diff options
Diffstat (limited to '409/CH3')
| -rwxr-xr-x | 409/CH3/EX3.1/Example3_1.sce | 26 | ||||
| -rwxr-xr-x | 409/CH3/EX3.2/Example3_2.sce | 29 | ||||
| -rwxr-xr-x | 409/CH3/EX3.3/Example3_3.sce | 17 | ||||
| -rwxr-xr-x | 409/CH3/EX3.4/Example3_4.sce | 41 | ||||
| -rwxr-xr-x | 409/CH3/EX3.5/Example3_5.sce | 33 |
5 files changed, 146 insertions, 0 deletions
diff --git a/409/CH3/EX3.1/Example3_1.sce b/409/CH3/EX3.1/Example3_1.sce new file mode 100755 index 000000000..e0321fcbd --- /dev/null +++ b/409/CH3/EX3.1/Example3_1.sce @@ -0,0 +1,26 @@ +clear ;
+clc;
+
+// Example 3.1
+printf('Example 3.1\n\n');
+//Page no. 79
+// Solution
+
+// Let component 1 be Ce and component 2 be O
+// Basis 2kg mol CeO
+mol1 = 1.0 ;//[kg mol]
+mol2 = 1.0 ;//[kg mol]
+total = mol1+mol2 ;//[kg mol]
+mol_fr1 = mol1/total ;//mole fraction of Ce
+mol_fr2 = mol2/total ;//mole fraction of O
+mw1 = 140.12; //molecular weight of Ce
+mw2 = 16.0 ;//molecular weight of O
+m1 = mw1*mol1;
+m2 = mw2*mol2;
+m_fr1 = m1/(m1+m2) ;//mass fraction of Ce
+m_fr2 = m2/(m1+m2) ;//mass fraction of O
+
+printf('Component kg mol Mole fraction Mol.Wt. kg. Mass fraction\n')
+printf('\n Ce %.2f %.3f %.2f %.3f %.2f\n',mol1,mol_fr1,mw1,m1,m_fr1);
+printf(' O %.2f %.3f %.2f %.3f %.2f\n',mol2,mol_fr2,mw2,m2,m_fr2);
+printf(' Total %.2f %.3f %.2f %.3f %.2f',mol1+mol2,mol_fr1+mol_fr2,mw1+mw2,m1+m2,m_fr1+m_fr2);
\ No newline at end of file diff --git a/409/CH3/EX3.2/Example3_2.sce b/409/CH3/EX3.2/Example3_2.sce new file mode 100755 index 000000000..42facbecb --- /dev/null +++ b/409/CH3/EX3.2/Example3_2.sce @@ -0,0 +1,29 @@ +clear ;
+clc;
+
+// Example 3.2
+printf('Example 3.2\n\n');
+//Page no. 80
+// Solution
+
+// Basis 100kg mol gas
+ml1 = 20.0 ;//[kg mol]
+ml2 = 30.0 ;//[kg mol]
+ml3 = 40.0 ;//[kg mol]
+ml4 = 10.0 ;//[kg mol]
+mw1 = 44.0 ;//molecular weight of CO2
+mw2 = 28.0 ;//molecular weight of CO
+mw3 = 16.04 ; //molecular weight of CH4
+mw4 = 2.02 ;//molecular weight of H2
+m1 = mw1*ml1;
+m2 = mw2*ml2;
+m3 = mw3*ml3;
+m4 = mw4*ml4;
+printf(' Component kg mol Mol.Wt. kg. \n')
+printf(' CO2 %.2f %.2f %.0f \n',ml1,mw1,m1);
+printf(' CO %.2f %.2f %.0f \n',ml2,mw2,m2);
+printf(' CH4 %.2f %.2f %.0f \n',ml3,mw3,m3);
+printf(' H2 %.2f %.2f %.0f \n',ml4,mw4,m4);
+printf('\n Total %.2f %.2f %.0f \n',ml1+ml2+ml3+ml4,mw1+mw2+mw3+mw4,m1+m2+m3+m4);
+av_m = (m1+m2+m3+m4)/100 ;//[kg]
+printf('\nAverage molecular mass of gas is %.1f kg.\n',av_m);
\ No newline at end of file diff --git a/409/CH3/EX3.3/Example3_3.sce b/409/CH3/EX3.3/Example3_3.sce new file mode 100755 index 000000000..f225f9ca6 --- /dev/null +++ b/409/CH3/EX3.3/Example3_3.sce @@ -0,0 +1,17 @@ +clear ;
+clc;
+
+// Example 3.3
+printf('Example 3.3\n\n');
+//Page no. 81
+// Solution
+
+// Basis 1 hour
+rc = 5000 ;//[cpm-counts per minute]
+cg = 10000/24 ;//[cells/hr]
+k = cg/rc ;//[cells/cpm]
+n_rc = 8000 ;//[cpm]
+n_cg = k*n_rc ;//[cells/hr]
+printf('New average cell growth rate is %.0f cells/hr.\n',n_cg);
+in_p = ((n_cg-cg)/cg)*100 ;//[increase percent]
+printf(' Increase percent of cell growth rate is %.1f %% .\n',in_p);
\ No newline at end of file diff --git a/409/CH3/EX3.4/Example3_4.sce b/409/CH3/EX3.4/Example3_4.sce new file mode 100755 index 000000000..33e97518f --- /dev/null +++ b/409/CH3/EX3.4/Example3_4.sce @@ -0,0 +1,41 @@ +clear;
+clc;
+
+// Example 3.4
+printf('Example 3.4\n\n');
+//Page no. 82
+// Solution
+
+// Basis 100 g mol of Nd(4.5)Fe(77)B(18.5)
+//(a)
+n_Fe = 77-0.2;
+printf('(a) Molecular formula after adding Cu is Nd(4.5)Fe(%.1f)B(18.5)Cu(.2).\n',n_Fe);
+
+//(b)
+o_ml1 = 4.5 ;//[kg mol]
+o_ml2 = 77.0 ;//[kg mol]
+o_ml3 = 18.5 ;//[kg mol]
+o_ml4 = 0.0 ;//[kg mol]
+f_ml1 = 4.5 ;//[kg mol]
+f_ml2 = 77.0-0.2 ;//[kg mol]
+f_ml3 = 18.5 ;//[kg mol]
+f_ml4 = 0.2 ;//[kg mol]
+mw1 = 144.24 ;//molecular weight of Nd
+mw2 = 55.85 ;//molecular weight of Fe
+mw3 = 10.81 ; //molecular weight of B
+mw4 = 63.55 ;//molecular weight of Cu
+m1 = mw1*f_ml1;
+m2 = mw2*f_ml2;
+m3 = mw3*f_ml3;
+m4 = mw4*f_ml4;
+f1 = f_ml1/100;
+f2 = f_ml2/100;
+f3 = f_ml3/100;
+f4 = f_ml4/100;
+tf = f1+f2+f3+f4;
+printf('\n (b) Component Original g mol Final g mol Mol.Wt. g. Mass fraction\n')
+printf(' Nd %.2f %.2f %.2f %.2f %.3f\n',o_ml1,f_ml1,mw1,m1,f1);
+printf(' Fe %.2f %.2f %.2f %.2f %.3f\n',o_ml2,f_ml2,mw2,m2,f2);
+printf(' B %.2f %.2f %.2f %.2f %.3f\n',o_ml3,f_ml3,mw3,m3,f3);
+printf(' Cu %.2f %.2f %.2f %.2f %.3f\n',o_ml4,f_ml4,mw4,m4,f4);
+printf('\n Total 100.0 100.0 %.2f %.3f\n',m1+m2+m3+m4,tf);
diff --git a/409/CH3/EX3.5/Example3_5.sce b/409/CH3/EX3.5/Example3_5.sce new file mode 100755 index 000000000..9cd3fe09c --- /dev/null +++ b/409/CH3/EX3.5/Example3_5.sce @@ -0,0 +1,33 @@ +clear ;
+clc;
+
+// Example 3.5
+printf('Example 3.5\n\n');
+//Page no. 84
+// Solution
+
+// Basis 100 kg coal
+ml_r = 9;
+wt_r = (9*1.008)/(1*12) ;//conversion of mole ratio to wt.ratio
+m1 = 2 ;//[kg] wt.of sulphur
+m2 = 1 ;//[kg] wt. of nitrogen
+m3 = 6 ;//[kg] wt. of oxygen
+m4 = 11 ;//[kg] wt. of ash
+m5 = 3 ;//[kg] wt. of water
+m6 = (1*77)/(wt_r+1) ;//[kg] wt. of carbon
+m7 = wt_r*m6 ;//[kg] wt. of hydrogen
+wc = 100-(m4+m5) ;//[kg] wt. of coal excluding ash and water
+wf1 = m1/wc;
+wf2 = m2/wc;
+wf3 = m3/wc;
+wf4 = m4/wc;
+wf6 = m6/wc;
+wf7 = m7/wc;
+tf = wf1+wf2+wf3+wf6+wf7;
+printf(' Component kg. Mass fraction');
+printf('\n C %.2f %.2f\n',m6,wf6);
+printf(' H %.2f %.2f\n',m7,wf7);
+printf(' S %.2f %.2f\n',m1,wf1);
+printf(' N %.2f %.2f\n',m2,wf2);
+printf(' O %.2f %.2f\n',m3,wf3);
+printf('\n Total %.2f %.2f\n',wc,tf);
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