Dit is nie altyd maklik om die steenkoolverbruik van 'n steenkoolaangedrewe ketel te bepaal nie; daarom word verskillende metodes gebruik wat met mekaar vergelyk word om die mees waarskynlike verbruik te bepaal.
Een metode om die steenkoolverbruik te bepaal, is om die lugvloei en die suurstofinhoud in die rookgas te gebruik. Indien die steenkoolsamestelling bekend is, kan die rookgassamestelling bepaal word en uit die suurstof in die rookgas kan die steenkoolvloei bereken word.
Steenkool verbranding
wysig
Die volgende is 'n lys van simbole wat gebruik word:
c, h, o, n, s = massa fraksie van C, H2 , O2 , N2 en S in die steenkool volgens die uitgebreide steenkoolanalise gebaseer op luggedroogde steenkool.
m = totale vog in steenkool = oppervlakvog (ms ) + inherente vog (mi ) gebaseer op luggedroogde steenkool. Die totale vog (m) kan ook soos volg geskryf word in terme van ms and mi :
m
l
u
g
d
r
o
o
g
=
m
s
1
−
m
s
+
m
i
{\displaystyle m^{lug\ droog}={m_{s} \over 1-m_{s}}+m_{i}}
nC , nH2 , nO2 , nN2 , nS , nH2O = aantal mol van die komponente in die steenkool wat opeindig in die ketel (kmol/h). Neem kennis dat beide oppervlakvog en inherente vog in die ketel opeindig.
Steenkool = Massa vloei van luggedroogde steenkool wat volledig verbrand (kg/h)
Lug = Lugvloei (kmol/h)
wa = mol fraksie van waterdamp in die lug (weens lughumiditeit)
Xi = omsetting van komponent i tydens ontbranding. Byvoorbeeld, ongeveer 90% van S word verbrand na SO2 . Dus, XS ≈ 0.9. Baie min N2 sal omskakel na NO2 . Dus sal XN2 baie klein wees. Die volgende tabel wys die aantal mol van elke komponent voor verbranding:
Komponent
Naam
MW (kg/kmol)
Mol voor verbranding
Lug
Lug
Lug
Lugdroog
Lug droog
(1-wa)×Lug
C
Koolstof
12
nC = c×Steenkool/12
H2
Waterstof
2
nH2 = h×Steenkool/2
O2
Suurstof
32
nO2 = o×Steenkool/32 + 0.21×(1-wa)×Lug
N2
Stikstof
28
nN2 = n×Steenkool/28 + 0.781×(1-wa)×Lug
S
Swael
32
nS = s×Steenkool/32
Ar
Argon
40
nAr = 0.009×(1-wa)×Lug
H2 O
Water
18
nH2O = m×Steenkool/18 + wa×Lug
Komponentbalans voor en na verbranding
wysig
Die volgende tabel wys die aantal mol van elke komponent voor en na verbranding van die steenkool met lug:
Kort
Komponent
Reaksie
Mol voor
Verandering
Mol na verbranding
C
Koolstof
C + O2 → CO2
nC
-nC
0
H2
Waterstof
H2 + ½O2 → H2 O
nH2
-nH2
0
O2
Suurstof
All combustion reactions
nO2
-(nC + ½nH2 + 2XN2 nN2 + XS nS )
nC + ½nH2 + 2XN2 nN2 + XS nS )
N2
Stifstof
N2 + 2O2 → 2NO2
nN2
-XN2 nN2
(1-XN2 )nN2
S
Swael
S + O2 → SO2
nS
-XS nS
(1-XS )nS
Ar
Argon
nAr
0
nAr
CO2
Koolstofdioksied
C + O2 → CO2
0
+nC
nC
H2 O
Water
H2 + ½O2 → H2 O
nH2O
0
nH2O + nH2
NO2
Stikstofdioksied
N2 + 2O2 → 2NO2
0
+2XN2 nN2
2XN2 nN2
SO2
Swaeldioksied
S + O2 → SO2
0
+XS nS
XS nS
Die suurstofinhoud in die rookgas kan dus soos volg uitgedruk word:
x
O
2
=
m
o
l
s
u
u
r
s
t
o
f
i
n
r
o
o
k
g
a
s
t
o
t
a
l
e
m
o
l
r
o
o
k
g
a
s
{\displaystyle x_{O2}={\frac {mol\ suurstof\ in\ rookgas}{totale\ mol\ rookgas}}}
x
O
2
=
n
O
2
−
(
n
C
+
1
2
n
H
2
+
2
X
N
2
n
N
2
+
X
S
n
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 x_{O2}={n_{O2}-\left(n_{C}+{\frac {1}{2}}n_{H2}+2X_{N2}n_{N2}+X_{S}n_{S}\right) \over n_{O2}-\left(n_{C}+{\frac {1}{2}}n_{H2}+2X_{N2}n_{N2}+X_{S}n_{S}\right)+(1-X_{N2})n_{N2}+n_{c}+n_{H2O}+n_{H2}+2X_{N2}n_{N2}+X_{S}n_{S}+n_{Ar}}}
x
O
2
=
n
O
2
−
n
C
−
1
2
n
H
2
−
2
X
N
2
n
N
2
−
X
S
n
S
n
O
2
+
1
2
n
H
2
+
(
1
−
X
N
2
)
n
N
2
+
n
H
2
O
+
n
A
r
{\displaystyle x_{O2}={n_{O2}-n_{C}-{\frac {1}{2}}n_{H2}-2X_{N2}n_{N2}-X_{S}n_{S} \over n_{O}2+{\frac {1}{2}}n_{H2}+(1-X_{N2})n_{N2}+n_{H2O}+n_{Ar}}}
x
O
2
n
O
2
+
1
2
x
O
2
n
H
2
+
(
1
−
X
N
2
)
n
N
2
x
O
2
+
x
O
2
n
H
2
O
+
x
O
2
n
A
r
=
n
O
2
−
n
C
1
2
n
H
2
−
2
X
N
2
n
N
2
−
X
S
n
S
{\displaystyle x_{O2}n_{O2}+{\frac {1}{2}}x_{O2}n_{H2}+(1-X_{N2})n_{N2}x_{O2}+x_{O2}n_{H2O}+x_{O2}n_{A}r=n_{O}2-n_{C}{\frac {1}{2}}n_{H}2-2X_{N2}n_{N2}-X_{S}n_{S}}
Vervang die molfraksies i met die vergelykings in die tabel hierbo:
Laat eers die humiditeitsterm,
(
1
−
w
a
)
{\displaystyle (1-wa)}
(
o
S
t
e
e
n
k
o
o
l
32
+
0.21
L
u
g
)
x
O
2
+
1
2
(
h
S
t
e
e
n
k
o
o
l
2
)
x
O
2
+
(
1
−
X
N
2
)
(
n
S
t
e
e
n
k
o
o
l
28
+
0.781
L
u
g
)
x
O
2
+
(
m
S
t
e
e
n
k
o
o
l
18
+
w
a
.
L
u
g
)
x
O
2
+
(
0.009
L
u
g
)
x
O
2
{\displaystyle \left(o{\frac {Steenkool}{32}}+0.21Lug\right)x_{O2}+{\frac {1}{2}}\left(h{\frac {Steenkool}{2}}\right)x_{O2}+(1-X_{N2})\left(n{\frac {Steenkool}{28}}+0.781Lug\right)x_{O2}+\left(m{\frac {Steenkool}{18}}+wa.Lug\right)x_{O2}+(0.009Lug)x_{O2}}
=
(
o
S
t
e
e
n
k
o
o
l
32
+
0.21
L
u
g
)
−
(
c
S
t
e
e
n
k
o
o
l
12
)
−
1
2
(
h
S
t
e
e
n
k
o
o
l
2
)
−
2
X
N
2
(
n
S
t
e
e
n
k
o
o
l
28
+
0.781
L
u
g
)
−
X
S
(
s
S
t
e
e
n
k
o
o
l
32
)
{\displaystyle =\left(o{\frac {Steenkool}{32}}+0.21Lug\right)-\left(c{\frac {Steenkool}{12}}\right)-{\frac {1}{2}}\left(h{\frac {Steenkool}{2}}\right)-2X_{N2}\left(n{\frac {Steenkool}{28}}+0.781Lug\right)-X_{S}\left(s{\frac {Steenkool}{32}}\right)}
Maal elke term binne die hakies met die term aan die buitekant:
S
t
e
e
n
k
o
o
l
o
32
x
O
2
+
L
u
g
0.21
x
O
2
+
S
t
e
e
n
k
o
o
l
h
4
x
O
2
+
S
t
e
e
n
k
o
o
l
(
1
−
X
N
2
)
n
28
x
O
2
+
L
u
g
(
1
−
X
N
2
)
0.781
x
O
2
+
S
t
e
e
n
k
o
o
l
m
18
x
O
2
+
L
u
g
.
w
a
.
x
O
2
+
L
u
g
0.009
x
O
2
{\displaystyle Steenkool{\frac {o}{32}}x_{O2}+Lug0.21x_{O2}+Steenkool{\frac {h}{4}}x_{O2}+Steenkool(1-X_{N2}){\frac {n}{28}}x_{O2}+Lug(1-X_{N2})0.781x_{O2}+Steenkool{\frac {m}{18}}x_{O2}+Lug.wa.x_{O2}+Lug0.009x_{O2}}
=
S
t
e
e
n
k
o
o
l
o
32
+
L
u
g
0.21
−
S
t
e
e
n
k
o
o
l
c
12
−
S
t
e
e
n
k
o
o
l
h
4
−
S
t
e
e
n
k
o
o
l
2
X
N
2
n
28
−
L
u
g
2
X
N
2
0.781
−
S
t
e
e
n
k
o
o
l
X
S
s
32
{\displaystyle =Steenkool{\frac {o}{32}}+Lug0.21-Steenkool{\frac {c}{12}}-Steenkool{\frac {h}{4}}-Steenkool2X_{N2}{\frac {n}{28}}-Lug2X_{N2}0.781-SteenkoolX_{S}{\frac {s}{32}}}
L
K
=
S
t
e
e
n
k
o
o
l
o
32
x
O
2
+
S
t
e
e
n
k
o
o
l
h
4
x
O
2
+
S
t
e
e
n
k
o
o
l
(
1
−
X
N
2
)
n
28
x
O
2
+
S
t
e
e
n
k
o
o
l
m
18
x
O
2
−
S
t
e
e
n
k
o
o
l
o
32
+
S
t
e
e
n
k
o
o
l
c
12
+
S
t
e
e
n
k
o
o
l
h
4
+
S
t
e
e
n
k
o
o
l
2
X
N
2
n
28
+
S
t
e
e
n
k
o
o
l
X
S
s
32
{\displaystyle LK=Steenkool{\frac {o}{32}}x_{O2}+Steenkool{\frac {h}{4}}x_{O2}+Steenkool(1-X_{N2}){\frac {n}{28}}x_{O2}+Steenkool{\frac {m}{18}}x_{O2}-Steenkool{\frac {o}{32}}+Steenkool{\frac {c}{12}}+Steenkool{\frac {h}{4}}+Steenkool2X_{N2}{\frac {n}{28}}+SteenkoolX_{S}{\frac {s}{32}}}
L
K
=
S
t
e
e
n
k
o
o
l
[
o
32
(
x
O
2
−
1
)
+
h
4
(
x
O
2
+
1
)
+
(
1
−
X
N
2
)
n
28
x
O
2
+
2
X
N
2
n
28
+
m
18
x
O
2
+
c
12
+
X
S
s
32
]
{\displaystyle LK=Steenkool\left[{\frac {o}{32}}(x_{O2}-1)+{\frac {h}{4}}(x_{O2}+1)+(1-X_{N2}){\frac {n}{28}}x_{O2}+2X_{N2}{\frac {n}{28}}+{\frac {m}{18}}x_{O2}+{\frac {c}{12}}+X_{S}{\frac {s}{32}}\right]}
L
K
=
S
t
e
e
n
k
o
o
l
[
o
32
(
x
O
2
−
1
)
+
h
4
(
x
O
2
+
1
)
+
n
28
[
(
1
−
X
N
2
)
x
O
2
+
2
X
N
2
]
+
m
18
x
O
2
+
c
12
+
X
S
s
32
]
{\displaystyle LK=Steenkool\left[{\frac {o}{32}}(x_{O2}-1)+{\frac {h}{4}}(x_{O2}+1)+{\frac {n}{28}}[(1-X_{N2})x_{O2}+2X_{N2}]+{\frac {m}{18}}x_{O2}+{\frac {c}{12}}+X_{S}{\frac {s}{32}}\right]}
R
K
=
L
u
g
0.21
−
L
u
g
2
X
N
2
0.781
−
L
u
g
0.21
x
O
2
−
L
u
g
(
1
−
X
N
2
)
0.781
x
O
2
−
L
u
g
.
w
a
.
x
O
2
−
L
u
g
0.009
x
O
2
{\displaystyle RK=Lug0.21-Lug2X_{N2}0.781-Lug0.21x_{O2}-Lug(1-X_{N2})0.781x_{O2}-Lug.wa.x_{O2}-Lug0.009x_{O2}}
R
K
=
L
u
g
[
0.21
−
2
X
N
2
0.781
−
0.21
x
O
2
−
(
1
−
X
N
2
)
0.781
x
O
2
−
w
a
.
x
O
2
−
0.009
x
O
2
]
{\displaystyle RK=Lug[0.21-2X_{N2}0.781-0.21x_{O2}-(1-X_{N2})0.781x_{O2}-wa.x_{O2}-0.009x_{O2}]}
R
K
=
L
u
g
[
0.21
−
0.21
x
O
2
−
(
1
−
X
N
2
)
0.781
x
O
2
−
2
X
N
2
0.781
−
w
a
.
x
O
2
−
0.009
x
O
2
]
{\displaystyle RK=Lug[0.21-0.21x_{O2}-(1-X_{N2})0.781x_{O2}-2X_{N2}0.781-wa.x_{O2}-0.009x_{O2}]}
R
K
=
L
u
g
[
0.21
−
0.21
x
O
2
−
0.781
x
O
2
+
X
N
2
0.781
x
O
2
−
2
X
N
2
0.781
−
w
a
.
x
O
2
−
0.009
x
O
2
]
{\displaystyle RK=Lug[0.21-0.21x_{O2}-0.781x_{O2}+X_{N2}0.781x_{O2}-2X_{N2}0.781-wa.x_{O2}-0.009x_{O2}]}
R
K
=
L
u
g
[
0.21
−
x
O
2
−
w
a
.
x
O
2
+
X
N
2
0.781
x
O
2
−
2
X
N
2
0.781
]
{\displaystyle RK=Lug[0.21-x_{O2}-wa.x_{O2}+X_{N2}0.781x_{O2}-2X_{N2}0.781]}
R
K
=
L
u
g
[
0.21
−
x
O
2
(
1
+
w
a
)
+
X
N
2
0.781
(
x
O
2
−
2
)
]
{\displaystyle RK=Lug[0.21-x_{O2}(1+wa)+X_{N2}0.781(x_{O2}-2)]}
(
1
−
w
a
)
{\displaystyle (1-wa)}
R
K
=
L
u
g
[
0.21
(
1
−
w
a
)
−
x
O
2
(
1
+
w
a
)
+
X
N
2
0.781
(
1
−
w
a
)
(
x
O
2
−
2
)
]
{\displaystyle RK=Lug[0.21(1-wa)-x_{O2}(1+wa)+X_{N2}0.781(1-wa)(x_{O2}-2)]}
R
H
=
L
u
g
[
(
1
−
w
a
)
[
0.21
+
0.781
X
N
2
(
x
O
2
−
2
)
]
−
x
O
2
(
1
+
w
a
)
]
{\displaystyle RH=Lug[(1-wa)[0.21+0.781X_{N2}(x_{O2}-2)]-x_{O2}(1+wa)]}
S
t
e
e
n
k
o
o
l
100
%
v
e
r
b
r
a
n
d
i
n
g
A
D
=
L
u
g
[
(
1
−
w
a
)
[
0.21
+
0.781
X
N
2
(
x
O
2
−
2
)
]
−
x
O
2
(
1
+
w
a
)
]
o
32
(
x
O
2
−
1
)
+
h
4
(
x
O
2
+
1
)
+
n
28
[
(
1
−
X
N
2
)
x
O
2
+
2
X
N
2
]
+
m
18
x
O
2
+
c
12
+
X
S
s
32
{\displaystyle Steenkool_{100\%\ verbranding}^{AD}={Lug[(1-wa)[0.21+0.781X_{N2}(x_{O2}-2)]-x_{O2}(1+wa)] \over {\frac {o}{32}}(x_{O2}-1)+{\frac {h}{4}}(x_{O2}+1)+{\frac {n}{28}}[(1-X_{N2})x_{O2}+2X_{N2}]+{\frac {m}{18}}x_{O2}+{\frac {c}{12}}+X_{S}{\frac {s}{32}}}}
Omdat die lugvloei (
L
u
g
{\displaystyle Lug}
x
O
2
{\displaystyle x_{O2}}
Bereken waterdamp in lug weens humiditeit
wysig
As:
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}}
Die maksimum hoeveelheid waterdamp in lug is:
x
w
a
t
e
r
,
m
a
k
s
=
p
s
a
t
P
{\displaystyle x_{water,maks}={\frac {p_{sat}}{P}}}
Die werklike molfraksie van water in lug is:
x
w
a
t
e
r
=
R
H
×
x
w
a
t
e
r
,
m
a
k
s
{\displaystyle x_{water}=RH\times x_{water,maks}}
Die volgende Buck vergelyking is 'n baie goeie benadering om die dampdruk van water onder 100°C te bereken:
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}}}
Dus, die molfraksie van water in lug is:
x
w
a
t
e
r
=
R
H
100
×
P
×
0.61121
×
e
17.368
T
T
+
238.88
{\displaystyle x_{water}={\frac {RH}{100\times P}}\times 0.61121\times e^{\frac {17.368T}{T+238.88}}}
Totale vog in steenkool in terme van oppervlakvog en inherente vog
wysig
Oppervlakvog (ms ) is gebaseer op steenkool soos ontvang (SO) en inherente vog (mi ) is gebaseer op luggedroogde steenkool (LD). Die totale vog (m) kan egter soos volg uitgedruk word in terme van ms en mi :
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{\displaystyle m^{ld}={\frac {M}{Steenkool^{ld}}}={M_{s}+M_{i} \over Steenkool^{ld}}={(Steenkool^{so}m_{s}+Steenkool^{ld}m_{i}) \over Steenkool^{ld}}}
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{\displaystyle Steenkool^{ld}=(1-m_{s})Steenkool^{so}\qquad \Rightarrow \qquad Steenkool^{so}={Steenkool^{ld} \over 1-m_{s}}}
Dan is:
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{\displaystyle m^{lug\ droog}={m_{s} \over 1-m_{s}}+m_{i}}
![{\displaystyle LK=Steenkool\left[{\frac {o}{32}}(x_{O2}-1)+{\frac {h}{4}}(x_{O2}+1)+(1-X_{N2}){\frac {n}{28}}x_{O2}+2X_{N2}{\frac {n}{28}}+{\frac {m}{18}}x_{O2}+{\frac {c}{12}}+X_{S}{\frac {s}{32}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8dc669771b91258dcb18f6c30b65749b7cbf7246)
![{\displaystyle LK=Steenkool\left[{\frac {o}{32}}(x_{O2}-1)+{\frac {h}{4}}(x_{O2}+1)+{\frac {n}{28}}[(1-X_{N2})x_{O2}+2X_{N2}]+{\frac {m}{18}}x_{O2}+{\frac {c}{12}}+X_{S}{\frac {s}{32}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9d9a5c341c3511f647d921ceae836d2ed4a1970)
![{\displaystyle RK=Lug[0.21-2X_{N2}0.781-0.21x_{O2}-(1-X_{N2})0.781x_{O2}-wa.x_{O2}-0.009x_{O2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/310f91bc33daf6d393eb9a5d7520f10960fed7e7)
![{\displaystyle RK=Lug[0.21-0.21x_{O2}-(1-X_{N2})0.781x_{O2}-2X_{N2}0.781-wa.x_{O2}-0.009x_{O2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e12a2b664193fdeee98ab9854f5b1ddcdedea937)
![{\displaystyle RK=Lug[0.21-0.21x_{O2}-0.781x_{O2}+X_{N2}0.781x_{O2}-2X_{N2}0.781-wa.x_{O2}-0.009x_{O2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c2415d4ef0f4bcf2344493d9d816b818e43eb162)
![{\displaystyle RK=Lug[0.21-x_{O2}-wa.x_{O2}+X_{N2}0.781x_{O2}-2X_{N2}0.781]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f512fc932a735fbf8f5ad9f0953dd62f91a26beb)
![{\displaystyle RK=Lug[0.21-x_{O2}(1+wa)+X_{N2}0.781(x_{O2}-2)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c28888677bad1179f7d6c2e22524e181c3361087)
![{\displaystyle RK=Lug[0.21(1-wa)-x_{O2}(1+wa)+X_{N2}0.781(1-wa)(x_{O2}-2)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8092f14d9b7745ed89a20eee9a69ceab4524ac27)
![{\displaystyle RH=Lug[(1-wa)[0.21+0.781X_{N2}(x_{O2}-2)]-x_{O2}(1+wa)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d8c8a4b72b90de4541161a84ef1bca1e2577c05)
![{\displaystyle Steenkool_{100\%\ verbranding}^{AD}={Lug[(1-wa)[0.21+0.781X_{N2}(x_{O2}-2)]-x_{O2}(1+wa)] \over {\frac {o}{32}}(x_{O2}-1)+{\frac {h}{4}}(x_{O2}+1)+{\frac {n}{28}}[(1-X_{N2})x_{O2}+2X_{N2}]+{\frac {m}{18}}x_{O2}+{\frac {c}{12}}+X_{S}{\frac {s}{32}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2eee99d3047d64dab63174b7de41c82ec93a7694)
