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-100 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..................................................................................... 10
Reaction 1:
......................................................................................................... 12
Reaction 2:
................................................................................................... 14
Reaction 3:
........................................................................................ 17
Reaction 4:
.............................................................................................................. 20
Reaction 5:
........................................................................................................... 22
Reaction 6:
............................................................................................................. 24
Reaction 7:
................................................................................................................. 26
Reaction 8:
..................................................................................... 29
Reaction 9: (
.................................................................................................. 31
Reaction 10:
........................................................................................... 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
𝐹𝑒
𝐹𝑒𝑂
𝐹𝑒
0 5𝑂2
𝑆𝑖
2 𝐹𝑒𝑂
𝑆𝑖
𝑂2
𝐶𝑂
2 𝑀𝑛𝑂
𝐹𝑒𝑂
2 𝐹𝑒
𝑆𝑖𝑂2
𝑆𝑖
𝐶
(𝑀𝑛𝑂
𝑆𝑖𝑂2
2 𝑀𝑛
𝑀𝑛
𝑀𝑛
𝐹𝑒𝑂
𝐹𝑒
3 𝐹𝑒𝑂
2 𝐶𝑟
3 𝐹𝑒
𝐶𝑂
0 5𝑂2
𝐶𝑂2
2 𝐶𝑟
𝐶
0 5𝑂2
𝐶𝑂
5 𝐹𝑒𝑂
𝐶
𝑂2
𝐶𝑂2
𝐶
1 5𝑂2
𝑆𝑖𝑂2
𝐶𝑂
𝑀𝑛𝑂
𝐶𝑟2𝑂3
𝐶𝑟2𝑂3
2𝑃
5 𝐹𝑒
𝐹𝑒𝑂
𝐹𝑒
𝑃2𝑂5
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
ΔE
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
ΔE
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
Enthalpy can be defined by the equation: H
PVp – Σ Er
E
5
PV
(4)
PVr
(3)
Therefore Qp
Σ Hp – Σ Hr so: Qp
ΔH
(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
T2:
eE
H
→
f
(1) An increase/decrease of the temperature of the reactants (
the enthalpy
eE， … …) from T1 to T2, gives
.
(2) At T2, the chemical reaction occurs , which gives the enthalpy
(3) An decrease/increase of the temperature of the products (f
the enthalpy as
T2
， … …) from T2 to T1, gives
.
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 …
(10)
∫
𝐶
T
∫
𝐶
T …
(11)
T2
Then, we can have:
T1
∫
𝐶 T
(12)
The above equation is named “Kirchhoff’s law”. For a chemical reaction, if the values of
T1 and all 𝐶
The values of 𝐶
T2 can be calculated based on Kirchhoff’s law.
are given, then
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 T0 [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
conditions, so
𝑅𝐸𝑆
𝑅𝐸𝑆
[8]
. As enthalpies of reactions depend significantly on temperature
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 [10], we can calculate the enthalpies of chemical reactions one by one.
H T   H 298( products )   H 298(reac tan ts ) 
0
0
0
 C
T
p

( products )  C P (reac tan ts ) dT
298
(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, I0, is used to
estimate the resultant thermal effect of the reaction, ΔHRES.
H
RES
0
   I ( products ) 
TP

 I T (reac tan ts ) * 3600
0
R
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 ΔHRES
10
depends on temperature conditions. The resultant thermal effect of the reaction, ΔHRES 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:
H
0
T
 H T ( products )  H T (reac tan ts )
0
0
(14)
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:
0
H T

 H (products )   H
0
0
25
25
  H (products )   H
(reac tan ts ) 
0
0
T
T
 (15)
(reac tan ts )
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.
11
Reaction 1:
Method 1: PERRY-NIST-JANAF method
Heat capacities of substances
(
)
13 0 00 3
12 0 0033
𝐶 𝐹𝑒
2 3 10 1
10 1 11
11
1
1
1 00
{10 0
𝐶 𝐶𝑂
0
𝐶 𝐶 𝑟
𝑛
0 00120
{
2
3
2 3
𝐶 𝐹𝑒𝑂
{
00
31
0 001
12 50
1 3
11
0 002 1
0 000
12 2
2500
200
2
0 001 3
2 3
11 3
11 3
1 00
2 3
11 3
11 3
1 50
1 50
1 00
Enthalpies of formation used in the model. (
H
)
( FeO )  249.5, H C S  27, H 298(CO)  110.5
0
0
298
H
H
1
∫
(𝐶 𝐹𝑒

𝐶 𝐹𝑒𝑂 )
𝐶 𝐶
𝐶 𝐶𝑂
0
0
𝐶 𝐶𝑂
𝐶 𝐶
+∫
(𝐶 𝐹𝑒
𝐶 𝐹𝑒𝑂 )
+∫
𝐶 𝐶
110 5
2
0 00120
)
12 50
∫
0 001 3
𝐶 𝐹𝑒𝑂 )
𝐶 𝐶𝑂
𝐶 𝐶
(𝐶 𝐹𝑒
(𝐶 𝐹𝑒
+∫
𝐶 𝐹𝑒𝑂 )
𝐶 𝐶𝑂
𝐶 𝐶
𝐶 𝐶𝑂
𝐶 𝐶
(𝐶 𝐹𝑒
+∫
𝐶 𝐹𝑒𝑂 )
𝐶 𝐶𝑂
(𝐶 𝐹𝑒
∫
𝐶 𝐹𝑒𝑂 )
2
5
0 002 1
0

(CO)  H C S  H 298( FeO ) 
298
(
∫
13
12 2
0 001
2
3
0 002 1
12
0 0033
0 00 3
2
0
)
+∫
12 2
0
0 00120
0
∫
12
0 00120
0 00120
(
12
0 001
2
0 0033
2
0 000
0 000
31
31
3
12 50
1 3
0 001 3
∫
1000
1
1
)
∫
10 0
0
(
5 3
∫
2
∫
5
0 000 1
1
5 3
5
1000
1
1
1000
0 003
0 000
1
)
1
0 000
31
31
1 3
( 25 3
∫
1
∫
∫
0 000 51
0 00131 1
0 000 1
1
1000
0 003
1 05
25 3
0 000 51
2
0 0020 1 0 5
1
0 00131 1
0 000 1
1201 21
0 00120
0 00120
0 0020 1
05
05
1
0
∫
05
252
1
11 2 3
1 20 35
0 000 1
122 331
05
3
150 1
Method 2: Total enthalpy method
When the chemical reaction happens at 1800K, 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 1800K, so
the chemical reaction energy can be calculated as the following formula:
H
RES
 (I
0
 I CO(1800)  I FeO (1800)  I C (1800)) * 3600
0
Fe (1800)
0 01 21
0
0 01
0
0 0 02
0 00 53
3 00
5 2
Method 3: HT enthalpy method
To use HT enthalpy method, the formula can be made when the chemical reaction happens at
1800K. In addition, the energy of carbon dissolved should also be taken into account in this
chemical reaction.
H  
(CO)  H C S  H 1800( FeO )
'
0
1
1800
0
11
2
2 3
13
153
Method 4: Atomic energy method
E press o for e h p of p re s
s
es
H ( Fe)  (0,196 *T  5,03) * 3600 /1000
Ws
(25-1527℃)
 3* 2
T  4 * T  0,0117  * 3600 * 22.4
H (CO)  
 100000000 10000

28


(25-1527℃)
H (C)  (0,572 *T 174) * 3600 /1000
(1400-1724℃)
H ( FeO )  0,226 *T  6,08* 3600 /1000
(25-1527℃)
He
of for
o
Ws
:
H ( Fe) f  0, H (CO) f  1,055 * 3600, H ( FeO ) f  1,057 * 3600, H (C) f  0
H1  H ( Fe) f  H (CO) f  H ( FeO ) f  H (C) f   H ( Fe)  H (CO)  H ( FeO)  H (C)
 (0  1,055 * 3600 * 28  1,057 * 3600 * 72  0)  0 1
(
0 011 ) 3 00 22
0 5 2 152
1
0 22
12]
1 2 15
Reaction 2:
Method 1: PERRY-NIST-JANAF method
Heat capacities of substances
13
12
𝐶 𝐹𝑒
(
)
0 00 3
0 0033
{10 0
𝐶 𝑂2
2
0 00025
1
00
2 3
10 1
11
1
10 1
11
1
1 00
300
5000
14
152
152
0
5 03
5
2
𝐶 𝐹𝑒𝑂
{
12 2
0 001
12 50
1 3
200
2
0 001 3
2 3
11 3
11 3
1 50
1 50
1 00
Enthalpies of formation used in the model. (
H
0
298
( FeO )  249.5, H Fe S  0
H
H
2
∫
(𝐶 𝐹𝑒𝑂

0
298
𝐶 𝑂2 )
𝐶 𝐹𝑒

( FeO )  H Fe S +
𝐶 𝐹𝑒
5
2
(𝐶 𝐹𝑒𝑂
+∫
0
0 5 𝐶 𝑂2 )
𝐶 𝐹𝑒
( 12 2
∫
0 00025
0 0033
12
)
05
2
0 0033
𝐶 𝐹𝑒
05
(𝐶 𝐹𝑒𝑂
∫
0 5 𝐶 𝑂2 )
2
+∫
2
05
2
𝐶 𝐹𝑒
(𝐶 𝐹𝑒𝑂
∫
13
( 12 2
0 001
)
)
+∫
10 0
0 001 3
( 12 50
+∫
)
(1 3
05
12
( 12 50
)
0 00025
0 00 3
2
+∫
0 00025
2
0 00025
)
0 001
0 00025
05
0 001 3
(1 3
+∫
05
2
05
0 00025
]*4.184/1000
2
5
1
2
5
0 00501
0 001 53
(3
∫
)
( 355
∫
(2 25
+∫
)
5
)
)
0 00012
)
(2 3 5
∫
( 0 02
+∫
∫
0 001
0 001 0
(2 1 5
0 00012
1000℃
355
05
05
𝐶 𝐹𝑒
(𝐶 𝐹𝑒𝑂
∫
0 5 𝐶 𝑂2 )
2
05
0 5 𝐶 𝑂2 )
(𝐶 𝐹𝑒𝑂
∫
0 5 𝐶 𝑂2 )
)
)
0 00501
2 25
3
1
5
05
0 001 53 0 5
0 00012
05
1000
15
23 5
0 02
21 5
0 001
0 001 0
0 00012
2
5
23
22 2
15 2
10 1 11
03
2
2 02
Method 2: Total enthalpy method
When the chemical reaction happens at 1800 K, 1 mole iron reacts with 0.5 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 1800 K. The chemical reaction energy can be calculated as the
following formula
H
RES
 (I
0
 I Fe (1800)  0.5 * I O 2( 298)) * 3600
0
FeO (1800)
0 0 02
0
0 01 21
05 0
3 00
1
Method 3: HT enthalpy method
To use HT enthalpy method, the formula is calculated for the chemical reaction at 1800 K. In
addition, the enthalpy of dissolved iron should also be taken into account in this chemical
reaction.
H  H
2
( FeO )  H Fe S
0
1800
2 3
0
2 3
Method 4: Atomic energy method
E press o for e h p of p re s
s
es
Ws
H ( FeO )  0,226 *T  6,08* 3600 /1000
(25-1527℃)
H ( Fe)  (0,196 *T  5,03) * 3600 /1000
(25-1527℃)
 3* 2
T  4 *T  0,0142  * 3600 * 22,4
H (O2)  
 100000000 10000

32


He
of for
o
Ws
:
16
(25-2000℃)
H ( FeO ) f  1,057 * 3600, H ( Fe) f  0, H (O2) f  0
H 2  H ( FeO) f  H ( Fe) f  0.5 * H (O2) f   H ( FeO )  H ( Fe)  0,5 * H (O2)
1 05
01
152
3 00
2
0
5 03
0
0 22
152
0
05 (
5
2
0 01 2) 3 00 22
2 22
Reaction 3:
Method 1: PERRY-NIST-JANAF method
Heat capacities of substances
13
12
𝐶 𝐹𝑒
(
)
0 00 3
0 0033
2 3
10 1
11
1
{10 0
𝐶 𝑆𝑖𝑂2
12 0
𝐶 𝐹𝑒𝑂
{
12 2
𝐶 𝑆𝑖
0 001
12 50
1 3
5
{
302000
0 00
0 001 3
101000
0 000 1
53
5
200
2
0 001
10 1
11
1
1 00
2 3
1
2 3
11 3
11 3
1 50
1 50
1 00
2 3
11
1 5
11
1 5
1 00
Enthalpies of formation used in the model. (
H
0
3
)
( FeO )  249.5, H 298(SiO 2)  910.9, H SiS  132, H SiO2S  45
0
298
H
H
3
∫
(2𝐶 𝐹𝑒
𝐶
𝑖

𝐶 𝑆𝑖𝑂2
2𝐶 𝐹𝑒𝑂 )
2𝐶 𝐹𝑒𝑂 )
+∫

 H 298(SiO 2)  2*H 298( FeO )  H SiS +
0
SiO2 S
+∫
(2𝐶 𝐹𝑒
0
𝐶
𝑖
2𝐶 𝐹𝑒𝑂 )
(2𝐶 𝐹𝑒
𝐶 𝑆𝑖𝑂2
+∫
𝐶 𝑆𝑖𝑂2
𝐶
17
𝑖
𝐶
(2𝐶 𝐹𝑒
𝐶 𝑆𝑖𝑂2
𝑖
2𝐶 𝐹𝑒𝑂 )
+∫
(2𝐶 𝐹𝑒
𝐶 𝑆𝑖𝑂2
𝐶
𝑖
2𝐶 𝐹𝑒𝑂 )
2𝐶 𝐹𝑒𝑂 )
(2𝐶 𝐹𝑒
+∫
10
5
5
2 2
5
(2𝐶 𝐹𝑒
𝐶 𝑆𝑖𝑂2
𝐶
132 + ∫
0 000 1
12
0 0033
12 2
0 001
53
)
2
2
53
0 00
53
0 001
0 00
5
2 1 3)
)
∫
( 52
2
1
)
30
0 003
)
)
∫
(
30
0 00
)
]
[ 52
0 013 3 0 5
0 00
(2
+∫
(2 10 0
12 0
12 0
1
∫
( 0 1
∫
( 5 10
]
]
05
]
05
]
[
30
0 00
05
]
5 53
330 3
Method 2: Total enthalpy method
18
0 000002
0 003
[ 5
05
2
0 00 5
]*4.184/1000
0 003
1
( 5
∫
30
32
0 00
12 0
(2 10 0
[
31
)
∫
∫
]
0
(2
0 001 3
05
32
2
12 0
+∫
)
(
2
(2
+∫
*4.184/1000
∫
{[
)
0 00
0 000 1
)
2 1 3)
0 013 3
0 00
2
2 1 3)
)
32
05
(
12 0
0 0033
12 50
0 001
𝑖
0 00 3
0 001
12
0 001 3
0 001
0 00
∫
12 2
(2
12 50
53
32
13
5
+∫
𝐶
2𝐶 𝐹𝑒𝑂 )
0 00
2
0 001
𝐶 𝑆𝑖𝑂2
𝑖
(2
2
12 0
0 00
12 0
+∫
12 3
0 00 5
[ 0 1
[ 5 10
0 000002
0 003
}
3 13
2
15
When the chemical reaction happens at 1800K, 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 1800K, so the
chemical reaction energy can be calculated as the following formula:
H
 (2*I Fe (1800)  I SiO2(1800)  2 * I FeO (1800)  I Si(1800)) * 3600
0
RES
0
2 0 01 21
0
0 22 32
0
2 0 0 02
0 02
3 00
5 32
Method 3: HT enthalpy method
According to the HT enthalpy method, the formula is written for the chemical reaction at 1800 K.
In addition, the enthalpy of the dissolved carbon should also be taken into account in this
chemical reaction.
H  H
(SiO 2)  H SiO2S  2*H 1800( FeO )  H SiS
'
0
3
1800
0
0
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)  (0,196 *T  5,03) * 3600 /1000
(25-1527℃)
H (SiO 2)  0,316 *T  8,09* 3600 /1000
(25-2000℃)
H (Si)  0,272 *T  7,30* 3600 /1000
(25-1527℃)
H ( FeO )  0,226 *T  6,08* 3600 /1000
(25-1527℃)
He
of for
o
Ws
:
H ( Fe) f  0, H (Si) f  0, H ( FeO ) f  1,057 * 3600, H (SiO 2) f  4,071* 3600
H 2  2H ( Fe) f  H (SiO 2) f  2H ( FeO ) f  H (Si) f   2H ( Fe)  H (SiO 2)  2H ( FeO )  H (Si)
0
0 1 3 00
0
2 1 05
3 00
19
2
0
2
01
152
5 03
5
0 31
152
0 2 2 152
0
30
331 3
0
2
0 22
152
0
2
2
2
32 1 1
Reaction 4:
Method 1: PERRY-NIST-JANAF method
Heat capacities of substances
𝐶 𝑆𝑖𝑂2
12 0
5
{
𝐶 𝑆𝑖
𝐶 𝑂2
(
302000
0 00
101000
0 000 1
53
5
2
)
0 001
1
0 00025
00
2 3
2 3
1
11
11
1 5
1 5
1 00
300
5000
Enthalpies of formation used in the model. (
H
3
)
(SiO 2)  910.9, H SiS  132, H SiO2S  45
0
298
H
H
4
∫
(𝐶 𝑆𝑖𝑂2
𝐶
𝐶
𝑖 𝐶 𝑂2 )
+∫

10
0 000 1
𝑖
5
)
)
53
)
5

 H 298(SiO 2)  H SiS 
0
SiO2 S
0 001
( 2
𝐶 𝑂2 )
(𝐶 𝑆𝑖𝑂2
+∫
(𝐶 𝑆𝑖𝑂2
𝐶
𝑖
𝐶 𝑂2 )
132
((12 0
0 00
( 2
[∫
0 00025
( 2
0 00025
)
))
0 00025
))
20
]
((12 0
∫
))
(5
∫
0 00
((12 0
0 00
23
)
( 1
∫
23
05
( 1 21
∫
122 5
1 23 2
1
0 003212
1000
]
0 00 212 0 5
( 0
∫
)
0 0035 5 0 5
[ 1
23
)
0 00 212
{[ 1 21
]
0 0035 5
[ 0
]
0 003212
}
13 15
0 1
Method 2: Total enthalpy method
When the chemical reaction happens at 1800K, 1 mole silicon reacts with 0.5 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 1800K. The chemical reaction
energy can be calculated as the following formula:
HRES  (I
0
 I Si(1800)  I O 2( 298)) * 3600
0
SiO2 (1800)
0 223 2
1
0
0 02
0
3 00
2
Method 3: HT enthalpy method
To use HT enthalpy method, the formula can be made when the chemical reaction happens at
1800K. In addition, the energy of silicon dissolved and silicon oxide dissolved should also be
taken into account in this chemical reaction.
H  H
(SiO 2)  H SiO2S  H SiS
'
0
4
1800
0
5
132
53
Method 4: Atomic energy method
E press o for e h p of p re s
s
es
Ws
21
H (SiO 2)  0,316 *T  8,09* 3600 /1000
(25-2000℃)
H (Si)  0,272 *T  7,30* 3600 /1000
(25-1527℃)
 3* 2
T  4 *T  0,0142  * 3600 * 22,4
H (O2)  
 100000000 10000

32


He
of for
o
Ws
(25-2000℃)
:
H (Si) f  0, H (O2) f  0, H (SiO 2) f  4,071* 3600
H 4  H (SiO 2) f  H (Si) f  H (O2) f   H (SiO 2)  H (Si)  H (O2)
0 1 3 00
152
30
2
0
0
0
3 152 2
0 31
152
10000
(100000000
152
0
0
02 2
0 01 2) 3 00 22
1 3
Reaction 5:
Method 1: PERRY-NIST-JANAF method
Heat capacities of substances
(
1 5500
10 3
0 002
{
11 0 0 00133
𝐶 𝐶𝑂2
𝐶 𝐶𝑂
0
𝐶 𝑂2
2
0 00120
2 3
1
0 00025
)
2 3
1200
1200
1 00
2500
00
300
5000
Enthalpies of formation used in the model. (
H
0
)
(CO)  110.5, H 298(CO2)  393.5
0
298
H
H
5
∫
(𝐶 𝐶𝑂2

0
298

(CO2)  H 298(CO) 
0
𝐶 𝐶𝑂
0 5 𝐶 𝑂2 )
+∫
𝐶 𝑂2 )
22
(𝐶 𝐶𝑂2
𝐶 𝐶𝑂
05
3 35
05
2
110 5
∫
0 00025
0 00120
)
05 ( 2
2 3
)
1
{[ 0 3 5
]
0
))
0 001 11
0 00133
1
)
0 00120
0
1000
(1 125
∫
0 00000
0 001 11 0 5
]
[1 125
0 00000
}
2 3
2
( 11
∫
0
1000
2 3
05
0 002
0 00025
( 03 5
∫
( 10 3
3 05
0 1
0
Method 2: Total enthalpy method
When the chemical reaction happens at 1800K, 1 mole carbon monoxide reacts with 0.5 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 1800K. The
chemical reaction energy can be calculated as the following formula:
H
RES
 (I
0
 I CO(1800)  0.5 * I O 2( 298)) * 3600
0
CO 2 (1800)
00
31
0
0 01
05 0
3 00
253 2
Method 3: HT enthalpy method
To use HT enthalpy method, the formula can be made when the chemical reaction happens at
1800K:
H  H
(CO2)  H 1800(CO)
'
0
5
1800
3
0
11
2
23
Method 4: Atomic energy method
E press o for e h p of p re s
s
es
Ws
:
 6* 2
T  6 *T  0,0308  * 3600 * 22,4
H (CO 2)  
 100000000 10000

44


(25-2000℃)
 3* 2
T  4 *T  0,0117  * 3600 * 22,4
H (CO)  
 100000000 10000

28


(25-2000℃)
 3* 2
T  4 *T  0,0142  * 3600 * 22,4
H (O2)  
 100000000 10000

32


(25-2000℃)
He
of for
o
Ws
:
H (CO) f  1,055 * 3600, H (O2) f  0, H (CO2) f  2,476 * 3600
H 5  H (CO2) f  H (CO) f  0,5 * H (O2) f   H (CO2)  H (CO)  0,5 * H (O2)
2
3 00
3 00 22
1 055 3 00 2
(
(
0 011 ) 3 00 22
0 030 )
05 (
0 01 2) 3 00 22
2 5 5
2
1 53
0
Reaction 6:
Method 1: PERRY-NIST-JANAF method
Heat capacities of substances
𝐶 𝐶𝑂
𝐶 𝐶 𝑟
0
𝑛
0 00120
{
2
3
(
2 3
0 002 1
0 000
𝐶 𝑂2
2
)
0 00025
2500
11
00
31
1
00
300
24
2 3
11 3
11 3
1 00
5000
Enthalpies of formation used in the model. (
H
0
298
)
(CO)  110.5, H C S  27
H
H
6
∫
(𝐶 𝐶𝑂

0
298

(298)  H C S 
𝐶 𝐶
110 5
05
2
0 000
31
0 5 𝐶 𝑂2 )
2
(
[∫
0 00025
0 00120
)
05
35
0
2
2
(
∫
0 00025
0 20
∫
(𝐶 𝐶𝑂
+∫
0
)
𝐶 𝐶
3
0 5 𝐶 𝑂2 )
0 002 1
0 00120
]*4.184/1000
0 0015
∫
2 025
0 0002
]*4.184/1000
35
05
{[ 0 20
]
0 0015
05
]
[ 2 025
0 0002
}
35
3
10 2
Method 2: Total enthalpy method
When the chemical reaction happens at 1800K, 1 mole carbon reacts with 0.5 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 1800K. The chemical
reaction energy can be calculated as the following formula:
H
RES
 (I
0
CO(1800)
0 01
 I C (1800)  0.5 * I O 2( 298)) * 3600
0
0
0 00 53
05 0
3 00
30 3
Method 3: HT enthalpy method
25
To use HT enthalpy method, the formula can be made when the chemical reaction happens at
1800K. In addition, the energy of carbon dissolved should also be taken into account in this
chemical reaction.
H  H
6
0
(CO)  H C S
1800
11
2
0
Method 4: Atomic energy method
E press o for e h p of p re s
s
es
Ws
:
 3* 2
T  4 *T  0,0117  * 3600 * 22,4
H (CO)  
 100000000 10000

28


(25-2000℃)
 3* 2
T  4 *T  0,0142  * 3600 * 22,4
H (O2)  
 100000000 10000

32


(25-2000℃)
H (C)  (0,572 *T 174) * 3600 /1000
He
of for
o
Ws
(1400-1724℃)
:
H (CO) f  1,055 * 3600, H (O2) f  0, H (C) f  0
H 6  H (CO) f  H (C ) f  0,5 * H (O2) f   H (CO)  H (C )  0,5 * H (O2)
1 055 3 00 2
0 5 2 152
1
10
3
0
12
0
(
0 011 ) 3 00 22
05 (
0 01 2) 3 00 22
Reaction 7:
Method 1: PERRY-NIST-JANAF method
Heat capacities of substances
(
)
26
1 5500
10 3
0 002
{
11 0 0 00133
𝐶 𝐶𝑂2
𝐶 𝐶 𝑟
𝑛
{
2
3
𝐶 𝑂2
2
1200
1 00
00
31
1
0 00025
1200
11
0 002 1
0 000
2 3
00
2 3
11 3
11 3
1 00
300
5000
Enthalpies of formation used in the model. (
H
0
298
(CO2)  393.5, H C S  27
H
H
7
∫
(𝐶 𝐶𝑂2

𝐶 𝑂2 )
0
298
)
𝐶 𝐶
+∫
( 2
))
𝐶 𝐶
2
((10 3
∫
∫
)
05
]
3
)
(2
( 10 3
∫
0 00025
3
0 002 1
0 002
)
0 000
( 0 03
0 000135
( 0
0 0002
∫
{[ 0 03
[ 0
5
0 002
31
( 2
0 00025
)
∫
1000
5
5
𝐶 𝐶
𝐶 𝑂2 )
0 00133
0 001
(𝐶 𝐶𝑂2
))
2
1
3
+∫
(𝐶 𝐶𝑂2
31
0
3
𝐶 𝑂2 )
0 00025
0 000
( 11

(CO2)  H C S +
3 35
∫
)
3 1 0
0 000135 0 5
0 0002
11 0 5
05
220 51
3
Method 2: Total enthalpy method
27
)
]
]
( 2 2
1
[ 2 2
}
1000
0 001
When the chemical reaction happens at 1800K, 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 1800 K. The chemical reaction energy
can be calculated as the following formula:
HRES  (I
0
 I C (1800)  I O 2( 298)) * 3600
0
CO 2 (1800)
00
31
0
0 00 53
0
3 00
2 3
Method 3: HT enthalpy method
To use HT enthalpy method, the formula can be made when the chemical reaction happens at
1800 K. In addition, the energy of carbon dissolved should also be taken into account in this
chemical reaction.
H  H
(CO2)  H C S
'
0
7
1800
3
2
3
Method 4: Atomic energy method
E press o for e h p of p re s
s
es
Ws
:
 6* 2
T  6 *T  0,0308  * 3600 * 22,4
H (CO 2)  
 100000000 10000

44


 3* 2
T  4 *T  0,0142  * 3600 * 22,4
H (O2)  
 100000000 10000

32


H (C)  (0,572 *T 174) * 3600 /1000
He
of for
o
Ws
(25-2000℃)
(25-2000℃)
(1400-1724℃)
:
H (CO2) f  2,476 * 3600, H (O2) f  0, H (C) f  0
H 6  H (CO2) f  H (C) f  H (O2) f   H (CO2)  H (C)  H (O2)
28
2
3 00
0 5 2 152
1
3 22
0
0
12
(
(
0 030 ) 3 00 22
0 01 2) 3 00 22
12
3 3
Reaction 8:
Method 1: PERRY-NIST-JANAF method
Heat capacities of substances
3
50
𝐶 𝑀𝑛
12 0
𝐶 𝑀𝑛𝑂
𝐶 𝑆𝑖
2 3 110
110
131
131
1 3
1 3 1 00
0 00
3
5
{
)
0 00
0 003 5
0 00 22
0
{ 11 0
𝐶 𝑆𝑖𝑂2
(
(273-1973K)
0 0103
0 000003 2
101000
0 000 1
53
5
0 001
2 3
2 3
11
1 5
Enthalpies of formation used in the model. (
H
0
1 23
11
1 5
1 00
)
(MnO)  385, H 298(SiO 2)  910.9, H SiS  132, H SiO2S  45
0
298
H
H
8
∫
(2𝐶 𝑀𝑛

2𝐶 𝑀𝑛𝑂
𝐶 𝑆𝑖 )
𝐶 𝑆𝑖𝑂2
𝐶 𝑆𝑖𝑂2
𝐶 𝑆𝑖 )
+∫

 H 298(SiO 2)  2H 298(MnO)  H SiS 
0
SiO2 S
+∫
(2𝐶 𝑀𝑛
2𝐶 𝑀𝑛𝑂
0
2𝐶 𝑀𝑛𝑂
𝐶 𝑆𝑖 )
(2𝐶 𝑀𝑛
𝐶 𝑆𝑖𝑂2
𝐶 𝑆𝑖𝑂2
𝐶 𝑆𝑖 )
2𝐶 𝑀𝑛𝑂
∫
(2𝐶 𝑀𝑛
𝐶 𝑆𝑖 )
29
+∫
(2𝐶 𝑀𝑛
𝐶 𝑆𝑖𝑂2
2𝐶 𝑀𝑛𝑂
𝐶 𝑆𝑖 )
𝐶 𝑆𝑖𝑂2
∫
(2𝐶 𝑀𝑛
2𝐶 𝑀𝑛𝑂
10
5
0 00
)
)
(12 0
0 001
3
(12 0
0 001
5)
53
0 00 00
1
53
0 00
)
{[ 0 2
0 000003 2
)
5
(2
∫
53
2
3
)
0 00 3
0 00000 2
0 0103
(2 32
∫
0 00000 2
)
(13
∫
0 001
05
05
∫
0 01 2
(1
0 00000 2
]
0 01 2
13
0 00000 2
0 00000 2
05
2
(2 11 0
0 00000 2
)
[1
53
50
0 000003 2
∫
0 00
)
0 00000 2
0 01 2
0 0103
3
1000
0 00 00
]
)
0 001
(2
∫
0 0103
2
1000
0 00000 2
)
3
53
0 001
0 00
)
)
2
0 000 1
0 00
(12 0
(12 0
(12 0
5
0 00 22
3
1
(2 2
∫
0 01 2
[13
2
0 00000 2
)
05
0
( 02
∫
0 00
0 000 1
(2
(2 11 0
∫
(12 0
)
)
0 000003 2
[2 32
5
3
0 000003 2
0 003 5
0 000003 2
0 00
)
0 0103
50
∫
(2
∫
0 00
)
0 0103
132
3
0 000003 2
0 003 5
53
2
(2
∫
0 0103
2 3 5
[2 2
05
05
]
[2
0 00
5 05
0 00 3
0 00000 2
0 00000 2
0 00000 2
2
]
]
302 15
35
Method 4: Atomic energy method
30
51
]
}
102
2
3
E press o for e h p of p re s
s
es
Ws
H (Mn)  (0,213 *T  5,71) * 3600 /1000
(25-1527℃)
H (SiO 2)  0,316 *T  8,09* 3600 /1000
(25-2000℃)
H (Si)  0,272 *T  7,30* 3600 /1000
(25-1527℃)
H (MnO)  0,226 *T  5,31* 3600 /1000
(25-1527℃)
He
of for
o
Ws
:
H (Mn) f  0, H (Si) f  0, H (MnO) f  1,517 * 3600, H (SiO 2) f  4,071* 3600
H 8  2H (Mn) f  H (SiO 2) f  2H (MnO) f  H (Si) f   2H (Mn)  H (SiO 2)  2H (MnO)  H (Si)
0
0 1 3 00
55
0 31
0
152
0 2 2 152
2 1 51
0
30
3 00
0
2
1
0
0 22
152
2
1
Reaction 9: (
Method 1: PERRY-NIST-JANAF method
Heat capacities of substances
𝐶 𝑀𝑛
3
50
0
𝐶 𝑀𝑛𝑂
𝐶 𝐶 𝑟
0 00120
3
𝑛
{
)
0 00
0 003 5
0 00 22
0
{ 11 0
𝐶 𝐶𝑂
(
2
2 3 110
110
131
131
1 3
1 3 1 00
2 3
0 0103
3
0 000003 2
0 002 1
0 000
2500
11
2 3
00
31
Enthalpies of formation used in the model. (
2 3
11 3
11 3
1 00
)
31
1 23
2
0 213 152
5 31
5 1
1
H
0
(MnO)  385, H 298(CO)  110.9, H MnS  20, H C S  27
0
298
H
H
9
∫
(𝐶 𝑀𝑛

𝐶 𝐶 )
0
298
+∫
𝐶 𝑀𝑛𝑂
𝐶 𝐶𝑂
𝐶 𝑀𝑛𝑂
𝐶 𝐶 )
(𝐶 𝑀𝑛
𝐶 𝐶𝑂
𝐶 𝑀𝑛𝑂
𝐶 𝐶 )
110
3
20
0 003 5
3
0 00120
0 000
31
1
(0 25
∫
0 00
0 00 32
0 52
∫
0 000003 2
31
2 11
[1 55
{[0 25
0 00
]
0 002 1
31
( 50
∫
0 000003 2
0
2
3
0 00120
(
∫
0 0103
0 00120
)
0
3
0 00 22
0 000
31
)
0 000003 2
)
)
∫
02
0 005
31
0 000003 2
1
0 000003 2
]
0 00
31 0 5
2 11
2 5 3
11 25
(1 55
5
∫
1000
0 000003 2
]
∫
0 00 0131
05
[5
2
𝐶 𝐶𝑂
0
0 000003 2
0 00 32
05
0 000003 2
(𝐶 𝑀𝑛
1000
0 000003 2
0 000003 2
𝐶 𝑀𝑛𝑂
𝐶 𝐶 )
0 000003 2
3
𝐶 𝐶𝑂
0 00
0 003 5
0 0103
+∫
𝐶 𝑀𝑛𝑂
0 0103
50
0 00120
)
3
0 000
3
0
2 11
05
0 00120
∫
𝐶 𝐶 )
( 3
∫
2
)
(11 0
0 00
2
0 000003 2
0
∫
3 5
(𝐶 𝑀𝑛
+∫
𝐶 𝐶𝑂
0 000003 2
0
0 002 1
(𝐶 𝑀𝑛
∫
0 0103
0 0103

(CO)  H MnS  H 298(MnO)  H C S 
0
[0 52
0 005
105
35
Method 4: Atomic energy method
32
[0 2
31 0 5
0 000003 2
13
]
0 000003 2
]
2
0 00 0131
}
2 0 1
E press o for e h p of p re s
s
es
Ws
H (Mn)  (0,213 *T  5,71) * 3600 /1000
(25-1527℃)
 3* 2
T  4 *T  0,0117  * 3600 * 22,4
H (CO)  
 100000000 10000

28


(25-2000℃)
H (MnO)  0,226 *T  5,31* 3600 /1000
(25-1527℃)
H (C)  (0,572 *T 174) * 3600 /1000
He
of for
o
Ws
(1400-1724℃)
:
H (Mn) f  0, H (C) f  0, H (MnO) f  1,517 * 3600, H (CO) f  1,055 * 3600
H 9  H (Mn) f  H (CO) f  H (MnO) f  H (C) f   H (Mn)  H (CO)  H (MnO)  H (C )
0
55
1 055 3 00 2
(
1 51
3 00
1
0 011 ) 3 00 22
0 5 2 152
1
0
0 22
0 213 152
152
12
2 1 55
Reaction 10:
Method 1: PERRY-NIST-JANAF method
Heat capacities of substances
13
12
𝐶 𝐹𝑒
(
0 00 3
0 0033
2 3
10 1
11
1
{10 0
𝐶 𝑀𝑛𝑂
𝐶 𝐹𝑒𝑂
3
{
12 2
12 50
1 3
0 0103
0 001
)
10 1
11
1
1 00
0 000003 2
2
0 001 3
200
2 3
2 3
11 3
11 3
1 50
1 50
1 00
33
1 23
5 31
5 1
1
3
50
𝐶 𝑀𝑛
0 00
0 003 5
0 00 22
0
{ 11 0
2 3 110
110
131
131
1 3
1 3 1 00
Enthalpies of formation used in the model. (
H
H
0
(MnO)  385, H 298( FeO )  249.5, H MnS  20
0
298
10

H
0
298
𝐶 𝑀𝑛𝑂
𝐶 𝐹𝑒𝑂
𝐶 𝑀𝑛 )
𝐶 𝑀𝑛 )
+∫
𝐶 𝑀𝑛𝑂
𝐶 𝐹𝑒𝑂
𝐶 𝑀𝑛 )
∫
𝐶 𝑀𝑛𝑂
𝐶 𝐹𝑒𝑂
3 5
(
12
)
3
𝐶 𝐹𝑒𝑂
𝐶 𝑀𝑛𝑂
𝐶 𝐹𝑒𝑂
(𝐶 𝐹𝑒
𝐶 𝑀𝑛𝑂
𝐶 𝐹𝑒𝑂
𝐶 𝑀𝑛 )
2
𝐶 𝑀𝑛 )
20
𝐶 𝐹𝑒𝑂
𝐶 𝑀𝑛𝑂
𝐶 𝐹𝑒𝑂
𝐶 𝑀𝑛 )
+∫
(𝐶 𝐹𝑒
𝐶 𝑀𝑛𝑂
𝐶 𝐹𝑒𝑂
∫
(
)
∫
0 001
2
0 001 3
50
0 003 5
∫
0 001 3
0
0 00 22
0 000003 2
1 3
11 0
115 5
0 00
∫
0 0103
0 001 3
1 3
11 0
1
1000
(
2
0 000003 2
12
0 00
)
∫
34
(
13
12
∫
50
12 50
0 0103
0 000003 2
0 0103
3
∫
0 000003 2T
2
0 0103
)
3
3
0 001
0 001 3
∫
10 0
T
0 003 5
0 000003 2
11 0
∫
(12 2
3
12 50
𝐶 𝑀𝑛 )
)
0 0033
3
(𝐶 𝐹𝑒
0 0103
0 00
50
∫
12 50
3
0 000003 2
3
3
0 000003 2
(
0 003 5
0 000003 2
0 00 3
0 0103
)
0 0103
13
2
3
12 2
3
𝐶 𝑀𝑛𝑂
0 001
0 00
(𝐶 𝐹𝑒
+∫
(𝐶 𝐹𝑒
∫
0 0033
𝐶 𝑀𝑛𝑂
∫
𝐶 𝑀𝑛 )
5
0 000003 2
12 50
(𝐶 𝐹𝑒
+∫
(𝐶 𝐹𝑒
(12 2
0 0033
𝐶 𝑀𝑛 )
+∫
(𝐶 𝐹𝑒
0 000003 2
∫

(MnO)  H 298( FeO )  H MnS 
0
(𝐶 𝐹𝑒
∫
)
0 0103
) T
0 00 2
0 0103
0 000003 2
∫
( 2 3
0 000003 2
)
(
∫
0 00
01
0 00 05
0 000003 2
0 000003 2
115 5
05
2
0 00
05
05
]
[ 1 3
[ 1
0 00
05
0 000003 2
]
[
115 5
2
0 000003 2
1
[
01
05
0 0103
[ 11
05
3
1
13
0 0103
3 00
1000
210 22
0 000003 2
]
0 00
05
]
0 00 2
05
[
0 000003 2
121 32
3
2
[
0 00 05
]
1000
]
0 000003 2
0 000003 2
]
051
101 23
0 0103
]
0 00
05
0 000003 2
0 000003 2
]
1 3
∫
11
0 000003 2
0 000003 2
2
∫
0 0103
0 00
)
0 00
0 000003 2
∫
{[
[ 2 3
1
∫
0 00
∫
0 000003 2
0 000003 2
}
25
3
3 355
125
Method 4: Atomic energy method
E press o for e h p of p re s
s
es
Ws
:
H ( Fe)  (0,196 *T  5,03) * 3600 /1000
(25-1527℃)
H (MnO)  0,226 *T  5,31* 3600 /1000
(25-1527℃)
H ( FeO )  0,226 *T  6,08* 3600 /1000
(25-1527℃)
H (Mn)  (0,213 *T  5,71) * 3600 /1000
(25-1527℃)
He
of for
o
Ws
:
H (Mn) f  0, H ( Fe) f  0, H (MnO) f  1,517 * 3600, H ( FeO ) f  1,507 * 3600
H10  H ( Fe) f  H (MnO) f  H (Mn) f  H ( FeO ) f   H ( Fe)  H (MnO)  H (Mn)  H ( FeO )
35
1 51
0 22
152
3 00
1
1 50
5 31
0
1
3 00
2
0 213 152
01
152
5 1
55
5 03
0 22
5
152
2
21
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
13
12
𝐶 𝐹𝑒
(
)
0 00 3
0 0033
2 3
10 1
11
1
{10 0
𝐶 𝐶𝑟2𝑂3
𝐶 𝐹𝑒𝑂
2 0
{
0 00 00
12 2
12 50
1 3
𝐶 𝐶𝑟
0 001
2 3
22 3
200
2
0 001 3
0 002 5
10 1
11
1
1 00
2 3
2 3
11 3
11 3
1 50
1 50
1 00
1 23
Enthalpies of formation used in the model. (
H
H
∫
)
(Cr 2O3)  1018.4, H 298( FeO )  249.5, H CrS  42
0
0
298
11

H
0
298
(3𝐶 𝐹𝑒
3 p e
𝐶 𝐶𝑟2𝑂3
0
𝐶 𝐶𝑟2𝑂3
2 p r ) T +∫
2 p r ) T +∫

(Cr 2O3)  3 H 298( FeO )  H CrS 
(3𝐶 𝐹𝑒
3𝐶 𝐹𝑒𝑂
3𝐶 𝐹𝑒𝑂
2𝐶 𝐶𝑟 )
(3 p e
p r2 3
𝐶 𝐶𝑟2𝑂3
3𝐶 𝐹𝑒𝑂
2𝐶 𝐶𝑟 )
(3𝐶 𝐹𝑒
∫
2𝐶 𝐶𝑟 )
36
+∫
(3 p e
p r2 3
3 p e
2𝐶 𝐶𝑟 )
𝐶 𝐶𝑟2𝑂3
+∫
(3𝐶 𝐹𝑒
3𝐶 𝐹𝑒𝑂
101
(12 2
3 2
2
[∫
2
)
2
2 0
0 00 00
0 001
0 0033
0 002 5
) T
2
12 50
0 001 3
) T
(
∫
( 2
{[
]
15
)
0 001
0 0033
) T
)
∫
)
2 0
)
(3 10 0
12
3
12 50
0 00 00
(3
∫
3
2 0
2 0
3
2
0 00 00
(3
∫
0 00 00
(3
∫
2
2 0
0 012
)
0 002
2
25
∫
0 012
[ 2
0 00 00
0 00 00
0 001
( 31
∫
) T
3
∫
3 1 3
1
]
[ 31
2 05
]
05
3
25
0 003 0
3
0 001
05
0 002
3
0 002 5
2 0
]
15
05
0 00 3
0 002 5
0 001
22
1
)
∫
3
∫
12
2
0 002 5
13
3 (12 2
0 002 5
22
(3
0 002 5
2
2
(3
∫
0 001 3
3 1 3
5
0 00 10
1000
0 003 0
3
0 001
05
0 00 10
05
}
1000
22
3 3
1
1 5
2 5
5
252 0
0 1
1
1000
2 2 5
Method 2: Total enthalpy method
When the chemical reaction happens at 1800K, 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 1800K.
The chemical reaction energy can be calculated as the following formula:
H
RES
 (3*I Fe (1800)  I Cr 2O3(1800)  3 * I FeO (1800)  2*I Cr (1800)) * 3600
0
3 0 01 21
5
0
0
02
0
0
3 0 0 02
1
37
2 0 01 25
3 00
Method 3: HT enthalpy method
To use HT enthalpy method, the formula can be made when the chemical reaction happens at
1800K. In addition, the energy of chromium dissolved should also be taken into account in this
chemical reaction.
H
'
 H 1800(Cr 2O3)  3*H 1800( FeO )  H CrS
0
0
11
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)  (0,196 *T  5,03) * 3600 /1000
(25-1527℃)
H (Cr 2O3)  0,231*T  5,84* 3600 /1000
(25-1527℃)
H ( FeO )  0,226 *T  6,08* 3600 /1000
(25-1527℃)
H (Cr )  (0,173 *T  4,43) * 3600 /1000
He
of for
o
Ws
(25-1527℃)
:
H (Cr ) f  0, H ( Fe) f  0, H ( FeO) f  1,507 * 3600, H (Cr 2O3) f  2,101* 3600
H11  H (Cr 2O3) f  3H ( Fe) f  3H ( FeO ) f  H (Cr ) f   H (Cr 2O3)  3H ( Fe)  3H ( FeO )  2H (Cr )
2 101 3 00 152
152
3
01
0 1 3 152
22 1
152
3
5 03
3 1 05
5
3 00
3
2
0 22
0 231 152
152
0
5
2
2
52
2
2 0
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 0
𝐶 𝐶𝑟
(
)
0 00 00
0 002 5
𝐶 𝑂2
2
2 3
22 3
2 3
1 23
1
0 00025
00
300
5000
Enthalpies of formation used in the model. (
H
H
)
(Cr 2O3)  1018.4, H CrS  42
0
298
12

H
0
298

(Cr 2O3)  2 * H CrS +∫
101
( 2
2
2
0 00025
3
∫
3
(𝐶 𝐶𝑟2𝑂3
( 2 0
∫
))
1
(3 15
0 0022
[3 15
0 00 00
2𝐶 𝐶𝑟
1 5𝐶 𝑂2 )
2
0 002 5
15
1000
0 0022
)
05
]
1
1000
1
1000
21 5 5
Method 2: Total enthalpy method
When the chemical reaction happens at 1800K, 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 1800K. The chemical reaction energy
can be calculated as the following formula:
H
RES
 (I
0
0
Cr 2O 3(1800)
02
 2*I Cr (1800)  1.5 * I O 2( 298)) * 3600
0
0 01 25
0
0
3 00
31
39
Method 3: HT enthalpy method
To use HT enthalpy method, the formula can be made when the chemical reaction happens at
1800K. In addition, the energy of chromium dissolved should also be taken into account in this
chemical reaction.
H
'
12
 H 1800(Cr 2O3)  2 * H CrS
0
1125
2
2
10 1
Method 4: Atomic energy method
E press o for e h p of p re s
s
es
Ws
H (Cr 2O3)  0,231*T  5,84* 3600 /1000
(25-1527℃)
H (Cr )  (0,173 *T  4,43) * 3600 /1000
(25-1527℃)
 3* 2
T  4 *T  0,0142  * 3600 * 22,4
H (O2)  
 100000000 10000

32


He
of for
o
Ws
(25-2000℃)
:
H (Cr ) f  0, H (O2) f  0, H (Cr 2O3) f  2,101* 3600
H12  H (Cr 2O3) f  2H (Cr ) f  1,5 * H (O2)  H (Cr 2O3)  2H (Cr )  1,5 * H (O2)
2 101 3 00 152
3
52
113
3 152
0 231 152
2
1 5 (100000000
152
10000
5
152
0 01 2) 3 00 22
1
40
2
0 1 3 152
Reaction 13:
Method 1: PERRY-NIST-JANAF method
Heat capacities of substances
(
)
13 0 00 3
12 0 0033
𝐶 𝐹𝑒
2 3 10 1
10 1 11
11
1
1
1 00
{10 0
15 2
{
3
𝐶 𝑃 𝑂10
𝐶 𝐹𝑒𝑂
{
0 10 2
12 2
0 001
12 50
1 3
200
2
0 001 3
5 5
{ 30
𝐶 𝑃
2 3
31
31 1 00
2 3
11 3
11 3
1 50
1 50
1 00
2 3 31
31
11 0
11 0 1 00
Enthalpies of formation used in the model. (
H
H
∫
0
)
( P2O5)  3009.9, H 298( FeO )  249.5, H CrS  29
0
298
13

H
0
(5𝐶 𝐹𝑒
5𝐶 𝐹𝑒𝑂
2𝐶 𝑃 )
𝐶 𝑃2𝑂5
2𝐶 𝑃 )
𝐶 𝑃2𝑂5
300
5 (12 2

( P2O5)  5 * H 298( FeO )  2 * H PS 
0
298
𝐶 𝑃2𝑂5
5𝐶 𝐹𝑒𝑂
2𝐶 𝑃 )
+∫
(5𝐶 𝐹𝑒
𝐶 𝑃2𝑂5
5𝐶 𝐹𝑒𝑂
(5𝐶 𝐹𝑒
𝐶 𝑃2𝑂5
5𝐶 𝐹𝑒𝑂
2𝐶 𝑃 )
2𝐶 𝑃 )
+∫
5𝐶 𝐹𝑒𝑂
∫
(5𝐶 𝐹𝑒
5𝐶 𝐹𝑒𝑂
5 2
0 001
2𝐶 𝑃 )
5
2
∫
𝐶 𝑃2𝑂5
(5𝐶 𝐹𝑒
+∫
(5𝐶 𝐹𝑒
+∫
(5𝐶 𝐹𝑒
𝐶 𝑃2𝑂5
5𝐶
2𝐶 𝑃 )
𝐹𝑒𝑂
𝐶 𝑃2𝑂5
5𝐶 𝐹𝑒𝑂
∫
(5𝐶 𝐹𝑒
2𝐶 𝑃 )
2 2
[∫
)
(5
13
2 5 5)
∫
41
0 00 3
(5
15 2
13
0 00 3
0 10 2
15 2
0 10 2
13
5 (12 2
0 00 3
∫
(5
∫
5
30
3
12
3
0 0033
3
5 1 3
2
)
0 001
5
5
2
∫
30 )
)
0 001
12 50
12 50
2
2
5 (12 2
3
5
∫
1
5 (12 2
0 0033
12
3
0 001
2
0 001 3
5 10 0
30 )
2
)
0 001 3
(5
∫
2
2
2
3
30 )
5
∫
5 1 3
2
1000
1 0
∫
( 3 23
)
∫
(1 55
0 02
2 055
0 00 12
∫
∫
33 1
)
∫
0 00
]
{[ 3 23
[1 55
]
0 133
0 02
2 055
25 1
131 052
1 0
1
1 0
1
(2 5
∫
1
]
05
( 3 33
∫
0 133
00 3
135
05
1000
[ 3 33
]
0 00 12 0 5
33 1
1 0
0 32
)
)
25 1
∫
1 0
0 133
[2 5
135
0 133
00 3
0 00
05
05
05
}
13 12
1000
1 220 3
1
51 1
2 0 25
1
0 1
25 2 2
5 12
Method 3: HT enthalpy method
To use HT enthalpy method, the formula can be made when the chemical reaction happens at
1800K. In addition, the energy of phosphorus dissolved should also be taken into account in this
chemical reaction.
H
'
13
 H 1800( P2O5)  5 * H 1800( FeO )  2 * H PS
0
3115
0
5 2 3
2 2
42
1
2
Reaction 14:
Method 1: PERRY-NIST-JANAF method
Heat capacities of substances
(
)
13 0 00 3
12 0 0033
𝐶 𝐹𝑒
2 3 10 1
10 1 11
11
1
1
1 00
{10 0
𝐶 𝐶𝑂
0
𝐶 𝐶 𝑟
𝑛
0 00120
{
2
3
2 3
𝐶 𝐹𝑒𝑂
{
11
0 002 1
0 000
12 2
2500
0 001
12 50
1 3
00
31
200
2
0 001 3
2 3
11 3
11 3
1 00
2 3
11 3
11 3
1 50
1 50
1 00
Enthalpies of formation used in the model. (
H
)
( FeO )  249.5, H 298(CO)  110.5
0
0
298
H
H
14
∫
(𝐶 𝐹𝑒

𝐶 𝐹𝑒𝑂 )
𝐶 𝐶
𝐶 𝐶𝑂
0
𝐶 𝐶𝑂
𝐶 𝐶
+∫
(𝐶 𝐹𝑒
𝐶 𝐹𝑒𝑂 )
+∫
𝐶 𝐶
110 5
2
5
0 00120
)
12 50
𝐶 𝐹𝑒𝑂 )
𝐶 𝐶𝑂
𝐶 𝐶
(𝐶 𝐹𝑒
(𝐶 𝐹𝑒
+∫
𝐶 𝐹𝑒𝑂 )
𝐶 𝐶𝑂
𝐶 𝐶
𝐶 𝐶𝑂
𝐶 𝐶
(𝐶 𝐹𝑒
+∫
𝐶 𝐹𝑒𝑂 )
𝐶 𝐶𝑂
(𝐶 𝐹𝑒
∫
𝐶 𝐹𝑒𝑂 )
∫
0 001 3
(
∫
0 002 1
0

(CO)  H 298( FeO ) 
0
298
13
0 00 3
12 2
0 001
2
3
0 002 1
12
0 0033
0
2
)
0 00120
+∫
12 2
0
0 00120
0
∫
43
0 00120
(
2
12
0 001
3
0 0033
2
0 000
0 000
31
31
12 50
1 3
0 001 3
∫
1000
1
13
)
0
(
5 3
2
∫
5
∫
10 0
∫
5
1
13
1000
1
)
1
0 000
31
31
1 3
( 25 3
0 000 51
0 00131 1
0 000 1
1
1000
0 003
1 05
25 3
0 000 51
2
0 0020 1 0 5
1
0 00131 1
0 000 1
1201 21
∫
1
∫
∫
05
1000
0 003
0 000
0 0020 1
5 3
0 00120
0 00120
0 000 1
13
05
1
0
∫
05
252
1
11 2 3
1 20 35
0 000 1
122 331
05
3
122 1
Method 2: Total enthalpy method
When the chemical reaction happens at 1800K, 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 1800K. The
chemical reaction energy can be calculated as the following formula:
H
 (I
RES
0
 I CO(1800)  I FeO (1800)  I C ( 298)) * 3600
0
Fe (1800)
0 01 21
0
0 01
0
0 0 02
3 00
25 5
Method 3: HT enthalpy method
To use HT enthalpy method, the formula can be made when the chemical reaction happens at
1800:
H
'
14
 H 1800(CO)  H 1800( FeO )
0
0
44
110
2 3
133
Method 4: Atomic energy method
E press o for e h p of p re s
s
es
H ( Fe)  (0,196 *T  5,03) * 3600 /1000
Ws
:
(25-1527℃)
 3* 2
T  4 *T  0,0117  * 3600 * 22,4
H (CO)  
 100000000 10000

28



(25-2000℃)

2


1,02 * T  273
2,1*100000
3600

H (C )  4,1* T  273 

 1972 *

 860 *12
2 *1000
T  273


H ( FeO )  0,226 *T  6,08* 3600 /1000
He
of for
o
Ws
(25-2727℃)
(25-1527℃)
:
H ( Fe) f  0, H (CO) f  1,055 * 3600, H ( FeO ) f  1,057 * 3600, H (C) f  0
H14  H ( Fe) f  H (CO) f  H ( FeO ) f  H (C) f   H ( Fe)  H (CO)  H ( FeO)  H (C)
 (0  1,055 * 3600 * 28  1,057 * 3600 * 72  0)  { 0 1
(
[ 1
0 011 ) 3 00 22
152
1
0 22
2 3
3
152
1
2
1 2
45
2]
152
5 03
0
5
2
12}
Table 3.Comparison of enthalpies calculated by four methods (KJ/mole)
PERRY-NIST-JANAF
[
Atomic energy method
Heat of
formation
Increasing
temperature
Total
Total
enthalpy
method
HT
Enthalpy
method
Heat of
formation
Increasing
temperature
Total
167
-16,086
150,914
56,268
153
167.63
-4.815
162.815
-249,5
9,3
-240,2
86.616
-243
-273.974
1.676
-272.298
-324.9
-5,53
-330.43
543,24
-367
-324.81
-2,331
-327.141
-823,9
15,736
-808,164
716,472
-853
-831.399
-40,339
-871.738
-283
4,392
-278,608
253,26
-279
-239.338
-48,369
-287.707
-83,5
-20,79
-104,29
30,348
-90
-92.311
-17,172
-109.483
-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
-7,704
281.556
-115,,5
-10,47
-125,97
2.753
-4,853
-2.1
-227,9
-15,059
-242,959
-354
86.541
-57,445
29.096
-934,4
12,825
-921,575
-1041
1301.043
-163,325
-1137.718
-1704,4
259,272
1445,128
139
-16,086
122,914
215.496
-52,508
162.988
(
[
)
)
]
)
46
588,168
899,316
-1842
25 5
133
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
𝐶
0 5𝑂2
𝐶
𝑂2
𝐶𝑂
𝐶
𝑀𝑛
𝐹𝑒𝑂
𝐹𝑒
(
𝐶𝑂
𝐶
𝐶𝑂
𝐶𝑂
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
𝐹𝑒𝑂
𝐹𝑒
0 5𝑂2
𝐶𝑂, 𝐶
𝑂2
𝐶𝑂, 𝐶
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
0 5𝑂2
[𝑆𝑖
𝑂2
𝐶𝑂
0 5𝑂2
𝐹𝑒𝑂
𝑆𝑖𝑂2
𝐶𝑂2
𝐶
𝐶
2 𝐶𝑟
0 5𝑂2
𝑂2
1 5𝑂2
𝐶𝑂
𝐶𝑂2
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
e
e
(3) [
2
(2) 3 e
2 r
(4) 5 e
2 P]
3 e
5 e
r2 3)
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
2𝑃
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”
PERRY-NIST-JANAF
[
Atomic energy method
Heat of
formation
Increasing
temperature
EC/HF
Heat of
formation
Increasing
temperature
EC/HF
167
-16,086
-0.096
167.63
-4.815
-0.0287
-249,5
9,3
-0.037
-273.974
1.676
-0.00612
-324.9
-5,53
-0.017
-324.81
-2,331
0.00712
-823,9
15,736
-0.019
-831.399
-40,339
0.0485
-283
4,392
-0.0053
-239.338
-48,369
0.202
-83,5
-20,79
0.248
-92.311
-17,172
0.186
-366,5
-2,398
0.0065
-319.29
-74,196
0.232
-53.9
17,954
-0.333
-95.255
5,594
-0.0587
281,1
16,835
0.06
289.26
-7,704
-0.266
-115,,5
-10,47
0.091
2.753
-4,853
-1.763
-227,9
-15,059
0.066
86.541
-57,445
-0.664
(
[
)
49
)
-934,4
12,825
-0.0137
-1704,4
259,272
-0.152
139
-16,086
-0.116
1301.043
-163,325
-0.126
215.496
-52,508
-0.244
]
)
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 (0.005 - 0.35), 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 -0.664, 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 “PERRYNIST-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.
50
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 (50 MW, 70 MW and 90 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 =
30 min (1800 s), at power of 90 MW, solid scrap mass almost reach zero which means the scrap
totally melt into liquid metal. At t = 50 min (3000 s), for the power of 70 MW, scrap mass totally
melt into liquid metal. However, for the power of 50 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 90
MW. The equilibrium is reached faster than that at the power of 70 MW, but the equilibrium
cannot be reached even after one hour at 50 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 70 MW, solid group melt and the temperature of liquid group
56
increases faster at 90 MW, but the solid group cannot even reach its melting temperature (1800 K)
at 50 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.10 Comparison of the changes of solid scrap mass at different heat transfer coefficient
57
Fig.11 Comparison of the changes of liquid metal mass at different heat transfer coefficient
Fig.12 Comparison of the changes of solid slag mass at different heat transfer coefficient
58
Fig.13 Comparison of the changes of liquid slag mass at different heat transfer coefficient
Fig.14 Comparison of the changes of FeO mass at different heat transfer coefficient
59
Fig.15 Comparison of the changes of liquid metal temperature at different heat transfer coefficient
Fig.16 Comparison of the changes of solid group temperature at different heat transfer coefficient
60
In Fig.10, the amount of solid scrap mass decreases with time at three investigated heat transfer
coefficients. In the beginning, the decreasing rates of three curves can be roughly measured:
Scrap mass: for kther-ems = 4*kther, Vd (1) = 200kg/s; for kther-ems = 2*kther, Vd (2) = 100kg/s;
for kther, Vd (3) = 40kg/s.
Then, it is obvious that the scrap mass decreases faster with higher heat transfer coefficient when
other conditions are equal. Due to higher decreasing rate, it is also found scrap is totally melt
earliest when kther-ems = 4*kther.
In Fig.11, Fig.13 and Fig.14, it can be seen that the amount of liquid metal mass, liquid slag mass
and FeO mass increase as time increases. For the increasing rates of these three substances in the
beginning, the rates of liquid metal and liquid slag can be measured, but the differences of FeO at
three heat transfer coefficients are not obvious.
Liquid metal: for kther-ems = 4*kther, Vi (1) = 140kg/s; for kther-ems = 2*kther, Vi (2) = 70kg/s;
for kther, Vi (3) = 35kg/s.
Liquid slag: for kther-ems = 4*kther, Vi (1’) = 180kg/s; for kther-ems = 2*kther, Vi (2’) = 90kg/s;
for kther, Vi (3’) = 45kg/s.
It can be seen that the amount of liquid metal and liquid slag increase faster at higher heat
transfer coefficient in the beginning. For FeO, although the increasing rates of three curves
cannot be distinguished, it still can also be found that FeO mass is a bit more at higher heat
transfer coefficient in the beginning which also means the increasing rate of FeO mass is a bit
faster at higher heat transfer coefficient.
In Fig.12, the final result of solid slag mass is similar to that of solid scrap mass. For the
decreasing rates of solid slag in the beginning:
Solid slag: for kther-ems = 4*kther, Vd (1’) = 120kg/s; kther-ems = 2*kther, Vd (2’) = 60kg/s; for
kther, Vd (3’) = 30kg/s.
From the rates of solid slag at three heat transfer coefficients, it is also found that solid slag mass
decreases faster at higher heat transfer coefficient. Due to higher decreasing rate, solid slag mass
reach equilibrium earliest when kther-ems = 4*kther.
61
Fig.15 and Fig.16 show the effects of three heat transfer coefficients on the temperature of liquid
group and solid group:
In Fig.15, It shows higher heat transfer coefficient, representative of higher force convection in
the melt due to EMS, leads to a lower melt temperature. This is due to the higher heat loss to the
solid scrap when the heat transfer coefficient is higher. It means, at higher heat transfer
coefficient, the effect on transferring heat from liquid metal to solid group is better than that at
lower heat transfer coefficient. Fig.16 shows that the temperature of solid group reach the
melting point faster at higher heat transfer coefficient which means that the melting rate of solid
group is faster at higher heat transfer coefficient.
To quantify the changing rates of liquid metal temperature and solid group temperature, the rates
in the beginning can also be measured to compare:
Liquid metal temperature: for kther-ems = 4*kther, Vd (1’’) = 10K/s; kther-ems = 2*kther, Vd (2’’) =
5K/s; for kther, Vd (3’’) = 2K/s.
Solid group temperature: for kther-ems = 4*kther, Vi (1’’) = 8K/s; for kther-ems = 2*kther, Vi (2’’) =
4K/s; for kther, Vi (3’’) = 2K/s.
The data of changing rates also show that temperature of solid group and liquid metal change
faster at higher heat transfer coefficient.
Above all, EAF process goes faster at higher heat transfer coefficient.
5. Summary
In this study, the calculation of EAF chemical reaction energies is studied and four methods are
applied to compare: Through the comparison, it is obvious that the calculation process of
“PERRY-NIST-JANAF method” is the most complicated, because it is necessary to conclude the
expressions of related heat capacities before calculation. However, this method is also the only
one which can obtain all the values of EAF chemical reaction energies. Moreover, as this method
is conducted totally based on the thermodynamics of chemical reactions in EAF, the calculation
results should be accurate. Investigating on the values of “EC/HF”, it is also shown that the range
62
of “EC/HF” resulted by “PERRY-NIST-JANAF method” is more reasonable compared to that
obtained by “Atomic energy method”.
Except the “PERRY-NIST-JANAF method”, other three methods are able to conduct the
calculation even without the expressions of heat capacities of the substances. As some basic data
are not available for “Total enthalpy method” and “HT enthalpy method”, some final results of
EAF chemical energies cannot be obtained, but it is also possible to make a rough estimation of
EAF chemical energies. In engineering field, these two methods can also be applied to conduct
the calculation if the accuracy of the results is not necessary. “Atomic energy method” is a
method used in nuclear power field, and is not suitable to be used for the calculation of EAF
chemical reaction energies. Even investigating on the values of “EC/HF”, it is found that the
range of “EC/HF” of this method is not reasonable.
Besides the calculation of EAF chemical energies, the effects of electric power and EMS on EAF
process are also studied through simulation based on Bekker’s EAF dynamic model:
Using “50MW”, “70MW” and “90MW” as EAF power respectively, it is obvious that EAF
process goes faster at higher electric power in the given conditions. Meanwhile, “kther”, “2*kther”
and “4*kther” are applied to study the effects of stirring intensity of EMS, and it finds that EAF
process goes faster at higher heat transfer coefficient.
However, this conclusion of the effects of these two factors can only be made in the general
conditions. If the conditions change, the results might have some difference.
6. Future works
When the values of enthalpies of chemical reaction obtained, further researches can be studied on
EAF process. Based on the values of chemical energies, it is possible to investigate the EAF
dynamic model which is helpful to find a way to improve the productivity and reduce energy
waste.
63
7. References
[1] Logar V., Dovžan D. and Škrjanc I.: Modeling and Validation of an Electric Arc Furnace: Part 2, Thermochemistry, ISIJ Int., 52 (2012), No. 3, 413-423.
[2] Narayanan K. V.: A Textbook of Chemical Engineering Thermodynamics, Prentice-Hall of India, New Delhi,
(2006), 62-63
[3] Achuthan M.: Engineering Thermodynamics, PHI learning, New Delhi, (2009), 85-106
[4] Halder G.: Introduction to Chemical Engineering Thermodynamics, PHI learning, New Delhi, (2009), 119
[5] Bekker J. G., Craig I. K. and Pistorius P. C.: Modeling and Simulation of an Electric Arc Furnace Process,
ISIJ Int., 39 (1999), No. 1, 23-32.
[6] Perry R.H. and Green D.W., Perry’s Chemical Engineers’ Handbook, McGraw-Hill, 1997
[7] American Institute of Phy, Nist-janaf Thermochemical Tables, Fourth Edition, 1998
[8] Toulouevski Y. N. and Zinurov I. Y.: Innovation in electric arc furnaces, Chapter 4: Energy (Heat)
Balances of Furnace, Springer Heidelberg Dordrecht London New York, (2010), 75-79
[9] Stål R. and Carlsson C., Electromagnetic stirring in electric arc furnace, Innovation in EAF and in
steelmaking processes Conference, 2009.
[10] Turkdogan E. T. and Frueham R. J.: The Making, Shaping and Treating of Steel, 10th ed., Chapter 2:
Fundamentals of Iron and Steelmaking, The AISE Steel Foundation, Pittsburgh, PA, USA, (1998), 13
[11] U.S. Atomic Energy Commission Report: ANL-5750
[12] Bulletin 542, U.S. Bureau of Mines, Washington DC, 1954
64
Appendix 1 – Abbreviation Index
EAF
Electric Arc Furnace
EMS
Electromagnetic Stirring
EAF-EMS
Electromagnetic stirring in electric arc furnace
EC/HF
Enthalpy change by increasing temperature/ Heat of formation
Appendix 2 – Symbol List
p
Heat capacity
The total enthalpy of a chemical substance at the temperature T
Σ Ep
Internal energy of the products
Σ Er
Internal energy of the reactants
Vp
Volume of the products
Vr
Volume of the reactants
Σ Hp
Enthalpy of the products
Σ Hr
Enthalpy of the reactants
(T1)
Enthalpy of the reaction at T1
(T1)
Enthalpy of the reaction at T2
𝑅𝐸𝑆
Kther
Resultant thermal effect
Heat transfer coefficient
Kther-ems
Heat transfer coefficient when using electromagnetic stirring
Vd
Decreasing rate
Vi
Increasing rate
Appendix 3 – Heat capacities of the substances
Heat capacities of substances are closely related to the calculation of chemical reaction enthalpies
in “PERRY-NIST-JANAF method”. According to the data of PERRY-NIST-JANAF, the curves
of heat capacities can be generated as following:
𝐶 𝐶 𝑟
𝑛
{
2
3
0 002 1
0 000
31
65
11
00
2 3
11 3
11 3
1 00
Fig.A-1 Heat capacity of carbon at the temperature from 273K to 1800K
𝐶 𝐶𝑂
0
0 00120
2 3
2500
Fig.A-2 Heat capacity of CO at the temperature from 273K to 1800K
𝐶 𝐶𝑂2
{
10 3
11
1 5500
0 002
0
0 00133
66
2 3
1200
1200
1 00
Fig.A-3 Heat capacity of CO2 at the temperature from 273K to 1800K
𝐶 𝐶𝑟
0 002 5
2 3
1 23
Fig.A-4 Heat capacity of Cr at the temperature from 273K to 1800K
𝐶 𝐶𝑟2𝑂3
2 0
0 00 00
67
2 3
22 3
Fig.A-5 Heat capacity of Cr2O3 at the temperature from 273K to 1800K
𝐶 𝐹𝑒
13 0 00 3
12 0 0033
2 3 10 1
10 1 11
11
1
1
1 00
{10 0
Fig.A-6 Heat capacity of Fe at the temperature from 273K to 1800K
68
𝐶 𝐹𝑒𝑂
{
12 2
12 50
1 3
0 001
2
0 001 3
200
2 3
11 3
11 3
1 50
1 50
1 00
Fig.A-7 Heat capacity of FeO at the temperature from 273K to 1800K
𝐶 𝑀𝑛
3
50
0
{ 11 0
0 00
0 003 5
0 00 22
2 3 110
110
131
131
1 3
1 3 1 00
Fig.A-8 Heat capacity of Mn at the temperature from 273K to 1800K
69
𝐶 𝑀𝑛𝑂
3
0 0103
0 000003 2
2 3
1 23
Fig.A-9 Heat capacity of MnO at the temperature from 273K to 1800K
𝐶 𝑂2
2
1
0 00025
00
300
5000
Fig.A-10 Heat capacity of O2 at the temperature from 300K to 1800K
70
5 5
{ 30
𝐶 𝑃
2 3 31
31
11 0
11 0 1 00
Fig.A-11 Heat capacity of P at the temperature from 273K to 1800K
𝐶 𝑃 𝑂10
{
15 2
3
0 10 2
2 3
31
31 1 00
Fig.A-12 Heat capacity of P2O5 at the temperature from 273K to 1800K
71
𝐶 𝑆𝑖
5
{
101000
0 000 1
53
5
0 001
2 3
11
1 5
11
1 5
1 00
Fig.A-13 Heat capacity of Si at the temperature from 273K to 1800K
𝐶 𝑆𝑖𝑂2
12 0
302000
0 00
2 3
1
3
Fig.A-14 Heat capacity of SiO2 at the temperature from 273K to 1800K
72