flux of cuso4 from crystal surface to bulk solution
print " the rate at which crystal dissolves :%f *10**-5 kmol/m**2*s"%(Na/10**-5)
#end
Example 2.5.a¶
In [6]:
ya1=0.8;
ya2=0.1;
T=(273+35); #temperature in kelvin
pt=1*1.013*10**5; #total pressure in pascal
z=0.3*10**-3; #gas film thickness in m
Dab=.18*10**-4; #diffusion coefficient in m**2/s
R=8314; #universal gas constant
# Calculation
Na=Dab*pt*(ya1-ya2)/(z*R*T) #diffusion flux in kmol/m**2*s
rate=Na*100*10**-4*3600*46; #since molecular weight of mixture is 46
# Result
print "\n rate of diffusion of alcohol-water vapour :%f kg/hr "%rate
#end
Example 2.5.b¶
In [35]:
ya1=0.8;
ya2=0.1;
T=(273+35); #temperature in kelvin
pt=1*1.013*10**5; #total pressure in pascal
z=0.3*10**-3; #gas film thickness in m
Dab=.18*10**-4; #diffusion coefficient in m**2/s
R=8314; #universal gas constant
import math
#diffusion through stagnant film
# Calculation
Na=Dab*pt*math.log((1-ya2)/(1-ya1))/(z*R*T); #diffusion flux in kmol/m**2*s
rate=Na*100*10**-4*3600*46; #since molecular weight of mixture is 46
# Result
print "\n rate of diffusion if water layer is stagnant :%f *10**-3 kg/s "%(rate/(3600*10**-3))
#end
Example 2.6¶
In [36]:
T=298.; #temperature in kelvin
pt=1.*1.013*10**5; #total pressure in pascal
ID=25.*10**-3; #internal diameter in m of unvulcanised rubber in m
OD=50.*10**-3; #internal diameter in m of unvulcanised rubber in m
Ca1=2.37*10**-3; #conc. of hydrogen at the inner surface of the pipe in kmol/m**3
Ca2=0; #conc. of hydrogen at 2
Dab=1.8*10**-10; #diffusion coefficient in cm**2/s
l=2; #length of pipe in m
import math
# Calculation
# Va=Da*Sa*(pa1-pa2)/z;
z=(50-25)*10**-3/2.; #wall thickness in m
Va=Dab*(Ca1-Ca2)/z; #diffusion through a flat slab of thickness z
Sa=2*3.14*l*(OD-ID)/(2*math.log(OD/ID)); #average mass transfer area of
rate=Va*Sa; #rate of loss of hydrogen by diffusion
# Result
print "\n rate of loss hydrogen by diffusion through a pipe of 2m length :%f*10**-12kmol/s"%(rate/10.**-12)
#end
Example 2.7¶
In [37]:
pa1=(1.5)*10**4; #vapour pressure of ammonia at pt.1
pa2=(0.5)*10**4; #vapour pressure of ammonia at pt.2
Dab=2.3*10**-5 #molecular diffusivity in m**2/s
z=0.15; #diffusion path in m
R=8314.; #universal gas constant
#ammonia diffuses through nitrogen under equimolar counter diffusion
T=298.; #temperature in kelvin
pt=1.013*10**5; #total pressure in pascal
# Calculation
Na=Dab*(pa1-pa2)/(z*R*T); #flux in kmol/m**2*S
# Result
print "the ammonia diffusion through nitrogen under equimolar \
counter diffusion:%f *10**-7 kmol/m**2*s"%(Na/10**-7);
#end
Example 2.8¶
In [39]:
import math
z=0.4*10**-2; #film thickness sorrounding the crystal
xa1=0.0453; #mole fraction of ethanol at pos.2
xa2=0.02775; #mole fraction of ethanol at pos.1
w1=46; #molecular weight of ethanol
w2=18; #molecular weight of water
Dab=74*10**-5*10**-4; #diffusivity of ethanol water sol.in m**2/s
#av=d/m
# Calculation
Mavg1=xa2*w1+(1-xa2)*w2; #average molecular wght of solution at pos 1
d1=0.9881*10**3; # density of 6.8 wt% solution
av1=d1/Mavg1; #value of (d/m) of copper solution
#for pure water
d2=972.8; # density of 10.8 wt% solution
Mavg2=xa1*w1+(1-xa1)*w2; #average molecular wght of solution at pos.2
av2=d2/Mavg2; #value of (d/m) of water
allavg=(av1+av2)/2; #average value of d/m
Na=Dab*(allavg)*math.log((1-xa2)/(1-xa1))/z; #steady state flux in kmol/m**2*s of ethanol water sol.
# Result
print "\n the rate at which crystal dissolves :%f *10**-5 kmol/m**2*s"%(Na/10**-5)
#end
Example 2.9¶
In [41]:
import math
# variable declaration
#basis : 100kg of mixture
z=2*10**-3; #film thickness sorrounding the water
xa1=0.0323; #mole fraction of ethanol at pos.2
xa2=0.0124; #mole fraction of ethanol at pos.1
w1=60.; #molecular weight of acetic acid
w2=18.; #molecular weight of water
Dab=0.000095; #diffusivity of acetic water sol.in m**2/s
#av=d/m
# Calculation
Mavg1=xa1*w1+(1-xa1)*w2; #average molecular wght of solution at pos 1
d1=1013.; # density of 10 % acid
av1=d1/Mavg1; #value of (d/m) of copper solution
#for pure water
d2=1004; #density of 4% acid
Mavg2=xa2*w1+(1-xa2)*w2; #average molecular wght of solution at pos.2
av2=d2/Mavg2; #value of (d/m) of water
allavg=(av1+av2)/2.; #average value of d/m
#assuming water to be non diffusing
Na=Dab*(allavg)*math.log((1-xa2)/(1-xa1))/z; #diffusion rate of acetic acid aacross film of non diffusing water sol.
# Result
print "\n diffusion rate of acetic acid aacross film of non diffusing water sol. :%f kmol/m**2*s"%Na
#end
Example 2.10.a¶
In [43]:
r=(50./2)*10**-3; #radius pf circular tube
pa1=190; #vapour pressure of ammonia at pt.1
pa2=95; #vapour pressure of ammonia at pt.2
Dab=2.1*10**-5 #molecular diffusivity in m**2/s
z=1;
R=760*22.414/273; #universal gas constant in mmHg*m**3*K*kmol
#carbondioxide and oxygen experiences equimolar counter diffusion
T=298.; #temperature in kelvin
# Calculation
pt=(10.0/780)*1.013*10**5; #total pressure in pascal
Na=Dab*(pa1-pa2)/(z*R*T); #flux in kmol/m**2*S
rate=Na*(3.14*r**2); #rate of mass transfer..(3.14*r**2)-is the area
# Result
print "\n the rate of mass transfer.:%f *10**-10 kmol/s"%(rate/10**-10)
#end
Example 2.10.b¶
In [44]:
r=(50./2)*10**-3; #radius pf circular tube
pa1=(190.); #vapour pressure of ammonia at pt.1
pa2=(95.); #vapour pressure of ammonia at pt.2
Dab=2.1*10**-5 #molecular diffusivity in m**2/s
R=760.*22.414/273; #universal gas constant in mmHg*m**3*K*kmol
#carbondioxide and oxygen experiences equimolar counter diffusion
T=298.; #temperature in kelvin
# Calculation
pt=(10./780)*1.013*10**5; #total pressure in pascal
#part (ii)
#(ya-ya1)/(ya2-ya1)=(z-z1)/(z2-z1);
z2=1; #diffusion path in m at pos.2
z1=0; #diffusion path in m at pos.1
z=.75; #diffusion at general z
#pa=poly([0],'pa'); #calc. of conc. in gas phase
x= 118.750000 #roots((pa-pa1)/(pa2-pa1)-(z-z1)/(z2-z1));
# Result
print "\n partial pressure of co2 at o.75m from the end where partial pressure is 190mmhg is:%f mmHg"%x
#end
Example 2.11.a¶
In [45]:
ya1=0.2; #initial mole fraction
ya2=0.1; #final mole fraction
T=298 #temperature in kelvin
pt=1*1.013*10**5; #total pressure in pascal
z=0.2*10**-2; #gas film thickness in m
Dab=.215*10**-4; #diffusion coefficient in m**2/s
R=8314.; #universal gas constant
#part (i)when N2 is non diffusing
import math
# Calculation
Na=Dab*pt*math.log((1-ya2)/(1-ya1))/(z*R*T); #diffusion flux in kmol/m**2*s
# Result
print " diffusion flux if N2 is non diffusing :%f *10**-5 kmol/m**2*s "%(Na/10**-5);
#end
Example 2.11.b¶
In [46]:
ya1=0.2;
ya2=0.1;
T=(298); #temperature in kelvin
pt=1*1.013*10**5; #total pressure in pascal
z=0.2*10**-2; #gas film thickness in m
Dab=.215*10**-4; #diffusion coefficient in m**2/s
R=8314; #universal gas constant
#part (ii) equimolar counter diffusion
# Calculation
Na=Dab*pt*(ya1-ya2)/(z*R*T) #diffusion flux in kmol/m**2*s
# Result
print " diffusion flux of oxygen during equimolar counter-diffusion :%f *10**-5 kmol/m**2*s "%(Na/10**-5)
#end
Example 2.12¶
In [15]:
ya1=0.1;
ya2=0;
T=293. #temperature in kelvin
pt=1*1.013*10**5; #total pressure in pascal
z=0.2*10**-2; #gas film thickness in m
Dab=.185*10**-4; #diffusion coefficient in m**2/s
R=8314.; #universal gas constant
#part (i)when air is assumed to be stagnant and non-diffusing
import math
# Calculation
Na=Dab*pt*math.log((1-ya2)/(1-ya1))/(z*R*T); #diffusion flux in kmol/m**2*s
mw=17; #molecular weight of ammonia
massflux=Na*mw; #mass flux of given NH3
print " diffusion flux when total presssure is 1atm and air \
is non-diffusing :%f *10**-4 kg/m**2*s "%(massflux/10**-4);
#part (ii) when pressure is increased to 10atm
#Dab_1/Dab_2=pt_2/pt_1
pt_2=10; #final pressure in atm
pt_1=1; #initially pressure was 1atm
Dab_1=.185; #initially diffusion coefficient was.185
Dab_2=Dab_1*pt_1/pt_2; #for gases Dab is proportional to 1/pt
Dab=Dab_2*10**-4; #new diffusion coefficient
pt=pt_2*1.013*10**5; #new total pressure
Na=Dab*pt*math.log((1-ya2)/(1-ya1))/(z*R*T); #diffusion flux in kmol/m**2*s
# Result
print "diffusion flux when pressure is increased to 10atm :%f *10**-5 kmol/m**2*s "%(Na/10**-5);
print "so the rate of diffusion remains same on increasing the pressure"
#end
Example 2.13¶
In [47]:
# variable declaration
T=290.0; #temperature in kelvin
z=2*10**-3; #film thickness sorrounding the water
xa2=0.0092; #mole fraction of ethanol at pos.2
xa1=0.0288; #mole fraction of ethanol at pos.1
w1=60.0; #molecular weight of acetic acid
w2=18.0; #molecular weight of water
Dab=0.95*10**-9; #diffusivity of acetic water sol.in m**2/s
#av=d/m
# Calculation
Mavg1=xa1*w1+(1-xa1)*w2; #average molecular wght of solution at pos 1
d1=1012.0; # density of 10 % acid
av1=d1/Mavg1; #value of (d/m) of copper solution
#for position 2
d2=1003.0; #density of 4% acid
Mavg2=xa2*w1+(1-xa2)*w2; #average molecular wght of solution at pos.2
av2=d2/Mavg2; #value of (d/m) of water
allavg=(av1+av2)/2; #average value of d/m
#assuming water to be non diffusing
import math
Na=Dab*(allavg)*math.log((1-xa2)/(1-xa1))/z; #diffusion rate of acetic acid aacross film of non diffusing water sol.
# Result
print "diffusion rate of acetic acid aacross film of non diffusing water sol. :%f *10**-7 kmol/m**2*s"%(Na/10.0**-7)
#end
Example 2.14¶
In [48]:
ya1=0.2; #molefraction at pos.1
ya2=0.1; #molefraction at pos.2
T=(293); #temperature in kelvin
pt=1*1.013*10**5; #total pressure in pascal
z=0.2*10**-2; #gas film thickness in m
Dab=.206*10**-4; #diffusion coefficient in m**2/s
R=8314; #universal gas constant
#for ideal gases volume fraction =mole fraction
#part (i)when N2 is non diffusing
import math
# Calculation
Na=Dab*pt*math.log((1-ya2)/(1-ya1))/(z*R*T); #diffusion flux in kmol/m**2*s
print "\n diffusion flux if N2 is non diffusing :%f *10**-5 kmol/m**2*s "%(Na/10**-5);
#part (ii) equimolar counter diffusion
Na=Dab*pt*(ya1-ya2)/(z*R*T) #diffusion flux in kmol/m**2*s
# Result
print "\n diffusion flux of nitrogen during equimolar counter-diffusion :%f *10**-5 kmol/m**2*s "%(Na/10**-5)
#end
Example 2.15¶
In [49]:
pa1=0.2*10**5; #partial pressure at pos.1
pa2=0; #partial pressure at pos.2
r=10./2; #radius of tank in which benzene is stored
T=298.; #temperature in kelvin
pt=1*1.013*10**5; #total pressure in pascal
z=10*10**-3; #gas film thickness in m
Dab=.02/3600; #diffusion coefficient in m**2/s
R=8314.; #universal gas constant
#benzene is stored in atank of dia 10m
#part (i)when air is assumed to be stagnant
import math
# Calculation
Na=Dab*pt*math.log((pt-pa2)/(pt-pa1))/(z*R*T); #diffusion flux in kmol/m**2*s
rate=Na*(3.14*r**2); #rate of loss of benzene if air is stagnant
# Result
print "diffusion rate of loss of benzene :%f *10**-4 kmol/s "%(rate/10**-4);
#end
Example 2.16¶
In [50]:
ya2=0.1; #molefraction at pos.2
ya1=0.8; #molefraction at pos.1
T=(370.); #temperature in kelvin
pt=1*1.013*10**5; #total pressure in pascal
z=0.1*10**-3; #gas film thickness in m
Dab=.15*10**-2; #diffusion coefficient in m**2/s
R=8314; #universal gas constant
Area=10; #area of the film is 10m**2
#for gase Dab=T**3/2
#Dab1/Dab2=(T1/T2)**3/2
import math
T2=370.; #final temperature in kelvin
T1=298.; #initial temperature in kelvin
# Calculation
Dab1=.15*10**-2; #initial diffusion coefficient
Dab2=((T2/T1)**(3./2))*Dab1; #final diffusion coefficient
Na=Dab2*pt*math.log((1-ya2)/(1-ya1))/(z*R*T); #diffusion flux in kmol/m**2*s
rate=Na*3600*46*Area; #rate of diffusion of alcohol-water vapour in kg/hour
# Result
print " rate of diffusion of alcohol-water vapour :%f *10**6 kg/hour "%(rate/10**6)
#end
Example 2.17¶
In [52]:
ya2=0; #molefraction at pos.2
ya1=0.1; #molefraction at pos.1
T=(273); #temperature in kelvin
pt=1*1.013*10**5; #total pressure in pascal
z=2*10**-3; #gas film thickness in m
Dab=.198*10**-4; #diffusion coefficient in m**2/s
R=8314; #universal gas constant
#for gase Dab=T**3/2
#Dab1/Dab2=(T1/T2)**3/2
import math
T2=293; #final temperature in kelvin
T1=273; #initial temperature in kelvin
# Calculation
Dab1=0.198*10**-4; #initial diffusion coefficient
Dab2=((T2/T1)**(3.0/2))*Dab1; #final diffusion coefficient
Na=Dab2*pt*math.log((1-ya2)/(1-ya1))/(z*R*T2); #diffusion flux in kmol/m**2*s
print " flux of diffusion of ammonia through inert film :%f *10**-5 kmol/m**2*s "%(Na/10**-5)
#if pressure is also incresed from 1 to 5 atm
#for gases Dab=(T**3/2)/pt;
#Dab1/Dab2=(T1/T2)**3/2*(p2/p1)
T2=293.; #final temperature in kelvin
T1=273.; #initial temperature in kelvin
pa2=5.; #final pressure in atm
pa1=1; #initial pressure in atm
p=pa2*1.013*10**5;
Dab1=.198*10**-4; #initial diffusion coefficient
Dab2=((T2/T1)**(3.0/2))*Dab1*(pa1/pa2); #final diffusion coefficient
Na=Dab2*p*math.log((1-ya2)/(1-ya1))/(z*R*T2); #diffusion flux in kmol/m**2*s
# Result
print "flux of diffusion of ammonia if temp. is 20 and pressure is 5 atm :%f*10**-5 kmol/m**2*s "%(Na/10**-5)
print "so there is no change in flux when pressure is changed"
#end
Example 2.18¶
In [54]:
# variable declaration
pa1=0.418*10**5; #partial pressure initially
pa2=0; #partial pressure of pure air
r=10/2; #radius of tank in which benzene is stored
T=(350); #temperature in kelvin
pt=1*1.013*10**5; #total pressure in pascal
z=2*10**-3; #gas film thickness in m
Dab=.2*10**-4; #diffusion coefficient in m**2/s
R=8314; #universal gas constant
r=0.2/2; #radius of open bowl is 0.2
#when air layer is assumed to be stagnant of thickness 2mm
import math
# Calculation
Na=Dab*pt*math.log((pt-pa2)/(pt-pa1))/(z*R*T); #diffusion flux in kmol/m**2*s
rate=Na*(3.14*r**2)*18; #rate of loss of evaporation
# Result
print "\n diffusion rate loss of evaporation :%f *10**-4 kg/s "%(rate/10**-4)
#end
Example 2.19¶
In [55]:
# variable declaration
#stefan tube experiment
import math
Ml=92.; #molecular weight of toluene
T=(312.4); #temperature in kelvin
pt=1*1.013*10**5; #total pressure in pascal
R=8314.; #universal gas constant
t=275.*3600; #after 275 hours the level dropped to 80mm from the top
zo=20.*10**-3; #intially liquid toluene is at 20mm from top
zt=80.*10**-3; #finally liquid toluene is at 80mm from top
#air is assumed to be satgnant
d=850.; #density in kg/m**3
pa=7.64*10**3; #vapour pressure of toluene in at 39.4degree celcius
# Calculation
cal=d/Ml; #conc. at length at disxtance l
ca=pt/(R*T); #total conc.
xa1=pa/pt; #mole fraction of toluene at pt1 i.e before evaporation
xb1=1-xa1; #mole fraction of air before evaporation i.e at pt1
xb2=1.; #mole fraction of air after evaporation i.e at pt.2
xa2=0.; #mole fraction of toluene at point 2
xbm=(xb2-xb1)/(math.log(xb2/xb1));
#t/(zt-zt0) = (xbm*cal*(zt+zo))/(2*c*(xa1-xa2)*t);
Dab=(xbm*cal*(zt**2-zo**2))/(2*ca*t*(xa1-xa2));
# Result
print "\n the diffusivity of the mixture in stefan tube of toluene in air is :%f*10**-5 m**2/s"%(Dab/10**-5)
#end
Example 2.20¶
In [1]:
# variable declaration
T=(360); #temperature in kelvin
pt=372.4/760; #total pressure in atm
R=82.06; #universal gas constant
Dab=0.0506; #diffusion coefficient in cm**2/s
z=0.254; #gas layer thickness in cm
vp=368.0/760; #vapour pressure of toluene in atm
xtol=.3; #mole fractoin of toluene in atm
pb1=xtol*vp; #partial pressure of toluene
#since pb1 is .045263 bt in book it is rounded to 0.145
# Calculation
pb2=xtol*pt; #parial pressure of toluene in vapour phase
Na=Dab*(pb1-pb2)/(z*R*T); #diffusion flux
# Result
print "\n the diffusion flux of a mixture of benzene and toluene %f*10**-8 gmol/cm**2*s\n"%(Na/10**-8)
print "\nthe negative sign indicates that toluene is getting transferred from gas phase \
to liquid phase(hence the transfer of benzene is from liquid to gas phase)"
#end
Example 2.21¶
In [57]:
# variable declaration
Ml=92.; #molecular weight of toluene
T=(303.); #temperature in kelvin
pt=1.*1.013*10**5; #total pressure in pascal
R=8314.; #universal gas constant
t=275.*3600; #after 275 hours the level dropped to 80mm from the top
zo=20.*10**-3; #intially liquid toluene is at 20mm from top
zt=77.5*10**-3; #finally liquid toluene is at 80mm from top
#air is assumed to be satgnant
import math
# Calculation
d=820.; #density in kg/m**3
pa=(57.0/760)*1.0135*10**5; #vapour pressure of toluene in at 39.4degree celcius
cal=d/Ml; #conc. at length at disxtance l
ca=pt/(R*T); #total conc.
xa1=pa/pt; #mole fraction of toluene at pt1 i.e before evaporation
xb1=1.-xa1; #mole fraction of air before evaporation i.e at pt1
xb2=1.; #mole fraction of air after evaporation i.e at pt.2
xa2=0; #mole fraction of toluene at point 2
xbm=(xb2-xb1)/(math.log(xb2/xb1));
#t/(zt-zt0) = (xbm*cal*(zt+zo))/(2*c*(xa1-xa2)*t);
Dab=(xbm*cal*(zt**2-zo**2))/(2*ca*t*(xa1-xa2));
# Result
print "\n the diffusivity of the mixture in stefan tube of toluene in air is :%f*10**-5 m**2/s"%(Dab/10**-5)
#end
Example 2.22¶
In [2]:
#variation in liquid level with respect to time is given below
%matplotlib inline
t=[26,185,456,1336,1958,2810,3829,4822,6385]
# let Zt-Zo= x;
x=[.25,1.29,2.32,4.39,5.47,6.70,7.38,9.03,10.48]
i=0;
y = [0,0,0,0,0,0,0,0,0] #looping starts
# Calculation
while(i<9):
y[i]=t[i]/x[i]; #for calculating the t/Zt-Zo value
i=i+1;
from matplotlib.pyplot import *
plot(x,y,"o-");
show()
#xtitle(" Fig.2.2 Example 22 ","X--(zi-zo),cm --->","Y-- vs (t/(zi-zo))min/cm ---->");
slope=51.4385*60 *10**4; #slope of the curve in 1/sec*m**2
#slope = Cal *(xblm)/(2*Dab*C*(xa1-xa2))
d=1540.; #density in kg/m**3
Ml=154.; #molecular weight of toluene
Cal=d/Ml ; #conc. at length at disxtance l in mol/m**3
T=(321.); #temperature in kelvin
pt=1.; #total pressure in atm
R=82.06; #universal gas constant
C=pt/(R*T) *10**3; #total conc. in kg mol/m**3
pa=(282./760); #vapour pressure of toluene
xa1=pa/pt; #mole fraction of toluene at pt1 i.e before evaporation
xb1=1.-xa1; #mole fraction of air before evaporation i.e at pt1
xb2=1.; #mole fraction of air after evaporation i.e at pt.2
xa2=0.; #mole fraction of toluene at point 2
xblm=(xb2-xb1)/(math.log(xb2/xb1)); #log mean temp. difference
Dab = Cal *(xblm)/(2*slope*C*(xa1-xa2)); #diffusivity coefficient
# Result
print "\n the diffusivity of the mixture by winklemann method of toluene in air is :%f*10**-6 m**2/s"%(Dab/10**-6)
#end
Example 2.23¶
In [59]:
d=0.001;
area=3.14*(d/2)**2; #area of the bulb
T=298.; #temperature in kelvin
p=1.013*10**5; #total pressure of both the bulbs
R=8314.; #universal gas constant
# Calculation
c=p/(R*T); #total concentration
Dab=.784*10**-4; #diffusion coefficient in m**2/s
xa1=0.8; #molefraction of nitrogen gas at the 1 end
xa2=0.25; #molefraction of nitrogen gas at the 2nd end
z=.15; #distance between the bulbs
#rate=area*Na;
rate=area*Dab*c*(xa1-xa2)/z; #rate of transfer of hydrogen and hydrogen
# Result
print "\n the rate of transfer from 1 to 2 of nitrogen and 2 to 1of hydrogen is :%f *10**-11kmol/s"%(rate/10.0**-11)
#end
Example 2.24¶
In [60]:
T=288; #temperature in kelvin
Mb=32; #molecular weight of methanol
phi=1.9; #association factor for solvent
# Calculation
va=(14.8+(4*24.6))*10**-3 #solute(CCl4) volume at normal BP in m**3/kmol
u=.6*10**-3; #viscosity of solution in kg/m*s
Dab=(117.3*10**-18)*(phi*Mb)**0.5*T/(u*va**0.6); #diffusion coefficient in m**2/s
# Result
print "\ndiffusivity of methanol in carbon tetrachloride is :%f*10**-9 m**2/s"%(Dab/10.0**-9)
#end
Example 2.25¶
In [28]:
T=288.0; #temperature in kelvin
Mb=18.0; #molecular weight of methanol
phi=2.26; #association factor for solvent
# Calculation
va=(2*14.8+(6*3.7)+7.4)*10**-3 #solute(water) volume at normal BP in m**3/kmol
u=1*10**-3; #viscosity of solution in kg/m*s
Dab=(117.3*10**-18)*(phi*Mb)**0.5*T/(u*va**0.6); #diffusion coefficient in m**2/s
# Result
print "\ndiffusivity of methanol in water is :%f*10**-9 m**2/s"%(Dab/10**-9)
#end
Example 2.26¶
In [61]:
u=20.*10**-6; #viscosity in Ns/m**2
pt=2666.; #total pressure in N/m**2
pa1=pt; #pressure at 1
pa2=0; #pressure at 2
mw=32.; #molecular weight of oxygen
R=8314.; #universal law constant
T=373.; #temp. in kelvin
gc=1.;
# Calculation
l=(3.2*u/pt)*((R*T)/(2*3.14*gc*mw))**0.5;#mean free path
d=.2*10**-6; #pore diameter
s=d/l; #value of dia/l
#hence knudsen diffusion occurs
Na=0.093*20*273/(760*373*22414*10**-1); #diffusion coefficient in kmol/m**2*s
Dka=(d/3)*((8*gc*R*T)/(3.14*mw))**0.5;
l1=Dka*(pa1-pa2)/(R*T*Na); #length of the plate
# Result
print "\n the length of the plate is :%f m "%l1
#for diffusion with hydrogen
u=8.5*10**-6; #viscosity in Ns/m**2
pt=1333; #total pressure in N/m**2
pa1=pt; #pressure at 1
pa2=0; #pressure at 2
mw=2; #molecular weight of oxygen
R=8314; #universal law constant
T=298; #temp. in kelvin
gc=1;
l=(3.2*u/pt)*((R*T)/(2*3.14*gc*mw))**0.5;#mean free path
d=.2*10**-6; #pore diameter
s=d/l; #value of dia/l
#hence knudsen diffusion occurs
Dka=(d/3)*((8*gc*R*T)/(3.14*mw))**0.5;
Na=Dka*(pa1-pa2)/(R*T*l1); #diffusion coefficient in kmol/m**2*s
print "\n the diffusion coefficient is :%f *10**-4 kmol/m**2*s"%(Na/10.0**-6)
#end