FUNDAMENTALS OF STEELMAKING METALLURGY CHAPTER 9 THERMODYNAMICS AS TOOL FOR PROCESS ANALYSIS

FUNDAMENTALS OF STEELMAKING METALLURGY CHAPTER 9 THERMODYNAMICS AS TOOL FOR PROCESS ANALYSIS Brahma De Visiting Prfessr Schl f Minerals, Metallurgical and Materials Engineering IIT Bhubaneswar, India bde@iitk.ac.in; bde@iitbbs.ac.in ABSTRACT Thermdynamics is a pwerful tl t analyze the cnditins required fr a reactin t g in a frward r backward directin. But when the chemical reactins are written, the students and researchers ften make errrs in chice f standard state f different reacting cmpnents and reactin prducts. The steps invlved in cnversin f ne standard t anther are explained thrugh wrked ut examples. In steelmaking, the activity f irn xide in slag plays a key rle in xidatin f elements acrss slag metal interface. Over past five decades, a large number f appraches have been published n calculatin f irn xide activity in slag thrugh several mdels, like inic thery f slag, ptical basicity, regular slutin mdel, cell mdel f slag, and a number f empirical relatins develped thrugh actual experiments. Hwever, each f these methds result in different activity f irn xide in slag, smetimes differing by a factr f tw r mre, and the researcher is at a lss t chse which ne. A wrked ut example is given t bring ut this aspect. Similarly, several appraches can be used t predict phsphrus in steel, like inic thery f slag, Healy s mdel, mlecular thery f slag, ptical basicity, and quasiregular slutin mdel. Again, each f these appraches result in different phsphrus cntent f liquid steel fr the same slag cmpsitin. A wrked ut example is given t bring ut these differences t shw that in all cases, with ne mdel r anther, the calculated results indicate nly a pssible trend. What is required in an actual applicatin t steelmaking prcess is t tune the mdel t a practical situatin. The usefulness f simple thermdynamic calculatins in selecting a prper desulphurizatin agent, like CaO, CaO+ Al, Ca+ C+ Al, is shwn thrugh a wrked ut examples. A wrked ut example is als given n dexidatin f steel with Al in rder t shw the pwer f thermdynamic analysis thrugh simple thermdynamic calculatins. The strength f present wrk is essentially f its educative nature in analyzing and applying thermdynamic fundamentals t steelmaking. The students, researchers and teachers can use it t understand r t explain the cncepts invlved. 1.0 Intrductin Thermdynamics is fundamental tl fr prcess analysis f steelmaking. The students and researchers ften make errr in calculatin f apprpriate equilibrium cnstant because the standard states are nt prperly chsen fr the prcess invlved. Further, irn xide activity plays a crucial rle in steel refining. But several appraches t calculate irn xide activity are available. When we use these appraches different values are btained, smetimes differing by a 1

factr f tw r mre. This is demnstrated thrugh a wrked ut example by using inic thery f slag, regular slutin mdel, ptical basicity mdel, and empirical relatin based n experiments. Similarly, wrked ut example is given t shw hw different mdels can be used t predict phsphrus cntent f liquid steel, but each mdel giving a different result. The strength f thermdynamics in chsing a desulphurizer is als demnstrated thrugh wrked ut example. An example f deep thermdynamic analysis f dexidatin f steel with aluminum is als given. A detailed backgrund reading material n thermdynamic fundamentals related t dilute slutins in steel, ptical basicity, inic thery f slag, cell mdel f slag, and mlecular thery f slag can be fund in Chapter f [1]. Details f quasi regular slutin and it applicatin t calculatin f activity f irn xide in slag are given in [,3]. The set f wrked ut examples have been chsen t cver a wide spectrum and are essentially f educative value in prcess analysis f steelmaking. Otherwise, t a first time student learner, researcher, teacher, and steelmaker alike, wanting t learn and apply thermdynamics, it appears like a maize and ne des nt where t start frm and fcus upn, which mdel t chse r rely upn. Fr actual applicatin n shp flr, nly after learning the basics ne can prceed t tune the chsen mdel t prcess data, because mdel by itself is inadequate and different mdels may predict widely different results..0 Selectin and change f standard state in chemical reactins The first step in equilibrium calculatins is t chse standard state r reference state f reactants and prducts. The reference states are: are pure slid (<>, with crystal structure mentined ), pure liquid ({ }), gaseus ( ( )g), disslved state f 1 Mass% and ( [ ]1 mass %), and disslved state f infinite dilutin( [ ]ID ]. Fr example, in the reactin given belw, the reference state f Al, by sign cnventin, is pure Al liquid (using { } brackets), f xygen is pure gas (using ( )g), and f AlO3 is pure slid (using <>). {Al} + 3/ (O)g = <AlO3> (9.1) It is t be nted that AlO3can be present in different crystalline states and that has als t be mentined, depending upn temperature, say α AlO3, r β AlO3, etc. Fr cncentrated slutins the Raults is stated as ai= γi* i where i is mle fractin, ai is Raultian activity and γi is Raultian activity cefficient. Fr ideal slutins γi = 1. Fr dilute slutins in steel a i= γ i i. Henery s law has tw frms:hi= fi*mass% (i) r h i =fi* i.here,bth hi and h i are Henerian activity, but with different reference states. Fr slutins beying Henery s law, fi=1. And hence hi=1 fr 1 mass% in disslved state. h i is Henerian activity defined at infinite dilutin when cncentratin is expressed in mle fractin and when fi=1 at infinite dilutin, and h i is equal t i and thus h i is never unity whereas hi=1 iffi=1

Fr dilute slutins in steel [1], the general relatinship between Rault s and Henrey s law is in terms f equality a i γ i i = h i = h i mass% i i = f i (9.) The Henrian activity cefficient fi can be mre than ne (psitive deviatin), equal t 1 (ideal cnditin) and less than 1 (negative deviatin). Since, mathematically, ai= γi* i= f i γ i i the prduct γ i f i is equal t γi which is Raultian activity cefficient when reference state is pure substance, and γ i is defined fr infinite dilutin. Obviusly, γi = γ i when fiis unity. γiis always psitive and can be equal t 1 (ideal slutin) mre than 1 (psitive deviatin) and less than 1 fr negative deviatin. S, activity and activity cefficients will always be psitive.activity itself cannt be mre than ne, except fr the case f Henrian activity hi, (reference state 1 mass%); fr example if Henery s law is beyed up t mass % then hi=. By cnventin, the disslved state is represented by square brackets. In case f 1 mass % as reference state, [Al]1mass% + 3 [O]1mass% = <AlO3> (9.3) If the square brackets d nt have a subscript then it is assumed by default that the standard state f the disslved element is 1 mass%. In the case f infinite dilutin as reference state, Eqn (9.1) is written as [Al]ID + 3 [O]ID = <AlO3> (9.3) The selectin f standard state (r reference state)f reactants and prducts shuld be dne depending upn the actual physical state f the reactants in the system. Otherwise the equilibrium cnstant calculated will nt crrespnd t the cnditins which actually exist in a system. Fr example, if a particular cmpnent in a system exists in liquid state then the standard state fr that cmpnent shuld be chsen as pure liquid, and if it is present is in a disslved state then reference state shuld be disslved state [ ]1 mass %,r [ ]ID. If the reference state is pure liquid then activity in pure liquid state wuld be unity, and similarly the activity in pure slid state will als be unity. But it des nt mean that, all ther things remaining same, the equilibrium cnstant with reference state chsen as pure liquid will be same as the equilibrium cnstant with reference state chsen as pure slid. This is because a free energy change is invlved when slid state is changed t liquid state and this is reflected in the equilibrium cnstant value.in fact equilibrium cnstant is slely dependent n the particular reference states chsen fr the set f reactants and prducts in a particular reactin. Each f the reactants can have a different reference state in a given reactin r same state if they are actually present in same state. It is clear that the standard state can be changed (i.e. frm ne state t anther, say frm pure slid t pure liquid) nly by including the free energy invlved, say frm sld t liquid, r liquid t gas, r frm [ ]1 mass % t [ ]ID etc. 3

It is imprtant t nte that fr the cmpnents disslved in slag nly Rault s law is applicable. Bth Rault s and Henery law are applicable nly fr the elements disslved in the metal and difference is the chice f standards state. The examples given belw clarify hw ne can change the standard state f reactants and prducts..1 Change f standard state fr Al disslved in metal If the free energy data fr the reactin is (frm Table 1) {Al} + 3/ (O)g = <AlO3> is = =-168567 + T*11.5368, j/ml G 1 then the prcedure fr finding ut the free energy fr the reactin [Al]1 mass% + 3 [O]1 mass% = <AlO3> is as fllws. Sultin In the reactin {Al} + 3/ (O)g = <AlO3> G 1 ==-168567 + T*11.5368 As indicated by apprpriate symbls, the chsen standard state f Al is pure liquid Al, standard state f xygen is pure xygen gas and standard state f alumina is pure slid alumina; ne can define in additin the crystallgraphic structure f alumina, α, β, etc. We cannt straight away use the abve equatin t calculate the disslved xygen cntent f liquid steel cntaining, fr example, 0.0% Al in equilibrium with slid alumina. It is necessary t g thrugh a series f steps t finally get the equatin [Al]1 mass% + 3 [O]1 mass% = <AlO3> The steps t arrive at abve equatin are as fllws. The crrespnding free energy values are taken frm Table 1. [Al]1 mass%= {Al} G = * (63118+ T*7.8806) This represents the free energy invlved in change f standard state f Al frm 1 mass% t {Al}. Similarly fr xygen, frm Table 1 3 3[ O ] 1mass% ( O ) g G 3 3*(11740 T *.884) 4

G 3 where is the verall equatin representing free energy f disslutin and als the free energy invlved in the change f standard state f pure gaseus xygen t 1 mass%. Nw we can write all the steps tgether {Al}pure liquid + 3 (O)g = <AlO3> G 1 ==-168567 + T*11.5368 [Al]1 mass% = {Al}pure liquid G = *(63118 T* 7.8806) 3 3[O] 1 mass % (O ) g G 3 3*(11740 T *.884) G 4 = G 1 + G + [Al]1 mass% + 3 [O] 1 mass% = <AlO3> G 3 = -154171 +75.9606 *T. Change f standard state f manganese frm pure liquid t Mn disslved in metal Suppse the reactin is {Mn} + [O]1 mass% = (MnO). It is required t change the standard state f Mn frm {Mn}, ie. pure liquid, t [Mn]1 mass% s as t find ut the free energy fr the reactin [Mn]1mass% + [O]1mass% = (MnO). Assume that Mn fllws ideal Rault s law when disslved in liquid steel. Slutin The reference state f disslved Mn in the equatin {Mn} + [O]1 mass% = (MnO) is pure Mn liquid. Nw, if we want t change the standard state frm {Mn} t [Mn]1 mass% then the steps needed are {Mn}pure liquid + [O]1 mass% = (MnO) G 5 Mn [Mn]1mass% {Mn}pure liquid G RT ln 0.5585 / M [Mn]1mass% + [O]1 mass% = (MnO) G i 7 G 5 G i Mn i Mn In the equatin [Mn]1mass% {Mn} the value fr change f standard state is (see Chapter )is M partial mlar free energy f slutin f Mn in liquid irn = G RT ln 0.5585 / M MMn is atmic mass f Mn (= 55.5). i i Mn. 5

In principle, whenever a free energy equatin is written it must be clearly mentined as t what the chsen reference states are. As per cnventin, the term (MnO) indicates that MnO is present in disslved frm liquid slag and in writing the free energy f disslutin f MnO in slag has been taken int accunt. The example given belw illustrates the pint. G 5.3 Change f standard state fr MnO in slag and Mn in metal fr the given reactin {[Mn} + [O]1 mass% = <MnOα> Slutin The steps invlved are as fllws {Mn} + [O]1 mass% = <MnOα> <MnOα>= {MnO} {MnO}= (MnO) [Mn]1mass%={Mn}pure liquid G 8 G 9 G 10 G 11 [Mn]1 mass% + [O]1 mass% = (MnO) = G 1 G 8 + G G 8 9 G10 G 11 The reference state f MnO is nw liquid MnO and the free energy f slutin f MnO in slag is incrprated int G 1 thrugh G 10 Care must be taken in writing the equilibrium cnstants as well. The equilibrium cnstant fr the reactin {Mn} + [O]1 mass% = <MnOα>, is t be written as a MnO G 13 =-RTlnK where K a *[h Mn ] G 7 It is t be nted that if since MnO is present as pure slid amno = 1.If Rault s law is beyed by metal then amn=mn because the reference state f Mn is pure Mn liquid. if Mn did nt fllw ideal behavir (Rault s law) in irn then amn=γmnmn where γmn is the activity cefficient f Mn. Similarly, in case f dilute slutin, h= mass%o if xygen fllws ideal Henery s law, therwise h= fo*mass%o where fo is the Henrian activity cefficient. Nw cnsider the reactin [Mn]1 mass% + [O]1 mass% = (MnO), G 14 The equilibrium cnstant is t be written as a MnO G 14 =-RTlnK where K [h ]*[h ] Mn 6

Here amno cannt be assumed t be unity because MnO is present in the disslved frm, in fact amno= MnO if it fllws Rault s law therwise amno=γmno* MnO where γmno is the Raultian activity cefficient f MnO in slag. Similarly, hmn=mass% Mn and ho=mass %O if bth Mn and O fbey Henery s law, therwise hmn=fmn*mass%mn and ho=fo*mass%o, where fmn and fo are the Hnerian activity cefficients fr Mn and O in metal, respectively. 3. Dexidatin f steel with aluminum Liquid steel (1873 K) weighs 300 tns and cntains 600 ppm disslved xygen and 0.03% carbn. It is required t reduce the disslved xygen t 6 ppm at 1873 K by additin f aluminum. Assume that the equilibrium cnstant (KAl, at 1873 K) fr the reactin, [Al] + 3 [O] = <AlO3>, is lg KAl =13.888, where the cncentratins f [Al] and [O] are expressed in mass%. (a) Calculate the amunt f aluminum required t reduce the disslved xygen t 6 ppm and find the crrespnding equilibrium mass % f disslved aluminum in steel. (b) What is the lwest cntent f disslved xygen that can be achieved at 1873 K by adding aluminum and what will be the crrespnding equilibrium cncentratin f disslved aluminum. Assume that activity f alumina is unity. Is the assumptin f unit activity f alumina always a valid ne? (c) If the AlO3 frmed is trapped in steel in the frm f 10 micrn spherical particles (inclusins) then what will be the number f such particles; assume that density f AlO3is 300 kg/m 3 and density f steel is 700 kg/m 3. If the dexidized slid sample f steel is bserved under the scanning electrn micrscpe (SEM) then what will be fractinal area cvered by the AlO3 particles. (d) What will happen if the cmputed amunt f aluminum is added t liquid steel ( as calculated in Part (a)) but temperature f steel drps t 1853 K due t heat lsses. Assume that the equilibrium cnstant at 1853 lg K=13.653 and the same interactin parameters can be used as given belw. (e) What is the effect f disslved carbn n the amunt f aluminum required t bring dwn the xygen, say the carbn cntent f steel is 0.04% instead f 0.03% in Part (a ) f the prblem. (f) What wuld happen if disslved nitrgen is als present in steel? The interactin parameters are: = -0.. Slutin C e Al = 0.091, Al e Al =0.043, O e Al = -4.77, Al e O = -.81, C e O = -.35, Part (a) Let the basis f calculatin be: 100 g f metal f liquid metal at 1873 K, cntaining 600 ppm f disslved xygen and 0.03% f disslved carbn t start with. In all equilibrium calculatins it is best t d mass balance in terms f mles f material reacted r frmed. Let x mles f pure alumina (AlO3) frm due t the reactin [Al] + 3 [O] = O e O 7

<AlO3>, fr which lg KAl =13.0. Let the mass percent f [Al] which is needed at the start (s that the final xygen, [O], at equilibrium is 0.0006%) be y. Frm stichimetry, we can write [Al] + 3 [O] = <AlO3> a AlO3 1 KAl= (A) 3 3 h ] [ h ] [ h ] [ h ] [ Al O Al O Initial mles: y 7 [Al] 0.06 16 [O] 0 (AlO3) Final mles: y x 7 [Al] 0.06 3x 16 [O] x (AlO3) Final mass%: y-54x [Al] 0.06-48x [O] B) Since irn is the slvent and the relative amunts f [Al] and [O] are negligible, nly as a first rder apprximatin we can write the final mass% as shwn abve; therwise, fr example, the final mass percentage f [Al] will be given by a rather cmplicated equatin, y [( ) x]*7 7 *(100) y 0.06 [( ) x]*7 (100 y 0.06) [( ) 3x]*16 7 16 Nw, the final xygen cntent is given, [O]=0.0006%, hence, 0.06-48x= 0.0006 r x=1.375*10-3 mles. Frm Eqn (B), the final (unknwn) mass% f [Al] is, y-54x=y-54*1.375*10-3 =y- 6.685*10 -. On writing the equilibrium cnstant in terms f activity cefficients and interactin parameters, Lg KAl = 13.0 = - lg [hal ] 3 lg [O] = - lg [Al]- lg [fal] -3 lg [O] - 3 lg [fo] (C) The activity cefficients, lg fal and lg fo, can be written as lg fal= Al O C e Al * [Al] + e Al * [O] + e Al * [C] = =0.043*(y 6.685*10 - ) -4.77 * 0.0006 + 0.091*0.03 (D) lg fo= e Al O * [Al] + O e O * [O] + C e O * [C] 8

= -.81*(y 6.685*10 - ) - 0. * 0.0006-0.35 * 0.03 (E) On inserting the abve values int eq. (C), we get a nnlinear equatin in ne unknwn y, which n slving by Newtn-Raphsn methd gives y=0.086%. Hence, fr a 300 tn bath we need t add 300*10*0.086=58.6 kg f Al in rder t reduce the disslved xygen cntent f liquid steel frm 600 ppm t 6 ppm. Fully killed steels require disslved xygen cntents t be even lwer than 6 ppm. The amunt f Al needed in an actual dexidatin practice n the shp flr is usually tw t three times f the theretically calculated (equilibrium) amunt. This is because a majr part f Al is used up (r wasted) in reactin with carried ver slag lying n tp f ladle and als in reactin with xygen in pen atmsphere. Since carried ver slag frm furnace may cntain high percentage f FeO (especially in the case f BOF), slag stppers are used t minimize the carryver f slag during tapping peratin (puring f liquid steel frm furnace t ladle). Only when the variatins in perating practice are minimized (minimum slag carryver etc.), the recvery f Al becmes mre predictable and then the amunt f Al can be calculated mre precisely, by fllwing the same prcedure f calculatin as explained abve. Befre the additin f aluminum, the disslved xygen cntent and temperature f metal can be measured by using a dispsable xygen sensr. Hwever, xygen sensrs are expensive and therefre add t the cst f steel prduced. It becmes ecnmical t use xygen sensrs nly when the recvery factr f aluminum is apprximately knwn. Otherwise, we end up adding either mre, r less, than the amunt f aluminum actually required and thereby the disslved xygen cntent f metal als becmes uncertain. The mass % f Al in equilibrium with 6 ppm f xygen will be [Al] = y-6.685*10 - = 0.086-6.685*10-0.0194 %, r 194 ppm. Part (b) The disslved xygen cntent at 1873 K can be calculated easily frm Eqn (C) fr the knwn cntent f disslved aluminum. On pltting a graph between [Al] versus [O] it wuld be bserved that the minima, [O]= 0.00033% (r 3.3 ppm), lies smewhere in the range [Al]= 0.10-0.107%. The actual lcatin f the minima wuld depend partly upn the accuracy f the numerical prcedure used t slve the nnlinear Eqn (C). At this pint, it als necessary t nte that the free energy equatin itself, which is used t calculate the equilibrium cnstant in Eqn (C) has sme errr (due t the nature f equilibrium experiments cnducted at high temperature). Thus, ne can nt be very critical f the small differences in the results btained by using ne numerical methd r the ther. Pure alumina (AlO3) inclusins can frm in liquid steel and therefre, theretically, it may be crrect t assume the activity f alumina as unity. If liquid metal is cvered with slag then the activity f alumina in this slag can be lwer than unity (viz. in calcium aluminate slag). Thus, if cmplete (simultaneus) thermdynamic equilibrium is nt established between liquid metal, slag, refractry, and entrapped gas bubbles, then xygen cntent f metal wuld vary frm ne lcatin t anther. Under such a cnditin, we can talk f nly an average disslved xygen cntent f metal. The average disslved xygen cntent f fully killed steel (say at 1873 K) may 9

vary frm.5-3.5 ppm. If the slag cvering the metal has high FeO cntent then the disslved xygen cntent culd be much higher, in spite f the presence f pure alumina particles in steel. Calescence and grwth f alumina particles in liquid steel, althugh present in ppm levels, results in sme very interesting mrphlgies (see Figures* in Chapter*) and is a subject f great theretical and practical interest (discussed in Chapter 8) in quality steel making. Part (c): Disslved xygen is reduced frm 600 ppm t 6 ppm. Thus, ttal amunt f alumina frmed in 300 tn f steel wuld be 0.0594*0.01* 300*1000*10/48= 378.675 kg. Ttal cntent f alumina inclusins is therefre 378.675/300/1000*10= 0.163 % r 163 ppm. With the passage f time, the inclusins being lighter than metal, flat up, aided by inert gas stirring etc. The final metal may cntain 30-100 ppm f inclusins. Let us assume that steel cntains 100 ppm f alumina particles. Mass f ne particle f 5 micrmeter radius and a density f 300 kg /m 3 will be Mass f ne particle= 6 4 (5*10 ) 3 3 *300 =.45*10-11 kg Ttal mass f particles (crrespnding t 100 ppm) = 0.0100/100*300*1000= 3 kg Hence ttal number f alumina particles = 3/(.45*10-11 ) = 1.*10 11 11 1.*10 *700 Vlume fractin f particles will be = =.9*10 9 precipitates /m 3 300*1000 Since vlume fractin f spherical particles is same as area fractin, the number f precipitates (f size 10 micrmeter diameter) per mm f the surface f steel sample will be.9 r apprximately 3. In this way, frm image analysis we can get a quantitative idea f the inclusin cntent f steel. These estimates are ften as accurate as the chemical analysis, but, f curse, the truble f sample preparatin, selecting a representative area fr image analysis, etc, pse a challenge. Part (d): The equilibrium [Al] and xygen [O] at 1873 K, as calculated in part (a) are, respectively, 0.0194% and 0.0006%, respectively. Fllwing the same prcedure as in part (a), let x mles f alumina be prduced n lwering f temperature frm 1873 K t 1853 K. Thus, similar t eq. (B), the final percentage f [Al] and [O], respectively, wuld be 0.0194-54*x and 0.0006-48x. Substituting these int Eqn (C), alng with the Eqns(D,E) fr the activity cefficients, fal and fo, and then slving fr x, we can calculate the equilibrium amunts f disslved aluminum and xygen at 1853 K as [Al] = 0.019% and [O] = 0.0004%, respectively. Thus, n cling frm 1873 K t 1853 K, which des take place during the perid f transfer f liquid metal frm ladle t muld, the disslved xygen may further reduce frm 6 ppm at 1873 K t 4. ppm at 1853 K. Very fine alumina inclusins may frm at this stage. At these lw xygen levels, hwever, we begin t apprach the virtual lwer limit f the accuracy f measurement f disslved xygen with the help f xygen sensrs. The slw kinetics (mass transfer, diffusin f species) at such lw cncentratins actually gvern bth the frmatin and the grwth rate f inclusins. Special kinetic aspects f frmatin and grwth f inclusins will be discussed in Chapter 8. 10

Part (e): The interactin parameter between carbn and xygen atms in steel is strngly negative ( = + 0.041). Thus the verall effect n equilibrium is such that the cmputed value f y in Part (a) increases t 0.0864% (and nt 0.086%). Hence we will require slightly greater amunt f aluminum t bring dwn the xygen t 6 ppm! This wuld nt be the case if the initial xygen cntent is high (say 0.08%) because then the disslved carbn and xygen can react and remve bth carbn and xygen as CO gas. C e O = -0.35) cmpared t the repulsin between aluminum and carbn ( The apprximate (equilibrium) value f the slubility prduct f carbn and xygen in liquid steel at 1873 K and 1 bar pressure f CO gas is 0.005. In BOF practice (withut argn stirring) it varies frm 0.0030-0.0036. With argn stirring it reduces t apprximately 0.00-0.006 under the influence f argn stirring (due t lwering f the partial pressure f CO gas). In the given prblem, the carbn-xygen prduct fr 0.08% disslved xygen wuld be 0.03*0.08=0.004. If the carbn cntent is increased t 0.04% then the prduct will be 0.03. Srategically, in the latter case, the reactin between carbn and xygen can be enhanced by purging the steel with argn gas s as t bring dwn the prduct t <0.006 and then add aluminum fr dexidatin. This is ne f the advantages f stirring with argn gas in steelmaking. It helps t reduce the disslved xygen and, therefre, the cnsumptin f aluminum. Reductin f aluminum cnsumptin als lesser inclusins t be flated ut frm steel. It is, hwever, necessary t evaluate the cst f argn stirring versus cst f aluminum additin and inclusin cntent. Argn stirring reduces the temperature f steel (i.e. additinal heat lss frm liquid steel) by 10-30 K, depending upn the gas flw rate and the amunt f slag cver n tp f steel. In a fair evaluatin, the cst f energy lss has t be included, alng with cst f stirring elements, refractry life etc. Part (f): If disslved nitrgen is als present in steel then, depending f relative amunts f aluminum, xygen, and nitrgen, bth AlN and AlO3 can frm. The slutin f the prblem wuld invlve simultaneus slutin f tw nnlinear equatins in tw unknwns. Prcedure f develping a mathematical mdel is presented in the next prblem. 4. Simultaneus frmatin f xide and nitride The initial mass percentages f disslved aluminum, nitrgen and xygen in steel are WAL, WAN and WO. Assuming that bth ALN and AlO3 frm in liquid steel, develp equatins t calculate the equilibrium mass percents, [Al], [O] and [N], and als the amunts f AlO3 and AlN that wuld frm. Assume that equilibrium cnstants are knwn and that all disslved elements fllw Henrian behaviur in metal and AlN and AlO3 are slid prducts (unit activity) and there is n slubility f AlN in AlO3. Al e O Slutin Initial mass WAL WN WO 11

At equilibrium [Al] [N] [O] Let the basis f calculatin be 100 g f liquid steel cntaining WAL, WN, and WO mass% f elements, rest Fe. Fllwing a similar prcedure as in Pblem 1, let the number f mles f AlO3 and AlN frmed be, respectively, x and y. Frm mass balance, as a first apprximatin (see prblem 1), the mass f Al lst due t frmatin f x mles f AlO3 is 54x and y mles f AlN is 7y. Hence the final mass f Al left in steel is [Al] = WAL - 54x - 7y Similarly, the amunt f nitrgen left after frming y mles f AlN is [N] = WN - 14y and the amunt f xygen left after frming x mles f AlO3 is [O] = WO - 48x Since the ttal mass f [Al], [N] and [O] in metal is very small cmpared t the mass f irn, they may be taken as mass% (100 g basis). Equilibrium cnstants fr AlO3, and AlN, assuming unit activity f alumina and Henerian behavir f [Al], [N] and [O], can be written as 1 KAl = 3 [ Al] [ O] (A) KN = 1 [ Al][ N] (B) The equatins (A) and (B) cnstitute tw equatins in tw unknwns (x, y) and can be slved by a suitable numerical prcedure. Alternatively, it can be seen that Eqn (B) is a quadratic in y and the value f y, calculated in terms f x frm Eqn (B), can be substituted int Eqn (A) and then Eqn (A) can be slved fr x by a simple iterative search prcedure. If yet anther element is present in steel, say Ti, which may react with xygen and/r nitrgen, then, fr each cmpund (TiO, TiN etc) there will be an additinal nnlinear equatin slve (arising frm the crrespnding equilibrium cnstant). The slutin prcedure invlving mre than tw nnlinear equatins, hwever, becmes mre and mre cmplicated because f the difficulty in making initial guesses f the slutin. All numerical prcedures fr slving nnlinear equatins are iterative. It is ur experience that genetic algrithm (GA) is the best methd fr such prblems because the search space (in which the slutin may exist) can be frmulated rather easily and the rate f cnvergence t the final slutin is nt a prblem because f the very evlutinary nature f GA (survival f the fittest individual in the ppulatin chsen frm within the search space). 1

Only as a simplificatin, it has been assumed abve that the elements bey Henery s law. The reactive elements like Al, Ca, and Ti can significantly influence the activity cefficient f xygen. If the effect f interactin parameters is t be taken in t accunt, then it is best t first determine the Henerian activities and then perfrm a separate calculatin t find ut the respective mass percentages, using the set f simultaneus equatins. 5. Desulphurizatin f liquid irn ally with lime (a) What are the cnditins fr btaining a high sulphur distributin between slag and metal? (b) Mlten irn cntaining 3.5% carbn, 0.5% manganese, 0.5% phsphrus and 1% silicn is in equilibrium with pure lime at 1583K. Calculate the equilibrium sulphur cntent f the metal. Slutin Part (a) In the absence f carbn, silicn and aluminum in the metal, the desulphurizatin reactin can be written as [S] + [Fe] + (CaO) = (CaS) + (FeO) (A) a a h a a CaS FeO K (B) S Fe CaO The mle fractin f CaS can be directly related t wt% sulphur in slag, f S (wt % sulphur in metal) and a FeO =1.With this simplificatin, h S can be written as % S inslag a k` f (C) [% S] inmetal a CaO Distributin rati = S Accrding t equatin ( C) the sulphur distributin rati will be higher prvided that: FeO high, afeo is lw, i.e. reducing slag; and k`is high. In Equatin ( C) k`increases with temperature. Nte again that a high basicity des nt necessarily me high lime activity. a CaO is Part (b) In the presence f silicn, the reactin can be written as 1 1 CaO [ Si] [ S] CaO. SiO CaS Fr which the free energy equatin is 13

G O 465 78.0 T J/ml Since G O RT ln K, hence, at 1583K, lg K=4.0469 It can be assumed, fr the sake f simplicity, that the activity f lime, the activity f dicalcium silicate ( acao. SiO ) and the activity f calcium sulphide ( a CaS ) pure slid state, is unity. The equilibrium cnstant can then be written as 1 1 K 1 h h f wt% Si 1 f wt S Si S Si S % The Henrian activity cefficient f silicn can be btained frm j lg f e ( wt%) Si = e S Si Si [ wt% S] j e Mn Si + + e C Si [ wt% C] + [ wt% Mn] e P Si [ wt% P] e Si Si a CaO, all assumed t be the + [ wt% Si] lgfsi= = e S S [ wt% S] + e C S [ wt% C] e Mn S + + [ wt% Mn] e P S [ wt% P] e Si S + [ wt% Si] On substituting the fllwing interactin parameter values: P e Si =0.11, S e S Si e S C e S Si e Si =0.11, S e Si = 0.056, C e Si = 0.18, =0.033, = -0.08, = 0.063, =0.11, =-0.05, =0.9 and the given mass percentages f C, Mn, P and Si, as well as the equilibrium cnstant value (lg K=4.0469) lg f Si 0.056*[ wt% S] 0.775 Similarly the activity cefficient f sulphur f can be btained frm lg f S e Mn = e S S *[ wt% S] + e c S *[ wt% C] + S *[ wt% Mn] + e P S *[ wt% P] + e Si S *[ wt% Si] S Mn e S On substituting the interactin parameter values and the cmpsitin: P e S Mn e Si lg f S 0.08*[ wt% S] 0.57 lg f S On substituting the value f lg fsiand int the expressin fr the equilibrium cnstants K, we can calculate the equilibrium sulphur cntent f metal fr the given silicn cntent f metal as 0.00001 wt% r 0.1ppm. Such lw cntents are nt btained in industrial practice due t the fact that the equilibrium is nt achieved because f the very lw reactin kinetics at lw sulphur cntents. 6.Use f lime (CaO) as an agent t remve sulphur frm mlten irn in presence f C, Si, Al It is prpsed t evaluate the use f lime (CaO) as an agent t remve sulphur frm mlten irn. Suppse mlten irn cntains 4% carbn, at 1583 K. In additin, the silicn r aluminum cntent f the metal can be made t vary. In which ne f the fllwing three cases, lwest sulphur cntent can be btained: 14

Case (a): Aluminum is nt present in metal (r is present in negligible amunt) and the silicn cntent is varied frm 0.0-0.35%. Only the fllwing reactin is assumed t ccur <CaO> + ½ [Si] + [S] = ½ < CaO.SiO> + <CaS>G = -465 +78.0 T (A) This case is essentially same as in prblem.5, except that the silicn cntent is made t vary. Case (b): Silicn is nt present in the metal (r is present in negligible amunt). The aluminum cntent is varied frm 0.0-0.35% and nly the fllwing reactin ccurs 6 <CaO> + [Al] + 3 [S] = <3 CaO.AlO3> + 3 <CaS>G = -899304 +7.8 T (B) Case(c): Neither silicn nr aluminum is present in the metal but the partial pressure f CO gas is varied in the range 0.01-1bar and nly the fllwing reactin ccurs <CaO> + <C> + [S] = <CaS> + (CO)g G = -110993 +114.4 T (C) The free energy values, given abve, is in j/ml and the interactin parameters S e Si = 0.056, S e Al =-0.03 Slutin C e Si = 0.18, S e S = -0.08, Si e S = 0.063, C e Si =0.11, AL e Si =0.035, AL e Al Si e Si =0.11, =0.043, C e Al =0.091, Let the equilibrium sulfur cntent f metal be represented as [S]. The activity cefficients fr the elements can be written in terms f the interactin parameters as: Lg fsi = Lg fs = Lg fal = Si e Si * [Si] + S e S Al e Al * [S] + Si e S * [Al] + S e Si [S]+ * [Si] + C e Al C e Si * 4.0 C e S * [C] + S e Al * [C]+ * [S] Al e S * [al] In the given prblem, [Al] = 0.01-0.35%, [C] = 4.0%, [Si] = 0.01-0.35%. Assuming that the activity f all the reactants/prducts in slid state is unity and partial pressure f CO gas is 1 bar, the equilibrium cnstants can be written as K Si [ h 1 ; K Al ; K C 1/ 3 Si ] * hs [ hal ] *[ hs ] 1 1 h s The respective values f the equilibrium cnstants fr the three reactins (A), (B) and (C) at 1583 K, respectively, can be calculated frm the free energy equatins. Further, the equatins fr Henrian activities, viz lg hal = fal * [Al], etc. can be written using the activity cefficient values given abve. On substituting the respective values f activity cefficient and mass% int 15

the equilibrium cnstant and slving fr the nly unknwn [S] in the nnlinear we can get the equilibrium sulfur cntents fr reactins (A), (B) and (C), as dne in previus example. Calculatins will reveal that at lw partial pressures f CO gas (say 0.01 bar) the best ptin (frm the pint f view f thermdynamics) is reactin (C), fllwed by reactin (B) when metal cntains 0.35% Al, and then reactin (A) when metal cntains 0.35% Si. Frm the practical pint f view, when we inject lime pwder with nitrgen gas then the cnditins f lw partial pressure f CO gas can be fulfilled because the bubbles f nitrgen will have lw partial pressure f CO gas, but the kinetics f reactin with pure slid lime is extremely slw. IN the ld industrial practices, apprximately 3-5% f CaF was mixed with lime t bring dwn the melting pint f lime. The use f CaF, hwever, is nt permitted any mre fr envirnmental cnsideratins. In the place f CaF even pre-fused mixture f calcium aluminate can be used, but then the activity f lime in calcium aluminate is much lwer. In such a case, it may be better t pt fr reactin (B), prvided the cst f adding aluminum (r disslving aluminum in bath befre hand) is nt prhibitive. Bth the reactins (A) and reactin (C) have the prblem that, that an un-reactive layer f di-calcium silicate may frm n the surface f lime which in turn culd slw dwn the reactin rates. In all the three reactins, thermdynamics predicts very lw equilibrium sulfur cntent but, in actual practice, because f the slw kinetics f reactins the equilibrium is never attained and the sulphur cntents is always higher than that expected frm thermdynamics. 7. Determinatin f activity f FeO in slag by different mdels A slag cntains 50% CaO, 0% SiO, 0% FeO, 5% AlO3 and 5% PO5 at 1873 K. Calculate the activity f FeO in this slag frm: (a) regular slutin mdel (b) inic thery f slag (c) IRSID mdel based n cell mdel f slag (d) experimental determinatin in a recent wrk (e) cmpare the values btained and cmment upn the differences in the calculated activity valuesf FeO in the fur appraches. Assume that in the inic thery mdel (similar t f () term, the deviatin frm nn-ideal behavir is accunted fr by the equatin Lg ()FeO = (-0.51 1.1 S + 1.16110-3 T ) + (6.13 +.85 S 5.93 10-3 T )NO - + (- 6.1 1.6 S + 4.58 10-3 T)(NO - ) where S Slutin N N 4 SiO4 4 SiO4 N 3 PO4 The mle fractin f FeO in the bulk slag can be calculated as 0.181. 16

Part (a): Regular slutin mdel In the case f the regular slutin mdel [,3], the activity cefficient f FeO can be calculated frm RT ln Fe + = Fe + - Ca + (NCa + ) + Fe + - si 4+ (NSi 4+ ) + Fe + - P 5+ (N P 5+ ) + Fe + - Al 3+ (N Al 3+ ) + ( Fe + -Ca + + Fe + - Si 4+ + Ca + - Si 4+ ) NCa + NSi 4+ + ( Fe + -Ca + + Fe + - Al 3+ + Ca + - Al 3+ ) NCa + N Al 3+ + ( Fe + -Ca + + Fe + - P 5+ + Ca + - P 5+ ) NCa + NP 5+ + ( Fe + -Si 4+ + Fe + - Al 3+ + Si 4+ - Al 3+ ) N Si 4+ N Al 3+ + ( Fe + - Si 4+ + Fe + - P 5+ + Si 4+ - P 5+ ) N Si 4+ NP 5+ + ( Fe + - Al 3+ + Fe + - P 5+ + Al 3+ - P 5+ ) N Al 3+ NP 5+ + ( -8540 + 7.14 T ) The last term in Eqn (A), i.e. ( -8540 + 7.14 T ), represents the free energy change f cnversin f standard state frm regular slutin t nrmal liquid slutin. The individual values f catinic fractins can be calculated as fllws. The basis f calculatin is 100 g f slag. Slag Cmpsitin Ml.Wt. Mles f Oxide Mles f Catins Inic Fractin %CaO = 50 56 (50/60) (50/60) =Ca + NCa + = 0.533 %FeO= 0 7 (0/7) (0/7) = Fe + NFe + = 0.1659 %SiO = 0 60 (0/60) (0/60) = Si 4+ NSi 4+ = 0.199 %AlO3 = 5 10 (5/10) (5/10)=Al 3+ NAl 3+ = 0.058 %PO5 = 5 140 (5/140) (5/140)=P 5+ NP 5+ = 0.04 Fr calculating the respective inic fractins, as shwn abve, NR n+ = nr n+ / ncatins In the given case, ncatins = 1.6734. The value f different ij terms cab btained frm Table (*) and n substituting int Eqn (A), RT ln FeO =-31380)(0.5335) + (-41840)(0.199) + (-41000)(0.058) + 31380)(0.04) + (-31380-41840 + 133890 ) (0.5335) (0.199) + (-31380-41000 + 154810 ) (0.5335) (0.058) + (-31380 31380 + 51040 ) (0.5335) (0.04) + (-41840 41000 + 17610 ) (0.199 ) (0.0558) + (-41840 31380 83680 ) (0.199) (0.04) + (-41000 31380 + 61500 ) (0.058) (0.04) + (-8540 + 7.14 1873 ) Since, R = 8.314,T = 1873 17

ln FeO = 6931.9395/(8.314 1873) FeO = 1.56 Since the mle fractin f FeO in slag is.181, the activity f FeO will be 0.181*1.56=0.83. Part (b) Inic thery f slag mdel Accrding t inic thery f slag [1] nly the basic xides give xygen ins (O - ins), fr example, CaOCa + + O - Similarly, FeOFe + + O - MgO Mg + + O - Fr the given slag cmpsitin the Ca + ins wuld be =0.56, Fe + = 0.1748, and ttal mle fractin f O - given by Ca + and Fe + wuld be =0.56 + 0.1748 = 0.7368. These xygen ins are cnsumed in frming anins as fllws: SiO + O - SiO4 4- PO5 + 3O - PO4 3- AlO3 + 3O - AlO3 3- Therefre, the number f mles f O - cnsumed = + SiO 3 P 3 O 5 AlO 3 = (0.098) + 3(0.00 + 3(0.0308) = 0.578 Hence the number f mles f O - remaining = 0.7368 0.578 = 0.1588 N f mles f SiO4 4- frmed = SiO = 0.098 N f mles f PO4 3- frmed = P 5 = (0.0) = 0.044 N f mles f AlO3 3- frmed = Al O = (0.0308) = 0.0616 3 18

N O 0.1588 0.1588 0.3348 0.1588 0.098 0.044 0.0616 0.474 n n Fe Fe 0.1748 N 0.37 Fe n n n 0.1748 0.56 N N 4 SiO4 3 PO4 S N 4 SiO4 O N n n n n anins catins 4 SiO4 n 3 PO4 n 4 SiO4 N anins anins n 3 PO4 O Fe n 0.098 0.444 0.474 0.044 0.474 SiO4 4 Ca n O n PO4 3 0.0978 n AlO3 3 0.444 0.866 0.444.0978 Lg ()FeO = (-0.51 1.1 S + 1.16110-3 T ) + (6.13 +.85 S 5.93 10-3 T )NO - + (-6.1 1.6 S + 4.58 10-3 T)(NO - ) At T = 1873 K Lg ()FeO = (-0.51 1.1 0.866+ 1.16110-3 1873 ) + (6.13 +.85 0.866 5.93 r ()FeO =0.9476 10-3 1873)0.3348 + (-6.1 1.6 0.866 + 4.58 10-3 1873) (0.3348) = (0.738761) (0.8775) + (0.1153) = -0.0337 Since the mle fractin f FeO in slag is 0.181, the activity f FeO= 0.181*0.9476 = 0.171 Part (c): IRSIDmdel IRSID mdel is based n cell thery f slag [1] invlving large and cmplex calculatins. It is necessary t use a cmputer prgram t slve the set f nn-linear equatins. It first calculates the cmpsitin f liquid slag frm the phase diagram and then the crrespnding activity f CaO, FeO, etc. in the liquid slag. Fr the given bulk slag cmpsitin in this prblem, the results frm the IRSID cell mdel shw that fr the given cmpsitin the slag is nt cmpletely liquid at 1873K and will cntain a slid phase. Due t the precipitatin f di-calcium silicate the cmpsitin f liquid is nt the same as that f the bulk slag and due t precipitatin f dicalcium silicate the FeO is rejected int the liquid slag and hence the liquid slag has higher mle fractin f FeO than the bulk slag. Fr the given liquid slag cmpsitin and temperature the calculated activity f FeO frm IRSID mdel is apprximately 0.396. Part (d): Experimental data n activity f FeO in CaO-FeO-SiO 19

Accrding t a recent wrk [4], the activity f irn xide in steelmaking CaO-FeO-SiO slag can be calculated frm lg γ (FeO) = 16 T 1.130 FeO 0.96 SiO + 0.13 CaO -0.4198 If the given slag cmpsitin in the prblem is apprximated t a quasi-ternary slag cntaining 50% CaO, 0% SiO and 0% FeO, then crrespnding mle fractins wuld be as fllws: Mles f CaO = 50/56, mles f FeO=0/7, mles f silica =0/60. Hence mle fractin f CaO wuld be (50/56)/(50/56 + 0/7 + 0/60) = CaO= 0.4181; similarly fr SiO=0.457, FeO = 0.1561. Nw frm Eqn(3), the activity cefficient f FeO wuld be lg10 γfeo = 16/1873 1.130*0.16 + 0.96* 0.4 +0.13*0.4-0.4198 = 0.54 Hence, γfeo= 3.44 and afeo= 3.44*0.1561= 0.53 Several such relatins have been reprted in literature and a summary f all references is prvided in [4]. Part (e): Cmparisn f different mdels Summarizing all the values: estimatin by Regular slutin mdel: afeo=0.83; estimatin by inic thery f slag, afeo= 0.171; estimatin by IRSID cell mdel,afeo= 0.396, and frm experimental data,afeo= 0.53 There are many ther empirical relatins and mdels reprted in literature frm time t time, fr activity f FeO in a variety f slags. FACTSAGE, if purchased, can als be used fr cmparisn. A researcher, while develping a mathematical mdel, is thus always faced with the dilemma: which slag mdel r equatin t use fr activity calculatins because each mdel gives a different result. It is clear frm the calculatins that in cmplex slags it is always a prblem t chse an apprpriate mdel as it can be a surce f errr in equilibrium calculatins. In the calculatin prcedure adpted in the Parts (a) and (b) the entire slag is assumed t be liquid. This assumptin may nt be crrect. The value f 0.396 calculated frm the IRSID mdel is mre apprpriate as it takes int accunt the change in FeO activity due t pssible precipitatin f a slid phase. This example is given t demnstrate that when using a particular slag mdel, care shuld be taken t first check whether slag is cmpletely liquid, and if nt then the liquid cmpsitin must be fund ut. Fr the present, it is advised t resrt t IRSID mdel as it has been successfully tested fr a wide variety f slag cmpsitins in literature. If IRSID mdel is nt available, then the next simpler ptin is t use the Regular Slutin Mdel, with an understanding that it predicts a slightly lwer activity value in mst cases, when cmpared with results f IRSID mdel. 0

The impact f using different slag mdels fr phsphrus predictin is described in the next example, fr the same slag cmpsitin. 8. Determinatin f equilibrium phsphrus cntent f steel by using different mdels Slag cntaining CaO = 55%, SiO = 5%, FeO = 15%, PO5 =5% and weighing 0 tns is kept in cntact with 300 tns f metal at temperature f 193 K. (a) What are the cnditins fr btaining lw phsphrus cntent in steel (b) Calculate equilibrium phsphrus and xygen cntent f metal frm Healy s equatin; If the temperature is lwered t 1913 K, then what will be the new phsphrus and xygen cntent f metal at the new temperature (c) Calculate equilibrium phsphrus and xygen cntent f metal frm ptical basicity mdel (d) Calculate equilibrium phsphrus and xygen cntent f metal frm quadratic frmalism mdel, (e) Calculate equilibrium phsphrus and xygen cntent f metal frm mlecular thery f slag (f) Finally, cmpare and cmment upn the values btained frm fur different mdels Assume that, due t the xidizing nature f slag, CaO and SiO d nt dissciate and thus the disslved xygen cntent f metal depends nly n phsphrus cntent f metal and activity f FeO in slag. Equilibrium cnstant fr the reactin {Fe} + [O] = (FeO) is given by: lg K FeO 637.73 T The atmic masses f different elements are: O=15.9994, Fe=55.847, Ca=40.08, Si=8.086 and P= 30.9738. Slutin Part (a)what are the cnditins fr btaining lw phsphrus cntent in steel Phsphrus distributin rati is defined as the rati f mass percentage f phsphrus in slag t that in metal. During steelmaking phsphrus disslved in metal [P] reacts with disslved xygen [O] in metal t prduce PO5 [P] + 5[O] (PO5) Disslved xygen is related t the activity f FeO in slag 1

Hence, the net reactin is (FeO) {Fe} + [O] [P] + 5 (FeO) (PO5) + 5{Fe} The PO5 thus frmed reacts with lime in slag, 3(CaO) + (PO5) 3CaO.PO5 (r smetimes 4CaO.PO5) The verall reactin cab be written as 5 [ P] 5( FeO) 3( CaO) (3CaO. PO ) 5{ Fe} The equilibrium cnstant fr the abve equatin is knwn t decrease with temperature and hence temperature has an adverse effect n xidatin f phsphrus. The FeO in the slag is required t prvide xygen fr xidatin f phsphrus t PO5 and CaO is required t arrest the PO 5 in the slag as 3 CaO.P O5. High FeO cntent f slag means an xidizing slag (r high xidizing pwer f slag) and high CaO cntent means a basic slag. Frm the pint f view f thermdynamics we thus require high activity f FeO and high activity f CaO in the slag t remve phsphrus frm metal t slag. Frm the pint f view f irn yield and refractry lining life f the vessel, hwever, high mass % f FeO in slag is undesirable. The simultaneus effects f temperature, lime and FeO cntents f slag n phsphrus distributin rati are described by Healy s equatin [1] which is based n inic thery f slag % P lg % P where, 350 16.0 0.08(% CaO).5lg(% Fe) T [%P] = mass % phsphrus in metal, (% P) = mass % phsphrus in slag, (%Fe) = mass % Fe in slag, (% CaO) = mass % CaO in slag Slag basicity is defined as the ratin f mass percent f CaO and SiO in the slag. It is clear that we need t ptimize, simultaneusly, the three parameters, mass % FeO and CaO and temperature s as t btain maximum pssible value f phsphrus distributin rati. Since, fr a given mass f slag, high FeO % in slag means lw % CaO in slag, there is an ptimum % FeO cntent fr a given basicity, r vice versa.

Besides Healy s equatin, as described abve, the phsphrus cntent can als be predicted frm regular slutin mdel (quadratic frmalism), ptical basicity and empirical relatin based n the mlecular thery f slag. Part (b)calculate equilibrium phsphrus and xygen cntent f metal frm Healy s equatin. If the temperature is lwered t 1913 K, then what will be the new phsphrus and xygen cntent f metal at the new temperature. Predictins using Healy s equatin: the initial mass balance f cnstituents in the slag will yield, Mass f CaO in slag = 11 tns, Mass f SiO in slag = 5 tns Mass f PO5 in slag = 1 tn Mass f FeO in slag = 3 tns Mass percent f phsphrus in slag (%P) = (%PO5) (61.9476 / 141.9446) =.180 % Mass percent f irn in slag (%Fe) = (% FeO )(55.847 / 71.8464 ) = that all irn in slag is present as Fe + nly. 11.6597 % ; it is assumed Accrding t Healy s equatin, based n inic thery f slag, inserting he given values, ( P) lg [ P].180 lg [ P] 350 193 16.0 + 0.08(55) +.5 lg(11.6597) The initial phsphrus cntent f slag is = 5*6/14=.689 Inserting he given values int Healy s equatin, ( P) lg [ P].180 lg [ P] 350 193 16.0 + 0.08(55) +.5 lg(11.6597) Hence, lg [%P] = -.351 Or, [% P] = 0.0045 %, i.e. metal initially cntains 0.0045 mass % P. Equilibrium cnstant fr the reactin {Fe} + [O] = (FeO) is given by: 637 lg K FeO.73 T At 193 K, KFeO = 3.8333. At 1913 K = 3.9893. Nw, K = afeo /h = 3.8333 Assuming ideal Henrian behaviur f xygen, 3

h= [wt %O] Als, afeo = FeO * FeO where FeOis the activity cefficient and FeO is mle fractin f FeO in slag. The initial mle fractins f different cmpnents in slag are, FeO = 0.17, CaO = 0.5977, SiO = 0.536, PO5 = 0.015 On applying Regular Slutin Mdel t calculate the activity cefficient f FeO in slag, RT ln P O + FeO FeOCaO CaO FeOSiO SiO FeOP O ( FeO SiO ( FeO O 5 5 SiO ) FeOP O 5 CaOSiO CaO SiO FeO P O 5 SiO P ) O 5 SiO P 5 + ( 5 FeO FeOCaO CaO P O5 CaO P O 5 P ) 8540 7. 17T On substituting the respective interactin parameter () values frm literature and the mle fractins different cnstituents as calculated abve, Or, RT ln FeO = -31380(0.5977) + (-41840)(0.536) + (-31380)(0.015) + (-31380-41840+133890)0.5977*0.536+ (-41840-31380-83680)0.4 x 0.015 + (-31380-31380+51040) * 0.5977 * 0.015-8540 + 7.17 * 193. = 038.588 FeO = 1.136 Therefre, afeo = 1.136 x 0.17 = 0.1445 Hence, [wt % O] = afeo / K =0.1445 / 3.8333 = 0.0377 The initial xygen cntent f metal is therefre 0.0%. Nw we can calculate the phsphrus and xygen cntent f metal at new temperature, 1913 K. Mass Balances: The first step is t calculate ttal mass f irn, xygen, phsphrus, lime and silica. -Irn balance: Irn in metal + irn in slag =300 x ( ( 100-0.0377-0.0045) /100 ) + 3 x 55.847 / 71.8464 = 30.0533 tn -Oxygen balance: Disslved xygen in metal + xygen in FeO and PO5 in slag (lime and silica d nt dissciate) 4

= Initial mass f FeO x (15.994 / 71.8464) + Initial mass f PO5 x (79.997/141.9666) + disslved mass % xygen in metal * (300 /100) =1.3447 tn -Phsphrus balance: mass f phsphrus in slag + mass f phsphrus in metal = mass f PO5 in slag x ( 61.9476 / 141.9466) +( mass %P in metal /100) x 300 =0.44991 tn -Lime and silica balance: ttal initial mass f silica and lime is 5+ 11 = 16 tn The next step is t write dwn mass cnservatin equatins in terms f final unknwn quantities. Let the mass % f phsphrus in metal at 1913 K be 1, mass percent f FeO in slag is, mass percent f phsphrus in slag be 3, final mass f slag be 4, final mass f metal be 5 and mass percent f disslved xygen in metal be 6. Cnservatin equatins fr different cmpnents can nw be written as fllws. -Phsphrus mass balance at new temperature x 100 3x 100 1 5 4 0.44991 On further simplificatin we get, 15 + 34 = 44.991 -----------(1) -Oxygen mass balance at new temperature, 100 ( 15.9994 718464 79.997 4 )( ) 141.9466 100 6 5 3 1.3447 () -Irn balance equatin at new temperature, 5 55.847 100 1 6) ( ) 30.0533 (3) 100 71.8464x100 ( 4 -Lime and silica balance at new temperature, 141.9466 4 [ 100 3( )]( ) 16 61.9476 100 On further simplificatin we get, 5

141.9466 [ 100 3( )] 4 1600 (4) 61.9476 The last step is t write equilibrium distributin equatins fr phsphrus and xygen. -Equilibrium phsphrus distributin rati at new temperature Applying Healy s equatin at new temperature f 1913 K, lg ( (%P) / [%P] ) = 350 1913-16.0 + 0.08 (%CaO) +.5 lg (% Fe) Or, 350 11 lg( ) 16.0 0.08( x100).5lg( 1913 3 x 1 4 55.847 71.8464 ) On further simplificatin we get, -Equilibrium disslved xygen in metal at new temperature 3 88 lg( ) 4.5903.5lg( ) 1 4 (5) At new temperature the xygen cntent f metal can be calculated frm equilibrium cnstant, {Fe} + [O] = (FeO) ; at 1913 K =3.9893 (calculated earlier), where, K = a FeO h 0 FeO [ ] f FeO and the activity cefficient f xygen f can be assumed t be unity and FeO can be assumed t be nearly same as calculated earlier at 193 K (i.e.1.1316). The mle fractin f FeO in slag, FeO, can be calculated as shwn belw. FeO [ 100 4 100x71.8464 4 3 4 11 x71.8464 100x61.9476 56.0794 5 ] 60.0848 On substituting the value f equilibrium cnstant and the mle fractin f FeO and n further simplificatin we get, 4 3.9893 100x71.8464 1.1358 4 3 4 11 [ 100x71.8464 100x61.9476 56.0794 6 5 ] 60.0848 (6) 6