initial commit / add all books

7 files changed, 226 insertions, 0 deletions

diff --git a/409/CH8/EX8.1/Example8_1.sce b/409/CH8/EX8.1/Example8_1.sce
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index 000000000..9dcc08689
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+clear ;

+clc;

+// Example 8.1

+printf('Example 8.1\n\n');

+//Page no. 197

+// Solution

+

+// Basis : 1 min

+d_w =  1.0 ;// Density of aqueous solution-[g/cubic metre]

+d_sol =  0.6 ;// Density of organic solvent-[g/cubic metre]

+

+n_un = 8 ;// Number of unknowns in the given problem

+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);

+

+// Material balance of Strep.

+x =  (200*10+10*0-200*0.2)/10;//[g]

+printf('Strep per litre of solvent is  %.1f g .\n',x);

+

+cnc = x/(1000*d_sol) ;//[g Strep/g of S]

+printf('Strep per gram of solvent is  %.4f g Strep/g of S .\n',cnc);

+

+m_fr = cnc/(1+cnc) ;//Mass fraction

+printf('Mass fraction of Strep is  %.3f g .\n',m_fr);
\ No newline at end of file
diff --git a/409/CH8/EX8.2/Example8_2.sce b/409/CH8/EX8.2/Example8_2.sce
new file mode 100755
index 000000000..4cf04244e
--- /dev/null
+++ b/409/CH8/EX8.2/Example8_2.sce
@@ -0,0 +1,24 @@
+clear ;

+clc;

+// Example 8.2

+printf('Example 8.2\n\n');

+// Page no. 199

+// Solution Fig. E8.2b

+

+F_O2 = 0.21 ;// fraction of O2 in feed(F) 

+F_N2 = 0.79 ;// fraction of N2 in feed(F) 

+P_O2 = 0.25 ;// fraction of O2 in product(P)

+P_N2 = 0.75 ;// fraction of N2 in  product(P)

+F = 100 ;// Feed - [g mol]

+w = 0.80 ;// Fraction of waste

+W = w*F ;// Waste -[g mol]

+

+// By analysis for degree of freedom , DOF comes to be zero 

+P = F - W ;// By overall balance - [g mol]

+W_O2 = (F_O2*F - P*P_O2)/100 ;// Fraction of O2 in waste stream by O2 balance 

+W_N2 = (W - W_O2*100)/100 ;//Fraction of N2 in waste stream

+ 

+printf('Composition of Waste Stream\n' );

+printf('\n  Component             Fraction in waste stream\n' );

+printf('  O2                    %.2f\n',W_O2 );

+printf('  N2                    %.2f\n',W_N2 );
\ No newline at end of file
diff --git a/409/CH8/EX8.3/Example8_3.sce b/409/CH8/EX8.3/Example8_3.sce
new file mode 100755
index 000000000..5e37c0351
--- /dev/null
+++ b/409/CH8/EX8.3/Example8_3.sce
@@ -0,0 +1,31 @@
+clear;

+clc;

+// Example 8.3

+printf('Example 8.3\n\n');

+// Page no. 202

+// Solution

+

+// Basis : 1 hr so F  = 1000 kg

+F = 1000 ;// feed rate-[kg/hr]

+P = F/10 ;// product mass flow rate -[kg/hr]

+

+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);

+

+//  Overall mass balance: F  =  P+B

+B =  F-P ;// bottom mass flow rate -[kg/hr]

+printf('\n Bottom mass flow rate -              %.1f kg \n',B);

+

+// Composition of bottoms by material balances

+m_EtOH = 0.1*F-0.6*P ;// By EtOH balance-[kg]

+m_H2O = 0.9*F - 0.4*P ;// By H2O balance-[kg]

+total  = m_EtOH+m_H2O ;//[kg]

+f_EtOH = m_EtOH/total ;// Mass fraction of EtOH

+f_H2O = m_H2O/total ;// Mass fraction of H2O

+

+printf(' Mass of EtOH in bottom -             %.1f kg \n',m_EtOH);

+printf(' Mass of H2O in bottom -              %.1f kg \n',m_H2O);

+printf(' Mass fraction of EtOH in bottom -    %.3f \n',f_EtOH);

+printf(' Mass fraction of H2O in bottom -     %.3f \n',f_H2O);
\ No newline at end of file
diff --git a/409/CH8/EX8.4/Example8_4.sce b/409/CH8/EX8.4/Example8_4.sce
new file mode 100755
index 000000000..2fbdc4b94
--- /dev/null
+++ b/409/CH8/EX8.4/Example8_4.sce
@@ -0,0 +1,26 @@
+clear ;

+clc;

+// Example 8.4

+printf('Example 8.4\n\n');

+// Page no. 205

+// Solution Fig E8.4

+

+// Given

+A = 200 ;// Mass of added solution [kg] 

+P_H2SO4 = .1863 ;//Fraction of H2SO4 in P(Final solution)

+P_H2O = .8137 ;//Fraction of H2O in P(Final solution)

+A_H2SO4 = .777 ;//Fraction of H2SO4 in A(Added solution)

+A_H2O = .223 ;//Fraction of H2O in A(Added solution)

+F_H2SO4 = .1243 ;//Fraction of H2SO4 in F(Original solution)

+F_H2O = .8757 ;//Fraction of H2O in F(Original solution)

+

+// By analysis for degree of freedom , DOF comes to be zero 

+// Solve following equations simultaneously for F and P,

+// P*P_H2O - F*F_H2O = A*A_H2O - By H2O balance

+// P - F = A - By overall balance

+a = [P_H2O -F_H2O;1 -1] ;// Matrix of coefficient

+b = [A*A_H2O;A] ;// Matrix of contants

+x = a\b ;// Matrix of solutions- P = x(1) and F = x(2)

+

+printf(' Original solution taken-                                         %.0i kg\n',x(2) );

+printf('  Final solution or kilograms of battery acid formed-              %.0i kg\n',x(1) );
\ No newline at end of file
diff --git a/409/CH8/EX8.5/Example8_5.sce b/409/CH8/EX8.5/Example8_5.sce
new file mode 100755
index 000000000..e068a6ec0
--- /dev/null
+++ b/409/CH8/EX8.5/Example8_5.sce
@@ -0,0 +1,23 @@
+clear ;

+clc;

+// Example 8.5

+printf('Example 8.5\n\n');

+// Page no. 207

+// Solution Fig E8.5

+

+// Given

+W = 100 ;// Water removed - [kg]

+A_H2O = 0.80 ;// Fraction of water in A(intial fish cake)

+A_BDC = 0.20 ;//  Fraction of BDC(bone dry cake) in B(final dry fish cake)

+B_H2O = 0.40 ;// Fraction of water in A(intial fish cake)

+B_BDC = 0.60 ;//  Fraction of BDC(bone dry cake) in B(final dry fish cake)

+

+// By analysis for degree of freedom , DOF comes to be zero 

+// Solve following equations simultaneously for A and B,

+// A*A_H2O = B*B_H2O + W - By H2O balance

+// A = B + W - By overall balance

+a = [A_H2O -B_H2O;1 -1] ;// Matrix of coefficient

+b = [W;W] ;// Matrix of contants

+x = a\b; // Matrix of solutions- A = x(1) and B = x(2)

+

+printf('Weight of the fish cake originally put into dryer -     %.0i kg\n',x(1) );
\ No newline at end of file
diff --git a/409/CH8/EX8.6/Example8_6.sce b/409/CH8/EX8.6/Example8_6.sce
new file mode 100755
index 000000000..692d69a3b
--- /dev/null
+++ b/409/CH8/EX8.6/Example8_6.sce
@@ -0,0 +1,42 @@
+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);
\ No newline at end of file
diff --git a/409/CH8/EX8.7/Example8_7.sce b/409/CH8/EX8.7/Example8_7.sce
new file mode 100755
index 000000000..2ee55f226
--- /dev/null
+++ b/409/CH8/EX8.7/Example8_7.sce
@@ -0,0 +1,55 @@
+clear ;
+clc;
+// Example 8.7
+printf('Example 8.7\n\n');
+// Page no. 213
+// Solution
+
+// Write given data
+B_in =  1.1 ;// Flow rate in  of blood -[L/min]
+B_out = 1.2;// Flow rate out of blood -[L/min]
+S_in = 1.7;// Flow rate in  of solution -[L/min]
+
+// Composition of input blood
+B_in_CR = 2.72 ;//[g/L]
+B_in_UR = 1.16 ;//[g/L]
+B_in_U = 18 ;//[g/L]
+B_in_P = 0.77 ;//[g/L]
+B_in_K = 5.77 ;//[g/L]
+B_in_Na = 13.0 ;//[g/L]
+B_in_water = 1100 ;//[mL/min]
+
+// Composition of output blood
+B_out_CR = 0.120 ;//[g/L]
+B_out_UR = 0.060;//[g/L]
+B_out_U = 1.51 ;//[g/L]
+B_out_P = 0.040 ;//[g/L]
+B_out_K = 0.120 ;//[g/L]
+B_out_Na = 3.21 ;//[g/L]
+B_out_water = 1200 ;//[mL/min]
+
+n_un = 7 ;// Number of unknowns in the given problem
+n_ie  = 7 ;// 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);
+
+// Water balance in grams, assuming 1 ml is equivalent to 1 g
+S_in_water = 1700 ;//[ml/min]
+S_out_water =  B_in_water+ S_in_water - B_out_water;
+S_out = S_out_water/1000 ;//[L/min]
+printf(' Flow rate of water in output solution is  %.2f L/min.\n\n',S_out);
+
+// The component balance in grams for CR,UR,U,P,K and Na are
+S_out_CR  =  (B_in*B_in_CR - B_out*B_out_CR)/S_out;
+S_out_UR  =  (B_in*B_in_UR - B_out*B_out_UR)/S_out;
+S_out_U  =  (B_in*B_in_U - B_out*B_out_U)/S_out;
+S_out_P  =  (B_in*B_in_P - B_out*B_out_P)/S_out;
+S_out_K  =  (B_in*B_in_K - B_out*B_out_K)/S_out;
+S_out_Na  =  (B_in*B_in_Na - B_out*B_out_Na)/S_out;
+printf(' Component     Concentration(g/L) in output Dialysis solution   \n');
+printf(' UR            %.2f    \n',S_out_UR);
+printf(' CR            %.2f    \n',S_out_CR);
+printf(' U             %.2f    \n',S_out_U);
+printf(' P             %.2f    \n',S_out_P);
+printf(' K             %.2f    \n',S_out_K);
+printf(' Na            %.2f    \n',S_out_Na);
\ No newline at end of file