# Chapter 4:Equipment for Gas-Liquid Mass-Transfer Operations¶

### Example 4.2,Page number:227¶

In :
#Variable declaration
u = 3*10**-6  				# [Kinematic viscosity, square m/s]
v = 0.01  				# [Superficial liquid velocity, m/s]
g = 9.8  				# [square m/s]

# From table 4.1
# For metal pall rings
a_pr = 112.6  				# [ square m/cubic m]
e_pr = 0.951
Ch_pr = 0.784
# For Hiflow rings
a_hr = 92.3  				# [square m/cubic m]
e_hr = 0.977
Ch_hr = 0.876

#Calculations

# Renoylds and Froude's number for metal pall rings
Rel_pr = v/(u*a_pr)
Frl_pr = v**2*a_pr/g
# From equation 4.5 since Rel is greater than 5, for pall rings
# ah/a = x_pr
x_pr = 0.85*Ch_pr*Rel_pr**0.25*Frl_pr**0.1
# From equation 4.3
hl_pr = (12*Frl_pr/Rel_pr)**(1.0/3.0)*(x_pr)**(2.0/3.0)

# Renoylds and Froude's number for Hiflow rings
Rel_hr = v/(u*a_hr)
Frl_hr = v**2*a_hr/g
# From equation 4.5 since Rel is greater than 5, for pall rings
# ah/a = x_pr
x_hr = 0.85*Ch_hr*Rel_hr**0.25*Frl_hr**0.1
# From equation 4.3
hl_hr = (12*Frl_hr/Rel_hr)**(1.0/3.0)*(x_hr)**(2.0/3.0)

#Result

print"The specific liquid holdup for Metal pall ring and Hiflow ring are",round(hl_pr,3),"cubic m holdup/cubic m packed bed and",round(hl_hr,3),"cubic m holdup/cubic m packed bed respectively"


The specific liquid holdup for Metal pall ring and Hiflow ring are 0.066 cubic m holdup/cubic m packed bed and 0.064 cubic m holdup/cubic m packed bed respectively


### Example 4.3,Page number:233¶

In :
#Variable declaration
# a-ammonia   b-air   c-water
P = 101.3  					# [kPa]
T = 293  					# [K]
R = 8.314
Vb = 20  					# [kmole/h]
xab = 0.05
Vc = 1500  					# [kg/h]
d = 0.9  					# [ammonia absorbed]
Ma = 17  					# [gram/mole]
Mb = 29  					# [gram/mole]
Mc = 18  					# [gram/mole]
g = 9.8  					# [square m/s]

#Calculation

import math
# For Inlet gas
Mg = (1-xab)*Mb+xab*Ma  			# [gram/mole]
V = Vb*Mg/3600  				# [kg/h]
rowg = P*Mg/(R*T)  				# [kg/cubic m]
Qg = V/rowg  					# [cubic m/s]

# For exiting liquid
b = Vb*xab*Ma*d  				# [ammonia absorbed in kg/h]
L = (Vc+b)/3600  				# [kg/s]
rowl = 1000  					# [kg/cubic m]

X = (L/V)*(math.sqrt(rowg/rowl))
# From equation 4.8
Yflood = math.exp(-(3.5021+1.028*math.log(X)+0.11093*(math.log(X))**2))

#Illustration 4.3(a)
# Solution(a)
# For 25-mm ceramic Raschig rings
Fp = 179  					# [square ft/cubic ft]
ul = 0.001  					# [Pa.s]
# From equation 4.6
Csflood = math.sqrt(Yflood/(ul**0.1*Fp))  	# [m/s]
# From equation 4.7
vgf = Csflood/(math.sqrt(rowg/(rowl-rowg)))  	# [m/s]
# From equation 4.9
deltaPf = 93.9*(Fp)**0.7  			# [Pa/m of packing]

# For operation at 70% of the flooding velocity
f = 0.7
# From equation 4.10
vg = f*vgf  					# [m/s]
D = math.sqrt(4*Qg/(vg*math.pi))

# From Table 4.1, for 25 mm ceramic Raschig rings
a_c = 190  					# [square m/cubic m]
Ch_c = 0.577
e_c = 0.68
Cp_c = 1.329

# From equation 4.13
dp = 6*(1-e_c)/a_c 				 # [m]
# From equation 4.12
Kw = 1/(1+(2*dp/(3*D*(1-e_c))))

# The viscosity of the gas phase is basically that of air at 293 K and 1 atm
ug = 1.84*10**-5  				# [kg/m.s]
# From equation 4.15
Reg = vg*rowg*dp*Kw/(ug*(1-e_c))
# From equation 4.14
sia_o = Cp_c*((64/Reg)+(1.8/(Reg**0.08)))

# From equation 4.11
# deltaP_o/z = T
T = sia_o*a_c*rowg*vg**2/(2*Kw*e_c**3)  	# [Pa/m]

# Now
Gx = L/((math.pi*D**2.0)/4)  			# [kg/square m.s]
Rel = Gx/(a_c*ul)
Frl = Gx**2*a_c/(rowl**2*g)

# From equation 4.5
# ah/a = x_pr
x = 0.85*Ch_c*Rel**0.25*Frl**0.1
# From equation 4.3
hl = (12*Frl/Rel)**(1.0/3.0)*(x)**(2.0/3.0)

# From equation 4.16
# daltaP/deltaP_o = Y
Y = (e_c/(e_c-hl))**1.5*math.exp(Rel/200)
# Therefore
# deltaP/z = H
H = Y*T  					# [Pa/m]

#Result

print"The superficial velocity is",round(vgf,3)," m/s\n"
print"The pressure drop at flooding is",round(deltaPf),"Pa/m\n"
print"The superficial velocity at 70 percent of flooding is",round(vg,4),"m/s\n"
print"The column inside diameter at 70 percent of flooding is" ,round(D,3)," m\n"
print"The pressure drop for operation at 70 percent of flooding is",round(H),"Pa/m\n\n"

#Illustration 4.3(b)
# Solution (b)
# Similarly for 25 mm metal Hiflow rings above quantities are determined

#Calcualtion

Fp1 = 42  						# [square ft/cubic ft]
Csflood1 = math.sqrt(Yflood/(ul**0.1*Fp1))  		# [m/s]
vgf1 = Csflood1/(math.sqrt(rowg/(rowl-rowg)))  		# [m/s]
# From equation 4.9
deltaPf1 = 93.9*(Fp1)**0.7  				# [Pa/m of packing]

# For operation at 70% of the flooding velocity
f = 0.7
# From equation 4.10
vg1 = f*vgf1  						# [m/s]
D1 = math.sqrt(4*Qg/(vg1*math.pi))

# For Hiflow rings
a_h = 202.9  						# [square m/cubic m]
e_h = 0.961
Ch_h = 0.799
Cp_h = 0.689

# From equation 4.13
dp1 = 6*(1-e_h)/a_h  					# [m]
# From equation 4.12
Kw1 = 1/(1+(2*dp1/(3*D1*(1-e_h))))

# The viscosity of the gas phase is basically that of air at 293 K and 1 atm
ug = 1.84*10**-5  					# [kg/m.s]
# From equation 4.15
Reg1 = vg1*rowg*dp1*Kw1/(ug*(1-e_h))
# From equation 4.14
sia_o1 = Cp_h*((64/Reg1)+(1.8/(Reg1**0.08)))

# From equation 4.11
# deltaP_o/z = T
T1 = sia_o1*a_h*rowg*vg1**2/(2*Kw1*e_h**3)  		# [Pa/m]

# Now
Gx1 = L/(math.pi*D1**2/4)  				# [kg/square m.s]
Rel1 = Gx1/(a_h*ul)
Frl1 = Gx1**2*a_h/(rowl**2*g)

# From equation 4.5
# ah/a = x_pr
x1 = 0.85*Ch_h*Rel1**0.25*Frl1**0.1
# From equation 4.3
hl1 = (12*Frl1/Rel1)**(1.0/3.0)*(x1)**(2.0/3.0)

# From equation 4.16
# daltaP/deltaP_o = Y
Y1 = (e_h/(e_h-hl1))**1.5*math.exp(Rel1/200)
# Therefore
# deltaP/z = H
H1 = Y1*T1  # [Pa/m]

#Result

print"The superficial velocity is",round(vgf1,3),"m/s\n"
print"The pressure drop at flooding is",round(deltaPf1)," Pa/m\n"
print"The superficial velocity at 70 percent of flooding is",round(vg1,2),"m/s"
print"The column inside diameter at 70 percent of flooding is",round(D1,3)," m\n"
print"The pressure drop for operation at 70 percent of flooding is",round(H1),"Pa/m\n\n"

The superficial velocity is 1.326  m/s

The pressure drop at flooding is 3545.0 Pa/m

The superficial velocity at 70 percent of flooding is 0.9283 m/s

The column inside diameter at 70 percent of flooding is 0.428  m

The pressure drop for operation at 70 percent of flooding is 505.0 Pa/m

The superficial velocity is 2.738 m/s

The pressure drop at flooding is 1285.0  Pa/m

The superficial velocity at 70 percent of flooding is 1.92 m/s
The column inside diameter at 70 percent of flooding is 0.298  m

The pressure drop for operation at 70 percent of flooding is 443.0 Pa/m



### Example 4.4,Page number:237¶

In :
#Variable declaration
# a-ethanol   b- gas(CO2 rich vapor)   c-liquid water
P = 110  						# [kPa]
T = 303 	 					# [K]
R = 8.314
Vb = 180  						# [kmole/h]
xab = 0.02  						# [molar composition of ethanol in gas]
Vc = 151.5  						# [kmole/h]
d = 0.97  						# [ethanol absorbed]
Ma = 46  						# [gram/mole]
Mb = 44  						# [gram/mole]
Mc = 18  						# [gram/mole]
g = 9.8  						# [square m/s]

#Calculation

import math
# For Inlet gas
Mg = (1-xab)*Mb+xab*Ma  				# [gram/mole]
V = Vb*Mg/3600  					# [kg/h]
rowg = P*Mg/(R*T)  					# [kg/cubic m]
Qg = V/rowg  						# [cubic m/s]

# For exiting liquid
b = Vb*xab*Ma*d  					# [ethanol absorbed in kg/h]
L = (Vc*Mc+b)/3600  					# [kg/s]
rowl = 986  						# [kg/cubic m]

X = (L/V)*(math.sqrt(rowg/rowl))
# From equation 4.8
Yflood = math.exp(-(3.5021+1.028*math.log(X)+0.11093*(math.log(X))**2))
#Illustration 4.4(a)
# Solution(a)

# For 50 mm metal Hiflow rings
Fp = 16  						# [square ft/cubic ft]
ul = 6.31*10**-4  					# [Pa.s]
# From equation 4.6
Csflood = math.sqrt(Yflood/(ul**0.1*Fp))  		# [m/s]
# From equation 4.7
vgf = Csflood/(math.sqrt(rowg/(rowl-rowg)))  		# [m/s]
# From equation 4.9
deltaPf = 93.9*(Fp)**0.7  				# [Pa/m of packing]

# For operation at 70% of the flooding velocity
f = 0.7
# From equation 4.10
vg = f*vgf  						# [m/s]
D = math.sqrt(4*Qg/(vg*math.pi))

# From Table 4.1, for 50 mm metal Hiflow rings
a = 92.3 						 # [square m/cubic m]
Ch = 0.876
e = 0.977
Cp = 0.421

# From equation 4.13
dp = 6*(1-e)/a  					# [m]

# From equation 4.12
Kw = 1/(1+(2*dp/(3*D*(1-e))))

# The viscosity of the gas phase is basically that of air at 303 K and 110 kPa
ug = 1.45*10**-5  					# [kg/m.s]
# From equation 4.15
Reg = vg*rowg*dp*Kw/(ug*(1-e))
# From equation 4.14
sia_o = Cp*((64/Reg)+(1.8/(Reg**0.08)))

# From equation 4.11
# deltaP_o/z = I
I = sia_o*a*rowg*vg**2/(2*Kw*e**3)  			# [Pa/m]

# Now
Gx = L/(math.pi*D**2.0/4.0)  				# [kg/square m.s]
Rel = Gx/(a*ul)
Frl = Gx**2*a/(rowl**2*g)

# From equation 4.5
# ah/a = x
x = 0.85*Ch*Rel**0.25*Frl**0.1
# From equation 4.3
hl = (12*Frl/Rel)**(1.0/3.0)*(x)**(2.0/3.0)

# From equation 4.16
# daltaP/deltaP_o = Y
Y = (e/(e-hl))**1.5*math.exp(Rel/200)
# Therefore
# deltaP/z = H
H = Y*I  						# [Pa/m]

#Result
print"PArt-a\n"

print"Since the pressure drop is too high, we must increase the tower diameter to reduce the pressure drop"
# The resulting pressure drop is too high  therefore, we must increase the tower diameter 	to reduce the pressure drop. Appendix D presents a Mathcad computer
# program designed to iterate automatically until the pressure drop criterion is 			satisfied.
# From the Mathcad program we get
D1 = 0.738  						# [m]

print"The tower diameter for pressure drop of 300 Pa/m of packed height is",D1,"m"

#Illustration 4.4(b)
# Solution(b)

print"\nPart-(b)\n"
# For the tower diameter of D = 0.738 m, the following intermediate results were obtained 	from the computer program in Appendix D:
vg1 = 2.68  						# [m/s]
vl1 = 0.00193  						# [m/s]
hl1 = 0.017
ah1 = 58.8  						# [square m/cubic m]
Reg1 = 21890
Rel1 = 32.6
Kw1 = 1/(1+(2*dp/(3*D1*(1-e))))

#Calculation

f1 = vg1/vgf
print"The fractional approach to flooding conditions is ",round(f1,2)

#Illustration 4.4(c)
# Solution(c)
# For ethanol

print "\nPart-(c)\n"
Vc_a = 167.1  						# [cubic cm/mole]
sigma_a = 4.53*10**-10  				# [m]
# E/k = M
M_a = 362.6  						# [K]

# For carbon dioxide
sigma_b = 3.94*10**-10  				# [m]
M_b = 195.2  						# [K]

# From equation 1.48
Vb_a = 0.285*Vc_a**1.048  				# [cubic cm/mole]

e1 = (9.58/(Vb_a)-1.12)
# From equation 1.53
Dl = 1.25*10**-8*((Vb_a)**-0.19 - 0.292)*T**1.52*(ul*10**3)**e1 	 # [square cm/s]

# From equation 1.49
Dg = 0.085  						# [square cm/s]

# From Table 4.2, for 50 mm metal Hiflow rings
Cl = 1.168
Cv = 0.408
# From equation 4.17
kl = 0.757*Cl*math.sqrt(Dl*a*vl1*10**-4/(e*hl1))  	# [m/s]
mtcl = kl*ah1  						# [s**-1]

Sc = ug/(rowg*Dg*10**-4)
# From equation 4.18
ky = 0.1304*Cv*(Dg*10**-4*P*1000/(R*T))*(Reg1/Kw1)**(3.0/4.0)*Sc**(2.0/3.0)*(a/(math.sqrt(e*(e-hl1)))) 							 # [mole/square m.s]
mtcg = ky*ah1*10**-3  					 # [kmole/cubic m.s]

#Result

print"The gas and liquid volumetric mass transfer coefficients are",round(mtcg,3),"kmole/cubic m.s and",round(mtcl,4),"s**-1 respectively"

PArt-a

Since the pressure drop is too high, we must increase the tower diameter to reduce the pressure drop
The tower diameter for pressure drop of 300 Pa/m of packed height is 0.738 m

Part-(b)

The fractional approach to flooding conditions is  0.58

Part-(c)

The gas and liquid volumetric mass transfer coefficients are 0.192 kmole/cubic m.s and 0.0074 s**-1 respectively


### Example 4.5,Page number:245¶

In :
#Variable declaration
#  a-chloroform    b-water    c-air
T = 298  						# [K]
Dv = 1  						# [vessel diameter, m]
Vb = 10  						# [kg/s]
ca = 240*10**-6  					# [gram/l]
xr = 0.9  						# [chloroform which is to be removed]
m = 220
Ds = 0.5  						# [diameter of sparger, m]
no = 90  						# [number of orifices]
Do = 3*10**-3  						# [diameter of orifice, m]
nm = 0.6  						# [mechanical efficiency]
rowb = 1000  						# [kg/cubic m]
R = 8.314
Mc = 29  						# [gram/mole]
Mb = 18  						# [gram/mole]
g = 9.8 	 					# [square m/s]

#Calculation

import math
from scipy.optimize import fsolve
Vair = 0.1  						# [kg/s as calculated in chapter 3]
mg = Vair/no  						# [mass flow rate through each orifice, 							kg/s]
ug = 1.8*10**-5  					# [kg/m.s]
Reo = 25940  						# [Renoylds number]
# From equ. 4.20
dp = 0.0071*Reo**-0.05  				# [m]

# Since the water column height is not known, therefore an iterative procedure must be 		implemented.
# Assuming column height, Z = 0.5 m
Z = 0.5  						# [m]
# For Z = 0.5 m
rowl = 1000  						# [kg/cubic m]
Ps = 101.3  						# [kPa]
Po = Ps+(1000*9.8*0.5/1000)  				# [kPa]
Pavg = (Po+Ps)/2  					# [kPa]
rowg = Pavg*Mc/(R*T)  					# [kg/cubic m]

area = math.pi*Dv**2.0/4.0  				# [square m]
vg = Vair/(rowg*area)  					# [m/s]
# In this case rowl = rowg and sigma = sigmaAW
# From equation 4.22
# Vg = vg
# vg/vs = 0.182
vs = vg/0.182  						# [m/s]
vl = -Vb/(rowl*area)  					# [negative because water flows downward, 							m/s]
# From equ 4.21

def f12(phig):
return(vs - (vg/phig)-(-vl/(1-phig)))
phig = fsolve(f12,0.1)
# Now in this case
S = vl/(1-phig)
# Value of 'S' comes out to be less than 0.15 m/s
# Therefore
dp = (dp**3*Po/Pavg)**(1.0/3.0)  			# [m]
# From equ 4.23
a = 6*phig/dp  						# [m**-1]
# Now we calculate diffusivity of chloroform
Vba = 88.6  						# [cubic cm/mole]
u = 0.9*10**-3  					# [Pa-s]
e = (9.58/(Vba)-1.12)
# From equation 1.53
Dl = 1.25*10**-8*((Vba)**-0.19 - 0.292)*T**1.52*(u*10**3)**e  	# [square cm/s]

# And Schmidt number is
Scl = 833  						# [Schmidt Number]

# Now we calculate  dp*g**(1/3)/Dl**(2/3) = J
J = dp*g**(1.0/3.0)/(Dl*10**-4)**(2.0/3.0)
Reg = dp*vs*rowl/u  					# [Gas bubble Renoylds number]
# From equ 4.25
Shl = 2 + 0.0187*Reg**0.779*Scl**0.546*J**0.116

# For dilute solution xbm = 1 or c = 55.5 kmole/cubic m
# Then for Nb = 0
c = 55.5  						# [kmole/cubic m]
kx = Shl*c*Dl*10**-4/dp  				# [kmole/square m.s]
mtc = kx*a  						# [kmole/cubic m.s]

L = Vb/Mb  						# [kmole/s]
Gmx = L/area  						# [kmole/square m.s]
V = Vair/Mc  						# [kmole/s]
A = L/(m*V)  						# [absorption factor]

# From equ 4.28
# For, xin/xout = x = 10
x = 10
Z = (Gmx/(kx*a*(1-A)))*math.log(x*(1-A)+A)

# With this new estimated Z ,we again calculate average pressure in the   # column of 	water
Po1 = 110.1  						# [kPa]
Pavg1 = 105.7  						# [kPa]
rowg1 = Pavg1*Mc/(R*T)
# Now value of rowg1 obtained is very close to value used in the first   # iteration. 	Therefore on three iteractions we achieve a value of 'Z'
Z1 = 0.904  						# [m]

rowgo = Po1*Mc/(R*T)  					# [kg/cubic m]
vo1 = 4*mg/(math.pi*Do**2*rowgo)  			# [m/s]
# Therefore,    vo1**2/(2*gc) = F
gc = 1
F = vo1**2/(2*gc)  					# [J/kg]
# And R*T*math.log(Po/Ps)/Mc = G
G = R*T*1000*math.log(Po1/Ps)/Mc  			# [J/kg]
Zs = 0
# And   (Z1-Zs)*g/gc = H
H = (Z1-Zs)*g/gc  					# [J/kg]
# From equ 4.27
W = F+G+H  						# [J/kg]
# Now the air compressor power is
W1 = W*Vair*10**-3/nm  					# [kW]

#Result

print"The depth of the water column required to achieve the specified 90percent removal efficiency is ",Z1,"m"

print"The power required to operate the air compressor is",round(W1,2),"kW"

The depth of the water column required to achieve the specified 90percent removal efficiency is  0.904 m
The power required to operate the air compressor is 2.43 kW


### Example 4.6,Page number:255¶

In :
#Variable declaration
Ff = 0.9  					# [foaming factor]
sigma = 70  					# [liquid surface tension, dyn/cm]
Do = 5  					# [mm]
#From Example 4.4
# X = 0.016
p = 15 						# [pitch, mm]
# From equ 4.35
# Ah/Aa	 = A
A = 0.907*(Do/p)**2  				# [ratio of vapor hole area to tray active area]

#Calculations

import math
from scipy.optimize import fsolve
# Assume
t = 0.5 					 # [m]
# From equ 4.32
alpha = 0.0744*t+0.01173
beeta = 0.0304*t+0.015

# Since X<0.1, therefore
X = 0.1
# From equ 4.31
Cf = alpha*math.log10(1.0/X) + beeta
# Since Ah/Aa > 0.1, therefore
Fha = 1
Fst = (sigma/20)**0.2  				# [surface tension factor]
# From equ 4.30
C = Fst*Ff*Fha*Cf

# From Example 4.4
rowg = 1.923  					# [kg/cubic m]
rowl = 986  					# [kg/cubic m]
Qg = 1.145  					# [cubic m/s]
# From equation 4.29
vgf = C*(math.sqrt((rowl-rowg)/rowg))  		# [m/s]
# Since X<0.1
# Equ 4.34 recommends Ad/At = B = 0.1
B = 0.1
# For an 80% approach to flooding, equation 4.33 yields
f = 0.8
D =math.sqrt((4*Qg)/(f*vgf*math.pi*(1-B)))  	# [m]
# At this point, the assumed value of tray spacing ( t = 0.5 m) must be   # checked 	against the recommended values of Table 4.3. Since the calculated
# value of D < 1.0 m, t = 0.5 m is the recommended tray spacing, and no
# further iteration is needed.

def f14(Q):
return(B-((Q-math.sin(Q))/(2*math.pi)))
Q = fsolve(f14,1.5)
Lw = D*math.sin(Q/2)  				# [m]
rw = D/2*math.cos(Q/2)  			# [m]

At = math.pi/4*D**2  				# [total cross sectional area, square m]
Ad = B*At  					# [Downcomer area, square m]
Aa = At-2*Ad  					# [ Active area over the tray, square m]
Ah = 0.101*Aa  					# [Total hole area, square m]

#Result

print"Summarizing, the details of the sieve-tray design are as follows"

print"Diameter =",round(D,3),"m\n Tray spacing =",t," m\n Total cross-sectional area=",round(At,3),"square m\n Downcomer area =",round(Ad,3)," square m\n Active area over the tray = ",round(Aa,3)," square m\n Weir length =",round(Lw,3)," m\n Distance from tray center to weir = ",round(rw,2)," m\n Total hole area =",round(Ah,3)," square m\n Hole arrangement: 5 mm diameter on an equilateral-triangular pitch 15 mm between hole centers, punched in stainless steel sheet metal 2 mm thick"

Summarizing, the details of the sieve-tray design are as follows
Diameter = 0.989 m
Tray spacing = 0.5  m
Total cross-sectional area= 0.768 square m
Downcomer area = 0.077  square m
Active area over the tray =  0.615  square m
Weir length = 0.719  m
Distance from tray center to weir =  0.34  m
Total hole area = 0.062  square m
Hole arrangement: 5 mm diameter on an equilateral-triangular pitch 15 mm between hole centers, punched in stainless steel sheet metal 2 mm thick


### Example 4.7,Page number:257¶

In :
#Variable declaration
Do = 5  						# [mm]
g = 9.8  						# [square m/s]
hw = 50  						# [mm]
# From example 4.4
Qg = 1.145  						# [cubic m/s]
# From example 4.6
Ah = 0.062 						 # [square m]
# Do/l = t = 5/2 = 2.5
t = 2.5
# Ah/Aa = A = 0.101
A = 0.101
rowg = 1.923  						# [kg/cubic m]
rowl = 986  						# [kg/cubic m]
roww = 995  						# [kg/cubic m]

#Calculation

import math
vo = Qg/Ah  						# [m/s]
# From equation 4.39
Co = 0.85032 - 0.04231*t + 0.0017954*t**2 	 	# [for t>=1]
# From equation 4.38
hd = 0.0051*(vo/Co)**2*rowg*(roww/rowl)*(1-A**2)  	# [cm]

# From example 4.6
Aa = 0.615  						# [square m]
va = Qg/Aa  						# [m/s]

# From equation 4.41
Ks = va*math.sqrt(rowg/(rowl-rowg))  			# [m/s]
phie = 0.274

# From equation 4.4
ql = 0.000815  						# [cubic m/s]

# From example 4.6
Lw = 0.719  						# [m]
Cl = 50.12 + 43.89*math.exp(-1.378*hw)
sigma = 0.07  						# [N/m]
# From eqution 4.40
hl = phie*(hw*10**(-1)+Cl*(ql/(Lw*phie))**(2.0/3.0))

# From equation 4.42
ho = 6*sigma/(g*rowl*Do*10**-3)*10**2  			# [cm]
# From equation 4.37
ht = hd+hl+ho  						# [cm of clear liquid/tray]
deltaPg = ht*g*rowl*10**-2  				# [Pa/tray]

#Result
print"The tray gas-pressure drop for the ethanol is ",round(deltaPg),"Pa/tray"

The tray gas-pressure drop for the ethanol is  811.0 Pa/tray


### Example 4.8,Page number:259¶

In :
#Variable declaration
Do = 5*10**-3  						# [m]
rowg = 1.923  						# [kg/cubic m]
rowl = 986.0  						# [kg/cubic m]
g = 9.8  						# [square m/s]
hl = 0.0173  						# [m]
vo = 18.48  						# [m/s]
phie = 0.274
Ks = 0.082  						# [m]
A = 0.101  						# [Ah/Aa]
t = 0.5  						# [m]

#Calculation

import math
Fr = math.sqrt(rowg*vo**2/(rowl*g*hl))  		# [Froude number]
if Fr>0.5:
print"Weeping is not significant"
else:
print"Significant weeping occurs"

# From above weeping is not a problem under this circumstances
# From equation 4.47
k = 0.5*(1-math.tanh(1.3*math.log(hl/Do)-0.15))

# From equation 4.46
h2q = (hl/phie) + 7.79*(1+6.9*(Do/hl)**1.85)*(Ks**2/(phie*g*A))  	# [m]
# From equation 4.45
E = 0.00335*(h2q/t)**1.1*(rowl/rowg)**0.5*(hl/h2q)**k
# From Example 4.4, the gas mass flow rate is V = 2.202 kg/s
V = 2.202  								# [kg/s]
Le = E*V 								 # [kg/s]

#Result

print"The entrainment flow rate for the ethanol absorber is",round(Le,2),"kg/s"

Weeping is not significant
The entrainment flow rate for the ethanol absorber is 0.11 kg/s


### Example 4.9,Page number:264¶

In :
#Variable declaration
Do = 5*10**-3  						# [m]
Ml = 18.63  						# [molecular weight of water, gram/mole]
Mg = 44.04  						# [molecular weight of carbon dioxide, 							gram/mole]
rowg = 1.923  						# [kg/cubic m]
rowl = 986  						# [kg/cubic m]
vo = 18.48  						# [m/s]
hl = 0.0173  						# [m]
ug = 1.45*10**-5  					# [kg/m.s]
phie = 0.274
A = 0.101  						# [Ah/Aa]
Dg = 0.085  						# [square cm/s]
Dl = 1.91*10**-5  					# [square cm/s]
Aa = 0.614  						# [square m]
Qg = 1.145  						# [cubic m/s]
t = 0.5  						# [m]
h2q = 0.391  						# [m]
rw = 0.34  						# [m]
ql = 0.000815  						# [cubic m/s]
g = 9.8  						# [square m/s]
G = 2.202/44.04  					# [kg/s]
L = 0.804/18.63  					# [kg/s]

Refe = rowg*vo*hl/(ug*phie)

#Calculation

import math
cg =rowg/Mg  						# [kmole/cubic m]
cl = rowl/Ml  						# [kmole/cubic m]

# For the low concentrations prevailing in the liquid phase, the ethanol-   # water 	solution at 303 K obeys Henry's law, and the slope of the equilibriu# m curve is m = 0.57
m = 0.57
# From equation 4.53
a1 = 0.4136
a2 = 0.6074
a3 = -0.3195
Eog = 1-math.exp(-0.0029*Refe**a1*(hl/Do)**a2*A**a3/((math.sqrt(Dg*(1-phie)/(Dl*A)))*m*cg/cl+1))
# From equation 4.62
Deg = 0.01  						# [square m/s]
Peg = 4*Qg*rw**2/(Aa*Deg*(t-h2q))  			# [Peclet number]
# Since Peclet number is greater than 50, therefore vapor is unmixed
# From equation 4.60
Del = 0.1*math.sqrt(g*h2q**3)  				# [square m/s]
# From equation 4.59
Pel = 4*ql*rw**2/(Aa*hl*Del)
N = (Pel+2)/2
lamda = m*G/L
# From equation 4.58
Emg = ((1+lamda*Eog/N)**N -1)/lamda*(1-0.0335*lamda**1.073*Eog**2.518*Pel**0.175)
# From example 4.8
E = 0.05
# Substituting in equation 4.63
Emge = Emg*(1-0.8*Eog*lamda**1.543*E/m)

#Result

print"The entrainment corrected Murphree tray efficiency for the ethanol is",round(Emge,3)

The entrainment corrected Murphree tray efficiency for the ethanol is 0.812