# Variables
Qa = -393.51 ; # Heat of reaction of reaction (a) - [kJ/g mol C]
Qb = -282.99 ; # Heat of reaction of reaction (b) - [kJ/g mol CO]
del_Ha = Qa ; # Change in enthalpy of reaction A - [kJ/g mol C]
del_Hb = Qb ; # Change in enthalpy of reaction B - [kJ/g mol CO]
# Calculations
del_Hfc = del_Ha - del_Hb ; # Standard heat of formation of CO - [kJ/g mol C]
# Results
print 'Standard heat of formation of CO is %.2f kJ/g mol C.'%del_Hfc
# Variables
H_H2 = 0 ; # Standard heat of formation of H2 -[kJ/ g mol H2]
H_Cl2 = 0 ; # Standard heat of formation of Cl2 -[kJ/ g mol Cl2]
H_HCl = -92.311 ; # Standard heat of formation of HCl -[kJ/ g mol HCl]
# Calculations
H_f = 1*H_HCl - (1./2)*(H_H2 + H_Cl2) ; # Standard heat of formation of HCl by reaction - [kJ/ g mol HCl]
# Results
print 'Standard heat of formation of HCl(g) is %.3f kJ/g mol HCl.'%H_f
# Variables
H_fNH3 = -46.191 ; # Standard heat of formation of NH3 -[kJ/ g mol]
H_fO2 = 0 ; #Standard heat of formation of O2 -[kJ/ g mol]
H_fNO = 90.374 ; # Standard heat of formation of NO -[kJ/ g mol]
H_fH2O = -241.826 ; # Standard heat of formation of H2O -[kJ/ g mol]
# Calculations
H_rxn = ((4*H_fNO + 6*H_fH2O) - (4*H_fNH3 + 5*H_fO2))/4. ; # Heat of above reaction-[kJ/ g mol NH3]
# Results
print 'Heat of above reaction is %.3f kJ/g mol NH3.'%H_rxn
# Variables
P1 = 1. ; # Initial pressure - [atm]
P2 = 1. ; # Final pressure - [atm]
T1 = 500. ; # Initial temperature -[degree C]
T2 = 500. ; # Final temperature -[degree C]
m_CO2 = 1. ; # Moles of CO2 - [ g mol]
m_H2 = 4. ; # Moles of H2 - [ g mol]
m_H2O = 2. ; # Moles of H2O - [ g mol]
m_CH4 = 1. ; # Moles of CH4 - [ g mol]
H_fCO2 = -393.250 ; # Heat of formation of CO2 - [kJ/g mol]
H_fH2 = 0. ; # Heat of formation of H2 - [kJ/g mol]
H_fH2O = -241.835 ; # Heat of formation of H2O - [kJ/g mol]
H_fCH4 = -74.848 ; # Heat of formation of CH4 - [kJ/g mol]
H_CO2 = 21.425 ; # Change in enthalpy during temperature change from 25 to 500 degree C of CO2 - [kJ/g mol]
H_H2 = 13.834 ; # Change in enthalpy during temperature change from 25 to 500 degree C of H2 - [kJ/g mol]
H_H2O = 17.010 ; # Change in enthalpy during temperature change from 25 to 500 degree C of H2O - [kJ/g mol]
H_CH4 = 23.126 ; # Change in enthalpy during temperature change from 25 to 500 degree C of CH4 - [kJ/g mol]
# Calculations
H_rxn_25 = (m_CH4*H_fCH4 + m_H2O*H_fH2O) - (m_CO2*H_fCO2 + m_H2*H_fH2) ; # Heat of reaction at 25 C
sum_H_rct = m_CO2*H_CO2 + m_H2*H_H2 ; # sum of heat of formation of reactant - [kJ]
sum_H_pdt = m_CH4*H_CH4 + m_H2O*H_H2O ; #sum of heat of formation of product - [kJ]
# Heat of above reaction is calculated by eqn. 25.4
H_rxn_500 = sum_H_pdt - sum_H_rct + H_rxn_25 ; # Heat of reaction at 500 C
# Results
print 'Heat of above reaction at 500 degree C and 1 atm is %.1f kJ.'%H_rxn_500
# Variables
m_CO2 = 1. ; # Moles of CO2 - [ g mol]
m_H2 = 4. ; # Moles of H2 - [ g mol]
m_H2O = 2. ; # Moles of H2O - [ g mol]
m_CH4 = 1. ; # Moles of CH4 - [ g mol]
P1 = 1. ; # Initial pressure - [atm]
P2 = 1. ; # Final pressure - [atm]
T1_CO2 = 800. ; # Initial temperature of entering CO2 -[K]
T1_H2 = 298. ; # Initial temperature of entering H2 -[K]
T2 = 1000. ; # Temperature of exiting product - [K]
m1_CO2 = 1. ; # Moles of entering CO2 - [ g mol]
m1_H2 = 4. ; # Moles of entering H2 - [ g mol]
f_con = 70./100 ; # Fractional conversion of CO2
m2_H2O = 2*f_con ; # Moles of H2O in product - [ g mol]
m2_CH4 = 1*f_con ; # Moles of CH4 in product - [ g mol]
m2_CO2 = m1_CO2*(1-f_con) ; # Moles of CO2 in product - [ g mol]
m2_H2 = m1_H2*(1-f_con) ; # Moles of CO2 in product - [ g mol]
H_fCO2 = -393.250 ; # Heat of formation of CO2 - [kJ/g mol]
H_fH2 = 0 ; # Heat of formation of H2 - [kJ/g mol]
H_fH2O = -241.835 ; # Heat of formation of H2O - [kJ/g mol]
H_fCH4 = -74.848 ; # Heat of formation of CH4 - [kJ/g mol]
H1_CO2 = 22.798 ; # Change in enthalpy during temperature change from 298K to 800 K of CO2 - [kJ/g mol]
H1_H2 = 0 ; # Change in enthalpy during temperature change from 298K to 298 K of H2 - [kJ/g mol]
H2_H2O = 25.986 ; # Change in enthalpy during temperature change from 298K to 1000 K of H2O - [kJ/g mol]
H2_CH4 = 38.325 ; # Change in enthalpy during temperature change from 298K to 1000 K of CH4 - [kJ/g mol]
H2_CO2 = 33.396; # Change in enthalpy during temperature change from 298K to 1000 K of CO2 - [kJ/g mol]
H2_H2 = 20.620; # Change in enthalpy during temperature change from 298K to 1000 K of H2 - [kJ/g mol]
# Calculations
H_rxn_25 = (m_CH4*H_fCH4 + m_H2O*H_fH2O) - (m_CO2*H_fCO2 + m_H2*H_fH2) ; # Standard heat of reaction at 25 C-[kJ]
H_rxn_ac = f_con*H_rxn_25 ; # Heat of reaction actual - [kJ]
sum_H_rct = m1_CO2*H1_CO2 + m1_H2*H1_H2 ; # sum of heat of formation of reactant - [kJ]
sum_H_pdt = m2_CH4*H2_CH4 + m2_H2O*H2_H2O + m2_CO2*H2_CO2 + m2_H2*H2_H2 ; #sum of heat of formation of product - [kJ]
H_rxn = sum_H_pdt - sum_H_rct + H_rxn_ac ; # Heat of reaction -[kJ/ g mol CO2]
Q = H_rxn ; # Heat transfer to/from the reactor - [kJ]
# Results
print 'Heat transfer to/from the reactor is %.3f kJ.Since Q is negative , the reactor losses heat.'%Q
# Variables
H_EtOH =-1330.51 ; # Change in enthalpy of ethanol -[kJ/g mol]
H_Ac = -887.01 ; # Change in enthalpy of acetate -[kJ/g mol]
H_Fr = -221.75 ; # Change in enthalpy of formate -[kJ/g mol]
H_Lc = -1330.51 ; # Change in enthalpy of lactate -[kJ/g mol]
H_Mn = -2882.78 ; # Change in enthalpy of mannitol -[kJ/g mol]
mol_EtOH =1.29 ; #ethanol produced / g mol mannitol -[g mol]
mol_Ac = 0.22 ; #acetate produced / g mol mannitol -[g mol]
mol_Fr = 1.6 ; #formate produced / g mol mannitol-[g mol]
mol_Lc = 0.4 ; #lactate produced / g mol mannitol-[g mol]
mol_Mn = 1.0 ; #mannitol produced / g mol mannitol-[g mol]
B_growth = 40.5 ; # Biomass growth -[g cells/g mol mannitol]
# Calculations and Results
del_H1 = H_EtOH*mol_EtOH +H_Ac*mol_Ac + H_Fr*mol_Fr + H_Lc*mol_Lc - H_Mn*mol_Mn ; # Net enthalpy change for several products (metabolites) per g mol mannitol consumed -[kJ]
printnt ' (a) Net enthalpy change for several products (metabolites) per g mol mannitol consumed is %.2f kJ.'%del_H1
#(b)
del_H2 = del_H1 / B_growth ; #Net enthalpy change for several products (metabolites) per g cells produced-[kJ]
print ' (b) Net enthalpy change for several products (metabolites) per g cells produced is %.2f kJ.'%del_H2
# Solution
# Variables
H_Cb = -26. ; #Standard heat of formation of carbaryl(C12H11O2N) -[kJ/ g mol]
H_HCl = -92.311 ; #Standard heat of formation of HCl -[kJ/ g mol]
H_Ma = -20.0 ; #Standard heat of formation of methyl amine(CH3NH2) -[kJ/ g mol]
H_Mi = -9*10**4 ; #Standard heat of formation of methyl isocynate(C2H3NO) -[kJ/ g mol]
H_Nc = -17.9 ; #Standard heat of formation of 1-Napthalenyl chloroformate(C11H7O2Cl) -[kJ/ g mol]
H_N = 30.9 ; #Standard heat of formation of napthol(C10H8O) -[kJ/ g mol]
H_P = -221.85 ; #Standard heat of formation of phosgene(COCl2) -[kJ/ g mol]
# Calculations
H_rxn_a = (2*H_HCl + 1*H_Mi) - (1*(H_Ma) + 1*H_P ) ; # Heat of reaction (A)-[kJ]
H_rxn_b = (1*H_Cb ) - (1*(H_Mi) + 1*H_N ) ; # Heat of reaction (B)-[kJ]
H_rxn_c = (1*H_Nc) - (1*(H_N) + 1*H_P ) ; # Heat of reaction (C)-[kJ]
H_rxn_d = (1*H_Cb + 1*H_HCl) - (1*(H_Nc) + 1*H_Ma ) ; # Heat of reaction (D)-[kJ]
# Results
#Bhopal Process
print ' Bhopal process .'
print ' (a) Heat of reaction (A) is %.1e kJ.'%H_rxn_a
print ' (b) Heat of reaction (B) is %.1e kJ.'%H_rxn_b
#Alternate process
print ' Alternate process .'
print ' (c) Heat of reaction (C) is %.2f kJ.'%H_rxn_c
print ' (d) Heat of reaction (D) is %.2f kJ.'%H_rxn_d
print ' The above data show that capital cost of Bhopal process could be higher than alternate process.'
# Variables
P1 = 1. ; # Initial pressure - [atm]
P2 = 1. ; # Final pressure - [atm]
T1 = 500. ; # Initial temperature -[degree C]
T2 = 500. ; # Final temperature -[degree C]
m_CO2 = 1. ; # Moles of CO2 - [ g mol]
m_H2 = 4. ; # Moles of H2 - [ g mol]
m_H2O = 2. ; # Moles of H2O - [ g mol]
m_CH4 = 1. ; # Moles of CH4 - [ g mol]
H_fCO2 = -393.250; # Heat of formation of CO2 - [kJ/g mol]
H_fH2 = 0 ; # Heat of formation of H2 - [kJ/g mol]
H_fH2O = -241.835 ; # Heat of formation of H2O - [kJ/g mol]
H_fCH4 = -74.848 ; # Heat of formation of CH4 - [kJ/g mol]
H_CO2 = 21.425 ; # Change in enthalpy during temperature change from 25 to 500 degree C of CO2 - [kJ/g mol]
H_H2 = 13.834 ; # Change in enthalpy during temperature change from 25 to 500 degree C of H2 - [kJ/g mol]
H_H2O = 17.010 ; # Change in enthalpy during temperature change from 25 to 500 degree C of H2O - [kJ/g mol]
H_CH4 = 23.126 ; # Change in enthalpy during temperature change from 25 to 500 degree C of CH4 - [kJ/g mol]
# Calculations
H_in = (H_fCO2 + H_CO2)*m_CO2 + (H_fH2 + H_H2)*m_H2 ; # Enthalpy change for inputs -[kJ]
H_out = (H_fH2O + H_H2O)*m_H2O + (H_fCH4 + H_CH4)*m_CH4 ; # Enthalpy change for outputs -[kJ]
del_H = H_out - H_in ; # Net enthalpy change of process -[kJ]
# Results
print 'Heat of above reaction at 500 degree C and 1 atm is %.1f kJ.'%del_H
# Solution
# Variables
m_CO = 1. ; # Moles of CO input- [g mol]
m1_O2 = 1.5 ; # Moles of O2 input - [g mol]
m_CO2 = 1. ; # Moles of CO2 output - [g mol]
m2_O2 = 1. ; # Moles of O2 output - [g mol]
T_in_CO = 298. ; # Temperature of entering CO -[K]
T_in_O2 = 400. ; #Temperature of entering O2 -[K]
T_out_CO2 = 300. ; # Temperature of exiting CO2 -[K]
T_out_O2 = 300. ; # Temperature of exiting O2 -[K]
H_fCO = -110.520 ; # Heat of formation of CO - [kJ/g mol]
H_fO2 = 0 ; # Heat of formation of O2 - [kJ/g mol]
H_fCO2 = -393.250 ; # Heat of formation of CO2 - [kJ/g mol]
H_CO = 0 ; # Change in enthalpy during temperature change from 298K to 298 K of CO - [kJ/g mol]
H1_O2 = 11.619 ; # Change in enthalpy during temperature change from 298K to 400 K of input O2 - [kJ/g mol]
H_CO2 = 11.644 ; # Change in enthalpy during temperature change from 298K to 300 K of CO2 - [kJ/g mol]
H2_O2 = 8.389 ; # Change in enthalpy during temperature change from 298K to 300 K of output O2 - [kJ/g mol]
# Calculations
H_in = (H_fCO + H_CO)*m_CO + (H_fO2 + H1_O2)*m1_O2 ; # Enthalpy change for inputs -[kJ]
H_out = (H_fCO2 + H_CO2)*m_CO2 + (H_fO2 + H2_O2)*m2_O2 ; # Enthalpy change for inputs -[kJ]
del_H = H_out - H_in ; # Net enthalpy change of process -[kJ]
# Results
print 'Heat of above reaction is %.1f kJ.'%del_H
# Solution
# Given
Ex_hv = 29770.0 ; # Experimental heating value of given coal - [kJ/kg]
C = 71.0/100 ; #Fraction of C in coal
H2 = 5.6/100 ; # Fraction of H2 in coal
N2 = 1.6/100 ; # Fraction of N2 in coal
S = 2.7/100 ; # Fraction of S in coal
ash = 6.1/100 ; # Fraction of ash in coal
O2 = 13.0/100 ; #Fraction of O2 in coal
# Calculations
HHV = 14544*C + 62028*(H2 - O2/8) + 4050*S ; # Higher heating value (HHV) by Dulong formula -[Btu/lb]
HHV_SI = HHV *1.055/0.454 ; # HHV in SI unt - [kJ/kg]
# Results
print 'The experimental heating value - %.0f kJ.'%Ex_hv
print ' Higher heating value (HHV) by Dulong formula - %.0f kJ.'%HHV_SI
print ' The two values are quite close.'
# Variables
H_req = 10**6 ; # Heat requirement - [Btu]
d_N6 = 60.2 ; # Density of fuel no. 6-[lb/ft**3]
d_N2 = 58.7 ; # Density of fuel no. 2-[lb/ft**3]
S_N6 = 0.72/100 ; # Sulphur content in fuel no. 6
S_N2 = 0.62/100; #Sulphur content in fuel no. 2
lhv_N6 = 155000 ; #Lower heating value of No.6 -[Btu/gal]
lhv_N2 = 120000 ; #Lower heating value of No.2 -[Btu/gal]
# Calculations
S1 = H_req*d_N6*S_N6/lhv_N6 ; # Sulphur emmited when we use fuel NO. 6-[lb]
S2 = H_req*d_N2*S_N2/lhv_N2 ; # Sulphur emmited when we use fuel NO. 2-[lb]
# Results
print ' Sulphur emmited when we use fuel NO. 6 is %.2f lb.'%S1
print ' Sulphur emmited when we use fuel NO. 2 is %.2f lb.'%S2
print 'Clearly fuel no. 6 should be selected because of its low SO2 emmission.'