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+clear ;
+clc;
+// Example 8.6
+printf('Example 8.6\n\n');
+// Page no. 209
+// Solution
+
+// Composition of initial solution at 30 degree C
+s_30 = 38.8 ;//  solublity of Na2CO3 at 30 degree C, by using the table for solublity of Na2CO3-[g Na2CO3/100 g H2O]
+If_Na2CO3 =  s_30/(s_30+100) ;// Initial mass fraction of Na2CO3
+If_H2O = 1-If_Na2CO3 ;// Initial mass fraction of H2O
+
+// Composition of crystals
+// Basis : 1g mol Na2CO3.10H2O
+n_mol_Na2CO3 = 1 ;// Number of moles of Na2CO3
+n_mol_H2O = 10 ;// Number of moles of H2O
+mwt_Na2CO3 = 106 ;// mol. wt of Na2CO3
+mwt_H2O = 18 ;// mol. wt of H2O
+m_Na2CO3 = mwt_Na2CO3*n_mol_Na2CO3 ;// Mass of Na2CO3
+m_H2O = mwt_H2O*n_mol_H2O ;// Mass of H2O
+Cf_Na2CO3 =  m_Na2CO3/(m_Na2CO3+m_H2O) ;// mass fraction of Na2CO3 
+Cf_H2O = 1-Cf_Na2CO3 ;// mass fraction of H2O
+
+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);
+
+// Final composition of tank
+//Basis :I = 10000 kg
+// Material balance reduces to Accumulation  =  final -initial = in-out(but in = 0)
+I = 10000 ;//initial amount of saturated solution-[kg]
+amt_C = 3000 ;// Amount of crystals formed-[kg]
+Fm_Na2CO3 = I*If_Na2CO3-amt_C*Cf_Na2CO3 ;// Mass balance of Na2CO3
+Fm_H2O = I*If_H2O-amt_C*Cf_H2O ;// Mass balance of H2O
+
+//To find temperature,T
+s_T =  (Fm_Na2CO3/Fm_H2O)*100 ;// Solublity of Na2CO3 at temperature T
+s_20 = 21.5 ;//Solublity of Na2CO3 at temperature 20 degree C ,from given table-[g Na2CO3/100 g H2O]
+// Find T by interpolation
+T =  30-((s_30-s_T)/(s_30-s_20))*(30-20) ;// Temperature -[degree C]
+printf(' Temperature to which solution has to be cooled to get 3000 kg crystals is %.0f degree C .\n',T);
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