Page 31 - Handbook of Energy Engineering Calculations

3
3
ft (1.158 m ), as computed in step 4, and the total volume of the combustion
3
3
products is 303.85 ft (8.604 m ). Therefore, the percent CO on a wet basis
2
3
3
(i.e., including the moisture in the combustion products) = ft CO /total ft =
2
53.6/303.85 = 0.1764, or 17.64 percent.
The percent CO on a dry, or Orsat, basis is found in the same manner,
2
except that the weight of H O in the products of combustion, 17.6 lb (7.83
2
kg) from step 4, is subtracted from the total gas weight. Or, percent CO , dry,
2
or Orsat basis = (53.6)/(303.85 − 17.6) = 0.1872, or 18.72 percent.

6. Compute the air required with the stated excess flow
With 20 percent excess air, the air flow required = (0.20 + 1.00)(air flow with
no excess) = 1.20 (10.9672) = 13.1606 lb (5.970 kg) of air per pound
(kilogram) of coal burned. The air flow with no excess is obtained from step

2.

7. Compute the weight of the products of combustion
The excess air passes through the furnace without taking part in the
combustion and increases the weight of the products of combustion per

pound (kilogram) of coal burned. Therefore, the weight of the products of
combustion is the sum of the weight of the combustion products without the
excess air and the product of (percent excess air)(air for perfect combustion,
lb); or, given the weights from steps 3 and 2, respectively, = 11.9139 + (0.20)

(10.9672) = 14.1073 lb (6.399 kg) of gas per pound (kilogram) of coal burned
with 20 percent excess air.

8. Compute the volume of the combustion products and the percent CO 2

The volume of the excess air in the products of combustion is obtained by
converting from the weight analysis 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 (10.9672)(0.20)(359/28.95)
3
3
(2.15) = 58.5 ft (1.656 m ). In this calculation the value 28.95 is the
molecular weight of air. The total volume of the products of combustion is
the sum of the column for perfect combustion, step 4, and the excess-air

3
3
volume, above, or 303.85 + 58.5 = 362.35 ft (10.261 m ).