A: Analysis by Volume Per Cent B: Molecular Weight C: Volume times Molecular Weight D: Analysis by Weight Per Cent _____________________________________________________________________ | | | | | | | Gas | A | B | C | D | |________________________|_______|___________|________|_______________| | | | | | | | | | | | | | | | | | 536.8 | | Carbon Dioxide CO_{2} | 12.2 | 12+(2×16) | 536.8 | ------ = 17.7 | | | | | | 3022.8 | | | | | | | | | | | | 11.2 | | Carbon Monoxide CO | .4 | 12+16 | 11.2 | ------ = .4 | | | | | | 3022.8 | | | | | | | | | | | | 220.8 | | Oxygen O | 6.9 | 2×16 | 220.8 | ------ = 7.3 | | | | | | 3022.8 | | | | | | | | | | | | 2254.0 | | Nitrogen N | 80.5 | 2×14 | 2254.0 | ------ = 74.6 | | | | | | 3022.8 | |________________________|_______|___________|________|_______________| | | | | | | | Total | 100.0 | | 3022.8 | 100.0 | |________________________|_______|___________|________|_______________| Application of Formulae and Rules--Pocahontas coal is burned in the furnace, a partial ultimate analysis being: _Per Cent_ Carbon 82.1 Hydrogen 4.25 Oxygen 2.6 Sulphur 1.6 Ash 6.0 B. t. u., per pound dry 14500 The flue gas analysis shows: _Per Cent_ CO_{2} 10.7 O 9.0 CO 0.0 N (by difference) 80.3 Determine: The flue gas analysis by weight (see Table 33), the amount of air required for perfect combustion, the actual weight of air per pound of fuel, the weight of flue gas per pound of coal, the heat lost in the chimney gases if the temperature of these is 500 degrees Fahrenheit, and the ratio of the air supplied to that theoretically required. Solution: The theoretical weight of air required for perfect combustion, per pound of fuel, from formula (11) will be,