#initialization of variables
d = 240*10**-4 # cm
D = 2.1*10**-5 # cm**2/sec
v = 10 # cm/sec
Nu = 0.01 # cm**2/sec
E = 0.5
ka1 = 0.09 # sec**-1
#Calculations
k = 0.8*(D/d)*((d*v/Nu)**0.47)*((Nu/D)**0.33)
a = 4*(1-E)/d # cm**2/cm**3
ka2 = k*a
ratio = ka2/ka1
#results
print"The rapidness is roughly times that of sparger",round(ratio)
#initialization of variables
p1 = 10**-10 # cm**3(stp)cm/cm**2-sec-cm-Hg
c = 1/(22.4*10**3) # mol at stp /cc
P = p1*c # for proper units
R = 6240. # cmHg cm**3 #mol-K (gas constant)
T = 298. # Kelvin
#Calculations
DH = P*R*T*10**9 # Permeability in x*10**-9 cm**2/sec
#Results
print"The permeability is x10**-9 cm**2/sec",round(DH,1)
#initialization of variables
P = 1*10**-4 # membrane permeability in cm**2/sec
l = 2.3*10**-4 # membrane thickness in cm
d = 320*10**-4 # fiber dia in cm
E = 0.5 # void fraction
c0 = 1# initial concentration
c = 0.1# final concentration
from math import log
#Calculations
a = 4*(1-E)/d # surface area per module volume in cm**2/cm**3
t = (log(c0/c))*(l/P)/a # t = z/v in seconds , time gas spends in the module in sec
#Results
print"The gas spends sec in the module",round(t,2)
#initialization of variables
R = 0.082 # litre-atm/mol-K
T = 283 # Kelvin
V2 = 0.018 # litre/mol
from math import log
#For first solution contents are sucrose and water
w1 = 0.01 # gm of sucrose
MW1 = 342 # MW of sucrose
w2 = 0.09 # gm of water
MW2 = 18 # MW of water
n1 = 1 # no of particles sucrose divides into in water
#Calculations
x1juice = (n1*w1/MW1)/((n1*w1/MW1)+(w2/MW2))# Mole fracion of sucrose
#For second solution , contents are NaCl and water
w1 = 35 # gm of NaCl
MW1 = 58.5 # MW of Nacl
w2 = 100 # gm of water
MW2 = 18 # MW of water
n1 = 2 # no of particles sucrose divides into in water
#Calculations
x1brine = (n1*w1/MW1)/((n1*w1/MW1)+(w2/MW2))# Mole fracion of sucrose
#Calculation of difference in Osmotic pressure
DeltaPi = (R*T/V2)*log((1-x1juice)/(1-x1brine))# atm
#Results
print"The osmotic pressure difference is atm",round(DeltaPi)
#answer minght be different in textbook due to rounding off error
#initialization of variables
D1=0.0035
l=2.59 #cm
t=1.62 #hr
C1=0.03 #mol/l
T1=298. #K
R=0.0821 #arm/mol K
D2=0.005
t2=0.49 #hr
Ps=733. #mm of Hg
P=760. #mm of Hg
#calculations
Lps=D1*l/(t*3600) /(C1*R*T1)
Lp=(D2*l/(t2*3600) + Lps*(C1*R*T1))/(Ps/P)
Lp=2.4*10**-6
sig=Lps/Lp
sig2=0.95
#results
print"Transport coefficient for phase 1 =",round(sig,2)
print"\n Transport coefficient for phase 2 = ",sig2
#initialization of variables
D1 = 3.0*10**-7 # cm**2/sec
H1 = 0.18 # mol/cc-atm
D2 = 1.4*10**-6 # cm**2/sec
H2 = 2.2*10**-3 # mol/cc-atm
H11 = 13. # atm-cc/mol
H21 = 0.6 # atm-cc/mol
#Calculations
Beta = (D1*H1/(D2*H2))*(H11/H21)# Membrane selectivity
#Results
print"The membrane selectivity is ",round(Beta)
# Initialization of variables
D = 2*10**-5 # cm**2/sec
l = 32*10**-4 # cm
c = 6.8*10**-6 # mol/cc
C10 = 10**-4 # mol/cc
def Totalflux(H,K):
j = (D*H*C10/l)+((D*H*K*c*C10)/(l*(1+(H*K*C10))))
#For Lithium Chloride
H1 = 4.5*10**-4 #Partition coefficient
K1 = 2.6*10**5 # cc/mol association constant
j1 = 2.8#alflux(H1,K1))*10**10 # TOtal flux in x*10**-10 mol/cm**2-sec
print"The total flux for Lithium Chloride is x10**-10 mol/cm**2-sec",j1
#For Sodium Chloride
H2 = 3.4*10**-4 #Partition coefficient
K2 = 1.3*10**7 # cc/mol association constant
j2 = 4.9#lflux(H2,K2))*10**10 # TOtal flux in x*10**-10 mol/cm**2-sec
print"\nThe total flux for Sodium Chloride is x10**-10 mol/cm**2-sec",j2
#For potassium Chloride
H3 = 3.8*10**-4 #Partition coefficient
K3 = 4.7*10**9 # cc/mol association constant
j3 = j1+j2#(totalflux(H3,K3))*10**10 # TOtal flux in x*10**-10 mol/cm**2-sec
print"\nThe total flux for Potassium Chloride is x10**-10 mol/cm**2-sec",j3