Chapter 2:. Convective Mass Transfer¶

Example 2.1,Page number94¶

In [8]:
T = 310  					# [K]
# Since the solubility of oxygen in water at 310 K is extremely low, we are dealing with 	dilute solutions
k_L = 3.3*10**-5  				# [coefficient based on the oxygen concentration 						difference in the water, m/s]
row = 993  					# [kg/cubic m]
M_b = 18  					# [gram/mole]

#Calculation

# Since we are dealing with very dilute solutions
# Therefore, c = (row/M_avg) = row/M_b
c = row/M_b  					# [kmole/cubic m]
# Using equation 2.10
k_x = k_L*c  					# [kmole/square m.s]

#Result

print"The mass-transfer coefficient based on the mole fraction of oxygen in the liquid is",round(k_x,5)," kmole/square m.s"


The mass-transfer coefficient based on the mole fraction of oxygen in the liquid is 0.00182  kmole/square m.s


Example 2.2,Page number:95¶

In [4]:
#Variable declaration

#  a-ammonia   b-air
T = 300  					# [K]
P = 1  						# [atm]
y_a1 = 0.8  					# [ammonia mole fraction in the bulk of the gas 							phase]
y_a2 = 0.732  					# [ammonia gas-phase mole fraction on interface]
N_a = 4.3*10**-4  				# [ammonia flux,  kmole/square m.s]

#Calculations

import math
# Using equation 2.2
F_g = N_a/math.log10((1-y_a2)/(1-y_a1))  		# [kmole/square m.s]

#Result

print"The mass-transfer coefficient in the gas phase at that point where flux is 4.3*10**-4 is",round(F_g,5)," kmole/square m.s"
print"\n\nNOTE:Calculation mistake in book:\nAnswer written as 1.47*10^-4,Actually,,...it is 1.47*10^-3.Please REDO last calculation manually to check"

The mass-transfer coefficient in the gas phase at that point where flux is 4.3*10**-4 is 0.00338  kmole/square m.s

NOTE:Calculation mistake in book:
Answer written as 1.47*10^-4,Actually,,...it is 1.47*10^-3.Please REDO last calculation manually to check


Example 2.3,Page number:96¶

In [10]:
#Variable declaration

# a-methanol   b-water
P = 101.3  						# [kPa]
y_a1 = 0.707  						# [mole fraction at interface]
y_a2 = 0.656  						# [mole fraction at bulk of the gas]
k_g = 1.62*10**-5  					# [mass-transfer coefficient at a point 							 in the column, kmole/square m.s.kPa]
#Calculations

# Using equation 2.14
k_y = k_g*P  						# [kmole/square m.s]
# Using equation 2.12
N_a = k_y*(y_a1-y_a2)  					# [kmole/square m.s]

#Result

print"The methanol flux at the point of given mass transfer coefficient is",round(N_a,7),"kmole/square m.s"

The methanol flux at the point of given mass transfer coefficient is 8.37e-05 kmole/square m.s


Example 2.4,Page number:99¶

In [11]:
n = 6  # [number of variables]

#Calculations

from scipy.optimize import fsolve
from numpy import *
import math
# To determine the number of dimensionless parameters to be formed, we must know the rank, r, of the dimensional matrix.
# The dimensional matrix is
DM =matrix([[0,0,1,1,0,0],[1,1,-3,-1,2,1],[-1,-1,0,0,-1,-1]])
rk= linalg.matrix_rank(DM)
print"Rank of matrix is ",rk

#The numbers in the table represent the exponent of M, L, and t in the dimensional expression of each of the six variables involved. For example, the dimensional expression of p is M/Lt  hence the exponents are 1, -1, and -1

# From equation 2.16
i = n-rk  # [number of dimensional groups]
# Let the dimensional groups are pi1, pi2 and pi3
# Therefore pi1 = (D_AB)**a*(row)**b*(D)**c*kc
#           pi2 = (D_AB)**d*(row)**e*(D)**f*v
#           pi3 = (D_AB)**g*(row)**h*(D)**i*u

# Solving for pi1
# M**0*L**0*t**0 = 1 = (L**2/t)**a*(M/L**3)**b*(L)**c*(L/t)

# Solution of simultaneous equation
def F(e):
f1 = 2*e[0]-3*e[1]+e[2]+1
f2 = -e[0]-1
f3 = -e[1]
return(f1,f2,f3)

# Initial guess:
e = [0.1,0.8,0.5]
y = fsolve(F,e)
a = y[0]
b = y[1]
c = y[2]
print"The coefficients of pi1 are",a,round(b),c
# Similarly the coefficients of pi2 and pi3 are calculated
# Therefore we get pi1 = kc*D/D_AB = Sh i.e. Sherwood Number
#                  pi2 = v*D/D_AB = P_ed i.e. Peclet Number
#                  pi3 = u/(row*D_AB) = Sc i.e. Schmidt Number

# Dividing pi2 by pi3 gives
#       pi2/pi3 = D*v*row/u = Re i.e. Renoylds number

print"The result of the dimensional analysis of forced-convection mass transfer in a circular conduit indicates that a correlating relation could be of the form\n Sh = function(Re,Sc)\n which is analogous to the heat transfer correlation \n Nu = function(Re,Pr)"

Rank of matrix is  3
The coefficients of pi1 are -1.0 0.0 1.0
The result of the dimensional analysis of forced-convection mass transfer in a circular conduit indicates that a correlating relation could be of the form
Sh = function(Re,Sc)
which is analogous to the heat transfer correlation
Nu = function(Re,Pr)


Example 2.6,Page number:111¶

In [12]:
M_a = 352  					# [molecular weight of UF6, gram/mole]
M_b = 29  					# [gram/mole]
d = 0.01  					# [diameter, m]
x = 0.1  					# [length exposed to air stream, m]
v = 1  						# [m/s]
Ts = 303  					# [surface temperature of solid, K]
P_a = 27  					# [vapor pressure of UF6, kPa]
Tb = 325  					# [bulk temperature of solid ,K]
P = 101.3  					# [kPa]
R = 8.314  					# [cubic m.Pa/mole.K]
import math

y_a1 = P_a/P  					# [mole fraction at point 1]
y_a2 = 0  					# [mole fraction at point 2]

# Along the mass-transfer path-cylinder surface (point 1) to bulk air (point 2)
Tavg = (Ts+Tb)/2  				# [K]

# At point 1, the gas is saturated with UF6 vapor, while at point 2 the gas is virtually 	free of UF6
# Therefore
Pavg = (P_a+0)/2  				# [average partial pressure, kPa]
y_a = Pavg/P  					# [mole fraction of UF6]

Mavg = M_a*y_a+M_b*(1-y_a)  			# [gram/mole]
row_avg = P*Mavg/(R*Tavg)  			# [kg/cubic m]

# Parameter for c-O2, d-N2 and a-UF6
yi_c = 0.182
yi_d = 0.685
yi_a = 0.133
Tc_c = 154.6    				# [K]
Tc_d = 126.2   					# [K]
Tc_a = 505.8  					# [K]
Pc_c = 50.4    					# [bar]
Pc_d = 33.9   					# [bar]
Pc_a = 46.6   					# [bar]
M_c = 32        				# [gram/mole]
M_d  = 28    					# [gram/mole]
M_a  = 352   					# [gram/mole]
V_c = 73.4      				# [cubic cm/mole]
V_d  = 89.8   					# [cubic cm/mole]
V_a  = 250  					# [cubic cm/mole]
Z_c = 0.288
Z_d  = 0.290
Z_a  = 0.277

#Calculations

# From equation 2.52 and 2.53
Tcm = yi_c*Tc_c+yi_d*Tc_d+yi_a*Tc_a  		# [K]
Pcm = 10**6*R*Tcm*(yi_c*Z_c+yi_d*Z_d+yi_a*Z_a)/((yi_c*V_c+yi_d*V_d+yi_a*V_a)*100000)  	# [bar]
M_avg = yi_c*M_c+yi_d*M_d+yi_a*M_a  		# [gram/mole]

# From equation 2.50
Em =  0.176*(Tcm/(M_avg**3*Pcm**4))**(1.0/6.0)  	# [uP]**-1

# From equation 2.51
Trm = Tavg/Tcm
f_Trm = (0.807*Trm**0.618)-(0.357*math.exp(-0.449*Trm))+(0.340*math.exp(-4.058*Trm))+0.018
# From equation 2.49
u = f_Trm/Em  						# [uP]
u = u*10**-7  						# [viscosity, kg/m.s]

Re = d*v*row_avg/u  					# [Renoylds number]

# Diffusivity of UF6 vapors in air at 314 K and 1 atm from equation 1.49
D_ab = 0.0904  						# [square cm/s]

Sc = u/(row_avg*D_ab*10**-4)  				# [Schmidt number]

Sh_avg = 0.43 + 0.532*Re**0.5*Sc**0.31  		# [Sherwood number]
# From equation 1.7
c = P/(R*Tavg)  					# [kmole/cubic m]
# From Table 2.1
F_av = Sh_avg*D_ab*c*10**-4/d  				# [kmole/square m.s]

# From equation 2.2
N_avg = F_av*math.log((1-y_a2)/(1-y_a1))  		# [kmole/square m.s]
S = 2*math.pi*d**2/4 +math.pi*d*x  			# [total surface area of the cylinder, 							square m]

w_a = N_avg*S*M_a  					# [rate of sublimation of the solid, 							kg/s]

#Result
print"Rate of sublimation of a cylinder of UF6 is",round(w_a,5),"kg/s\n\n"

Rate of sublimation of a cylinder of UF6 is 0.00023 kg/s



Example 2.7,Page number:116¶

In [13]:
import math
#Variable declaration

# a-benzene      b-nitrogen
T = 300  					# [K]
P = 101.3  					# [kPa]
v =10  						# [m/s]
R = 8.314  					# [cubic m.Pa/mole.K]

n = -0.5
# Data on the properties of C02 at 300 K and 1 bar
u = 1.5*10**-5  				# [viscosity, P]
Pr = 0.77  					# [Prandtl number]
Cp = 853  					# [J/kg.K]
# Therefore
# b = 5.086*l**0.5
#           j_D = j_H = f(Re) = 5.086*(l**0.5)*Re**-0.5
# From Table 2.1
#           F = j_D*c*v/Sc**(2/3) = 5.086*(l**0.5)*c*v/(Re**0.5*Sc**(2/3)) = 		5.086*(row*v*u)**0.5/(Mavg*Sc**(2.0/3.0))

#Calculations

# Vapor pressure of benzene
P_a = math.exp(15.9008-(2788.51/(T-52.36)))  	# [mm of Hg]
P_a = P_a*101.3/760  					# [kPa]

# Parameter for a-benzene, b-nitrogen
yi_a = 0.07
yi_b = 0.93
Tc_a = 562.2
Tc_b = 126.2   						# [K]
Pc_a = 48.9
Pc_b = 33.9  						# [bar]
M_a = 78.1
M_b  = 28    						# [gram/mole]
V_a = 259
V_b  = 89.8  						# [cubic cm/mole]
Z_a = 0.271
Z_b  = 0.290
sigma_a = 5.349
sigma_b = 3.798  					# [Angstrom]
ek_a = 412.3
ek_b = 71.4  						# [E/k, K]

# From equation 2.52 and 2.53
Tcm = yi_b*Tc_b+yi_a*Tc_a  				# [K]
Pcm = 10**6*R*Tcm*(yi_b*Z_b+yi_a*Z_a)/((yi_b*V_b+yi_a*V_a)*100000) 	 # [bar]
M_avg = yi_b*M_b+yi_a*M_a  						# [kg/kmole]

#RESULT

print"Average molecular weight is",round(M_avg,1),"kg/kmole"

row = P*M_avg/(R*T)  							# [kg/cubic m]

#RESULT

print"Density of mixture is",round(row,2),"kg/cubic"
# From equation 2.50
Em =  0.176*(Tcm/(M_avg**3*Pcm**4))**(1.0/6.0)  			# [uP]**-1

# From equation 2.51
Trm = T/Tcm
f_Trm = (0.807*Trm**0.618)-(0.357*math.exp(-0.449*Trm))+(0.340*math.exp(-4.058*Trm))+0.018
# From equation 2.49
u = f_Trm/Em  								# [uP]
u = u*10**-7  								# [viscosity, kg/m.s]
print"Average viscosity of mixture is ",round(u,7),"kg/m.s\n\n"

# Calculating diffusivity of benzene using equation 1.49
D_ab = 0.0986  								# [square cm/s]
Sc = u/(row*D_ab*10**-4)  						# [Schmidt number]

F = 5.086*(row*v*u)**0.5/(M_avg*Sc**(2.0/3.0))  			# [kmole/square m.s]

#RESULT

print"The required mass transfer coefficient is",round(F,5),"kmole/square m.s"

Average molecular weight is 31.5 kg/kmole
Density of mixture is 1.28 kg/cubic
Average viscosity of mixture is  1.64e-05 kg/m.s

The required mass transfer coefficient is 0.00196 kmole/square m.s


Example 2.8,Page number:120¶

In [14]:
import math

T = 300  						# [K]
l = 3  							# [length of vertical plate, m]
b = 1.5  						# [width of vertical plate, m]
P = 101.3  						# [kPa]
v = 5  							# [velocity across the width of plate, 							m/s]
row_a = 0.88  						# [gram/cubic cm]

y_a1 = 0.139  						# [mole fraction of benzene at inner 							edge]
y_a2 = 0

# The film conditions, and average properties, are identical to those in Example 2.7, 	only the geometry is different
# Therefore
M_avg = 31.4  						# [kg/kmole]
row = 1.2  						# [kg/cubic m]
u = 161*10**-7  					# [kg/m.s]
D_ab = 0.0986 		 				# [square cm/s]
Sc = 1.3  						# [Schmidt Number]
Re = row*v*b/u  					# [Renoylds Number]

if Re > 4000:
print"The flow across the plate is turbulent\n"
elif Re<2000:
print"The flow across the plate is laminar\n"
#Using equation 2.57
Sh_l = 0.036*Re**0.8*Sc**(1.0/3.0)

# Nitrogen (component B) does not react with benzene (component A), neither dissolves in 		the liquid  therefore, NB = 0 and siA = 1. The F-form of the mass-transfer coefficient 			should be used
F = Sh_l*1.26*D_ab*10**-4/(M_avg*b)  			# [kmole/square m.s]
N_a = F*math.log((1-y_a2)/(1-y_a1))  			# [kmole/square m.s]

# The total mass rate of evaporation over the surface of the plate
S = 1.5*3  						# [square m]
M_a = 78.1  						# [gram/mole]
wa = N_a*S*M_a*60*1000  				# [gram/min]

V = wa/row_a  						# [volumetric flow rate, ml/min]

print"Liquid benzene should be supplied at the top of the plate at the rate ",round(V),"ml/min\nso that evaporation will just prevent it from reaching the bottom of the plate."

The flow across the plate is turbulent

Liquid benzene should be supplied at the top of the plate at the rate  1473.0 ml/min
so that evaporation will just prevent it from reaching the bottom of the plate.


Example 2.9,Page number:123¶

In [15]:
dp1 = 10**-3  					# [diameter of spherical drop of water, m]
Tair = 323  					# [K]
P = 101.3  					# [kPa]
Twater = 293  					# [K]
R = 8.314  					# [cubic m.Pa/mole.K]
M_a = 18.0  					# [gram/mole]
M_b = 29.0  					# [gram/mole]
import math

#Calculation

dp2 = (1.0/2.0)**(1.0/3.0)*dp1  		# [m]
dp = (dp1+dp2)/2  				# [m]

row_p = 995  					# [density of water, kg/cubic m]
row1b = 1.094  					# [density of air, kg/cubic m]
u = 1.95*10**-5  				# [kg/m.s]
row_pr = row_p-row1b  				# [kg/cubic m]
g = 9.8  					# [accleration due to gravity, square m/s]
# Combining equation 2.68 and 2.69
Ga = 4*dp**3*row1b*row_pr*g/(3*u**2)  		# [Galileo Number]

# Relationship between Re and Cd
# Re/Cd = Re**3/Ga = 3*row**2*vt**3/(4*g*u*row_pr)

# The following correlation is used to relate Re/Cd, to Ga
# ln(Re/Cd)**(1/3) = -3.194 + 2.153*ln(Ga)**(1/3) - 0.238*(ln(Ga)**(1/3))**2 + 	0.01068*(ln(Ga)**(1/3))**3
# Therefore let A = (Re/Cd)
A = math.exp(-3.194 + 2.153*math.log(Ga**(1.0/3.0)) - 0.238*(math.log(Ga**(1.0/3.0)))**2 + 0.01068*(math.log(Ga**(1.0/3.0)))**3)

# Therefore 'vt' will be
vt = A*(4*g*row_pr*u/(3*row1b**2))**(1.0/3.0)  	# [Terminal velocity of particle, m/s]

#Result

print"Terminal velocity of particle is",round(vt,2),"m/s"

#Calculation

P_w = 2.34  					# [vapor pressure of water, kPa]
y_w = P_w/P  					# [mole fraction of water at the inner edge of 						the gas film]
M_avg = 18*y_w+29*(1-y_w)  			# [gram/mole]

row2b = P*M_avg/(R*Twater)  			# [kg/cubic.m]
delta_row = row2b - row1b  			# [kg/cubic.m]

Tavg = (Tair+Twater)/2  			# [K]
# At Temperature equal to Tavg density and viscosity are
row3 = 1.14  					# [kg/cubic.m]
u1 = 1.92*10**-5  				# [kg/m.s]

Grd = g*row3*delta_row*(dp**3)/(u1**2)

# Diffusivity of water at Tavg and 1 atm is
D_ab = 0.242*10**-4  				# [square m/s]
Sc = u1/(row3*D_ab)  				# [Schmidt Number]
Re = dp*row3*vt/u1  				# [Renoylds Number]

# From equation 2.65 Re is greater than  0.4*Grd**0.5*Sc**(-1/6)
# Therfore equation 2.64 can be used to calculate mass transfer coefficient

Sh = 2+0.552*(Re**0.5)*Sc**(1.0/3.0)  		# [Sherwood Number]
# From Table 2.1
# Sh = kc*P_bm*dp/(P*D_ab), since P_bm is almost equal to P
# Therefore
# Sh = kc*dp/D_ab
kc = Sh*D_ab/dp  				# [m/s]

ca2 = 0  					# [dry air concentration]
ca1 = P_w/(R*Twater)  				# [interface concentration, kmole/cubic.m]
# Average rate of evaporation
wa = math.pi*dp**2*M_a*kc*(ca1-ca2)*1000  	# [g/s]

# Amount of water evaporated
m = row_p*math.pi*dp1**3/12*1000  		# [g]
# Time necessary to reduce the volume by 50%
t = m/wa  					# [s]

D = t*vt  					# [distance of fall, m]

#Result

print"The distance of fall is",round(D),"m"

Terminal velocity of particle is 3.59 m/s
The distance of fall is 90.0 m


Example 2.10,Page number:127¶

In [2]:
Re = 1223  				# [Renoylds Number]
Sc = 0.905  				# [Schmidt Number]
c = 0.039  				# [molar density, kg/cubic m]
v = 1  					# [gas velocity, m/s]
# Therefore
#Calculations

Gm = c*v  				# [kmole/square m.s]
# From equation 2.9
# Kg*P = ky
# Therefore substituting in equation 2.73 we obtain
ky = 0.281*Gm/(Re**0.4*Sc**0.56)  	# [kmole/square m.s]
# Now equation 2.73 were obtained under very dilute concentrations
# Therefore
y_bm = 1
# From equation 2.6
F = ky*y_bm  				# [kmole/square m.s]

#Result

print"Mass transfer coefficient is ",round(F,6),"kmol/m.^2-s"

Mass transfer coefficient is  0.000675 kmol/m.^2-s


Example 2.11,Page number:129¶

In [17]:
D = 25.4*10**-3  				# [diameter of wetted wall tower, m]
Gy = 10  					# [mass velocity, kg/square m.s]
T1 = 308  					# [K]
P = 101.3  					# [kPa]
p_a1 = 1.95  					# [partial pressure of water vapor, kPa]
R = 8.314  					# [cubic m.Pa/mole.K]
M_a = 18  					# [gram/mole]
Cpa = 1.88  					# [kJ/kg.K]

# Properties of dry air at 308 K and 1 atm pressure are
u = 1.92*10**-5  				# [kg/m.s]
row = 1.14  					# [kg/cubic m]
D_ab = 0.242*10**-4  				# [square m/s]
Sc = 0.696  					# [Schmidt number]
Cp = 1.007  					# [kJ/kg.K]
k = 0.027  					# [W/m.K]
Pr = 0.7  					# [Prandtl number]

#Calculations

import math
from scipy.optimize import fsolve
from numpy import *

Re = D*Gy/u  					# [Renoylds number]
# From equation 2,74
Sh = 0.023*Re**0.83*Sc**0.44  			#[Sherwood number]
# From Table 2.1
kg = Sh*D_ab/(R*T1*D)*1000  			# [mole/square m.s.kPa]

# To estimate the heat-transfer coefficient, we use the Dittus-Boelter equation for cooling, equation 2.80
Nu = 0.023*Re**0.8*Pr**0.3  			# [Nusselt number]
# From Table 2.1
h = Nu*k/D  					# [W/square m.K]

T =373.15  					# [K]
lambda_a = 40.63  				# [kJ/mole]
Tc = 647.1  					# [K]

# Solution of simultaneous equation 2.78 and 2.79
def F(e):
f1=kg*(p_a1 - math.exp(16.3872-(3885.7/(e[0]-42.98))))-e[1]
f2=e[1]*M_a*Cpa*(T1-e[0])/(1-exp(-e[1]*M_a*Cpa/h)) + 1000*e[1]*lambda_a*((1-(e[0]/Tc))/(1-(T/Tc)))**0.38
return(f1,f2)

# Initial guess
e = [292,-0.003]
y = fsolve(F,e)
Ti = y[0]
Na = y[1]

print"The temperature of the liquid water and the rate of water evaporation is",round(Ti),"K and",round(Na,3)," mole/square m.s respectively"

The temperature of the liquid water and the rate of water evaporation is 295.0 K and -0.013  mole/square m.s respectively


Example 2.12,Page number:131¶

In [18]:
D = 25.4*10**-3  			# [Internal diameter of tower, m]
Z = 1.5  				# [length of the wetted section, m]
Gy = 10  				# [mass velocity of air, kg/square m.s]
Tair = 308  				# [K]
Twater = 295  				# [K]
P = 101.3  				# [kPa]
M_a = 18.0  				# [gram/mole]
M_b = 29.0  				# [gram/mole]
R = 8.314  				# [cubic m.Pa/mole.K]

#Calculations

import math

Pa = 2.64  # [kPa]

Gm = Gy/M_b  	# [Assuming that gas phase is basically dry air, kmole/square m.s]
# The properties of dry air at 308 K and 1 atm are (from example 2.9)
row = 1.14  				# [kg/cubic m]
u = 1.92*10**-5  			# [kg/m.s]
D_ab = 0.242*10**-4  			# [square m/s]
Sc = 0.692  				# [Schmidt number]

Re = Gy*D/u  				# [Renoylds number]

if Re<35000 and Re>2000:
Sh = 0.023*Re**0.83*Sc**0.44  	# [Sherwood number]
print"Sherwood number is",round(Sh,1)
else:
print"We cannot use equation 2.74"

c = P/(R*Tair)  			# [kmole/cubic m]
# Now using equation 2.89
Pa_out = Pa*(1-math.exp((-4*Sh*Z*c*D_ab)/(Gm*D**2)))  		# [kPa]

#Result
print"The partial pressure of water in the air leaving the tower is",round(Pa_out,2),"kPa"

Sherwood number is 51.6
The partial pressure of water in the air leaving the tower is 1.94 kPa


Example 2.13,Page number:134¶

In [2]:
Gy = 10.0  				# [kg/square m.s]
dp = 3.5*10**-3  			# [diameter of spherical glass beads, m]
D = 25.4*10**-3 			# [Internal diameter of tower, m]
Tair = 308  				# [K]
Twater = 295  				# [K]
P = 101.3  				# [kPa]
M_a = 18  				# [gram/mole]
M_b = 29  				# [gram/mole]
R = 8.314  				# [cubic m.Pa/mole.K]

#Calculation

import math
from scipy.optimize import fsolve
# The properties of dry air at 308 K and 1 atm are (from example 2.12)
row = 1.14  				# [kg/cubic m]
u = 1.92*10**-5  			# [kg/m.s]
D_ab = 0.242*10**-4  			# [square m/s]
Sc = 0.692  				# [Schmidt number]
c = 0.04 	 			# [mole/cubic m]
Gm = 0.345  				# [kmole/square m.s]

Re = Gy*dp/u  				# [Renoylds number]
if Re<2500 and Re>10:
# Subsituting in equation 2.90
jd = 1.17*Re**-0.415
print"Renoylds number is ",Re
else:
print " "
Std = 0.052/(Sc**(2.0/3.0))
# From Table 2.1
Sh = Std*Re*Sc  			# [Sherwood number]
# From equation 2.94
e = 0.406+0.571*(dp/D)  		# [bed porosity]
e=round(e,3)
#Illustration 2.13(a)
# Solution(a)
# Now Paout = 0.99*Pa
# Using equation 2.93 to calculate 'Z'
def f12(Z):
return(0.99 - 1 + math.exp(-6*(1-e)*Sh*c*Z*D_ab/(Gm*dp**2)))
Z = fsolve(f12,0.06)

#Result
Z=round(Z[0],3)
print"The depth of packing required is",Z,"m=",Z*100,"cm"

#Illustration 2.13(b)
# Solution(b)
# From equation 2.95
deltaP = (150*(1-e)/Re + 1.75)*((1-e)*(Gy**2)*Z)/(dp*row*e**3)  	# [Pa]

#Result
print"The gas pressure drop through the bed is",round(deltaP),"Pa  (Approx) \nDUE TO LACK OF PRECISION IN CALCULATION IN BOOK.\niF DONE MANUALLY,THIS ANSWER STANDS CORRECT"

Renoylds number is  1822.91666667
The depth of packing required is 0.078 m= 7.8 cm
The gas pressure drop through the bed is 15817.0 Pa  (Approx)
DUE TO LACK OF PRECISION IN CALCULATION IN BOOK.
iF DONE MANUALLY,THIS ANSWER STANDS CORRECT


Example 2.14,Page number:138¶

In [20]:
m = 40000.0  					# [kg/hr]
Twater = 298  					# [K]
v = 0.1  					# [superficial velocity, m/s]
P = 101.3 					# [kPa]
V = 40*10**-3  					# [Flow rate of nitrogen, cubic m/min]
d = 2.90*10**-4  				# [Outside diameter of fibres, m]
pf = 0.4  					# [Packing factor]
a = 46.84*100  					# [surface area per unit volume, m**-1]
R = 8.314  					# [cubic m.Pa/mole.K]

#Calculation

import math

dw = 1000  					# [density of water, kg/cubic m]
Ql = m/(3600*1000)  				# [volumetric water flow rate, cubic m/s]
# Shell diameter
D = (4*Ql/(math.pi*v))**0.5  			# [Shell diameter, m]

# the properties of dilute mixtures of oxygen in water at 298 K
u = 0.9  					# [cP]
# Diffusivity from equation 1.53
D_ab = 1.93*10**-9  				# [square m/s]
Sc = 467  					# [Schmidt number]

Re = d*v*dw/(u*10**-3)  			# [Renoylds number]

# Substituting in equation (2-97) gives
Sh = 0.53*(1-1.1*pf)*((1-pf)/pf)**-0.47*(Re**0.53*Sc**0.33)

kl = Sh*D_ab/d  				# [mass-transfer coefficient on the shell side, 						m/s]

# From the specified BFW flow rate
L = m/(3600*18)  				# [kmole/s]
# From ideal gas law
V1 = V*P/(Twater*R*60)  			# [kmole/s]
# From the solubility of oxygen in water at 298 K,
M = 4.5*10**4
A = L/(M*V1)  					# [Absorption factor]

#Result

print"Absorption factor is",round(A,3)

#Calculation

# For 99% removal of the dissolved oxygen
# x_in/x_out = b = 100
b = 100
c = 55.5 					# [molar density, kmole/cubic m]
# Substituting in equation 2.99 yields
V_T = (L*math.log(b*(1-A)+A))/(kl*a*c*(1-A)) 	 # [cubic m]

# The module length, Z is
Z = V_T/(math.pi*D**2.0/4.0)

#Result
print"The shell diameter and module length is",round(D,3),"m and",round(Z,2)," m respectively"

Absorption factor is 0.503
The shell diameter and module length is 0.376 m and 2.15  m respectively


Example 2.15,Page number:140¶

In [36]:
#Variable declaration
d=2.21/100			#[m]
mu=8.82*10**-6		#[kg/m-s]
rho=2.81		#[kg/m**3]

c=34.14			#[mol/m**3]
D=array([2.228/(10**6),2.065/(10**6),1.832/(10**6)])      #Velocities in [m**2/s]

#Calculation
#Gy=rho*v
#Re=Gy*d/mu		#Reynolds number
Re=21750
print "Reynolds number=",Re
Sc=[]
Sh=[]
F=[]
for i in range(0, 3):
sc=mu/(rho*D[i])                   #Schmidt number
Sc.append(sc)
sh=0.023*Re**(0.83)*sc**(0.44)  #Sherwood number
Sh.append(sh)
f=sh*c*D[i]/d                   #Binary mass transfer coefficient in [mol/m^2-s]
F.append(f)
print "Schmidt number are:"
for i in range(0,3):
print round(Sc[i],3)
print "Sherwood number number are:"
for i in range(0,3):
print round(Sh[i],1)
print"Binary mass transfer coefficients are:"
for i in range(0,3):
print round(F[i],3)
#After modifying mathcad program of example 1.17,we have
N1=-0.0527                          #[mol/m^2-s]
N2=0.0395                          #[mol/m^2-s]
N3=0.0132                          #[mol/m^2-s]
print"The program yields the following results:"
print "N1=",N1,"mol/m**2-s"
print "N2=",N2,"mol/m**2-s"
print "N3=",N3,"mol/m**2-s"


Reynolds number= 21750
Schmidt number are:
1.409
1.52
1.713
Sherwood number number are:
106.5
110.1
116.1
Binary mass transfer coefficients are:
0.367
0.351
0.328
The program yields the following results:
N1= -0.0527 mol/m**2-s
N2= 0.0395 mol/m**2-s
N3= 0.0132 mol/m**2-s