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+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);