Page 278 - Materials Chemistry, Second Edition

CAT3525_C09.qxd 2/8/2005 10:11 AM Page 249
Incineration of MSW 249
EXAMPLE 9.2
A carbonaceous waste given by the empirical formula C H O N is to be incinerated.
65.5 102.3 40.8 1.1
Proximate and elemental analyses of the waste are as follows:

Proximate Analysis % Elemental Analysis %
Moisture 4.8 Carbon 47.36
Noncombustibles 6.2 Hydrogen 6.25
Oxygen 39.25
Nitrogen 0.85
Sulfur 0.19
Ash 6.10
Calculate the following: (a) the gross heat value and net heat value of this waste as received; (b) the
volume of air needed for the complete combustion of 1000 kg (i.e., 1 metric ton) of the input material.

SOLUTION
(a) The higher heat value (HHV) and lower heat value (LHV) of the waste can be calculated
using Equations 4.6 and 4.7:
HHV 0.339 (C ) 1.44 (H) – 0.139 (O) 0.105 (S) MJ/kg
HHV 0.339 (47.36) 1.44 ( 6.25) – 0.139 (39.25) 0.105 (0.19) MJ/kg,
HHV 19.61 MJ/kg
LHV HHV (in MJ/kg ) – 0.0244 (W 9H) MJ/kg
LHV 19.61 MJ/kg – 0.0244 (4.8 9 (6.25) MJ/kg
LHV 18.12 MJ/kg
(b) When computing the oxygen requirements, the chlorine and sulfur components may be
neglected. Given that a 65.5, b 102.3, c 40.8, d 0, e 0, f 0.85, g 0, the
combustion equation is as follows:
C H O M 71.1O → 65.5CO 51.2H O 1.1NO
65.5 102.3 40.8 0.85 2 2 2
The formula mass of the waste is then calculated:

Carbon 12 65.5 786
Hydrogen 1 102.3 102.3
Oxygen 16 40.8 652.8
Nitrogen 14 0.85 11.9

Total 1553
Therefore the molar mass of the material is 1553 or 1.55kg.
Of the 1000 kg of the material, 890 kg (i.e., 1000 kg minus 48 kg moisture and 62 kg inert mate-
rial) is combustible. This quantity corresponds to 890 kg/1553 kg/mol 573 mol.
From the equation, we see that 1 mol of the material requires 71.1 mol of O . Therefore 573
2
mol of material requires 573 71.1 40,746 mol of O .
2
At standard temperature and pressure (i.e., 0ºC and 1atm), 1 mole of a gas occupies 22.4
–3
3
10 m . Consequently,
–3
3
volume 40,746 mol of O 22.4 10 m /mol of O
2 2
3
volume 913 m of O
2