# Chapter 17 : Two Phase Gas Liquid Systems Saturation Condensation and Vaporization¶

### Example 17.1 Page no. 511¶

In :
# Variables
F = 1. ;			#H2C2O4- [mol]
ex_O2 =  248. ;		#Excess air- [%]
f_C = 65/100. ;		# Fraction of Carbon which convert to CO2
P = 101.3 ;			# Atmospheric pressure-[kPa]

# Calculations
O2_req = F*0.5 ;			# O2 required by the above reaction-[mol]
O2_in = (1. + ex_O2*F/100)*0.5 ;			# Mol. of O2 entering

# Use Elemental balance moles of species in output
n_CO2 = f_C*2 ;			# [mol]
n_H2O = (2*F)/2. ;		# From 2H balance-[mol]
n_N2 = ((O2_in*0.79)/(0.21)) ;			#  From 2N balance-[mol]
n_CO = 2-n_CO2 ;			             # From C balance-[mol]
n_O2 = ((4 + O2_in*2)-(n_H2O + n_CO + 2*n_CO2))/2 ;			# From O2 balance-[mol]
total_mol = n_CO2 + n_H2O + n_N2 + n_CO + n_O2 ;			# Total moles in output stream-[mol]
y_H2O = n_H2O/total_mol ;	# Mole fraction of H2O
pp_H2O = y_H2O*P ;			# Partial pressure of H2O-[kPa]

# Results
print 'Partial pressure of H2O %.2f kPa.'%pp_H2O
print 'Use partial pressure of H2O to get dew point temperature T from steam table: T  = 316.5 K'

Partial pressure of H2O 9.10 kPa.
Use partial pressure of H2O to get dew point temperature T from steam table: T  = 316.5 K


### Example 17.2 Page no. 517¶

In :
# Variables
gas =  1. ;			# Entering gas-[g mol]
T = 26. ;			# Temperature (for isothermal process)-[degree C]
vp =  99.7 ;			# vapour pressure of benzene at 26 C-[mm of Hg]

# Analysis of entering gas
f_C6H6 = 0.018 ;			# Mol fraction of benzene
f_air  =  0.982 ;			# Mol fraction of air
mol_C6H6 = 0.018*gas ;		# Moles  of benzene-[g mol]
mol_air  =  0.982*gas ;		# Moles  of air-[g mol]

# Calculations
# Analysis of exit gas
C6H6_rec =  95./100 ;			# Fraction of benzene recovered
C6H6_out  = 1-C6H6_rec ;		#Fraction of benzene in exit stream
C6H6_out = mol_C6H6*C6H6_out ;	#Moles of benzene in exit stream-[g mol]
air_out = mol_air ;			    #Moles of air in exit stream-[g mol]
total_mol = C6H6_out+air_out ;	# Total moles in exit stream
y_C6H6_out = C6H6_out/total_mol ;			# Mole fraction of benzene in exit
P = vp/y_C6H6_out ;			                # Pressure total of exit

# Results
print ' Pressure total at exit of compressor %.2e mm of Hg.'%P

 Pressure total at exit of compressor 1.09e+05 mm of Hg.


### Example 17.3 Page no. 519¶

In :
# Variables
ex_air = .4 ;			# Fraction of excess air required
w_C = 12. ;			    # Mol. wt. of C-[g]
mol_C = 71./w_C ;		#[kg mol]
w_H2 = 2.016 ;			# Mol. wt. of H2 - [g]
mol_H2 = 5.6/w_H2;
air_O2 = 0.21;			# Fraction of O2 in air
air_N2 = 0.79;			# Fraction of N2 in air

# Calculations
CO2_1 = 1. ;			#  By Eqn. (a) CO2 produced -[kg mol]
H2O_1 = 2. ;			# By Eqn. (a) H2O produced -[kg mol]
Req_O2_1 = 2. ;			# By Eqn. (a) -[kg mol]
ex_O2_1 = Req_O2_1*ex_air  ;			# Excess O2 required -[kg mol]
O2_1 = Req_O2_1 + ex_O2_1 ;			    # Total O2 required - [kg mol]
N2_1 = O2_1*(air_N2/air_O2) ;			#Total N2 required - [kg mol]
Total_1 = CO2_1 + H2O_1 + N2_1 + ex_O2_1 ;			# Total gas produced- [kg mol]

CO2_2 = 1 ;			#  By Eqn. (a) CO2 produced -[kg mol]
H2O_2 = mol_H2/mol_C ;			# By Eqn. (a) H2O produced -[kg mol]
Req_O2_2 = 1 + (mol_H2/mol_C)*(1./2) ;			# By Eqn. (b) and (c) -[kg mol]
ex_O2_2 = Req_O2_2*ex_air  ;			# Excess O2 required -[kg mol]
O2_2 = Req_O2_2 + ex_O2_2; 			# Total O2 required - [kg mol]
N2_2 = O2_2*(air_N2/air_O2); 			#Total N2 required - [kg mol]
Total_2 = CO2_2 + H2O_2 + N2_2 + ex_O2_2 ;			# Total gas produced- [kg mol]

P = 100. ;			# Total pressure -[kPa]
p1 = P*(H2O_1/Total_1) ;			# Partial pressure of water vapour in natural gas - [kPa]
Eq_T1 = 52.5  ;			# Equivalent temperature -[degree C]
p2 = P*(H2O_2/Total_2) ;			# Partial pressure of water vapour in coal - [kPa]
Eq_T2 = 35  ;			# Equivalent temperature -[degree C]

# Results
print '                                Natural gas                         Coal'
print '                            ----------------------             --------------------'
print 'Partial pressure:                %.1f kPa                            %.1f kPa'%(p1,p2 )
print 'Equivalent temperature:          %.1f C                              %.1f C'%(Eq_T1,Eq_T2 );

                                Natural gas                         Coal
----------------------             --------------------
Partial pressure:                14.0 kPa                            5.5 kPa
Equivalent temperature:          52.5 C                              35.0 C


### Example 17.4 Page no. 522¶

In :
# Variables
F = 30. ;			# Volume of initial gas-[m**3]
P_F =  98.6 ;			# Pressure of gas-[kPa]
T_F =  273.+100 ;			# Temperature of gas-[K]
P_p = 109. ;			#[kPa]
T_p = 14.+273 ;			# Temperature of gas-[K]
R = 8.314 ;			# [(kPa*m**3)/(k mol*K)]
vpW_30 = 4.24 ;			#Vapour pressure-[kPa]
vpW_14 = 1.60 ;			#Vapour pressure-[kPa]
n_F = (P_F*F)/(R*T_F) ;			# Number of moles in F

# Calculations
# Material balance to calculate  P & W
P = (n_F*((P_F-vpW_30)/P_F))/((P_p-vpW_14)/P_p) ;			# P from mat. bal. of air -[kg mol]
W = (n_F*(vpW_30/P_F))- P*(vpW_14/P_p); 			# W from mat. bal. of water -[kg mol]
iW = n_F*(vpW_30/P_F) ;			# Initial amount of water -[kg mol]
fr_con = W/iW ;			#Fraction of water condenseed

# Results
print ' Fraction of water condenseed %.3f.'%fr_con

 Fraction of water condenseed 0.668.


### Example 17.5 Page no. 527¶

In :
# Variables
P =  100. ;			# Pressure of air-[kPa]
T =  20. + 273 ;			# Temperature of air-[K]
R = 8.314 ;			# [(kPa*m**3)/(k mol*K)]
EOH =  6 ;			# Amount of ethyl alcohol to evaporate-[kg]
mw_EOH = 46.07 ;			# Mol.wt. of 1 k mol ethyl alcohol-[kg]

# Calculations
# Additional data needed
vp_EOH = 5.93 ;			# Partial pressure of alcohol at 20 C-[kPa]
vp_air = P-vp_EOH ;			# Partial pressure of air at 20 C-[kPa]
n_EOH  = EOH/mw_EOH ;			#Moles of ethyl alcohol -[kg mol]
n_air = (n_EOH*vp_air)/vp_EOH ;			# Moles of air -[kg mol]
V_air = n_air*R*T/P ;			# Volume of air required

# Results
print ' Volume of air required to evaporate 6 kg of ethyl alcohol is %.1f cubic metre . '%V_air

 Volume of air required to evaporate 6 kg of ethyl alcohol is 50.3 cubic metre .


### Example 17.6 Page no. 529¶

In :
import math

# Variables
P =  760. ;			# Pressure -[ mm of Hg]
vp = 40. ;			# vapour pressure of n-heptane-[mm of Hg]

# Calculations
K = 10**((math.log10(vp/P)-0.16)/1.25) ;
x = 0.5 ;			# mole fraction after  t_half
x0 = 1. ;			# initial mole fraction
t_half = (math.log(x/x0))/(-K);			# Time required to reduce the concentration to one-half-[min]

# Results
print 'Time required to reduce the concentration to one-half is %.1f min. '%t_half

Time required to reduce the concentration to one-half is 9.8 min.

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