Investigation of Electric Arc Furnace Chemical Reactions and stirring effect

Investigation of Electric Arc Furnace Chemical Reactions and stirring effect Lei Deng Supervisor:Dr. Xiaojing Zhang (ABB) Niloofar Arzpeyma (KTH) Master Degree Thesis School of Industrial Engineering and Management Department of Materials Science and Engineering Royal Institute of Technology SE-1 44 Stockholm Sweden 1

Abstract Chemical energy plays a big role in the process of modern Electric Arc Furnace (EAF). The objective of this study is to compare the results of chemical reaction enthalpies calculated by four different methods. In general, the PERRY-NIST-JANAF method is used to calculate the chemical energies. However, this method heavily depend on heat capacities of the substances which have to be deduced from Perry s Chemical Engineers Handbook and NIST-JANAF Thermochemical Tables, even the calculation process is complicated. Then, some other methods are introduced: Total enthalpy method, HT (High Temperature) enthalpy method and Atomic energy method. In this thesis, the above four methods have been used to calculate the enthalpies of chemical reactions in EAF process. Both of Total enthalpy method and HT enthalpy method are not complicated, but some basic data are not available. The calculation for chemical reaction enthalpies cannot be completely made by these two methods. Atomic energy method is more complicated than Total enthalpy method and HT enthalpy method, even almost all data are available, but some results of these methods are far from those of the other three methods. The results show that values of enthalpies obtained by PERRY-NIST-JANAF method are more reasonable, though the calculation process is more complicated. In this study, it is also discussed two influencing factors on EAF process: electric power and electromagnetic stirring (EMS). 2

Contents 1. Introduction... 4 2. Literature Review... 5 2.1 Enthalpy of a Reaction... 5 2.2 Hess s law... 6 2.3 Kirchhoff s law... 6 2.4 Total enthalpy method... 7 2.4.1 Total enthalpy... 7 2.4.2 Calculation of the thermal effects... 8 2.5 Electromagnetic Stirring in Electric Arc Furnace (EAF EMS)... 9 3. Calculation of chemical reaction energy in EAF... 1 Reaction 1:... 12 Reaction 2:... 14 Reaction 3:... 17 Reaction 4:... 2 Reaction 5:... 22 Reaction 6:... 24 Reaction 7:... 26 Reaction 8:... 29 Reaction 9: (... 31 Reaction 1:... 33 Reaction 11:... 36 Reaction 12:... 38 Reaction 13:... 41 Reaction 14:... 43 Comparison of enthalpies calculated by four methods... 46 4. Influencing Factors on EAF process... 52 4.1 Power on EAF process... 52 4.2 Effects of EMS... 57 5. Summary... 62 6. Future works... 63 7. References... 64 Appendix 1 Abbreviation Index... 65 Appendix 2 Symbol List... 65 Appendix 3 Heat capacities of the substances... 65 3

1. Introduction EAF process is an essential part of steelmaking. Melting is the main task of EAF process, and it is accomplished by energy supply which includes both electric power and chemical energy. The electric power is supplied by the graphite electrodes and is the main energy supplier. Chemical energy is supplied via several sources which mainly include oxy-fuel burners and oxygen injection. Fig. 1 Schematic diagram of Electric Arc Furnace (EAF) In electric arc furnace, some oxidation and reduction reactions happen subsequently [1] : C FeO Fe CO 2 MnO Si 2 Mn SiO2 Fe 5O2 FeO (MnO C Mn CO Si 2 FeO 2 Fe SiO2 Mn FeO Fe MnO Si O2 SiO2 3 FeO 2 Cr 3 Fe Cr2O3 CO 5O2 CO2 2 Cr 1 5O2 Cr2O3 C 5O2 CO 5 FeO 2 P 5 Fe P2O5 C O2 CO2 C FeO Fe CO These chemical reactions generate a lot of chemical energy which can also improve the efficiency and reduce the time of EAF process. In EAFs, chemical reactions can affect the temperature of steel which is one of the most important factors for steelmaking. However, there are several ways 4

to calculate the EAF chemical reaction energy. In this study, four methods are applied to calculate EAF chemical energy as follows: 1) PERRY-NIST-JANAF method 2) Total enthalpy method 3) HT (High-Temperature) enthalpy method 4) Atomic energy method In this study, the enthalpies of chemical reactions are calculated by these four methods. Besides, the effects of two factors on EAF process are also studied through simulation: 1) Electric power 2) Electromagnetic stirring The results showed that the changing of these factors influence EAF process. 2. Literature Review 2.1 Enthalpy of a Reaction Enthalpy of a Reaction is defined as the amount of heat absorbed or evolved in the transformation of the reactants at a given temperature and pressure into the products at the same temperature and pressure [2]. It depends on the conditions under which the reaction is carried out. When a chemical reaction happens at constant pressure, the thermal change will not only involve the change of internal energy of the system but also the work performed either in expansion or contraction of the system. According to the First law of thermodynamics [3], it can be obtained that the heat of reaction is equal to the enthalpy of reaction at constant pressure as follows: Based on First law of thermodynamics : Qp Δ E W, (1) Where Qp is heat, W is work and E is internal energy. When the pressure p is constant, it can be expressed as: Qp Δ E PΔV (2) Heat of the reaction is the internal energy difference between products and reactants: Qp Σ Ep Σ Er P Vp Vr Σ Ep PVp Σ Er PVr (3) Enthalpy can be defined by the equation: H E PV (4) 5

Therefore Qp Σ Hp Σ Hr so: Qp Δ H (6) 2.2 Hess s law Experiments show that thermal effects of a chemical reaction are not influenced by number of reaction steps. In other words, the thermal effect of a chemical reaction is only related to the initial and final state, which is stated as Hess s law [4]. Hess s law can be used only when pressure or volume is constant. Based on the definition, Hess s law can be expressed as the following equation: (7) Bekker [5] calculated the chemical reaction power using the enthalpy of formation of substances which are assumed at high temperature. The law of his method also lies in Hess s law. 2.3 Kirchhoff s law When a chemical reaction is carried out in two different temperatures T1 and T2 at constant pressure, the thermal effects normally will not be the same. The enthalpy of the reaction at T1, (T1), can be related to that at T2, (T2), by using the following procedure: T1: ee f H H T2: ee f (1) An increase/decrease of the temperature of the reactants ( ee, ) from T1 to T2, gives the enthalpy. (2) At T2, the chemical reaction occurs, which gives the enthalpy T2 (3) An decrease/increase of the temperature of the products (f, ) from T2 to T1, gives the enthalpy as. As enthalpy is a state function, the sum of the enthalpies of the three reactions is equal to the enthalpy of the reaction at the temperature of T1. Therefore: 6

T1 T2 (8) Or T2 T1 (9) Also, the following formulas are given: C T ec T (1) C T C T (11) Then, we can have: T2 T1 C T (12) The above equation is named Kirchhoff s law. For a chemical reaction, if the values of T1 and all C are given, then T2 can be calculated based on Kirchhoff s law. The values of C can be found from Perry s Chemical Engineers Handbook [6] and NIST- JANAF Thermochemical Tables [7]. Vito [1] obtained the enthalpies of EAF chemical reactions by the difference between enthalpies of products and those of reactants as well as by the temperature integral of the difference between the heat capacities of the products and reactants. The method totally lies in Kirchhoff s law. 2.4 Total enthalpy method 2.4.1 Total enthalpy The total enthalpy of the chemical compound is the enthalpy of its formation at the temperature T from the elements in their standard state at the temperature T [8]. 7

Table 1.Total enthalpies of some chemical compounds Table 2.Total enthalpies of some individual elements Tables 1 and 2 [8] show the values of the total enthalpies of some chemical compounds and individual elements. 2.4.2 Calculation of the thermal effects 8

In traditional method, it is necessary to know the value of heat capacities when calculating the enthalpy of a chemical reaction. However, the data of heat capacities which are temperature dependent are not always available, especially for high temperature steelmaking processes. In total enthalpy method, the quantity of heat released is also named the resultant thermal effects, and designated as RES [8]. As enthalpies of reactions depend significantly on temperature conditions, so RES depends on temperatures of original substances and those of final products. Using total enthalpy method to calculate the resultant thermal effect, RES can be determined as the difference between the absolute values of total enthalpies of products of the reaction and total enthalpies of original substances. Compared to the traditional method to calculate the enthalpies, total enthalpy method is more convenient and now is used in different fields of technology, but not widespread in steelmaking field, perhaps because metallurgists are more familiar with the traditional method [8]. 2.5 Electromagnetic Stirring in Electric Arc Furnace (EAF EMS) Since 1947, ABB has supplied various electromagnetic stirring (EMS) systems for steel industries. In the early days, EMS system was used for secondary refining for EAF process. Because of the strong electromagnetic force in the melt, stirring intensity can be enhanced which will reduce the melting time. See figures [9] below. Fig. 2 Electromagnetic Stirring (EMS) in EAF Nowadays, EAF-EMS is still used for making high alloyed tool steel where some alloys with high melting points are added. For example, Uddeholm, which works on processing tool steel, 9

still takes advantage of EAF-EMS for steelmaking processes. EAF-EMS has several benefits for steelmaking, such as reducing the electric energy consumption and decreasing disturbances, both of which improve the productivity. In fact, EAF-EMS can stabilize the EAF process. 3. Calculation of chemical reaction energy in EAF In EAFs, chemical reactions can affect the temperature of steel which is one of the most important factors for steelmaking. However, there are several ways to calculate the chemical reaction energy in EAFs. In this study, four methods are applied to calculate EAF chemical energy as follows: 1) PERRY-NIST-JANAF method: Based on the formula below [1], we can calculate the enthalpies of chemical reactions one by one. H products reac ts products T ( ) ( tan ) C ( ) 298 298 p T 298 C P ( reac tants) dt (13) Before calculation, it is necessary to find the values of standard enthalpies of products and reactants as well as the expressions of heat capacities of the substances. From Perry s Chemical Engineers Handbook and NIST-JANAF Thermochemical Tables, heat capacities and standard enthalpies of substances can be found, so the method is named PERRY-NIST-JANAF method. This principle of this method lies in Kirchhoff s law, which calculate the chemical reaction energies by the difference between enthalpies of products and those of reactants as well as by the temperature integral of the difference between the heat capacities of the products and reactants. The curves of heat capacities of the substances are shown in Appendix 3. 2) Total enthalpy method: The value of the total enthalpy of a chemical compound at a certain temperature, I, is used to estimate the resultant thermal effect of the reaction, ΔH RES. H RES ( ) ( tan ) I products reac ts T I P T R * 36 Compared with PERRY-NIST-JANAF method, it is not necessary to conclude the expressions of heat capacities for Total enthalpy method. To calculate the chemical reaction enthalpies, it is important to clear the temperatures of original substances and those of final products as ΔH RES 1

depends on temperature conditions. The resultant thermal effect of the reaction, ΔH RES can be obtained by the difference between the absolute values of total enthalpies of products and total enthalpies of original substances. 3) HT (High-Temperature) enthalpy method: The calculation of this method is totally conducted by the enthalpies at high temperature. The calculation formula [5] as following shows: T T ( products ) 11 T ( reac tants) Similar to Total enthalpy method, this method is also not related to heat capacities of the substances. In order to conduct the calculation, it is necessary to find the values of enthalpies of reactants and products at temperature T. Then, the chemical reaction enthalpies can be determined by the difference between the values of enthalpies of products at T and enthalpies of reactants at T. This method is treated as the simplified method of PERRY-NIST-JANAF method. 4) Atomic energy method [11, 12] : This calculation formula is consisted by standard enthalpies of formation and enthalpies at temperature T. Similar to PERRY-NIST-JANAF method, the chemical reaction enthalpies can be determined by the difference between standard enthalpies of products and those of reactants as well as the enthalpy change by increasing temperature. However, the way to calculate is different from PERRY-NIST-JANAF method. Atomic energy method is no related to heat capacities, but to calculate the enthalpy changes of increasing temperature by using the formulas of all substances which are functions of final temperatures of the reactions. The calculation formula can be concluded as following: (14) ( products) ( reac tan ts) ( products) ( reac tan ts) H H H H T 25 25 T T (15) Based on the above methods, chemical energies of listed reactions (1-14) can be obtained. To be mentioned, since the basic data of some substances are not sufficient, some methods in several reactions cannot be conducted. However, for most reactions, chemical reaction energies can be calculated by these four methods. Abbreviations used in this thesis can be found in Appendix 1 and symbol list can be found in Appendix 2.

Reaction 1: Method 1: PERRY-NIST-JANAF method Heat capacities of substances ( ) C Fe 13 3 2 3 1 1 12 33 1 1 11 11 1 { 1 1 1 C CO 12 2 3 25 C C r n { 2 3 2 1 11 2 3 11 3 31 11 3 1 12 2 1 2 2 2 3 11 3 C FeO { 12 5 1 3 11 3 1 5 1 3 1 5 1 Enthalpies of formation used in the model. ( ) ( FeO ) 249.5, 27, ( ) 11.5 298 CO C S 298 ( ) ( ) 1 H CO FeO 298 H C S H 298 (C Fe C CO C C C FeO ) + (C Fe C CO C C C FeO ) + (C Fe C CO C C C FeO ) + (C Fe C CO C C C FeO ) + (C Fe C CO C C C FeO ) (C Fe C CO C C C FeO ) 11 5 2 2 5 ( 13 3 12 2 3 2 1 12 2 1 2 ) + ( 12 33 12 2 3 2 1 12 2 1 2 ) 12 33 12 31 12 5 1 3 12 31 12

12 5 1 3 12 31 1 3 1 12 31 1 3 1 1 1 ( 5 3 3 1 ) ( 2 5 3 51 ) 2 2 1 1 131 1 5 1 1 1 1 1 1 5 3 3 1 5 2 5 3 51 5 2 2 1 5 1 131 1 5 5 1 5 1 1 5 1 1 1 121 21 252 11 2 3 1 2 35 122 331 3 1 1 15 1 Method 2: Total enthalpy method When the chemical reaction happens at 18K, 1mole iron oxide reacts with 1 mole carbon to form 1 mole iron and 1 mole carbon monoxide. All the reactants and products are at 18K, so the chemical reaction energy can be calculated as the following formula: H RES ( I Fe(18) I CO(18) I FeO (18) I C(18) )*36 1 21 1 2 53 3 5 2 Method 3: HT enthalpy method To use HT enthalpy method, the formula can be made when the chemical reaction happens at 18K. In addition, the energy of carbon dissolved should also be taken into account in this chemical reaction. ' ( CO) ( FeO ) 1 18 C S 18 11 2 2 3 13

153 Method 4: Atomic energy method E press o for e h p of p re s s es Ws H ( Fe) (,196* T 5,3)*36/1 (25-1527 ) 2 3* 4* 36*22.4 ( ) T H CO T,117 * 1 1 28 (25-1527 ) H ( C) (,572* T 174)*36/1 (14-1724 ),226* T 6,8*36/ 1 H ( FeO ) (25-1527 ) He of for o Ws : H ( Fe) f, H( CO) f 1,55*36, H( FeO ) f 1,57*36, H( C) f ( Fe) f H( CO) f H( FeO ) f H( C) f ( Fe) H( CO) H( FeO ) H( ) H1 C ( 1,55*36* 28 1,57*36*72 ) 1 152 5 3 5 ( 11 ) 3 22 22 152 2 5 2 152 1 12] 1 2 15 Reaction 2: Method 1: PERRY-NIST-JANAF method Heat capacities of substances ( ) C Fe 13 3 2 3 1 1 12 33 1 1 11 11 1 { 1 1 1 C O2 2 25 1 3 5 14

12 2 1 2 2 2 3 11 3 C FeO { 12 5 1 3 11 3 1 5 1 3 1 5 1 Enthalpies of formation used in the model. ( ) ( FeO ) 249.5, 298 FeS ( FeO ) + 2 298 FeS (C FeO C Fe 5 C O2 ) (C FeO C Fe 5 C O2 ) (C FeO C Fe 5 C O2 ) (C FeO C Fe 5 C O2 ) + (C FeO C Fe 5 C O2 ) (C FeO C Fe 5 C O2 ) 2 5 ( 12 2 1 2 13 3 5 2 25 ) + ( 12 2 1 2 12 33 5 2 25 ) + ( 12 5 1 3 12 33 5 2 25 ) + ( 12 5 1 3 5 2 25 ) + (1 3 5 2 25 ) + (1 3 1 5 2 25 ) ]*4.184/1 2 5 ( 355 51 ) (2 3 5 1 ) + (2 25 1 53 ) + ( 2 1 ) (3 5 12 ) (2 1 5 12 ) 1 1 2 5 355 51 5 2 3 5 1 5 2 25 1 53 5 2 1 5 3 5 12 5 2 1 5 12 5 1 1 15

2 5 2 3 22 2 1 5 2 1 1 11 3 2 2 2 Method 2: Total enthalpy method When the chemical reaction happens at 18 K, 1 mole iron reacts with.5 mole oxygen to form 1 mole iron oxide. The temperature of oxygen is 298 K since it is from the ambient medium, but iron and iron oxide are at 18 K. The chemical reaction energy can be calculated as the following formula H RES ( I FeO (18) I Fe(18).5* I O2(298) )*36 2 1 21 5 3 1 Method 3: HT enthalpy method To use HT enthalpy method, the formula is calculated for the chemical reaction at 18 K. In addition, the enthalpy of dissolved iron should also be taken into account in this chemical reaction. 2 ( FeO ) 18 FeS 2 3 2 3 Method 4: Atomic energy method E press o for e h p of p re s s es Ws,226* T 6,8*36/ 1 H ( FeO ) (25-1527 ) H ( Fe) (,196* T 5,3)*36/1 (25-1527 ) 2 3* 4* 36*22,4 ( 2) T H O T,142* (25-2 ) 1 1 32 He of for o Ws : 16

H ( FeO ) f 1,57*36, H( Fe) f, H( O2) f ( FeO ) f H( Fe) f.5* H( O2) f ( FeO ) H( Fe),5* H( 2) H2 O 1 5 3 2 22 152 2 1 152 5 3 5 5 ( 1 2) 3 22 2 2 2 Reaction 3: Method 1: PERRY-NIST-JANAF method Heat capacities of substances ( ) C Fe 13 3 2 3 1 1 12 33 1 1 11 11 1 { 1 1 1 C SiO2 12 32 2 3 1 3 12 2 1 2 2 2 3 11 3 C FeO { 12 5 1 3 11 3 1 5 1 3 1 5 1 5 1 11 2 3 11 C Si { 5 3 1 11 1 5 5 1 5 1 Enthalpies of formation used in the model. ( ) ( FeO ) 249.5, ( SiO2) 91.9, 132, 45 298 298 SiS SiO 2 S ( 2* 17 SiO2) ( FeO ) + 3 SiO2S 298 298 SiS (2C Fe C SiO2 C i 2C FeO ) + (2C Fe C SiO2 C i 2C FeO ) + (2C Fe C SiO2 C i 2C FeO ) + (2C Fe C SiO2 C i 2C FeO ) + (2C Fe

C SiO2 C i 2C FeO ) + (2C Fe C SiO2 C i 2C FeO ) + (2C Fe C SiO2 C i 2C FeO ) 1 5 2 2 5 132 + (2 13 3 12 5 1 2 12 2 1 2 ) + (2 12 33 12 5 1 2 12 2 1 2 ) + (2 12 33 12 5 3 1 2 12 5 1 3 ) + (2 12 5 3 1 2 12 5 1 3 ) + (2 12 5 3 1 2 1 3) (2 1 12 5 3 1 2 1 3) (2 1 12 5 2 1 3) *4.184/1 32 ( 2 13 3 ) ( 5 5 ) ( 5 2 1 ) ( 1 2 ) ( 3 3 ) ( 5 1 3 ) ( 3 ) ]*4.184/1 32 {[ 2 13 3 5 ] [ 5 5 5 ] [ 5 2 1 5 ] [ 1 2 5 ] [ 3 3 5 ] [ 5 1 3 5 ] [ 3 5 ] } 32 31 1 2 12 3 3 13 2 15 32 5 53 33 3 Method 2: Total enthalpy method 18

When the chemical reaction happens at 18K, 2mole iron oxide reacts with 1 mole silicon to form 2 mole iron and 1 mole silicon oxide. All the reactants and products are at 18K, so the chemical reaction energy can be calculated as the following formula: H ( RES 2* I Fe(18) I SiO2(18) 2* I FeO (18) I Si(18) )*36 2 1 21 22 32 2 2 2 3 5 3 2 Method 3: HT enthalpy method According to the HT enthalpy method, the formula is written for the chemical reaction at 18 K. In addition, the enthalpy of the dissolved carbon should also be taken into account in this chemical reaction. ' 3 ( SiO2) * SiO2S 2 ( FeO ) 18 18 5 2 2 3 132 3 Method 4: Atomic energy method E press o for e h p of p re s s es Ws H ( Fe) (,196* T 5,3)*36/1 (25-1527 ),316* T 8,9*36/ 1 H ( SiO2) (25-2 ),272* T 7,3*36/ 1 H ( Si) (25-1527 ),226* T 6,8*36/ 1 H ( FeO ) (25-1527 ) He of for o Ws : H ( Fe) f, H( Si) f, H( FeO ) f 1,57*36, H( SiO2) f 4,71*36 2H ( Fe) f H( SiO2) f 2H( FeO ) f H( Si) f 2H ( Fe) H( SiO2) 2H( FeO ) H( ) 2 Si 1 3 2 1 5 3 2 2 1 152 5 3 SiS 19

5 31 152 2 22 152 2 2 2 152 3 2 331 3 2 32 1 1 Reaction 4: Method 1: PERRY-NIST-JANAF method Heat capacities of substances ( ) C SiO2 12 32 2 3 1 3 5 1 11 2 3 11 C Si { 5 3 1 11 1 5 5 1 5 1 C O2 2 25 1 3 5 Enthalpies of formation used in the model. ( ) ( SiO 2) 91.9, 132, 45 298 SiS SiO 2 S 4 ( SiO 2) SiO2S 298 SiS (C SiO2 C i C O2 ) + (C SiO2 C i C O2 ) + (C SiO2 C i C O2 ) 1 5 132 [ ((12 ) (5 1 ) ( 2 25 )) ((12 ) 5 3 1 ( 2 25 )) ((12 ) 5 ( 2 25 )) ] 2

23 ( 1 21 35 5 ) ( 3212 ) ( 1 212 ) 1 1 23 {[ 1 21 35 5 5 ] [ 3212 5 ] [ 1 212 5 ] } 23 122 5 1 23 2 13 15 1 Method 2: Total enthalpy method When the chemical reaction happens at 18K, 1 mole silicon reacts with.5 mole oxygen to form 1 mole silicon oxide. Because oxygen is injected from normal environment, so the temperature of oxygen is 298K, but silicon and silicon oxide are at 18K. The chemical reaction energy can be calculated as the following formula: H RES ( I SiO2(18) I Si(18) I O2(298) )*36 223 2 2 3 1 2 Method 3: HT enthalpy method To use HT enthalpy method, the formula can be made when the chemical reaction happens at 18K. In addition, the energy of silicon dissolved and silicon oxide dissolved should also be taken into account in this chemical reaction. ' ( SiO2) 4 18 SiO 2 S SiS 5 132 53 Method 4: Atomic energy method E press o for e h p of p re s s es Ws 21

,316* T 8,9*36/ 1 H ( SiO2) (25-2 ),272* T 7,3*36/ 1 H ( Si) (25-1527 ) 2 3* 4* 36*22,4 ( 2) T H O T,142* (25-2 ) 1 1 32 He of for o Ws : H ( Si) f, H( O2) f, H( SiO2) f 4,71*36 ( SiO2) f H( Si) f H( O2) f ( SiO2) H( Si) H( 2) H4 O 1 3 31 152 2 2 152 3 2 ( 3 152 2 Reaction 5: 1 3 1 Method 1: PERRY-NIST-JANAF method 152 1 Heat capacities of substances ( ) 1 2) 3 22 C CO2 { 1 3 2 1 55 2 3 12 11 133 12 1 C CO 12 2 3 25 C O2 2 25 1 3 5 Enthalpies of formation used in the model. ( ) ( CO) 11.5, ( CO2) 393.5 298 298 H ( 2) ( ) 5 H CO CO 298 H 298 (C CO2 C CO 5 C O2 ) + (C CO2 C CO 5 C O2 ) 22

3 3 5 11 5 ( 1 3 2 12 5 2 25 ) ( 11 133 12 5 ( 2 25 )) 1 1 2 3 ( 3 5 1 11 ) (1 125 ) 1 1 5 ] } 2 3 {[ 3 5 1 11 5 ] [1 125 2 3 3 5 1 2 Method 2: Total enthalpy method When the chemical reaction happens at 18K, 1 mole carbon monoxide reacts with.5 mole oxygen to form 1 mole carbon dioxide. Because oxygen is injected from normal environment, so the temperature of oxygen is 298K, but carbon monoxide and carbon dioxide are at 18K. The chemical reaction energy can be calculated as the following formula: H RES ( I CO2(18) I CO(18).5* I O2(298) )*36 31 1 5 3 253 2 Method 3: HT enthalpy method To use HT enthalpy method, the formula can be made when the chemical reaction happens at 18K: ' ( CO2) ( CO) 5 18 18 3 11 2 23

Method 4: Atomic energy method E press o for e h p of p re s s es Ws : 2 6* 6* 36*22,4 ( 2) T H CO T,38* 1 1 44 (25-2 ) 2 3* 4* 36*22,4 ( ) T H CO T,117* 1 1 28 (25-2 ) 2 3* 4* 36*22,4 ( 2) T H O T,142* 1 1 32 (25-2 ) He of for o Ws : H ( CO) f 1,55*36, H( O2) f, H( CO2) f 2,476*36 ( CO2) f H( CO) f,5* H( O2) f ( CO2) H( CO),5* H( 2) H5 O 2 3 1 55 3 2 ( 3 ) 3 22 ( 11 ) 3 22 5 ( 1 2) 3 22 2 5 5 1 53 2 Reaction 6: Method 1: PERRY-NIST-JANAF method Heat capacities of substances ( ) C CO 12 2 3 25 C C r n { 2 3 2 1 11 2 3 11 3 31 11 3 1 C O2 2 25 1 3 5 24

Enthalpies of formation used in the model. ( ) ( CO ) 11.5, 27 298 CS 6 298 (298) CS (C CO C C 5 C O2 ) + (C CO C C 5 C O2 ) 11 5 2 [ ( 12 2 3 2 1 5 2 25 ) ( 12 31 5 2 25 ) ]*4.184/1 3 5 2 15 2 25 2 ]*4.184/1 5 ] } 3 5 {[ 2 15 5 ] [ 2 25 2 3 5 3 1 2 Method 2: Total enthalpy method When the chemical reaction happens at 18K, 1 mole carbon reacts with.5 mole oxygen to form 1 mole carbon monoxide. Because oxygen is injected from normal environment, so the temperature of oxygen is 298K, but carbon and carbon monoxide are at 18K. The chemical reaction energy can be calculated as the following formula: H RES ( I CO(18) I C(18).5* I O2(298) )*36 1 53 5 3 3 3 Method 3: HT enthalpy method 25

To use HT enthalpy method, the formula can be made when the chemical reaction happens at 18K. In addition, the energy of carbon dissolved should also be taken into account in this chemical reaction. 6 ( CO) 18 CS 11 2 Method 4: Atomic energy method E press o for e h p of p re s s es Ws : 2 3* 4* 36*22,4 ( ) T H CO T,117* 1 1 28 (25-2 ) 2 3* 4* 36*22,4 ( 2) T H O T,142* 1 1 32 (25-2 ) H ( C) (,572* T 174)*36/1 (14-1724 ) He of for o Ws : H ( CO) f 1,55*36, H( O2) f, H( C) f ( CO) f H( C) f,5* H( O2) f ( CO) H( C),5* H( 2) H6 O 1 55 3 2 ( 11 ) 3 22 5 2 152 1 12 5 ( 1 2) 3 22 1 3 Reaction 7: Method 1: PERRY-NIST-JANAF method Heat capacities of substances ( ) 26

C CO2 { 1 3 2 1 55 2 3 12 11 133 12 1 C C r n { 2 3 2 1 11 2 3 11 3 31 11 3 1 C O2 2 25 1 3 5 Enthalpies of formation used in the model. ( ) ( CO 2) 393.5, 27 298 CS ( CO2) + 7 298 CS (C CO2 C C C O2 ) + (C CO2 C C C O2 ) + (C CO2 C C C O2 ) 3 3 5 2 ((1 3 2 ) (2 3 2 1 ) ( 2 25 )) ( 1 3 2 31 2 25 ) ( 11 133 31 ( 2 25 )) 1 1 3 5 ( 3 135 ) ( 2 2 1 ) ( 2 ) 1 1 3 5 {[ 3 135 5 ] [ 2 2 1 5 ] [ 2 5 ] } 3 5 3 1 11 5 22 51 3 Method 2: Total enthalpy method 27

When the chemical reaction happens at 18K, 1 mole carbon reacts with 1 mole oxygen to form 1 mole carbon dioxide. Because oxygen is injected from normal environment, so the temperature of oxygen is 298 K, but carbon and carbon dioxide are at 18 K. The chemical reaction energy can be calculated as the following formula: H RES ( I CO2(18) I C(18) I O2(298) )*36 31 53 3 2 3 Method 3: HT enthalpy method To use HT enthalpy method, the formula can be made when the chemical reaction happens at 18 K. In addition, the energy of carbon dissolved should also be taken into account in this chemical reaction. ' 7 ( CO2) 18 CS 3 2 3 Method 4: Atomic energy method E press o for e h p of p re s s es Ws : 2 6* 6* 36*22,4 ( 2) T H CO T,38* 1 1 44 (25-2 ) 2 3* 4* 36*22,4 ( 2) T H O T,142* 1 1 32 (25-2 ) H ( C) (,572* T 174)*36/1 (14-1724 ) He of for o Ws : H ( CO2) f 2,476*36, H( O2) f, H( C) f ( CO2) f H( C) f H( O2) f ( CO2) H( C) H( 2) H6 O 28

2 3 ( 3 ) 3 22 5 2 152 1 12 ( 1 2) 3 22 3 2 2 1 2 3 3 Reaction 8: Method 1: PERRY-NIST-JANAF method Heat capacities of substances ( ) C Mn 3 2 3 11 5 3 5 11 131 22 131 1 3 { 11 1 3 1 C SiO2 12 (273-1973K) C MnO 3 13 3 2 2 3 1 23 5 1 11 2 3 11 C Si { 5 3 1 11 1 5 5 1 5 1 Enthalpies of formation used in the model. ( ) ( MnO ) 385, ( SiO2) 91.9, 132, 45 298 298 SiS SiO 2 S 8 SiO2S 298 ( SiO 2) 2 ( MnO) 298 (2C Mn C SiO2 2C MnO C Si ) + (2C Mn C SiO2 SiS 2C MnO C Si ) + (2C Mn C SiO2 2C MnO C Si ) + (2C Mn C SiO2 2C MnO C Si ) (2C Mn C SiO2 2C MnO C Si ) (2C Mn C SiO2 2C MnO C Si ) 29

1 5 2 3 5 132 (2 3 (12 ) 2 3 13 3 2 5 1 ) (2 5 3 5 (12 ) 2 3 13 3 2 5 1 ) (2 5 3 5 (12 ) 2 3 13 3 2 5 3 1 ) (2 22 (12 ) 2 3 13 3 2 5 3 1 ) (2 11 (12 ) 2 3 13 3 2 5 3 1 ) (2 11 (12 ) 2 3 13 3 2 5) 1 1 53 ( 2 1 2 ) (2 32 2 ) (2 3 2 ) (2 2 5 2 ) (1 1 2 2 ) (13 1 2 2 ) 1 1 53 {[ 2 1 5 2 ] [2 32 5 2 ] [2 3 5 2 ] [2 2 5 5 2 ] [1 1 2 5 2 ] [13 1 2 5 2 ] } 53 13 2 5 32 15 51 12 2 3 35 Method 4: Atomic energy method 3

E press o for e h p of p re s s es Ws H ( Mn) (,213* T 5,71)*36/1 (25-1527 ),316* T 8,9*36/ 1 H ( SiO2) (25-2 ),272* T 7,3*36/ 1 H ( Si) (25-1527 ),226* T 5,31*36/ 1 H ( MnO) (25-1527 ) He of for o Ws : H ( Mn) f, H( Si) f, H( MnO) f 1,517*36, H( SiO2) f 4,71*36 2H ( Mn) f H( SiO2) f 2H( MnO) f H( Si) f 2H ( Mn) H( SiO2) 2H( MnO) H( ) 8 Si 1 3 2 1 51 3 1 2 213 152 5 1 55 31 152 2 22 152 5 31 1 2 2 152 3 2 1 Reaction 9: ( Method 1: PERRY-NIST-JANAF method Heat capacities of substances ( ) C Mn 3 2 3 11 5 3 5 11 131 22 131 1 3 { 11 1 3 1 C CO 12 2 3 25 C MnO 3 13 3 2 2 3 1 23 C C r n { 2 3 2 1 11 2 3 11 3 31 11 3 1 Enthalpies of formation used in the model. ( ) 31

( MnO ) 385, ( CO) 11.9, 2, 27 298 298 MnS CS 9 298 ( CO ) ( MnO) MnS 298 (C Mn C CO C MnO C C ) + (C Mn C CO C MnO C C ) + (C Mn C CO C MnO C C ) + (C Mn C CO C MnO C C ) (C Mn C CO C MnO C C ) 11 2 3 5 2 ( 3 12 3 13 3 2 2 3 2 1 ) ( 5 3 5 12 3 13 3 2 2 3 2 1 ) 5 3 5 12 3 13 3 2 31 ( 22 12 3 13 3 2 31 ) (11 12 3 13 3 2 31 ) 1 1 CS 2 1 1 ( 25 32 3 2 ) (1 55 3 2 ) 2 131 3 2 52 5 31 3 2 5 31 3 2 1 1 2 1 1 {[ 25 32 5 3 2 ] [1 55 5 3 2 ] [ 2 131 5 3 2 ] [ 52 5 31 5 3 2 ] [5 31 5 3 2 ] } 2 1 1 2 5 3 11 25 13 15 2 2 1 2 35 Method 4: Atomic energy method 32

E press o for e h p of p re s s es Ws H ( Mn) (,213* T 5,71)*36/1 (25-1527 ) 2 3* 4* 36*22,4 ( ) T H CO T,117* 1 1 28 (25-2 ),226* T 5,31*36/ 1 H ( MnO) (25-1527 ) H ( C) (,572* T 174)*36/1 (14-1724 ) He of for o Ws : H ( Mn) f, H( C) f, H( MnO) f 1,517*36, H( CO) f 1,55*36 ( Mn) f H( CO) f H( MnO) f H( C) f ( Mn) H( CO) H( MnO) H( ) H9 C 1 55 3 2 1 51 3 1 213 152 5 1 55 ( 11 ) 3 22 22 152 5 31 1 5 2 152 1 12 2 1 55 Reaction 1: Method 1: PERRY-NIST-JANAF method Heat capacities of substances ( ) C Fe 13 3 2 3 1 1 12 33 1 1 11 11 1 { 1 1 1 C MnO 3 13 3 2 2 3 1 23 12 2 1 2 2 2 3 11 3 C FeO { 12 5 1 3 11 3 1 5 1 3 1 5 1 33

C Mn 3 2 3 11 5 3 5 11 131 22 131 1 3 { 11 1 3 1 Enthalpies of formation used in the model. ( ) ( MnO ) 385, ( FeO ) 249.5, 2 298 298 MnS 1 298 ( MnO ) ( FeO ) 298 (C Fe C MnO C FeO C Mn ) + (C Fe C MnO C FeO C Mn ) + (C Fe C MnO C FeO C Mn ) + (C Fe C MnO C FeO C Mn ) + (C Fe C MnO C FeO C Mn ) (C Fe C MnO C FeO C Mn ) (C Fe C MnO C FeO C Mn ) + (C Fe C MnO C FeO C Mn ) (C Fe C MnO C FeO C Mn ) MnS 3 5 2 5 2 ( 13 3 3 13 3 2 (12 2 1 2 ) 3 ) ( 12 33 3 13 3 2 (12 2 1 2 ) 3 ) ( 12 33 3 13 3 2 12 2 1 2 5 3 5 ) 12 33 3 13 3 2 12 5 1 3 5 3 5 3 13 3 2 12 5 1 3 5 3 5 3 13 3 2 12 5 1 3 22 3 13 3 2 12 5 1 3 11 3 13 3 2 1 3 11 1 3 13 3 2 1 3 11 1 1 115 5 ( 2 T 3 2T ) T ( 2 3 3 2 ) ( 13 2 3 2 34

) ( 1 5 3 2 ) 1 3 3 2 1 2 3 2 3 2 11 13 3 2 13 3 2 1 1 115 5 {[ 2 5 3 2 ] [ 2 3 5 3 2 ] [ 13 2 5 3 2 ] [ 1 5 5 3 2 ] [ 1 3 5 3 2 ] [ 1 2 5 3 2 ] [ 5 3 2 ] [ 11 13 5 3 2 ] [ 13 5 3 2 ] } 115 5 51 121 32 3 3 21 22 25 3 3 355 11 23 3 2 1 1 125 Method 4: Atomic energy method E press o for e h p of p re s s es Ws : H ( Fe) (,196* T 5,3)*36/1 (25-1527 ),226* T 5,31*36/ 1 H ( MnO) (25-1527 ),226* T 6,8*36/ 1 H ( FeO ) (25-1527 ) H ( Mn) (,213* T 5,71)*36/1 (25-1527 ) He of for o Ws : H ( Mn) f, H( Fe) f, H( MnO) f 1,517*36, H( FeO ) f 1,57*36 ( Fe) f H( MnO) f H( Mn) f H( FeO ) f ( Fe) H( MnO) H( Mn) H( ) 1 FeO 35

1 51 3 1 1 5 3 2 1 152 5 3 5 22 152 5 31 1 213 152 5 1 55 22 152 2 2 1 Reaction 11: (Only when the steel contains chromium, this chemical reaction will be taken into account.) Method 1: PERRY-NIST-JANAF method Heat capacities of substances ( ) C Fe 13 3 2 3 1 1 12 33 1 1 11 11 1 { 1 1 1 C Cr2O3 2 2 3 22 3 12 2 1 2 2 2 3 11 3 C FeO { 12 5 1 3 11 3 1 5 1 3 1 5 1 C Cr 2 5 2 3 1 23 Enthalpies of formation used in the model. ( ) ( Cr 2O3) 118.4, ( FeO ) 249.5, 42 298 298 CrS 11 298 ( Cr 2O3) 3 ( FeO ) 298 (3C Fe C Cr2O3 3C FeO 2C Cr ) + (3 p e p r2 3 CrS 3 p e 2 p r ) T + (3 p e p r2 3 3 p e 2 p r ) T + (3C Fe C Cr2O3 3C FeO 2C Cr ) + (3C Fe C Cr2O3 3C FeO 2C Cr ) (3C Fe C Cr2O3 3C FeO 2C Cr ) 36

11 3 2 5 2 [ (3 13 3 2 3 (12 2 1 2 ) 2 2 5 ) (3 12 33 2 3 (12 2 1 2 ) 2 2 5 ) T (3 12 33 2 3 12 5 1 3 2 2 5 ) T (3 2 3 12 5 1 3 2 2 5 ) (3 2 3 1 3 2 2 5 ) (3 1 2 3 1 3 2 2 5 ) ] 22 ( 15 12 ) ( 3 1 3 ) T ( 2 2 2 ) T 3 3 1 3 1 2 5 1 1 1 22 {[ 15 12 5 ] [ 3 1 3 5 ] [ 2 2 2 5 ] 3 3 1 5 3 1 5 2 5 1 5 } 1 1 22 3 3 1 1 5 2 5 5 252 1 1 1 2 2 5 Method 2: Total enthalpy method When the chemical reaction happens at 18K, 3 mole iron oxides react with 2 mole chromium to form 3 mole iron and 1 mole chromium oxide. All the reactants and products are at 18K. The chemical reaction energy can be calculated as the following formula: H RES 3* I ( 3* (18) I 2 3(18) I Fe Cr O FeO (18) 2* I Cr(18) )*36 3 1 21 2 3 2 2 1 25 3 5 1 37

Method 3: HT enthalpy method To use HT enthalpy method, the formula can be made when the chemical reaction happens at 18K. In addition, the energy of chromium dissolved should also be taken into account in this chemical reaction. ' 11 ( Cr2O3) 3* ( FeO ) 18 18 CrS 1125 3 2 3 2 35 Method 4: Atomic energy method E press o for e h p of p re s s es Ws : H ( Fe) (,196* T 5,3)*36/1 (25-1527 ),231* T 5,84*36/ 1 H ( Cr2O3) (25-1527 ),226* T 6,8*36/ 1 H ( FeO ) (25-1527 ) H ( Cr) (,173* T 4,43)*36/1 (25-1527 ) He of for o Ws : H ( Cr) f, H( Fe) f, H( FeO ) f 1,57*36, H( Cr2O3) f 2,11*36 ( Cr2O3) f 3H( Fe) f 3H( FeO ) f H( Cr) f ( Cr2O3) 3H( Fe) 3H( FeO ) 2H( ) 11 Cr 2 11 3 152 3 1 5 3 2 231 152 5 152 3 1 152 5 3 5 3 22 152 2 2 1 3 152 3 52 22 1 2 2 Reaction 12: (Only when the steel contains chromium, this chemical reaction will be taken into account.) 38

Method 1: PERRY-NIST-JANAF method Heat capacity of substances ( ) C Cr2O3 2 2 3 22 3 C Cr 2 5 2 3 1 23 C O2 2 25 1 3 5 Enthalpies of formation used in the model. ( ) ( Cr 2O3) 118.4, 42 298 CrS ( Cr2O3) 2* + (C Cr2O3 2C Cr 1 5C O2 ) 12 298 CrS 11 2 2 ( 2 2 2 5 1 5 ( 2 25 )) 1 1 3 (3 15 22 ) 1 1 3 [3 15 22 5 ] 1 1 21 5 5 Method 2: Total enthalpy method When the chemical reaction happens at 18K, 2 mole iron reacts with 1.5 mole oxygen to form 1 mole chromium oxide. Because oxygen is from normal environment, so the temperature of oxygen is 298K, but chromium and chromium oxide are at 18K. The chemical reaction energy can be calculated as the following formula: H ( I * RES Cr2O3(18) 2 I Cr(18) 1.5* I O2(298) 2 1 25 3 31 )*36 39

Method 3: HT enthalpy method To use HT enthalpy method, the formula can be made when the chemical reaction happens at 18K. In addition, the energy of chromium dissolved should also be taken into account in this chemical reaction. ' 12 ( Cr2O3) 2* 18 1125 2 2 1 1 Method 4: Atomic energy method CrS E press o for e h p of p re s s es Ws,231* T 5,84*36/ 1 H ( Cr2O3) (25-1527 ) H ( Cr) (,173* T 4,43)*36/1 (25-1527 ) 2 3* 4* 36*22,4 ( 2) T H O T,142* (25-2 ) 1 1 32 He of for o Ws : H ( Cr) f, H( O2) f, H( Cr2O3) f 2,11*36 ( Cr2O3) f 2H( Cr) f 1,5* H( O2) ( Cr2O3) 2H( Cr) 1,5* H( 2) H12 O 2 11 3 152 231 152 5 152 2 1 3 152 3 52 1 5 ( 3 152 2 113 1 1 152 1 1 2) 3 22 4

Reaction 13: Method 1: PERRY-NIST-JANAF method Heat capacities of substances ( ) C Fe 13 3 2 3 1 1 12 33 1 1 11 11 1 { 1 1 1 15 2 1 2 2 3 31 C P O1 { 3 31 1 12 2 1 2 2 2 3 11 3 C FeO { 12 5 1 3 11 3 1 5 1 3 1 5 1 C P { 5 5 2 3 31 3 31 11 11 1 Enthalpies of formation used in the model. ( ) ( P 2O5) 39.9, ( FeO ) 249.5, 29 298 298 CrS 13 298 ( P 2O5) 5* ( FeO ) 2* 298 (5C Fe C P2O5 5C FeO 2C P ) + (5C Fe C P2O5 5C FeO 2C P ) + (5C Fe C P2O5 5C FeO 2C P ) + (5C Fe C P2O5 5C FeO 2C P ) + (5C Fe C P2O5 5C FeO 2C P ) (5C Fe C P2O5 5C FeO 2C P ) (5C Fe C P2O5 5C FeO 2C P ) (5C Fe C P2O5 5C FeO 2C P ) 3 5 2 5 2 2 [ (5 13 3 15 2 1 2 5 (12 2 1 2 ) 2 5 5) (5 13 3 PS 41

15 2 1 2 5 (12 2 1 2 ) 2 3 ) (5 13 3 3 5 (12 2 1 2 ) 2 3 ) (5 12 33 3 5 (12 2 1 2 ) 2 3 ) 5 12 33 3 5 12 5 1 3 2 3 5 3 5 12 5 1 3 2 5 3 5 1 3 2 5 1 3 5 1 3 2 1 1 1 ( 3 23 133 ) ( 3 33 133 ) (1 55 2 ) (2 5 3 ) 2 55 12 135 25 1 33 1 1 1 1 {[ 3 23 133 5 ] [ 3 33 133 5 ] [1 55 2 5 ] [2 5 3 5 ] 2 55 12 5 135 5 25 1 33 1 } 1 131 52 13 12 1 22 3 1 51 1 2 25 1 1 32 1 1 1 1 25 2 2 1 5 12 Method 3: HT enthalpy method To use HT enthalpy method, the formula can be made when the chemical reaction happens at 18K. In addition, the energy of phosphorus dissolved should also be taken into account in this chemical reaction. ' 13 ( P2O5) 5* ( FeO ) 2* 18 18 3115 5 2 3 2 2 42 PS

1 2 Reaction 14: Method 1: PERRY-NIST-JANAF method Heat capacities of substances ( ) C Fe 13 3 2 3 1 1 12 33 1 1 11 11 1 { 1 1 1 C CO 12 2 3 25 C C r n { 2 3 2 1 11 2 3 11 3 31 11 3 1 12 2 1 2 2 2 3 11 3 C FeO { 12 5 1 3 11 3 1 5 1 3 1 5 1 Enthalpies of formation used in the model. ( ) ( FeO ) 249.5, ( CO) 11.5 298 298 H ( ) ( ) 14 H CO FeO 298 H 298 (C Fe C CO C C C FeO ) + (C Fe C CO C C C FeO ) + (C Fe C CO C C C FeO ) + (C Fe C CO C C C FeO ) + (C Fe C CO C C C FeO ) (C Fe C CO C C C FeO ) 11 5 2 5 ( 13 3 12 2 3 2 1 12 2 1 2 ) + ( 12 33 12 2 3 2 1 12 2 1 2 ) 12 33 12 31 12 5 1 3 12 31 43

12 5 1 3 12 31 1 3 1 12 31 1 3 1 1 13 ( 5 3 3 1 ) ( 2 5 3 51 ) 2 2 1 1 131 1 5 1 1 1 1 1 13 5 3 3 1 5 2 5 3 51 5 2 2 1 5 1 131 1 5 5 1 5 1 1 5 1 1 13 121 21 252 11 2 3 1 2 35 122 331 3 1 1 122 1 Method 2: Total enthalpy method When the chemical reaction happens at 18K, 1mole iron oxide reacts with 1 mole carbon to form 1 mole iron and 1 mole carbon monoxide. Because carbon is injected from normal environment, so the temperature of carbon is 298K, but other substances are at 18K. The chemical reaction energy can be calculated as the following formula: H RES ( I Fe(18) I CO(18) I FeO (18) I C(298) )*36 1 21 1 2 3 25 5 Method 3: HT enthalpy method To use HT enthalpy method, the formula can be made when the chemical reaction happens at 18: ' ( CO) ( FeO ) 14 18 18 44

11 2 3 133 Method 4: Atomic energy method E press o for e h p of p re s s es Ws : H ( Fe) (,196* T 5,3)*36/1 (25-1527 ) 2 3* 4* 36*22,4 ( ) T H CO T,117* (25-2 ) 1 1 28 H( C) 4,1* 1,2* T T 273 2*1 273 2 2,1*1 1972 * T 273 36 86*12 (25-2727 ),226* T 6,8*36/ 1 H ( FeO ) (25-1527 ) He of for o Ws : H ( Fe) f, H( CO) f 1,55*36, H( FeO ) f 1,57*36, H( C) f ( Fe) f H( CO) f H( FeO ) f H( C) f ( Fe) H( CO) H( FeO ) H( ) H14 C ( 1,55*36* 28 1,57*36*72 ) { 1 152 5 3 5 ( 11 ) 3 22 22 152 2 [ 1 152 2 3 1 2] 12} 1 3 2 1 2 45

Table 3.Comparison of enthalpies calculated by four methods (KJ/mole) Heat of formation PERRY-NIST-JANAF Increasing temperature Total Total enthalpy method HT Enthalpy method Heat of formation Atomic energy method Increasing temperature Total 167-16,86 15,914 56,268 153 167.63-4.815 162.815-249,5 9,3-24,2 86.616-243 -273.974 1.676-272.298-324.9-5,53-33.43 543,24-367 -324.81-2,331-327.141 [ -823,9 15,736-88,164 716,472-853 -831.399-4,339-871.738-283 4,392-278,68 253,26-279 -239.338-48,369-287.77-83,5-2,79-14,29 3,348-9 -92.311-17,172-19.483-366,5-2,398-368,898 283,6-369 -319.29-74,196-393.486-53.9 17,954-35.946-95.255 5,594-89.661 ( 281,1 16,835 297,935 289.26-7,74 281.556 [ -115,,5-1,47-125,97 2.753-4,853-2.1 ) -227,9-15,59-242,959 588,168-354 86.541-57,445 29.96 ) -934,4 12,825-921,575 899,316-141 131.43-163,325-1137.718 ] ) -174,4 259,272-1445,128-1842 139-16,86 122,914 25 5 133 215.496-52,58 162.988 46

Through the above table, it s obvious that the calculation results of these four methods show some difference for all the reactions. Besides, because some basic data are not available, so the corresponding final results cannot be obtained. In EAF process, carbon plays a major role in the process of EAF and contributes lots of chemical energy in reactions: C FeO Fe CO ( C Mn CO C 5O2 CO C FeO Fe CO C O2 CO2 Obviously, dissolved carbon participates in most of the chemical reactions, but the injected carbon only take part in the reaction which happens with iron oxide. In general, C FeO Fe CO is the main reaction in EAF process and contributes to most of the chemical energy. The Comparison of the results of four methods on this reaction shows that the values calculated from all methods, except the Total enthalpy method, are very close. This also happens on the other three reactions ( C 5O2 CO, C O2 CO, C FeO Fe CO ) which also shows the results of Total enthalpy method is a bit far from the calculation results of the other three methods. For the reaction, manganese oxide reacts with carbon, since some basic data are not available for Total enthalpy method and HT enthalpy method, only the results of the other two methods can be obtained which are quite close. Besides carbon, oxygen injected during EAF process also contributes a lot to chemical energy. In the process, oxygen mainly reacts with iron, but also reacts with some other elements in the metal through the following reactions: Fe 5O2 FeO C 5O2 CO [Si O2 SiO2 C O2 CO2 CO 5O2 CO2 2 Cr 1 5O2 Cr2O3 For the main reaction, oxygen reacts with iron oxide, it can be found that the result of chemical energy calculated by Total enthalpy method is different from the values of the other three 47

methods. However, for the other five reactions, the values of chemical energies calculated by four methods are quite close. To be mentioned, the oxidation reaction of chromium is very important for stainless steel making process. During EAF process, although carbon plays a major role in reducing iron oxide, some other alloying elements can also reduce iron oxides through the following reactions: (1) 2 e 2 e 2 (2) 3 e 2 r 3 e r2 3) (3) [ e e (4) 5 e 2 P] 5 e P2 5) Investigating on the values of chemical energies of these four reactions, it was found that the data regarding on the calculation are not sufficient. For the reaction which happens between iron oxide and manganese, the basic data of Total enthalpy method and HT enthalpy method are not available. Besides, for iron oxide reaction with phosphorous, the data of Total enthalpy method and Atomic energy method are not available. Besides the insufficient data, for reaction (3), it is obvious that the results of the other two methods have a big difference. For reaction (4), the results of the other two methods are relatively similar. For the reaction which happens between silicon and iron oxide, the results calculated by four methods are close, except for the result of Total enthalpy method which shows a bit difference. For reaction (2), the four results are different, especially for the one obtained by Atomic energy method which is far from the other three. Manganese oxide, generated due to oxidation of dissolved Mn in steel, can react with carbon and silicon: (5) (MnO C Mn CO (6) 2 MnO Si 2 Mn SiO2 In table 3, it can be seen that basic data of both of Total enthalpy method and HT enthalpy method are not available. The Comparison of the results of the other two methods shows that for reaction (5), the results are very close; but for reaction (6), the results are different. Considering the background of metallurgy, some chemical reactions listed might contribute little to chemical energy. For example, the reaction between iron oxide and phosphorous:5 FeO 48

2 P 5 Fe P2O5, since the content of P is very low in steel, the chemical reaction energy is quite limited. Compare PERRY-NIST-JANAF method and Atomic energy method, it is found that the values of the chemical reaction enthalpies are composed by two parts: heat of formation and enthalpy change by increasing temperature. In order to distinguish the two methods, Enthalpy change by increasing temperature/ Heat of formation (EC/HF) is used to study: Table 4.Comparison between PERRY-NIST-JANAF method and Atomic energy method Heat of formation PERRY-NIST-JANAF Increasing temperature EC/HF Heat of formation Atomic energy method Increasing temperature EC/HF 167-16,86 -.96 167.63-4.815 -.287-249,5 9,3 -.37-273.974 1.676 -.612-324.9-5,53 -.17-324.81-2,331.712 [ -823,9 15,736 -.19-831.399-4,339.485-283 4,392 -.53-239.338-48,369.22-83,5-2,79.248-92.311-17,172.186-366,5-2,398.65-319.29-74,196.232-53.9 17,954 -.333-95.255 5,594 -.587 ( 281,1 16,835.6 289.26-7,74 -.266 [ -115,,5-1,47.91 2.753-4,853-1.763 ) -227,9-15,59.66 86.541-57,445 -.664 49

) -934,4 12,825 -.137 131.43-163,325 -.126 ] ) -174,4 259,272 -.152 139-16,86 -.116 215.496-52,58 -.244 Investigating on the values of EC/HF of the two methods, it finds that values of EC/HF in PERRY-NIST-JANAF method are in a stable range (.5 -.35), but those in Atomic energy method show some special values: For the reaction [ e e, the EC/HF value in Atomic energy method is -1.763, which means enthalpy change by increasing temperature is almost twice of the heat of formation. Considering the EC/HF values of other chemical reactions, this value seems not reasonable. In addition, another EC/HF value of in Atomic energy method is -.664, which also seems out of reasonable range. Through the comparison, it is clear that the results of PERRY-NIST-JANAF method are more reasonable than those of Atomic energy method. Above all, although the calculation process is more complicated to that of other methods, it is still recommended to use PERRY-NIST-JANAF method to obtain the values of chemical reaction energies in order to reach more reasonable final results. From the Table.3 and Table.4, the comparison of the four methods can be made as above. Moreover, the comparison can also be made through the principles of each method: In general, the chemical reaction enthalpies are obtained by PERRY-NIST-JANAF method, we have to search the basic data to conclude the expressions of heat capacities before conducting the calculation. Investigated on the calculation process of these four methods, it finds that PERRY- NIST-JANAF method is a kind of calculation which totally based on the thermodynamics of the EAF chemical reactions. Therefore, the calculation results of the reactions should be the most accurate. 5

To make a simplification, HT enthalpy method and Total enthalpy method are generated: HT enthalpy method which does not concern the heat capacities is an approximate method of PERRY-NIST-JANAF method. This method simply treats the enthalpy change by increasing temperature is equal to the changes of enthalpies formation from standard state to the temperature T. Therefore, the final calculation results are different from the results of PERRY-NIST-JANAF method. In engineering field, it is possible to conduct the work if the approximate values of chemical enthalpies can be obtained. Besides HT enthalpy method, Total enthalpy method is another way to get the approximate results of chemical enthalpies. The method concludes the total enthalpy values of the substances which show in Table.1 and Table.2, then conduct the calculation based on the total enthalpy values of the reactants and products at their temperatures when the reactions happen. Similar to HT enthalpy method, Total enthalpy method also makes some changes for simplification: first, the enthalpies of formation are changed to the form of total enthalpies; second, thermal effects of dissolution of the substances are ignored; third, thermal effects of evolution of the substances are also ignored [8]. In fact, the calculation of this method is totally based on the states of the reactants and products. That s why the calculation results show some differences from the results of the other methods. Atomic energy method is a kind of calculation which is usually used in nuclear power field. It simplifies the changes of enthalpies of increasing temperature by using some particular formulas. However, from the point of view of thermodynamics, the enthalpy change by increasing temperature should be calculated by using the heat capacities of the substances. In addition, as this method is usually used in nuclear power field, the values of standard enthalpies of formation are also different from those of PERRY-NIST-JANAF method. Therefore, the final calculation results show the difference. Through the investigation of the principles of the four methods, the same conclusion can also be made: PERRY-NIST-JANAF method can obtain more reasonable calculation results, but other methods can also be used in suitable situations if it is necessary to simplify the calculation process. 51

4. Influencing Factors on EAF process In EAFs, besides chemical energy, there are some other factors can affect the process of EAF. In this chapter, two influencing factors are discussed: electric power and electromagnetic stirring (EMS). Given the initial conditions and operating actions, it is possible to simulate the process of EAF. If change one variable, but keep other conditions same as before, then the effects of the changed variable can be shown. In order to study the two factors, some related items are used to show the effects on EAF process: (1) Change of scrap mass (2) Change of liquid metal mass (3) Change of solid slag mass (4) Change of liquid slag mass (5) Change of FeO mass (6) Change of liquid metal temperature (7) Change of solid group temperature Based on Bekker s EAF dynamic model [5], it is possible to simulate the above research items in some particular condition. Varying the value of investigated factor, then the original curves might be changed. In order to discuss the effects of electric power and electromagnetic stirring (EMS), three values of the power and three different heat transfer coefficients will be given for making comparison. 4.1 Power on EAF process In EAFs, the electric power melts the scrap into liquid steel during the EAF process. If we vary the level of power and assume the other conditions are fixed, then it is possible to find the effects of electric power on EAF process. In order to make the EAF process more efficient, it is necessary to investigate the effects of EAF power. Through the simulation, the effects of EAF power (5 MW, 7 MW and 9 MW) on scrap melting are shown by the following figures: 52

Fig.3 Comparison of the changes of solid scrap mass at different power Fig.4 Comparison of the changes of liquid metal mass at different power 53

Fig.5 Comparison of the changes of solid slag mass at different power Fig.6 Comparison of the changes of liquid slag mass at different power 54

Fig.7 Comparison of the changes of FeO mass at different power Fig.8 Comparison of the changes of liquid metal temperature at different power 55

Fig.9 Comparison of the changes of solid group temperature at different power In Fig.3, the amount of solid scrap mass decreases with time at three investigated powers. At t = 3 min (18 s), at power of 9 MW, solid scrap mass almost reach zero which means the scrap totally melt into liquid metal. At t = 5 min (3 s), for the power of 7 MW, scrap mass totally melt into liquid metal. However, for the power of 5 MW, scrap mass cannot be totally melted even in one hour. In Fig.4, Fig.6 and Fig.7, it can be seen that liquid metal mass, liquid slag mass and FeO mass show the same tendency. In general, the amount of these three substances increase as time goes by. Investigating on each of them, it is found that the amount of each substance can reach higher point in one hour at higher power which means that the melting process goes faster at higher power. In Fig.5, the result of solid slag mass is similar to that of solid scrap mass: at the power of 9 MW. The equilibrium is reached faster than that at the power of 7 MW, but the equilibrium cannot be reached even after one hour at 5 MW. Fig.8 and Fig.9 show the effects of three powers on the temperature of liquid group and solid group: compared to the power of 7 MW, solid group melt and the temperature of liquid group 56

increases faster at 9 MW, but the solid group cannot even reach its melting temperature (18 K) at 5 MW even in one hour. Above all, EAF process goes faster at higher electric power. 4.2 Effects of EMS Since 1947, ABB has supplied various electromagnetic stirring (EMS) systems for steel industries [9]. Until now, EMS is still applied to improve the efficiency of steelmaking process. When changing the electric current, the stirring intensity can be changed and the heat transfer coefficient can also be changed indirectly. If using kther as the original heat transfer coefficient (without EMS), and kther-ems as the heat transfer coefficient with EMS. Taking kther, kther-ems (2*kther, 4*kther) into account, the results can be obtained through simulation: Fig.1 Comparison of the changes of solid scrap mass at different heat transfer coefficient 57