Chapter 11: An introduction to mass transfer¶
Example 11.1, Page number:603¶
In [14]:
from __future__ import division
#Variables
Mn2=0.7556; #mass fraction of nitrogen
Mo2=0.2315; #mass fraction of oxygen
Mar=0.01289; #mass fraction of argon gas
M1=28.02; #molar mass of N2,kg/kmol
M2=32; #molar mass of O2,kg/kmol
M3=39.95 ; #molar mass of Ar,kg/kmol
p=101325; #Atmospheric pressure in Pascal(Pa)
R=8314.5; #Gas constant, J/kmol-K
T=300; #Approximate room temperature, K
#Calculations
Mair=(Mn2/M1+Mo2/M2+Mar/M3)**-1; #molar mass of air,kg/kmol
Xo2=Mo2*Mair/M2; #mole fraction of O2
PO2=Xo2*p; #partial pressure of O2,Pa
Co2=PO2/(R*T); #molar volume of O2,kmol/m**3
ao2=Co2*M2; #density of O2,kg/m**3
#Result
print "Mole fraction of O2 is :",round(Xo2,4),"\n"
print "Partial pressure of O2 is :",round(PO2,4),"\n"
print "Molar volume of O2 is :",round(Co2,4)," kmol/m^3\n"
print "Density of O2 is :",round(ao2,4)," kg/m^3\n"
#end
Example 11.2, Page number: 606¶
In [15]:
from __future__ import division
#Variables
r=0.00241; #rate of consumption of carbon,kg/(m**2*s)
Mo2=0.2; #concentration of oxygen at surface s
Mco2=0.052; #concentration of CO2 at surface s
sd=0.29; #density of surface s,kg/m**3
#since carbon flows through a second imaginary surface u, the mass fluxes are relatedd by Ncu=-12/32*No2s=12/44*Nco2s
#the minus sign arises because the O2 flow is opposite the C and CO2 flows.in steady state if we apply mass conservation to the control volume between the u and s surface, wee find that the total mass flux entering the u surface equals that leaving the s surface
Ncu=r; #mass fluxes of carbon in u surface,kg/m**2/s
#Calculations
No2s=-32/12*Ncu; #mass flux of O2 in surface s,kg/(m**2*s)
Nco2s=44/12*Ncu; #mass flux of CO2 in surface s,kg/(m**2*s)
Vo2s=No2s/(Mo2*sd); #mass average speed,m/s
Vco2s=Nco2s/(sd); #mass average speed,m/s
Vs=(Nco2s+No2s)/sd; #effective mass average speed,m/s
j1=sd*Mo2*(Vo2s-Vs); #diffusional mass flux,kg/(m**2*s)
j2=sd*Mco2*(Vco2s-Vs); #diffusional mass flux,kg/(m**2*s)
#the diffusional mass fluxes are very nearly equal to the species m ss fluxes. tha is because the mass average speed is much less than species speeds.
N1 = Ncu/12; #mole flux of carbon through the surface s,kmol/(m**2*s)
N2 = -N1; #mole flux of oxygen through the surface s,kmol/(m**2*s)
#Result
print "Mass flux of O2 through an imaginary surface is :",round(j1,5),"kg/(m^2*s)\n"
print "Mass flux of CO2 through an imaginary surface is :",round(j2,5),"kg/(m^2*s)\n"
print "Mole flux of Co2 through an imaginary surface is :",round(N1,5),"kmol/(m^2*s)\n"
print "Mole flux of O2through an imaginary surface is :",round(N2,5),"kmol/(m^2*s)\n"
print "The two diffusional mole fluxes sum to zero themselves because ther is no convective mole flux for other species to diffuse against. the reader may find the velocity of the interface.that calculation shows the interface to be receding so slowly that the velocities are equal to those that would be seen by a stationary observer."
#end
Example 11.3, Page number: 617¶
In [2]:
from __future__ import division
import math
#Variables
T1=276; #temp.of air,K
aa=3.711; #lennard jones constant or collision diameter,A
ab=2.827; #lennard jones constant or collision diameter,A
b1=78.6; #lennard jones constant,K
b2=59.7; #lennard jones constant,K
Ma=28.97; #Molecular mass of air, kg/kmol
Mh=2.016; #Molecular mass of hydrogen, kg/kmol
#Calculations
a=(aa+ab)/2; #effective molecular diameter for collisions of hydrogen and air,m
b=math.sqrt(b1*b2); #effective potential well depth,K
c=T1/b;
d=0.8822; #potential well function, from table 11.3
Dab=(1.8583*math.pow(10,-7))*math.pow(T1,1.5)/(math.pow(a,2)*d)*math.sqrt(1/Mh+1/Ma); #diffusion coefficient of hydrogen in air,m**2/s
#Result
print "Diffusion coefficient of hydrogen in air is :",round(Dab,6),"m^2/s an experimental value from table is 6.34*10^-5 m^2/s,so the prediction is high by 5 percent.\n"
#end
Example 11.4, Page number: 625¶
In [1]:
from __future__ import division
import math
from numpy import array
#Variables
T1=373.15; #temp.of tea,K
XN2=0.7808; #mole fraction of nitrogen
XO2=0.2095; #mole fraction of oxygen
Xar=0.0093; #mole fraction of
Cp=1006 #mixture diffusivity,j/(kg*K)
#Calculations
a=array(([3.798, 3.467, 3.542])); #collisin diameter,m
b=array(([71.4, 106.7, 93.3])); #lennard jones constant,K
M=array(([28.02, 32, 39.95])); #molar masses,kg/kmol
c=array(([0.9599, 1.057, 1.021])); #potential well function
d=array(([1.8*10**-5, 2.059*10**-5, 2.281*10**-5])); #calculated viscosity,kg/(m*s)
e=array(([1.8*10**-5, 2.07*10**-5, 2.29*10**-5])); # theoritical viscosity,kg/(m*s)
f=array(([0.0260, 0.02615, 0.01787])); #theoritical thermal conducitvity,W/(m*K)
u=2.6693*10**-6*(M*T1)**0.5/((a**2*c)); #viscosity,kg/(m*s)
k=0.083228/((a**2*c*(T1/M**0.5))) #thermal conductivity,W/(m*s)
umc = XN2*u.item(0)/0.9978+XO2*u.item(1)/1.008+Xar*u.item(2)/0.9435 ; #calculated mixture viscosity,kg/(m*s)
umc1=1.857*10**-5;
umd=XN2*e.item(0)/0.9978+XO2*e.item(1)/1.008+e.item(2)*Xar/0.9435; #theoritical mixture viscosity,kg/(m*s)
kmc=XN2*k.item(0)/0.9978+XO2*k.item(1)/1.008+Xar*k.item(2)/0.9435; #calculated thermal conducitvity,W/(m*K)
kmc1=0.02623;
kmd=XN2*f.item(0)/0.9978+XO2*f.item(1)/1.008+Xar*f.item(2)/0.9435; #theoritical thermal conductivity, W/(m*K)
pr=umd*Cp/kmd; #prandtl no.
#Result
print "Theoritical mixture viscosity is :",round(umc1,6),"kg/(m*s)\n"
print "Calculated mixture viscosity is :",round(umd,6),"kg/(m*s)\n"
print "Theoritical thermal conducitvity is :",round(kmc1,4)," W/(m*K)\n"
print "Calculated thermal conducitvity is :",round(kmd,4),"W/(m*K)\n"
print "Prandtl no. is :",round(pr,4),"\n"
Example 11.5, Page number: 632¶
In [2]:
from __future__ import division
import math
from numpy import array
import matplotlib.pyplot as plt
%matplotlib inline
#Variables
Patm=101.325; #Atmospheric pressure in kPa.
Mh20=18.02; #Molecular mass of H20 in kg/kmol.
Mair=28.96; #Molecular mass of air in kg/kmol.
Psat =array(([0.6113, 1.2276, 2.3385, 4.2461, 7.3837, 12.35, 19.941, 31.188, 47.39, 70.139, 101.325])); #Saturated pressure of watrer in kPa
T=array(([0.01, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100])); #Temperature of air in in degree C
#Calculations
xw=Psat/Patm;
mw=(xw*Mh20)/(xw*Mh20+(1-xw)*Mair); #Mass fraction of water vapour.
#Result
plt.plot(T,mw);
plt.xlabel("Temperature(Degree C)")
plt.ylabel("Mass fraction of water vapour");
plt.show()
Example 11.6, Page number: 634¶
In [26]:
from __future__ import division
import math
#Variables
T1=263.15; #temp.of ice,K
p=101.325; #Atmospheric pressure, KPa
Molw=18.02; #Molecular mass of water vapour, g/mol
Mola=28.96; #Molecular mass of air, g/mol
#Calculations
Pv=math.exp(21.99-6141/(T1)); #vapor pressure,KPa
xw=Pv/p; #mole fraction of water
mw=xw*Molw/(xw*Molw+(1-xw)*Mola); #mass fraction
#Result
print "Mass fraction of watervapor above the surface of ice is :",round(mw,5),"\n"
#end
Example 11.10, Page number:¶
In [16]:
from __future__ import division
#Variables
T1=303; # isothermal temp.,K
v=5; #air speed,m/s
l=0.05; #length of naphthalene model that is flat, m
Mnap=128.2; #molar mass of naphthalene,kg/kmol
nu=1.867*10**-5; #Dynamic viscocity, m^2/s
#Calculations
D=0.86*(10**-5); #diffusion coefficient of naphthalene in air,m/s
Pv=10**(11.45-3729.3/T1)*133.31; #vapor pressure of napthalene, Pa
xn=Pv/101325; #mole fraction of naphthalene
mn=xn*Mnap/(xn*Mnap+(1-xn)*28.96); #mass fraction of naphthalene
mnp=0; #mass fraction of naphthalene in free stream is 0
Rel=v*l/nu; #reynolds no.
Sc=nu/D; #schimidt no.
Nul=0.664*math.sqrt(Rel)*(Sc**(1/3)); #mass transfer nusselt no.
Gmn=D*Nul*1.166/l; #gas phase mass transfer coefficient,kg/(m**2*s)
n=Gmn*(mn-mnp); #average mass flux,kg/(m**2*s)
n1=n*1000*3600;# average mass flux, g/m**2.h
#Result
print "Average rate of loss of naphthalene from a part of model is :",round(n,7),"kg/(m^2*s) or ",round(n1),"g/(m^2*h)\n"
print "Naphthalene sublimatin can be used to infer heat transfer coefficient by measuring the loss of naphthalene from a model over some length of time.since the schimidt no. of naphthalene is not generally equal to prandtl no. under the conditions of interest, some assumption about the dependence of nusselt no. on the prandtl no. must usually be introduced.\n"
#end
Example 11.11, Page number:¶
In [17]:
from __future__ import division
#Variables
T1=300; #temp. of helium-water tube,K
h=0.4; #height of vertical wall,m
m=0.087*10**-3; #flow rate of helium,kg/(m**2*s)
#this is a uniform flux natural convection problem.
Mhes=0.01; # assuming the value of mass fraction of helium at the wall to be 0.01
Mhef=Mhes/2; #film composition
af=1.141; #film density,kg/m**3
wd=1.107; #wall density,kg/m**3
Dha=7.119*10**-5; #diffusion coefficient,m**2/s
u=1.857*10**-5; #film viscosity at 300K,kg/(m*s)
aa=1.177; #air density,kg/m**3
g=9.8; #Gravity constant, m/s**2
#Calculations
Sc=(u/af)/Dha; #schimidt no.
Ra1=g*(aa-wd)*m*h**4/(u*af*Dha**2*Mhes); #Rayleigh no.
Nu=6/5*(Ra1*Sc/(4+9*math.sqrt(Sc)+10*Sc))**(1/5); #approximate nusselt no.
s=m*h/(af*Dha*Nu); #average concentration of helium at the wall
#thus we have obtained an average wall concentration 14 oercent higher than our initial guess of Mhes.we repeat this calclations with revised values of densities to obtain Mhes = 0.01142
#Result
print "Average conentration of helium at the wall is ",round(s,5),",since the result is within 0.5 percent of our second guess, a third iteration is not needed"
#end
Example 11.14, Page number:¶
In [18]:
from __future__ import division
import math
#Variables
T1=325; #temp. of helium-water tube,K
l=0.2; #length of tube,m
x=0.01; # mole fraction of water
R=8314.472; #gas constant,J/(kmol*K)
Mw=18.02;#Molecular mass of water, g/mol
#the vapor pressure of the liquid water is approximately the saturation pressure at the water temp.
#Calculations
p=1.341*10000 ; #vapor pressure using steam table,Pa
x1=p/101325; #mole fraction of saturated water
c=101325/(R*T1); #mole concentration in tube,kmol/m**3
D12=1.067*math.pow(10,-4); #diffusivity ofwater with respect to helium,m**2/s
Nw=c*D12*math.log(1+(x-x1)/(x1-1))/l ; #molar evaporation rate,kmol/(m**2*s)
nw=Nw*Mw; # mass evaporation rate,kg/(m**2*s)
#S=1+(x1-1)*math.exp(Nw*y/(c*D12)) conentration distribution of water-vapor
#Result
print "Conentration distribution of water-vapor is : x1(y)=1-0.8677*exp(0.6593*y) where y is expressed in meters.\n"
Example 11.15, Page number:¶
In [23]:
from __future__ import division
#Variables
T1=1473; #suraface temp.of hot water,K
x=0.05; #mass fraction of water
Gm=0.017; #average mass transfer coefficient,kg/(m**2*s)
A=0.04; #suraface area of pan,m**2
#only water vapour passes through the liquid surface, since air is not strongly absorbed into water under normal conditions.
#Calculations
p=38.58*1000; #saturation pressure of water,kPa
Xwater=p/101325; #mole fraction of saturated water
Mwater=Xwater*18.02/(Xwater*18.02+(1-Xwater)*28.96); #mass fraction of saturated water
B=(x-Mwater)/(Mwater-1); #mass transfer driving force
m=Gm*B*A; #evaporation rate,kg/s
#Result
print "Evaporation rate is:",round(m,6)," kg/s"
Example 11.16, Page number:¶
In [28]:
from __future__ import division
import math
#Variables
T1=298; #temp.of air,K
T2=323.15; #film temp.,K
x=0.05; #mass fraction of water at 75 C
Gm=0.017; #average mass transfer coefficient,kg/(m**2*s)
A=0.04; #suraface area of pan,m**2
l=0.2; #length of pan in flow direction,m
v=5; #air speed,m/s
m=(x+0.277)/2; #film composition of water at 50 C
Mf=26.34; #mixture molecular weight,kg/kmol
p=101325; #Atmospheric pressure, Pa
R=8314.5; #Gas constant, J/kmol-K
Uf=1.75*math.pow(10,-5); #film viscosity,kg/(m*s)
B=0.314; #mass transfer driving force
D=2.96*math.pow(10,-5); #diffusivity of water in air,m**2/s
df=0.993; #Density of ideal air, kg/m**#
#Calculations
af=p*Mf/(R*T2); #film density from ideal gas law,kg/m**3
Vf=Uf/af; #kinematic viscosity,m**2/s
Rel=v*l/Vf; #reynolds no. comes out to be 56,800 so the flow is laminar.
Sc=Vf/D; #scimidt no.
Nu=0.664*math.sqrt(Rel)*math.pow(Sc,1/3); #nussselt no.
Gmw1=Nu*(D*df/l); #appropriate value of mass transfer gas phase coeffficient of water in air,kg/(m**2*s)
Gmw=Gmw1*(math.log(1+B)/B); #mass transfer gas phase coeffficient of water in air,kg/(m**2*s)
#Results
print "Mass transfer gas phase coeffficient of water in air is :",round(Gmw,4),"kg/(m^2*s)\nIn this caes, the blowing factor is 0.870. Thus the mild blowing has reduced the mass transfer coefficient by about 13 percent"