# variables
# Given
#Main eqn. C6H12O6 + aO2 ---> bCO2 + cH2O
# By carbon balance
b = 6 ;
#By hydrogen balance
c = 6;
# calculation
#Balancing oxygen in reaction
a = (c*1+b*2-6)/2.0;
#result
print 'Value of a is %i'%a
print 'Value of b is %i'%b
print 'Value of c is %i'%c
# Variables
m_CO2 = 44.0 ; #molecular wt-[g]
m_C7H16 = 100.1 ; #molecular wt-[g]
p_con = 50. ; # percentage conversion of CO2 to dry ice
amt_di = 500. ; # amount of dry ice to be produce per hour-[kg]
# Calculation
# By using the given equation
amt_C7H16 = (amt_di*m_C7H16)/((p_con/100.)*m_CO2*7) ;# [kg]
# Result
print 'Amount of heptane required per hour to produce 500kg dry ice per hour is %.1f kg.'%amt_C7H16
# Variables
m_CaCO3 = 100.1 ; #molecular wt-[g]
m_MgCO3 = 84.32 ; #molecular wt-[g]
m_CaO = 56.08 ; #molecular wt-[g]
m_MgO = 40.32 ; #molecular wt-[g]
m_CO2 = 44.0 ; #molecular wt-[g]
# Limestone analysis
p_CaCO3 = 92.89 ; # percentage of CaCO3
p_MgCO3 = 5.41 ; # percentage of MgCO3
inrt = 1.7 ; #percentage of inert
# Calculation and Results
#(a)
amt_CaO = (((p_CaCO3/100)*m_CaO)/m_CaCO3)*2000 ; #Pounds of CaO produced from 1 ton(2000lb) of limestone
print ' Amount of CaO produced from 1 ton(2000lb) of limestone is %.0f lb.'%amt_CaO
#(b)
mol_CaCO3 = (p_CaCO3/100)/m_CaCO3 ; # lb mol of CaCO3
mol_MgCO3 = (p_MgCO3/100)/m_MgCO3 ; # lb mol of MgCO3
total_mol = mol_CaCO3+mol_MgCO3;
amt_CO2 = total_mol*m_CO2 ; # Amount of CO2 recovered per pound of limestone-[lb]
print ' Amount of CO2 recovered per pound of limestone is %.3f lb.'%amt_CO2
#(c)
amt_CaO = m_CaO*mol_CaCO3 ; # since lb mol of CaO = CaCO3
amt_MgO = m_MgO*mol_MgCO3 ; # since lb mol of MgO = MgCO3
total_lime = amt_CaO+amt_MgO+(inrt)/100 ; # total amount of lime per pound limestone
amt_lmst = 2000/total_lime ; # Amount of limestone required to make 1 ton(2000lb) of lime
print ' Amount of limestone required to make 1 ton(2000lb) of lime %.1f lb.'%amt_lmst
# Variables
f_NH3 = 5. ; # NH3 in feed-[g]
f_N2 = 100. ; # N2 in feed-[g]
f_H2 = 50. ; # H2 in feed-[g]
p_NH3 = 90. # NH3 in product-[g]
m_NH3 = 17. ; # Molecular wt. of NH3-[g]
m_N2 = 28. ; # Molecular wt. of N2-[g]
m_H2 = 2. ; # Molecular wt. of H2-[g]
# Calculations
# Extent of reaction can be calculated by using eqn. 9.3
# For NH3
ni = p_NH3/m_NH3 ; #[g mol NH3]
nio = f_NH3/m_NH3 ; #[g mol NH3]
vi = 2 ; # coefficint of NH3
ex_r = (ni-nio)/vi # Extent of reaction - moles reacting
#Determine H2 and N2 in product of reaction by Eqn. 9.4
# For N2
nio_N2 = f_N2/m_N2 ; #[g mol N2]
vi_N2 = -1 ; # coefficint of N2
ni_N2 = nio_N2 + vi_N2*ex_r ; #N2 in product of reaction-[g moles ]
m_N2 = ni_N2*m_N2 ; # mass of N2 in product of reaction-[g]
# Results
print ' N2 in product of reaction is %.2f g moles '%ni_N2
print ' Mass of N2 in product of reaction is %.2f g '%m_N2
# For H2
nio_H2 = f_H2/m_H2 ; #[g mol H2]
vi_H2 = -3 ; # coefficint of H2
ni_H2 = nio_H2 + vi_H2*ex_r ; #H2 in product of reaction-[g moles ]
m_H2 = ni_H2*m_H2 ; # mass of H2 in product of reaction-[g]
print ' H2 in product of reaction is %.2f g moles '%ni_H2
print ' Mass of H2 in product of reaction is %.2f g '%m_H2
# ARP
m_SO2 = 64. ; # Molecular wt.of SO2-[g]
mol_SO2 = 2. ; # moles of SO2
ARP = (1./m_NH3)/(mol_SO2/m_SO2);
print ' ARP is %.2f '%ARP
# Variables
f_N2 = 10. ; # N2 in feed-[g]
f_H2 = 10. ; # H2 in feed-[g]
m_NH3 = 17.02 # Molecular wt. of NH3-[g]
m_N2 = 28. ; # Molecular wt. of N2-[g]
m_H2 = 2. ; # Molecular wt. of H2-[g]
# Calculations
# Extent of reaction can be calculated by using eqn. 9.3
# Based on N2
nio_N2 = f_N2/m_N2 #[g mol N2]
vi_N2 = -1 ; # coefficint of N2
ex_N2 = -(nio_N2)/vi_N2 ; # Max. extent of reaction based on N2
# Based on H2
nio_H2 = f_H2/m_H2 ; #[g mol H2]
vi_H2 = -3 ; # coefficint of H2
ex_H2 = -(nio_H2)/vi_H2 # Max. extent of reaction based on H2
#(a)
vi_NH3 = 2 ; # coefficint of NH3
mx_NH3 = ex_N2*vi_NH3*m_NH3 ; # Max. amount of NH3 that can be produced
# Results
print ' (a) Max. amount of NH3 that can be produced is %.1f g'%mx_NH3
#(b) and (c)
if (ex_H2 > ex_N2 ):
print ' (b) N2 is limiting reactant '
print ' (c) H2 is excess reactant '
ex_r = ex_N2
else:
print ' (b) H2 is limiting reactant '
print ' (c) N2 is excess reactant '
ex_r = ex_H2 ;
# variables
#(a)
mol_bms = 0.59 ; # Biomass produced per g mol of glucose-[g mol biomass/ g mol glucose]
mw_bms = 23.74 ; # molecular wt. of biomass -[g]
mw_gls = 180.0 ; # molecular wt. of glucose -[g]
# calculations and Results
ms_bms = (mol_bms*mw_bms)/mw_gls ; # Biomass produced per gram of glucose-[g biomass/ g glucose]
print '(a) Biomass produced per gram of glucose is %.4f g biomass/ g glucose.'%ms_bms
#(b)
mol_etol = 1.3 ; #Ethanol produced per g mol of glucose-[g mol ethanol/ g mol glucose]
mw_etol = 46.0 ; # molecular wt. of ethanol -[g]
ms_etol = (mol_etol*mw_etol)/mw_gls ; # Ethanol produced per gram of glucose-[g ethanol/ g glucose]
print ' (b) Ethanol produced per gram of glucose is %.3f g ethanol/ g glucose.'%ms_etol
# Variables
# By using reaction (a)
H2_a = 3-0.50 # H2 produced in reaction (a)
# Calculations
C_a = (2./3)*H2_a ; # Nanotubes(the C) produced by reaction (a)
sel = C_a/0.50 ; # Selectivity of C reletive to C2H4-[g mol C/ g mol C2H4]
# Results
print 'Selectivity of C reletive to C2H4 is %.2f g mol C/ g mol C2H4.'%sel
# Variables
m_C3H6 = 42.08 # molecular wt. of propene-[g]
m_C3H5Cl = 76.53 ; # molecular wt. of C3H5Cl-[g]
m_C3H6Cl2 = 112.99 ; # molecular wt. of C3H6Cl2-[g]
# Product analysis
pml_Cl2 = 141.0 ; # [g mol]
pml_C3H6 = 651.0 ; #[g mol]
pml_C3H5Cl = 4.6 ; # [g mol]
pml_C3H6Cl2 = 24.5 ; # [g mol]
pml_HCL = 4.6 ; #[g mol]
# Calculation & Results
#(a)
a_Cl = pml_C3H5Cl; # Chlorine reacted by eqn.(a)
b_Cl = pml_C3H6Cl2 ; # Chlorine reacted by eqn.(b)
fed_Cl = pml_Cl2+a_Cl+b_Cl ; # Total chlorine fed to reactor-[g mol]
#by analysing reaction (a) and (b)
a_C3H6 = a_Cl+b_Cl ; # C3H6 reacted by reaction (a)
fed_C3H6 = pml_C3H6+a_C3H6 ; #Total C3H6 fed to reactor-[g mol]
print '(a) Total chlorine fed to reactor is %.2f g mol '%fed_Cl
print ' Total C3H6 fed to reactor is %.2f g mol '%fed_C3H6
#(b) and (c)
# Extent of reaction can be calculated by using eqn. 9.3
# Based on C3H6
nio_C3H6 = fed_C3H6 ; #[g mol C3H6]
vi_C3H6 = -1 ; # coefficint of C3H6
ex_C3H6 = -(nio_C3H6)/vi_C3H6 ; # Max. extent of reaction based on C3H6
# Based on Cl2
nio_Cl2 = fed_Cl; #[g mol Cl2]
vi_Cl2 = -1 ; # coefficint of Cl2
ex_Cl2 = -(nio_Cl2)/vi_Cl2 ;# Max. extent of reaction based on Cl2
if (ex_Cl2 > ex_C3H6 ):
print ' (b) C3H6 is limiting reactant '
print ' (c)Cl2 is excess reactant '
ex_r = ex_C3H6;
else:
print ' (b) Cl2 is limiting reactant '
print ' (c) C3H6 is excess reactant '
ex_r = ex_Cl2;
#(d)
fr_cn = pml_C3H5Cl/fed_C3H6 ; #Fractional conversion of C3H6 to C3H5Cl
print ' (d) Fractional conversion of C3H6 to C3H5Cl is %.2e '%fr_cn
#(e)
sel = pml_C3H5Cl/pml_C3H6Cl2 ; # Selectivity of C3H5Cl relative to C3H6Cl2
print ' (e) Selectivity of C3H5Cl relative to C3H6Cl2 is %.2f g mol C3H5Cl/g mol C3H6Cl2 '%sel
#(f)
yld = (m_C3H5Cl*pml_C3H5Cl)/(m_C3H6*fed_C3H6) # Yield of C3H5Cl per g C3H6 fed to reactor
print ' (f) Yield of C3H5Cl per g C3H6 fed to reactor is %.3f g C3H5Cl/g C3H6 '%yld
#(g)
vi_C3H5Cl = 1 ; # coefficint of C3H5Cl
vi_C3H6Cl2 = 1 ; # coefficint of C3H6Cl2
ex_a = (pml_C3H5Cl-0)/vi_C3H5Cl ; # Extent of reaction a as C3H5Cl is produced only in reaction a
ex_b = (pml_C3H6Cl2-0)/vi_C3H6Cl2 ; # Extent of reaction b as C3H6Cl2 is produced only in reaction b
print ' (g) Extent of reaction a as C3H5Cl is produced only in reaction a is %.1f '%ex_a
print ' Extent of reaction b as C3H6Cl2 is produced only in reaction b %.1f '%ex_b
#(h)
in_Cl = fed_Cl*2 ; #Entering Cl -[g mol]
out_Cl = pml_HCL ; # Exiting Cl in HCl-[g mol]
ef_w = out_Cl/in_Cl ; # Mole efficiency of waste
ef_pr = 1-ef_w ; # Mole efficiency of product
print ' (h) Mole efficiency of product is %.3f '%ef_pr