Fundamentals and Science-Based Foundations of Steel Making

so we now will start talking for few next lectures on the science based foundation of steel making

because steel making involves a lot of fundamentals

and no course in skill making i think can be appreciated unless the

fundamentals are known of course i'm not going to give you a complete coverage of the topics but it is

important for us to revise a few key concepts and show you that what are the fundamental components

or the science based foundation of steel making which we need to remember all the time

as we go and address still making problems be it process analysis process design optimization and so on

now steel making is more than 150 years old uh when sir henry bessemer first

made steel but by blowing air uh into a malt into into a clay crucible containing molten pig iron

and since the days of sir henry vissimar just making and treating making treating and

shaping up steel has undergone remarkable improvements today if you go to the bof shop for

example you are talking about 92 to 94 percentage of yield

in bof so that means if you charge 100 kg of uh iron

into pig iron in into bof you are going to get something like 92 to 94 kg extremely efficient process

in the bof the refining indus ld converter and combination blown reactors the refining

is so fast so rapid that you can get something like 2 to 3 kg of carbon per second that is the rate of

carbon removal we are talking about if you go and visit the

casting bay for example in a thin slab casting

shop today you will find that you are talking about 6.5 to about 7 metric tons of

steel being produced per minute per strand

so enormously efficient process a lot of developments went in the last 150 years and we have been able to reach here

achieve all these marks because we have relied at all stages heavily on fundamentals the application of

fundamental knowledge is extremely important in steering the technology further because

there is going to be many many challenges coming up in the future days for still making

for example stricter environmental regulations burning of fossil fuels

specific energy consumption these are recycling these are some of the fronts for the steel makers will be more and

more concerned to improve and there we will see that

the science-based foundation will come to our rescue and we will have to increasingly apply the fundamental

knowledges in order to find the ways out now there is a huge knowledge based

foundation of skill making this i call as the knowledge base foundation i am still making i will tell

you what is the components with the top down approach so if your knowledge base foundation

of steel making is good what you can do you can do very well

process analysis design control

optimize optimization and if you can do all these things

then what is the net result the net result is enormous you are talking of you can

develop new technology people are looking for direct still making technology

people are looking for technology where they don't have to use coke to produce iron

you can have new products every day you see new alloy systems are

being developed new products are being marketed and what was not possible 20 30 years

back these are possible today i mean for example i have great steel who heard

about i have great skills 30 40 years back today you can

roll shifts i have great steels and make thin automobile thin sheets ready for automobile usage

and these have been the development of sustained application of knowledge you can reduce

specific energy consumption specific energy consumption

you can improve you can decrease emission and make the process more

environmentally friendly today steel industries look down open in view of the global warming that we are

one of the biggest polluters as far as the greenhouse gases are concerned and we got to

tighten our belt and there we will find that in the context of direct emission decreasing emission

reducing specific energy consumption process analysis design will play a very big role and this is based on the

knowledge base foundation and what does this knowledge base foundation comprises of

it has four different components these are very important

still making we say that this is the manufacturing still making or the alternative

that i can put here in the box is the manufacturing then we have

the science base foundation i have

modeling and i have measurements so manufacturing process

science based foundation modeling and measurements

they constitute our knowledge base foundation you cannot omit any of this so you have

to know the steel making technology the process is still making processes very well you have to know the science based

foundation of steel making when you top up the science based foundation let us elucidate what does it contain the

science based foundation of steel making contains for example metallurgical thermodynamics

that is why before one a student or one takes a steel making course it is so important for him or her to know

about metallurgical thermodynamics which is introduced at the second year level or third third year beginning of the

third year level in engineering program we have to know about metallurgical kinetics

and kinetics basically comprises of two components will sub divide kinetics into red

phenomena and chemical reaction rate phenomena again can be

divided into three sections we can say fluid flow heat transfer

and mass transfer so you see the science based foundation of steel making is extremely vast and on

top of that i can say that multiphase or gas liquid interaction

basically i am talking of gas metal interaction because

these interactions as in the case of a basic oxygen furnace these are reacting interactions on the other hand in the

case of an argon molten steel interaction in a ladle these are inner interaction there is no

chemical reaction so we have to know thermodynamics well we have to know

metallurgical kinetics well and there we see that it contains of various subjects and each of these

can be a topic and their own and in many engineering disciplines like mechanical engineering chemical engineering we have

full-fledged one semester courses running on this so you can imagine the breadth of the science

and it is here that you get the largest inside unless your science base from the foundation is good you will not

be able to build models and if you don't build models you will not be able to make any

meaningful interpretation of the measurements that you might intend to carry out so

therefore we have to know these subjects well and most of you during your second year

stages of metallurgical engineering program have been exposed to thermodynamical thermodynamics also a

subject titled differently maybe transport phenomena or heat transfer and then

many of you may not have really done an exhaustive treatment of metallurgical kinetics and

metallurgical kinetics which comprises of so many things but in an isolated way you are already exposed to it so i

intend to revise this very briefly because as i said they are subjects on their own

and when i'm delivering lecture a few lectures on the topic of steel making perhaps it will beyond

any possibility to address these subjects on their entirety in the context of

steel making but nevertheless it is important for us to review certain key concepts because we can before we can

advance into the discussion of various other topics basic oxygen skill making little metallurgy continuous casting

etcetera because they will there will see extensive application of the fundamentals so before we can address

all those advanced topics or relevant topics it is worthwhile for us to briefly examine the key concepts of

thermodynamics metallurgical kinetics that are relevant to skill making so let us

talk very briefly about the key concepts in metallurgical finite kinetics first methodological thermodynamics

this is the first subject as i have mentioned but in the context of still making we want we apply metallurgical

thermodynamics particularly to compute or calculate the state of equilibrium of the system so in many instances we would

like to see whether this process has reached equilibrium or not for example we can say carbon oxygen reaction is

taking place in a basic oxygen still making converter is the reaction close at equilibrium how far it is from the

equilibrium if the reaction when the reaction reaches equilibrium at a certain temperature and pressure what is

the composition of the system in terms of its carbon concentration and oxygen concentration so all these equilibrium

calculations we wish to calculate we wish to perform in the context of steel making for example i have

a slag and i want to find out that how much of iron oxide will be there in the slag in

equilibrium with a certain given concentration of oxygen so i will carry out an equilibrium calculation and i am

going to find out that what is the level of iron oxide content based on that equilibrium concentration it is not

necessary that the prediction is going to calculation is going to match the

measurements it is only going to match the measure calculation is going to match measurements provide the process

has truly reached the state of equilibrium thermodynamics for example as you all know it predicts about the

feasibility of a process okay thermodynamics it depends on the initial in the final

stage it does not depend on the path okay so to study the feasibility of a process we apply thermodynamic

fundamentals so whether carbon will oxygen will react with carbon or not the feasibility of that process can we can

examine but if the process is occurring then it is away from the equilibrium because

at equilibrium there is no net rate of the process it is stand still okay so thermodynamics does not cannot talk

about rate it can only talk about the feasibility of a process it can talk about the state of equilibrium of a

system when you are when the process is occurring spontaneously from one direction to from left to right for

example carbon and oxygen reacting to form carbon monoxide there comes the question of rate that if it is reacting

and the reaction is proceeding in one direction if it is if it is moving towards equilibrium but not yet

at equilibrium then the question comes at what rate it is happening you know i have quoted that 2 to 3 kg of carbon per

second that is the rate of carbon elimination that means if you are talking of carbon oxygen reaction

okay the reaction obviously is going in the forward direction and certainly it is not

you know close to equilibrium which we represent basically as a reversible sign now when

you are talking of two to three kg of carbon per second of removal that means we are talking about the reaction being

going in the forward reaction now what is the condition for the process to go from the left or the reactant will be

converted into a product or in terms of a system generalized statement we can say that

free energy change is going to be less than 0 for the reaction

in that case we can say this is going to be going from spontaneously from left to

right so delta g essentially means that the product has a lesser free energy than the reactant and therefore this is

a negative quantity and the free energy becoming negative gives a essential condition for spontaneity of a process

so this criteria we have applied in many many many cases now let us see look at

that when you have a reaction like for example and let us say that the reaction

is occurring reversibly that means it is it is close operating close to equilibrium

and then we say we have so this is the chemical reaction that we wish to represent

so we have a forward reaction we have a backward reaction they are competing with each other and soon the process is

going to be under equilibrium condition in this particular case okay

the expression that we have seen or we know formula knowledge of thermodynamics is

delta g the free energy change for dxm is the standard free energy change plus the gas constant

logarithmic and then we say it is activity of l raised to the power l activity of m raised to the power m

activity of a raise to the power a and activity of b

and we know that if the process is occurring close to equilibrium in

that case this that is the product on the reactant will have

free energies which will be identical so there will be this quantity on the left hand side will be close to is equal to

0. now this quantity this is this is called as you have known mind it i am not trying to introduce all

these concepts in great detail because i am expecting that you know the subject you have been exposed to thermodynamics

we are just revising it very quickly so that we find these concepts to be later relevant and we can remember them often

as we advance or move on to different topics in still making so this is as you know this is

activity quotient and when the process is at equilibrium

the chemical reaction is in equilibrium as i have tried to equip then we say that

delta g naught is equal to this we have applied frequently to solve a variety of problems

in thermodynamics particularly you know inst for steel making systems for solutions and so on

and this term these are now we must denote that this this this

concentrations refer to the concentrations at this equilibrium state so i one should normally represent this

with a suffix suffix like a l capital l comma e a small m capital m comma e and so on to indicate

specifically that this when you when you write delta g is equal to zero at equilibrium delta g being equal to zero

so the equation translates to this particular form and this now refers to the equilibrium

concentrations and as i have written here these are essentially activities okay now

this concept the delta g and delta g naught have been extensively applied and what is the

delta g not essentially implies delta g implies it is the standard free energy change okay

when the elements or the reactants and the products are assumed to be in the in their standard state okay which we say

as a one atmosphere pressure and that is the condition at which ah delta g naught is

going to be evaluated in terms of this now in order to evaluate delta g naught we

must understand that we have to have proper activity versus composition relationship

okay so we must know because we will not be able to directly measure activity okay it is a activity is a measure of

chemical potential in the system so what we can do we can measure the concentration

based on that we will be able to evaluate these activities and therefore there are

various models which are available to us and which will be which will allow us

for example to convert activity in terms of relevant known parameters for example if i say that well the system that i am

talking about you know in the context of particularly solutions because we talk of steel

making molten steel and slag can be visualized as solutions of multi-component system and if i say that

activity is roughly equal to mole fraction okay in that case these activities can

be converted into corresponding mole fractions and mole fractions can be

easily converted into weight fractions which are measurable quantities through any standard chemical devices okay

before i talk about this activity composition relationships whereby we can convert these equations

or expressions in a more meaningful form let us explain

the context you know the relevance of a parameter called the free energy of formation

so many times you will find that

in the literature free energy of formation of compounds are extensively used for example if i start

with a compound and you will require in order to carry out that what is the difference

in the free energies you will you will have to know that you know the compound that you are taking uh what is the free

energy of formation for example i can have i can write down say a metal

plus oxygen and then i say it is mx

y so therefore it is going to be 2 by y and this is going to be 2 x by y

so this is a stoichiometrically balanced equation as you can see here ok let us denote it

now so this oxide has i can use for example what is the

free energy of formation of these particular oxides what is the relevance of this because i

may use these oxides for in other calculations also for example when i say that well i take iron

oxide and i want to find out the free energy change for this reaction co2 plus fe

then ratio will come that what is the free energy of formation of this compound

then only i will be able to find out the free energy so i will have to find out the free energy of formation of iron

oxide i have to find out the free energy of formation of carbon monoxide i have to find out because there are compounds

so only when i know the free energy of formation of these compounds then only i can calculate the net free energy change

for this particular reaction so we have free energy of formation of the sulphides we have free energy formation

of the chlorides with the free energy formation of oxides so these are very standard informations which are needed

for carrying out many pyro metallurgical calculations and i am going to just explain ah you

already know i am going to highlight this free energy formation in the context of oxides

which is known as an where it is stimulated is known as an oxide element diagram to which all of you have been

exposed with now here for example i have written this x and y are the stoichiometric compound so we you can

imagine that if i am talking about aluminum then what is the value of x so for aluminum we know the corresponding

oxides is l2 o3 okay so it turns out that x is is equal to two and y is is equal to three that's what i am talking

about if it is iron oxide in that case you can also clearly see that what will be the value of x and y for all other

oxides you should be able to visualize now free energy of formation standard free energy of formation standard state

that means the metal is in the pure state the oxide is in the pure state as a function of oxygen partial pressure

has been measured by various investigators and for the sake of our convenience so that

we can carry out thermodynamic calculations very easily these have been tabulated before us and

in the lingam diagram if one assumes that well delta g naught

formation can be represented in terms of a and b t now you can see that

this is a gibbs free energy expression gibbs free energy functional is

expressed in terms of combined expression of first and second law as enthalpy temperature and entropy and if

you assume that so therefore i can say that well if you assume that h naught standard

enthalpy and standard entropy of formations are not independent not dependent on temperature in that case

the free energy of formations of oxide materials can be expressed conveniently

in terms of this particular equation so that means it indicates that the free energy of

formation plotted against temperature will yield a straight line in which a and b are constants and that's what you

can you actually see when you look at the lingam diagram where you have a number of straight lines drawn and i am

going to now draw it very briefly and show you that what are the relative positions of the lines in the figure and

what does that really indicate so at equilibrium when delta g is equal to 0 i have shown that delta g naught is equal

to this parameter and this as you all know is called the equilibrium coefficient or equilibrium

constant and if i look at now this reaction

operating close to equilibrium i should also be similarly write down the equilibrium

coefficient and that is going to be activity of metal oxide raised to the power

2 by y and then activity of oxygen and if you assume ideal behavior then i am going to

say that this is equal to partial pressure of oxygen in the system and then we have activity of metal

raised to the power 2x since we are talking of standard free energy change so that is why we will the

standard state is the pure for metal and metal oxide so therefore their activity is going to go to be equal to

one so this is equation will going to come out to be partial pressure resistance

so therefore now one can measure the partial pressure of

oxygen and delta g naught is exactly going to be is equal to

rt ln po2 the negative sign cancels out and that is what it is and therefore we can

now plot study the equilibrium of this reaction at 800 degree centigrade 900 degree

centigrade 1000 degree centigrade measure the partial pressure of oxygen and on the basis of that we can compute

the delta g naught value because r is known temperature is known and

partial pressure is the equilibrium partial pressure that we have measured through a adequate

measuring device so we should be able to now at various temperature we should be able

to plot so temperature is measured temperature at which the equilibrium has

been obtained corresponding delta g naught is estimated based on the measured po2 value and we should be able

to draw now a straight line between the two because

of the fact that delta h naught f and delta s naught f are largely independent of temperature and if you look at it in

such plots you will find that there are straight lines and only the slope of the straight line changes when there is the

phase change okay for example we may have started with a solid but then at certain some

point of time the solid may melt so otherwise when the phase is maintained continuously unique in that case you

will see that there is a straight line and this straight line behavior for different metal is going to be

different in the system they need not be parallel as i have tried to indicate in fact there is a

straight line there is a line which is horizontal in the lingam diagram and there is a line which is actually with a

negative slope these are the two peculiar lines in the oxide lingam diagram so this i say as delta

standard free energy of formation of oxide which is actually

equal to rtl and po2 so now these are the negative values for

example this could be something like 1200 so kilo calories per mole and this could

be about minus 200 200 so more negative value is the standard free energy formation so

therefore this line indicates that the values are here standard free energy change values are far more negative than

this value which essentially indicates that a very little amount of oxygen will be able to oxidize this metal then it is

possible for this particular metal so a very low lines in the figure for example will be for

two calcium plus oxygen and note it so that we can compare the relative affinity of elements or the stability of

oxides at a given temperature all this is done for per mole of oxygen and this is

2 cao which essentially indicates that calcium oxide is very stable

oxide on the other hand if you look at iron oxide for example fe 2 fe plus o2 it is going to be

relatively so this are all at equilibrium so therefore i should

maintain and calcium there will come magnesium there will come aluminum there will become

titanium then will become silicon and if you go up and then you will see lead oxide tin oxide etcetera which are

relatively weak oxides and this line is a cco line and this is the coco 2 line we must you must be knowing that why the

slope of the lines are different because of simply the difference in the entropy values for this reactions and

with the standard entropy that causes a opposite slope for carbon monoxide and carbon dioxide lines

this you must be knowing so we have similar figures which are tabulated and these figures give us important

guidelines as far as pyro metallurgical processes are concerned for example

at this particular point when carbon carbon monoxide line intersects this particular line for example it tells us

that it should be possible to below this line the dotted line carbon carbon monoxide

line cook by smaller position so therefore lower position so therefore above this temperature carbon monoxide

is far more stable than the oxide which is corresponding to this particular line so therefore it indicates that beyond

this temperature if you go beyond this temperature carbon monoxide will be able to reduce this particular line for the

oxide itself and now we see that well carbon monoxide line intersects the fe fu line at 800 900

degree centigrade but it intersects the alumina line for example at more than 2000 degree centigrade and that is why

by looking at this diagram we can immediately conclude whether carbohydrate reduction of a metal oxide

is possible or not similarly we have sulphide diagrams also per mole of sulfide we can

find out that what is the standard free energy change where we can take metal take sulphur sulfur gas and then form

metal sulphide and exactly the same sort of graph can be produced and which is known as the sulfide illinois diagram

which gives us important clue regarding the pyrometallurgical or the smelting sulfate smelting

processes now coming back to this is a for example you see how

smartly i have been able to convert i made one assumption that well activity of oxygen is equal to partial pressure

of oxygen okay and ideal behavior has been ideal gas law has been assumed and thereby i have been able to convert the

oxygen activity into partial pressure so i have assumed an activity composition relationship here okay and now

we have various models particularly for steel melts and slag one model for particularly solutions when you are

talking of activity you know we are talking about activity

of sulfur and steel melt so basically we are talking about activity of a species in a solution so we have various for

various solutions we have various models available and as you all know

raoult's law is there so if you have concentrated solution in that case you can say activity is is equal to actually

activity of i is is equal to mole fraction of i

and then if you have departure from raoult's law you introduce what is

known as activity of y is equal to h introduce an activity coefficient and into x i

okay so also you all know for example iron and manganese iron and manganese they occupy very

close position in the periodic table the atomic numbers are similar crystal structures are similar so when

you mix iron and manganese they form ideal solution and there you can say that look if

that if the wet person if the mole fraction of manganese is 0.5 in that case its activity is also equal

to 0.5 because iron and manganese are ways but there are only few solutions

few systems which are truly ideal in nature most of the systems are non-ideal and departure from the raoult's law is

expressed by what is known as the hendricks law where we say that activity is equal to activity coefficient and

this can enhance the

activity can get enhanced bigger than the mole fraction or can get smaller than the mole fraction depending on the

value of gamma either activity coefficient if activity coefficient is greater than one then we say it is a

positive deviation that means a mole fraction is five in that case the activity will feel like it is point

seven point eight or something like that because activity coefficient is greater than one so attribute equation greater

than one causes positive deviation and activity coefficient less than one so we have

this is the raoult's line rolls law line activities is equal to mole fraction with an angle of 45 degree what is there

it is activity of i it is mole fraction and this is the negative deviation for example this is the

negative deviation and this is the positive deviation from the raoult's law

and that deviation behavior is explained on the basis of this

expression now we can have

different type of expressions also depending on what we choose as the standard state in this

case the standard state refers to the pure element okay but we can also define a standard state for example uh we can

use henry's law in the white percentage standard scale so one weight percentage standard scale

this is also very popular scale and finds extensive application in ah dilute

solutions and there we say that activity of a species i is

approximately equal to what percentage i and therefore activity becomes equal to unity when but percentage i is so so

when weight percentage i is is equal to 1 then the activity is going to be is equal to so this is the

henry an activity in one weight percentage scale this is with the interpreter as the henry activity in

concentrated or in the pure state when the elements are in their original solid state or

impure state and this is with reference to one weight percentage standard state so there are so the free energy is

standard free energy of formation if you calculate for example with respect to 100 standard state or with respect to a

pure element as the standard state these values are not going to be related so they are going to be some difference by

some constant amount when the thermodynamic parameters are evaluated with respect to different standard

states so one standard state it is like a frame of reference you know you have you can do the coordinate transformation

from one frame of reference you can shift the origin and go to another frame of reference so exactly the same thing

is applicable here it is based on only our convenience that we use sometimes you know pure element as the standard

state or sometimes ah you know pure compound is the standard state or sometimes one bit percentage as this one

so this will essentially tell that well you know that this activity is equal to one so this is

mole fraction is equal to one and this is activity is is equal to one so this is one and this is one so one

one activity so mole fraction is equal to or you can say it is not mole fraction but weight percentage so this

point is actually one weight percentage and this is actually activity is equal to

one so at the pure state

we have reference with the with with pure elements or pure

compounds we can have one standard state defined and with respect to one weight percent standard state we can have again

the same kind of an evaluation now if weight percentage is equal to 2 in that case what happens if henry's

lowest or weight up to that the activity doesn't become equal to 2 it becomes equal to 1 because that is the maximum

value of activity that is actually possible now this assumes that up to 1 weight

percentage henry's law is weighed in that one percentage standard scale but if it deviates from that in that case we

can introduce a parameter which is weight percentage of high

and this f i actually given in terms of the composition of the

mult and this is multi-component dilute solutions which have lot of

impurities and lot of elements present this sort of an analysis is very very meaningful in the sense that where we

can take into account uh the you know activity of ah influence of other elements on the activity of

the species for example we can say that well there is one e i

b into weight percentage b plus e i

k with percentage of k so this activity coefficient

is given in terms of some e i's which are known as the interaction parameters again these interaction parameters are

tabulated in thermodynamic books this tells us e i b into what percentage of b so therefore it essentially tells us the

influence of b on the activity coefficient of i now you must understand that when i have iron

and sulphur for example okay i have an activity i can measure the activity of sulfur

now when i introduce carbon into it so it becomes in a free body problem so there is carbon sulphur interaction

carbon iron sulphur iron interaction and carbon iron interaction so as a result of which in a

two component system iron and sulphur whatever is the activity that i measure in the presence of carbon the same

activity will not be measured so therefore we understand that the introduction of a

third element will either increase or decrease the activity of the second element and therefore all the solutes

present in steel melt will have for example influence on the activity of a given species for example if you take

the activity of oxygen okay so the activity of oxygen will be dependent on the activity of sulphur on the

activity of phosphorus how much phosphorus is there how much silicon is there how much manganese is there so on

all these things and that is dictated by f i and the dependence of f i in terms of

the weight percentage of various spaces so you can interpret this to be as oxygen i is oxygen here and this b may

be silicon k may be aluminum and so we have a multifaceted influence of all the solute elements

over any element and thereby its activity can be either increased or decreased now these are these

essentially tells the influence of b on i now we also have

parameters called ei which is called as the self interaction parameter okay and the self

interaction parameters sometimes may become zero and sometimes may not become is equal to zero but they nevertheless

come into the picture you know when you formulate fi in this particular form so

this parameter can be known okay but

for example if it is if there are self interaction parameters then the end i can know that how many how

much percentage of manganese for example is there and then i can say that this these

values are known from these are experimental values available to us

values which are available to us so these are known and these compositions b k

etcetera can be known but as you see here that f i itself is a function of w i so therefore when you solve this

kind of an equation we get an equation of this particular form

in which a naught is is equal to a constant because the log term is there log f i so

therefore we can convert the equation you know in terms of this unknown parameter so x we can interpret

as the weight percentage of i x plus log x i which is known as a transcendental equation you cannot solve these

equations analytically so you have to use computational routines and you can imagine that so many such equations are

going to be there in multi component system one equation will be there for oxygen another equation will be there

for sulphur another equation will be there for manganese and you know lot of activities that you are going to

calculate and then as a result of which you know you will you may carry out many other calculations for example you can

say i have aluminum the deoxidation calculation aluminium plus

oxygen giving you l2o3 now if you assume that there is no

formation of any slag just pure alumina forms ah there is no silica there is no

calcium oxide so pure alumina is there okay there is no reaction taking place simultaneously so aluminum reacts with

oxygen dissolved oxygen and produces so we can say dissolved oxygen two times

dissolved oxygen and three atoms of aluminum producing one ah molecule of l two o three and if

this is formed in the pure phase there is no ah

formation of any compound then we can say that well the activity is is equal to

l two o three is equal to one and then i can write down that the k equilibrium is actually

activity so this is equal to 1 and then i can say this is going to be hendrian activity because it is dilute solution

okay it does not obey so this is h a l and then we have h o

and then this is equal to three and this is two if there is an indication

that up to you know whatever oxygen is there below one weight percentage that you all

know okay as i have indicated it is about the saturation limit is about 0.20 percentage so we can say that well as an

approximation we can say that this is equal to weight percentage aluminum

cube and weight percentage oxygen square now free energy of formation

of l two o three is available to us note that this free energy formation and the free energy of formation in oxide in in

the aluminum diagram are not going to be identical because here the stage standard state is different they are the

standard state was pure aluminum but here the standard state is one weight percentage so the value of free energy

of formation for this equation cannot really be directly obtained from the lingam oxidal income diagram this is the

point because we have changed the standard state so this sign is very important for us this is an approximate

sign otherwise if there is a departure we should have introduced the parameter f i into this particular calculation and

it should have made the calculation it little bit more tedious if i substitute in this particular form in that case i

should be able to use a simple calculator and solve it but the moment i introduce

h i values here h a l and h o the conclusion becomes complex we have to take into account manganese

concentration sulphur concentration silicon concentration and as a result of which we will require

you know computational routines to solve because we generate you know equations of this particular form which are not

normally which one cannot really solve by analytical means so therefore we have to have today you know large computer

routines in order to calculate the state of equilibrium give us the composition give us the phases concentration

etcetera many computational packages are available so maybe you have heard

about this software which is called the fact a canada based software for thermodynamic calculations

ah facilities for ah chemical thermodynamics ah

calculation of chemical thermodynamics it is chemical thermodynamics then we have thermodope

this is another software which is available thermo calc

which is another software there are lot of software platforms which are now available where we can perform you know

many complex thermodynamic uh calculations ah considering multiphases uh various types of solutions

and you know innumerable number of possibilities to

predict the various states of equilibrium that you may be interested to know now

beyond this one important parameter that we often talk in oxy on oxygen steel making is a

term called oxygen chemical potential or oxygen chemical potential of oxygen

potential this is a very important term which we frequently use and chemical potential is

our notation for chemical potential is mu and mu is basically written for a

species i is is equal to mu i naught plus r t l n a i

that is the chemical potential of this standard formulation accepted well accepted formulation this is the

standard chemical potential of the species this is the activity and this is

the chemical potential under consideration at a given under a given condition now

transport of species from one phase to another phase is driven by chemical potential so we have

thermal potential thermal potential is created because of a temperature gradient heat will flow when there is a

temperature difference now mechanical potential is going to be created by head difference or pressure

differences similarly mass transport is going to be created because of

gradient in chemical potential which is directly related with activities we must understand that we

have studied this famous darkens couple there we have seen that in terms of

concentration it is possible for us to transfer carbon from a region of lower

concentration to a region of higher concentration because of the simple fact that in the

lower concentration region carbon actually the activity of carbon is much more because of the presence of many

other solute elements which have influenced through the activity coefficients okay and enhance the

activity of carbon so it is not the absolute value of concentration that dictates the rate of mass transport

rather the possibilities of mass transport or transfer of a species from one phase

to another phase is going to be dictated by the chemical potential so therefore chemical potential is a very very

important term in the context of mass transfer studies and also oxygen potential is a very

important term it then tells us that what is the level of oxygen potential in the system it tells us the oxidizing

power of the system for example oxidation oxygen potential in the blast furnace is extremely small okay 10

raised to the power minus 19 atmosphere this tells us that the environment within the blast furnace

is extremely reducing on the other hand in in oxygen still making converter the

oxygen potential because of the presence of oxygen is is tremendously large now if you say that

therefore if i have a slag here and if i have metal here and a species i

is being transferred from here to here and in equilibrium is going to be achieved

so you visualize it like this temperature is higher here temperature is lower here

so heat will flow in this direction and heat flow will come to a standstill where the temperature will become

identical now if the species i has a higher chemical potential in slag lower chemical potential in slab the transport

is going to take in the direction of slag to metal and a chemical equilibrium so that was the

thermal equilibrium when we are talking about temperature now a chemical equilibrium is going to be established

at the interface that means there is going to be no net movement across the interface

okay provided the activity is identical or chemical potential is identical here now

therefore if you assume that the same one is say this is metal which i denote

as this and this is slag and if i assume that

the same standard state condition applies so i am going to denote by the same bracket although

these are first brackets this is a third bracket because i am making an assumption that the same standard

chemical potential holds good or applies to both slag in the metal phase and then we can say that

rt ln activity of ai in the slack phase now since these two quantities have been assumed to be

identical so equality of chemical potential essentially implies that when there is a chemical equilibrium we must

have activity in the slag phase of i is exactly equal to the activity of i in the

metal phase and that is the essential condition of chemical potential now we can find out chemical potential

for example chemical potential of oxygen will be defined as vo2 is equal to mu to

zero standard plus rtln po2 that is the fundamental definition of chemical potential and this term

because it is in the pure state so if you assume that there is only oxygen available then this term is going to be

equal to zero and therefore the chemical potential is going to be directly related with the partial pressure of

oxygen so we can calculate the chemical potential for oxygen potential with respect to

this oxygen also okay and then translate it convert it to this particular scale because here the

standard state and here the standard states are different so this is the fundamental definition of

oxygen potential and we can calculate the activity of oxygen and oxygen potential here and then translate it to

this particular format so wherever there is oxygen it is possible for us to calculate the chemical potential of

oxygen and thereby tell that what is the oxidizing power of the system itself

okay you