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+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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