Page 34 - Handbook of Energy Engineering Calculations

w = 0.748 × [(4 × 14.2) + 4.7 + 704)]/[3(14.2 + 0.3)] = 13.16 lb/lb (5.97
dg
kg/kg).

4. Compute the amount of dry air supplied per lb (kg) of coal
The lb (kg) of dry air supplied per lb (kg) of coal, w = w − C + 8[H′ –
da
1
2
dg
(O′ /8)] − (N′ /N), where the percentage by weight of nitrogen in the fuel, N′ 2
2
2
= 1.56, and “atmospheric nitrogen” in the supply air, N = 0.768; other values
2
are as given or calculated. Then, w = 13.16 − 0.748 + 8[0.0572 –
da
(0.1382/8)] – (0.0156/0.768) = 12.65 lb/lb (5.74 kg/kg).

5. Compute the percent of excess air used
Percent excess air = (w − w )/w = (12.65 – 10.03)/10.03 = 0.261, or 26.1
da
ta
ta
percent.

Related Calculations. The percentage by weight of nitrogen in “atmospheric
air” in step 4 appears in Principles of Engineering Thermodynamics, 2nd

edition, by Kiefer et al., John Wiley & Sons, Inc.

FUEL OIL COMBUSTION IN A FURNACE

A fuel oil has the following ultimate analysis: C = 0.8543; H = 0.1131; O =
2
2
0.0270; N = 0.0022; S = 0.0034; total = 1.0000. This fuel oil is burned in a
2
steam-boiler furnace. Determine the weight of air required for theoretically
perfect combustion, the weight of gas formed per pound (kilogram) of oil
burned, and the volume of flue gas, at the boiler exit temperature of 600°F

(316°C), per pound (kilogram) of oil burned; the air required with 20 percent
excess air, and the volume of gas formed with this excess; the CO 2
percentage in the flue gas on a dry and wet basis.

Calculation Procedure:

1. Compute the weight of oxygen required per pound (kilogram) of oil
The same general steps as given in the previous calculation procedure will be
followed. Consult that procedure for a complete explanation of each step.