from __future__ import division
# given data
P1=4 # in bar
P2=2 # in bar
T=25 # in degree C
Dhp=9*10**-8 # in m**2/s
S=3*10**-3 # in kg mole/m**3 bar
del_x=0.5*10**-3 # thickness in m
#(a) The molar concentration of a gas in terms of solubility
CH1=S*P1 # in kg mole/m**3
CH2=S*P2 # in kg mole/m**3
#(b) Molar diffusion flux of hydrogen through plastic memberence is given by Fick's law of diffision
#N_H= N_h/A = Dhp*(CH1-CH2)/del_x#
N_H= Dhp*(CH1-CH2)/del_x # in kg mole/s-m**2
print "Molar diffusion flux of hydrogen through the membrane = %0.3e kg mole/s-m**2" %N_H
#Mass_d_Flux= N_H*Molecular_Weight
Molecular_Weight=2#
Mass_d_Flux= N_H*Molecular_Weight
print "Molar diffusion flux = %0.3e kg/s-m**2" %Mass_d_Flux
# given data
T=25 # in degree C
T=T+273 # in K
P=1#
V1=12 #Molecular volume of H2 in cm**3/gm mole
V2=30 #Molecular volume of Air in cm**3/gm mole
M1=2 # Molecular weight of H2
M2=29 # Molecular weight of Air
#The diffusion coefficient for gases in terms of molecular volumes may be express as
D_AB= .0043*T**(3/2)/(P*(V1**(1/3)+V2**(1/3)))*(1/M1+1/M2)**(1/2)#
print "The diffusion coefficient for gases in terms of molecular volumes = %0.3f cm**2/sec" %D_AB
# given data
T=300 # temp of gas mixture in K
D_HN2=18*10**-6 # in m**2/s at 300 K, 1 bar
T1=300 # in K
D_HO2=16*10**-6 # in m**2/s at 273 K, 1 bar
T2=273 # in K
O_2=0.2#
N_2=0.7#
H_2=0.1#
#The diffusivity at the mixture temperature and pressure are calculated as
# D1/D2 = (T1/T2)**(3/2)*(P2/P1)
D_HO2= (T/T2)**(3/2)*1/4*D_HO2#
D_HN2= (T/T1)**(3/2)*1/4*D_HN2#
#The composition of oxygen and nitrogen on a H2 free basis is
x_O= O_2/(1-H_2)#
x_N= N_2/(1-H_2)#
# The effective diffusivity for the gas mixture at given temperature and pressure is
D= 1/(x_O/D_HO2+x_N/D_HN2) # in m**2/s
print "Effective diffusivity = %0.3e m**2/s" %D
from numpy import pi
# given data
d=3 # in mm
d=d*10**-3 # in meter
T=25 # in °C
T=T+273 # in K
D= 0.4*10**-4 # in m**2/s
R= 8314#
P_A1=1 # in atm
P_A1=P_A1*10**5 # in w/m**2
P_A2=0#
C_A2=0#
x2= 15 # in meter
x1= 0#
A= pi/4*d**2#
M_A= D*A/(R*T)*(P_A1-P_A2)/(x2-x1) # in kg mole/sec
N_B= M_A#
M_B= M_A*29 # in kg/sec
print "Value of N_B = %0.3e kg mole/sec" %N_B
print "Value of M_B = %0.3e kg/sec" %M_B
# Note : The value of M_B in the book is wrong
from math import log
# given data
P=3 # in atm
P=P*10**5 # in N/m**2
r1=10 # in mm
r1=r1*10**-3 # in m
r2=20 # in mm
r2=r2*10**-3 # in m
R=4160 # in J/kg-K
T=303 # in K
D=3*10**-8 # in m**2/s
S=3*0.05# # Solubility of hydrogen at a pressure of 3 atm in m**3/m**3 of rubber tubing
del_x=r2-r1 # in m
L=1 # in m
Am=2*pi*L*del_x/log(r2/r1)#
#Formula P*V= m*R*T
V=S#
m=P*V/(R*T) # in kg/m**3 of rubber tubing at the inner surface of the pipe
C_A1=m#
C_A2=0#
#Diffusion flux through the cylinder is given
M=D*(C_A1-C_A2)*Am/del_x#
print "Diffusion flux through the cylinder = %0.1e kg/sm" %M
from numpy import pi
# given data
R=4160 # in J/kg-K
M=2#
D_AB=1.944*10**-8 # in m**2/s
R_H2=R/M#
S=2*0.0532# # Solubility of hydrogen at a pressure of 2 atm in cm**3/cm**3 of pipe
P=2 # in atm
P=P*1.03*10**5 # N/m**2
T=25 # in degree C
T=T+273 # in K
r1=2.5 # in mm
r1=r1*10**-3 # in m
r2=5 # in mm
r2=r2*10**-3 # in m
del_x=r2-r1 # in m
L=1 # in m
#Formula P*V= m*R*T
V=S#
m=P*V/(R*T) # in kg/m**3 of pipe
# So, Concentration of H2 at inner surface of the pipe
C_A1=0.0176 # in kg/m**3
# The resistance of diffusion of H2 away from the outer surface is negligible i.e.
C_A2=0#
Am=2*pi*L*del_x/log(r2/r1)#
# Loss of H2 by diffusion
M_A= D_AB*(C_A1-C_A2)*Am/del_x#
print "Loss of H2 by diffusion = %0.3e kg/s" %M_A
#Note: In the book , they put wrong value of C_A1 to calculate M_A, so the answer in the book is wrong
from numpy import pi
from math import log
# given data
Px1= 0.14 # in bar
Px2= 0#
P=1.013 # in bar
Py1=P-Px1# # in bar
Py2=P-Px2# # in bar
D=8.5*10**-6 # in m**2/s
d=5 # diameter in meter
L=1 # in mm
L=L*10**-3 #in meter
M=78 # molecular weight
Am_x= 1/4*pi*d**2*M#
R=8314#
del_x=3 # thickness in mm
del_x=del_x*10**-3 # in m
T=20 # in degree C
T=T+273 # in K
P=P*10**5 # in N/m**2
m_x= D*Am_x*P*log(Py2/Py1)/(R*T*del_x)#
# The mass of the benzene to be evaporated
mass= 1/4*pi*d**2*L#
density=880 # in kg/m**3
m_b= mass*density#
toh=m_b/m_x # in sec
print "Time taken for the entire organic compound to evaporate = %0.3f seconds" %toh
# Note: Answer in the book is wrong
# given data
A=0.5 # in m**2
Pi=2.2 # in bar
Pi=Pi*10**5 # in N/m**2
Pf=2.18 # in bar
Pf=Pf*10**5 # in N/m**2
T=300 # in K
S=0.072 # in m**3
V=0.028 # in m**3
L=10 # in mm
L=L*10**-3 # in meter
R=287#
# Diffusivity of air in rubber D
# Initial mass of air in the tube
mi= Pi*V/(R*T) # in kg
#final mass of air in the tube
mf= Pf*V/(R*T) # in kg
# Mass of air escaped
ma = mi-mf #in kg
# Formula Na = ma/A = mass of air escaped / Time elapsed * area
A=6*24*3600*0.5#
Na = ma/A #in kg/sm**2
# Solubility of air should be calculated at mean temperature
S_meanTemperature=(2.2+2.18)/2 # in bar
#Solubility of air at the mean inside Pressure is
S=S*S_meanTemperature # in m**3/m**3 of rubber
V1=S#
V2=0.072#
T1=T#
T2=T#
P1=2.19*10**5 # in N/m**2
P2=1*10**5 # in N/m**2
# The corresponding mass concentration at the inner and outer surface of the tube, from gas equation are calculated as
Ca1= P1*V1/(R*T1) # in kg/m**3
Ca2= P2*V2/(R*T2) # in kg/m**3
# The diffusion flux rate of air through the rubber is given by
# Na = ma/A = D*(Ca1-Ca2)/del_x, here
del_x=L#
D=Na*del_x/(Ca1-Ca2)#
print "Diffusivity of air in rubber = %0.3e m**2/s" %D