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//Chapter 13: Fuel and Combustions
//Problem: 9
clc;
//Declaration of Variables
wt_C = 3 // kg
// Solution
wt_a = wt_C * 32 * 100 / 12.0 / 23.0
vol_a = wt_a * 1000 * 22.4 / 28.94
mprintf("H2(g) + 1/2 O2(g) --> H20(l)\n")
mprintf(" 1 0.5 1\t\t(By Vol.)\n")
mprintf(" CO(g) + 1/2 O2(g) --> CO2(g)\n")
mprintf(" 1 0.5 1\t\t(By Vol.)\n")
mprintf(" CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(l)\n")
mprintf(" 1 2 1\t\t(By Vol.)\n")
mprintf(" Weight of air for the combustion of 3kg carbon %.3f kg\n",wt_a)
mprintf(" Vol. of air required for combustion of 3kg carbon %.3e L (or) %.2f metre cube",vol_a,vol_a / 1000)
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