#initiation of variable Ca=1.2 #Concentration at A in kg/m^3 Cb=0.8 #Concentration at B in kg/m^3 xa=5.0*10**-3#Position 1 in m xb=10.0*10**-3#Position 2 in m D=3.0*10**-11 #Diffusion coefficient in m^2/s #calcualtion J=-D*(Ca-Cb)/(xa-xb) #result print" Diffusion flux is %.1e kg/m^2-s" %J
Diffusion flux is 2.4e-09 kg/m^2-s
#initiation of variable from math import sqrt x=5*10**-4 #Position in m D=1.6*10**-11 #Diffusion coefficient in m^2/s Co=0.25#Initial Concentration in wt% Cs=1.2 #Surface concentration in wt% Cx=0.8 #Concentration at any x in wt% z1 = 0.35 # tabular data z2 = 0.4 # tabular data erf_z1 = 0.3794 # tabular data erf_z2 = 0.4284 # tabular data #calcualtion C=1-((Cx-Co)/(Cs-Co)) z = (C-erf_z1)/(erf_z2-erf_z1) * (z2-z1) + z1 # Calculation by interpolation t= ((x/(2*sqrt(D)))/z)**2 # calculation of time #result print" Time required is %d" %t,"s or %.2f h" %(t/3600); print "Answer in book is 25400 sec or 7.1 h. It is because of considering different number of significant figure"
Time required is 25355 s or 7.04 h Answer in book is 25400 sec or 7.1 h. It is because of considering different number of significant figure
#initiation of variable D_500=4.8*10**-14 #Diffusion coefficient at 500 C D_600=5.3*10**-13 #Diffusion coefficient at 600 C t_600=10.0 #Time in hours to diffuse #calculation t_500=D_600*t_600/D_500 #result print" Time to diffuse at 500 degree Celsius is %.1f h" %(t_500)
Time to diffuse at 500 degree Celsius is 110.4 h
#initiation of variable from math import exp T=550+273.0 # temperature of aluminium in K D_0=1.2*10**-4 #Temperature independent preexponential in m^2/s Q_d=131000.0 #Activation energy in J/mol-K R=8.31 #Universal Gas constant #calculation D=D_0*exp(-Q_d/(R*T)); #result print" Diffusion coefficient is %.1e m^2/s" %D;
Diffusion coefficient is 5.8e-13 m^2/s
#initiation of variable #From graph log D ad 1/T are deducted inv_T1=0.8*10**-3 #Reciprocal of temp. in K^-1 inv_T2=1.1*10**-3 #Reciprocal of temp. in K^-1 logD1=-12.4 logD2=-15.45 R=8.31 #Gas law Constant in J/mol-K #calculation Q_d=-2.3*R*(logD1-logD2)/(inv_T1-inv_T2) D_0=10**(logD2+(Q_d*inv_T2/(2.3*R)))#For calculating Preexponential factor #result print" Activation energy is %d kJ/mol" %(Q_d/1000) print" Preexponential factor D_0 is %.1e m^2/s" %D_0 print "Answer in book is 5.2e-5 m^2/s. It is because of consideration of different number of significant figures"
Activation energy is 194 kJ/mol Preexponential factor D_0 is 5.4e-05 m^2/s Answer in book is 5.2e-5 m^2/s. It is because of consideration of different number of significant figures
#initiation of variable P_m = 2.3e-14 # permissibility coefficient of CO2 through PET P_1 = 400.0 # Pressure inside bottle in KPa P_2 = 0.4 # Pressure outside bottle in KPa A = 500.0 # Surface area of bottle in cm^2 x = 0.05 # wall thickness of bottle in cm V = 750.0 # volume in cm^3 #part A #calculation J = -P_m*(P_2-P_1)*1e3/x # calculation of diffusion flux print" Diffusion flux is %0.1e cm^3 STP/(cm^2-s)" %J #partB V_co2 = J*A t = V/V_co2 # calculation of self life print" Self life for bottle of pop is %d" %(t/(60.0*60*24)), "days (or about %d months)." %(t/(60*60*24*30)) print "Answer in book is 97 days. It is because of considering different number of significant figure"
Diffusion flux is 1.8e-07 cm^3 STP/(cm^2-s) Self life for bottle of pop is 94 days (or about 3 months). Answer in book is 97 days. It is because of considering different number of significant figure