From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001 From: priyanka Date: Wed, 24 Jun 2015 15:03:17 +0530 Subject: initial commit / add all books --- 409/CH10/EX10.1/Example10_1.sce | 56 +++++++++++++++++++ 409/CH10/EX10.2/Example10_2.sce | 46 ++++++++++++++++ 409/CH10/EX10.3/Example10_3.sce | 69 +++++++++++++++++++++++ 409/CH10/EX10.4/Example10_4.sce | 48 ++++++++++++++++ 409/CH10/EX10.5/Example10_5.sce | 118 ++++++++++++++++++++++++++++++++++++++++ 409/CH10/EX10.6/Example10_6.sce | 30 ++++++++++ 409/CH10/EX10.7/Example10_7.sce | 19 +++++++ 409/CH10/EX10.8/Example10_8.sce | 48 ++++++++++++++++ 409/CH10/EX10.9/Example10_9.sce | 85 +++++++++++++++++++++++++++++ 9 files changed, 519 insertions(+) create mode 100755 409/CH10/EX10.1/Example10_1.sce create mode 100755 409/CH10/EX10.2/Example10_2.sce create mode 100755 409/CH10/EX10.3/Example10_3.sce create mode 100755 409/CH10/EX10.4/Example10_4.sce create mode 100755 409/CH10/EX10.5/Example10_5.sce create mode 100755 409/CH10/EX10.6/Example10_6.sce create mode 100755 409/CH10/EX10.7/Example10_7.sce create mode 100755 409/CH10/EX10.8/Example10_8.sce create mode 100755 409/CH10/EX10.9/Example10_9.sce (limited to '409/CH10') diff --git a/409/CH10/EX10.1/Example10_1.sce b/409/CH10/EX10.1/Example10_1.sce new file mode 100755 index 000000000..383a7bd8d --- /dev/null +++ b/409/CH10/EX10.1/Example10_1.sce @@ -0,0 +1,56 @@ +clear; +clc; +// Example 10.1 +printf('Example 10.1\n\n'); +//Page no. 264 +// Solution + +F = 100 ;// feed to the reactor-[g mol] +// Composition of feed +CH4 = 0.4*F ;// [g mol] +Cl2 = 0.5*F ;// [g mol] +N2= 0.1*F ;//[g mol] + +// Extent of reaction can be calculated by using eqn. 9.3 +// Based on CH4 +nio_CH4 = CH4 ;//[g mol CH4] +vi_CH4 = -1 ;// coefficint of CH4 +ex_CH4 = -(nio_CH4)/vi_CH4 ;// Max. extent of reaction based on CH4 + +// Based on Cl2 +nio_Cl2 = Cl2 ;//[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_CH4 ) + printf(' \n CH4 is limiting reactant \n'); + else +printf(' \n (b) Cl2 is limiting reactant \n'); +end +// By execution of above block its clear that CH4 is limiting reactant, therefore extent of reaction is +cn_CH4 = 67/100 ;// percentage conversion of CH4 +ex_r = (-cn_CH4)*CH4/vi_CH4 ;// extent of reaction +printf(' extent of reaction is %.1f g moles reacting \n',ex_r); + +n_un = 11 ;// Number of unknowns in the given problem +n_ie = 11 ;// Number of independent equations +d_o_f = n_un-n_ie ;// Number of degree of freedom +printf(' Number of degree of freedom for the given system is %i \n',d_o_f); + +// Product composition using species balance using eqn.10.2 +vi_CH3Cl = 1; +vi_HCl = 1; +vi_N2 = 0; +p_CH4 = CH4+(vi_CH4*ex_r);// [g mol] +p_Cl2 = Cl2+(vi_Cl2*ex_r);// [g mol] +p_CH3Cl = 0+(vi_CH3Cl*ex_r);// [g mol] +p_HCl = 0+(vi_HCl*ex_r);// [g mol] +p_N2 = N2+(vi_N2*ex_r);// [g mol] +// As we have taken F = 100 so answers we are getting can be directly used as percentage composition +printf('\n\nComposition of product stream in %% g mol of products\n'); +printf('\nProduct Percentage g mol\n'); +printf('\nCH4 %.1f%% g mol\n',p_CH4); +printf('\nCl2 %.1f%% g mol\n',p_Cl2); +printf('\nCH3Cl %.1f%% g mol\n',p_CH3Cl); +printf('\nHCl %.1f%% g mol\n',p_HCl); +printf('\nN2 %.1f%% g mol\n',p_N2); \ No newline at end of file diff --git a/409/CH10/EX10.2/Example10_2.sce b/409/CH10/EX10.2/Example10_2.sce new file mode 100755 index 000000000..8387a9b46 --- /dev/null +++ b/409/CH10/EX10.2/Example10_2.sce @@ -0,0 +1,46 @@ +clear ; +clc; +// Example 10.2 +printf('Example 10.2\n\n'); +// Page no. 266 +// Solution + +S = 5000 ;// Sulphur [lb] +// Composition of feed +CH4 = 80 ;// [%] +H2S = 20 ;// [%] + +n_un = 11 ;// Number of unknowns in the given problem +n_ie = 11 ;// Number of independent equations +d_o_f = n_un-n_ie ;// Number of degree of freedom +printf('Number of degree of freedom for the given system is %i \n',d_o_f); +m_S = 32.0 ;//molecular wt. of S -[lb] +mol_S = S/32.0; +// Extent of reaction can be calculated by using eqn. 9.3 +// Based on S +nio_S = 0 ;//[g mol S] +ni_S = mol_S ;//[g mol S] +vi_S = 3 ;// coefficint of S -from given reaction +ex_r = (ni_S-nio_S)/vi_S ;// Extent of reaction based on S +printf(' Extent of reaction is %.1f g moles reacting \n',ex_r); + +// Product composition +vi_H2O = 2 ;// coefficint of H2O +vi_H2S = -2 ;// coefficint of H2S +vi_SO2 = -1 ;//coefficint of SO2 +vi_CH4 = 0 ;//coefficint of CH4 +P_H2O = 0+(vi_H2O*ex_r);// [lb mol] +P_H2S = P_H2O/10 ;//[lb mol] +P_SO2 = 3*P_H2S ;//[lb mol] + +F = (P_H2S-vi_H2S*ex_r)/(H2S/100) ;// total feed-[lb mol] +F_SO2 = P_SO2-(vi_SO2*ex_r);// feed rate of SO2- [lb mol] +F_CH4 = (CH4/100)*F+vi_CH4*ex_r ;//feed rate of CH4- [lb mol] +F_H2S = ((H2S/100)*F) ;// feed rate of H2S-[lb mol] + +// We can see from situation that H2S is limiting reagent as ratio of SO2 to H2S in the product gas(3/1) is greater than their molar ratio in chemical reaction(2/1) +f_cn = -(vi_H2S*ex_r)/((H2S/100)*F) ;// Fractional conversion of limiting reagent + +printf('\n(1)Feed rate of H2S- %.1f lb mol\n',F_H2S); +printf('(2)Feed rate of SO2- %.1f lb mol\n',F_SO2); +printf('(3)Fractional conversion of limiting reagent- %.2f \n',f_cn); \ No newline at end of file diff --git a/409/CH10/EX10.3/Example10_3.sce b/409/CH10/EX10.3/Example10_3.sce new file mode 100755 index 000000000..5bf1e29c8 --- /dev/null +++ b/409/CH10/EX10.3/Example10_3.sce @@ -0,0 +1,69 @@ +clear ; +clc; +// Example 10.3 +printf('Example 10.3\n\n'); +// Page no. 270 +// Solution + +F = 1 ;//CH3OH -[gmol] +// Extent of reactions can be calculated by using eqn. 10.5 +// For reaction 1 based on CH3OH is limiting reagent +f_cn = 90 ;//[%] +vi_CH3OH = -1 ;//coefficint of CH3OH +ex_r1 = (-90/100)/vi_CH3OH ;// Extent of reaction based on CH3OH +printf(' Extent of reaction 1 is %.2f g moles reacting \n',ex_r1); +//For reaction 2 +yld = 75 ;//[%] +ex_r2 = ex_r1-(F*(yld/100)); +printf(' Extent of reaction 2 is %.2f g moles reacting \n',ex_r2); + +// For amount of air +// Entering O2 is twice the O2 required by reaction 1,therefore +f_O2 = 0.21 ;// mol. fraction of O2 +f_N2 = 0.79 ;// mol. fraction of N2 +n_O2 = 2*((1/2)*F) ;// entering oxygen -[g mol] +air = n_O2/f_O2 ;// Amount of air entering +n_N2 = air-n_O2 ;// entering nitrogen -[g mol] + +// Degree of freedom analysis +n_un = 11 ;// Number of unknowns in the given problem +n_ie = 11 ;// Number of independent equations +d_o_f = n_un-n_ie ;// Number of degree of freedom +printf(' Number of degree of freedom for the given system is %i \n',d_o_f); + +// Reaction 1 +v1_CH3OH = -1 ;//coefficint of CH3OH +v1_O2 = -1/2 ;//coefficint of O2 +v1_CH2O = 1 ;//coefficint of CH2O +v1_H2O = 1 ;//coefficint of H2O +v1_CO = 0 ;//coefficient of CO +//Reaction 2 +v2_O2 = -1/2 ;//coefficint of O2 +v2_CH2O = -1 ;//coefficint of CH2O +v2_H2O = 1 ;//coefficint of H2O +v2_CO = 1 ;//coefficient of CO +P = F+air +(v1_CH3OH+v1_O2+v1_CH2O+v1_H2O)*ex_r1 +(v2_O2+v2_CH2O+v2_H2O+v2_CO)*ex_r2 ;// Product -[g mol] + +no_CH3OH = F+(v1_CH3OH*ex_r1)+0 ;// [g mol] +no_O2 = n_O2+(v1_O2*ex_r1)+v2_O2*ex_r2 ;// [g mol] +no_CH2O = 0 + v1_CH2O*ex_r1 +v2_CH2O*ex_r2 ;//[g mol] +no_CO = 0+v1_CO*ex_r1 +v2_CO*ex_r2 ;//[g mol] +no_H2O = 0+v1_H2O*ex_r1+v2_H2O*ex_r2 ;// [g mol] +no_N2 = n_N2-0-0 ;// [g mol] + +// Composition of product +y_CH3OH = (no_CH3OH/P )*100 ;// mole % +y_O2 = (no_O2/P)*100 ;// mole % +y_CH2O = (no_CH2O/P)*100 ;// mole % +y_CO = (no_CO/P)*100 ;// mole % +y_H2O = (no_H2O/P)*100 ;// mole % +y_N2 = (no_N2/P )*100;// mole % + +printf('\nComposition of product\n'); +printf('Component mole percent\n'); +printf(' CH3OH %.1f %%\n',y_CH3OH); +printf(' O2 %.1f %%\n',y_O2); +printf(' CH2O %.1f %%\n',y_CH2O); +printf(' CO %.1f %%\n',y_CO); +printf(' H2O %.1f %%\n',y_H2O); +printf(' N2 %.1f %%\n',y_N2); \ No newline at end of file diff --git a/409/CH10/EX10.4/Example10_4.sce b/409/CH10/EX10.4/Example10_4.sce new file mode 100755 index 000000000..f0906d54c --- /dev/null +++ b/409/CH10/EX10.4/Example10_4.sce @@ -0,0 +1,48 @@ +clear; +clc; +// Example 10.4 +printf('Example 10.4\n\n'); +// Page no. 273 +// Solution + +F = 4000 ;//[kg] +m_H2O = 18.02 ;// molecular masss of water +m_C6H12O6 = 180.1 ;// molecular mass of glucose +m_CO2 = 44 ;//molecular mass of CO2 +m_C2H3CO2H = 72.03 ;// molecular mass of C2H3CO2H +m_C2H5OH = 46.05 ;// molecular mass of ethanol + +p_H2O = 88 ;// [%] +p_C6H12O6 = 12;// [%] +ni_H2O = (F*p_H2O/100)/m_H2O ;// initial moles of water +ni_C6H12O6 = (F*(p_C6H12O6/100))/m_C6H12O6 ;// initial moles of glucose + +// Degree of freedom analysis +n_un = 9 ;// Number of unknowns in the given problem +n_ie = 9 ;// Number of independent equations +d_o_f = n_un-n_ie ;// Number of degree of freedom +printf('Number of degree of freedom for the given system is %i \n',d_o_f); + +ur_C6H12O6 = 90 ;//[kg] +pr_CO2 = 120 ;//[kg] +nf_C6H12O6 = ur_C6H12O6/m_C6H12O6 ;// [kmoles] +nf_CO2 = pr_CO2/m_CO2 ;// [kmoles] + +// solve following eqn. (b) and (e)simultaneously +//(b): nf_C6H12O6 = ni_C6H12O6+-1*ex_r1+-1*ex_r2 +//(e): nf_CO2 = 0+2*ex_r1+ 0*ex_r2 +a = [-1 -1;2 0];// matrix formed by coefficients of unknowns +b = [(nf_C6H12O6-ni_C6H12O6);nf_CO2];//matrix formed by constant +x = a^(-1)*b;//matrix formed by solution +printf(' Extent of reaction 1 is %.3f kg moles reacting \n',x(1)); +printf(' Extent of reaction 2 is %.3f kg moles reacting \n',x(2)); + +nf_H2O = ni_H2O+0*x(1) +2*x(2);// from eqn. (a)-[kmoles] +nf_C2H5OH = 0+2*x(1)+0*x(2);// from eqn.(c)-[kmoles] +nf_C2H3CO2H = 0+0*x(1)+2*x(2) ;//from eqn.(d)-[kmoles] +total_wt = m_H2O*nf_H2O + m_C6H12O6*nf_C6H12O6 + m_CO2*nf_CO2 + m_C2H3CO2H*nf_C2H3CO2H + m_C2H5OH*nf_C2H5OH; +mp_C2H5OH = (m_C2H5OH*nf_C2H5OH*100)/total_wt ;// Mass percent of ethanol -[%] +mp_C2H3CO2H = (m_C2H3CO2H*nf_C2H3CO2H*100)/total_wt ;//Mass percent of propenoic acid -[%] + +printf(' \n Mass percent of ethanol in broth at end of fermentation process is %.1f %%\n',mp_C2H5OH); +printf(' Mass percent of propenoic acid in broth at end of fermentation process is %.1f %%\n',mp_C2H3CO2H); \ No newline at end of file diff --git a/409/CH10/EX10.5/Example10_5.sce b/409/CH10/EX10.5/Example10_5.sce new file mode 100755 index 000000000..091c95310 --- /dev/null +++ b/409/CH10/EX10.5/Example10_5.sce @@ -0,0 +1,118 @@ +clear ; +clc; +// Example 10.5 +printf('Example 10.5\n\n'); +// Page no. 279 +// Solution + +//(a)Solution of Example 10.1 using element balance +printf('(a)Solution of Example 10.1 using element balance\n'); +F = 100 ;// feed to the reactor-[g mol] +// Composition of feed +CH4 = 0.4*F ;// [g mol] +Cl2 = 0.5*F ;// [g mol] +N2 = 0.1*F ;//[g mol] + +n_un = 10 ;// Number of unknowns in the given problem(excluding extent of reaction) +n_ie = 10 ;// Number of independent equations +d_o_f = n_un-n_ie ;// Number of degree of freedom +printf(' Number of degree of freedom for the given system is %i \n',d_o_f); + +// Extent of reaction can be calculated by using eqn. 9.3 +// Based on CH4 +nio_CH4 = CH4 ;//[g mol CH4] +vi_CH4 = -1; // coefficint of CH4 +ex_CH4 = -(nio_CH4)/vi_CH4 ;// Max. extent of reaction based on CH4 + +// Based on Cl2 +nio_Cl2 = Cl2 ;//[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_CH4 ) + printf(' CH4 is limiting reactant \n'); + else +printf(' \n (b) Cl2 is limiting reactant \n'); +end +// By execution of above block its clear that CH4 is limiting reactant,therefore +cn_CH4 = 67/100 ;// percentage conversion of CH4(limiting reagent) +no_CH4 = CH4-(cn_CH4*CH4) ;//CH4 in product -[g mol] + +// Product composition using element balance +// By N2 balance +no_N2 = N2;//N2 in product -[g mol] + +C = CH4 ;//moles of CH4 = moles of C (by molecular formula) +H = 4*CH4 ;// moles of H = 4*moles of CH4 (by molecular formula) +Cl = 2*Cl2 ;// moles of Cl = 2* moles of Cl2 (by molecular formula) +// Solving following 3 eqn. obtained from balance of C,H,Cl for 3 unknowns +//1. C-no_CH4*1 = 1*no_CH3Cl +//2. H-4*no_CH4 = 3*no_CH3Cl+no_HCl*1 +//3. Cl = no_Cl2*2 + no_HCl*1+1*no_CH3Cl +a = [0 0 1;0 1 3;2 1 1] ;// matrix formed by coefficients of unknowns +b = [C-no_CH4*1;H-4*no_CH4;Cl] ;//matrix formed by constant +x = a^(-1)*b ;// matrix of solution + +// As we have taken F = 100 so answers we are getting can be directly used as percentage composition +printf('\nComposition of product stream in %% g mol of products\n'); +printf('Product Percentage g mol\n'); +printf('\nCH4 %.1f%% g mol\n',no_CH4); +printf('\nCl2 %.1f%% g mol\n',x(1)); +printf('\nCH3Cl %.1f%% g mol\n',x(3)); +printf('\nHCl %.1f%% g mol\n',x(2)); +printf('\nN2 %.1f%% g mol\n',no_N2); + +//(b)Solution of Example 10.3 using element balance +printf('______________________________________________________________________________'); +printf('\n\n(b)Solution of Example 10.3 using element balance\n'); +F = 1 ;//CH3OH -[gmol] +yld = 75 ;//[%] +cnv = 90 ;//conversion of methanol-[%] + +// For amount of air +// Entering O2 is twice the O2 required by reaction 1,therefore +f_O2 = 0.21 ;// mol. fraction of O2 +f_N2 = 0.79 ;// mol. fraction of N2 +n_O2 = 2*((1/2)*F) ;// entering oxygen -[g mol] +air = n_O2/f_O2 ;// Amount of air entering +n_N2 = air-n_O2 ;// entering nitrogen -[g mol] + +// Degree of freedom analysis +n_un = 9 ;// Number of unknowns in the given problem(excluding extent of reactions) +n_ie = 9 ;// Number of independent equations +d_o_f = n_un-n_ie ;// Number of degree of freedom +printf(' Number of degree of freedom for the given system is %i \n',d_o_f); + +// Product composition using element balance +// By N2 balance +no_N2 = n_N2 ;// inert ,terefore input = output +C = 1*F ;//moles of C = moles of CH3OH (by molecular formula) +H = 4*F ;//moles of H = 4*moles of CH3OH (by molecular formula) +O = 1*F +2*n_O2;// moles of O = 1*moles of CH3OH + O in air +no_CH2O = yld/100 ;//[g mol] +no_CH3OH = F-((cnv/100)*F);// [g mol] + +// Solving following 3 eqn. obtained from balance of C,H,O for 3 unknowns +a = [0 0 1;0 2 0;2 1 1] ;// matrix formed by coefficients of unknowns +b = [(C-(no_CH3OH*1+no_CH2O*1));(H-(4*no_CH3OH+2*no_CH2O));(O-(no_CH3OH*1+no_CH2O*1))] ;//matrix formed by constant +x = a\b ;// matrix of solution + +P = no_CH2O+no_CH3OH+no_N2+x(1)+x(2)+x(3); + +// Composition of product +y_CH3OH = (no_CH3OH/P )*100;// mole % +y_O2 = ((x(1))/P)*100;// mole % +y_CH2O = (no_CH2O/P)*100 ;// mole % +y_CO = (x(3)/P)*100 ;// mole % +y_H2O = (x(2)/P)*100 ;// mole % +y_N2 = (no_N2/P )*100;// mole % + + +printf('\nComposition of product\n'); +printf('Component mole percent\n'); +printf(' CH3OH %.1f %%\n',y_CH3OH); +printf(' O2 %.1f %%\n',y_O2); +printf(' CH2O %.1f %%\n',y_CH2O); +printf(' CO %.1f %%\n',y_CO); +printf(' H2O %.1f %%\n',y_H2O); +printf(' N2 %.1f %%\n',y_N2); \ No newline at end of file diff --git a/409/CH10/EX10.6/Example10_6.sce b/409/CH10/EX10.6/Example10_6.sce new file mode 100755 index 000000000..13228fe9c --- /dev/null +++ b/409/CH10/EX10.6/Example10_6.sce @@ -0,0 +1,30 @@ +clear ; +clc; +// Example 10.6 +printf('Example 10.6\n\n'); +// Page no. 281 +// Solution + +// Basis: P=100 // Product from the reactor-[g mol] +P=100 ;//Product from the reactor-[g mol] +// Composition of product +C3H8 = 0.195*P ;// [g mol] +C4H10 = 0.594*P ;// [g mol] +C5H12 = 0.211*P;// [g mol] + +n_un = 3 ;// Number of unknowns in the given problem(excluding extent of reaction) +n_ie = 3 ;// Number of independent equations +d_o_f = n_un-n_ie ;// Number of degree of freedom +printf('Number of degree of freedom for the given system is %i \n',d_o_f); + +C = C3H8*3+C4H10*4+C5H12*5 ;// moles of C on product side +H = C3H8*8+C4H10*10+C5H12*12 ;// moles of H on product side +// Solve following eqn.( obtained by element balance of C & H) for F and G +//8F+0G = C +//18F+2G = H +a = [8 0;18 2] ;// matrix formed by coefficients of unknowns +b = [C;H] ;//matrix formed by constant +x = a\b ;// matrix of solution + +R = x(2)/x(1) ;// Ratio of H2 consumed to C8H18 reacted = G/F +printf(' Molar ratio of H2 consumed to C8H18 reacted is %.3f \n',R); \ No newline at end of file diff --git a/409/CH10/EX10.7/Example10_7.sce b/409/CH10/EX10.7/Example10_7.sce new file mode 100755 index 000000000..3d90929e4 --- /dev/null +++ b/409/CH10/EX10.7/Example10_7.sce @@ -0,0 +1,19 @@ +clear ; +clc; +// Example 10.7 +printf('Example 10.7\n\n'); +// Page no. 286 +// Solution + +C3H8 = 20 ;//C3H8 burned in a test-[kg] +m_C3H8 = 44.09 ;// mol. wt . of 1 kmol C3H8 +cf_O2 = 5 ;// coefficient of O2 in given reaction +air = 400 ;// Air given -[kg] +m_air = 29 ;// molecular wt. of 1kmol air-[kg] +O2p = 21 ;// percentage of O2 in air-[%] +p_CO2 = 44 ;// CO2 produced -[kg] +p_CO = 12 ;// CO produced -[kg] +O2 = (air*O2p/100)/(m_air) ;// amount of entering O2-[k mol] +rqO2 = (C3H8*cf_O2)/(m_C3H8) ;// Required O2 for given reaction +ex_air = ((O2-rqO2)*100)/rqO2 ;// Excess air percent-[%] +printf('Excess air percent is %.0f %%.\n',ex_air); \ No newline at end of file diff --git a/409/CH10/EX10.8/Example10_8.sce b/409/CH10/EX10.8/Example10_8.sce new file mode 100755 index 000000000..939ec1e26 --- /dev/null +++ b/409/CH10/EX10.8/Example10_8.sce @@ -0,0 +1,48 @@ +clear; +clc; +// Example 10.8 +printf('Example 10.8\n\n'); +// Page no. 287 +// Solution + +F = 16 ;// feed of CH4 -[kg] +CH4p = 100 ;//[%] +m_CH4 = 16 ;// mass of kmol of CH4-[kg] +mol_CH4 = (F*CH4p/100)/m_CH4;//k moles of CH4 in feed-[kmol] +air = 300 ;// Air given -[kg] +m_air = 29 ;// molecular wt. of 1kmol air-[kg] +mol_air = air/m_air ;// kmoles of air-[kmol] +O2p = 21 ;// percentage of O2 in air-[%] +O2 = (mol_air*O2p/100) ;// amount of entering O2-[k mol] +N2 = mol_air-O2 ;// amount of entering N2-[k mol] + +// Degree of freedom analysis +n_un = 8 ;// Number of unknowns in the given problem(excluding extent of reactions) +n_ie = 8 ;// Number of independent equations +d_o_f = n_un-n_ie ;// Number of degree of freedom +printf('Number of degree of freedom for the given system is %i \n',d_o_f); + +// Product composition analysis using element balance of C,H,O and N +p_N2 = N2 ;// inert +C_in = 1*mol_CH4 ;// kmoles of carbon in input-[kmol] +H_in = 4*mol_CH4 ;// kmoles of hydrogen in input-[kmol] +O_in = 2*O2 ;// kmoles of oxygen in input-[kmol] +p_CO2 = C_in/1 ;//kmoles of CO2 in product obtained by carbon balance-[kmol] +p_H2O = H_in/2 ;//kmoles of H2O in product obtained by hydrogen balance-[kmol] +p_O2 = (O_in-(2*p_CO2+p_H2O))/2 ;//kmoles of O2 in product obtained by oxygen balance-[kmol] +p_CH4 = 0 ;// Complete reaction occurs +P = p_CH4 + p_N2+ p_CO2 + p_H2O + p_O2; + +y_N2 = p_N2*100/P ;//[mol %] +y_CO2 = p_CO2*100/P ;//[mol %] +y_H2O = p_H2O*100/P ;//[mol %] +y_O2 = p_O2*100/P ;//[mol %] +y_CH4 = p_CH4*100/P ;//[mol %] + +printf('\nComposition of product\n'); +printf('Component mole percent\n'); +printf(' CH4 %.1f %%\n',y_CH4); +printf(' O2 %.1f %%\n',y_O2); +printf(' CO2 %.1f %%\n',y_CO2); +printf(' H2O %.1f %%\n',y_H2O); +printf(' N2 %.1f %%\n',y_N2); \ No newline at end of file diff --git a/409/CH10/EX10.9/Example10_9.sce b/409/CH10/EX10.9/Example10_9.sce new file mode 100755 index 000000000..51b147e29 --- /dev/null +++ b/409/CH10/EX10.9/Example10_9.sce @@ -0,0 +1,85 @@ +clear; +clc; +// Example 10.9 +printf('Example 10.9\n\n'); +// Page no. 290 +// Solution + +F = 100 ;// feed of coal -[lb] +// given coal composition-given +C = 83.05 ;//[%] +H = 4.45 ;//[%] +O = 3.36 ;// [%] +N = 1.08 ;// [%] +S = 0.70 ;//[%] +ash = 7.36;//[%] +H2O = 3.9 ;//[%] +w_C = 12 ;// mol. wt. of C +w_H = 1.008;//mol. wt. of H +w_O = 16 ;// mol. wt. of O +w_N = 14 ;// mol. wt. of N +w_S = 32 ;//mol. wt. of S + +//given stack gas analysis-given +CO2 = 15.4 ;//[%] +CO = 0.0 ;//[%] +O2 = 4.0 ;// [%] +N2 = 80.6 ;//[%] +//given refuse analysis +ash_R = 86 ;//[%] +odr = 14 ;//[%] + +H2O_air = .0048 ;// [lb H2O/lb dry air] +m_air = 29 ;// mol. wt. of air +mf_O2 = 0.21 ;// mole fraction of O2 in air +mf_N2 = 0.79 ;//mole fraction of N2 in air +m_H2O = 18 ;// mol. wt. of H2O + +H_cl = (H2O*2)/m_H2O ;// lb mol of H in coal moisture +O_cl = H_cl/2 ;// lb mol of O in coal moisture + +H_air = (H2O_air*m_air )/m_H2O;// lb mol of H per lb mol air +O_air = H_air/2 ;// lb mol of O per lb mol air + +// Ash balance to get refuse(R) +R = ash/(ash_R/100) ;// Refuse-[lb] +//refuse composition +pub_cl = 14 ;// percentage of unburned coal in refuse-[%] +ub_cl = (14/100)*R ;// amount of unburned coal in refuse +C_p = (C/(100-ash))*ub_cl ;// C in unburned coal-[lb] +H_p = (H/(100-ash))*ub_cl ;// H in unburned coal-[lb] +O_p = (O/(100-ash))*ub_cl ;// O in unburned coal-[lb] +N_p = (N/(100-ash))*ub_cl ;// N in unburned coal-[lb] +S_p = (S/(100-ash))*ub_cl ;// S in unburned coal-[lb] +mol_C = C_p/w_C;// lb mol of C +mol_H = H_p/w_H ;// lb mol of H +mol_N = N_p/w_N ;// lb mol of N +mol_O = O_p/w_O ;// lb mol of O +mol_S = S_p/w_S ;// lb mol of S + +// Degree of freedom analysis +n_un = 4 ;// Number of unknowns in the given problem(excluding extent of reactions) +n_ie = 4 ;// Number of independent equations +d_o_f = n_un-n_ie ;// Number of degree of freedom +printf('Number of degree of freedom for the given system is %i \n\n',d_o_f); + +//Using element balance of C+S, N& H +P = (C/w_C + S/w_S - (mol_C+mol_S ))/.154 ;// mol of stack gas-[lb mol] +A = (2*P*.806 +2*mol_N-N/w_N)/(2*mf_N2) ;// mol of air -[lb mol] +W = (H/w_H +H_cl+H_air*A-mol_H)/2 ;// moles of exit water-[lb mol] +printf(' Moles of stack gas(P) - %.1f lb mol\n',P); +printf(' Moles of air (A) - %.1f lb mol \n',A); +printf(' Moles of exit water(W) - %.1f lb mol \n',W); +// by using P,W , A and O2 balance we get 19.8 = 20.3 , therefore difference is about 1% + +//Calculation of excess air +// For O2 required +C_req = (C/w_C)/1 ;// O2 required by entering C given by reaction C+O2--->CO2 -[lb mol] +H_req = (H/w_H)/4 ;// O2 required by entering H by given reaction H2+(1/2)*O2--->H20-[lb mol] +N_req = 0 ;// inert +O_req = (O/w_O)/2 ;// O2 required by entering O-[lb mol] +S_req = (S/w_S)/1 ;// O2 required by entering S-given by S+O2--->SO2 -[lb mol] +total_O2_req = C_req+H_req+N_req+O_req +S_req ;// Total oxygen required-[lb mol] +O2_in = A*mf_O2 ;// O2 entering in air +ex_air = 100*((O2_in-total_O2_req)/total_O2_req) ;//[% of excess air] +printf('\n Excess air is %.1f %%.\n',ex_air); \ No newline at end of file -- cgit