+clear ;

+clc;

+// Example 11.1

+printf('Example 11.1\n\n');

+// Page no. 311

+// Solution

+

+// Composition of each stream

+w_A1 = 1 ;//concentration of A in 1

+w_B2 = 1 ;// concentration of B in 2

+w_A3 = 0.8 ;// concentration of A in 3

+w_B3 = 0.2 ;// concentration of B in 3

+w_C4 = 1 ;// concentration of C in 4

+w_A5 = 0.571 ;//concentration of A in 5

+w_B5 = 0.143 ;//concentration of B in 5

+w_C5 = 0.286 ;//concentration of C in 5

+w_D6 = 1;// concentration of D in 6

+w_A7 = 0.714 ;// concentration of A in 7

+w_B7 = 0.286 ;// concentration of B in 7

+w_B8 = 0.333 ;//concentration of B in 8

+w_C8 = .667 ;//concentration of C in 8

+

+us1 = 2 ;// Species involved in unit 1

+us2 = 3 ;// Species involved in unit 2

+us3 = 4 ;// Species involved in unit 3

+total_sp = us1+us2+us3 ;// Total species in system

+

+// Element balance of all systems

+printf('Number of possible equations are 9, they are as follows- \n');

+printf(' Subsystem 1\n');

+printf(' A: F1*w_A1+F2*0 = F3*w_A3 (a)\n');

+printf(' B:F1*0 + F2*w_B2 = F3*w_B3 (b)\n');

+printf(' Subsystem 2\n');

+printf(' A: F3*w_A3+F4*0 = F5*w_A5 (c)\n');

+printf(' B:F3*w_B3 + F4*0 = F5*w_B5 (d)\n');

+printf(' C: F3*0+F4*w_C4 = F5*w_C5 (e)\n');

+printf(' Subsystem 3\n');

+printf(' A: F5*w_A5+F6*0 = F7*w_A7+F8*0 (f)\n');

+printf(' B:F5*w_B5 + F6*0 = F7*0+F8*w_B8 (g)\n');

+printf(' C: F5*w_C5+F6*0 = F7*0+F8*w_C8 (h)\n');

+printf(' D:F5*w_C5+F6*0 = F7*0+F8*w_C8 (i)\n');

+printf('\n The above equations do not form a unique set\n');

+

+// By inspection we can see that only 7 equations are independent

+//Independent Equations are:

+// Subsystem 1

+//A: F1*w_A1+F2*0 = F3*w_A3 (a)

+//B:F1*0 + F2*w_B2 = F3*w_B3 (b)

+//Subsystem 2

+//A: F3*w_A3+F4*0 = F5*w_A5 (c)

+// C: F3*0+F4*w_C4 = F5*w_C5 (e)

+// Subsystem 3

+//A: F5*w_A5+F6*0 = F7*w_A7+F8*0 (f)

+//B:F5*w_B5 + F6*0 = F7*0+F8*w_B8 (g)

+//D:F5*w_C5+F6*0 = F7*0+F8*w_C8 (i)

+

+printf('\n Number of independent equations are 7 \n'); \ No newline at end of file

+clear;

+clc;

+// Example 11.2

+printf('Example 11.2\n\n');

+// Page no.315

+// Solution

+

+//Basis:1 hr

+G = 1400 ;//[kg]

+//Unit 1

+// Degree of freedom analysis

+n_un = 16 ;// Number of unknowns in the given problem(excluding extent of reactions)

+n_ie = 16 ;// Number of independent equations

+d_o_f = n_un-n_ie ;// Number of degree of freedom

+printf('For unit 1 number of degree of freedom for the given system is %i .\n',d_o_f);

+//Given

+o1_air = 0.995 ;// Mass fraction of air at out of unit 1 in A

+i1_air = 0.95 ;// Mass fraction air at in of unit 1 in G

+i1_wtr = 0.02;// Mass fraction water at in of unit 1 in G

+F1_wtr = 0.81 ;// Mass fraction of water at out of unit 1 in F

+o1_wtr = 0.005 ;// Mass fraction of water at out of unit 1 in A

+o2_wtr = 0.96 ;// Mass fraction of water at out of unit 2 in B

+o3_wtr = 0.01;// Mass fraction of water at out of unit 3 in D

+i1_act = 0.03 ;// Mass fraction of acetone at in of unit 1 in G

+F1_act = 0.19 ;// Mass fraction of acetone at out of unit 1 in F

+o3_act = 0.99 ;// Mass fraction of acetone at out of unit 3 in D

+o2_act = 0.04 ;// Mass fraction of acetone at out of unit 2 in B

+

+//Mass balance to get A ,W & F

+A = G*i1_air/o1_air ;//air-[kg]

+F = G*i1_act/F1_act ;//[kg]

+W = (F*F1_wtr+A*o1_wtr-G*i1_wtr)/1 ;//Pure water in -[kg]

+// unit 2 and 3

+// 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(' For unit 2 and 3 number of degree of freedom for the given system is %i .\n',d_o_f);

+// Mass balance

+// solving eqn (d)& (e) simultaneously for D and B

+a = [o3_act o2_act;o3_wtr o2_wtr];// Matrix formed by coefficients of unknown

+b = [F*F1_act;F*F1_wtr];// Matrix formed by constant

+x = a\b ;// Solution matrix-x(1) = D and x(2) = B

+printf('\n W-Pure water in to unit 1 - %.2f kg/hr\n',W);

+printf(' A-Air out of unit 1 - %.2f kg/hr\n',A);

+printf(' F-out of unit 1 - %.2f kg/hr\n',F);

+printf(' B-out of unit 2 - %.2f kg/hr\n',x(2));

+printf(' D-out of unit 3 - %.2f kg/hr\n',x(1)); \ No newline at end of file

+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); \ No newline at end of file

+clear ;

+clc;

+// Example 11.4

+printf('Example 11.4\n\n');

+// Page no. 322

+// Solution fig E11.4

+

+//Basis : M = 1000 lb

+M = 1000 ;//[lb]

+//Given

+F_s = 16/100 ;// Fraction of sugar in F

+F_w = 25/100 ;// Fraction of water in F

+F_p = 59/100 ; // Fraction of pulp in F

+D_p = 80/100 ; // Fraction of pulp in D

+E_s = 13/100 ;// Fraction of sugar in E

+E_p = 14/100 ;// Fraction of pulp in E

+G_p = 95/100 ;// Fraction of pulp in G

+H_s = 15/100 ;// Fraction of sugar in H

+K_s = 40/100 ;// Fraction of sugar in K

+

+// For crystallizer equations are

+K_w = 1 - K_s ;// summation of wt. fraction is 1

+K = M/K_s ;// By sugar balance -[lb]

+L = K_w*K ;// By water balance -[lb]

+

+// For evaporator equations are

+H_w = 1- H_s ;//summation of wt. fraction is 1

+H = K_s*K/H_s ;//By sugar balance -[lb]

+J = H - K; //By overall balance -[lb]

+

+// For screen equations are

+E_w = 1 - (E_p + E_s) ; // summation of wt. fraction is 1

+// solve E and G by simultaneous eqn. obtained by overall and pulp balance

+a1 = [1 -1;E_p -G_p] ;// Matrix of coefficients of unknown

+b1 = [H;0] ;//Matrix of constants

+x1 = a1\b1 ;// Matrix of solutions ,x1(1) = E, x1(2) = G

+E = x1(1) ;//[lb]

+G = x1(2) ;//[lb]

+G_s = (E_s*E - H_s *H )/G ;// By sugar balance

+G_w = 1 -(G_s + G_p) ;// summation of wt. fraction is 1

+

+// For mill equations are

+// solve F and D by simultaneous eqn. obtained by overall and pulp balance

+a2 = [1 -1;F_p -D_p] ;// Matrix of coefficients of unknown

+b2 = [E;E_p*E] ;//Matrix of constants

+x2 = a2\b2 ;// Matrix of solutions ,x2(1) = F, x2(2) = D

+F = x2(1) ;//[lb]

+D = x2(2) ;//[lb]

+D_s = (F_s*F - E_s *E )/D ;// By sugar balance

+D_w = 1 -(D_s + D_p) ; // summation of wt. fraction is 1

+

+S_rec = M/(F*F_s) ; // Fraction of sugar recovered

+

+printf('\nFlow streams and their respective compositions.\n');

+printf('\n M = %.0f lb \n',M);

+printf(' Sugar: %.2f \n',1);

+

+printf('\n L = %.0f lb \n',L);

+printf(' Water: %.2f\n',1);

+

+printf('\n K = %.0f lb \n',K);

+printf(' Sugar: %.2f\n',K_s);

+printf(' Water: %.2f\n',K_w);

+

+printf('\n J = %.0f lb \n',J);

+printf(' Water: %.2f \n',1);

+

+printf('\n H = %.0f lb \n',H);

+printf(' Sugar: %.2f\n',H_s);

+printf(' Water: %.2f\n',H_w);

+

+printf('\n G = %.0f lb \n',G);

+printf(' Sugar: %.3f\n',G_s);

+printf(' Water: %.3f\n',G_w);

+printf(' Pulp : %.2f\n',G_p);

+

+printf('\n E = %.0f lb \n',E);

+printf(' Sugar: %.2f\n',E_s);

+printf(' Water: %.2f\n',E_w);

+printf(' Pulp : %.2f\n',E_p);

+

+printf('\n D = %.0f lb \n',D);

+printf(' Sugar: %.3f\n',D_s);

+printf(' Water: %.3f\n',D_w);

+printf(' Pulp : %.2f\n',D_p);

+

+printf('\n F = %.0f lb \n',F);

+printf(' Sugar: %.2f\n',F_s);

+printf(' Water: %.2f\n',F_w);

+printf(' Pulp : %.2f\n',F_p); \ No newline at end of file

+clear ;

+clc;

+// Example 11.5

+printf('Example 11.5\n\n');

+// Page no.324

+// Solution

+

+// Option 1

+F = 15 ;//[L/hr]

+cs_in = 10 ;//Nutrient conc. input vessel - [g nutrient/L substrate]

+V1 = 100 ;// [L]

+V2 = 50 ;//[L]

+Yxs = 0.2 ;// [cells/g]

+umax = 0.4 ;//[hr^ - 1]

+Ks = 2 ;//[g/L] - Monod constant

+// Use eqn. 10.1 for balances and Monod eqns. applies to each vessel

+//Cells: 0 - F/V * x_out + u * x_out - 0 = 0....(a)

+//Nutrient: F/V * cs_in - F/V * cs_out + 0 - (u * x_out)/(Yxs) = 0.....(b)

+//From eqn.(a) F/V = u(dilution rate)...(c)

+// From eqn. (b) x_out = Yxs(cs_in - cs_out)....(d)

+u1 = F/V1 ;//[hr^ - 1] //[hr^ - 1]

+cs_out = (Ks * u1/umax)/(1 - (u1/umax)) ;//Nutrient conc. output vessel - [g nutrient/L substrate]

+// Find x_out by eqn. (d)

+x_out = Yxs * (cs_in - cs_out) ;//[g cells / L substrate]

+

+//Option 2

+//For vessel 1

+u2 = F/V2;

+cs_out1 = (Ks * u2/umax)/(1 - (u2/umax)) ;//Nutrient conc. output vessel - [g nutrient/L substrate]

+x_out1 = Yxs * (cs_in - cs_out1) ;//[g cells / L substrate]

+// For vessel 2

+// Eqn. (a) is now F/V * x_out1 - F/V * x_out2 + u3 * x_out2 = 0...(e)

+// Eqn. (b) is now F/V * cs_out1 - F/V * cs_out2 + (u3 * x_out2)/Yxs = 0...(f)

+// u3 = (umax * cs_out2) / (Ks + cs_out2).. Monod Eqn...(g)

+// (e),(f) and (g) form a non - linear set of equations , solving them we get cs_out2 = 1.35 g nutrient/L substrate and x_out2 = 1.73 g cells/L

+x_out2 = 1.73 ;// From eqn. (e),(f) and (g) - [g cells / L substrate]

+

+printf('g cells/L from option 1 is %.2f.\n',x_out);

+printf(' g cells/L from option 2 is %.2f.\n',x_out2);

+printf(' By comparing option 1 and option 2 the respective answers are essentially the same.\n'); \ No newline at end of file