initial commit / add all books

9 files changed, 519 insertions, 0 deletions

diff --git a/409/CH10/EX10.1/Example10_1.sce b/409/CH10/EX10.1/Example10_1.sce
new file mode 100755
index 000000000..383a7bd8d
--- /dev/null
+++ b/409/CH10/EX10.1/Example10_1.sce
@@ -0,0 +1,56 @@
+clear;
+clc;
+// Example 10.1
+printf('Example 10.1\n\n');
+//Page no. 264
+// Solution
+
+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]
+
+// 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(' \n 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 extent of reaction is
+cn_CH4 = 67/100 ;// percentage conversion of CH4
+ex_r = (-cn_CH4)*CH4/vi_CH4 ;// extent of reaction
+printf(' extent of reaction is %.1f g moles reacting \n',ex_r);
+
+n_un = 11 ;// Number of unknowns in the given problem
+n_ie = 11 ;// 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 species balance using eqn.10.2
+vi_CH3Cl = 1;
+vi_HCl = 1;
+vi_N2 = 0;
+p_CH4 = CH4+(vi_CH4*ex_r);// [g mol]
+p_Cl2 = Cl2+(vi_Cl2*ex_r);// [g mol]
+p_CH3Cl = 0+(vi_CH3Cl*ex_r);// [g mol]
+p_HCl = 0+(vi_HCl*ex_r);// [g mol]
+p_N2 =  N2+(vi_N2*ex_r);// [g mol]
+// As we have taken F = 100 so answers we are getting can be directly used as percentage composition
+printf('\n\nComposition of product stream in %% g mol of products\n');
+printf('\nProduct            Percentage g mol\n');
+printf('\nCH4                %.1f%% g mol\n',p_CH4);
+printf('\nCl2                %.1f%% g mol\n',p_Cl2);
+printf('\nCH3Cl              %.1f%% g mol\n',p_CH3Cl);
+printf('\nHCl                %.1f%% g mol\n',p_HCl);
+printf('\nN2                 %.1f%% g mol\n',p_N2);
\ No newline at end of file
diff --git a/409/CH10/EX10.2/Example10_2.sce b/409/CH10/EX10.2/Example10_2.sce
new file mode 100755
index 000000000..8387a9b46
--- /dev/null
+++ b/409/CH10/EX10.2/Example10_2.sce
@@ -0,0 +1,46 @@
+clear ;
+clc;
+// Example 10.2
+printf('Example 10.2\n\n');
+// Page no. 266
+// Solution
+
+S = 5000 ;// Sulphur [lb]
+// Composition of feed
+CH4 =  80 ;// [%]
+H2S =  20 ;// [%]
+
+n_un = 11 ;// Number of unknowns in the given problem
+n_ie  = 11 ;// 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);
+m_S = 32.0 ;//molecular wt. of S -[lb]
+mol_S = S/32.0;
+// Extent of reaction can be calculated by using eqn. 9.3 
+// Based on S
+nio_S = 0 ;//[g mol S]
+ni_S = mol_S ;//[g mol S]
+vi_S = 3 ;// coefficint of S -from given reaction
+ex_r = (ni_S-nio_S)/vi_S ;//  Extent of reaction based on S
+printf(' Extent of reaction is %.1f g moles reacting \n',ex_r);
+
+// Product composition 
+vi_H2O = 2 ;// coefficint of H2O
+vi_H2S = -2 ;// coefficint of H2S
+vi_SO2 = -1 ;//coefficint of SO2
+vi_CH4 = 0 ;//coefficint of CH4
+P_H2O = 0+(vi_H2O*ex_r);// [lb mol]
+P_H2S = P_H2O/10 ;//[lb mol]
+P_SO2 = 3*P_H2S ;//[lb mol]
+
+F = (P_H2S-vi_H2S*ex_r)/(H2S/100) ;// total feed-[lb mol]
+F_SO2 = P_SO2-(vi_SO2*ex_r);// feed rate of SO2- [lb mol]
+F_CH4 = (CH4/100)*F+vi_CH4*ex_r ;//feed rate of CH4- [lb mol]
+F_H2S = ((H2S/100)*F) ;// feed rate of H2S-[lb mol]
+
+// We can see from situation that H2S is limiting reagent as ratio of SO2 to H2S in the product gas(3/1) is greater than their molar ratio in chemical reaction(2/1)
+f_cn = -(vi_H2S*ex_r)/((H2S/100)*F) ;// Fractional conversion of limiting reagent
+
+printf('\n(1)Feed rate of H2S-  %.1f lb mol\n',F_H2S);
+printf('(2)Feed rate of SO2-  %.1f lb mol\n',F_SO2);
+printf('(3)Fractional conversion of limiting reagent-  %.2f \n',f_cn);
\ No newline at end of file
diff --git a/409/CH10/EX10.3/Example10_3.sce b/409/CH10/EX10.3/Example10_3.sce
new file mode 100755
index 000000000..5bf1e29c8
--- /dev/null
+++ b/409/CH10/EX10.3/Example10_3.sce
@@ -0,0 +1,69 @@
+clear ;
+clc;
+// Example 10.3
+printf('Example 10.3\n\n');
+// Page no. 270
+// Solution
+
+F = 1 ;//CH3OH -[gmol]
+// Extent of reactions can be calculated by using eqn. 10.5
+// For reaction 1 based on CH3OH is limiting reagent
+f_cn = 90 ;//[%]
+vi_CH3OH = -1 ;//coefficint of CH3OH
+ex_r1 = (-90/100)/vi_CH3OH ;//  Extent of reaction based on CH3OH 
+printf(' Extent of reaction 1 is %.2f g moles reacting \n',ex_r1);
+//For reaction 2
+yld = 75 ;//[%]
+ex_r2 = ex_r1-(F*(yld/100));
+printf('  Extent of reaction 2 is %.2f g moles reacting \n',ex_r2);
+
+// 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 = 11 ;// Number of unknowns in the given problem
+n_ie  = 11 ;// 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);
+
+// Reaction 1
+v1_CH3OH = -1 ;//coefficint of CH3OH
+v1_O2 = -1/2 ;//coefficint of O2
+v1_CH2O = 1 ;//coefficint of CH2O
+v1_H2O = 1 ;//coefficint of H2O
+v1_CO = 0  ;//coefficient of CO
+//Reaction 2
+v2_O2 = -1/2 ;//coefficint of O2
+v2_CH2O = -1 ;//coefficint of CH2O
+v2_H2O = 1 ;//coefficint of H2O
+v2_CO = 1 ;//coefficient of CO
+P = F+air +(v1_CH3OH+v1_O2+v1_CH2O+v1_H2O)*ex_r1 +(v2_O2+v2_CH2O+v2_H2O+v2_CO)*ex_r2 ;// Product -[g mol]
+
+no_CH3OH = F+(v1_CH3OH*ex_r1)+0 ;// [g mol]
+no_O2 = n_O2+(v1_O2*ex_r1)+v2_O2*ex_r2 ;// [g mol]
+no_CH2O = 0 + v1_CH2O*ex_r1 +v2_CH2O*ex_r2 ;//[g mol]
+no_CO =  0+v1_CO*ex_r1 +v2_CO*ex_r2 ;//[g mol]
+no_H2O = 0+v1_H2O*ex_r1+v2_H2O*ex_r2 ;// [g mol]
+no_N2 =  n_N2-0-0 ;// [g mol]
+
+// Composition of product
+y_CH3OH = (no_CH3OH/P )*100 ;// mole %
+y_O2 = (no_O2/P)*100 ;// mole %
+y_CH2O = (no_CH2O/P)*100 ;// mole %
+y_CO = (no_CO/P)*100 ;// mole %
+y_H2O = (no_H2O/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
diff --git a/409/CH10/EX10.4/Example10_4.sce b/409/CH10/EX10.4/Example10_4.sce
new file mode 100755
index 000000000..f0906d54c
--- /dev/null
+++ b/409/CH10/EX10.4/Example10_4.sce
@@ -0,0 +1,48 @@
+clear;
+clc;
+// Example 10.4
+printf('Example 10.4\n\n');
+// Page no. 273
+// Solution
+
+F = 4000 ;//[kg]
+m_H2O = 18.02 ;// molecular masss of water
+m_C6H12O6 = 180.1 ;// molecular mass of glucose
+m_CO2 = 44 ;//molecular mass of CO2
+m_C2H3CO2H = 72.03 ;// molecular mass of C2H3CO2H
+m_C2H5OH = 46.05 ;// molecular mass of ethanol
+
+p_H2O = 88 ;// [%]
+p_C6H12O6 = 12;// [%] 
+ni_H2O = (F*p_H2O/100)/m_H2O ;// initial moles of water
+ni_C6H12O6 = (F*(p_C6H12O6/100))/m_C6H12O6 ;// initial moles of glucose
+
+// Degree of freedom analysis
+n_un = 9 ;// Number of unknowns in the given problem
+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);
+
+ur_C6H12O6 =  90 ;//[kg]
+pr_CO2 = 120 ;//[kg]
+nf_C6H12O6 = ur_C6H12O6/m_C6H12O6 ;// [kmoles]
+nf_CO2 = pr_CO2/m_CO2 ;// [kmoles]
+
+// solve following eqn. (b) and (e)simultaneously 
+//(b):  nf_C6H12O6 = ni_C6H12O6+-1*ex_r1+-1*ex_r2
+//(e):  nf_CO2 = 0+2*ex_r1+ 0*ex_r2
+a = [-1 -1;2 0];// matrix formed by coefficients of unknowns 
+b = [(nf_C6H12O6-ni_C6H12O6);nf_CO2];//matrix formed by constant
+x = a^(-1)*b;//matrix formed by solution
+printf(' Extent of reaction 1 is %.3f kg moles reacting \n',x(1));
+printf(' Extent of reaction 2 is %.3f kg moles reacting \n',x(2));
+
+nf_H2O = ni_H2O+0*x(1) +2*x(2);// from eqn. (a)-[kmoles]
+nf_C2H5OH = 0+2*x(1)+0*x(2);// from eqn.(c)-[kmoles]
+nf_C2H3CO2H = 0+0*x(1)+2*x(2) ;//from eqn.(d)-[kmoles]
+total_wt = m_H2O*nf_H2O + m_C6H12O6*nf_C6H12O6 + m_CO2*nf_CO2 + m_C2H3CO2H*nf_C2H3CO2H + m_C2H5OH*nf_C2H5OH;
+mp_C2H5OH = (m_C2H5OH*nf_C2H5OH*100)/total_wt ;// Mass percent of ethanol -[%]
+mp_C2H3CO2H = (m_C2H3CO2H*nf_C2H3CO2H*100)/total_wt ;//Mass percent of propenoic acid -[%]
+
+printf(' \n Mass percent of ethanol in broth at end of fermentation process is  %.1f %%\n',mp_C2H5OH);
+printf(' Mass percent of propenoic acid  in broth at end of fermentation process is  %.1f %%\n',mp_C2H3CO2H);
\ No newline at end of file
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
diff --git a/409/CH10/EX10.6/Example10_6.sce b/409/CH10/EX10.6/Example10_6.sce
new file mode 100755
index 000000000..13228fe9c
--- /dev/null
+++ b/409/CH10/EX10.6/Example10_6.sce
@@ -0,0 +1,30 @@
+clear ;
+clc;
+// Example 10.6
+printf('Example 10.6\n\n');
+// Page no. 281
+// Solution
+
+// Basis: P=100 // Product from the reactor-[g mol]
+P=100 ;//Product from the reactor-[g mol]
+// Composition of product
+C3H8 = 0.195*P ;// [g mol]
+C4H10 = 0.594*P ;// [g mol]
+C5H12 = 0.211*P;// [g mol]
+
+n_un = 3 ;// Number of unknowns in the given problem(excluding extent of reaction)
+n_ie  = 3 ;// 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);
+
+C = C3H8*3+C4H10*4+C5H12*5 ;// moles of C on product side
+H = C3H8*8+C4H10*10+C5H12*12 ;// moles of H on product side
+// Solve following eqn.( obtained by element balance of C & H) for F and G 
+//8F+0G = C
+//18F+2G = H
+a = [8 0;18 2] ;// matrix formed by coefficients of unknowns 
+b = [C;H] ;//matrix formed by constant
+x = a\b ;// matrix of solution
+
+R = x(2)/x(1) ;// Ratio of H2 consumed to C8H18 reacted  = G/F
+printf(' Molar ratio of H2 consumed to C8H18 reacted is %.3f  \n',R);
\ No newline at end of file
diff --git a/409/CH10/EX10.7/Example10_7.sce b/409/CH10/EX10.7/Example10_7.sce
new file mode 100755
index 000000000..3d90929e4
--- /dev/null
+++ b/409/CH10/EX10.7/Example10_7.sce
@@ -0,0 +1,19 @@
+clear ;
+clc;
+// Example 10.7
+printf('Example 10.7\n\n');
+// Page no. 286
+// Solution
+
+C3H8 = 20 ;//C3H8 burned in a test-[kg]
+m_C3H8 = 44.09 ;// mol. wt . of 1 kmol C3H8
+cf_O2 = 5 ;// coefficient of O2 in given reaction
+air = 400 ;// Air given -[kg]
+m_air = 29 ;// molecular wt. of  1kmol air-[kg]
+O2p = 21 ;// percentage of O2 in air-[%]
+p_CO2 =  44 ;// CO2 produced -[kg]
+p_CO = 12 ;//  CO produced -[kg]
+O2 = (air*O2p/100)/(m_air) ;// amount of entering O2-[k mol]
+rqO2 = (C3H8*cf_O2)/(m_C3H8) ;// Required O2 for given reaction
+ex_air = ((O2-rqO2)*100)/rqO2 ;// Excess air percent-[%]
+printf('Excess air percent is %.0f %%.\n',ex_air);
\ No newline at end of file
diff --git a/409/CH10/EX10.8/Example10_8.sce b/409/CH10/EX10.8/Example10_8.sce
new file mode 100755
index 000000000..939ec1e26
--- /dev/null
+++ b/409/CH10/EX10.8/Example10_8.sce
@@ -0,0 +1,48 @@
+clear;
+clc;
+// Example 10.8
+printf('Example 10.8\n\n');
+// Page no. 287
+// Solution
+
+F = 16 ;// feed of CH4 -[kg]
+CH4p = 100 ;//[%]
+m_CH4 = 16 ;// mass of kmol of CH4-[kg]
+mol_CH4 = (F*CH4p/100)/m_CH4;//k moles of CH4 in feed-[kmol]
+air = 300 ;// Air given -[kg]
+m_air = 29 ;// molecular wt. of  1kmol air-[kg]
+mol_air = air/m_air ;// kmoles of air-[kmol]
+O2p = 21 ;// percentage of O2 in air-[%]
+O2 = (mol_air*O2p/100)  ;// amount of entering O2-[k mol]
+N2 = mol_air-O2 ;// amount of entering N2-[k mol]
+
+// Degree of freedom analysis
+n_un = 8 ;// Number of unknowns in the given problem(excluding extent of reactions)
+n_ie  = 8 ;// 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 analysis using element balance of C,H,O and N
+p_N2 = N2 ;// inert 
+C_in = 1*mol_CH4 ;// kmoles of carbon in input-[kmol]
+H_in = 4*mol_CH4 ;// kmoles of hydrogen in input-[kmol]
+O_in = 2*O2 ;// kmoles of oxygen in input-[kmol]
+p_CO2 = C_in/1 ;//kmoles of  CO2 in product obtained  by carbon balance-[kmol]
+p_H2O = H_in/2 ;//kmoles of  H2O in product obtained  by hydrogen balance-[kmol]
+p_O2 = (O_in-(2*p_CO2+p_H2O))/2 ;//kmoles of  O2 in product obtained  by oxygen balance-[kmol]
+p_CH4 = 0 ;// Complete reaction occurs
+P = p_CH4 + p_N2+  p_CO2 + p_H2O + p_O2;
+
+y_N2 = p_N2*100/P ;//[mol %]
+y_CO2 = p_CO2*100/P ;//[mol %]
+y_H2O = p_H2O*100/P ;//[mol %]
+y_O2 = p_O2*100/P ;//[mol %]
+y_CH4 = p_CH4*100/P ;//[mol %]
+
+printf('\nComposition of product\n');
+printf('Component        mole percent\n');
+printf(' CH4             %.1f %%\n',y_CH4);
+printf(' O2              %.1f %%\n',y_O2);
+printf(' CO2             %.1f %%\n',y_CO2);
+printf(' H2O             %.1f %%\n',y_H2O);
+printf(' N2              %.1f %%\n',y_N2);
\ No newline at end of file
diff --git a/409/CH10/EX10.9/Example10_9.sce b/409/CH10/EX10.9/Example10_9.sce
new file mode 100755
index 000000000..51b147e29
--- /dev/null
+++ b/409/CH10/EX10.9/Example10_9.sce
@@ -0,0 +1,85 @@
+clear;
+clc;
+// Example 10.9
+printf('Example 10.9\n\n');
+// Page no. 290
+// Solution
+
+F = 100  ;// feed of coal -[lb]
+// given coal composition-given
+C = 83.05 ;//[%]
+H = 4.45 ;//[%]
+O = 3.36 ;// [%]
+N = 1.08 ;// [%]
+S = 0.70  ;//[%]
+ash = 7.36;//[%]
+H2O = 3.9 ;//[%]
+w_C = 12 ;// mol. wt. of C
+w_H =  1.008;//mol. wt. of H
+w_O = 16 ;// mol. wt. of O
+w_N = 14 ;// mol. wt. of N
+w_S = 32  ;//mol. wt. of S
+
+//given stack gas analysis-given
+CO2 =  15.4 ;//[%]
+CO = 0.0 ;//[%]
+O2 = 4.0 ;// [%]
+N2 = 80.6 ;//[%]
+//given refuse analysis
+ash_R = 86 ;//[%]
+odr = 14 ;//[%]
+
+H2O_air =  .0048 ;// [lb H2O/lb dry air]
+m_air = 29 ;// mol. wt. of air
+mf_O2 = 0.21 ;// mole fraction of  O2 in air
+mf_N2 = 0.79 ;//mole fraction of  N2 in air
+m_H2O =  18 ;// mol. wt. of H2O
+
+H_cl = (H2O*2)/m_H2O ;// lb mol of H in coal moisture
+O_cl = H_cl/2 ;// lb mol of O in coal moisture
+
+H_air = (H2O_air*m_air )/m_H2O;// lb mol of H per lb mol air
+O_air = H_air/2 ;// lb mol of O per lb mol  air 
+
+// Ash balance to get refuse(R)
+R = ash/(ash_R/100) ;// Refuse-[lb]
+//refuse composition
+pub_cl = 14 ;// percentage of unburned coal in refuse-[%]
+ub_cl =  (14/100)*R ;// amount of unburned coal in refuse
+C_p = (C/(100-ash))*ub_cl ;//  C in unburned coal-[lb]
+H_p = (H/(100-ash))*ub_cl ;//  H in unburned coal-[lb]
+O_p =  (O/(100-ash))*ub_cl ;//  O in unburned coal-[lb]
+N_p =  (N/(100-ash))*ub_cl ;//  N in unburned coal-[lb]
+S_p =  (S/(100-ash))*ub_cl ;//  S in unburned coal-[lb]
+mol_C =  C_p/w_C;// lb mol of C
+mol_H = H_p/w_H ;// lb mol of H
+mol_N = N_p/w_N ;// lb mol of N
+mol_O = O_p/w_O ;// lb mol of O
+mol_S = S_p/w_S ;// lb mol of S 
+
+// Degree of freedom analysis 
+n_un = 4 ;// Number of unknowns in the given problem(excluding extent of reactions)
+n_ie  = 4 ;// 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\n',d_o_f);
+
+//Using element balance of C+S, N& H
+P = (C/w_C + S/w_S - (mol_C+mol_S ))/.154 ;// mol of stack gas-[lb mol]
+A = (2*P*.806 +2*mol_N-N/w_N)/(2*mf_N2) ;// mol of air -[lb mol]
+W = (H/w_H +H_cl+H_air*A-mol_H)/2 ;// moles of exit water-[lb mol]
+printf(' Moles of stack gas(P)   -      %.1f  lb mol\n',P);
+printf(' Moles of air (A)        -      %.1f lb mol \n',A);
+printf(' Moles of exit water(W)  -      %.1f lb mol \n',W);
+// by using P,W , A and O2 balance we get 19.8 =  20.3 , therefore difference is about 1%
+
+//Calculation of excess air
+// For O2 required
+C_req =  (C/w_C)/1 ;// O2 required by entering C given by reaction C+O2--->CO2 -[lb mol]
+H_req = (H/w_H)/4 ;// O2 required by entering H by given reaction H2+(1/2)*O2--->H20-[lb mol]
+N_req = 0 ;// inert
+O_req = (O/w_O)/2 ;// O2 required by entering O-[lb mol]
+S_req = (S/w_S)/1 ;// O2 required by entering S-given by S+O2--->SO2 -[lb mol]
+total_O2_req =  C_req+H_req+N_req+O_req +S_req ;// Total oxygen required-[lb mol]
+O2_in = A*mf_O2 ;// O2 entering in air
+ex_air = 100*((O2_in-total_O2_req)/total_O2_req) ;//[% of excess air]
+printf('\n Excess air is %.1f %%.\n',ex_air);
\ No newline at end of file