diff options
Diffstat (limited to '409/CH25')
| -rwxr-xr-x | 409/CH25/EX25.1/Example25_1.sce | 18 | ||||
| -rwxr-xr-x | 409/CH25/EX25.10/Example25_10.sce | 25 | ||||
| -rwxr-xr-x | 409/CH25/EX25.11/Example25_11.sce | 23 | ||||
| -rwxr-xr-x | 409/CH25/EX25.2/Example25_2.sce | 16 | ||||
| -rwxr-xr-x | 409/CH25/EX25.3/Example25_3.sce | 18 | ||||
| -rwxr-xr-x | 409/CH25/EX25.4/Example25_4.sce | 38 | ||||
| -rwxr-xr-x | 409/CH25/EX25.5/Example25_5.sce | 55 | ||||
| -rwxr-xr-x | 409/CH25/EX25.6/Example25_6.sce | 27 | ||||
| -rwxr-xr-x | 409/CH25/EX25.7/Example25_7.sce | 38 | ||||
| -rwxr-xr-x | 409/CH25/EX25.8/Example25_8.sce | 36 | ||||
| -rwxr-xr-x | 409/CH25/EX25.9/Example25_9.sce | 34 |
11 files changed, 328 insertions, 0 deletions
diff --git a/409/CH25/EX25.1/Example25_1.sce b/409/CH25/EX25.1/Example25_1.sce new file mode 100755 index 000000000..dac0fae14 --- /dev/null +++ b/409/CH25/EX25.1/Example25_1.sce @@ -0,0 +1,18 @@ +clear ; +clc; +// Example 25.1 +printf('Example 25.1\n\n'); +//page no. 766 +// Solution Fig. E25.1 + +// Given +// C(s) + O2(g) --> CO2(g) (A) +// CO(g) + (1/2)(O2)(g) --> CO2 (g) (B) +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] + +// According to Hess's Law , subtract reaction (B) from reaction (A) , subtract corresponding del_H's to get enthalpy of formation of reaction (C)- C(s) + (1/2)*O2 --> CO(g) , therefore +del_Hfc = del_Ha - del_Hb ;// Standard heat of formation of CO - [kJ/g mol C] +printf('Standard heat of formation of CO is %.2f kJ/g mol C.',del_Hfc) ;
\ No newline at end of file diff --git a/409/CH25/EX25.10/Example25_10.sce b/409/CH25/EX25.10/Example25_10.sce new file mode 100755 index 000000000..51da40c19 --- /dev/null +++ b/409/CH25/EX25.10/Example25_10.sce @@ -0,0 +1,25 @@ +clear ; +clc; +// Example 25.10 +printf('Example 25.10\n\n'); +//page no. 788 +// Solution + +// Given +Ex_hv = 29770.0 ;// Experimental heating value of given coal - [kJ/kg] +// Coal analysis + +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 + +//Higher heating value (HHV) by Dulong formula +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] + +printf('The experimental heating value - %.0f kJ.\n',Ex_hv) ; +printf(' Higher heating value (HHV) by Dulong formula - %.0f kJ.\n',HHV_SI) ; +printf('\n The two values are quite close.' ) ;
\ No newline at end of file diff --git a/409/CH25/EX25.11/Example25_11.sce b/409/CH25/EX25.11/Example25_11.sce new file mode 100755 index 000000000..beeb8fce0 --- /dev/null +++ b/409/CH25/EX25.11/Example25_11.sce @@ -0,0 +1,23 @@ +clear ; +clc; +// Example 25.11 +printf('Example 25.11\n\n'); +//page no. 789 +// Solution + +// Given +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] + +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] + +printf('\n Sulphur emmited when we use fuel NO. 6 is %.2f lb.',S1 ) ; +printf('\n Sulphur emmited when we use fuel NO. 2 is %.2f lb.\n',S2 ) ; +printf('Clearly fuel no. 6 should be selected because of its low SO2 emmission.') ;
\ No newline at end of file diff --git a/409/CH25/EX25.2/Example25_2.sce b/409/CH25/EX25.2/Example25_2.sce new file mode 100755 index 000000000..3b3a81ce1 --- /dev/null +++ b/409/CH25/EX25.2/Example25_2.sce @@ -0,0 +1,16 @@ +clear ; +clc; +// Example 25.2 +printf('Example 25.2\n\n'); +//page no. 767 +// Solution + +// Given +// The main reaction is (1/2)*H2(g) +(1/2)* Cl2(g) --> HCl(g) (A) +// Look in Appendix F for heat of formation of H2 ,Cl2 and HCl +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] + +H_f = 1*H_HCl - (1/2)*(H_H2 + H_Cl2) ; // Standard heat of formation of HCl by reaction - [kJ/ g mol HCl] +printf('Standard heat of formation of HCl(g) is %.3f kJ/g mol HCl.',H_f) ;
\ No newline at end of file diff --git a/409/CH25/EX25.3/Example25_3.sce b/409/CH25/EX25.3/Example25_3.sce new file mode 100755 index 000000000..c74eaadf1 --- /dev/null +++ b/409/CH25/EX25.3/Example25_3.sce @@ -0,0 +1,18 @@ +clear ; +clc; +// Example 25.3 +printf('Example 25.3\n\n'); +//page no. 771 +// Solution + +// Given +// The main reaction is 4*NH3(g) + 5*O2(g) --> 4*NO(g) + 6*H2O (A) +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] + +// Heat of above reaction is calculated by eqn. 25.1 +H_rxn = ((4*H_fNO + 6*H_fH2O) - (4*H_fNH3 + 5*H_fO2))/4 ;// Heat of above reaction-[kJ/ g mol NH3] + +printf('Heat of above reaction is %.3f kJ/g mol NH3.',H_rxn) ;
\ No newline at end of file diff --git a/409/CH25/EX25.4/Example25_4.sce b/409/CH25/EX25.4/Example25_4.sce new file mode 100755 index 000000000..524c9eadd --- /dev/null +++ b/409/CH25/EX25.4/Example25_4.sce @@ -0,0 +1,38 @@ +clear ; +clc; +// Example 25.4 +printf('Example 25.4\n\n'); +//page no. 773 +// Solution + +// Given +P1 = 1 ;// Initial pressure - [atm] +P2 = 1 ;// Final pressure - [atm] +T1 = 500 ;// Initial temperature -[degree C] +T2 = 500 ;// Final temperature -[degree C] + +// The main reaction is CO2(g) + 4H2(g) + --> 2H2O(g) + CH4(g) (A) +// Data obtained from above reaction +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] + +// Additional required data are obtained from CD, according to book it is a follows- +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] + +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 + +printf('Heat of above reaction at 500 degree C and 1 atm is %.1f kJ.',H_rxn_500) ;
\ No newline at end of file diff --git a/409/CH25/EX25.5/Example25_5.sce b/409/CH25/EX25.5/Example25_5.sce new file mode 100755 index 000000000..2686fc6a5 --- /dev/null +++ b/409/CH25/EX25.5/Example25_5.sce @@ -0,0 +1,55 @@ +clear ; +clc; +// Example 25.5 +printf('Example 25.5\n\n'); +//page no. 775 +// Solution + +// Given +// The main reaction is CO2(g) + 4H2(g) + --> 2H2O(g) + CH4(g) (A) +// Data obtained from above reaction +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] + +// The main reaction is CO2(g) + 4H2(g) + --> 2H2O(g) + CH4(g) (A) +// Data obtained from above reaction +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] + +// Additional required data are obtained from CD, according to book it is a follows- +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] + +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] +// Heat of above reaction is calculated by eqn. 25.4 +H_rxn = sum_H_pdt - sum_H_rct + H_rxn_ac ;// Heat of reaction -[kJ/ g mol CO2] + +// By above conditions the energy balance reduces to Q = del_H , therefore +Q = H_rxn ;// Heat transfer to/from the reactor - [kJ] +printf('Heat transfer to/from the reactor is %.3f kJ.\nSince Q is negative , the reactor losses heat.',Q) ;
\ No newline at end of file diff --git a/409/CH25/EX25.6/Example25_6.sce b/409/CH25/EX25.6/Example25_6.sce new file mode 100755 index 000000000..a3d915985 --- /dev/null +++ b/409/CH25/EX25.6/Example25_6.sce @@ -0,0 +1,27 @@ +clear ; +clc; +// Example 25.6 +printf('Example 25.6\n\n'); +//page no. 776 +// Solution + +// Given +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] + +// (a) +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] +printf(' (a) Net enthalpy change for several products (metabolites) per g mol mannitol consumed is %.2f kJ.\n',del_H1) ; + +//(b) +del_H2 = del_H1 / B_growth ;//Net enthalpy change for several products (metabolites) per g cells produced-[kJ] +printf(' (b) Net enthalpy change for several products (metabolites) per g cells produced is %.2f kJ.',del_H2) ;
\ No newline at end of file diff --git a/409/CH25/EX25.7/Example25_7.sce b/409/CH25/EX25.7/Example25_7.sce new file mode 100755 index 000000000..5230ed2ec --- /dev/null +++ b/409/CH25/EX25.7/Example25_7.sce @@ -0,0 +1,38 @@ +clear ; +clc; +// Example 25.7 +printf('Example 25.7\n\n'); +//page no. 777 +// Solution + +// Given +//Bhopal Process +//CH3NH2 + COCl2 + --> C2H3NO +2HCl (A) +//C2H3NO + C10H8O --> C12H11O2N (B) +//Alternate process +//C10H8O + COCl2 --> C11H7O2Cl (C) +//C11H7O2Cl + CH3NH2 --> C12H11O2N + HCl (D) + +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] + +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] + +//Bhopal Process +printf(' Bhopal process .\n') ; +printf(' (a) Heat of reaction (A) is %.1e kJ.\n',H_rxn_a) ; +printf(' (b) Heat of reaction (B) is %.1e kJ.\n',H_rxn_b) ; + +//Alternate process +printf('\n Alternate process .\n') ; +printf(' (c) Heat of reaction (C) is %.2f kJ.\n',H_rxn_c) ; +printf(' (d) Heat of reaction (D) is %.2f kJ.\n',H_rxn_d) ; +printf(' \nThe above data show that capital cost of Bhopal process could be higher than alternate process.\n') ;
\ No newline at end of file diff --git a/409/CH25/EX25.8/Example25_8.sce b/409/CH25/EX25.8/Example25_8.sce new file mode 100755 index 000000000..f6d2efc43 --- /dev/null +++ b/409/CH25/EX25.8/Example25_8.sce @@ -0,0 +1,36 @@ +clear ; +clc; +// Example 25.8 +printf('Example 25.8\n\n'); +//page no. 782 +// Solution + +// Given +P1 = 1 ;// Initial pressure - [atm] +P2 = 1 ;// Final pressure - [atm] +T1 = 500 ;// Initial temperature -[degree C] +T2 = 500 ;// Final temperature -[degree C] + +// The main reaction is CO2(g) + 4H2(g) + --> 2H2O(g) + CH4(g) (A) +// Data obtained from above reaction +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] + +// Additional required data are obtained from CD, according to book it is a follows- +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] + +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] + +printf('Heat of above reaction at 500 degree C and 1 atm is %.1f kJ.',del_H) ;
\ No newline at end of file diff --git a/409/CH25/EX25.9/Example25_9.sce b/409/CH25/EX25.9/Example25_9.sce new file mode 100755 index 000000000..951b68555 --- /dev/null +++ b/409/CH25/EX25.9/Example25_9.sce @@ -0,0 +1,34 @@ +clear ; +clc; +// Example 25.9 +printf('Example 25.9\n\n'); +//page no. 783 +// Solution + +// Given +// The main reaction is CO(g,1 atm,298 K) + (1/2)O2(g,1 atm,400K) --> CO2(g,1at,300 K) (A) +// Conditions of input and output gases are shown in above reaction +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] + +// Additional data are obtained fro Appendix D and E , according to book it is as follows +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] + +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] + +printf('Heat of above reaction is %.1f kJ.',del_H) ;
\ No newline at end of file |