Chapter 9 : Theories of Mass Transfer

Example 9.1.1 pgno:277

In [1]:
#initialization of variables
p1 = 10. # pressure in atm
H = 600. # henrys constant in atm
c1 = 0 # gmol/cc
N1 = 2.3*10**-6 # mass flux in mol/cm**2-sec
c = 1./18. #total Concentration in g-mol/cc
D = 1.9*10**-5 # Diffusion co efficient in cm**2/sec
#Calculations
c1i = (p1/H)*c # Component concentration in gmol/cc
k = N1/(c1i-c1)#Mass transfer co efficient in cm/sec
l = D/k # Film thickness in cm
#Results
print"The film thickness is  cm",round(l,5)

The film thickness is  cm 0.00765

Example 9.2.1 pgno:281

In [2]:
#initialization of variables
D = 1.9*10**-5 #Diffusion co efficient in cm**2/sec
k = 2.5*10**-3 # M.T.C in cm/sec
from math import pi
#Calculations
Lbyvmax = 4*D/((k**2)*pi)#sec
tou = D/k**2 # sec
#Results
print"The contact time  sec",round(Lbyvmax,1)
print"\nThe surface resident time  sec",round(tou,1)

The contact time  sec 3.9

The surface resident time  sec 3.0

Example 9.3.1 pgno:35

In [3]:
#initialization of variables
const = 0.5 # The part of flow in the system which bypasses the region where the mass transfer occurs
v1 = 1. # cm/sec
al = 10**3
k = 10**-3 # cm/sec
v2 = 3. # cm/sec
from math import log
from math import exp
#Calculations
C1byC10first = const + (1-const)*(exp(-k*al/v1))# c1/c10
appk1 = (v1/al)*(log(1/C1byC10first))# Apparent m.t.c for first case in cm/sec
C1byC10second = const + (1-const)*(exp(-((3)**0.5)*k*al/v2))#c1/c10 in second case
appk2 = (v2/al)*log(1/C1byC10second)# apparent m.t.c for second case in cm/sec
power = log(appk2/appk1)/log(v2/v1)
#Results
print"The apparent m.t.c for the first case is  cm/sec",appk1
print"\nThe apparent m.t.c for the second case is  cm/sec",appk2
print"\nThe apparent is proportional to the power of  of the velocity",round(power,2)

The apparent m.t.c for the first case is  cm/sec 0.000379885493042

The apparent m.t.c for the second case is  cm/sec 0.000742723884992

The apparent is proportional to the power of  of the velocity 0.61

Example 9.4.1 pgno:283

In [2]:
#initialization of variables
D = 1*10**-5 #cm**2/sec
d = 2.3 # cm
L = 14 # cm
v0 = 6.1 # cm/sec
#gamma(4./3.)=0.8909512761;
#calculations
k = ((3**(1./3.))/(0.8909512761))*((D/d))*(((d**2)*v0/(D*L))**(1./3.))# cm/sec
#Results
print"The average mass transfer coefficient is  cm/sec",round(k,5)

The average mass transfer coefficient is  cm/sec 0.00043

Example 9.4.2 pgno:287

In [3]:
#initialization of variables
tn = 300000 # turbulence number
v0  = 10 # cm/sec
p = 1 # g/cc
mu = 0.01 # g/cm-sec
delta = 2.5 #cm
D = 1*10**-5 # cm**2/sec
#Calculations
x = tn*mu/(v0*p)# cm
delta = ((280/13)**(1/2))*x*((mu/(x*v0*p))**(1/2))#cm
deltac = ((D*p/mu)**(1/3))*delta#cm
k = (0.323*(D/x)*((x*v0*p/mu)**0.5)*((mu/(p*D))**(1/3)))*10**5# x*10**-5 cm/sec
#Results
print"The distance at which turbulent flow starts is  cm",x
print"\nThe boundary layer for flow at this point is  cm",delta
print"\nThe boundary layer for concentration at this point is  cm",deltac
print"\nThe local m.t.c at the leading edge and at the position of transistion is  x10**-5 cm/sec",round(k,3)

The distance at which turbulent flow starts is  cm 300.0

The boundary layer for flow at this point is  cm 300.0

The boundary layer for concentration at this point is  cm 300.0

The local m.t.c at the leading edge and at the position of transistion is  x10**-5 cm/sec 0.59