initial commit / add all books

10 files changed, 220 insertions, 0 deletions

diff --git a/409/CH2/EX2.1/Example2_1.sce b/409/CH2/EX2.1/Example2_1.sce
new file mode 100755
index 000000000..901009d62
--- /dev/null
+++ b/409/CH2/EX2.1/Example2_1.sce
@@ -0,0 +1,22 @@
+clear ;

+clc;

+

+// Example 2.1

+printf('Example 2.1\n\n');

+//Page no. 45

+// Solution

+

+// Count the number of each element from fig. E2.1.

+// Look for the atomic weights of elements from Appendix B

+// Assume the one cell is a molecule

+n_Ba = 2 ;// Number of atoms of Ba

+n_Cu = 16 ;// Number of atoms of Cu

+n_O = 24 ;// Number of atoms of O

+n_Y = 1 ;// Number of atoms of Y

+m_Ba = 137.34 ;//Atomic wt. -[g]

+m_Cu = 63.546 ;//Atomic wt.-[g]

+m_O = 16.00 ;//Atomic wt.-[g]

+m_Y = 88.905; //Atomic wt.-[g]

+mol_wt =  n_Ba*m_Ba + n_Cu*m_Cu + n_O*m_O + n_Y*m_Y ;//The molecular weight of given material-[g]

+

+printf('The molecular weight of given material is %1.1f g/g mol.\n',mol_wt);
\ No newline at end of file
diff --git a/409/CH2/EX2.10/Example2_10.sce b/409/CH2/EX2.10/Example2_10.sce
new file mode 100755
index 000000000..8fdb865c4
--- /dev/null
+++ b/409/CH2/EX2.10/Example2_10.sce
@@ -0,0 +1,38 @@
+clear;

+clc;

+

+// Example 2.10

+// Page no. 64

+// Solution

+

+// Given

+// Process a 

+// Let us take array of given values for compounds in following order 1- acetone, 2 - Hydrogen cyanide, 3- methanol, 4-Sulphuric acid , 5 - Methyl methacrylate

+Lb1  =  [0.68,0.32,0.37,1.63,1]  ;// Mass of compounds -[lb]

+Value1  =  [0.43, 0.67,0.064,0.04,0.78] ;// Cost of compounds -[$/lb]

+TLV1  =  [750,10,200,2,100] ;// TLV value of compounds -[ppm]

+OITF1  =  [0,1000,10,10000,10] ;// Note : (?) mark values are neglected as they are nearly equal to zero 

+

+// Process b

+// Let us take array of given values for compounds in following order 1- Isobutylene, 2 - Methanol, 3- Pentane, 4-Sulphuric acid , 5 - Methyl methacrylate

+Lb2  =  [1.12,0.38,0.03,0.01,1.00]  ;// Mass of compounds -[lb]

+Value2  =  [0.31,0.64,0.112,0.04,0.78] ;// Cost of compounds -[$/lb]

+TLV2  =  [200,200,600,2,100] ;// TLV value of compounds -[ppm]

+OITF2  =  [0,10,0,10000,10] ;// Note : (?) mark values are neglected as they are nearly equal to zero 

+

+NetV1  =  Lb1(5)*Value1(5) - Lb1(2)*Value1(2) - Lb1(3)*Value1(3) - Lb1(4)*Value1(4) - Lb1(1)*Value1(1); // Net Value for process (a) -[$]

+NetV2  =  Lb2(5)*Value2(5) - Lb2(2)*Value2(2) - Lb2(3)*Value2(3) - Lb2(4)*Value2(4) - Lb2(1)*Value2(1) ;// Net Value for process (b) -[$]

+

+printf('1.With respect to cost criteria\n');

+printf('  Net value for process (a) is %.2f $ and for process (b) is %.2f $.\n  Hence based on net value both process are equivalent.  \n',NetV1,NetV2);

+

+// With respect to two environmental criteria

+TLV_index1  =  Lb1(1)/TLV1(1) + Lb1(2)/TLV1(2)  + Lb1(3)/TLV1(3) + Lb1(4)/TLV1(4) + Lb1(5)/TLV1(5) ;// TLV index for process a

+OITF_index1  =  OITF1(1)*Lb1(1) +OITF1(2)*Lb1(2) + OITF1(3)*Lb1(3) + OITF1(4)*Lb1(4) + OITF1(5)*Lb1(5) ;// OITF index process a

+

+TLV_index2  =  Lb2(1)/TLV2(1) + Lb2(2)/TLV2(2)  + Lb2(3)/TLV2(3) + Lb2(4)/TLV2(4) + Lb2(5)/TLV2(5) ;// TLV index for process b

+OITF_index2  =  OITF2(1)*Lb2(1) +OITF2(2)*Lb2(2) + OITF2(3)*Lb2(3) + OITF2(4)*Lb2(4) + OITF2(5)*Lb2(5) ;// OITF index process b

+

+printf('\n 2.With respect to  two environmental criteria\n');

+printf('   Process (a)\n   TLV index for process a is %.2f .\n   OITF index process a is %.2f .  \n',TLV_index1,OITF_index1);

+printf('   \n   Process (b)\n   TLV index for process b is %.2f .\n   OITF index process b is %.2f .  \n',TLV_index2,OITF_index2);
\ No newline at end of file
diff --git a/409/CH2/EX2.2/Example2_2.sce b/409/CH2/EX2.2/Example2_2.sce
new file mode 100755
index 000000000..f94ee484d
--- /dev/null
+++ b/409/CH2/EX2.2/Example2_2.sce
@@ -0,0 +1,16 @@
+clear ;

+clc;

+

+// Example 2.2

+printf('Example 2.2\n\n');

+// Page no. 46

+// Solution

+

+//(a)

+m_NaOH = 40.0 ;//[lb]

+pnd_mol = 2*1/m_NaOH ;//[lb mol]

+printf('(a) The number of pound moles of NaOH in 2.00 lb NaOH is %.2f lb mol.\n',pnd_mol);

+

+//(b)

+grm_mol = pnd_mol*454 ;//[g mol]

+printf(' (b) The number of gram moles of NaOH in 2.00 lb NaOH is %.2f g mol.\n',grm_mol);
\ No newline at end of file
diff --git a/409/CH2/EX2.3/Example2_3.sce b/409/CH2/EX2.3/Example2_3.sce
new file mode 100755
index 000000000..c48f1ab31
--- /dev/null
+++ b/409/CH2/EX2.3/Example2_3.sce
@@ -0,0 +1,13 @@
+clear;

+clc;

+

+// Example 2.3

+printf('Example 2.3\n\n');

+//Page no. 46

+// Solution

+

+//Basis 7.5 g mol of NaOH

+m_NaOH = 40.0 ;//[lb]

+lb = 454 ;//[g mol]

+n = 7.50*1*m_NaOH/(lb*1);

+printf('Number of pounds of NaOH is %.3f lb.\n',n);
\ No newline at end of file
diff --git a/409/CH2/EX2.4/Example2_4.sce b/409/CH2/EX2.4/Example2_4.sce
new file mode 100755
index 000000000..498ad232a
--- /dev/null
+++ b/409/CH2/EX2.4/Example2_4.sce
@@ -0,0 +1,25 @@
+clear ;

+clc; 

+

+// Example 2.4

+printf('Example 2.4\n\n');

+//Page no. 53

+// Solution

+

+// (a)

+d_w1=1000 ;//[kg/cubic metre]

+

+d1=(1.184*d_w1*1000)/(10^6) ;//[g/cubic centimetre]

+printf('(a) Density in g/cubic centimetre is %.4f g/cubic centimetre.\n',d1);

+

+// (b)

+d_w2= 62.4 ;//[lbm/cubic feet]

+

+d2=1.184*d_w2/1 ;//[lbm/cubic feet]

+

+printf(' (b) Density in lbm/cubic feet is %.1f lbm/cubic feet.\n',d2);

+

+// (c)

+d3=1.184*d_w1 ;//[kg/cubic metre]

+

+printf(' (c) Density in kg/cubic metre is %.1f kg/cubic metre.\n',d3);
\ No newline at end of file
diff --git a/409/CH2/EX2.5/Example2_5.sce b/409/CH2/EX2.5/Example2_5.sce
new file mode 100755
index 000000000..5b5004888
--- /dev/null
+++ b/409/CH2/EX2.5/Example2_5.sce
@@ -0,0 +1,18 @@
+clear ;

+clc;

+

+// Example 2.5

+printf('Example 2.5\n\n');

+//Page no. 54

+// Solution

+

+m_wt=192 ;//[kg]

+d_sol=1.024*1000 ;//[kg/cubic metre]

+// 1000L=1 cubic metre

+c_sol=d_sol/1000 ;//[kg/L]

+c_drug=c_sol*.412 ;//[kg/L]

+printf('Concentration of drug in solution is %.3f kg/L .\n',c_drug);

+

+Q=10.5 ;//[L/min]

+Qmol=10.5*c_drug/m_wt ;//[kg mol/min]

+printf(' Flow rate of drug is %.3f kg mol/min. \n',Qmol); 
\ No newline at end of file
diff --git a/409/CH2/EX2.6/Example2_6.sce b/409/CH2/EX2.6/Example2_6.sce
new file mode 100755
index 000000000..c2d20532d
--- /dev/null
+++ b/409/CH2/EX2.6/Example2_6.sce
@@ -0,0 +1,25 @@
+clear ;

+clc;

+

+// Example 2.6

+printf('Example 2.6\n\n');

+//Page no.57

+// Solution

+

+// Let component 1 be water and component 2 be NaOH

+// Basis 10 kg total solution

+m1 = 5.0 ;//[kg]

+m2 = 5.0; //[kg]

+total = m1  + m2 ;//[kg] 

+m_fr1 = m1/total ;//mass fraction of water

+m_fr2 = m2/total ;//mass fraction of NaOH

+mw1 = 18.0 ;//molecular weight of water

+mw2 = 40.0 ;//molecular weight of NaOH

+mol1 = m1/mw1;

+mol2 = m2/mw2;

+mol_fr1 = mol1/(mol1  + mol2) ;//mol fraction of water

+mol_fr2 = mol2/(mol1  + mol2) ;//mol fraction of NaOH

+printf(' Component    kg        Mass fraction    Mol.Wt.    kg mol    Mole fraction\n');

+printf('n Water        %.2f      %.3f            %.1f       %.3f      %.2f\n',m1,m_fr1,mw1,mol1,mol_fr1);

+printf(' NaOH         %.2f      %.3f            %.1f       %.3f      %.2f\n',m2,m_fr2,mw2,mol2,mol_fr2);

+printf(' Total        %.2f     %.3f                       %.3f      %.2f',m1  + m2,m_fr1  + m_fr2,mol1  + mol2,mol_fr1  + mol_fr2);
\ No newline at end of file
diff --git a/409/CH2/EX2.7/Example2_7.sce b/409/CH2/EX2.7/Example2_7.sce
new file mode 100755
index 000000000..d51e030af
--- /dev/null
+++ b/409/CH2/EX2.7/Example2_7.sce
@@ -0,0 +1,16 @@
+clear ;

+clc;

+

+// Example 2.7

+printf('Example 2.7\n\n');

+//Page no.58

+// Solution

+

+// Basis 500 L solution containing 35g/L

+// (NH4)2SO4 is the only nitrogen source

+cn = 35 ;//[g/L]

+wt = 9  ;//[wt % N]

+m_wt1 = 132 ;//[g]

+m_wt2 = 14 ;//[g]

+amt = (500*(35)*.09*1*1*m_wt1)/(1*m_wt2*1*1);

+printf('Total amount of (NH4)2SO4 consumed  is  %.1f g.',amt);
\ No newline at end of file
diff --git a/409/CH2/EX2.8/Example2_8.sce b/409/CH2/EX2.8/Example2_8.sce
new file mode 100755
index 000000000..f60b0cd5b
--- /dev/null
+++ b/409/CH2/EX2.8/Example2_8.sce
@@ -0,0 +1,19 @@
+clear ;

+clc;

+

+// Example 2.8

+printf('Example 2.8\n\n');

+//Page no. 63

+// Solution

+

+// 1 kg of the air/HCN mixture

+// (a)

+m1 = 27.03 ;//[g]

+m2 = 29.0 ;//[g]

+cn = (10*m1*1000*1000)/(10^6*m2) ;//[mgHCN/kg air]

+printf('(a) 10.0 ppm HCN is %.2f mg HCN/kg air.\n',cn);

+

+// (b)

+ld = 300 ;//[mg/kg air]

+fr = cn/ld;

+printf(' (b) Fraction of lethal dose is 10.0 ppm is  %.3f.',fr);
\ No newline at end of file
diff --git a/409/CH2/EX2.9/Example2_9.sce b/409/CH2/EX2.9/Example2_9.sce
new file mode 100755
index 000000000..2e6ff7756
--- /dev/null
+++ b/409/CH2/EX2.9/Example2_9.sce
@@ -0,0 +1,28 @@
+clear ;

+clc;

+

+// Example 2.9

+printf('Example 2.9\n\n');

+//Page no. 64

+// Solution

+

+// Let component 1 be water and component 2 be HNO3

+// Basis 1L solution

+c = 15 ;//[g/L]

+sg = 1.10 ;

+L = 1000 ;//[cubic centimetre]

+m1 = 18.0 ;//[g]

+m2 = 63.02 ;//[g]

+cn2 = (15*1)/(L*sg) ;//[gHNO3/g soln]

+// Basis 1g soln

+cn1 = 1-cn2 ;// Mass of water in 1 g soln

+mg1 = cn1/m1;

+mg2 = cn2/m2;

+ml_fr1 = mg1/(mg1+mg2);

+ml_fr2 = mg2/(mg1+mg2);

+printf(' (a) Component   g(per 1g soln)           Mol.Wt.     g mol                Mole fraction\n')

+printf('     Water       %.4f                   %.2f       %.3f                %.2f\n',cn1,m1,mg1,ml_fr1);

+printf('     HNO3        %.4f                   %.2f       %e        %e\n',cn2,m2,mg2,ml_fr2);

+// (b)

+cpm = cn2*10^6 ;//[ppm]

+printf('\n (b)Ppm of HNO3 in soln. is %.2f ppm.',cpm);
\ No newline at end of file