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clear;
clc;
disp('Example 8.5');
// aim : To determine
// the volume of the air required
// Given values
H2 = .45;// volume fraction of H2
CO = .40;// volume fraction of CO
CH4 = .15;// volume fraction of CH4
// solution
V = 2.38*(H2+CO)+9.52*CH4;// stoichimetric volume of air, [m^3]
mprintf('\n The volume of air required is = %f m^3/m^3 fuel\n',V);
// Result in the book is misprinted
// End
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