It is not always easy to determine the coal consumption of a coal-fired boiler which is generally used at power plants; therefore, different methods are used which are compared to determine the most likely consumption.
One method to determine the coal consumption is to use the air flow and the oxygen content in the flue gas. If the coal composition is known, the flue gas composition can be determined and from the oxygen in the flue gas the coal flow can be calculated.
This method will be useful to:
Confirm the measured coal consumption
Calculate the carbon dioxide (CO2 ) and sulphur dioxide (SO2 ) emissions. Coal-fired boiler combustion Coal consists of C, H2 , O2 , N2 , S, H2 O and air consists of O2 , N2 and Ar.
The following combustion reactions take place in the boiler:
C
+
O
2
→
C
O
2
{\displaystyle C+O_{2}\rightarrow CO_{2}}
H
2
+
1
2
O
2
→
H
2
O
{\displaystyle H_{2}+{\frac {1}{2}}O_{2}\rightarrow H_{2}O}
N
2
+
2
O
2
→
2
N
O
2
{\displaystyle N_{2}+2O_{2}\rightarrow 2NO_{2}}
S
+
O
2
→
S
O
2
{\displaystyle S+O_{2}\rightarrow SO_{2}}
The following is a list of symbols which will be used:
c
{\displaystyle c}
h
{\displaystyle h}
o
{\displaystyle o}
n
{\displaystyle n}
s
{\displaystyle s}
2 , O2 , N2 and S in the coal according to the ultimate analysis based on air dry (ad) coal.
m
=
m
a
i
r
d
r
y
{\displaystyle m=m^{air\ dry}}
m
s
{\displaystyle m_{s}}
m
i
{\displaystyle m_{i}}
m
s
{\displaystyle m_{s}}
m
i
{\displaystyle m_{i}}
m
a
i
r
d
r
y
=
m
s
1
−
m
s
+
m
i
{\displaystyle m^{air\ dry}={m_{s} \over 1-m_{s}}+m_{i}}
(This formula is derived later in the article.)
n
C
{\displaystyle n_{C}}
n
H
2
{\displaystyle n_{H2}}
n
O
2
{\displaystyle n_{O2}}
n
N
2
{\displaystyle n_{N2}}
n
S
{\displaystyle n_{S}}
n
H
2
O
{\displaystyle n_{H2O}}
C
o
a
l
=
C
o
a
l
100
%
c
o
n
v
e
r
s
i
o
n
a
d
{\displaystyle Coal=Coal_{100\%\ conversion}^{ad}}
A
i
r
{\displaystyle Air}
w
a
{\displaystyle wa}
X
i
{\displaystyle X_{i}}
2 . Thus,
X
S
≈
0.9
{\displaystyle X_{S}\approx 0.9}
2 will convert to NO2 . Thus,
X
N
2
{\displaystyle X_{N2}}
The following air composition is assumed:
Component
Dry air
Humid air
N2
0.781
(1-wa) × 0.781
O2
0.21
(1-wa) × 0.21
Ar
0.009
(1-wa) × 0.009
H2 O
0
wa
Total 1.000 1.000
If another composition needs to be used, only change the 0.781, 0.21 and 0.009 constants throughout the document.
The following table shows the number of moles of each component before combustion:
Component
Name
MW (kg/kmol)
Moles of component before combustion
Air
Air (humid)
A
i
r
{\displaystyle Air}
Airdry
Air dry
(
1
−
w
a
)
A
i
r
{\displaystyle (1-wa)Air}
C
Carbon
12
n
C
=
c
⋅
C
o
a
l
12
{\displaystyle n_{C}={\frac {c\cdot Coal}{12}}}
H2
Hydrogen
2
n
H
2
=
h
⋅
C
o
a
l
2
{\displaystyle n_{H2}={\frac {h\cdot Coal}{2}}}
O2
Oxygen
32
n
O
2
=
o
⋅
C
o
a
l
32
+
0.21
(
1
−
w
a
)
A
i
r
{\displaystyle n_{O2}={\frac {o\cdot Coal}{32}}+0.21(1-wa)Air}
N2
Nitrogen
28
n
N
2
=
n
⋅
C
o
a
l
28
+
0.781
(
1
−
w
a
)
A
i
r
{\displaystyle n_{N2}={\frac {n\cdot Coal}{28}}+0.781(1-wa)Air}
S
Sulphur
32
n
S
=
s
⋅
C
o
a
l
32
{\displaystyle n_{S}={\frac {s\cdot Coal}{32}}}
Ar
Argon
40
n
A
r
=
0.009
(
1
−
w
a
)
A
i
r
{\displaystyle n_{Ar}=0.009(1-wa)Air}
H2 O
Water
18
n
H
2
O
=
m
⋅
C
o
a
l
18
+
w
a
⋅
A
i
r
{\displaystyle n_{H2O}={\frac {m\cdot Coal}{18}}+wa\cdot Air}
The following table shows the number of moles before and after combusion of the coal with air:
Short
Component
MW
Reaction
Moles before
Change
Moles after combustion
C
carbon
12
C
+
O
2
→
C
O
2
{\displaystyle C+O_{2}\rightarrow CO_{2}}
n
C
{\displaystyle n_{C}}
−
n
C
{\displaystyle -n_{C}}
0
{\displaystyle 0}
H2
hydrogen
2
H
2
+
1
2
O
2
→
H
2
O
{\displaystyle H_{2}+{\frac {1}{2}}O_{2}\rightarrow H_{2}O}
n
H
2
{\displaystyle n_{H2}}
−
n
H
2
{\displaystyle -n_{H2}}
0
{\displaystyle 0}
O2
oxygen
32
All combustion reactions
n
O
2
{\displaystyle n_{O2}}
−
(
n
C
+
1
2
n
H
2
+
2
X
N
2
n
N
2
+
X
S
n
S
)
{\displaystyle -\left(n_{C}+{\frac {1}{2}}n_{H2}+2X_{N2}n_{N2}+X_{S}n_{S}\right)}
n
O
2
−
(
n
C
+
1
2
n
H
2
+
2
X
N
2
n
N
2
+
X
S
n
S
)
{\displaystyle n_{O2}-\left(n_{C}+{\frac {1}{2}}n_{H2}+2X_{N2}n_{N2}+X_{S}n_{S}\right)}
N2
nitrogen
28
N
2
+
2
O
2
→
2
N
O
2
{\displaystyle N_{2}+2O_{2}\rightarrow 2NO_{2}}
n
N
2
{\displaystyle n_{N2}}
−
X
N
2
n
N
2
{\displaystyle -X_{N2}n_{N2}}
(
1
−
X
N
2
)
n
N
2
{\displaystyle (1-X_{N2})n_{N2}}
S
sulphur
32
S
+
O
2
→
S
O
2
{\displaystyle S+O_{2}\rightarrow SO_{2}}
n
S
{\displaystyle n_{S}}
−
X
S
n
S
{\displaystyle -X_{S}n_{S}}
(
1
−
X
S
)
n
S
{\displaystyle (1-X_{S})n_{S}}
CO2
carbon dioxide
44
C
+
O
2
→
C
O
2
{\displaystyle C+O_{2}\rightarrow CO_{2}}
0
{\displaystyle 0}
+
n
C
{\displaystyle +n_{C}}
n
C
{\displaystyle n_{C}}
H2 O
water
18
H
2
+
1
2
O
2
→
H
2
O
{\displaystyle H_{2}+{\frac {1}{2}}O_{2}\rightarrow H_{2}O}
n
H
2
O
{\displaystyle n_{H2O}}
n
H
2
{\displaystyle n_{H2}}
n
H
2
O
+
n
H
2
{\displaystyle n_{H2O}+n_{H2}}
NO2
nitrogen dioxide
46
N
2
+
2
O
2
→
2
N
O
2
{\displaystyle N_{2}+2O_{2}\rightarrow 2NO_{2}}
0
{\displaystyle 0}
+
2
X
N
2
n
N
2
{\displaystyle +2X_{N2}n_{N2}}
2
X
N
2
n
N
2
{\displaystyle 2X_{N2}n_{N2}}
SO2
sulphur dioxide
64
S
+
O
2
→
S
O
2
{\displaystyle S+O_{2}\rightarrow SO_{2}}
0
{\displaystyle 0}
+
X
S
n
S
{\displaystyle +X_{S}n_{S}}
X
S
n
S
{\displaystyle X_{S}n_{S}}
Ar
argon
40
n/a
n
A
r
{\displaystyle n_{Ar}}
0
{\displaystyle 0}
n
A
r
{\displaystyle n_{Ar}}
The total flue gas is the sum of all the products and all unreacted components, except sulphur (S) which will not leave with the flue gas but will leave with the ash.
F
l
u
e
g
a
s
=
n
O
2
−
(
n
C
+
1
2
n
H
2
+
2
X
N
2
n
N
2
+
X
S
n
S
)
+
(
1
−
X
N
2
)
n
N
2
+
n
c
+
n
H
2
O
+
n
H
2
+
2
X
N
2
n
N
2
+
X
s
n
s
+
n
A
r
{\displaystyle Flue\ gas=n_{O2}-(n_{C}+{\frac {1}{2}}n_{H2}+2X_{N2}n_{N2}+X_{S}n_{S})+(1-X_{N2})n_{N2}+n_{c}+n_{H2O}+n_{H2}+2X_{N2}n_{N2}+X_{s}n_{s}+n_{Ar}}
=
n
O
2
−
n
C
−
1
2
n
H
2
−
2
X
N
2
n
N
2
−
X
S
n
S
+
(
1
−
X
N
2
)
n
N
2
+
n
c
+
n
H
2
O
+
n
H
2
+
2
X
N
2
n
N
2
+
X
s
n
s
+
n
A
r
{\displaystyle =n_{O2}-n_{C}-{\frac {1}{2}}n_{H2}-2X_{N2}n_{N2}-X_{S}n_{S}+(1-X_{N2})n_{N2}+n_{c}+n_{H2O}+n_{H2}+2X_{N2}n_{N2}+X_{s}n_{s}+n_{Ar}}
=
n
O
2
+
1
2
n
H
2
+
(
1
−
X
N
2
)
n
N
2
+
n
H
2
O
+
n
A
r
{\displaystyle =n_{O2}+{\frac {1}{2}}n_{H2}+(1-X_{N2})n_{N2}+n_{H2O}+n_{Ar}}
Substitute the mol fractions ni with equations in table above:
F
l
u
e
g
a
s
=
(
o
.
C
o
a
l
32
+
0.21
(
1
−
w
a
)
A
i
r
)
+
1
2
(
h
.
C
o
a
l
2
)
+
(
1
−
X
N
2
)
(
n
.
C
o
a
l
28
+
0.781
(
1
−
w
a
)
A
i
r
)
+
(
m
.
C
o
a
l
18
+
w
a
.
A
i
r
)
+
(
0.009
(
1
−
w
a
)
A
i
r
)
{\displaystyle Flue\ gas=\left({\frac {o.Coal}{32}}+0.21(1-wa)Air\right)+{\frac {1}{2}}\left({\frac {h.Coal}{2}}\right)+(1-X_{N2})\left({\frac {n.Coal}{28}}+0.781(1-wa)Air\right)+\left({\frac {m.Coal}{18}}+wa.Air\right)+(0.009(1-wa)Air)}
=
C
o
a
l
o
32
+
A
i
r
0.21
(
1
−
w
a
)
+
0.5
C
o
a
l
h
2
+
C
o
a
l
n
28
(
1
−
X
N
2
)
+
A
i
r
0.781
(
1
−
w
a
)
(
1
−
X
N
2
)
+
C
o
a
l
m
18
+
A
i
r
.
w
a
+
A
i
r
0.009
(
1
−
w
a
)
{\displaystyle =Coal{\frac {o}{32}}+Air0.21(1-wa)+0.5Coal{\frac {h}{2}}+Coal{\frac {n}{28}}(1-X_{N2})+Air0.781(1-wa)(1-X_{N2})+Coal{\frac {m}{18}}+Air.wa+Air0.009(1-wa)}
=
C
o
a
l
(
o
32
+
0.5
h
2
+
n
28
(
1
−
X
N
2
)
+
m
18
)
+
A
i
r
[
w
a
+
0.781
(
1
−
w
a
)
(
1
−
X
N
2
)
+
0.009
(
1
−
w
a
)
+
0.21
(
1
−
w
a
)
]
{\displaystyle =Coal\left({\frac {o}{32}}+0.5{\frac {h}{2}}+{\frac {n}{28}}(1-X_{N2})+{\frac {m}{18}}\right)+Air\left[wa+0.781(1-wa)(1-X_{N2})+0.009(1-wa)+0.21(1-wa)\right]}
=
C
o
a
l
(
o
32
+
0.5
h
2
+
n
28
(
1
−
X
N
2
)
+
m
18
)
+
A
i
r
[
w
a
+
0.781
(
1
−
w
a
)
−
0.781
X
N
2
(
1
−
w
a
)
+
0.009
(
1
−
w
a
)
+
0.21
(
1
−
w
a
)
]
{\displaystyle =Coal\left({\frac {o}{32}}+0.5{\frac {h}{2}}+{\frac {n}{28}}(1-X_{N2})+{\frac {m}{18}}\right)+Air\left[wa+0.781(1-wa)-0.781X_{N2}(1-wa)+0.009(1-wa)+0.21(1-wa)\right]}
=
C
o
a
l
(
o
32
+
0.5
h
2
+
n
28
(
1
−
X
N
2
)
+
m
18
)
+
A
i
r
[
w
a
+
(
1
−
w
a
)
−
0.781
X
N
2
(
1
−
w
a
)
]
{\displaystyle =Coal\left({\frac {o}{32}}+0.5{\frac {h}{2}}+{\frac {n}{28}}(1-X_{N2})+{\frac {m}{18}}\right)+Air\left[wa+(1-wa)-0.781X_{N2}(1-wa)\right]}
F
l
u
e
g
a
s
=
C
o
a
l
(
o
32
+
0.5
h
2
+
n
28
(
1
−
X
N
2
)
+
m
18
)
+
A
i
r
[
1
−
0.781
X
N
2
(
1
−
w
a
)
]
{\displaystyle Flue\ gas=Coal\left({\frac {o}{32}}+0.5{\frac {h}{2}}+{\frac {n}{28}}(1-X_{N2})+{\frac {m}{18}}\right)+Air\left[1-0.781X_{N2}(1-wa)\right]}
Where
F
l
u
e
g
a
s
{\displaystyle Flue\ gas}
A
i
r
{\displaystyle Air}
C
o
a
l
{\displaystyle Coal}
c
,
h
,
o
,
n
,
s
{\displaystyle c,h,o,n,s}
w
a
{\displaystyle wa}
X
N
2
{\displaystyle X_{N2}}
The oxygen in the flue gas can be written as follows:
O
x
y
g
e
n
i
n
f
l
u
e
g
a
s
=
n
O
2
−
(
n
C
+
1
2
n
H
2
+
2
X
N
2
n
N
2
+
X
S
n
S
)
{\displaystyle Oxygen\ in\ flue\ gas=n_{O2}-\left(n_{C}+{\frac {1}{2}}n_{H2}+2X_{N2}n_{N2}+X_{S}n_{S}\right)}
=
n
O
2
−
n
C
−
1
2
n
H
2
−
2
X
N
2
n
N
2
−
X
S
n
S
{\displaystyle =n_{O2}-n_{C}-{\frac {1}{2}}n_{H2}-2X_{N2}n_{N2}-X_{S}n_{S}}
Substitute the mol fractions ni with equations in table above:
O
x
y
g
e
n
i
n
f
l
u
e
g
a
s
=
(
o
.
C
o
a
l
32
+
0.21
(
1
−
w
a
)
A
i
r
)
−
(
c
.
C
o
a
l
12
)
−
1
2
(
h
.
C
o
a
l
2
)
−
2
X
N
2
(
n
.
C
o
a
l
28
+
0.781
(
1
−
w
a
)
A
i
r
)
−
X
S
(
s
.
C
o
a
l
32
)
{\displaystyle Oxygen\ in\ flue\ gas=\left({\frac {o.Coal}{32}}+0.21(1-wa)Air\right)-\left({\frac {c.Coal}{12}}\right)-{\frac {1}{2}}\left({\frac {h.Coal}{2}}\right)-2X_{N2}\left({\frac {n.Coal}{28}}+0.781(1-wa)Air\right)-X_{S}\left({\frac {s.Coal}{32}}\right)}
=
C
o
a
l
o
32
+
A
i
r
0.21
(
1
−
w
a
)
−
C
o
a
l
c
12
−
0.5
C
o
a
l
h
2
−
C
o
a
l
.2
X
N
2
n
28
−
A
i
r
2
X
N
2
0.781
(
1
−
w
a
)
−
C
o
a
l
.
X
S
s
32
{\displaystyle =Coal{\frac {o}{32}}+Air0.21(1-wa)-Coal{\frac {c}{12}}-0.5Coal{\frac {h}{2}}-Coal.2X_{N2}{\frac {n}{28}}-Air2X_{N2}0.781(1-wa)-Coal.X_{S}{\frac {s}{32}}}
O
x
y
g
e
n
i
n
f
l
u
e
g
a
s
=
C
o
a
l
[
o
32
−
c
12
−
0.5
h
2
−
2
X
N
2
n
28
−
X
S
s
32
]
+
A
i
r
[
0.21
(
1
−
w
a
)
−
2
X
N
2
0.781
(
1
−
w
a
)
]
{\displaystyle Oxygen\ in\ flue\ gas=Coal\left[{\frac {o}{32}}-{\frac {c}{12}}-0.5{\frac {h}{2}}-2X_{N2}{\frac {n}{28}}-X_{S}{\frac {s}{32}}\right]+Air\left[0.21(1-wa)-2X_{N2}0.781(1-wa)\right]}
Where
O
x
y
g
e
n
{\displaystyle Oxygen}
A
i
r
{\displaystyle Air}
C
o
a
l
{\displaystyle Coal}
c
,
h
,
o
,
n
,
s
{\displaystyle c,h,o,n,s}
w
a
{\displaystyle wa}
X
s
,
X
N
2
{\displaystyle X_{s},X_{N2}}
The percentage oxygen in flue gas is given by the following formula:
x
O
2
=
%
o
x
y
g
e
n
i
n
f
l
u
e
g
a
s
=
O
x
y
g
e
n
i
n
f
l
u
e
g
a
s
T
o
t
a
l
f
l
u
e
g
a
s
{\displaystyle x_{O2}=\%\ oxygen\ in\ flue\ gas={\frac {Oxygen\ in\ flue\ gas}{Total\ flue\ gas}}}
C
o
a
l
[
o
32
−
c
12
−
0.5
h
2
−
2
X
N
2
n
28
−
X
S
s
32
]
+
A
i
r
[
0.21
(
1
−
w
a
)
−
2
X
N
2
0.781
(
1
−
w
a
)
]
{\displaystyle Coal\left[{\frac {o}{32}}-{\frac {c}{12}}-0.5{\frac {h}{2}}-2X_{N2}{\frac {n}{28}}-X_{S}{\frac {s}{32}}\right]+Air\left[0.21(1-wa)-2X_{N2}0.781(1-wa)\right]}
=
C
o
a
l
[
o
32
x
O
2
+
0.5
h
2
x
O
2
+
n
28
(
1
−
X
N
2
)
x
O
2
+
m
18
x
O
2
]
+
A
i
r
[
x
O
2
−
0.781
X
N
2
(
1
−
w
a
)
x
O
2
]
{\displaystyle =Coal\left[{\frac {o}{32}}x_{O2}+0.5{\frac {h}{2}}x_{O2}+{\frac {n}{28}}(1-X_{N2})x_{O2}+{\frac {m}{18}}x_{O2}\right]+Air[x_{O2}-0.781X_{N2}(1-wa)x_{O2}]}
C
o
a
l
[
o
32
−
c
12
−
0.5
h
2
−
2
X
N
2
n
28
−
X
S
s
32
]
−
C
o
a
l
[
o
32
x
O
2
+
0.5
h
2
x
O
2
+
n
28
(
1
−
X
N
2
)
x
O
2
+
m
18
x
O
2
]
{\displaystyle Coal\left[{\frac {o}{32}}-{\frac {c}{12}}-0.5{\frac {h}{2}}-2X_{N2}{\frac {n}{28}}-X_{S}{\frac {s}{32}}\right]-Coal\left[{\frac {o}{32}}x_{O2}+0.5{\frac {h}{2}}x_{O2}+{\frac {n}{28}}(1-X_{N2})x_{O2}+{\frac {m}{18}}x_{O2}\right]}
=
A
i
r
[
x
O
2
−
0.781
X
N
2
(
1
−
w
a
)
x
O
2
]
−
A
i
r
[
0.21
(
1
−
w
a
)
−
2
X
N
2
0.781
(
1
−
w
a
)
]
{\displaystyle =Air[x_{O2}-0.781X_{N2}(1-wa)x_{O2}]-Air\left[0.21(1-wa)-2X_{N2}0.781(1-wa)\right]}
C
o
a
l
[
o
32
(
1
−
x
O
2
)
−
c
12
−
0.5
h
2
(
1
+
x
O
2
)
−
n
28
(
2
X
N
2
+
(
1
−
X
N
2
)
x
O
2
)
−
X
S
s
32
−
m
18
x
O
2
]
{\displaystyle Coal\left[{\frac {o}{32}}(1-x_{O2})-{\frac {c}{12}}-0.5{\frac {h}{2}}(1+x_{O2})-{\frac {n}{28}}(2X_{N2}+(1-X_{N2})x_{O2})-X_{S}{\frac {s}{32}}-{\frac {m}{18}}x_{O2}\right]}
=
A
i
r
[
x
O
2
−
0.781
X
N
2
(
1
−
w
a
)
x
O
2
−
0.21
(
1
−
w
a
)
+
2
X
N
2
0.781
(
1
−
w
a
)
]
{\displaystyle =Air[x_{O2}-0.781X_{N2}(1-wa)x_{O2}-0.21(1-wa)+2X_{N2}0.781(1-wa)]}
C
o
a
l
[
o
32
(
1
−
x
O
2
)
−
c
12
−
0.5
h
2
(
1
+
x
O
2
)
−
n
28
(
2
X
N
2
+
(
1
−
X
N
2
)
x
O
2
)
−
X
S
s
32
−
m
18
x
O
2
]
{\displaystyle Coal\left[{\frac {o}{32}}(1-x_{O2})-{\frac {c}{12}}-0.5{\frac {h}{2}}(1+x_{O2})-{\frac {n}{28}}(2X_{N2}+(1-X_{N2})x_{O2})-X_{S}{\frac {s}{32}}-{\frac {m}{18}}x_{O2}\right]}
=
A
i
r
[
x
O
2
−
0.781
X
N
2
(
1
−
w
a
)
x
O
2
−
0.21
(
1
−
w
a
)
+
2
X
N
2
0.781
(
1
−
w
a
)
]
{\displaystyle =Air[x_{O2}-0.781X_{N2}(1-wa)x_{O2}-0.21(1-wa)+2X_{N2}0.781(1-wa)]}
R
H
=
A
i
r
[
−
0.781
X
N
2
(
1
−
w
a
)
x
O
2
−
0.21
(
1
−
w
a
)
+
2
X
N
2
0.781
(
1
−
w
a
)
+
x
O
2
]
{\displaystyle RH=Air[-0.781X_{N2}(1-wa)x_{O2}-0.21(1-wa)+2X_{N2}0.781(1-wa)+x_{O2}]}
R
H
=
A
i
r
[
(
1
−
w
a
)
[
−
0.781
X
N
2
x
O
2
+
2
X
N
2
0.781
−
0.21
]
+
x
O
2
]
{\displaystyle RH=Air[(1-wa)[-0.781X_{N2}x_{O2}+2X_{N2}0.781-0.21]+x_{O2}]}
R
H
=
A
i
r
[
(
1
−
w
a
)
[
0.781
X
N
2
(
2
−
x
O
2
)
−
0.21
]
+
x
O
2
]
{\displaystyle RH=Air[(1-wa)[0.781X_{N2}(2-x_{O2})-0.21]+x_{O2}]}
C
o
a
l
[
o
32
(
1
−
x
O
2
)
−
c
12
−
0.5
h
2
(
1
+
x
O
2
)
−
n
28
(
2
X
N
2
+
(
1
−
X
N
2
)
x
O
2
)
−
X
S
s
32
−
m
18
x
O
2
]
{\displaystyle Coal\left[{\frac {o}{32}}(1-x_{O2})-{\frac {c}{12}}-0.5{\frac {h}{2}}(1+x_{O2})-{\frac {n}{28}}(2X_{N2}+(1-X_{N2})x_{O2})-X_{S}{\frac {s}{32}}-{\frac {m}{18}}x_{O2}\right]}
=
A
i
r
[
(
1
−
w
a
)
[
0.781
X
N
2
(
2
−
x
O
2
)
−
0.21
]
+
x
O
2
]
{\displaystyle =Air\left[(1-wa)[0.781X_{N2}(2-x_{O2})-0.21]+x_{O2}\right]}
C
o
a
l
=
A
i
r
×
(
1
−
w
a
)
[
0.781
X
N
2
(
2
−
x
O
2
)
−
0.21
]
+
x
O
2
o
32
(
1
−
x
O
2
)
−
c
12
−
0.5
h
2
(
1
+
x
O
2
)
−
n
28
[
2
X
N
2
+
(
1
−
X
N
2
)
x
O
2
]
−
X
S
s
32
−
m
18
x
O
2
{\displaystyle Coal=Air\times {(1-wa)[0.781X_{N2}(2-x_{O2})-0.21]+x_{O2} \over {\frac {o}{32}}(1-x_{O2})-{\frac {c}{12}}-0.5{\frac {h}{2}}(1+x_{O2})-{\frac {n}{28}}[2X_{N2}+(1-X_{N2})x_{O2}]-X_{S}{\frac {s}{32}}-{\frac {m}{18}}x_{O2}}}
Where:
C
o
a
l
{\displaystyle Coal}
2 and SO2 emissions are calculated, this value can use used as is. However, if one wants to compare this calculated value with the actual coal received, on should take into consideration the uncombusted coal in the ash.
A
i
r
{\displaystyle Air}
A
i
r
=
a
i
r
m
e
a
s
u
r
e
d
+
a
i
r
l
e
a
k
a
g
e
{\displaystyle Air=air\ measured+air\ leakage}
The individual components in the flue gas is then the following (kmol/h):
O
2
i
n
f
l
u
e
g
a
s
=
n
O
2
−
(
n
C
+
1
2
n
H
2
+
2
X
N
2
n
N
2
+
X
S
n
S
)
{\displaystyle O_{2}\ in\ flue\ gas\quad =n_{O2}-\left(n_{C}+{\frac {1}{2}}n_{H2}+2X_{N2}n_{N2}+X_{S}n_{S}\right)}
=
C
o
a
l
[
o
32
−
c
12
−
0.5
h
2
−
2
X
N
2
n
28
−
X
S
s
32
]
+
A
i
r
[
0.21
(
1
−
w
a
)
−
2
X
N
2
0.781
(
1
−
w
a
)
]
{\displaystyle =Coal\left[{\frac {o}{32}}-{\frac {c}{12}}-0.5{\frac {h}{2}}-2X_{N2}{\frac {n}{28}}-X_{S}{\frac {s}{32}}\right]+Air\left[0.21(1-wa)-2X_{N2}0.781(1-wa)\right]}
N
2
i
n
f
l
u
e
g
a
s
=
(
1
−
X
N
2
)
n
N
2
{\displaystyle N_{2}\ in\ flue\ gas\quad =\quad (1-X_{N2})n_{N2}}
=
(
1
−
X
N
2
)
[
n
⋅
C
o
a
l
28
+
0.781
(
1
−
w
a
)
A
i
r
]
{\displaystyle =(1-X_{N2})\left[{\frac {n\cdot Coal}{28}}+0.781(1-wa)Air\right]}
C
O
2
i
n
f
l
u
e
g
a
s
=
n
C
{\displaystyle CO_{2}\ in\ flue\ gas\quad =\quad n_{C}}
=
C
o
a
l
c
12
{\displaystyle =Coal{\frac {c}{12}}}
H
2
O
i
n
f
l
u
e
g
a
s
=
n
H
2
O
+
n
H
2
=
(
C
o
a
l
m
18
+
w
a
.
A
i
r
)
+
(
C
o
a
l
h
2
)
{\displaystyle H_{2}O\ in\ flue\ gas\quad =\quad n_{H2O}+n_{H2}\quad =\quad \left(Coal{\frac {m}{18}}+wa.Air\right)+\left(Coal{\frac {h}{2}}\right)}
=
C
o
a
l
(
m
18
+
h
2
)
+
w
a
.
A
i
r
{\displaystyle =Coal\left({\frac {m}{18}}+{\frac {h}{2}}\right)+wa.Air}
N
O
2
i
n
f
l
u
e
g
a
s
=
2
X
N
2
n
N
2
{\displaystyle NO_{2}\ in\ flue\ gas\quad =\quad 2X_{N2}n_{N2}}
=
2
X
N
2
(
C
o
a
l
n
28
+
0.781
(
1
−
w
a
)
A
i
r
)
{\displaystyle =2X_{N2}\left(Coal{\frac {n}{28}}+0.781(1-wa)Air\right)}
S
O
2
i
n
f
l
u
e
g
a
s
=
X
s
n
S
=
X
S
(
C
o
a
l
s
32
)
{\displaystyle SO_{2}\ in\ flue\ gas\quad =\quad X_{s}n_{S}\quad =\quad X_{S}\left(Coal{\frac {s}{32}}\right)}
=
C
o
a
l
.
X
S
s
32
{\displaystyle =Coal.X_{S}{\frac {s}{32}}}
A
r
i
n
f
l
u
e
g
a
s
=
n
A
r
{\displaystyle Ar\ in\ flue\ gas\quad =\quad n_{Ar}}
=
0.009
(
1
−
w
a
)
A
i
r
{\displaystyle =0.009(1-wa)Air}
The CO2 , SO2 and NO2 emissions can be calculated as follows:
C
O
2
e
m
i
s
s
i
o
n
s
=
c
×
C
o
a
l
×
44
12
i
n
k
g
/
h
{\displaystyle CO_{2}\ emissions=c\times Coal\times {\frac {44}{12}}\ in\ kg/h}
S
O
2
e
m
i
s
s
i
o
n
s
=
X
s
×
c
×
C
o
a
l
×
64
32
i
n
k
g
/
h
{\displaystyle SO_{2}\ emissions=X_{s}\times c\times Coal\times {\frac {64}{32}}\ in\ kg/h}
N
O
2
e
m
i
s
s
i
o
n
s
=
2
X
s
×
(
n
⋅
C
o
a
l
28
+
0.781
(
1
−
w
a
)
A
i
r
)
×
46
i
n
k
g
/
h
{\displaystyle NO_{2}\ emissions=2X_{s}\times \left({\frac {n\cdot Coal}{28}}+0.781(1-wa)Air\right)\times 46\ in\ kg/h}
If:
T
{\displaystyle T}
R
H
{\displaystyle RH}
P
{\displaystyle P}
p
s
a
t
{\displaystyle p_{sat}}
x
w
a
t
e
r
{\displaystyle x_{water}}
The maximum amount of water vapour in air is:
x
w
a
t
e
r
,
m
a
x
=
p
s
a
t
P
{\displaystyle x_{water,max}={\frac {p_{sat}}{P}}}
The actual mol fraction of water in air is:
x
w
a
t
e
r
=
R
H
×
x
w
a
t
e
r
,
m
a
x
{\displaystyle x_{water}=RH\times x_{water,max}}
The following Buck equation is a very good approximation of the vapour pressure of water below 100°C:
p
s
a
t
=
0.61121
×
e
17.368
T
T
+
238.88
{\displaystyle p_{sat}=0.61121\times e^{\frac {17.368T}{T+238.88}}}
Thus, the mol fraction of water in air is:
x
w
a
t
e
r
=
R
H
P
×
0.61121
×
e
17.368
T
T
+
238.88
{\displaystyle x_{water}={\frac {RH}{P}}\times 0.61121\times e^{\frac {17.368T}{T+238.88}}}
Total coal moisture in terms of superficial and inherent moisture [ edit | edit source ] The superficial moisture fraction (ms ) is based on as received (AR) coal and the inherent moisture fraction (mi ) is based on air dry (AD) coal. However, the total moisture (m), can be expressed in terms of ms and mi on an air dry basis as follows.
m
a
d
=
M
C
o
a
l
a
d
=
M
s
+
M
i
C
o
a
l
a
d
=
C
o
a
l
a
r
m
s
+
C
o
a
l
a
d
m
i
C
o
a
l
a
d
{\displaystyle m^{ad}={\frac {M}{Coal^{ad}}}={M_{s}+M_{i} \over Coal^{ad}}={Coal^{ar}m_{s}+Coal^{ad}m_{i} \over Coal^{ad}}}
If:
C
o
a
l
a
d
=
(
1
−
m
s
)
C
o
a
l
a
r
⇒
C
o
a
l
a
r
=
C
o
a
l
a
d
1
−
m
s
{\displaystyle Coal^{ad}=(1-m_{s})Coal^{ar}\qquad \Rightarrow \qquad Coal^{ar}={Coal^{ad} \over 1-m_{s}}}
Then:
m
a
d
=
C
o
a
l
a
d
1
−
m
s
m
s
+
C
o
a
l
a
d
m
i
C
o
a
l
a
d
{\displaystyle m^{ad}={{Coal^{ad} \over 1-m_{s}}m_{s}+Coal^{ad}m_{i} \over Coal^{ad}}}
m
a
d
=
m
s
1
−
m
s
+
m
i
{\displaystyle m^{ad}={m_{s} \over 1-m_{s}}+m_{i}}
Mass flow rate of coal consumption and theoretical air mass
Application of mass and energy balances to determine coal, air required and flue gas flow rates in a power plant], Katende, Landry Mbangu, 2019
Calculation of combustion air required for burning solid fuels (coal / biomass / solid waste) and analysis of flue gas composition], Lizica Simona Paraschiv, Alexandru Serban, Spiru Paraschiv, 6 September 2019 
![{\displaystyle =Coal\left({\frac {o}{32}}+0.5{\frac {h}{2}}+{\frac {n}{28}}(1-X_{N2})+{\frac {m}{18}}\right)+Air\left[wa+0.781(1-wa)(1-X_{N2})+0.009(1-wa)+0.21(1-wa)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e8d012ad84da516da12bda5a888435de1224b9b)
![{\displaystyle =Coal\left({\frac {o}{32}}+0.5{\frac {h}{2}}+{\frac {n}{28}}(1-X_{N2})+{\frac {m}{18}}\right)+Air\left[wa+0.781(1-wa)-0.781X_{N2}(1-wa)+0.009(1-wa)+0.21(1-wa)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/45b5c16a752cfd867701d58de3fcbc370ba5c879)
![{\displaystyle =Coal\left({\frac {o}{32}}+0.5{\frac {h}{2}}+{\frac {n}{28}}(1-X_{N2})+{\frac {m}{18}}\right)+Air\left[wa+(1-wa)-0.781X_{N2}(1-wa)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ba89aea353b8e0d250014fe59c9b0002e5c9834)
![{\displaystyle Flue\ gas=Coal\left({\frac {o}{32}}+0.5{\frac {h}{2}}+{\frac {n}{28}}(1-X_{N2})+{\frac {m}{18}}\right)+Air\left[1-0.781X_{N2}(1-wa)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/407a7cfd1783f6cb69dfa28cea7bfc048d582e1e)
![{\displaystyle Oxygen\ in\ flue\ gas=Coal\left[{\frac {o}{32}}-{\frac {c}{12}}-0.5{\frac {h}{2}}-2X_{N2}{\frac {n}{28}}-X_{S}{\frac {s}{32}}\right]+Air\left[0.21(1-wa)-2X_{N2}0.781(1-wa)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e6d2924d8cc1bfc140983a3c9dfb29917e916d4)
![{\displaystyle Coal\left[{\frac {o}{32}}-{\frac {c}{12}}-0.5{\frac {h}{2}}-2X_{N2}{\frac {n}{28}}-X_{S}{\frac {s}{32}}\right]+Air\left[0.21(1-wa)-2X_{N2}0.781(1-wa)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3a04b1478cee61869ad39318bd940e9b446a7a6)
![{\displaystyle =Coal\left[{\frac {o}{32}}x_{O2}+0.5{\frac {h}{2}}x_{O2}+{\frac {n}{28}}(1-X_{N2})x_{O2}+{\frac {m}{18}}x_{O2}\right]+Air[x_{O2}-0.781X_{N2}(1-wa)x_{O2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79eb97272d330a19a86051d62dc4925d93c2a69b)
![{\displaystyle Coal\left[{\frac {o}{32}}-{\frac {c}{12}}-0.5{\frac {h}{2}}-2X_{N2}{\frac {n}{28}}-X_{S}{\frac {s}{32}}\right]-Coal\left[{\frac {o}{32}}x_{O2}+0.5{\frac {h}{2}}x_{O2}+{\frac {n}{28}}(1-X_{N2})x_{O2}+{\frac {m}{18}}x_{O2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/44879878500a128a7a94b256bca8ac7676bb03ef)
![{\displaystyle =Air[x_{O2}-0.781X_{N2}(1-wa)x_{O2}]-Air\left[0.21(1-wa)-2X_{N2}0.781(1-wa)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/835c5fddf84bb9848bbf362304f56b0e077a6201)
![{\displaystyle Coal\left[{\frac {o}{32}}(1-x_{O2})-{\frac {c}{12}}-0.5{\frac {h}{2}}(1+x_{O2})-{\frac {n}{28}}(2X_{N2}+(1-X_{N2})x_{O2})-X_{S}{\frac {s}{32}}-{\frac {m}{18}}x_{O2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed1de40d038de02a6e75dc4084fbe8bca70e602c)
![{\displaystyle =Air[x_{O2}-0.781X_{N2}(1-wa)x_{O2}-0.21(1-wa)+2X_{N2}0.781(1-wa)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d94efbe68c5cf559fedf3838eb180622127267ed)
![{\displaystyle RH=Air[-0.781X_{N2}(1-wa)x_{O2}-0.21(1-wa)+2X_{N2}0.781(1-wa)+x_{O2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5fddaea1f716b6d87c2802840bc43908180da49c)
![{\displaystyle RH=Air[(1-wa)[-0.781X_{N2}x_{O2}+2X_{N2}0.781-0.21]+x_{O2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49507fbd9d7a2d4e04c6652072094c0883df9947)
![{\displaystyle RH=Air[(1-wa)[0.781X_{N2}(2-x_{O2})-0.21]+x_{O2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9642fafa645efb19976107a50885bbb2a0968b07)
![{\displaystyle =Air\left[(1-wa)[0.781X_{N2}(2-x_{O2})-0.21]+x_{O2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/36ebc36d4374e6d51ed6d47b3249ae31ebafa7da)
![{\displaystyle Coal=Air\times {(1-wa)[0.781X_{N2}(2-x_{O2})-0.21]+x_{O2} \over {\frac {o}{32}}(1-x_{O2})-{\frac {c}{12}}-0.5{\frac {h}{2}}(1+x_{O2})-{\frac {n}{28}}[2X_{N2}+(1-X_{N2})x_{O2}]-X_{S}{\frac {s}{32}}-{\frac {m}{18}}x_{O2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0180e01d964383798dc5e647288953e2a5b82bf)
![{\displaystyle =Coal\left[{\frac {o}{32}}-{\frac {c}{12}}-0.5{\frac {h}{2}}-2X_{N2}{\frac {n}{28}}-X_{S}{\frac {s}{32}}\right]+Air\left[0.21(1-wa)-2X_{N2}0.781(1-wa)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7cd3065ba0274f764bf4fec1f7f1297fb452713)
![{\displaystyle =(1-X_{N2})\left[{\frac {n\cdot Coal}{28}}+0.781(1-wa)Air\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0bbb8217a361d0a21ca3f9c4ec2170f53d1ed5c3)
