Page 54 - Handbook of Energy Engineering Calculations

Air has some moisture because of its relative humidity. Estimate the amount
of moisture in dry air in M lb/lb (kg/kg) from M = 0.622 (p )/(14.7 – p ),
w
w
where 0.622 is the ratio of the molecular weights of water vapor and dry air;
p = partial pressure of water vapor in the air, psia (kPa) = saturated vapor
w
pressure (SVP) × relative humidity expressed as a decimal; 14.7 =

atmospheric pressure of air at sea level (101.3 kPa). From the steam tables, at
80°F (26.7°C), SVP = 0.5069 psia (3.49 kPa). Substituting, M = 0.622
(0.5069 × 0.65)/(14.7 – [0.5069 × 0.65]) = 0.01425 lb of moisture/lb of dry
air (0.01425 kg/kg).
The total flow rate of the wet air then = 1.0142 (101,142.5) = 102,578.7

lb/h (46,570.7 kg/h). To convert to a volumetric-flow basis, recall that the
density of air at 80°F (26.7°C) and 14.7 psia (101.3 kPa) = 39/(480 + 80) =
3
3
0.0722 lb/ft (1.155 kg/m ). In this relation, 39 = a constant and the
temperature of the air is converted to degrees Rankine. Hence, the volumetric
3
flow = 102,578.7/(60 min/h)(0.0722) = 23,679.3 actual cfm (670.1 m /min).

4. Estimate the rate of fuel firing and flue-gas produced
6
The rate of fuel firing = Q /HHV = (120.48 × 10 )/23,000 = 5238 lb/h (2378
f
kg/h). Hence, the total flue gas produced = 5238 + 102,578 = 107,816 lb/h

(48,948 kg/h).
If the temperature of the flue gas is 400°F (204.4°C) (a typical value for a
natural-gas fired boiler), then the density, as in step 3, is: 39/(400 + 460) =
3
3
0.04535 lb/ft (0.7256 kg/m ). Hence, the volumetric flow = (107,816)/(60
3
min/h × 0.04535) = 39,623.7 actual cfm (1121.3 m /min).

Related Calculations. Detailed combustion calculations based on actual fuel
gas analysis can be performed to verify the constants given in the list above.

For example, let us say that the natural-gas analysis was: methane = 83.4
percent; ethane = 15.8 percent; nitrogen = 0.8 percent by volume. First
convert the analysis to a percent weight basis.