Page 40 - Handbook of Energy Engineering Calculations

4. Convert the flue-gas weight to volume
The products of complete combustion of any fuel that does not contain sulfur
are CO , H O, and N . Using the combustion equation in step 1, compute the
2
2
2
products of combustion thus: CH + 2O = CO + H O; 16 + 64 = 44 + 36; or
2
2
2
4
the CH burns to CO in the ratio of 1 part CH to 44/16 parts CO . Since,
2
4
4
2
3
3
from step 1, there is 0.03896 lb CH per ft (0.624 kg/m ) of natural gas, this
4
forms (0.03896) (44/16) = 0.1069 lb (0.048 kg) of CO . Likewise, for C H ,
2 6
2
(0.003247)(88/30) = 0.00952 lb (0.004 kg). The total CO in the combustion
2
products = 0.00464 + 0.1069 + 0.00952 = 0.11688 lb (0.053 kg), where the
first quantity is the CO in the fuel.
2
Using a similar procedure for the H O formed in the products of
2
combustion by CH , we find (0.03896)(36/16) = 0.0875 lb (0.039 kg). For
4
C H , (0.003247)(54/30) = 0.005816 lb (0.003 kg). The total H O in the
2
2 6
combustion products = 0.0875 + 0.005816 = 0.093316 lb (0.042 kg).
Step 2 shows that 12.39 lb (5.58 kg) of N is required per lb (kg) of fuel.
2
3
3
Since 1 ft (0.028 m ) of the fuel weights 0.04517 lb (0.02 kg), the volume of
3
3
gas which weighs 1 lb (2.2 kg) is 1/0.04517 = 22.1 ft (0.626 m ). Therefore,
3
the weight of N per ft of fuel burned = 12.39/22.1 = 0.560 lb (0.252 kg).
2
This, plus the weight of N in the fuel, step 1, is 0.560 + 0.0025 = 0.5625 lb
2
(0.253 kg) of N in the products of combustion.
2
Next, find the total weight of the products of combustion by taking the sum
of the CO , H O, and N weights, or 0.11688 + 0.09332 + 0.5625 = 0.7727 lb
2
2
2
3
(0.35 kg). Now convert each weight to ft at 650°F (343°C), the temperature