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//Chapter 13: Fuel and Combustions
//Problem: 6
clc;
//Declaration of Variables
w1 = 2.5 // g
w2 = 2.415 // g
r = 1.528 // g
ma = 0.245 // Mass of ash, g
// Solution
m = w1 - w2 // Mass of moisture in coal
mv = w2 - r // Mass of volatile matter
moip = m * 100 / w1
vp = mv * 100 / w1
ap = ma * 100 / w1
cp = 100 - (moip + vp + ap)
mprintf("Percentage of moisture:%.1f percentage\n", moip)
mprintf(" Percentage of volatile matter:%.2f percentage\n", vp)
mprintf(" Percentage of ash:%.1f percentage\n", ap)
mprintf(" Percentage of fixed carbon:%.2f percentage", cp)
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