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Subtracting the enthalpy of the waste gas stream exiting the incinerator from the waste gas
stream entering the incinerator gives the heat that must be supplied by the fuel. This is referred to as
a change in enthalpy or heat content . Using Equation 16.19, the enthalpy entering ( T 1 ) is subtracted
from the enthalpy leaving ( T 2 ), giving
q = m H = mC p ( T 2 - T 1 )
(16.20)
where
q = Heat rate (Btu/hr).
m = Mass flow rate (lb/hr).
H = Enthalpy change.
C p = Specific heat (J/kg, °C; Btu/lb, °F).
T 2 = Exit temperature of the substance (°C or °F).
T 1 = Initial temperature (°C or °F).
16.5.2 i inCineration e xample C alCulations
EXAMPLE 16.10. RESIDENCE TIME (METRIC UNITS)
Problem: A thermal incinerator controls emissions from a paint-baking oven. The cylindrical unit
diameter is 1.5 m (5 ft) and it is 3.5 m (11.5 ft) long. The exhaust from the oven is 3.8 m 3 s (8050
scfm). The incinerator uses 300 scfm of natural gas and operates at a temperature of 1,400°F. If
all the oxygen necessary for combustion is supplied from the process stream, no outside air added,
what is the residence time in the combustion chamber (USEPA, 1981, p. 3-7)?
Solution: To solve this problem, we use the approximation that 11.5 m 3 of combustion products
are formed for every 1.0 m 3 of natural gas burned at standard conditions (16°C and 101.3 kPa).
Moreover, 10.33 m 3 of theoretical air is required to combust 1 m 3 of natural gas at standard condi-
tions. First determine the volume of combustion products from burning the natural gas:
3
) ×
11.5 mofproduct
1.0 mofgas
= 1.61 m/s
(
3
3
0.14 m/s
3
Determine the air required for combustion:
3
) ×
10.3 mofair
1.0 mofgas
= 1.45 m/s
(
3
3
0.14 m/s
3
Add the volumes and subtract the air required for combustion:
Flow from paint bake oven = 3.8
Products from combustion = 1.61
Air required for combustion = 1.45
Total volume = 3.8 + 1.61 - 1.45 = 3.96 m 3 /s
Convert the m 3 /s calculated under the standard conditions to actual conditions:
273°C 760°C
273°C
+
(
) ×
=
3
3
3.96 m/s
14.98 m
/s
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