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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);
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