Chapter 3:lnterphase Mass Transfer¶
Example 3.1,Page number:161¶
In [9]:
#Variable declaration
# a-benzene b-toluene
T = 300 # [K]
x_a = 0.4 # [mole fraction in liquid phase]
# Antoine constants for benzene and toluene are given
# For benzene
A_a = 15.9008
B_a = 2788.51
C_a = -52.36
# For toluene
A_b = 16.0137
B_b = 3096.52
C_b = -53.67
#Calculation
import math
# Using equation 3.5 vapor pressure of component 'a' and 'b'
P_a = math.exp(A_a-(B_a/(T+C_a))) # [mm of Hg]
P_b = math.exp(A_b-(B_b/(T+C_b))) # [mm of Hg]
P_a = P_a*101.3/760 # [kPa]
P_b = P_b*101.3/760 # [kPa]
# Partial pressure of component 'a' and 'b'
p_a = x_a*P_a # [kPa]
p_b = (1-x_a)*P_b # [kPa]
P_total = p_a+p_b # [kPa]
#Result
print"The total equilibrium pressure of the binary system of benzene and toluene is",round(P_total,2)," kPa"
y_a = p_a/P_total # [mole fraction in vapor phase]
print"The composition of the vapor in equilibrium is",round(y_a,3)
Example 3.2,Page number:162¶
In [10]:
#Variable declaration
# A-oxygen B-water
T = 298 # [K]
H = 4.5*10**4 # [atm/mole fraction]
P = 1 # [atm]
row_B = 1000.0 # [density of water, kg/cubic m]
M_B = 18.0 # [Molecular mass of water, gram/mole]
M_A = 32.0 # [,Molecular mass of oxygen, gram/mole]
# Dry air contains 21% oxygen then p_A = y*P = 0.21 atm
# Therefore using Henry's Law
p_A = 0.21 # [atm]
x_A = p_A/H # [mole fraction in liquid phase]
# Basis: 1L of saturated solution
# For 1 L of very dilute solution of oxygen in water, the total moles of solution, n_t, will be approximately equal to the moles of water
n_t = row_B/M_B
# Moles of oxygen in 1L saturated solution is
n_o = n_t*x_A # [mole]
# Saturation concentration
c_A = n_o*M_A*1000 # [mg/L]
#Result
print"The saturation concentration of oxygen in water exposed to dry air at 298 K and 1 atm is",round(c_A,2),"mg/L"
Example 3.3,Page number:162¶
In [11]:
#Variable declaration
# a-ammonia b-air c-water
T = 300 # [K]
P = 101.3 # [kPa]
R = 8.314 # [cubic m.Pa/mole.K]
V_b = 15.0 # [cubic m]
m_a = 10.0 # [kg]
m_c = 45.0 # [kg]
M_a = 17.0 # [molecular mass of ammonia, gram/mole
M_c = 18.0 # [molecular mass of water, gram/mole]
#Calculations
n_b = V_b*P/(R*T) # [kmole]
n_a = m_a/M_a # [kmole]
n_c = m_c/M_c # [kmole]
import math
from scipy.optimize import fsolve
def f12(L_a):
return(((n_a-L_a)/(n_b+n_a-L_a))-(10.51*(0.156+(0.622*(L_a/(n_c+L_a))*(5.765*(L_a/(n_c+L_a))-1)))*(L_a/(n_c+L_a))))
L_a = fsolve(f12,0.3) # [kmole]
x_a = L_a/(n_c+L_a)
y_a = (n_a-L_a)/(n_b+n_a-L_a)
gammma_a = 0.156+0.622*x_a*(5.765*x_a-1)
#Result
print"At equilibrium the ammonia content of the liquid phase will be",x_a[0]
print"At equilibrium the ammonia content of the gas phase will be",y_a[0]
print"The amount of ammonia absorbed by the water will be",round(L_a[0],3),"kmole"
Example 3.4,Page number:169¶
In [12]:
#Variable declaration
# a-ammonia
T = 300 # [K]
P = 101.3 # [kPa]
Kg = 2.75*10**-6 # [kmole/square m.s.kPa]
m = 1.64
res = 0.85 # [gas phase resistance]
xa_g = 0.115/100.0 # [mole fraction of NH3 in liquid phase at a point]]
#Calculations
ya_g = 8.0/100.0 # [mole fraction of NH3 in gas phase at a point]
import math
from scipy.optimize import fsolve
Ky = Kg*P # [kmole/square m.s]
# Using equation 3.24
ky = Ky/res # [kmole/square m.s]
# Using equation 3.21
def f12(kx):
return((m/kx)-(1/Ky)+(1/ky))
kx = fsolve(f12,0.0029) # [kmole/square m.s]
# Interfacial concentrations at this particular point in the column, using equation (3.15)
ystar_a = m*xa_g
# Using equation 3.12
N_a = Ky*(ya_g-ystar_a) # [kmole/square m.s]
# Gas-phase interfacial concentration from equation (3.9)
ya_i = ya_g-(N_a/ky)
# Since the interfacial concentrations lie on the equilibrium line, therefore
xa_i = ya_i/m
# Cross checking the value of N_a
N_a = kx*(xa_i-xa_g) # [kmole/square m.s]
#Result
print"The individual liquid film coefficient and gas film coefficient are",round(kx[0],5),"kmole/square m.s and ",round(ky,6),"kmole/square m.s respectively"
print"The gas phase and liquid phase interfacial concentrations are",round(ya_i,5),"and,",round(xa_i,5),"respectively"
Example 3.5,Page number:171¶
In [13]:
#Variable declaration
# a-ammonia
T = 300 # [K]
P = 101.3 # [kPa]
ya_g = 0.6 # [ammonia concentration in bulk gas]
xa_l = 0.12 # [ammonia concentration in bulk liquid]
Fl = 3.5*10**-3 # [kmole/square m.s]
Fg = 2*10**-3 # [kmole/square m.s]
#Calculations
import math
from scipy.optimize import fsolve
# Algebraic solution (a)
def f12(xa_i):
return(1-(1-ya_g)*((1-xa_l)/(1-xa_i))**(Fl/Fg) - 10.51*(0.156+0.622*xa_i*(5.765*xa_i-1))*xa_i)
xa_i = fsolve(f12,0.2)
ya_i = 1-(1-ya_g)*((1-xa_l)/(1-xa_i))**(Fl/Fg)
#Result
print"The local gas and liquid interfacial concentrations are",round(ya_i[0],3),"and",round(xa_i[0],3),"respectively"
# Using equation 3.28
N_a = Fg*math.log((1-ya_i)/(1-ya_g))
print"The local ammonia mass-transfer flux is",round(N_a,5),"kmole/square m.s"
Example 3.6,Page number:174¶
In [14]:
#Variable declaration
# a-methanol b-water
T = 360.0 # [K]
P = 101.3 # [kPa]
lambda_a = 33.3 # [MJ/kmole]
lambda_b = 41.3 # [MJ/kmole]
Fg = 0.0017 # [kmole/square m.s]
Fl = 0.0149 # [kmole/square m.s]
yag = 0.36 # [bulk gas phase concentration]
xag = 0.20 # [bulk liquid phase concentration]
R = 1.987
#Calculations
import math
from numpy import *
from scipy.optimize import fsolve
sia_ag= 1.0/(1-(lambda_a/lambda_b))
print"sia_g is",sia_ag
#sia_ag =5.155
sia_al = sia_ag
# Therefore equation 3.29 becomes
# yai = 5.155-4.795(4.955/(5.155-xai))**8.765
# Using equation 3.33, 3.34, 3.35
V2 = 18.07 # [cubic cm/mole]
V1 = 40.73 # [cubic cm/mole]
a12 = 107.38 # [cal/mole]
a21 = 469.5 # [cal/mole]
# Solution of simultaneous equation
def F(e):
x1,x2,y1,y2,e4=e
#####SIDE Quantities#####
a=(V1/V2)*math.exp(-a21/(R*e4)) #d2lT
a1=(V2/V1)*math.exp(-a12/(R*e4)) #dl2(T)
b=math.exp(16.2620-3800.0/(e4-47)) #P2(T)
b1=math.exp(16.5938-3644.3/(e4-33)) #P1(T)
c=math.exp(-log(x2+x1*a)-x1*(a1/(x1+x2*a1)-a/(x2+x1*a))) #gamma_2
d=math.exp(-log(x1+x2*a1)-x1*(a1/(x1+x2*a1)-a/(x2+x1*a))) #gamma_1
#########################
x1=1-x2
y2=1-y1
y1=5.155-4.795*(4.955/(5.155-x2))**(Fl/Fg)
y1=x1*d*b1/P
y2=x2*c*b/P
return(x1,x2,y1,y2,e4)
# Initial guess
y = fsolve(F,(0.1,0.9,0.2,0.8,300))
xai = y[0]
xbi = y[1]
yai = y[2]
ybi = y[3]
T = y[4] # [K]
print"yai is",yai
print"ybi is",ybi
print"xai is",xai
print"xbi is",xbi
print"Temperature is",T,"K"
# Local Methanol flux, using equation 3.28
Na = sia_ag*Fg*math.log((sia_ag-yai)/(sia_ag-yag)) # [kmole/square m.s]
# From energy balance
Nb = -(lambda_a/lambda_b)*Na
#Result
print"Local Methanol flux is",round(Na,4)," (APPROX) kmole/square m.s"
Example 3.7,Page number:183¶
In [1]:
#Variable declaration
# 1-benzene a-absorber s-steams
T = 300 # [K]
P = 101.3 # [kPa]
R = 8.314 # [gas constant]
v = 1 # [cubic m/s]
# Gas in
y1a = 0.074
# Liquid in
x2a = 0.0476
# Recovery is 85 %
# Calculations for absorber section
#Calculations
import matplotlib
%matplotlib inline
import math
from pylab import *
from numpy import *
V1a = P*v/(R*T) # [kmole/s]
# Inert gas molar velocity
Vsa = V1a*(1-y1a) # [kmole/s]
Y1a = y1a/(1-y1a) # [kmole of benzene/kmole of dry gas]
X2a = x2a/(1-x2a) # [kmole of benzene/kmole of oil]
# Since the absorber will recover 85% of the benzene in the entering gas, the concentration of the gas leaving it will be
r = 0.85
Y2a = (1-r)*Y1a # [kmole of benzene/kmole of dry gas]
# The benzene-wash oil solutions are ideal, and the pressure is low therefore, Raoult’s law applies. From equations 3.1, 3.44, and 3.45
# yia = 0.136*xia
# or Yia/(1+Yia) = 0.136*Xia/(1+Xia)
# Data_eqm = [Xia,Yia]
Data_eqm =matrix([[0,0],[0.1,0.013],[0.2,0.023],[0.3,0.032],[0.4,0.04],[0.6,0.054],[0.8,0.064],[1,0.073],[1.2,0.080],[1.4,0.086]])
# Here because of the shape of equilibrium curve, the operating line for minimum oil rate must be tangent to curve
# Therefore
# From the curve X1a_max = 0.91
X1a_max = 0.91 # [kmol benzene/kmol oil]
# For minimum operating line slope is
S = (Y1a-Y2a)/(X1a_max-X2a) # [kmol oil/kmol air]
# Therfore
Lsa_min = S*Vsa # [kmole oil/s]
Data_minSlope1 = matrix([[X2a,Y2a],[X1a_max,Y1a]])
# For Actual operating line, oil flow rate is twice the minimum
Lsa = 2*Lsa_min # [kmole oil/s]
M_oil = 198 # [molecular weight of oil, gram/mole]
Wsa = Lsa*M_oil # [mass flow rate of oil, kg/s]
# Using equation 3.47 to calculate the actual concentration of the liquid phase leaving the absorber
X1a = X2a + Vsa*(Y1a-Y2a)/Lsa # [kmol benzene/kmol oil]
Data_opline1 = matrix([[X2a,Y2a],[X1a,Y1a]])
figure(1)
a1=plot(Data_eqm[:,0],Data_eqm[:,1],label='$Equilibrium line for absorber$')
a2=plot(Data_minSlope1[:,0],Data_minSlope1[:,1],label='$Minimum Flow Rate Line for absorber$')
a3=plot(Data_opline1[:,0],Data_opline1[:,1],label='$Operating Line for absorber$')
legend(loc='lower right')
plt.grid()
xlabel("Xa,mole benzene/mole oil")
ylabel("Ya, mole benzene/mole air")
title("XY diagram for the absorber of Example 3.7")
show(a1)
show(a2)
show(a3)
# Calculations for stripping section
Lss = Lsa
X2s = X1a
X1s = X2a
Y1s = 0
T = 373 # [K]
# Applying Raoult’s law at this temperature gives us
# yis = 1.77*xis
# Yis/(1+Yis) = 1.77*Xis/(1+Xis)
# Equilibrium data
# Data_equm = [Xis,Yis]
Data_equm = matrix([[0,0],[0.05,0.092],[0.1,0.192],[0.15,0.3],[0.2,0.418],[0.25,0.548],[0.3,0.691],[0.35,0.848],[0.4,1.023],[0.45,1.219],[0.5,1.439]])
# Similar procedure as above is followed
# The operating line for minimum oil rate must be tangent to curve
# Therefore from the curve
Y2s_max = 1.175 # [kmol benzene/kmol steam]
S = (Y2s_max-Y1s)/(X2s-X1s) # [kmole oil/kmole steam]
Vss_min = Lss/S # [kmole/s]
Vss = 1.5*Vss_min # [kmole/s]
Mss = 18 # [molecular weight of steam, gram/mole]
Wss = Vss*Mss # [kg steam/s]
Data_minSlope2 = matrix([[X1s,Y1s],[X2s,Y2s_max]])
Y2s_act = Y1s + Lss*(X2s-X1s)/Vss # [kmol benzene/kmol steam]
Data_opline2 = matrix([[X1s,Y1s],[X2s,Y2s_act]])
figure(2)
a1=plot(Data_equm[:,0],Data_equm[:,1],label='$Equilibrium line for stripping$')
a2=plot(Data_minSlope2[:,0],Data_minSlope2[:,1],label='$Minimum Flow Rate for stripping Line$')
a3=plot(Data_opline2[:,0],Data_opline2[:,1],label='$Operating Line for stripping$')
#legend("Equilibrium line for stripping","Minimum Flow Rate for stripping Line","Operating Line #for stripping")
legend(loc='upper left')
xlabel("Xa, mole benzene/mole oil")
ylabel("Ya, mole benzene/mole air")
title("XY diagram for stripper of Example 3.7")
plt.grid()
show(a1)
show(a2)
show(a3)
#Result
print"The oil circulation rate and steam rate required for the operation is",round(Wsa,2),"kg/s",round(Wss,3)," kg steam/s respectively"
Example 3.8,Page number:190¶
In [1]:
import matplotlib
%matplotlib inline
from numpy import *
from pylab import *
#Variable declaration
# 1-Nitrogen dioxide 2-air
T = 298 # [K]
P = 101.3 # [kPa]
y1 = 0.015
V1 = 0.5 # [mass flow rate of the gas entering the adsorber, kg/s]
M1 = 46 # [gram/mole]
M2 = 29 # [gram/mole]
#Calculations
# Data_eqm1 = [P1 m] (where 'P1' is Partial pressure of NO2 in mm of Hg, 'm' is solid concentration in kg NO2/kg gel)
Data_eqm1 = matrix([[0,0],[2,0.4],[4,0.9],[6,1.65],[8,2.60],[10,3.65],[12,4.85]])
Y1 = y1*M1/((1-y1)*M2) # [kg NO2/kg air]
# For 85% removal of the NO2,
Y2 = 0.15*Y1 # [kg NO2/kg air]
# Since the entering gel is free of NO2,
X2 = 0
Data_eqm =matrix([[0,0],[0.4,0.42],[0.9,0.83],[1.65,1.26],[2.6,1.69],[3.65,2.11],[4.85,2.54]])
# The operating line for minimum slope is tangent to curve, from which we get
X1_max = 0.0375 # [kg NO2/kg gel]
wb1 = 1/(1+Y1)
Vs = V1*wb1 # [mass velocity of the air, kg/s]
Ls_min = Vs*(Y1-Y2)/(X1_max-X2) # [kg gel/s]
Data_minSlope = matrix([[X2,Y2],[X1_max,Y1]])*100
# Operating line
Ls = 2*Ls_min # [kg gel/s]
X1 = X2 + Vs*(Y1-Y2)/Ls # [kg NO2/kg gel]
#Result
a1=plot(Data_eqm[:,0],Data_eqm[:,1],label='$Equilibrium line$')
a2=plot(Data_minSlope[:,0],Data_minSlope[:,1],label='$Minimum Flow Rate Line$')
a3=plot((3.75,3.75),(0,2.45),'--')
a4=plot((0,3.75),(2.45,2.45),'r--')
legend(loc='upper left')
xlabel("$Xa*100, kg NO2/kg gel$")
ylabel("$Ya*100, kh NO2/kg air$")
title('X-Y diagram for the adsorber')
annotate("X1(max)", xy=(3.75, 2.45), xytext=(4,2.5),
arrowprops=dict(facecolor='black', shrink=0.01),
)
show(a1)
show(a2)
show(a3)
show(a4)
print"Mass flow rate of the and the composition of the gel leaving the absorber are",round(Ls,3),"kg gel/s and",round(X1,4),"kg No2/kg gel"
Example 3.9,Page number:194¶
In [2]:
#Variable declaration
# 1-Nitrogen dioxide 2-air
# From Example 3.8
Y1 = 0.0242 # [kg NO2/kg air]
Y2 = 0.0036 # [kg NO2/kg air]
Vs = 0.488 # [kg air/s]
M1 = 46 # [gram/mole]
M2 = 29 # [gram/mole]
# However here
import matplotlib
%matplotlib inline
import math
from pylab import *
from numpy import *
X1 = 0
#Calculations
# Data_eqm1 = [P1 m] (where 'P1' is Partial pressure of NO2 in mm of Hg, 'm' is solid concentration in kg NO2/kg gel)
Data_eqm1 =matrix([[0,0],[2,0.4],[4,0.9],[6,1.65],[8,2.60],[10,3.65],[12,4.85]])
# The equilibrium data are converted to mass ratios as follows:
# Yi = P1/(760-P1)*46/29 (kg NO2/kg air) Xi = m/100 (kg NO2/kg gel)
# Equilibrium data
# Data_eqm = [Xi*100 Yi*100]
Data_eqm=matrix([[0,0],[0.4,0.4185242],[0.9,0.8392629],[1.65,1.2622336],[2.6,1.6874541],[3.65,2.1149425],[4.85,2.5447169]])
# From the intersection of the minimum operating line and equilibrium curve
X2_max = 0.0034 # [kg NO2/kg gel]
S = (Y1-Y2)/(X1-X2_max) # [kg gel/kg air]
Ls_min = -S*Vs # [kg/s]
Ls = 2*Ls_min # [kg/s]
Data_minSlope =matrix([[X1,Y1],[X2_max,Y2]])*100
#Result
a1=plot(Data_eqm[:,0],Data_eqm[:,1],label='$Equilibrium line $')
a2=plot(Data_minSlope[:,0],Data_minSlope[:,1],label='$Minimum Flow Rate Line$')
legend(loc='upper right')
xlabel("Xa*100, kg NO2/kg gel")
ylabel("Ya*100, kh NO2/kg air")
title('X-Y diagram for the cocurrent adsorber')
annotate("M", xy=(0.34,0.36), xytext=(0.34,0.36)
)
annotate("X2(max)", xy=(0.34,0.0), xytext=(0.34,0.01)
)
show(a1)
show(a2)
print"The mass velocity of the silica gel required for cocurrent operation is",round(Ls,2),"kg/s which is 11 times that required for countercurrent operation"
Example 3.10,Page number:199¶
In [20]:
#Variable declaration
# From Example 3.7
X2a = 0.05
X0 = X2a # [kmole benzene/kmole oil]
Y2a = 0.012
Y1 = Y2a # [kmole benzene/kmole dry gas]
X1a = 0.480
Xn = X1a # [kmole benzene/kmole oil]
Y1a = 0.080
Yn1 = Y1a # [kmole benzene/kmole dry gas]
# Ideal stages for absorber section
m = 0.097 # [mole of oil/mole of dry gas]
Lsa = 0.006 # [kmole/s]
Vsa = 0.038 # [kmole/s]
#Calculations
A = Lsa/(m*Vsa) # [Absorption factor]
import math
# From equation 3.54 by Kremser equation
Nk = math.log((((Yn1-m*X0)*(1-1/A))/(Y1-m*X0))+1/A)/(math.log(A))
# Ideal stages from graph
# Stair case construction is being made between equilibrium curve and operating line from piont X2a,Y2a to X1a,Y1a
# A more precise estimate of stages
# From figure 3.25 or from graph made for absorber in Example 3.7
Xa = 0.283
Xb = 0.480
Xc = 0.530
Na = 3+(Xb-Xa)/(Xc-Xa)
print"ABSORBER"
print"The number of ideal stages from graph in the absorber is",round(Na,1)
print"Number of ideal stages from Kremser equation in the absorber is",round(Nk,2)
# Ideal satges for stripping section
X2s = 0.480
X0 = X2s # [kmol benzene/kmol oil]
Y2s = 0.784
Y1 = Y2s # [kmol benzene/kmol steam]
X1s = 0.05
Xn = X1s # [kmol benzene/kmol oil]
Y1s = 0
Yn1 = Y1s # [kmol benzene/kmol steam]
# Similarly here also stair case construction is being made between equilibrium curve and operating line from piont X0,Y1 to Xn,Yn1
# A more precise estimate of stages
# From figure 3.26 or from graph made for stripping section in Example 3.7
Ns = 5+(0.070-0.050)/(0.070-0.028)
#Result
print"STRIPPER"
print"The number of ideal stages from graph in the stripping section is",round(Ns,1)