#given data
from math import log
P = 900.3 #pressure in mmHg
R = 0.08317 #m^3.bar/kmol.k
T = 300 #in K
kG = 2.4E-6 #gas phase mass transfer coefficient
ky = P*kG
print '(a)'
print '(i) Mass transfer-coefficient is %0.3E'%ky,'kmol/(m^2.s.dy)'
pA1 = 0.0877 #vapor pressure of alochol in bar
pB1 = (1.2 - 0.0877)*760/1.013 #in mmHg
pA2 = 0
pB2 = 1.2*760/1.013 #in mmHg
kY = kG*pB1*pB2/P
print '(ii) Mass transfer-coefficient is %0.3E'%kY,'kmol/(m^2.s.dy)'
kc = kG*(760/1.013)*R*T
print '(iii) Mass transfer-coefficient is %0.4f'%kc,'m/s'
Fg = kG*(pB2-pB1)/(log(pB2/pB1))
print 'Coefficient of Fg: %0.4f'%Fg,'kmol/m^2.s'
print '(b)'
kGo = kG
kG1 = kG*2.2046*51.7/10.764
print '(i) kG: %0.3E'%kG,'lbmol/s.ft^2.psi'
kG2 = kG1*3600*14.69
print '(ii) kG: ',round(kG2,3),'lbmol/h.ft^2.atm'
kG3 = kG*2.2046*3600/(10.764*0.0393)
print '(iii) kG: %0.3E'%kG3,'lbmol/h.ft^2.inchHg'
pB2 = 1.2
pB1 = 1.2 - 0.0877
P = 1.2
Dab = 0.102*(1.013/1.185)*((300.0/273)**1.75)*1e-4 #Diffusivity value at 1.2bar and 300K
pBM = (pB2-pB1)/log(pB2/pB1)
kG4 = kG*(760/1.013)
d = Dab*P/(kG4*R*T*pBM)
print 'Thickness of the stagnant film is: %0.1E'%d,'m'
#Answers may vary due to round off error
#given data
Q = 13.2e-6 #liquid flow rate (pi*r^2*v)
l = 0.05 #length of the jet
W = 34 #Molecular weight of H2S
P = 1.03 #total pressure in the jet chamber
Cai = P*(0.1136) #kmol/m^3
Cab = 0
Rate = (4.42e-7)/W #in m^3/s
#let Dab = 1
Dab = 1
rate = 4*((Dab*Q*l)**0.5)*Cai
Dab = ((Rate/(4*Cai))**2)/(Q*l)
print 'Diffusivity of H2S in water at 25C Dab: %0.2E m^2/s'%Dab
#given data
from math import pi
tc = 1.0/20 #contact time of a liquid element with a gas bubble db/vb
Dab = 2.19e-5
kL = 2*(Dab/(pi*tc))**0.5 #Mass transfer coefficient
tr = 0.3/0.2 #residence time of a single bubble
r = 1.0/2 #radius of the bubble
vol = (4.0/3)*pi*(r**3) #volume of the bubble
area = 4*pi*(r**2) #surface area of the bubble
Cs = 1.493e-5 #solubility of CO2 water
Cb = 0
Amt = kL*area*(Cs-Cb)*tr #amount of CO2 absorbed from a single bubble during its residence time
Bub = 15/vol
rate = Amt*Bub
print 'Rate of absorption of CO2 from the bubbles per minute: %0.3E gmol/min'%rate
#Answer vary due to round off error
#given data
from math import pi
p = 2 #in atm
x = 2.0/1640 #mole fraction
W = 44*x + 18*(1-x) #Molecular wieght of the solution
sol = 997.0/W #Moles of solution per m^3
CO2 = 55.4*x #Moles of CO2 per m^3 solution
Cco = 2.3/44 #Concentration of the carbonated solution(kmol/m^3)
RoW = 1.0/(60*1000) #Volume rate of input of water(m^3/s)
RoA = RoW*Cco #rate of absorption of CO2 in the vessel(kmol/s)
V = 8e-3 #volume of the gas-liquid dispersion in the vessel(m^3)
a = 80 #specific interfacial area of contact between the dispersed gas snd the liquid(m^2/m^3)
#let kL = 1
kL = 1.0
Dab = 1.92e-9 #diffusivity of CO2
rateOA = V*a*(CO2-Cco)
kL = RoA/rateOA
#(a) part
delta = Dab/kL
print '(a)'
print 'Film thickness: %0.2e m'%delta
#(b) part
tc = (4*Dab)/(pi*(kL**2))
print '(b)'
print 'The contact time: %0.3f s'%tc
#(c) part
s = (kL**2)/(Dab)
print '(c)'
print 'The fractional rate of surface renewal: %0.2f per sec'%s
#given data
from math import exp
Cb1 = 0 #concentration of the benzoic acid at the inlet
Cs = 3.01 #concentration of the liquid in contact with the surface of a sphere
dp = 0.8 #diameter of the sphere
vo = 2.2 #superficial liquid velocity
nu = 0.0095 #kinematic viscosity
Dab = 1e-5 #diffusivity of the solute
e = 0.4 #bed voidage
h = 70 #height of the column
QbyA = 2.2 #liquid flow rate per unit area of the bed
Re = (dp*vo)/nu #Reynolds number
Sc = nu/Dab #Schmidt number
ejd = 0.25*((Re)**(-0.31)) #epsilon*jD
Sh = ejd*Re*(Sc**(1.0/3))/e #Sherwood number
kL = Sh*Dab/(dp)
a = 6*(1-e)/dp #specific interfacial velocity
Cb2 = Cs-((Cs)/exp(kL*a*h/(QbyA)))
print 'The effluent concentration: %i kg/m^3'%round(Cb2,1)