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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);
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