Control of Volatile Organic Compounds Emissions from Manufacturing

Tab1e E-5. GENERALIZED WASTE GAS COMBUSTION CALCULATIONS
Basis: per 100 l b waste gas (w.g.) Stream 1.0.:
Assumptions: (1) 18 percent excess air (1 02:3.76 N 2 by volume or by mole);
Reaction (in dry air):
(2) known waste gas composition of C, H, N, 0, (in lb-moles of atom per 100 lb w.g.);
(3) known amount of AIR (in lb-moles/100 lb w.g.) in waste gas stream (also included in N & 0 compositions);
(4) oxygen in waste gas (in air or hydrocarbons) thoroughly mixed with VOC so only stoichiometric requirement.
Cc Hh Nn Oo + (1.18) (C+ t - 3 (02 + 3.76 N2) c(co2) + 8 (H20) + # + (C t # - $)(1.18)(3.76) X(N2) + (0.18) (C + - $) (02)
Products of Combustion in lb1100 Ib w.g. (wet):
44.01 lb
Ib-mol e
Ib-mole C02
(
100 lb w.g.
= (44.01) =
H20: if air required 18.02 lb H
Ib
H 6 lb-mole (2+ [AIR + (C + 9- (4.76) (1.18)] (0.013)
a (29 i:ixeair)
i s( C ) 7 Ib dry air
= (2.118) C * (9.539) H - (1.059) 0 + (0.377) AIR =
If (B) + (AIR) (0.013)
H 0 '
i.e., (C 7 lb-mole 2
(29 -)
lb
lb a (9.01) H + (0.377) AIR =
dry air
=
ifno air required
0 : Ib-mole 28.02 lb (N) 2
= (14.01)
i,e., (c*$ 7
02: if air required 32.00 lb
6 lb-mole (0.18)
i.e., (C+$
(C + q - $!) = (5.760) C + (1.44) H - (2.88) 0 =
if no air required* 32.00 lb
i.e., lb-mole 2 - (C +
'!)I
2
Total: prod. = lb products (wet) = lb(C02 + H20 + N2 to2) =
w.g. 100 lb w.g. 100 lb w.g.
= (-32.00)C - (8.00) H + (16.00) 0 =
a~ounds of water per pound of dry air at 80°~.and 60% relative humidity.