COMBUSTION AND FUELS 77 



oxygen combine they form water, HzO. These are called the 

 products of combustion. The oxygen required for combustion 

 is usually obtained from the air, which is a mixture composed 

 of approximately 23 parts of oxygen and 77 parts of nitrogen 

 by weight. The nitrogen that enters the furnace with the- 

 oxygen takes no part in the combustion, but passes through 

 the furnace and up the chimney without any change in its 

 nature. 



Air Required for Combustion. When carbon is burned to 

 carbon dioxide, COz, 1 atom of carbon unites with 2 atoms of 

 oxygen. Carbon has an atomic weight of 12 and oxygen has 

 an atomic weight of 16, so that the molecular weight of COi is 

 (1 X 12) + (2 X 16) = 44 ; hence COi is composed of 12 -^ 44 = 27.27 

 per cent, of carbon and 32 -=-44 = 72. 73 % of oxygen. To 

 burn a pound of carbon to COz, therefore, requires 32 -hi 2 

 = 2f Ib. of oxygen. If the oxygen is taken from the air, it 

 will take 2f -h. 23 = 11.6 Ib. of air to supply the 2f Ib. of oxy- 

 gen. This is because only 23% of air is oxygen. The com- 

 bustion of a pound of carbon to COz may be represented as 

 follows: 



Mixture Elements Products 



in Pounds in Pounds in Pounds 



Carbon, 1.0 Carbon, l.OOl _ , 



Air 1,6 -I """' 2.6r)- Carbo " d '- de ' 3 ' 67 



I Nitrogen, 8.93 Nitrogen, 8.93 



12.6 12.60 12.60 



That is, 1 Ib. of carbon requires 11.6 Ib. of air for complete 

 combustion. Of this air, 2.67 Ib. is oxygen which combines 

 with the pound of carbon, forming 3.67 Ib. of carbon dioxide. 

 The 8.93 Ib. of nitrogen contained in the air passes off with 

 the CO-i as a product of combustion. 



Take, next, the complete combustion of 1 Ib. of hydrogen. 

 The product of the combustion is water, HiO. It has been 

 shown that HzO is composed by weight of 2 parts hydrogen 

 to 16 parts oxygen. Hence 1 Ib. of H requires 16 -=-2 = 8 Ib. 

 of O to unite with it. The air required to furnish 8 Ib. of O is 

 8 -T-. 23 = 34.8 Ib. The process of combustion is, therefore, as 

 follows: 



