Chapter 17: Transport Phenomena¶
Example Problem 17.1, Page Number 417¶
In [17]:
from scipy import constants
from math import sqrt,pi
#Variable Declaration
M = 0.040 #Molecualar wt of Argon, kh/mol
P, T = 101325.0, 298.0 #Pressure and Temperature, Pa, K
sigm = 3.6e-19 #
R = 8.314 #Molar Gas constant, mol^-1 K^-1
N_A = 6.02214129e+23 #mol^-1
#Calculations
DAr = (1./3)*sqrt(8*R*T/(pi*M))*(R*T/(P*N_A*sqrt(2)*sigm))
#Results
print 'Diffusion coefficient of Argon %3.1e m2/s'%DAr
Example Problem 17.2, Page Number 418¶
In [20]:
from math import sqrt
#Variable Declaration
DHebyAr = 4.0
MAr, MHe = 39.9, 4.0 #Molecualar wt of Argon and Neon, kg/mol
P, T = 101325.0, 298.0 #Pressure and Temperature, Pa, K
sigm = 3.6e-19 #
R = 8.314 #Molar Gas constant, mol^-1 K^-1
N_A = 6.02214129e+23 #mol^-1
#Calculations
sigHebyAr = (1./DHebyAr)*sqrt(MAr/MHe)
#Results
print 'Ratio of collision cross sections of Helium to Argon %4.3f'%sigHebyAr
Example Problem 17.3, Page Number 420¶
In [23]:
from math import sqrt
#Variable Declaration
D = 1.0e-5 #Diffusion coefficient, m2/s
t1 = 1000 #Time, s
t10 = 10000 #Time, s
#Calculations
xrms1 = sqrt(2*D*t1)
xrms10 = sqrt(2*D*t10)
#Results
print 'rms displacement at %4d and %4d is %4.3f and %4.3f m respectively'%(t1,t10,xrms1,xrms10)
Example Problem 17.4, Page Number 421¶
In [29]:
#Variable Declaration
D = 2.2e-5 #Diffusion coefficient of benzene, cm2/s
x0 = 0.3 #molecular diameter of benzene, nm
#Calculations
t = (x0*1e-9)**2/(2*D*1e-4)
#Results
print 'Time per random walk is %4.3e s or %4.2f ps'%(t,t/1e-12)
Example Problem 17.5, Page Number 424¶
In [34]:
from math import sqrt,pi
#Variable Declaration
P = 101325 #Pressure, Pa
kt = 0.0177 #Thermal conductivity, J/(K.m.s)
T = 300.0 #Temperature, K
k = 1.3806488e-23 #Boltzmanconstant,J K^-1
sigm = 3.6e-19 #
R = 8.314 #Molar Gas constant, mol^-1 K^-1
NA = 6.02214129e+23 #mol^-1
M = 39.9 #Molecualar wt of Argon and Neon, kg/mol
#Calculations
CvmbyNA = 3.*k/2
nuavg = sqrt(8*R*T/(pi*M*1e-3))
N = NA*P/(R*T)
labda = 3*kt/(CvmbyNA*nuavg*N)
sigm = 1/(sqrt(2)*N*labda)
#Results
print 'Mean free path %4.3e m and collisional cross section %4.2e m2'%(labda, sigm)
Example Problem 17.6, Page Number 427¶
In [48]:
from math import sqrt,pi
#Variable Declaration
eta = 227. #Viscosity of Ar, muP
P = 101325 #Pressure, Pa
kt = 0.0177 #Thermal conductivity, J/(K.m.s)
T = 300.0 #Temperature, K
k = 1.3806488e-23 #Boltzmanconstant,J K^-1
R = 8.314 #Molar Gas constant, mol^-1 K^-1
NA = 6.02214129e+23 #mol^-1
M = 39.9 #Molecualar wt of Argon and Neon, kg/mol
#Calculations
nuavg = sqrt(8*R*T/(pi*M*1e-3))
N = NA*P/(R*T)
m = M*1e-3/NA
labda = 3.*eta*1e-7/(nuavg*N*m) #viscosity in kg m s units
sigm = 1./(sqrt(2)*N*labda)
#Results
print 'Collisional cross section %4.2e m2'%(sigm)
Example Problem 17.7, Page Number 429¶
In [80]:
from math import sqrt,pi
#Variable Declaration
m = 22.7 #Mass of CO2, kg
T = 293.0 #Temperature, K
L = 1.0 #length of the tube, m
d = 0.75 #Diameter of the tube, mm
eta = 146 #Viscosity of CO2, muP
p1 = 1.05 #Inlet pressure, atm
p2 = 1.00 #Outlet pressure, atm
atm2pa = 101325 #Conversion for pressure from atm to Pa
M = 0.044 #Molecular wt of CO2, kg/mol
R = 8.314 #Molar Gas constant, J mol^-1 K^-1
#Calculations
p1 = p1*atm2pa
p2 = p2*atm2pa
F = pi*(d*1e-3/2)**4*(p1**2-p2**2)/(16.*eta/1.e7*L*p2)
nCO2 = m/M
v = nCO2*R*T/((p1+p2)/2)
t = v/F
#Results
print 'Flow rate is %4.3e m3/s'%(F)
print 'Cylinder can be used for %4.3e s nearly %3.1f days'%(t, t/(24*3600))
Example Problem 17.8, Page Number 431¶
In [54]:
from math import sqrt,pi
#Variable Declaration
eta = 0.891 #Viscosity of hemoglobin in water, cP
T = 298.0 #Temperature, K
k = 1.3806488e-23 #Boltzmanconstant,J K^-1
R = 8.314 #Molar Gas constant, mol^-1 K^-1
D = 6.9e-11 #Diffusion coefficient, m2/s
#Calculations
r = k*T/(6*pi*eta*1e-3*D)
#Results
print 'Radius of protein is %4.3f nm'%(r/1e-9)
Example Problem 17.9, Page Number 432¶
In [56]:
from math import sqrt,pi
#Variable Declaration
s = 1.91e-13 #Sedimentation constant, s
NA = 6.02214129e+23 #mol^-1
M = 14100.0 #Molecualr wt of lysozyme, g/mol
rho = 0.998 #Density of water, kg/m3
eta = 1.002 #Viscosity lysozyme in water, cP
T = 293.15 #Temperature, K
vbar = 0.703 #Specific volume of cm3/g
#Calculations
m = M/NA
f = m*(1.-vbar*rho)/s
r = f/(6*pi*eta)
#Results
print 'Radius of Lysozyme particle is %4.3f nm'%(r/1e-9)
Example Problem 17.10, Page Number 433¶
In [4]:
from numpy import arange,array,ones,linalg,log, exp
from matplotlib.pylab import plot,show
%matplotlib inline
#Variable Declaration
t = array([0.0,30.0,60.0,90.0,120.0,150.0]) #Time, min
xb = array([6.00,6.07,6.14,6.21,6.28,6.35]) #Location of boundary layer, cm
rpm = 55000. #RPM of centrifuge
#Calculations
nx = xb/xb[0]
lnx = log(nx)
A = array([ t, ones(size(t))])
# linearly generated sequence
[slope, intercept] = linalg.lstsq(A.T,lnx)[0] # obtaining the parameters
# Use w[0] and w[1] for your calculations and give good structure to this ipython notebook
# plotting the line
line = slope*t+intercept # regression line
#Results
plot(t,line,'-',t,lnx,'o')
xlabel('$ Time, min $')
ylabel('$ \log(x_b/x_{b0}) $')
show()
sbar = (slope/60)/(rpm*2*pi/60)**2
print 'Slope is %6.2e 1/min or %4.3e 1/s '%(slope, slope/60)
print 'Sedimentation factor is %4.3e s'%(sbar)
Example Problem 17.11, Page Number 439¶
In [59]:
#Variable Declaration
LMg = 0.0106 #Ionic conductance for Mg, S.m2/mol
LCl = 0.0076 #Ionic conductance for Cl, S.m2/mol
nMg, nCl = 1, 2 #Coefficients of Mg and Cl
#Calculations
LMgCl2 = nMg*LMg + nCl*LCl
#Results
print 'Molar conductivity of MgCl2 on infinite dilution is %5.4f S.m2/mol'%(LMgCl2)