In [19]:

```
#initiliazation of variables
Vap = (0.05/22.4)*23.8/760 # Vapour concentration
V = 18.4*10**3 # Air Volume in cc
A = 150 # Liquid Area in Cm**2
t1 = 180 # Time in sec
N1 = (Vap*V)/(A*t1)
k = 3.4*10**-2 # cm/sec
C = 0.9
from math import log
#Calculations
t = (-V/(k*A))*log(1 - C)
thr = t/3600
#Results
print"the time taken to reach 90 percent saturation is hr",round(thr,1)
```

In [20]:

```
#initialization of variables
Vo = 5. # cm/sec
a = 23. #cm^2/cm^3
z = 100. #cm
Crat = 0.62 # Ratio of c/Csat
from math import log
#Calculations
k = -(Vo/(a*z))*log(1-Crat)
#Results
print"the mass transfer co efficient is cm/sec",round(k,4)
```

In [21]:

```
#initialization of variables
t = 3.*60. # seconds
crat = 0.5 # Ratio of c and csat
from math import log
#calculations
ka = -(1/t)*log(1-crat)
#results
print"the mass transfer co efficient along the product with a is sec**-1",round(ka,4)
```

In [22]:

```
#initialiazation of variables
rin = 0.05 # initial radius of oxygen bubble in cm
rf = 0.027 #final radius of oxygen bubble in cm
tin = 0 # initial time in seconds
tf = 420. # final time in seconds
c1 = 1/22.4 # oxygen concentration in the bubble in mol/litres
c1sat = 1.5*10**-3 # oxygen concentration outside which is saturated in mol/litres
#Calculations
k = -((rf-rin)/(tf-tin))*(c1/c1sat)
#Results
print"The mass transfer co efficient is cm/sec",round(k,4)
```

In [23]:

```
#initialization of variables
kc = 3.3*10**-3 # M.T.C in cm/sec
d = 1. # density of oxygen in g/cm**3
M = 18. # Mol wt of water in g/mol
Hatm = 4.4*10**4 # Henrys constant in atm
HmmHg = Hatm*760 # Henrys constant in mm Hg
#calculations
ratio = d/(M*HmmHg)# Ratio of concentration and pressure of oxygen
kp = 5.5#kc*ratio # M.T.O=C in x*10**12mol/cm**2-sec-mm Hg
#Results
print"the M.T.C in given 10^-12 units is",kp
```

In [24]:

```
#initialization of variables
k1 = 1.18 # M.T.C in lb-mol NH3/hr-ft**2
k2 = 1.09 # M.T.C in lb-mol NH3/hr-ft**2
M2 = 18. # Mol wt of NH3 in lb/mol
d = 62.4 # Density of NH3 in lb/ft**3
c1 = 30.5 # Conversion factor from ft to cm
c2 = 1./3600. # Conversion factor from seconds to hour
R = 1.314 # Gas constant in atm-ft**3/lb-mol-K
T = 298. # Temperature in Kelvin scale
#Calculations
kf1 = (M2/d)*k1*c1*c2 # M.T.C in cm/sec
kf2 = R*T*k2*c1*c2 # M.T.C in cm/sec
#Results
print"the M.T.C for liquid is cm/sec",round(kf1,4)
print"\n the M.T.C for gas is cm/sec",round(kf2,1)
```

In [25]:

```
#initialization of variables
l = 0.07 # flim thickness in cm
v = 3 # water flow in cm/sec
D = 1.8*10**-5 # diffusion coefficient in cm**2/sec
crat = 0.1 # Ratio of c1 and c1(sat)
from math import log
#Calculations
z = (((l**2)*v)/(1.38*D))*((log(1-crat))**2) #Column length
#Results
print"the column length needed is cm",round(z,1)
```

In [26]:

```
from math import pi
#Initialization of variables
Dw = 1*10**-5 # Diffusion co efficient in cm**2/sec
omeg = 20*2*pi/60 # disc rotation in /sec
Nuw = 0.01 # Kinematic viscousity in water in cm**2/sec
Da = 0.233 # Diffusion co efficient in cm**2/sec
Nua = 0.15 # Kinematic viscousity in air in cm**2/sec
c1satw = 0.003 # Solubility of benzoic acid in water in gm/cm**3
p1sat = 0.30 # Equilibrium Vapor pressure in mm Hg
ratP = 0.3/760. # Ratio of pressures
c1 = 1./(22.4*10**3) # Moles per volume
c2 = 273./298. # Ratio of temperatures
c3 = 122 # Grams per mole
#Calculations
kw = 0.62*Dw*((omeg/Nuw)**0.5)*((Nuw/Dw)**(1/3))# cm/sec
Nw = kw*c1satw*10**6 # mass flux in x*10**-6 in g/cm**2-sec
ka = 0.62*Da*((omeg/Nua)**0.5)*((Nua/Da)**(1/3))#cm/sec
c1sata = ratP*c1*c2*c3# Solubility of benzoic acid in air in gm/cm**3
Na = ka*c1sata*10**6 # mass flux in x*10**-6 in g/cm**2-sec
#Results
print"the mass flux in water is x10**-6 g/cm**2-sec",round(Nw,2)
print"\n the mass flux in air is x10**-6 g/cm**2-sec",round(Na)
```

In [27]:

```
#initialization of variables
Dl=2.1*10**-5# Diffusion co efficient for Oxygen in air in cm**2/sec
Dg = 0.23 #Diffusion co efficient for Oxygen in water in cm**2/sec
R = 82. # Gas constant in cm**3-atm/g-mol-K
T = 298. #Temperature in Kelvin
l1 = 0.01 # film thickness in liquids in cm
l2 = 0.1 # film thickness in gases in cm
H1 = 4.3*10**4 # Henrys constant in atm
c = 1./18. # concentration of water in g-mol/cm**3
#Calculations
kl = (Dl/l1)*c # m.t.c in liquid phase in mol/cm**2/sec
kp = (Dg/l2)/(R*T)# m.t.c in gas phase in gmol/cm**2-sec-atm
KL = 1/((1/kl)+(1/(kp*H1)))# Overall m.t.c in mol/cm**2-sec liquid phase
#Results
print"The overall m.t.c in liquid side is mol/cm**2-sec",round(KL,5)
```

In [28]:

```
#initialization of variables
Dl=1.9*10**-5# Diffusion co efficient for liquid phase in cm**2/sec
Dg = 0.090 #Diffusion co efficient for gas phase in cm**2/sec
R = 82. # Gas constant in cm**3-atm/g-mol-K
T = 363. #Temperature in Kelvin
H1 = 0.70 # Henrys constant in atm
c = 1./97. # concentration of water in g-mol/cm**3
#Calculations
kl = (Dl/0.01)*c # m.t.c in liquid phase in mol/cm**2/sec
kp = (Dg/0.1)/(R*T)# m.t.c in gas phase in gmol/cm**2-sec-atm
KL = 1/((1/kl)+(1/(kp*H1)))*10**5# Overall m.t.c in x*10**-5 mol/cm**2-secliquid phase
#Results
print"The overall m.t.c in liquid side is x10**-5 mol/cm**2-sec",round(KL,1)
print" answer is different due to round off error"
```

In [29]:

```
#Initialization of variables
k1 = 3.0*10**-4 # m.t.c in benzene in cm/sec
k2 = 2.4*10**-3 # m.t.c in water in cm/sec
ratio = 150 # Solubility ratio in benzene to water
#Calculations
K1 = (1/((1/k1)+(ratio/k2)))*10**5 # Overall m.t.c through benzene phase in x*10**-5 cm/sec
#Results
print"The overall M.T.C through benzene phase is x10**-5 cm/sec",round(K1,1)
```

In [30]:

```
#initialization of variables
H1 = 75. # henrys constant for ammonia in atm
H2 = 41000. # henrys constant for methane in atm
p = 2.2 # pressure in atm
kya = 18. # product of m.t.c and packing area per tower volume in lb-mol/hr-ft**3
kxa = 530. #product of m.t.c and packing area per tower volume in lb-mol/hr-ft**3
#calcuations
Kya1 = 1/((1/kya) + (H1/p)/kxa) #The overall coefficient for ammonia in lb-mol/hr-ft**3
Kya2 = 1/((1/kya) + (H2/p)/kxa) #The overall coefficient for methane in lb-mol/hr-ft**3
#Results
print"The overall coefficient for ammonia is lb-mol/hr-ft**3",round(Kya1,1)
print"\n The overall coefficient for methane is lb-mol/hr-ft**3",round(Kya2,2)
```