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Mod-01 Lec-06 Lecture-06-Principles of Carbon Reduction — Transcript

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0:00[Music]
0:19now
0:22friends during my last lecture I started
0:25discussing free energy diagram for
0:28oxides which are also known know as Elam
0:31diagrams let me quickly recapitulate
0:35some of the basic features of eling gam
0:39diagrams these diagrams plot standard
0:42free energies of formation of various
0:48oxides from the metal and
0:50oxygen as a function of
0:54temperature and the values are always
0:58with reference to one
1:00molecule of
1:02oxygen so all the formation reactions
1:05are so
1:07written that it is always one molecule
1:11of oxygen in the
1:13reaction and there is a very interesting
1:16reason why we do that we do that because
1:19then we can easily deduct values of
1:22formation of one oxide from values of
1:26formation of another oxide we can
1:28calculate very easily
1:30the free energy change in the reaction
1:33when one metal reduces a
1:37lesser less stable oxide to produce a
1:41more stable oxide I'll give an example
1:43of that but first of all let us see the
1:48relevance of this these diagrams for
1:51carbon reduction of oxides which as you
1:55know is the basic reaction in pyrometer
2:01now carbon forms two
2:04oxides carbon dioxide and carbon
2:09monoxide the line
2:13for carbon reacting with oxygen to form
2:17CO2 is a horizontal
2:19line because there is no entropy change
2:22in this reaction the same number of
2:25moles of
2:27gas are involved in the left hand side
2:29as well as the right hand side one mole
2:31of oxygen giving you one mole of
2:34CO2 and as I mentioned earlier these
2:37lines represent the equation Delta G is
2:40equal to Delta H minus t Delta H so the
2:44slope is comes from Delta s not which is
2:47the entropy
2:48change in the case of Co reaction
2:52however two 2
2:56C plus O2 giving you 2
3:01Co we have the situation that one mole
3:05of
3:05oxygen reacts with carbon to produce 2
3:09moles of Co so the volume is increasing
3:13volume of gas the entropy is increasing
3:15and therefore this line has a negative
3:20slope this implies that with increasing
3:23temperature carbon monoxide becomes
3:27increasingly
3:28stable carbon dioxide stability does not
3:33change with increasing
3:35temperature on the other hand all oxides
3:39which are represented by the parallel
3:42lines show that as the temperature
3:46increases all oxides become less stable
3:51more prone to reduction and
3:54decomposition now these lines intersect
3:58carbon Cuts these lines
4:00as well as for formation of Co cuts the
4:03CO2 Cuts these lines as well as carbon
4:06going to carbon dioxide also cut these
4:08lines consider an
4:12intersection of this line and that line
4:15This oxide is becoming increasingly
4:18unstable with
4:20temperature and Co is becoming
4:23increasingly stable with temperature
4:26Beyond this intersection Co is
4:30comparatively more stable than the
4:33oxide there therefore carbon can reduce
4:37this oxide to form Co and liberate that
4:41metal now since the co line has a
4:45negative slope in theory it will cut all
4:49lines of course some intersections will
4:52be at a very high temperature some would
4:54be at low temperatures so those oxides
4:58which are not very stable will be
5:00reduced
5:01easily oxides that are far more stable
5:05will be reduced at high temperatures in
5:07theory it's quite possible that the
5:09temperatures required for reduction at
5:13for these Metals at such high
5:15temperatures what will be more
5:18stable some carbide will become more
5:21stable and not oxide because the metal
5:23would also react with carbon in the
5:25system then we have a different kind of
5:28problem but in theory
5:32carbon oxygen to form carbon monoxide
5:35this
5:37reaction can be the basis of reduction
5:41of any metal oxide at sufficiently high
5:44temperatures how do we calculate the uh
5:48thermodynamic quantities free energies
5:50of formation such reactions for such
5:52reactions is very simple actually
5:56consider the basic reactions here
6:002 C+ O2 2 Co call it Delta G1 C+ O2 CO2
6:06call it Delta
6:08G2 and formation reactions for oxide the
6:11simple oxide Mo we take
6:13Delta G
6:163 if you write
6:18it in the reverse manner this becomes
6:21minus Delta
6:23G now you have to add this
6:27reaction with this reaction
6:31to get the values for these like 2 m o +
6:352 C to give you 2 m + 2 Co will be
6:38obtained by Delta G1 minus Delta G3 and
6:43for 2 m mo+ c giving
6:47you 2
6:50m+
6:52CO2 will be given by Delta G2 minus
6:56Delta G 3 so you can calculate the free
6:59energy change for the reduction of uh
7:02these reduction reactions by
7:05carbon to form Co and
7:09CO2 these calculations are important in
7:12metery and there will be many examples
7:15of that similarly we can calculate the
7:18free energy
7:19change when a metal M Prime reduces a
7:24less
7:25stable metal oxide to produce release
7:29that metal and produce a more stable
7:31metal oxide Again by taking a
7:35difference of two free energy change for
7:38two
7:40reactions now another thing that I had
7:42mentioned is that in
7:45these plots the slope represent Delta s
7:49not the change in slope is not because
7:52of change in Delta is not but because of
7:55some melting of either metal or the uh
7:58metal oxide
8:04now let's take this
8:07example of reduction of
8:11al23 by
8:13carbon so we have to combine these two
8:19reactions 2 C + O2 2
8:23Co
8:25and 4x3 aluminum plus O2 2x3
8:30l23 you have to deduct the free energy
8:33values for this
8:35reaction from the free energy change for
8:38the reaction and then we get the desired
8:42reaction 2x3 l23 + 2 C 4 by + 2 Co and
8:48we can calculate the free energy change
8:50for that
8:51reaction you can see here that the
8:54carbon going Co that line is represent
8:59approxim by this line
9:02carbon oxygen reaction from CO2 is about
9:06this line This intersection takes place
9:08at around
9:10710 and for ION oxide the line is here
9:14so we can say that AO can be reduced by
9:18both
9:19carbon carbon to form either CO2 or Co
9:24at around here but the line for l23
9:29can be met only by the line for
9:34Co this line that represents
9:39carbon oxygen reaction to form CO2 does
9:43not cut this l23 line we can say that if
9:48we can have temperatures of around
9:511800° even a stable oxide like l23 will
9:55be reduced by carbon to form Co because
9:59Beyond this point Co becomes more stable
10:02as compared to
10:05al23 however as I
10:07mentioned although this is okay in
10:10theory we cannot ignore another reaction
10:13that will take place that will
10:16form aluminium carbide which will not be
10:20reduced by Co so this reaction actually
10:24it it cannot be exploited in the
10:28industry but for lesser less stable
10:31oxides where these lines are placed
10:34higher above we'll find many
10:38reactions many reactions are
10:41possible where the reduction is by
10:45carbon there's another thing that
10:48happens that Suppose there is a metal
10:52oxide which is reduced by
10:57carbon to form
11:01metal and carbon
11:03monoxide suppose the temperature is very
11:07high we can help the reaction in two
11:10ways we can apply
11:13vacuum if you can apply vacuum and
11:15release Co from the system then this
11:19reaction will be driven to the right and
11:21it can take place at lower temperatures
11:25the other way would be if you can lower
11:27the activity of M
11:30by dissolving in
11:39something for example suppose we produce
11:44not the metal but a Ferro
11:48alloy a metal dissolved in a pool of
11:51iron then the activity of metal will go
11:55down and the reaction will be driven to
11:57the right we'll be able to C carry out
11:59the reaction at a lower temperatures
12:02will
12:03also eliminate if not totally partially
12:08the tendency to form a carbide because
12:11again formation of a carbide will depend
12:14on activity of the metal so in the case
12:18of those oxides which are very stable
12:20and which will need very high
12:22temperatures for reduction by
12:24carbon we can help the reduction
12:27reaction by using by by producing Ferro
12:30Alloys this is the basis of production
12:33of Ferro chromium pherom manganese
12:37ferrovanadium
12:38Etc which are possible because the metal
12:43being produced is being dissolved in
12:45iron and we can drive the reaction to
12:48the
12:54right unfortunately we do not produce
12:57anything called Ferro aluminum
13:00otherwise if
13:01we could dissolve aluminum in Iron then
13:05this reaction would have occurred at a
13:08much lower temperatures because
13:10aluminium
13:12activity will go
13:17down okay let us now
13:23proceed
13:25the
13:28thermodynamics has been
13:30extensively applied in the case of
13:33roasting reactions also to understand
13:36the reactions now during
13:39roasting all kinds of reactions are
13:42possible I am writing some reactions
13:45here as you can
13:50see let's let's assume the sulfide is
13:52written as MS2 like
13:56F2 it can react with other sulfides like
14:01FES release sulfur this sulfur oxygen
14:05reaction is possible the metal sulfide
14:08can react with oxygen to produce a metal
14:11oxide and so SO2 there's reaction
14:14between s SO2 and there should be half
14:17O2 oxide and S3 can form uh
14:22M4 then there also this sort of products
14:27are possible in other words
14:29when you have a in a system you have
14:32metal sulfur and
14:34oxygen you can produce oxide you can
14:37produce different sulfides you can
14:39produce sulfates you can produce
14:43compounds which can be written as a
14:46combination of oxide and
14:49sulfate what will exist at a particular
14:52temperature would depend on the partial
14:54pressures of sulfur and
14:58oxygen
14:59or if you fix the pressures of sulfur
15:03and oxygen then what will exist what
15:06phases will be present will depend on
15:09the
15:10temperature this information is vital
15:13and necessary because if you want to
15:16control roasting reactions we would like
15:18to aim at certain products we must know
15:22what should be the value of partial
15:23pressure of oxygen partial pressure of
15:27um sulfur and what should be the
15:30temperature so let's see how we do
15:42that there
15:44are Elam diagrams for sulfites also
15:48where all the reactions are
15:51shown for formation of the
15:54sulfites for reactions between the metal
15:58and sulfur Always written as
16:02S2 sulfur Vapor so all the reactions as
16:06in the case of oxid it was with O2 it's
16:08written in terms of H2 and whatever we
16:11did with sulfides oxides we can do the
16:14same thing here
16:16but we do not consider these diagrams
16:19for reduction by carbon or not even so
16:22much for metallothermic reaction but
16:25there are other uses of these diagrams
16:27and I'll I'll I'll try to show you one
16:30or two
16:35uses people have been able to draw using
16:39thermodynamic
16:40data diagrams called predominance area
16:46diagrams now predominance area diagram
16:49show
16:50us
16:53that at a particular
16:56temperature for different values of
17:02P2 and PSO2 what are the phrases that
17:06are present this will be for a
17:09particular
17:11temperature which means that we can have
17:15this phase for a
17:18variety of
17:20combinations of partial pressure of
17:22oxygen and partial pressure of SO2 there
17:25are limits but if you have the partial
17:29pressure of oxygen at this and if you
17:32exceed the value of
17:36P SO2 Beyond this then you'll end up
17:39with is n
17:42io4 so how do we draw such diagrams I'll
17:45give you a an example of a simple
17:49example that suppose you want to study
17:52the uh
17:56NIS this diagram of
17:59roasting of NIS with there are many many
18:02reactions possible let us consider one
18:05particular
18:06reaction that is nickel
18:09sulfide giving you nio and
18:14SO2 now we know that the basic reaction
18:19for this would be Delta G minus rtln K
18:22Prime where K Prime is the equilibrium
18:24constant which is written like
18:27this
18:30now from this if we take log we can
18:33write log PSO2 3x2 log P sub2 plus log K
18:41Prime we are considering an equilibrium
18:45between n and niio and that is the line
18:50BC n and
18:57ni now similar
18:59for reaction ni3 S2 7x 22 3 N2 s O2 we
19:05will obtain another
19:08equation
19:10now how do you analyze these look at
19:13this
19:15uh look at these two equations 2 n + O2
19:192
19:20nio from Elam diagrams we can get the
19:23free energy change value for
19:26this 2 n i + S2 2N n we can get the free
19:31energy change for this from the Elum
19:34diagram from sulfides now you write it
19:37in the reverse this reaction so that you
19:40can add these two equations to get a
19:44equation that
19:46represents oxidation of n to
19:49nio so we can get the Delta G value for
19:53this Reaction 2 N plus O2 + 2 nio
19:58considering the free energies of
20:00formation of the oxide and
20:02sulfur from
20:04this the Delta G value for this has some
20:09value this will have another value these
20:13values are also available so if you
20:16substract this value from that value we
20:19will get the Delta G not for this
20:21Reaction 2 n i plus 3 by2 2 N plus 2 S2
20:26and finally if you have it we'll get the
20:30Delta G value for this now if you have
20:33the Delta G value for this then we can
20:38do
20:39the we can get
20:42the value of K because Delta
20:49G is equal to minus rtln K
20:54Prime and once we have that value of K
20:58Prime if you substitute there then you
21:01get an equation that relates PSO2 with
21:05log P2 for a particular temperature for
21:09different values of T we will get
21:12different relationships and in using
21:15that we can draw the boundaries between
21:17NIS and nio similarly you have to
21:20consider the various equilibria and we
21:23can draw these lines now obviously if
21:25you consider the equilibrium between
21:27nickel and Nel side it will depend only
21:32on P2 it has nothing to do
21:35with SO2 at all that's why this is a
21:39vertical line similarly NIS and N4 would
21:43also the boundary also would be a
21:45vertical
21:48line we have also this sort of uh phas
21:53stability diagrams or predominance area
21:56diagrams for other oxides sulfates and
21:59sulfites this is a very very important
22:02diagram because we need to know that
22:05when we take lead sulfide and we roast
22:09it to get PBO where should the furnace
22:14operate normally this is the range where
22:18the usual rooster gas composition is
22:21usually
22:22here now you know this sort of
22:26calculations gives us the limiting Valu
22:28that this is this is the equilibrium
22:29value obviously if suppose you get a
22:32equilibrium temperature of T the actual
22:35operation in the industry would be high
22:37higher temperatures for various reasons
22:40firstly to accelerate the rates or
22:42sometimes to melt different phases but
22:45we still want to know the limiting
22:47values that will tell us exactly where
22:52the operation should be carried
22:55out now let's go back to some
22:59things I mentioned earlier I had talked
23:01about roasting roasting is you take a
23:05sulfide and you convert it into an oxide
23:08or a sulfate or whatever it is there are
23:11so many
23:13reactions this is a roer which is very
23:17commonly used to very commonly used in
23:19the industry
23:20called he rooster yet there are many
23:24many hearts and the there's a central
23:27shaft on on which these
23:29hearts are circular hearts and the whole
23:33thing
23:35rotates the the the hearts rotate and
23:41the feed from the top actually goes from
23:45one he to
23:47another in this fashion and see these
23:50are the
23:51teeth which sort of go through the
23:54charge so the charge flows from one to
23:57the other one to the other other one is
24:00stationary this part is stationary these
24:02are the attached to the centrer shaft
24:05which are rotating and the these are the
24:06teeth the teeth kind of uh make the
24:10charge flow from one to the other so the
24:13by the time the
24:15sulfide has come from here to the bottom
24:19you get a calci you got the
24:22oxide now a lot of experiments on this
24:26showed that actual Ro in
24:29reactions took place when the particles
24:32are falling from one Earth to the other
24:35immediate by people had taken samples
24:38not so much as when they were on on the
24:40earth and they were being stirred or
24:43they were they going from one place to
24:45another so that gave an idea of this
24:49flash rooster which is what is used in
24:53the industry now there the idea is that
24:56finally divided sulfide concentrate
24:58air is allowed to drop through a
25:01combustion chamber maintain at a certain
25:04temperature and this is discharging
25:06device you get the roast straight
25:09away what will be the nature of the
25:11product will depend on of course the
25:14partial pressure of oxygen and of
25:16temperature etc etc but this is a very
25:19rapid
25:20process that the multiple he rooster
25:24will take very long time because the
25:26central shaft is rotating and the charge
25:29is coming very slowly from top to the
25:31bottom but here it is practically almost
25:35in instantaneous of course there's a
25:37height of the chamber they simply drop
25:40through a hot chamber and immediately
25:44roasted and they taken
25:46out now this give rise
25:49to another idea to which outc
25:54come that for that I have to come to the
25:57concept of Smith melting now I had
25:59mentioned in roasting there's no melting
26:02you're charging you're heating a solid
26:05charge sulfide and it becomes a solid
26:07oxide but in the system if you bring in
26:11a reducing agent like carbon and some
26:13fluxing agents like calcium oxide and um
26:18quartz you create a slag phase that
26:22operation is called melting where you
26:24create like in Blast Furnace you create
26:27Slack
26:29and you create metal and there is
26:31separation between one and the
26:33other we have smelting in case of
26:36sulfide do also but
26:40sulfides do not give metal straight away
26:44what they do is a phase called
26:48mat and we have an operation called
26:52smelting where the sulfides are first
26:56partly roasted and and then the whole
26:59charge reduced there is a slag phase and
27:02we do have a separation of slag from the
27:06metallic values but the metallic values
27:09stay basically as a mixture of sulfites
27:12in the case of copper ion sulfide and
27:14copper sulfide we call that matte so
27:17there is matte slag separation so
27:20smelting can be of two kinds metal slag
27:23separation or matte slag
27:26separation this discuss in detail in
27:29when we come to extraction of
27:32copper now this idea of flash smelting
27:37has come flash roasting has come into
27:40flash smelting also there the idea is
27:44the sulfide is particles are dropped
27:47into a chamber hot chamber controlled
27:51oxygen partial pressure along with the
27:54fluxing
27:56material and there is auxiliary Fuel and
27:58oxygen to maintain temperature so while
28:02in flight and then later on it not only
28:06produces the caline but caline also
28:08reacts with reducing agents and you end
28:11up with a slag and mat so this is also a
28:15very rapid process flash M flash
28:18roasting is where there's no reducing
28:20agent you simply caling it very quickly
28:24and Flash smelting is where in a Flash
28:28you are caling then you are also fluxing
28:31out on the gang and you're producing a
28:33slag and a mat two separate
28:40phes now I have been talking about this
28:44word slag quite frequently so I would
28:46like to say a few words about what is
28:49slag and how is slag made but before
28:52that let me give you one or two small
28:55examples of how thermodynamic calcul
28:58ations are applied in the case of
29:00reduction reactions now here is a small
29:03problem find the vacuum required to
29:06reduce
29:07nb205 by carbon at 1200° K now if you
29:12look at the lingam
29:14diagrams this
29:16reaction reduction of
29:19nb205 by
29:21carbon will need very high
29:25temperatures because if you write in
29:27this reaction
29:28nb205 Toc
29:31this normally if you look at the lingam
29:34diagrams what we are we are plotting
29:38standard free
29:40energies it is
29:42for a given values of partial pressure
29:45of Co it will need high temperatures but
29:48this reaction obviously can
29:51be sent forward if you find ways to
29:55reduce this carbon monoxide by applying
29:58vacuum so the problem is
30:01this can we reduce NB 203 to5 by carbon
30:07at a relatively low temperature of 1200°
30:11K which is 900°
30:13C normally it will not happen but
30:16suppose we apply vacuum what kind of
30:18vacuum would you need it's very easy we
30:22have the free energy change values for
30:24this
30:25reaction standard free energy change
30:28change we have the standard free energy
30:30change for formation of
30:33nb205 by difference we get the standard
30:36free energy change for the reaction that
30:39we are studying which is reduction of
30:42nb205 by carbon to form metal and
30:46Co at 1200° scale we'll get the value of
30:51delta G not as 68.85 Kil calories which
30:56is minus RTL and k k is the equilibrium
30:59constant the equilibrium constant we can
31:03obtain by putting the right
31:06values
31:0968.85 RT
31:11value and taking 2.33 log K if WR it
31:17comes to 4575 in
31:201200 K is represented by this
31:24expression and so we end up with an ex
31:28expression from which we can calculate
31:31that the equilibrium partial pressure of
31:33carbon monoxide would be 3 into 10us 3
31:38atmosphere for this
31:40reaction now obviously which is equal to
31:432.28 uh mm H this is the equilibrium
31:48partial pressure of Co for this
31:50reduction reaction so if we can maintain
31:52a vacuum better than 2.28 mm Mercury
31:57then we can make this reaction happen at
32:01temperatures as low as
32:07900° we use thermodynamic data for
32:12analysis of thermit reactions which
32:15refer to reduction of an oxide by
32:17another
32:18metal like right in the beginning I had
32:21said there is a process called thermit
32:23process where f23 is reduced by
32:26aluminium exothermically to produce
32:29liquid ion all it needs is you take f23
32:33powder and aluminium powder and
32:35ignite immediately the reaction starts
32:38the temperature is so high that
32:40everything becomes molten even l23
32:42becomes molten and the molten iron will
32:45go into the cracks in rails if you want
32:49to uh repair those
32:51rails now in this cases as I have shown
32:54here after initiation temperature rises
32:56melting Rises
32:58enormously to melt everything l23 can be
33:01easily slagged off means if you put some
33:03flux it will it will go out very easily
33:07and the
33:09rest are not volatile so the reaction is
33:12easy but sometimes even such reactions
33:15are okay in theory there are a lot of
33:17problems in practice for example suppose
33:19you want to reduce
33:21tio2 solid by
33:24calcium this reduction needs
33:28high temperatures where calcium becomes
33:30a gas so the reaction has to be written
33:33like this it has to be in in a closed
33:36chamber TM dioxide being reduced by
33:40calcium Vapors to produce Co which is
33:42solid which can be slagged up and
33:44titanium which is solid now titanium
33:48melts at 16
33:4970° calcium boils at 14 92° cenr calcium
33:54melts only at
33:56260 so it is a very complicated reaction
34:00because we are not able to get liquid
34:02phases very easily so thermodynamics
34:07gives us guidance about what should be
34:10the temperature uh etc etc but then in
34:14practice we need to do a lot more
34:17things this we would can discuss only
34:21when we come to uh extraction of
34:23individual Metals now before we before I
34:27end this
34:29lecture I want to say something about
34:32structure of
34:39slags generally in an
34:42ore we have metallic
34:48values means minerals and we have gang
34:52gang means things we do not want like
34:55silica alumino silicates other things
34:59now when we do
35:01smelting by Say by reducing agent and we
35:04add a flux the flux is
35:08limestone quartz Etc the whole idea is
35:12to produce a
35:16liquid silicate
35:21phase which takes out many impurities
35:24which separates out from the metal and
35:26so that we have a clean separation
35:29between slag and metal how do
35:32we create a fluid
35:36slag to that we have to go into a bit of
35:39discussion of silicate
35:43structures pure
35:45silica sio2 although it is written like
35:50this it is not made up of molecules of
35:54ao2 silicon actually is
35:58bonded
35:59to four oxygen
36:02atoms this is
36:05the basic unit and many such units
36:09attach themselves to one
36:12another like
36:23this so in
36:26si2 there
36:28are these tetrahedral things attached to
36:32each other so that the entire mass is
36:34actually one molecule in theory and that
36:37is
36:38why molten silica is very viscous
36:43because it it is the flow unit is very
36:45large of course if one raises the
36:49temperature too high then many of these
36:51bonds will break thermally so we make
36:54smaller and smaller flow units
36:58but there is a very clever way we can
37:01make silica less
37:04viscous and this
37:06is if let us
37:11represent the basic silica structure by
37:15two dimension in two Dimension these are
37:18oxygen
37:20atoms if we add to this silica which is
37:23an acid oxide a basic oxide like calcium
37:26oxide which gives
37:32calcium this
37:34oxygen goes and breaks a silicon oxygen
37:39Bond so it splits we from a big flow
37:44unit we create two smaller flow
37:47units and then it becomes less viscous
37:52so we can represent the
37:54reaction in a in a thing like this
38:05this will happen with metal oxide you
38:06have
38:14added we have broken it
38:18into and the metalon is hanging
38:22around now in
38:25silica the more more calcium oxide you
38:29add the more fluid it becomes because
38:32the more bonds are broken the flow units
38:36become smaller and smaller and
38:38smaller but there's a limit to that this
38:42once you have broken it down to the
38:45smallest unit which is
38:51Si you cannot break it any
38:54further so we have a long chain or a
38:58complicated thing you keep on breaking
39:00and finally this is the smallest unit
39:03and this happen when
39:072/3 calcium
39:10oxide and there is
39:131/3
39:14S2 it's very easy to understand why this
39:17is so because from stomri region if you
39:22have added sufficient amount of oxygen
39:25sio2 has to
39:28go down to the smallest unit now the lot
39:31of work has been done on structure of
39:33silicates and it it's a very vast
39:35subject I don't want to go into that but
39:38you should understand that there is a
39:41concept of
39:45acidity and basicity in
39:50slags acid slacks are where
39:55silica is on the highest side basic
39:58slids are where calcium oxide is on the
40:01high side why this is a base this is
40:04called a base because it
40:07donates oxygen it donates
40:13oxygen it is called an acid
40:16oxide because it
40:20accepts oxygen for breaking into smaller
40:24and smaller unit don't think only
40:26calcium oxide is the basic
40:30oxide feo
40:33CAO MGO they are all basic
40:38oxides al23 we call it is an OTC oxide
40:42it sometimes it acts as a
40:45base that it donates oxygen ion
40:48sometimes it acts as an acid it adds
40:51oxygen depending on the
40:54composition it will be enough for you to
40:57know that if the slag is
41:00viscous it's flowing it a large because
41:04there are polymeric silicon oxygen units
41:07in that it can be made fluid by adding
41:11basic oxides like calcium oxide this is
41:13the most
41:14common magnesium oxide also will make it
41:18fluid but not all basic oxides are
41:22equally effective in reducing the uh the
41:26the viscosity of a I2 in other words the
41:30basicity of different
41:33oxides are different so we actually have
41:36a basicity scale of different oxides
41:41calcium oxide is high PG MGO is not so
41:46strong a base weaker bases are
41:49a then there are even other oxides which
41:52are even weaker there are many
41:54definitions of
41:56basicity the
41:58commonest definition of basicity
42:01is CAO by
42:06S2 and very often you'll see in pyromet
42:09as operation they will say this slag is
42:12maintained with a basicity of so and
42:16so there there are modifications
42:18required of this if there are other
42:19oxides coo plus
42:23mg by S2 is another definition
42:27commonly used in the industry some
42:30people say since MJ is not as effective
42:32as coo it should be written as two two3
42:36some people say under certain conditions
42:39the basicity index is best written as Al
42:42203 plus si2 etc etc so in all
42:46pyrometric logical
42:49operations the operator wants to know
42:52what is the basicity of the slag or is
42:54often advised about the basicity of slag
42:57because proper basicity defines proper
43:02viscosity of slag so the viscosity of
43:06slag which is a
43:11crucial parameter in pomological
43:14operations depends on
43:18temperature because the higher the
43:20temperature lower will be the viscosity
43:23no matter what slag is it because at
43:26higher temperature
43:27bonds tend to break and the other will
43:33vity so
43:36temperature should be
43:38high basicity should be high to lower
43:44viscosity or increase
43:46fluidity there are also some fluidizing
43:48agents like calcium fluide which when
43:52added increases fluidity of slag now
43:56slag basicity control is very important
44:00in the industry and there is a common
44:03saying that any many pyral
44:07operations the aim is to look at the
44:10slag if the slag is right the metal
44:13would be right this was particularly so
44:16for Blast Furnace
44:17operation in Blast
44:20Furnace you have a slag layer covering
44:24the metal
44:26layer and the slag layer is has many
44:29functions first of all it is covering
44:31the metal layer that is whatever gaseous
44:35atmosphere is there on top of the slag
44:37is away from the
44:39metal between slag and metal all kinds
44:42of slag metal reactions are taking place
44:45depending on the uh chemistry of the
44:47slag so the slag chemistry and slack
44:52properties are of vital
44:54importance a simple example I have to to
44:57go for that to Iron and
44:59Steel perhaps you know that one of the
45:03problems
45:05of slags in blast furnace was because of
45:10alumina high
45:13alumina that came from Iron ORS Indian
45:17ORS are very good but aluminia comes
45:19from more as well as some
45:21from Coke and Indian slags used to be
45:26very highly VIs discuss because of Al
45:28203
45:29content many efforts have been made to
45:32remove Al 203 for my but it's very
45:35difficult so it will end up with slag
45:38with lot of alumina high
45:42viscosity many efforts
45:44were there to find out how to bring down
45:48this l23 content you cannot simply go on
45:51adding a lot of calcium oxide increase
45:54the basicity that will increase the um
45:57volume of slag and increasing the
46:00basicity will also have effect on slag
46:03metal reactions finally the adverse
46:06effect of
46:07l23 was met by adding
46:12mgu and many blastness operations were
46:17found to be Optimum with addition of 9%
46:22NGO that took care of high alumina in
46:25the slag which was going to increase is
46:27the basicity the viscosity viscosity was
46:30brought down by m now why I'm saying
46:33these things is that you often come
46:37across terms like basic
46:42slag neutral
46:46slag acid
46:50slag generally by basic slag it will
46:55mean
47:01CAO by
47:03sio2 more than
47:072 Mo in this case coo by sio2 neutal
47:13slag will be two acid slag will Mo
47:18coo by S2 less than two I'm writing coo
47:23in in case there is MGO that has to come
47:25here again so you can say m Mo by
47:31A2 basic oxides and acid oxide they
47:35ratio governs the basicity neutrality or
47:39acidity of
47:46SL I think with that it's time
47:50to wind up this
47:54lecture I have mainly discussed
47:56discussed here the use of eling gam
48:00diagrams to understand reduction of
48:03oxides by
48:04carbon and reduction of oxides by Metals
48:08which
48:10form more stable
48:13oxides I've have talked about
48:15calcination roasting
48:18smelting and when we talked about
48:21smelting I mentioned that in
48:24smelting you have to have a slap phase
48:28in contact with either a metal phase or
48:31a matte phase a matte phase is not a
48:34metal phas it's a mixture of
48:38sulfites but when you have a
48:40slag the first Criterion for this flag
48:44must be that it should be
48:47fluid that it should be
48:50easily taken out it should be easily
48:53Tapped Out it should separate out very
48:56easily
48:57from the metal phase so we have liquid
49:01liquid
49:03separation
49:05sometimes it is also important to
49:08control the chemistry of the slag just
49:11making it fluid is not enough it should
49:14have such
49:15a chemical
49:18composition that some reactions that are
49:21favorable to produce a purer metal are
49:25possible
49:27like maybe the composition will be such
49:32that it will this lag phase will absorb
49:34sulfur or phosphorus and other things
49:37that you do not want in the metal
49:39phas so not only fluidity is important
49:44but chemistry is also
49:48important but slag phase is a vital
49:50phase and slack
49:54chemistry physical properties of the
49:56slack
49:57they form a very important part of
50:00pomological
50:03operations I think I will end the
50:06lecture on pyromet reactions right now
50:10in the next lecture I will move into
50:14hydrology and try to discuss some of the
50:16basic principles of hydrology thank
50:25you
50:32[Music]

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