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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /409/CH11/EX11.3 | |
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diff --git a/409/CH11/EX11.3/Example11_3.sce b/409/CH11/EX11.3/Example11_3.sce new file mode 100755 index 000000000..4d704dcc5 --- /dev/null +++ b/409/CH11/EX11.3/Example11_3.sce @@ -0,0 +1,71 @@ +clear ; +clc; +// Example 11.3 +printf('Example 11.3\n\n'); +//Page no. 318 +// Solution + +P = 6205 ;//[lb mol/hr] +//Given +amt_F = 560 ;//[bbl] +// Fuel oil(F) analysis +C_F = 0.50 ;// [mol fraction] +H2_F = 0.47 ;//[mol fraction] +S_F = 0.03 ;//[mol fraction] +// Natural Gas(G) analysis +CH4_G = 0.96 ;//[mol fraction] +C2H2_G = 0.02 ;//[mol fraction] +CO2_G = 0.02 ;//[mol fraction] +// Analysis of air into Gas furnace(A) +O2_A = 0.21 ;//[mol fraction] +N2_A = 0.79 ;//[mol fraction] +// Analysis of air into Oil furnace(A1) +O2_A1 = 0.20 ;//[mol fraction] +N2_A1 = 0.76 ;//[mol fraction] +CO2_A1 = 0.04 ;//[mol fraction] +//Stack gas(P) analysis +N2_P = .8493 ;//[mol fraction] +O2_P = .0413 ;//[mol fraction] +SO2_P = .0010 ;// [mol fraction] +CO2_P = .1084 ;//[mol fraction] + +// Degree of freedom analysis +n_un = 5;// Number of unknowns in the given problem(excluding extent of reactions) +n_ie = 5 ;// 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); + +// Elemental mole balance for 2N,2H,2O,S and C +// Use S balance to get F +F = P* SO2_P/S_F ;// [lb mol/hr] +//Solve other four balances to get G +//2H: G*(2*CH4_G+C2H2_G)+F*H2_F-W*1 +//2N: A*N2_A+A1*N2_A1 = P*N2_P +//2O: A*(O2_A)+A1*(O2_A1+CO2_A1)+G*CO2_G-W*(1/2) = P*(O2_P+CO2_P+SO2_P) +//C: G*(CH4_G+2*C2H2_G+CO2_G)+F*C_F+A1*CO2_A1 = P*CO2_P +//Solving above eqns. by matrix method[G W A A1] +a = [2*CH4_G+C2H2_G -1 0 0;0 0 N2_A N2_A1;CO2_G -.5 O2_A O2_A1+CO2_A1;CH4_G+2*C2H2_G+CO2_G 0 0 CO2_A1];// matrix of coefficients +b = [-F*H2_F;P*N2_P;P*(O2_P+CO2_P+SO2_P);(P*CO2_P-F*C_F)];// matrix of constants +x = a\b ;// matrix of solutions x(1) = G,x(2) = W,x(3) = A & x(3) = A1 +G = x(1);//[lb mol/hr] +m_F = 7.91 ;// Molecular wt. of fuel oil-[lb] +Fc = (F*m_F)/(7.578*42);// Fuel gas consumed -[bbl/hr] +time = amt_F/Fc ;// Time for which available fuel gas lasts-[hr] +printf('(1) Fuel gas consumed(F) is %.2f bbl/hr .\n',Fc); +printf('(2) Time for which available fuel gas lasts is %.0f hr .\n',time); + +// For increase in arsenic and mercury level +F_oil = Fc*42; //[gal/hr] +Em_ars2 = (3.96 *10^(-4))/1000 ;// [lb/gal] +Em_Hg2 = (5.92 *10^(-4))/1000 ;// [lb/gal] +ars_F = F_oil*Em_ars2 ;// Arsenic produced on burning oil-[lb] +Hg_F = F_oil*Em_Hg2 ;//Mercury produced on burning oil-[lb] +G_gas = G*359 ;//[ft^3/hr] +Em_ars1 = (2.30 *10^(-4))/10^6 ;// [lb/ft^3] +Em_Hg1 = (1.34 *10^(-4))/10^6 ;// [lb/ft^3] +ars_G = G_gas*Em_ars1; // Arsenic produced on burning Natural gas-[lb] +Hg_G = G_gas*Em_Hg1 ;//Mercury produced on burning Natural Gas-[lb] +in_ars = ((ars_F-ars_G)/ars_G)*100 ;//[% increase in Arsenic emission] +in_Hg = ((Hg_F-Hg_G)/Hg_G)*100 ; //[% increase in Mercury emission] +printf('(3) Increase in Arsenic emission is %.1f %% .\n',in_ars); +printf('(4) Increase in Mercury emission is %.1f %% .\n',in_Hg);
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