University of
Portsmouth
Biomass Combustion Study Pack
for WEBCT Contract
by
Prof. M. R. I. Purvis
S. O. Santos
©
February 2002
Department of Mechanical and Manufacturing Engineering
University of Portsmouth
United Kingdom of Great Britain
Table of Content
Learning Objectives
1
Study Topic
1
1.1.
What is Biomass?
1
1.2.
Importance of biomass as energy source
1
1.3.
How do we convert biomass to energy
4
1.4.
Biomass Composition and Energy Content
5
1.5.
Advantages and Disadvantages of Using Biomass
8
1.6.
Principles of Biomass Combustion
1.6.1.
Devolatilisation
1.6.2.
Char Combustion
8
10
11
1.7.
17
17
18
20
Combustion Equipment
1.7.1.
Fixed Bed Combustion
1.7.2.
Fluidised Bed Combustion
1.7.3.
Suspension Burning
1.8.
Thermal Efficiency
21
1.9.
Combustion Air
22
1.10.
Products of Combustion
24
1.11.
Environmental Issues and Biomass Combustion
1.11.1.
Effects of Pollutants
1.11.2.
Quantification of Emissions
1.11.3.
Nitrogen Oxides
25
26
27
27
1
Learning Objectives
On the completion of this study pack, you should be able to:
” describe properties of biomass materials and their importance to the design of combustion plant
” understand principles of biomass combustion
” explain the importance of biomass as a sustainable source of energy
” recognise the environmental impact of biomass combustion
” identify significant literature to aid biomass combustion modelling
Study Topic
1.1.
What is Biomass?
Biomass is…
¾ plant and other growing species capable of being used as fuel.
¾ organic material mainly composed of carbohydrate and lignin compounds, the building blocks of which are the elements carbon, hydrogen and oxygen.
¾ stored form of solar energy relying on the process of photosynthesis.
Some examples of biomass are…
fuel wood
sugar cane bagasse
switch grass
rice hull
coconut shells
1.2.
tree barks
wheat straw
corn cobs
vineyard pruning
almond shell
Importance of biomass as energy source
As late as the mid 1800s, biomass supplied the vast majority of the world’s energy and
fuel needs (as shown in Figure 1). It only started to phase out in industrialised countries
as the fossil fuel era began, slowly at first and then at a rapid rate (Klass, 1998). But with
the onset of the first major oil crisis in the early 1970s and the increasing stringency of
regulations on pollution emissions, biomass was again realised by many governments and
policy makers to be a viable, domestic energy resource which has the potential for reducing oil consumption and imports and limiting various air pollutants such as CO2, SOx,
NOx and many trace emission. A case in point is shown in Figure 2 which indicates the
Page 1
Figures 3 and 4 describe world’s energy consumption. Figure 3 shows the share of the different fuels to
the world’s primary energy supply. It is seen that the
global utilisation of combustible renewables and
wastes, which includes biomass as a primary source
of energy has increased from 677 Mtoe to 1,063
Mtoe during the period of 1973 to 1998.
Energy Inputs (%)
energy distribution for Sweden for the year 1973 and
1998. The figure illustrates the contribution of combustible renewables and wastes (which also includes
biomass fuel) to the primary energy supply in Sweden, i.e. increases from 8.9% to 14.8% between 1973
and 1998. Likewise, it also shows the contribution of
oil to the primary energy supply, i.e. decreasing from
44.9% in 1973 to 35.7% in 1998.
Calendar Year
Figure 1: Evolution of the primary energy structure. This figure shows the contribution of various
types of fuel to the global energy input from
1860 to 1990 (Dupont-Roc et al, 1996)
Sweden’s Total Primary Energy Supply
1998
1973
Com bustible
Renew ables &
Waste
8.9%
Hydro
13.0%
Coal
4.1%
Others**
0.7%
Coal
5.0%
Oil
30.3%
Gas
1.3%
Hydro
12.0%
Nuclear
1.5%
Oil
72.4%
Com bustible
Renew ables &
Waste
14.8%
Nuclear
35.9%
39.3 Mtoe
** Others include geothermal, wind, solar, heat etc…
52.5 Mtoe
Figure 2: The share of various fuels to the total primary energy supply of Sweden between 1973 and 1998. The total
energy used is 39.3 and 52.5 million tonne of oil equivalent (Mtoe) for 1973 and 1998 respectively. (IEA, 2000a)
World’s Total Primary Energy Supply
1973
6,043 Mtoe
1998
** Others include geothermal, wind, solar, heat etc…
9,491 Mtoe
Figure 3: The share of various fuels to the world’s total primary energy supply between 1973 and 1998. The total
energy used is 6,043 and 9,491 million tonne of oil equivalent (Mtoe) for 1973 and 1998 respectively. (IEA, 2000b)
Page 2
Gross National Product per capita (US$ / capita-year)
Figure 4 presents the relationship between gross national product (GNP) per capita to energy consumption per capita. The figure shows a clear difference in the energy consumption of developed and developing countries. For many developing countries, fuel wood is
the primary source of energy particularly in the domestic sector. Figure 5 shows the contribution of wood energy in various Asian countries.
Energy consumption per capita (kg of oil equivalent / capita-year)
Figure 4: Gross national product vs energy consumption of selected countries (Klass, 1998)
Share of wood energy in the total energy consumption in various Asian countries
Figure 5: Share of wood energy in total energy consumption in various Asian countries (FAO United Nation, 1997)
Page 3
Question 1:
1.1. From Figure 2, calculate the energy consumption per capita of Sweden in 1998 assuming that the population is about 8.8 million.
1.2. From Figure 3, draw the histogram showing the changes in World’s energy supply between 1998 and
1973 in terms of the following categories: fossil fuel, non-fossil fuel and biomass fuel (Note that nonfossil fuel includes nuclear, wind, solar, geothermal and hydro).
1.3.
How do we convert biomass to energy
Biomass materials are processed in various ways to produce heat, chemicals and other
types of fuels. The different conversion processes involved are generally classified as either: (i.) thermal conversion process or (ii.) biochemical conversion process. Figure 6
shows a schematic diagram summarising how biomass is converted into energy and heat.
Direct Combustion
Thermal
Conversion
Gasification
Liquefaction
Pyrolysis
Various
Gaseous and
Liquid Fuels,
Tars and
Charcoal
Biomass
Energy
and Heat
Anaerobic
Processes
Methane
Fermentation
Ethanol
Biological
Conversion
Figure 6: Schematic diagram showing how biomass are converted to produce heat and energy.
Thermal conversion processes include combustion, liquefaction, gasification and pyrolysis, while biochemical conversion processes include anaerobic digestion and fermentation. In this study pack, the direct combustion of biomass is emphasised.
Table 1 presents some of the common products produced during gasification, liquefaction
or pyrolysis of biomass. Table 2 presents the product yield from the thermal decomposition of Birch, Pine and Spruce woods heated over an 8 hour period to a final temperature
of 400oC.
Table 1: Some of the common products produced during gasification, liquefaction
and pyrolysis of biomass
Solid Products
charcoal, ash
Liquid Products
water, tar, volatile acids, alcohols, aldehydes, esters, ketones
Gaseous Products
H2, CO, CO2, CH4, C2H4, C2H6
Page 4
Table 2: Product yields from thermal decomposition of Birch, Pine and Spruce woods
heated over an 8 hour period to a final temperature of 400oC. (Klass, 1998)a
Products
Birch (wt%)
Pine (wt%)
Spruce (wt%)
Gases
H2
CO
CO2
CH4
C2H4
0.03
4.12
11.19
1.51
0.21
0.03
4.10
11.17
1.49
0.14
.03
4.07
10.95
1.59
0.15
Charcoal
33.66
36.40
3.43
21.42
3.75
10.42
22.61
10.81
5.90
23.44
10.19
5.13
7.66
1.83
0.50
1.63
1.13
3.70
0.89
0.19
1.22
0.26
3.95
0.88
0.22
1.30
0.29
0.94
1.09
0.38
Liquid products
water
settled tar
soluble tar
volatile acids
alcohols
aldehydes
esters
ketones
Losses
a
Data are obtained by Klass (1998) from work of Nikitin et al (1962) & Bagrova and Kozlov (1958). Volatile acids are calculated as acetic acids; Alcohols as methanol; Aldehydes as formaldehydes; Esters as methyl acetate and Ketones as acetone.
1.4.
Biomass Composition and Energy Content
In a similar way to coal, biomass materials are also characterised according to their Ultimate Analysis, Proximate Analysis and their Gross Calorific Values. Table 3 presents the
Ultimate Analysis of some biomass materials and UK coal; Table 4 presents their Gross
Calorific Values and Proximate Analysis. Unlike coal, wood and other biomass contains
complex carbohydrates and lignin. Table 5 presents the chemical composition of softwood and hardwood. Figures 7A to 7C illustrate the chemical structure of lignin, cellulose and coal respectively. These figures show that coal consists mostly of complex aromatic compounds while most biomass, which consists of carbohydrates and lignin (the
only source of aromatics), would have less complex aromatic compounds and more oxygen and hydrogen atoms attached. This explains why biomass has higher atomic H/C and
O/C ratio than coal.
Table 3: Ultimate Analysis of Some Biomass Fuels and UK Coal (CRE, 1995; Rossi, 1984)
Fuel
Rice hulls
Gin Trash
Peach pits
Black Oak
Red Alder
Red Alder Bark
Douglas-fir
Douglas-fir Bark
W. Hemlock
Markham Main
Daw Mill
Cynheidre
Taff Merthyr
Fuel Types
Agricultural
Agricultural
Agricultural
Hardwood
Hardwood
Hardwood
Softwood
Softwood
Softwood
UK Coal
UK Coal
UK Coal
UK Coal
Weight Percent by Element (Dry Basis)
Atomic Ratio
C
H
O
N
S
Ash
H/C
O/C
38.0
42.8
49.1
49.0
49.6
50.9
50.6
54.1
50.4
78.7
76.8
93.2
88.4
4.4
5.1
6.4
6.0
6.1
5.5
6.2
6.1
5.8
5.0
4.5
2.8
4.0
35.5
35.4
43.5
43.5
43.8
40.7
43.0
38.8
41.4
8.9
10.9
0.3
1.2
0.8
1.5
0.5
0.2
0.1
0.4
0.1
0.2
0.1
1.7
1.2
1.0
1.4
0.1
0.6
0.0
0.0
0.1
-0.0
tr.
0.1
1.5
1.4
0.6
0.7
21.00
14.69
0.50
1.34
0.41
2.50
0.10
1.00
2.20
3.71
4.69
1.83
4.03
1.36
1.43
1.55
1.47
1.48
1.30
1.47
1.35
1.38
0.76
0.71
0.37
0.55
0.69
0.62
0.66
0.67
0.66
0.60
0.64
0.54
0.62
0.08
0.11
0.00
0.01
Page 5
Table 4: Proximate Analysis of Some Biomass Fuels and UK Coal (Rossi, 1984; CRE, 1995)
Fuel
Rice hulls
Gin Trash
Peach pits
Black Oak
Red Alder
Red Alder Bark
Douglas Fir
Douglas Fir Bark
W. Hemlock
Markham Main
Daw Mill
Cynheidre
Taff Merthyr
Fuel Types
Agricultural
Agricultural
Agricultural
Hardwood
Hardwood
Hardwood
Softwood
Softwood
Softwood
UK Coal
UK Coal
UK Coal
UK Coal
Gross Calorific
Value (MJ/kg)
14.89
15.58
19.42
18.65
19.30
19.44
20.37
21.93
19.89
33.60
32.82
35.64
36.50
Proximate Analysis (weight per cent, oven dry basis)
Volatile Matter
(VM)
Fixed Carbon
(FC)
Ash
VM/FC Ratio
63.60
75.40
79.10
85.60
87.10
87.00
87.30
73.60
87.00
36.14
38.02
4.67
12.80
15.80
15.40
19.80
13.00
12.50
19.70
12.60
25.90
12.70
60.15
57.29
93.50
83.17
20.60
9.20
1.10
1.40
0.40
3.00
0.10
0.50
0.30
3.71
4.69
1.83
4.03
4.00
4.90
4.00
6.60
7.00
3.90
6.90
2.80
6.80
0.60
0.66
0.05
0.15
Figure 7-A: Hypothetical representative structure of a
coniferous lignin (Antal, 1982)
Figure 7-B: Hypothetical representation of
cellulose structure (Antal, 1982)
Figure 7-C: Hypothetical representation of coal structure (Wender et al, 1981)
Page 6
Table 5: Chemical Composition of Wood (%)a (Tsoumis, 1991)
Components
Softwood
Hardwood
Holocellulose
59.8 - 80.9
71.0 - 89.1
Cellulose
30.1 - 60.7
31.1 - 64.4
12.5 - 29.1
18.0 - 41.2
4.5 - 17.5
12.6 - 32.3
Lignin
21.7 - 37.0
14.0 - 34.6
Extractives (hot water)
0.2 - 14.4
0.3 - 11.0
Extractives (cold water)
0.5 - 10.6
0.2 - 8.9
Extractives (ether)
0.2 - 8.5
0.1 - 7.7
Ash
0.02 - 1.1
0.1 - 5.4
Polyoses
b
Pentosans
b
a. Ranges of values are derived from 153 Temperate Zone species. As a rule, extractives are determined from oven dry basis and others are determined from oven dry and extractives free basis.
b. Hemicellulose
Another important property of biomass is the moisture content. Siau (1984) has noted that
water in wood can be contained in the pores or chemically bound within the structure of
the biomass. The moisture present in most biomass materials can be classified as:
o bound water
o free water
o water vapour
Table 6 summarises the total moisture content of some biomass fuels.
Table 6: Moisture Contents of Some Biomass Fuels (Rossi, 1984)
Fuel
Range of Moisture Content (%)
Woody Biomass
Bark
Chips
Cull material
Hog fuel
Planer shavings
Sander dust
Sawdust
30 - 60
40 - 50
40 - 70
30 - 60
8 - 19
2-6
40 - 55
Agricultural Wastes
Gin trash
Grape pomace
Nuts
Orchard pruning
Peach pits
Tomato pomace
Rice hulls
Vineyard pruning
7 - 12
50 - 60
10 - 35
20 - 40
30 - 40
50 - 75
7 - 10
20 - 40
Question 2:
2.1. Comment on whether ash or moisture would help or hinder the combustion of wood in a furnace.
2.2. From Table 3 and 4, tabulate the following for coal (not including lignite), agricultural biomass, wood, and tree
barks:
a.) volatile to fixed carbon ratio
b.) atomic hydrogen to carbon ratio
c.) atomic oxygen to carbon ratio
d.) gross calorific values
Page 7
1.5.
Advantages and Disadvantages of Using Biomass
Advantages
¾ lower sulphur and nitrogen contents as compared to coal and fuel oil results in
lower SOx and NOx emissions.
¾ sustainable management of biomass will results in a reduction of CO2 emission by
displacement of fossil fuels (CO2 produced during combustion are considered CO2
neutral.)
¾ provide savings for most developing countries by displacing imported fossil fuels
Disadvantages
¾ have low thermal intensity as compared to fossil fuels; therefore, for a specified
plant duty, biomass mass flows are greater than those for coal.
¾ most biomass has low density and bulk density and therefore requires larger
equipment for handling, storage and burning.
¾ biomass materials are normally high moisture content therefore reducing combustion efficiency.
Question 3:
3.1. Identify the main pollutants from biomass combustion and their related impact?
3.2. Describe the regulation for the control of smoke emission from stationary combustion plant in Sweden?
1.6.
Principles of Biomass Combustion
Burning of biomass to obtain heat and light is one of the oldest biomass conversion processes known to mankind. Complete combustion (i.e. incineration, direct firing) of biomass consists of (a.) rapid chemical reactions (oxidation) of biomass and oxygen, (b.) the
release of energy and (c.) the simultaneous formation of the ultimate oxidation products
of organic matter (i.e. CO2 and water). The basic stoichiometric equation for the combustion of wood, represented by the empirical formula of cellulose, [C6H10O5]n, is given by:
[C6H10O5]n + 6nO2 → 6nCO2 + 5nH2O
The combustion of lump biomass can be simply described by the burning of a single biowet core
pyrolysis front at r = rp
dry shell
volatiles from the particle
water from the particle
drying front at r = rd
char layer
flame region
Figure 8: Schematic diagram showing the combustion of a small biomass particle (i.e. with a 20 mm diameter)
Page 8
mass particle as shown in Figure 8.
¾ Biomass fuel follows a similar sequence of processes as applies to the combustion of
coal. These include the following:
o Drying
o Devolatilisation
o Char combustion
o Volatile combustion
¾ Assuming that a 20 millimetre sized particle is burned in a furnace, the combustion
mechanisms of the biomass particle can be classified into two phases namely: preignition phase and post-ignition phase. For the combustion of this biomass particle,
the following stages may occur simultaneously or concurrently. The basic steps are
described as follows:
o Step 1: Drying of the biomass particle immediately commences as the particle is
introduced into the hot combustion environment. Due to the higher permeability of the vapour phase, a wet-dry interface (as indicated by the
drying front at r = rd as shown in Figure 7) is formed within the particle.
o Step 2: As the dry shell starts heating up, devolatilisation commences with the
breaking up of the solids structure.
o Step 3: As the volatile release becomes more rapid, ignition of the volatiles are assumed to occur in the gas phase. Following ignition, volatiles burn in a
thin flame enveloping the particle surface. It is believed that this phenomenon prevents oxygen reaching the charred surface and the flame energy promotes the drying and devolatilisation of the particle.
o Step 4: As all the moisture is vaporised, volatilisation may still continue. Several
workers (Veras et al, 1999; Saastamoinen et al, 1993; Simmons, 1983)
have indicated that there is a possible simultaneous occurrence of devolatilisation and char combustion at this stage of the process.
o Step 5: As the volatiles become depleted, the volatile flame collapses and oxygen
is permitted to attack the charred surface therefore leading to the ignition
and burnout of the residual char.
¾ Among the complex processes involved during combustion, the devolatilisation of
the biomass materials and the subsequent combustion of the char produced during
devolatilisation are seen to be the most important process that define parameter for
the design of a furnace. This will be described in more detail in the succeeding sections.
Question 4:
4.1. How does the reactivity of the fuel particle vary with biomass properties?
4.2. What additional physical processes are involved if the particle is burned in a fuel bed?
Page 9
1.6.1.
Devolatilisation
The devolatilisation of biomass materials involves several complex processes of
thermal decomposition and can be generalised as:
¾ Removal of bound moisture and some volatiles
¾ Breakdown of hemicellulose; emission of CO and CO2
¾ Exothermic reaction causing the wood temperature to rise from 250 to 360oC;
emission of methane and ethane.
¾ External energy is required to continue the process (i.e. breakdown of cellulose and lignin).
The products of biomass devolatilisation are generally classified into gas, tar and char
(Di Blasi, 1993). The distribution of these products depends on various parameters
such as biomass composition, reaction temperature, heating rate, particle size and
catalyst used. Usually, under a fast devolatilisation condition yields more gases than
solids.
The rate of mass loss during devolatilisation can be determined using various models.
The simplest among these models assumes an Arrhenius form of equation with a first
order reaction represented as:
−
dm v
= Am v exp( − E / RT )
dt
(Eqn. 1)
where
mv
A
E
R
T
= mass of volatiles remaining (kg)
= Arrhenius constant
= activation energy (kJ/kmol)
= universal gas constant (kJ/kmol-K)
= temperature (K)
The activation energy (E) and pre-exponential constant (A) vary considerably depending upon the pyrolysis conditions and fuel type as shown in Table 7. Some other models assume two competing reactions or a series of competing and concurrent steps.
Figure 9 presents an outline of a commonly used reaction scheme describing the devolatilisation of biomass.
Table 7: Kinetic data for devolatilisation assuming first order reaction kinetics (Di Blasi, 1993)
Temperature (K)
E (kJ/mol)
A(s-1)
Reference
α - cellulose
Sample
550 - 1000
79.4
1.7 x 104
Kanury (1972)
Cellulose
600 - 850
100.5
1.2 x 106
Tabatabaie et al (1989)
-3
Beech sawdust
450 - 700
18 (T < 600)
5.3 x 10
Barooah and Long (1976)
Beech sawdust
450 - 700
84 (T > 600)
2.3 x 104
Barooah and Long (1976)
Cellulose
Lignin
520 – 1270
520 - 1270
166.4
141.3
3.9 x 10
11
Lewellen et al (1978)
8
Min (1977)
1.2 x 10
9
Hemicellulose
520 - 1270
123.7
1.45 x 10
Wood
321 - 720
125.4
1.0 x 108
Nolan et al (1973)
Min (1977)
Almond Shell
730 - 880
95 - 121
1.8 x 106
Font et al (1990)
Page 10
i.) Simple Single Step Reaction
k
Biomass 

→ Volatiles + Char
Biomass
k

→ a Gases + b Tars + c Char
ii.) Multiple Single Step Reactions
k1
Product (1)
k2
Product (2)
k3
Biomass
Product (3)
•
•
•
•
•
•
•
•
•
•
•
•
Product (i)
iii.) Multi Step Reaction
k1
Biomass
k2
Gases + Volatiles
Flaming Combustion
k5
Tar
k4
k3
Char + Gases
Glowing Combustion
Figure 9: A common reaction scheme used in biomass pyrolysis (Adapted from
Di Blasi, 1993)
Recent development in the modelling of biomass devolatilisation involves the use of
network models such as the bio-FG DVC [Functional Group – Depolymerisation, Vaporisation, and Crosslinking] (Chen et al, 1998) and bio-FLASHCHAIN (Niksa,
2000). These network models are developed based on an assumed structure of the
biomass and a set of kinetic data from a fuel library. These data are determined from
various experiments and modelled on the principle of distributed activation energy.
For example, bio-FG DVC describes the devolatilisation process in two steps. The
first step involves the evolution of gases and tars based on the breakaway of various
functional groups from their parent structure. The second step describes the reactions
of the solid materials based on a Monte Carlo simulation. The model was applied to
several types of biomass such as wheat, corn stalk and wood. Further details of this
model are described in the literature (Solomon et al, 1992; Chen et al, 1998).
Illustrative Example (#1):
A small wood particle has a temperature of 800 K. Find the time required to devolatilise 90% of the volatile mass, assuming
that it follows a first order reaction and with Arrhenius constant (A) = 7 x 107 and activation energy (E) = 125 kJ/mole.
Solution to Illustrative Example (#1):
Using the equation for a first order reaction
dm v
−
= Am v exp( − E / RT )
dt
Rearranging
m
t
dm v
∫m − m Aexp(-E / RT) = ∫t dt
v
v2
v1
2
1
Integrating
m 
− ln v2 
 m v1  = t − t
2
1
Aexp(- E/RT)
since
mv2 = 0.1 mv total & mv1 = mv total
Substituting
 0.1 m v total 

− ln

 m v total 
= t2 − 0
7 x10 7 exp(-125,000/8.314/800)
thus
time (t2) required is 4.78 sec.
Page 11
1.6.2.
Char Combustion
The final step in solid fuel burning is the char combustion. When devolatilisation is
complete, char and ash remain. Oxygen or other oxidisers (i.e. CO2, H2O) can diffuse
through the external boundary layer into the char particle and react with the char surface, (Generally, char is greater than 95% carbon). Basically, the char combustion
process involves the following gas-solid reaction processes (Laurendeau, 1978):
o Transport of the oxygen molecules or oxidisers to the surface by convection
and/or diffusion.
o Adsorption of the oxidiser molecule on the surface.
o Reaction of the adsorbed oxidiser molecule with the surface, reaction of the surface itself or reaction of the surface and gas phase molecules.
o Desorption of the combustion product from the surface.
The burning rate of char mainly depends on both chemical rate of the carbon-oxygen
(oxidiser) reaction at the surface and the rate of diffusion of the oxygen (oxidiser) into
the boundary layer and into the internal voids within the char. Likewise, burning rate
also depends on the oxygen (oxidiser) concentration, gas temperature, Reynolds Number, and char size and porosity.
Biomass char is known to be highly porous [i.e. wood char is about 94% porous]
(Williams et al, 2000). This therefore results in an internal surface area of around
10,000 m2/g (for wood char with 90% porosity). This property of biomass char
strongly affects the combustion rate of the char.
For engineering purposes, the global reaction rate of order ‘n’ with respect to oxygen
is used to calculate the char burning rate and is given by:
 M
dm c
= −i c
 MO
dt

2

A p k c ρ O (s ) n


(
2
)
(Eqn. 2)
where i , MC, MO2, Ap, kc, ρ O (s ) and n are the stoichiometric ratio (moles of carbon
per mole of oxygen; i = 2 if product is CO and 1 if product is CO2), molecular mass of
carbon, molecular mass of oxygen, total surface area, kinetic rate constant, concentration of oxygen at the char surface and reaction order respectively.
2
Assuming that reaction order (n) is equal to one, thus:
[
]
A p k c ρ O (s ) = A p hD ρ O (∞ ) − ρ O (s )
2
2
2
(Eqn. 3)
where hD and ρ O (s ) are the mass transfer coefficient and ambient oxygen concentration respectively. Therefore, burning rate can be simplified to:
2
 M
dm c
= −i c
 MO
dt

2

A p k e ρ O (s )


2
(Eqn. 4)
where ke is the effective kinetic rate constant that can be expressed as: k e =
hD k c
hD + k c
Page 12
To aid the understanding of the char combustion process, burning of non-porous pure
carbon is used as an example. There are two limiting conditions available in describing the combustion of a spherical carbon particle. These conditions are defined in:
o the one-film model and
o the two-film model
Figures 10 and 11 present the species and temperature profiles of a burning carbon
particle assumed to follow one-film and two-film models respectively.
Yi, T
Gas-Solid Interface
(Carbon Surface)
Yi – concentration of species i.
T – temperature
r - radius
YCO2 (s)
YO2 (∞)
YCO2 (r)
Tp
YO2 (r)
Ts
T(r)
T(∞)
YO 2 (s)
r
rs
0
Figure 10: Species and temperature profiles for one film model of carbon combustion
assuming that CO2 is the only product of combustion at the carbon surface (Turns, 1998).
Yi, T
Carbon
Surface
Yi – concentration of species i.
T – temperature
r - radius
Flame
Sheet
YCO (s)
YO 2 (∞)
T(r)
Tp
Ts
YCO (r)
T(∞)
YCO2 (r)
YO2 (r)
YCO2 (s)
0
rs
r
Figure 11: Species and temperature profiles for two-film model of a burning spherical
carbon particle (Turns, 1998)
Page 13
At the surface, carbon can be attacked by O2, CO2 or H2O as shown in the following
reactions:
C + O2 ⇒ CO2
C + ½O2 ⇒ CO
C + H2O ⇒ CO + H2
C + CO2 ⇒ 2CO
Then CO diffuses away from the surface through the boundary layer where it can react with the inward diffusing O2 as:
CO + O2 ⇒ CO2
For the one-film model (as shown in Figure 10), it can be seen that the CO2 concentration is a maximum at the carbon surface and is approximately zero far from the particle surface. Conversely, O2 concentration is at minimum at the surface. If the
chemical reaction rate is much greater than the oxygen diffusion rate, then the O2 concentration approaches zero; otherwise, if the diffusion rate is much faster than the
chemical reaction rate, then O2 concentration at the surface will be appreciable.
Since, for this model, it is assumed that there is no reaction involved in the gas phase,
the temperature is at a maximum on the carbon surface and then falls monotonically
to T∞.
On the other hand, for the two-film model (as shown in Figure 11), it can seen that a
flame sheet is present at some distance away from the surface; and the CO2 species attacking the carbon surface is reduced to yield CO. The maximum temperature is located around the interface of the flame sheet. Likewise, O2 and CO are a minimum at
the interface of the flame sheet and CO2 is a maximum.
In the one-film model the following assumption are usually made (Turns, 1998):
o Burning process is quasi-steady state.
o Spherical carbon particle burns in a quiescent, infinite ambient medium that
contains only oxygen and an inert gas such as nitrogen. There are no interactions with other particles and effects of convection are ignored.
o At the particle surface, the carbon particle reacts kinetically with oxygen to produce only CO2.
o The gas phase only consists of O2, CO2 and inert gas. The O2 diffuses inward,
reacts with the surface to form CO2 which then diffuses outward. The inert gas
forms a stagnant layer as in the Stefan problem.
o Gas phase thermal properties (i.e. conductivities, specific heat, product of the
density and mass diffusivity) are constant. Lewis number is equal to one (Le =
k/ρcpD = 1).
o Particle temperature is uniform and radiates as a gray body to the surroundings
without participation of the intervening medium.
Using a mass balance and assuming that diffusion of O2 and CO2 into and out of the
carbon particle follows Fick’s Law, the burning rate of carbon can be calculated as
(Turns, 1998):
Page 14
&c =
m
(Y
O2 ∞
−0
)
(Eqn. 5)
R kin + R diff
where
R kin ≡
1
k kin
ν s( 1− film ) RTs
=
4πrs2 MWmix k c P
(Eqn. 6)
where ν s( 1− film ) , R, Ts, rs, MWmix, kc and p are the stoichiometric ratio of the reacting
species at the surface for the one-film model (32/12), universal gas constant (8.314
J/mol-K), carbon surface temperature (K), radius of the particle, molecular mass of
gas mixture (kg/kg-mole) and pressure (Pa) respectively.
and
R diff ≡
(ν
s ( 1− film )
+ YO s
2
)
4πrs ρD
(Eqn. 7)
where YO s , ρ and D are oxygen concentration at the surface, gas mixture density and
diffusivity respectively.
2
Likewise, at the carbon surface, the burning rate is calculated as (Turns, 1998):
& c = k kin YO 2s
m
(Eqn. 8)
For the two-film model, the following assumptions are used:
o Burning process is quasi-steady state
o Spherical carbon particle burns in a quiescent, infinite ambient medium that
contains only oxygen and an inert gas such as nitrogen. There are no interactions with other particles and effects of convection are ignored.
o At the particle surface, the carbon particle reacts kinetically with CO2 to produce
only CO.
o The gas phase only consists of O2, CO2 and inert gas. The O2 diffuses inward,
reacts with CO along the first gas film (gas layer before the flame sheet interface) to form CO2 which then diffuses outward. The inert gas forms a stagnant
layer as in the Stefan problem.
o Gas phase thermal properties (i.e. conductivities, specific heat, product of the
density and mass diffusivity) are constant. Lewis number is equal to one (Le =
k/ρcpD = 1).
o Particle temperature is uniform and radiates as gray body to the surroundings
without participation of the intervening medium.
In this case, the burning rate of carbon surface is calculated as (Turns, 1998):
a.) At the carbon surface
& c = k kin YCO s
m
2
(Eqn. 9)
where
k kin(2-film) =
4πrs2 MWmix k c P
ν s( 2− film ) RTs
(Eqn. 10)
Page 15
b.) At the gas phase, the burning rate is calculated as
& c = 4πrs ρD ln(1 + B )
m
and
(
(Eqn. 11)
)
−1
ν

2YO ∞ −  s( 2− film )
 YCO s
ν
s ( 2 − film ) 

B=
−1
ν

ν s( 2 − film ) − 1 +  s( 2 − film )
 YCO s
ν
s ( 2 − film ) 

2
2
(
(Eqn. 12)
)
2
where ν s( 2− film ) , YO ∞ and YCO s are the stoichiometric ratio at the surface in the
two-film model (44/12), oxygen concentration at the free stream and carbon
dioxide concentration at the surface.
2
2
Illustrative Example (#2):
Using the one-film model, estimate the burning rate of a 90 µm carbon particle burning in a still air ( YO ∞ = 0.233) and 1 atm.
2
The particle temperature is 1750K and the kinetic rate constant is assumed to be 13.9 m/s. Assume that the mean molecular
mass of the gas mixture surrounding the particle is 30 kg/kmol and the mass diffusivity is estimated using the value for the mass
diffusivity of CO2 in N2 which is 1.6 x 10-5 m2/s at 120oC. Also what is the prevailing combustion regime?
Solution to Illustrative Example (#2)
Calculation Procedure:
1.5
Note: calculation for the onefilm or two-film model usually
involves iteration. To solve for
the burning rate, the following
procedure is used:
a.) Calculate Rdiff and initially
assuming that YO s = 0
2
b.) Calculate Rkin and then
solve for the burning rate.
c.) Using the equation for burning rate at the surface, calculate the value of YO s .
 1750 K 
D=
 1.6 x 10−5 m 2 /s = 1.19 x 10-4 m 2 /s
 393 K 
Assuming
Solution:
ρ =
=
P MWmix
RTs
(101,325Pa)(30 kg/kmol)
(8314 J/kmol - K)(1750 K)
= 0.209 kg/m3
for the time being.
(ν s( 1− film ) + YO s )
=
R diff
2
4πrsρD
4π( 45 x 10-6
R kin =
ν s( 1− film )RTs
4πrs2 MWmix k c P
(3212)(8314)(1750)
=
4π( 45 x 10-6 )2 (30) (13.9)(101,325)
= 3.61 x 107 s/kg
=
Recalculating Rdiff
 R diff - 2nd iter

R
 diff -1st iter.
(
(
(YO ∞ − 0)
2
R kin + R diff
(0.233 − 0)
= 1.25 x 10-9 kg/s
3.61 x 107 + 1.50 x 108
Using the equation for burning rate on the surface,
& c = k kin YO s
m
2
)
)
 ν s( 1− film ) + YO2 s 2nd iter.
=

ν s( 1− film ) + YO2 s

1st iter.
=
)(0.209)(1.19 x 10-4 )
= 1.50 x 108 s/kg
&c =
m
& c = (3.61 x 107 )(1.25 x 10-9 ) = 0.0452
YO 2s = R kin m
(3212 + 0)
=
2
d.) Recalculate Rdiff.and compare the values of previous
Rdiff
YO 2 s ≈ 0
rearranging
(3212 + 0.0452)
(3212 + 0)
2nd iter.
= 1.0169
1st iter.
∴ Rdiff (2nd iteration) = (1.50 x 108)(1.0169) = (1.52 x 108)
Recalculate burning rate (2nd iteration)
YO2 ∞ − 0
(0.233 − 0)
&c =
m
=
R kin + R diff
3.61 x 107 + 1.52 x 108
= 1.23 x 10-9 kg/s
(
)
(diffusion controlled combustion regime)
Since changes in the burning rate are less than
2%, no further iteration is required.
Challenge Question
Recalculate problem using the two-film model assuming
that kc = 4.016 x 108 exp (-29,790/Ts) [m/s] and compare
the result from one-film model. (Ans: 2.3784x 10-9 kg/s)
Page 16
1.7.
Combustion Equipment
Biomass materials can be burned in many ways. They can be pulverised, crushed,
chipped, cut, chopped and fed into one of the several types of furnaces or boilers available. Generally, biomass materials can be burned in a:
”
Fixed bed
”
Suspension fire
”
Fluidised bed
Fuel handling and feeding are some of the factors which need to be considered regarding
how the biomass materials are burned. The nature of combustion system dictates the required fuel preparation. Fixed bed combustion systems require the least amount of fuel
size reduction as compared to suspension burning or fluidised bed combustion. Typical
biomass materials are difficult to crush or pulverise due to their fibrous nature thus most
biomass materials are burned either in a fixed bed or a fluidised bed where uniform particle size is not a pressing requirement (Borman and Ragland, 1998).
1.7.1.
Fixed Bed Combustion
The traditional campfire is a classical example of the fixed bed combustion of biomass using the natural convection of air as the oxidiser. Normally this kind of combustion (simple pile burning method) is considered low intensity combustion (ie. low
heat output per unit volume) and smoky. In order to increase the combustion intensity, improve control of heat output, and reduced emissions, force draft and grates are
used.
Fixed beds can be operated using co-current, cross-current or counter current air flow.
This broad classification of fixed bed is based on how the fuel and air is introduced
into the combustion system. Figure 12 shows how these three types of fixed bed
combustion operate. It also illustrates the different reaction zones that can occur
within the bed.
Counter Current
Co-Current
Cross Current
ash
product gas
product gas
solid fuel
product gas
Ash
Drying
1
Devolatilisation
Gasification
ash
3
Combustion
4
5
Combustion
Devolatilisation
Drying
air
Ash
ash
Gasification
2
solid
fuel
air
1: Drying
2: Devolatilization
3: Combustion
4: Gasification
5: Ash
solid fuel
air
Figure 12: Schematic diagram of different fixed bed reactors
Page 17
(a.)
(b.)
Figure 13: Some examples of fixed bed combustion. (a.) spreader
stoker (b.) travelling grate stoker (Podolski et al, 1997)
An example of counter current fixed bed combustion is the spreader stoker. This type
of stoker operates in such a way that fuel is fed on top of the bed and air is introduced
through several ports under the fuel bed. Figure 13(a) shows how the spreader stoker
operates. For the counter current reactor, the fuel is generally fed on top of the bed
and moves downward under the influence of gravity and moves counter to the flow of
the gas stream.
An example of a cross current combustion system is the travelling grate stoker as
shown in Figure 13(b). In this type of stoker, fuel is fed through a hopper at one end
of the grate while air is introduced through the small gaps that are situated all over of
the grate.
A typical example for co-current combustion is the underfeed stoker. In an
underfeed stoker, both fuel and air are
fed more or less in the same direction.
This type of stoker is built with a single
or multiple retort configuration. In a
single retort, a ram or a screw pushes the
fuel to the end of the stoker and upwards
toward a tuyere block where air is introduced into the bed. On the other hand, a
multiple retort stoker consists of a series
of shallow, longitudinal retorts separated
by rows of stepped air tuyeres. Figure 14
shows a schematic diagram of the underfeed stoker.
1.7.2.
Heat Exchanger Tubes
Combustion Gas to Chimney
Figure 14: Schematic diagram of the underfeed
stoker (Purvis et al, 2000)
Fluidised Bed Combustion
A fluidised bed is a bed of solid particles which are put into motion by blowing air
upwards through the bed at a sufficient velocity to locally suspend the particles. The
air blown through the bed has a velocity that is not great enough to blow the particle
out of the bed and will make the bed appear like “boiling liquid”. The fluidised bed
therefore exhibits buoyancy and a hydrostatic head. Figure 15 presents the schematic
diagram of a fixed bed, a bubbling fluidised bed and a circulating fluidised bed.
Page 18
Fixed Bed
Bubbling Fluidised Bed
Circulating Fluidised
Bed
Figure 15: Schematic diagram of a fixed bed, bubbling fluidised bed and
circulating fluidised bed combustors (Borman and Ragland, 1998)
Fluid bed combustion has been given much attention in recent times because of its
advantages particularly in large scale systems. Combustion takes place in a cylindrical
reactor in which air is dispersed through an orifice or a sintered plate at the bottom of
the vessel. The air then passes through a bed of inert refractory pieces, particles of fuels, ash and other residual inorganic particles remaining from combustion therefore
causing the bed to expand and to become “fluidised”. Smaller fuel particles burn rapidly on top of the bed while large fuel particles filter into the bed where they are dried
and gasified. Most char is burned completely within the bed while the volatiles are
burned partially in and partially above the bed. The fuel is fed either by a ram or by
pneumatic means into the reactor clearance at around 900K. Fluidised bed combustion is suitable to burn high moisture content fuels because of its low heat input requirement and high thermal inertia. Materials such as limestone are often added to
the bed to minimise pollutants in the flue gases. The constant motion of the bed allows good mixing between air and fuel, this improves combustion, reduces emissions
and makes it possible to burn a wide range of fuels having different sizes, moisture
contents and calorific values. Figure 16 presents a schematic diagram of a circulating
fluidised bed combustion system.
Figure 16: Schematic diagram of circulating fluidised bed combustion system
(Borman and Ragland, 1998)
Page 19
The air velocity in a bubbling fluidised bed is maintained slightly lower that the air
velocity of a circulating fluidised bed to prevent particle carryover. Likewise, a bubbling fluidised bed requires a fuel feed point for approximately every 1 m2 of bed area
for good mixing. Circulating fluidised bed combustors were developed to overcome
particle carryover and facilitate fuel feeding.
Fluidised bed can be operated either atmospheric or pressurised. Pressurised fluidised
bed combustion systems are currently developed with an aim of directly powering a
gas turbine using various solid fuels.
1.7.3.
Suspension Burning
Suspension furnaces burn pulverised fuel particles that are fed through a nozzles into
the furnace volume which is large enough to allow burnout of the fuel chars. Figure
17 shows a schematic diagram of a pulverised fuel steam power plant.
Solid Fuels
Figure 17: Schematic diagram of a pulverised fuel combustion system (Borman and Ragland, 1998)
As received, solid fuel is fed into a pulveriser (ie. ball mill) or a shredder. From the
size reduction equipment, the pulverised fuel (coal or biomass) is piped with air to
burners, where the fuel is mixed with pre-heated air from an air plenum (windbox).
The volatile flames are stabilised by the burner while the chars are burned out in the
radiant section of the furnace.
The burner consists of a nozzle that delivers the fuel particle. The conveying air that
delivers the pulverised fuel into the furnace is called the primary air and the air is
about 20% of the required combustion air. The secondary or main air is supplied
through a swirl vane surrounding the fuel nozzle. Figure 18 shows the schematic diagram of a fuel nozzle burner. The flame shape is controlled by the extent of secondary
air swirl and the contour of the burner throat. The recirculation pattern which is set up
inside and extends several throat diameters into the furnace provides a stabilisation
zone for ignition and combustion of the volatiles.
The velocity of the primary air and pulverised fuel should be greater than the speed of
flame propagation so as to avoid flashback. The flame speed depends on the fuel-air
ratio, the amount of volatile matter, ash in the fuel particle, particle size distribution,
Page 20
Figure 18: Schematic diagram of a low NOx burner
air pre-heat temperature and nozzle diameter. Fuel rich mixtures have the highest
flame speeds. For low volatile and high ash fuel, the flame speed is low and the air
from the secondary nozzle should be mixed appropriately to avoid flame instability.
As mentioned earlier, wood and bark are more difficult to grind than coal because of
their fibrous nature which is less friable than coal particles. However, pulverised biomass burners are noted to be economically attractive for retrofitting oil burners when
an ample supply of pulverised biomass (i.e. sawdust) is available. Co-firing of biomass with coal in a pulverised combustor has been done to a limited extent (Sami et
al, 2001). Several issues remain to be resolved when co-firing biomass and coal and
some of these are:
a.) Fouling and corrosion problems due to the alkalinity of biomass ash.
b.) Burner stability and particle burnout should be considered due to differences in reactivity.
c.) Fuel feeding mechanisms should be taken into account due to differences
in physical properties.
1.8.
Thermal Efficiency
The thermal efficiency refers to the amount of heat recovered as useful heat divided by
the heat input. This can be determined in two ways:
a.) direct method
b.) indirect method
The Direct method is expressed as:
η=
amount of the energy absorbed by the working fluid
x 100%
energy input
The Indirect method is expressed as:
 sum of energy losses 
 x 100%
η = 1 −
energy input


Page 21
The sum of energy losses considered are:
a.) the amount of energy lost in the flue gas
b.) the amount of energy lost due to CO in the flue gas
c.) the amount of energy lost due to unburned fuel in ash
d.) the conduction and convection losses from the boiler structure
The overall energy balance for a boiler may be written from the first law of thermodynamics as:
Energy input = Energy to working fluid + Energy losses in the flue gas + unaccounted losses
Where
Unaccounted losses = radiation loss from boiler structure + conduction losses + energy loss to the
floor of the boiler house
Therefore from first law of thermodynamics:
Energy input = Energy to steam + Energy loss in the flue gases + unaccounted loss
Divide by the “Energy input”
1=
Energy to steam Energy loss in the flue gases unaccounted loss
+
+
Energy input
Energy input
Energy input
Rearranging and neglecting the unaccounted energy loss (usually the unaccounted loss
contributes only to around 4% of the total energy input):
Energy loss in the flue gases
Energy to steam
≈ 1−
Energy input
Energy input
(Direct Method)
1.9.
(Indirect Method)
Combustion Air
The oxygen required in any combustion processes is supplied by the atmospheric air and
the oxygen within the fuel itself. As shown in Table 3, biomass has a substantial amount
of oxygen within the fuel therefore requires lesser amount of oxygen from the atmosphere. Atmospheric air (as shown in Table 8) has approximately 78% N2 and 21% O2.
However, for simplicity in combustion calculation, N2 is taken as 79% and O2 is taken as
21%. With this assumption, the molar ratio of the nitrogen to oxygen is 0.79/0.21 = 3.76.
Thus every mole of oxygen supplied from air is always accompanied by 3.76 moles of nitrogen. Also for calculation simplicity, nitrogen in the combustion process does not undergo chemical reaction. (i.e. nitrogen is inert).
In most combustion calculation, the parameters that are frequently used to quantify the
amounts of fuel and air in a particular combustion process are the air-fuel ratio, theoretical or stoichiometric amount of air and percent excess air.
Page 22
Table 8: Typical composition of dry air
Components
Mole Fraction (%)
Nitrogen (N2)
78.08
Oxygen (O2)
20.95
Argon (Ar)
0.93
Carbon Dioxide (CO2)
0.03
Others (Neon, Helium, Methane etc…)
0.01
The air-fuel ratio can be written on a molar basis (moles of air / moles of fuel) or on a
mass basis (kg of air / kg of fuel). Conversion between these values is accomplished by
using the molecular weight of air (MWair) and fuel (MWfuel).
Air - Fuel Ratio (AF) =
(moles of air)(MWair )
mass of air
=
mass of fuel (moles of fuel)(MWfuel )
The molecular mass of air is normally taken as 28.84 (29.0), while the molecular mass of
biomass may be calculated from the Ultimate Analysis of the fuel.
The minimum amount of air necessary to supply sufficient amount of oxygen to complete
the combustion process is called the theoretical or stoichiometric amount of air. To calculate the stoichiometric amount of air, the following global reactions are considered:
C + O2 → CO2
2H + ½ O2 → H2O
S + O2 → SO2
The products of combustion would then consist only of carbon dioxide, sulphur dioxide,
water, nitrogen from air and fuel. (nb: there will be no free oxygen appearing in the combustion product).
Note that 1 kmol of O2 is required to burn 1 kmol of carbon atom, ½ kmole of O2 is required to burn 2 kmol of H atoms and 1 kmol of O2 is required to burn 1 kmol of sulphur
atom. It should be noted that oxygen in the fuel is used first before oxygen in air is used.
Thus the theoretical amount of air is calculated as:
stoichiometric amount of O2 = kmol C + ¼ kmol H + kmol S – ½ kmol O in the fuel
Likewise,
(stoichiometric amount of O2)(0.21)-1 = stoichiometric amount of air
Finally, the percent excess air is defined as follows:
Percent Excess Air =
amount of air supplied
x100%
stoichiometric amount of air
Page 23
Illustrative Example (#3):
Determine the air to fuel ratio on mass basis for the complete combustion of woodchips with the following ultimate analysis of woodchips:
Carbon
48.6%
Hydrogen
5.2%
Oxygen
46.0%
Nitrogen
0.2%
Sulphur
0.0%
and burned under the following conditions:
a.) only the theoretical amount of air is supplied.
b.) When 50% excess air is supplied
Solution to Illustrative Example (#3)
Calculation Procedure:
Calculate the ultimate
analysis in dry ash free
(daf) basis to as fired basis.
Calculate the moles
b.)
of C, H, O, N and S atoms
per kg of fuel.
Calculate
the
c.)
stoichiometric amount of
air needed per kg of fuel.
Basis 100 kg of
d.)
woodchips and assuming
complete combustion.
a.)
mass of C in woodchip as fired
= % C(daf)/100*(100 – moisture – ash)
= 48.6/100 * (100 – 12.7 – 0.79)
= 42.04 kg C / 100 kg wood
Theoretical (stoichiometric) amount of oxygen
= kmol of C + ¼ kmol of H – ½ kmol O
= 3.503 + 4.50/4 – 2.487/2
= 3.385 kmol of O2
mass of H in woodchip as fired
= % H (daf)/100*(100 – moisture – ash)
= 5.2/100 * (100 – 12.7 – 0.79)
= 4.50 kg H / 100 kg wood
Theoretical amount of air on mass basis
= 3.385 kmol of O2 / 0.21 * 28.84
= 464.81 kg of air
mass of O in woodchip as fired
= % O (daf)/100*(100 – moisture – ash)
= 46.0/100 * (100 – 12.7 – 0.79)
= 39.79 kg O / 100 kg wood
moles C (per 100 kg woodchips)
= 42.04 / 12 = 3.503 kmol C atoms
Solution:
Mass of ash (as fired)
= % ash/100 x (100 – moisture)
= (0.9/100) (100 - 12.7)
= 0.79 kg of ash /100 kg wood
moles H (per 100 kg woodchips)
= 4.50 / 1 = 4.50 kmol H atoms
moles O (per 100 kg woodchips)
= 39.79 / 16 = 2.487 kmol O atoms
a.) air to fuel ratio on mass basis if theoretical
amount of air is supplied
= 464.81 kg of air / 100 kg of woodchips
= 4.65 kg of air / kg of woodchips
b.) air to fuel ratio on mass basis if 50% excess
air is supplied
= 4.65 kg of air / kg of woodchips * 1.5
= 6.97 kg of air / kg of woodchips
Challenge Question
(i.)
Calculate the composition of CO2,
O2, N2 and H2O in the flue gas. (assuming that nitrogen in the fuel is inert).
1.10. Products of Combustion
Illustrative example #3 assumes that combustion is complete and nitrogen in the fuel is
inert. Thus the product of combustion consists only of CO2, H2O, N2 from air, N2 from
fuel and SO2. However, in reality, combustion is seldom complete and CO in small quantity is one of the products.
Several factors affect the overall combustion process and these include (i.) amount of air
supplied, (ii.) degree of mixing between air and fuel, (iii.) temperature and pressure of the
combustion surroundings (i.e. furnace), (iv.) overall kinetics of the combustion process.
(v.) residence time of the air.
For example, when the amount of air supplied is less than theoretical amount required, the
products of combustion may include CO2 and CO; and there may also include unburned
fuel in the products. Unlike the case of complete combustion process, the products of an
incomplete combustion process can only be determined by experiment.
Page 24
Most commonly used devices for the experimental determination of the composition of
products of combustion are the gas chromatograph, infrared analyser, flame ionisation detector. Measurements obtained from these devices are generally reported as a dry flue gas
analysis (ie. moisture is removed from the combustion products).
Illustrative Example (#4):
(see next page for solution)
Woodchips, fed into an underfeed stoker at a rate of 5.8 kg/hr, are burned with dry air. The molar analysis of the flue
gas on a dry basis is CO2, 7.3%; CO, 0.3% and O2, 12.7%. Assuming that there are no unburned woodchips in the ash,
determine the following:
a.) Thermal efficiency of the plant
b.) Measured excess air in the flue gas
Properties of Woodchips
Ultimate Analysis (daf)
48.6 %
Carbon
5.2 %
Hydrogen
46.0 %
Oxygen
0.2 %
Nitrogen
0.0 %
Sulphur
Proximate Analysis and GCV
12.1 %
Fixed Carbon (dry)
87.0 %
VM (dry)
0.9 %
Ash (dry)
12.7 %
Moisture
17,756 kJ/kg
GCV (as received)
Operating Parameters of the boiler
Fuel Feed Rate
Cooling Water Flow Rate
Temperature Water In
Temperature Water Out
5.8 kg/hr
7.4 litre/min.
12.6 oC
55.9 oC
1.11. Environmental Issues and Biomass Combustion
The control of pollutant emissions is a major factor in the design of modern combustion
systems. This is also one of the main driving forces in the development and use of biomass as a renewable energy source. Pollutants of concern related to biomass combustion
are:
o particulate matter – soot, fly ash, fumes and various aerosols
o sulphur oxides – SO3, SO2
o nitrogen oxides –NOx (which consist of NO & NO2), N2O
o unburned and partially burned hydrocarbons (PAH, aldehydes, CH4)
o dioxin and furan
o carbon monoxide
o greenhouse gases – CO2, N2O
Page 25
Solution to Illustrative Example (#2)
Calculation Procedure:
a.) Calculate the total energy input to
the boiler and the energy absorbed
by the cooling water.
b.) Calculate the stoichiometric amount
of air required for combustion.
c.) Calculate the total amount of DFG
per 100 kg of fuel.
d.) Calculate the amount of O2, CO in
the DFG per 100 kg of fuel.
e.) Calculate the amount of O2 required
to burn the CO in DFG completely
to CO2.
f.) Basis 100 kg of woodchips and assuming that no unburned fuel in ash.
Solution:
Thermal Efficiency of the plant (direct)
amount of energy absorbed by CW
=
amount of energy input
22.34 kW
* 100%
=
28.60 kW
= 78.11%
Mass of ash (as fired)
= % ash/100 x (100 – moisture)
= (0.9/100) (100 - 12.7)
= 0.79 kg of ash /100 kg wood
mass of C in woodchip as fired
= % C(daf)/100*(100 – moisture – ash)
= 48.6/100 * (100 – 12.7 – 0.79)
= 42.04 kg C / 100 kg wood
Amount of energy input
= (fuel flow rate ) (fuel’s GCV)
kg  1 hr 
kJ 

=  5.8 
17756 
hr
3600
s
kg




mass of H in woodchip as fired
= % H (daf)/100*(100 – moisture – ash)
= 5.2/100 * (100 – 12.7 – 0.79)
= 4.50 kg H / 100 kg wood
= 28.60 kW
mass of O in woodchip as fired
= % O (daf)/100*(100 – moisture – ash)
= 46.0/100 * (100 – 12.7 – 0.79)
= 39.79 kg O / 100 kg wood
Mass flow rate of cooling water
3
 7.4 L

0.001 m

min.
L. 1000 kg 
=

m3 


60 s
.
min.


= 0.1233 kg/s
Amount of energy absorbed by water
= (water flow rate ) (Cp H2O ) (Tout – Tin)
 0.123 kg  4.184 kJ 
(55.9 − 12.6 )o C
=

o
s
kg
C



moles C (per 100 kg woodchips)
= 42.04 / 12 = 3.503 kmol C atoms
moles H (per 100 kg woodchips)
= 4.50 / 1 = 4.50 kmol H atoms
Theoretical amount of oxygen
= moles C + ¼ moles H – ½ moles O
= 3.503 + 4.50/4 – 2.487/2
= 3.385 kmol of O2
Total Dry Flue Gas (DFG)/100 kg fuel
kmol of C in the fuel per 100 kg fuel
=
kmol of C in DFG
kmol DFG
3.503 kmol C per 100 kg fuel
=
(0.3 + 7.3) kmol C
100 kmol DFG
= 46.09 kmol DFG per 100 kg fuel
moles of O2 in DFG per 100 kg fuel
= (0.127)(46.09)
= 5.854 kmol of O2 per 100 kg fuel
moles of CO in DFG per 100 kg fuel
= (0.003)(46.09)
= 0.138 kmol of CO per 100 kg fuel
moles of O2 required to burn CO
= 0.138 kmol / 2
= 0.069 kmol O2 per 100 kg fuel
Measured excess air in flue gas
amt. of O 2 - amt. O 2 req. to burn CO
=
theoretical amt. of O 2
(5.854 − 0.069)* 100%
=
3.385
= 170.9 %
moles O (per 100 kg woodchips)
= 39.79 / 16 = 2.487 kmol O atoms
= 22.34 kW
1.11.1.
Effects of Pollutants
Pollutants can be classified as primary or secondary. Priimary pollutants are emitted
directly from source while secondary pollutants are formed via reaction with the primary pollutants in the atmosphere. Pollutants primarily affect the environment and
human health. Seinfield (1986) indentify four principal effects of air pollutants in the
troposphere:
o Altered properties of the atmosphere and precipitation
o Harm to vegetation
o Soiling and deterioration of materials
o Potential increase of morbidity and mortality in humans
Page 26
1.11.2.
Quantification of Emissions
Emission levels are expressed in many ways, which can make comparisons difficult
and ambiguous. Some examples of these expressions are pounds per million BTUs
(lb/million BTU), gram per kilojoules (g/kJ), parts per million (volume basis) at a
specific oxygen level (ppm @ certain % of O2) and many others.
Concentrations, corrected to a particular level of O2 in the products stream are frequently used in the literature and in practice. The purpose of correcting to a specific
O2 level is to remove the effect of various degree of dilution so that a true comparison
of emission levels can be made.
To correct a concentration measured at a certain level of O2 the following equation
can be used:
 21% − %O 2[desired level] 

χ [corrected to a desired O2 level] = χ [at a given O2 level] 
 21% − %O 2[given level] 


Illustrative Example (#5):
75 ppm NOx was measured in an exhaust stream containing 1.5% O2. What is the reported NOx value when reference O2
level is .set at 6%.
Solution to Illustrative Example (#5):
To correct the NOx concentration from a level of 1.5% O2 to a level of 6% O2 the following formula is:
 21% − %O 2[desired level] 

χ [corrected to a desired O2 level] = χ [at a given O2 level] 
 21% − %O 2[given level] 


 21% − 6% 
= 75 ppm @ 1.5% O 2 

 21% − 1.5% 
= 57.7 ppm @ 6% O2
1.11.3.
Nitrogen Oxides
Nitrogen oxides are one of the many pollutants that cause the acid rain. NO and NO2
are the most common oxides of nitrogen produced during combustion. Both are usually lumped together and reported as NOx. Oxides of nitrogen are formed through
several mechanisms. Bowman (1992) classifies these into the following three categories:
o Thermal NO
o Prompt NO
o Fuel NO
Page 27
The Thermal NO mechanism involves the reaction between atmospheric nitrogen and
atmospheric oxygen to form NO. This overall reaction is described by the extended
Zeldovich mechanism. It is noted that this mechanisms is highly dependent on temperature and slightly dependent on the O2 concentration. This route of NO formation
is only significant when the gas temperature is greater than 1500K (Purvis et al,
2000).
The Prompt NO mechanism involves the reaction of hydrocarbon fragments with molecular nitrogen. In this mechanism, NO is formed more rapidly than the thermal NO.
Prompt NO is formed either by one of the following mechanisms:
o Fenimore CN and HCN pathways.
o N2O intermediate route
o As a result of superequilibrium concentration of O and OH radicals in conjunction with the extended Zeldovich mechanisms
The Fuel NO mechanism involves the formation of NO from fuel bound nitrogen. This is
the most significant source of oxides of nitrogen during the combustion of biomass. Generally, NOx from bound nitrogen evolves in
two pathway namely: NOx released during devolatilisation as HCN and NH3 species and the
NOx released from the bound nitrogen in the
char fraction during combustion. Figure 19
shows a simplified mechanism of fuel nitrogen conversion to NO. Figure 20 presents a
more detailed nitrogen oxide formation and
reduction pathway during biomass combustion.
NO
Roxid
Fuel N
HCN
O
NH
O2
RH
Rred.
NH2
NO
N2
NH3
Figure 19: Simplified chemistry for fuel nitrogen conversion in combustion processes
(Nimmo et al, 1991).
Figure 20: Simplified reaction pathway of the fuel nitrogen during biomass combustion (main path
indicated by thick arrows) [Nussbaumer, 1997].
Page 28
NOx emissions can be reduced in many ways. Figure 21 shows various strategies employed to abate NOx emissions. Most of these technologies have been applied in coal
combustion. These technologies have also been applied to biomass combustion.
Combustion
Modification
Staged
Combustion
Reburn
Technology
Low NOx
burner
Post Combustion
Control
Temperature
Reduction
Selective NonCatalytic Reduction
Selective Catalytic
Reduction
Figure 21: NOx control technologies applied to solid fuel combustion (Adapted from Turns, 1998)
Staged combustion: In this method of NOx control, operation of existing burners are
applied in such a way to create a rich-lean stage of combustion. Normally this
method is applied to suspension burning. In fixed bed combustion, staged combustion
is achieved by reducing the air passing through the grate and increasing the amount of
secondary or overfire air to achieve rich-lean burning.
Reburn Technology: This method is a type of staged combustion. In this method,
about 15% of the total fuel is introduced downstream of the main fuel lean combustion zone. Within the reburning zone, NO is reduced via reactions of HCN species
with hydrocarbons and hydrocarbon intermediates. Additional air is supplied after the
reburning zone to provide the final burnout of the reburn fuel.
Low NOx burner: This is a burner designed for low NOx emission by employing fuel
or air staging within the flame. Fuel staging creates a sequential lean-rich combustion
process while air staging creates a rich-lean process.
Temperature Reduction: In this method, the main objective is to reduce the contribution of the thermal NO mechnism. Examples of the reducing temperature technique are flue gas recirculation, water injection and reduced air pre-heating.
Selective non-catalytic reduction (SNCR): In this post-combustion control technology, a nitrogen containing additive, either ammonia, urea or cyanuric acid, is injected
and mixed with flue gases to effect chemical reduction of NO to N2 without the aid of
a catalyst. Temperature is a critical variable and operation within a relatively narrow
range of temperatures is necessary to achieve significant NO reduction.
Selective catalytic reduction (SCR): In this technique, a catalyst is used in conjunction with ammonia injection to reduce NO to N2. The temperature window for effective reduction depends upon the catalyst used but is contained within the range of
about 480K. to 780K (Turns, 1998). The advantage of SCR over SNCR is that greater
NOx reduction is possible when operating temperature is lower. However, the disadvantage is cost.
Page 29
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