1 October 1997
7,219,575/ (1195.5-158.0) = 6,959 lb/h of steam
B-3. THEORETICAL COMBUSTION AIR. There are several ways to calculate the theoretical
air requirement for complete combustion of the waste.
a. Alternative Methods.
(1) If the waste has been carefully characterized by component, tables 3-7 and 3-8
give the ultimate analysis of most of the major components found in waste. A composite
ultimate analysis may be calculated by assuming a lot of 100 lb and adding the amount of
each element that each component contributes to the lot. If a waste stream was 2%
newspaper, there would be 0.9828 lb of carbon (0.02 x 49.14) from the newspaper in the lot.
The amount of carbon contributed by each component would be calculated and added to
determine the total amount of carbon in the lot. The same would then be done for hydrogen,
oxygen, nitrogen, sulfur, and ash. Table 3-5 lists the ultimate analyses for some typical waste
streams. Once a composite ultimate analysis is obtained, table B-1 may be used to calculate
the theoretical air and flue gases produced. Careful attention should be given to the note on
table B-1 regarding chlorine being an oxidizer and the consequent reduction in air required.
This also reduces the amount of water and nitrogen produced in the flue gas.
(2) If a less complete characterization has been done, or a less tedious method is
desired to calculate the theoretical air requirements, table 3-9 gives values for several major
waste stream components on a moisture and ash free (MAF) basis. These values are based
on the MAF HHV of each component. The procedure described in the first note on the table
may be reversed in order to determine the MAF HHV of the waste stream from the "as
received" data developed during the waste survey.
(3) The same criteria may be used to determine the theoretical air requirements for
the entire waste stream based on its MAF HHV. The MAF value may also be converted back
to the "as received" condition.
b. Use of Tables. Table 3-2 gives an estimated value of theoretical air for a waste
stream and compares it to the values for wood and paper. The amount of theoretical air also
has to be adjusted for the moisture (humidity) in it. A typical value can be assumed based
upon average local climactic data. Using the data from table 3-2 and the above example, the
theoretical dry air requirement would be calculated as follows:
(2,917) (6.53) = 19,048 lb air/h
B-4. EXCESS AIR. Because of problems with incomplete mixing and insufficient reaction
times, supplying only the theoretically required amount of air will not result in complete
combustion. Additional air must be supplied. Different types of fuels (more or less reactive)
and different types of combustion systems will require different amounts of combustion air as
listed in table 3-10. The third note on table 3-11 gives instructions on how to compute the air
mass flow requirements for whatever amount of excess air is required. For the above
example and 130% excess air (from table 3-10) the results would be as follows:
(19,048) 100+130) / (100) = 43,810 lbs dry air/hr