Page 41 - Handbook of Energy Engineering Calculations

of the combustion products, or:

In this calculation, the value of 379 is used in the molecular-weight ratio
2
because at 60°F (15.6°C) and 14.7 lb/in (abs) (101.3 kPa), the volume of 1
lb (0.45 kg) of any gas = 379/gas molecular weight. The fuel gas used is
2
initially at 60°F (15.6°C) and 14.7 lb/in (abs) (101.3 kPa). The ratio 2.255 =
(650 + 460)/(32 + 460).

5. Compute the CO content of the flue gas
2
CO , wet basis = 2.265/23.88 = 0.947, or 9.47 percent. CO , dry basis =
2
2
2.265/(23.88 − 4.425) = 0.1164, or 11.64 percent.

6. Compute the air required with the stated excess flow
With 20 percent excess air, (1.20)(16.132) = 19.3584 lb of air per lb (8.71
3
3
kg/kg) of natural gas, or 19.3584/22.1 = 0.875 lb of air per ft (13.9 kg/m ) of
natural gas. See step 4 for an explanation of the value 22.1.

7. Compute the weight of the products of combustion
Weight of the products of combustion = product weight for perfect
combustion, lb + (percent excess air) (air for perfect combustion, lb) = 16.80

+ (0.20)(16.132) = 20.03 lb (9.01 kg).

8. Compute the volume of the combustion products and the percent CO 2
The volume of excess air in the products of combustion is found by

converting from the weight to the volumetric analysis and correcting for
temperature as in step 4, using the air weight from step 2 for perfect
combustion and the excess-air percentage, or (16.132/22.l)(0.20)(379/28.95)
3
3
(2.255) = 4.31 ft (0.122 m ). Add this to the volume of the products of
3
3
combustion found in step 4, or 23.88 + 4.31 = 28.19 ft (0.798 m ).