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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 | |
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initial commit / add all books
Diffstat (limited to '409/CH11')
-rwxr-xr-x | 409/CH11/EX11.1/Example11_1.sce | 57 | ||||
-rwxr-xr-x | 409/CH11/EX11.2/Example11_2.sce | 48 | ||||
-rwxr-xr-x | 409/CH11/EX11.3/Example11_3.sce | 71 | ||||
-rwxr-xr-x | 409/CH11/EX11.4/Example11_4.sce | 90 | ||||
-rwxr-xr-x | 409/CH11/EX11.5/Example11_5.sce | 40 |
5 files changed, 306 insertions, 0 deletions
diff --git a/409/CH11/EX11.1/Example11_1.sce b/409/CH11/EX11.1/Example11_1.sce new file mode 100755 index 000000000..a0779b072 --- /dev/null +++ b/409/CH11/EX11.1/Example11_1.sce @@ -0,0 +1,57 @@ +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 diff --git a/409/CH11/EX11.2/Example11_2.sce b/409/CH11/EX11.2/Example11_2.sce new file mode 100755 index 000000000..9a7ea8821 --- /dev/null +++ b/409/CH11/EX11.2/Example11_2.sce @@ -0,0 +1,48 @@ +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 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);
\ No newline at end of file diff --git a/409/CH11/EX11.4/Example11_4.sce b/409/CH11/EX11.4/Example11_4.sce new file mode 100755 index 000000000..32547d31e --- /dev/null +++ b/409/CH11/EX11.4/Example11_4.sce @@ -0,0 +1,90 @@ +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 diff --git a/409/CH11/EX11.5/Example11_5.sce b/409/CH11/EX11.5/Example11_5.sce new file mode 100755 index 000000000..75ef0e680 --- /dev/null +++ b/409/CH11/EX11.5/Example11_5.sce @@ -0,0 +1,40 @@ +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');
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