Slag formation
The composition of slag is changed as stated in the table below
–Estimate B2, B3 och B4
B2: CaO/SiO2=32,4/33,1=0,98; 33,6/33,7=1,00
B3: (CaO+MgO)/SiO2=(32,4+17,6)/33,1=1,51;
(33,6+16,0)/33,7=1,47
B4: (CaO+MgO)/(SiO2+Al2O3)
=(32,4+17,6)/(33,1+11,9)=1,11;
(33,6+16,0)/(33,7+11,9)= 1,09
Discuss based on the chemical composition how the
sulphur and alkali refining properties of slag can be
affected by the change in composition
Higher basicity promotes S refining because S can be
bound to both Ca and Mg. A higher basicity, especially
when it is caused by Ca is negative for alkali refining. Kat
ions of Ca and K have approximately the same diameter
and compete for the same positions in the silica network.
weight%
CaO
SiO2
Al2O3
MgO
Alt. 1
32,4
33,1
11,9
17,6
Alt. 2
33,6
33,7
11,9
16,0
Slag formation
The slag composition is changed according to the table below.
–The change in slag composition is caused by changeover
from 60%MPBO/40%KPBO to100%KPBO. KPBO has
higher P content, which limits the use of BOF slag as flux..
BOF slag is replaced by limestone. What effects can this
have on the slag formation in the BF? The slag volume is
unchanged because KPBO has higher SiO2 content
compared to MPBO.
BOF slag has a low melting point compared to limestone
(in reality burnt lime at this level of the BF) an it therefore
reacts completely with slag from pellets and ash from coal
and coke in the cohesive zone of the BF. The content of
other oxides as for example MnO, MgO etc contributes to
decreased melting point and viscosity in formed bosh slag.
Residual burnt lime, that is formed from limestone, has a
high melting point and is only partly dissolved in the bosh
slag. This is shown by residual lime in the lower part of
the BF found during excavation and core drilling. As a
result a bosh slag with lower basicity is formed, which is
negative for sulphur refining but positive for alkali
refining.
weight%
CaO
SiO2
Al2O3
MgO
Alt. 1
32,4
33,1
11,9
17,6
Alt. 2
33,6
33,7
11,9
16,0
Slag
•At SSAB BOF slag and limestone is used as
basic fluxes. In the table below the chemical
composition of these materials can be seen
together with the BF slag. The slag volume is
160kg/tHM. The aim for B2 is 1,12.
a)
Estimate B2 och B4 of the BOF slag.
B2=32,38/34,47=0,94
B4=(32,38+16,25)/(34,47+11,51)=1,06
b)
Estimate the amount that has to be
added to reach the aim of B2 if you only can
use limestone or BOF slag
- Limestone
{(160*32,30+x*53,49)/(160*34,47+x*1,12)}=1,1
2
The eq. is solved for x, x= 19,3 kg limestone
-BOF-slag
{(160*32,30+x*41,42)/160*34,47+x*8,10)}=1,12
The eq. is solved for x, x=31,2 kg BOF slag
CaO MgO SiO2 Al2 O3 TiO2 V2 O5 Na2 O K2 O S
BOF slag 41,42 10,38 8,10
1,55 5,18 4,81 0,013 0,03
Limestone 53,49 0,96 1,12
0,50 0,02 0,04
0,02 0,08
BF slag 32,38 16,25 34,47 11,51 2,46 0,070 0,77 0,51 1,37
* MnO
P
0,286
0,013
0,010
Mn
3,07
0,02
0,48*
Fetot CO2
20,07
0,22 41,6
0,23
Slag
How much is the slag volume increased in each case if
Fe, Mn, P och V is supposed to reach HM to 100% and
other elements
-For the elements that is reduced and dissolved in the HM
the O is supposed to be transferred to the gas phase.
%MnO= %Mn*[(54,9380+15,9994)/54,9380]=1,29*%Mn
%P2O5=%P *[(30,9738*2+15,9994*5)/
(2*30,9738)]=2,29%P
%Fe2O3=
%Fe*[(55,847*2+15,9994*3)/(2*55,847)=1,43*%Fe
In BOF slag Fe is present as Femet, Fe2+ and Fe3+ and
chemical analyses indicates 6-7% O Bound to Fe.
-Limestone 19,3 kg increased charging
To HM and gas phases:
19,3*{0,01*(41,6+0,04+2,29*0,013+1,29*0,02+1,43*0,22)}=
8,11 kg
To slag
19,3-8,11=11,2 kg, the increase in slag volume
-BOF-slag 31,2 kg BOF slag
To HM and gas 31,2*{0,01*[4,81+2,29*0,286+1,29*3,07+
(20,07+6,5)]=11,2
To slag
31,2-11,2=20,0 kg, the increase in slag volume
CaO MgO SiO2 Al2 O3 TiO2 V2 O5 Na2 O K2 O S
BOF slag 41,42 10,38 8,10
1,55 5,18 4,81 0,013 0,03
Limestone 53,49 0,96 1,12
0,50 0,02 0,04
0,02 0,08
BF slag 32,38 16,25 34,47 11,51 2,46 0,070 0,77 0,51 1,37
* MnO
P
0,286
0,013
0,010
Mn
3,07
0,02
0,48*
Fetot CO2
20,07
0,22 41,6
0,23
HM Si content
• Write the reaction for dissolution of Si in HM.
Dissolution of Si in HM occurs through reduction
of SiO2 to SiO(g) that is dissolved into C
containing HM
ƒSiO(g)+C' Si +CO
•
Important sources for Si dissolution in the BF?
Sources for reduction of SiO2 toSiO(g) are
-Coke
-Slag via slag/metal or slag/coke reaction
- PC
• What factors has effect on the
distribution of Si between HM and slag?
Explain.
Factors that has effect on SiO(g) generation are
-Temperature
- SiO2 activity
-Wettability
- Endothermal reactions
-Oxygen potential
- Basicity
MgO in the slag changes wettability, activity of
SiO2 activity and is reduced and evaporated in an
energy consuming reaction.
Temperature has effect on K for the reaction
High oxygen potential as for example the
presence of FeO
HM Si content
• Why can the %Si of HM be used for estimation
of the heat level of the BF.
The reactions for dissolution of Si into HM are
strongly endothermic. I other parameters
affecting the Si content are kept constant the Si
content can be used for heat level control.
• %Si in HM can increase/decrease without change
of heat level, explain.
Changes of basicity and/or composition in bosh
slag, tuyere slag and final slag.
Change of PC type, coke type with changed ratio
of SiO2/Al2O3 in the ash
The level of the cohesive zone has effect on the
volume with high temperatures in the BF. The
level of the cohesive zone can be changed for
example if PC type or tuyere parameters are
changed.
Describe state and actions
•
HM -%S is high but %C, %Si and HM
temperature are low
The heat level of the BF is low. This is counter
acted primarily by increased injection rate and if
this is not enough the production can be
decreased by decreasing the blast.
•
HM - %C is low but %S, %Si and HM
temperature are high
High %Si and temperature indicates normal or
high heat level. High %S and low %C corresponds
to the low heat level. This can be explained by bad
drainage/inactive dead man that prohibit the HM
from flowing through the coke bed. The
temperature of the coke bed is lower compared
to the temperature close to the tuyeres.
•
The content of MnO och FeO in the slag
increases
Increased amount of unreduced material reaches
the hearth. Increased direct reduction may result
in low heat level of the BF. Badly reduced
material can consist of.
– Sculls/scaffolds is scaled off
– Uncompletly reducd ferrous burden
Describe state and action
•
(S)/[S] increases
Increased ration means improved S refining by
the slag. The sulphur distribution is affected
positively by
– Low oxygen potential, complete reduction of
FeO and MnO
– Increased basicity of slag
– Low viscosity of slag
No action is necessary
•
Alkali output has for a while been only 75% and
the cooling effects has decreased below the
critical limit.
The probable reason is sculls and/or scaffold
formed on the BF wall.
Possible counter actions
– Decreased slag basicity
– Changed burden distribution to get increased
gas flow on the wall
– Increased addition of reducing agents
S in the BF
•
What conditions are beneficial to S
refining?
– Low oxygen potential, optimal pre-reduction of
of MnO, FeO etc
– Incerased slag basicity
– Low slag viscosity
– Increased slag volume
•
Explain with a formula why FeO in the
slag will retard the S refining.
(CaO)+S '(CaS)+(FeO)
•
High S load in the BF results in high S
content of the slag. What will the effect
on Si content be?
– The slag basicity has to be adjusted to
higher values (lower Si)
– The wettability of slag to coke
increases (higher Si)
Recirculation
•
Some elements shows recirculation in the BF.
Describe the recirculation of alkali.
•Alkali is reduced and evaporated from
–Charged material when i melts. The alkali content is
highest in the cohesive zone and distributed in
correlation with the cohesive zone shape.
– Coal and coke ash raceway (increases with increased
temperature and decreased PCO)
–2K2SiO3 + 2C⇄4K(g) + 2SiO2 +2CO
–2Na2SiO3 + 2C⇄4Na(g) + 2SiO2 +2CO
•1300-1600°C, high activity of C and N2
–K(g)+C+1/2N2 ⇄ KCN(g, liq)
– Na(g)+C+1/2N2 ⇄ NaCN(g, liq)
•700-1300°C, CO2oxidises alkalis
–K(g)+2CO2(g) ⇄ K2CO3 + CO(g)
–Na(g)+2CO2(g) ⇄ Na2CO3 + CO(g)
–2KCN(l)+4CO2(g) ⇄ K2CO3+5CO(g) + N2(g)
–2NaCN(l)+4CO2(g) ⇄ Na2CO3+5CO(g) + N2(g)
•
The oxidation of alkalis in the shaft is affected
by the atmosphere? How?
– The reduction potential of the gas
– The temperature
•
Mention negative effects caused by alkali
accumulation?
–Scull and scaffold formation
–Increased recirculation
• Increases the consumption of reducing agents
• Has negative effect on coke and pellets leading to increased
dust generation
Estimate the direct reduction degree
• HM 6307 t/24h, slag 168 kg/tHM, BF flue
dust 14 kg/tHM
• Coke 2050 t/24h, PC 851 t/24h, pellets 8482
t/24h, limestone 164 t/24h, BOF-slag 296
t/24h
• Blast flow 239 kNm3/h, moisture 12,5
g/Nm3, O2 flow 9,39 kNm3/h
• HM composition, weight%
C = 4,75,
Si = 0,39,
Mn = 0,33
P = 0,035,
S = 0,047,
V = 0,35
Ti=0,13,
(S)/[S] = 27
• Operation time 24h
• Top gas CO=20,57 %, CO2=24,94 %,
H2=3,31 %
CPC =85,0 %
• Ccoke = 88,0 %
CO2limestone= 41,6 %
CBF flue dust= 45,4 %
Opellets = 28,7 %
Estimation of direct reduction degree
blast volume
•
•
•
•
•
•
•
•
Amounts corresponds to 1000 kg HM, blast volume
C in HM = 4,75%*1000=47,5 kg
C consumption for reduction of Si, Mn, P, S, V, Ti
– Si: (1000*0,01*0,39*2/28,09)*12=3,332 kg
– Mn: (1000*0,01*0,33/54,94)*12=0,721 kg
– P: (1000*0,01*0,035*2,5/30,97)*12=0,339 kg
– S: (168*0,01*1,269/32)*12=0,799 kg
– V: (1000*0,01*0,35*2,5/50,94)*12=2,061 kg
– Ti: (1000*0,01*0,13*2/47,9)*12=0,651 kg
– Totally:7,903 kg
C in BF flue dust 45,4*0,01*14=6,356 kg
Tot C in (coke, PC, briquette)
325*87,96+135*85+42*17,84=408,11 kg
Consumption of C by CO2 in limestone
(26*41,6*0,01*0,50/44)*12=1,475 kg
Consumption of C to CO is estimated based on O
– Specifik blast:[245,7/(6307/24)]=935,0Nm3
inclusive O2 enrichment and moisture
– O2 enrichment 9,39 Nm3
– {20,9%*(935-9,39 /(6307/24)]) +9,39
/(6307/24)]}/22,41=20,0 kmol
– H2O 12,5 g/Nm3*935Nm3/trj*0,001=11,69 kg
11,69/18 kmol O=0,667 kmol
– O iin PC 135*2,1%/16=0,177 kmol
– Totally O 20,04+0,667+0,177=20,81 kmol
– 20,81*12=249,7 kg C is combusted to CO
408,11-(47,5+6,636+7,03+1,475+249,7)=95,17 kg C for
direct reduction of Fe, 95,17/12=7,91 kmol
Estimation of direct reduction degree
blast volume
• Direct reduction based on total amount reduced Fe
{1000*(1-0,01*(0.39+0,33+0,035+0,047+0,35+0,13)}/55,847
=16,83 kmol
–Ratio direct reduction : 7,91/16,83=47,1%
–Ratio indirect reduction: 1-47,1%=52,9%
• Direct reduction based on O content of pellets
–O in pellets:
{(28,7%*1345/15,9994)-0,24/(55,847+15,9994)}
=24,09 kmol
- Ratio direct reduction :7,91/24,09=32,9%
- Ratio indirect reduction : 1-32,9%=67,1%
• Direct reduction based on O content of pellets and O
bound to leg elements
–O in pellets 24,09 kmol
–O in Si, Mn, P, S, V, Ti
• Si: (1000*0,01*0,39*2/28,09)
• Mn: (1000*0,01*0,33/54,94)
• P: (1000*0,01*0,035*2,5/30,97)
• S: (168*0,01*1,269/32)
• V: (1000*0,01*0,35*2,5/50,94)
• Ti: (1000*0,01*0,13*2/47,9)
–Totally:0,660 kmol
• O i ”raw material”: 24,09+0,66=24,75 kmol
• C red+comb =[ Ctot-(CHM+Cdust+Climestone)]/12
[408,11-(47,5+6,36+1,48]/12=29,40 kmol
• Direktred=(Cred-Oblast)/Opellets +leg= (29,420,81)/24,75=34,7%
• Indirekt reduktion 1-34,7%=65,3%
Estimation of direct reduction degree
C-N balance
• Amounts corresponds to 1000 kg HM, blast volume
• C in HM = 4,75%*1000=47,5 kg
• C consumption for reduction of Si, Mn, P, S, V, Ti
– Si: (1000*0,01*0,39*2/28,09)*12=3,332 kg
– Mn: (1000*0,01*0,33/54,94)*12=0,721 kg
– P: (1000*0,01*0,035*2,5/30,97)*12=0,339 kg
– S: (168*0,01*1,269/32)*12=0,799 kg
– V: (1000*0,01*0,35*2,5/50,94)*12=2,061 kg
– Ti: (1000*0,01*0,13*2/47,9)*12=0,651 kg
– Totally:7,903 kg
• C in BF flue dust 45,4*0,01*14=6,356 kg
• Tot C in (coke, PC, briquette)
325*87,96+135*85+42*17,84=408,11 kg
• Consumption of C by CO2 in limestone
(26*41,6*0,01*0,50/44)*12=1,475 kg
• Estimation of blast volume based on på C-N balance
– Total amount of C in top gas:Ctot-Cdust-CHM=408,1-6,35647,50=354,2kg
– [(24,94%CO2+20,57%CO)/22,41]*12=0,2437kg/Nm3
– Top gas volume:354,2/0,223,7=1454 Nm3
– %N2 in top gas=1-(%CO+%CO2+%H2)=51,18%
– %N2 i blast=1-%O2-%H2O=
1(223,7/935+0,667*22,41/935)=74,48%
– Vblast= Vtop gas*(%N2top gas/%N2blast)=
1454*(51,18/74,48)=998,9Nm3
• Combustion of C to CO estimated based on O in blast=
{[998,9*23,9/(22,41*16)]+(998,9*0,001*12,5/18)+0,177}kmol
=22,20 kmol
• C combustion in raceway 22,22*12 kg=266,4 kg C
• 408,11-(47,5+6,636+7,03+1,475+266,4)=78,48 kg C for direct
reduction of Fe, 78,48/12=6,54 kmol
Estimation of direct reduction degree
C-N balance
• Direct reduction based on totally reduced Fe
{1000*(1-0,01*(0.39+0,33+0,035+0,047+0,35+0,13)}/55,847
=16,83 kmol
–Ratio direct reduction : 6,54/16,83=38,9%
–Ratio indirect reduction : 1-47,1%=61,1%
• Direct reduction based O content in pellets
–O in pellets (FeO in slag is discounted):
{(28,7%*1345/15,9994)-0,24/(55,847+15,9994)}
=24,09 kmol
- Ratio direct reduction :6,54/24,09=27,1%
- Ratio indirect reduction : 1-32,9%=72,9%
• Direct reduction based on O content of pellets and O
bound to leg elements
–O in pellets 24,09 kmol
–O in leg. elements Si, Mn, P, S, V, Ti
• Si: (1000*0,01*0,39*2/28,09)
• Mn: (1000*0,01*0,33/54,94)
• P: (1000*0,01*0,035*2,5/30,97)
• S: (168*0,01*1,269/32)
• V: (1000*0,01*0,35*2,5/50,94)
• Ti: (1000*0,01*0,13*2/47,9)
–Totally:0,660 kmol
• O pellets+leg 24,09+0,66=24,75 kmol
• C red+comb =[ Ctot-(CHM+Cdust+Climestone)]/12
[408,11-(47,5+6,36+1,48]/12=29,40 kmol
• Direct red=(Cred+comb-Oblast)/Opellets +leg= (29,422,2)/24,75=29,1%
• Indirect reduction 1-29,1%=70,9%
CDRR diagram
decreased demand of reducing agents
• How can the position of the chemical
limitation line be improved?
• Reduction degree/oxygen content in
charged ferrous burden
• Composition of BF gas (H2, CO)
• Control of the thermal reserve zone
temperature
• How can the position of the
thermal limitation line be improved?
•
•
•
•
Increased blast temperature
Decreased blast moisture
Decreased slag volume
Decreased heat losses
Questions about the CDRRdiagram on next page
• Mark in the figure
– The chemical and thermal limitation line.(see next
–
–
page)
The operational area of the BF in the figure. (see next
page)
Were reduction is possible? (See next page)
• Draw the operational triangle based on data
in the CDRR rapport. (se following three
pages)
• Give the formula for the gas efficiency
(ÎCO)?
• How much is the lowest possible C rate
changed (kg/tHM), according to the diagram,
if the heat loss decreases from 400 till
200MJ? (see the three following pages)
• Write the formula for indirect and direct
reduction of FeO, respectively. State if the
reaction is exotherm or endotherm. Which of
the reactions consumes most C? (See the
following three pages)
Chemical
Limitation
line
Thermal
limitation
line
The operational area is the area above the
limitation lines.
Reduction is possible above the chemical
limitation line
Eta CO = %CO2/(%CO+%CO2)
C rate 400 MJ
C rate 200 MJ
• Indirect reduction
–
–
–
FeO +nCO 'Fe+CO2+(n-1) CO
900°C, 70%CO, 30%CO2 ⇒
FeO +3,3CO 'Fe+CO2+ 2,3CO
Exotherm reaction
High C demand via CO consumption;
0,709 kg/kg Fe
• Direct reduction
– FeO+C ' Fe+CO
– Endotherm reaction
– C consumption 0,215kgC/kgFe
Estimation based on process data week
37 2005
•
How much does the briquette contribute to
the slag volume.
•The distribution factor tells the ratio of a
charged element that is transferred to HM
Amount Factor
Briquett- of
Oxide/ele Amount
comp oxide
ment
element
%
kg
kg
CaO
18.13
7.61
-----------
-----------
MgO
SiO2
2.48
6.69
1.04
2.81
----------0.47
----------1.31
Al2O3
TiO2
V2O5
2.14
0.41
0.65
0.90
0.17
0.27
----------0.60
0.56
----------0.10
0.15
Na2O
0.08
0.03
-----------
-----------
K2O
S
P
Mn
Ni
Fe tot
0.26
0.406
0.053
0.64
0.11
-----------
----------0.406
0.053
0.64
0
Charg.
weight,
kg/tHM
Distri.
butionfactor
100%to
slag
100% to
slag
0.13
100% tol
slag
0.30
0.96
100% to
slag
100% to
slag
0.16
0.84
0.81
0.97
Amount
Factor
element element/o
slag
xid
kg
amount
oxide
slagg
kg
-----------
-----------
7.61
----------1.14
----------2.14
1.04
2.81
----------0.072
0.006
----------1.67
1.79
0.90
0.12
0.01
-----------
-----------
0.03
----------0.341
0.008
0.122
0.000
-----------
0.11
0.17
0.02
0.16
0.00
2.29
1.29
32.56
42
Total contribution to slag
kg
12.99
Questions to process data week 37
2005
•
Eta CO is relatively high for BF-s operating with
100% pellets
– State the formula for Eta CO and estimate
the value for the actual period
•
EtaCO=ŋCO=%CO2/(%CO+%CO2)
– How can a high EtaCO in BF-s with 100%
pellet operation be explained?
• High ratio of Fe2O3 in pellets
• Low slag volume
– How much should EtaCO be changed if the the
limestone was replaced by BOF slag? Estimate
with the assumptions that 100% and 60%,
respectively of the limestone is calcinated at
temperatures <900°C.
•26 kg limeston/tHM corresponds to 11 k g CO2
•100% calcination at temperatures <900°C 11kg CO2 corr.
to (11/44)*22,41 Nm3=5,6025Nm3
•Top gas volume=1456 Nm3, %CO2=24,94, %CO=20,57
•Corrected CO2=1456*24,94%-5,6025=357,239
CO=1456*20,57%=299,499
•Corrected EtaCO=0,544 which can be compared with not
corrected of 0,550
•60% of CO2 calcinates at temperatures <900°C corr. to
(0,6*11/44)*22,41Nm3 CO2 = 3,3615 Nm3och
(0,4*11/44)*22,41Nm3 CO=2,241 Nm3 CO
•Corrected EtaCO: (1456*24,94%3,3615)/{(1456*24,94%-3,3615)+ (1456*20,57%2,241)}=0,545
Some more questions based on process
data week 37 2005
•
In the drawn CDRR diagram is the
operational triangle large. Try to explain
why.
Incorrect
•
•
•
data concerning
Specific blast
Consumption of reducing agents
EtaCO
Which can be caused by
– Large variations in process conditions during
the period
– Incorrect moisture content of coke
– Incorrect analysed gas composition
– Inconsistent weights
– Inconsistent measurements of gas flow
RIST diagram –describes schematically the
change of o content in the burden and reduction
gas as it descends and ascends, respectively.
Which interval on the line AE corresponds to
− Indirect reduction of FeO
− C combustion in the raceway
− Direct reduction of FeO
− Pre-heating
− Direct reduction of P, Si etc
Explain the term
Shaft efficiency.
•
The answer of these
questions can easliy be
found in in the ppt
presentation.
Limestone used as slag former (flux) in the BF i
mainly consistent of CaCO3.
•Write the formula for calcination of CaCO3.
•CaCO3 ' CaO+CO2
•How is the temperature for calcination changed if the
top pressure of the blast furnace is increased from
atmospheric pressure to 1,5 bars over pressure?
•The temperature for calcination increases.
•How will the calcination effect the gas efficiency if
CaCO3 is calcinated below or above the thermal
reserve zone, respectively?Above the thermal reserve
zone.
•CO2 released above the thermal reserve zone
leaves the BF and gives a small increase of an
uncorrected EtaCO. CO2 released below the
thermal reserve zone can react with C in coke.
CO2 +C' 2CO (Boudoard reaction)