weight% Alt. 1 Alt. 2 CaO 32,4 33,6 SiO2 33,1 33,7 Al2O3 11,9 11,9 MgO 17,6 16,0

Transcription

1 Slag formation The composition of slag is changed as stated in the table below Estimate B2, B3 och B4 B2: CaO/SiO 2 =32,4/33,1=0,98; 33,6/33,7=1,00 B3: (CaO+MgO)/SiO 2 =(32,4+17,6)/33,1=1,51; (33,6+16,0)/33,7=1,47 B4: (CaO+MgO)/(SiO 2 +Al 2 O 3 ) =(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% Alt. 1 Alt. 2 CaO 32,4 33,6 SiO2 33,1 33,7 Al2O3 11,9 11,9 MgO 17,6 16,0

2 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 SiO 2 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% Alt. 1 Alt. 2 CaO 32,4 33,6 SiO2 33,1 33,7 Al2O3 11,9 11,9 MgO 17,6 16,0

3 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 SiO 2 Al 2 O 3 TiO 2 V 2 O 5 Na 2 O K 2 O S P Mn Fe tot CO 2 BOF slag 41,42 10,38 8,10 1,55 5,18 4,81 0,013 0,03 0,286 3,07 20,07 Limestone 53,49 0,96 1,12 0,50 0,02 0,04 0,02 0,08 0,013 0,02 0,22 41,6 BF slag 32,38 16,25 34,47 11,51 2,46 0,070 0,77 0,51 1,37 0,010 0,48* 0,23 * MnO

4 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, ,9994)/54,9380]=1,29*%Mn %P 2 O 5 =%P *[(30,9738*2+15,9994*5)/ (2*30,9738)]=2,29%P %Fe 2 O 3 = %Fe*[(55,847*2+15,9994*3)/(2*55,847)=1,43*%Fe In BOF slag Fe is present as Fe met, Fe 2+ and Fe 3+ 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 SiO 2 Al 2 O 3 TiO 2 V 2 O 5 Na 2 O K 2 O S P Mn Fe tot CO 2 BOF slag 41,42 10,38 8,10 1,55 5,18 4,81 0,013 0,03 0,286 3,07 20,07 Limestone 53,49 0,96 1,12 0,50 0,02 0,04 0,02 0,08 0,013 0,02 0,22 41,6 BF slag 32,38 16,25 34,47 11,51 2,46 0,070 0,77 0,51 1,37 0,010 0,48* 0,23 * MnO

5 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 SiO(g)+C Si +CO Important sources for Si dissolution in the BF? Sources for reduction of SiO 2 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 - SiO 2 activity -Wettability - Endothermal reactions -Oxygen potential - Basicity MgO in the slag changes wettability, activity of SiO 2 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

6 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 SiO 2 /Al 2 O 3 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.

7 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

8 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

9 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)

10 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 P CO ) 2K 2 SiO 3 + 2C 4K(g) + 2SiO 2 +2CO 2Na 2 SiO 3 + 2C 4Na(g) + 2SiO 2 +2CO C, high activity of C and N 2 K(g)+C+1/2N 2 KCN(g, liq) Na(g)+C+1/2N 2 NaCN(g, liq) C, CO 2 oxidises alkalis K(g)+2CO 2 (g) K 2 CO 3 + CO(g) Na(g)+2CO 2 (g) Na 2 CO 3 + CO(g) 2KCN(l)+4CO 2 (g) K 2 CO 3 +5CO(g) + N 2 (g) 2NaCN(l)+4CO 2 (g) Na 2 CO 3 +5CO(g) + N 2 (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

11 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/nm 3, O 2 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 %, CO 2 =24,94 %, H 2 =3,31 % C coke = 88,0 % C PC =85,0 % CO 2limestone = 41,6 % C BF flue dust = 45,4 % O pellets = 28,7 %

12 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 CO 2 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)]=935,0nm 3 inclusive O 2 enrichment and moisture O 2 enrichment 9,39 Nm 3 {20,9%*(935-9,39 /(6307/24)]) +9,39 /(6307/24)]}/22,41=20,0 kmol H 2 O 12,5 g/nm 3 *935Nm 3 /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, ,7)=95,17 kg C for direct reduction of Fe, 95,17/12=7,91 kmol

13 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-(C HM +C dust +C limestone )]/12 [408,11-(47,5+6,36+1,48]/12=29,40 kmol Direktred=(C red -O blast )/O pellets +leg = (29,4-20,81)/24,75=34,7% Indirekt reduktion 1-34,7%=65,3%

14 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 CO 2 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:c tot -C dust -C HM =408,1-6,356-47,50=354,2kg [(24,94%CO2+20,57%CO)/22,41]*12=0,2437kg/Nm3 Top gas volume:354,2/0,223,7=1454 Nm3 %N 2 in top gas=1-(%co+%co 2 +%H 2 )=51,18% %N 2 i blast=1-%o2-%h 2 O= 1- (223,7/935+0,667*22,41/935)=74,48% V blast = V top gas *(%N 2top gas /%N 2blast )= 1454*(51,18/74,48)=998,9Nm 3 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, ,4)=78,48 kg C for direct reduction of Fe, 78,48/12=6,54 kmol

15 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-(C HM +C dust +C limestone )]/12 [408,11-(47,5+6,36+1,48]/12=29,40 kmol Direct red=(c red+comb -O blast )/O pellets +leg = (29,4-22,2)/24,75=29,1% Indirect reduction 1-29,1%=70,9%

16 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 (H 2, 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

17 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 ( CO )? 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)

18 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 = %CO 2 /(%CO+%CO 2 )

19 C rate 400 MJ C rate 200 MJ

20 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

21 Estimation based on process data week 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 Distri. Amount Factor amount Briquettcomp of oxide Oxide/ele ment Amount element butionfactor element slag element/o xid oxide slagg % kg kg kg kg CaO %to slag MgO % to slag SiO Al2O % tol slag TiO V2O Na2O % to slag K2O % to slag S P Mn Ni Fe tot Charg. weight, kg/thm 42 Total contribution to slagkg 12.99

22 Questions to process data week 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=%CO 2 /(%CO+%CO 2 ) How can a high EtaCO in BF-s with 100% pellet operation be explained? High ratio of Fe 2 O 3 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 CO 2 100% calcination at temperatures <900 C 11kg CO 2 corr. to (11/44)*22,41 Nm 3 =5,6025Nm 3 Top gas volume=1456 Nm3, %CO2=24,94, %CO=20,57 Corrected CO 2 =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 CO 2 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

23 Some more questions based on process data week 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

24 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.

25 Limestone used as slag former (flux) in the BF i mainly consistent of CaCO 3. Write the formula for calcination of CaCO 3. CaCO 3 CaO+CO 2 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 CaCO 3 is calcinated below or above the thermal reserve zone, respectively?above the thermal reserve zone. CO 2 released above the thermal reserve zone leaves the BF and gives a small increase of an uncorrected EtaCO. CO 2 released below the thermal reserve zone can react with C in coke. CO 2 +C 2CO (Boudoard reaction)