# 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'
# 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
# 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 );
# 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)]
# Additional condition
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
# 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
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