diff options
Diffstat (limited to '409/CH10/EX10.5/Example10_5.sce')
| -rwxr-xr-x | 409/CH10/EX10.5/Example10_5.sce | 118 |
1 files changed, 118 insertions, 0 deletions
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 |