# Illustration 2.1

# solution
import math

#***Data***#
# a = O2 & b = CO
Dab = 1.87*10**(-5);#[square m/s]
Pt = 10**5;#[N/square m]
z = 0.002;#[m]
R = 8314;#[Nm/kmol]
T = 273;#[K]
Pa1 = 13*10.0**(3);#[N/square m]
Pb1 = 10**(5)-13*10**(3);#[N/square m]
Pa2 = 6500;#[N/square m]
Pb2 = 10**(5)-6500.0;#[N/square m]
#********#

# Calculation from Eqn. 2.30
Pbm = (Pb1-Pb2)/math.log(Pb1/Pb2);#[N/square m]
Na = Dab*Pt*(Pa1-Pa2)/(R*T*z*Pbm);#[kmol/square m.s]
print" Rate of diffusion of oxygen is",round(Na,7),"kmol/square m.sec "


# Illustration2.2

# solution
import math

#***Data***#
Pt = 10**5.0;#[N/square m]
z = 0.002;#[m]
R = 8314.0;#[Nm/kmol]
T = 273;#[K]
#a = O2 b = CH4 c = H2
Pa1 = 13*10**(3);#[N/square m]
Pb1 = 10**(5)-13*10**(3);#[N/square m]
Pa2 = 6500.0;#[N/square m]
Pb2 = 10.0**(5)-6500;#[N/square m]
Dac = 6.99*10**(-5);#[N/square m]
Dab = 1.86*10.0**(-5);#[N/square m]
#*******#

# Calculation from Eqn. 2.30
Pbm = (Pb1-Pb2)/math.log(Pb1/Pb2);#[N/square m]
Yb_prime = 2.0/(2+1);
Yc_prime = 1-Yb_prime;
Dam = 1.0/((Yb_prime/Dab)+(Yc_prime/Dac));#[square m.s]
Na = Dam*(Pa1-Pa2)*Pt/(R*T*z*Pbm);#[kmol/square m.s]
print "Rate of diffusion is",round(Na,7),"kmol/square m.sec"


# Illustration2.3

import math

# solution

#***Data***#
# a = C2H5OH b = air
Pt = 101.3*10**(3);#[N/square m]
T = 273.0 ;#[K]
#********#

Ma = 46.07;# [kg/kmol]
Mb = 29.0;# [kg/kmol]
#For air from Table 2.2 (Pg 33)
Eb_by_k = 78.6;# [K]
rb = 0.3711; # [nm]
# For C2H5OH using Eqn. 2.38 & 2.39
# From Table 2.3
Va = (2*0.0148)+(6*0.0037)+(0.0074);# [cubic m/kmol]
Tba = 351.4;# [K]
ra = 1.18*(Va**(1/3.0));#[nm]
Ea_by_k = 1.21*Tba;# [K]
rab = (ra+rb)/2.0;# [nm]
Eab_by_k = math.sqrt(Ea_by_k*Eb_by_k);# [K]
Collision_value = T/Eab_by_k;
#From Fig. 2.5 (Page: 32) f(collision value)
Collision_func = 0.595;
Dab = (10**(-4)*(1.084-(0.249*math.sqrt((1/Ma)+(1/Mb))))*T**(3.0/2)*math.sqrt((1/Ma)+(1/Mb)))/(Pt*(rab**2)*Collision_func);#[square m/s]
print" The diffusivity of ethanol through air at 1 atm. & 0 degree C is",round(Dab,7),"m^2/s"
print" The observed value from (Table 2.1) is 1.02*10^(-5) square m/s'"


# Illustration 2.4
import math

# solution

#***Data****#
# a = acetic acid b = H2O
z = 0.001;# [m]
Dab = 0.95*10**(-9);#[square m/s]
#************#

Ma = 60.03;# [kg/kmol]
Mb = 18.02;# [kg/kmol]
#At 17 C & 9% solution
density1 = 1012; #[kg/cubic m]
Xa1 = (0.09/Ma)/((0.09/Ma)+(0.91/Mb));
Xb1 = 1-Xa1;
M1 = 1/((0.09/Ma)+(0.91/Mb));# [kg/kmol]
#At 17 C & 3% solution
density2 = 1003.2; #[kg/cubic m]
Xa2 = (0.03/Ma)/((0.03/Ma)+(0.97/Mb));
Xb2 = 1-Xa2;
M2 = 1/((0.03/Ma)+(0.97/Mb));# [kg/kmol]
avg_density_by_M = ((density1/M1)+(density2/M2))/2;#[kmol/cubic m]
# From Eqn. 2.42
Xbm = (Xb2-Xb1)/math.log(Xb2/Xb1);
# From Eqn. 2.41
Na = Dab*(avg_density_by_M)*(Xa1-Xa2)/(Xbm*z); #[square m/s]
print" The rate of diffusion is",round(Na,9),"square m/s"


# Illustration 2.5


# solution

#***Data****#
# a = mannitol b = H2O
T = 293; # [K]
#*****#

Mb = 18.02;# [kg/kmol]
# From Table 2.3 (Pg 33)
Va = (0.0148*6)+(0.0037*14)+(0.0074*6); # [cubic m/kmol]
viscosity = 0.001005; # [kg/m.s]
association_factor = 2.26; # [water as a solvent]
Dab = (117.3*10**(-18))*((association_factor*Mb)**0.5)*T/(viscosity*Va**0.6); # [square m/s]
print" Diffusivity of mannitol is",round(Dab,12),"square m/s"
print" Observed value is 0.56*10^(-9) square m/s"


# Illustration 2.6


# solution

#****Data****#
T2 = 70+273;# [K]
#**********#

# a = mannitol b = H2O
# From Illustration 2.5 at 20 C
viscosity1 = 1.005*10**(-3); # [kg/m.s]
Dab1 = 0.56*10**(-9); #[m^2/s]
T1 = 273+20;# [K]
# At 70 C
viscosity2 = 0.4061*10**(-3); # kg/m.s
# Eqn. 2.44 indicates Dab*viscocity/T  =  constnt
Dab2 = Dab1*(T2)*(viscosity1)/(T1*viscosity2);# [square m/s]
print" Diffusivity of mannitol at 70 degree C is",round(Dab2,11),"square/s "
print" Observed value at 70 degree C is 1.56*10^(-9) square m/s"