Charge Mix Preparation during Steel Making

Composition of Charge Mix

Charge

Mix

C % Mn

%

S % P % Si

%

N.

Mag

Moist Fem Fet Net

Fem

Charge

Qty

Liquid

Metal

DRI(GB) 1.2 0.015 0.040 1.00 88.00 93.80 90.29 4.500 4.022

DRI(CB) 0.12 0.040 0.040 1.00 82.00 92.00 87.87 8.500 7.394

CI Mold 4.00 0.30 0.100 0.350 1.50 94.02 5.50 5.156

CIBoring 3.20 0.40 0.100 0.300 1.50 1.50 4.00 89.27 1.000 0.890

HMS 0.50 0.80 0.025 0.025 0.40 1.00 97.50 5.000 4.910

Skull 0.50 0.80 0.025 0.025 0.30 5.00 93.65 .0500 0.468

MSTB 0.45 0.80 0.025 0.025 0.25 1.50 4.00 93.65 1.000 0.932

Shredded 0.20 0.60 0.015 0.015 0.20 4.00 95.20

Pig Iron 3.50 0.60 0.040 0.15 1.50 0.10 94.37

Total 26.000 23.772

91.43 %

1

Main reactions involved in the steel making

Carbon reaction

[C] + [O] {CO}

12 16 28

Mn reaction

[Mn] + [O] [MnO]

55 16 71

S reaction

[S] + (CaO) (CaS) + [O]

32 56 72 16

P reaction

2[P] + 5 [O] [P2O5]

62 80 142

Si reaction

[Si] + 2[O] [SiO2]

28 32 60

2

DRI Gas Based

Fe T = 93.8%, Fe M = 88.0%, C = 1.20%, S = 0.015%, P = 0.040%,

N. Mag. = 1 – 2%

FeO% in DRI = (Fe T – Fe M) X 72 / 56 = (93.8 – 88.0) X 72 / 56 = 7.46%

Out of 7.46% FeO, 60 – 70% FeO gets reduced and 30 – 40% FeO remains

unreduced and goes into slag.

So 60% of 7.46 = 4.48% FeO will be reduced to Fe = 4.48 X 56 / 72 =

3.49% Fe

Total Fe M = 3.49% Fe + 88% Fe M = 91.49% Fe M

4.5 Ton X 7.46% X 35% = 0.117 Ton unreduced FeO

1.2% carbon will escape, hence, net Fe M = 90.29%

1% Non Mag is present.

For 4.5 Ton we have 45 kg of non Mag

Hence total Fe M = (4500 – 45) X 90.29% Fe M = 4.022T

3

DRI Coal Based

Fe T = 92.0%, Fe M = 82.0%, C = 0.12%, S = 0.030%, P = 0.040%,

N. Mag. = 1 – 2%

FeO% in DRI = (Fe T – Fe M) X 72 / 56 = (92.0 – 82.0) X 72 / 56 = 12.85%

Out of 12.85% FeO, 60 – 70% FeO gets reduced and 30 – 40% FeO remains

unreduced and goes into slag.

So 60% of 12.85% = 7.71% FeO will be reduced to Fe = 7.71 X 56 / 72 =

5.99% Fe

Total Fe M = 5.99% Fe + 82% Fe M = 87.99% Fe M

8.5 Ton X 12.85% X 35% = 0.382 Ton unreduced FeO

0.12% carbon will escape, hence, net Fe M = 87.87%

1% Non Mag is present.

For 8.5 Ton we have 85 kg of non mag

Hence total Fe M = (8500 – 85) X 87.87% Fe M = 7.394T

C. I. Mould

4

100 – (4.0 – 0.10 C %) – (0.30 – 0.05 Mn %) – (0.100 – 0.050 S %) – (0.350 – 0.025 P

%) – (1.50 – 0.05 Si %)

= 100 – 3.9 C % - 0.25 Mn % - 0.05 S % - 0.325 P % - 1.45 Si %

= 94.02 % Fe M

C. I. Boring

100 – (3.2 – 0.10 C %) – (0.40 – 0.05 Mn %) – (0.100 – 0.050 S %) – (0.30 – 0.025 P %)

– (1.50 – 0.05 Si %) – 1.5 Non Mag % - 4.0 Moisture %

= 100 – 3.1 C % - 0.35 Mn % - 0.05 S % - 0.275 P % - 1.45 Si % - 1.5 Non Mag. % - 4.0

Moisture %

= 89.275 % Fe M

H.M.S.

100 – (0.50 – 0.10 C %) – (0.80 – 0.05 Mn %) – (0.4 – 0.05 Si %) – 1.0 Non Mag %

= 100 – 0.40 C % - 0.75 Mn % - 0.35 Si % - 1 Non Mag. %

= 97.50 % Fe M

M. S. T. B.

100 – (0.45 – 0.10 C %) – (0.80 – 0.05 Mn %) – (0.25 – 0.05 Si %) – 1.5 Non Mag % -

4.0 Moisture %

= 100 – 0.35 C % - 0.75 Mn % - 0.20 Si % - 1.5 Non Mag. % - 4.0 Moisture %

= 93.25 % Fe M

Shreaded

100 – (0.20 – 0.10 C %) – (0.60 – 0.05 Mn %) – (0.20 – 0.05 Si %) – 4.0 Dust %

= 100 – 0.10 C % - 0.55 Mn % - 0.15 Si % – 4.0 Dust %

= 95.20 % Fe M

Pig Iron

5

100 – (3.5 – 0.10 C %) – (0.60 – 0.05 Mn %) – (0.15 – 0.025 P %) - (1.5 – 0.05 Si %) –

0.10 Non Mag %

= 100 – 3.4 C % - 0.55 Mn % - 0.125 P % - 1.45 Si % - 0.1 Non Mag. %

= 94.37 % Fe M

Yield from L.M. To Billet Slag volume in E. A. F.

Scrap = 26.000 Ton

Lime = 1.400 Ton

SOF = 0.200 Ton

DPX = 0.100 Ton

R. Mass = 0.161 Ton (7 Kg / T)

E. A. F. Brick = 0.004 Ton (0.2 Kg / T)

Total Input = 27.865 Ton

Liquid Metal = 23.772 Ton

Slag Weight = 4.093 Ton

Liquid Metal produced 23.772 out of scrap charged 26.000 Ton i.e. 91.43 % and Net Slag

Produced = 4.093 Ton

At the time of injection (CPC) in different grades, FeO loss i.e. Fe loss does not

remain same. Approximately 1.5 Ton slag comes out while injection.

Out of 4.093 Ton, Slag volume having 10 % FeO (Average of different stages) =

0.409 Ton

C % C % C % C % C %

0.04 0.05 0.08 0.10 0.12

FeO % 35 30 22.50 20.0 18.33

Fe 408 349 262 233 213

% 1.69 1.45 1.09 0.97 0.88

i.e. C % at the time of injection contributes Fe loss as per grades to be made.

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Assuming 10 % FeO in slag (average of different stages)

4.093 Ton slag will contain FeO = 409 Kgs or Fe = 409 X 0.777 = 318 Kgs

Net Liquid Metal = 23.772 – 0.318 = 23.454 MT = 90.21 %

Yield in Different GradesNet Liquid Metal = 23.454 out of 26.000 Ton = 90.21 %

16MnCr5 / 20MnCr5 En-8DCr / C-45Cr Spring Steel

C = 0.10 (0.16 – 0.06) C = 0.35 (0.45 – 0.10) C = 0.50 (0.60 – 0.10)

Mn = 1.10 Mn = 0.80 Mn = 0.80

Si = 0.28 Si = 0.020 Si = 1.75

Cr = 1.00 Cr = 0.25 Cr = 0.30

2.48 % 1.60 % 3.35 %

+90.21 +90.21 +90.21

92.69 91.81 93.56

- 2.50 % ST. Loss - 2.50 % ST. Loss - 2.50 % ST. Loss

90.19 % 89.31 % 91.06 %

Average of 92.69 %, 91.81 %, 93.56 %, is 92.68 %

This stands 24.097 of 26 Ton charge, 2.5 % ST. loss of 24.097 = 602 Kgs.

Slag analysis of samples taken from EAF / LRF / VD is done to know the Feo %,

MnO %, SiO2 %, Cao %, MgO %, Al2O3 % present with the help of Volumetric

Methods of Titration with different reagents like Potassium Dichromate and

EDTA.

7

Calculation of Flux addition needed for 25 T Scrap Mix

Break up of the 25 T charge mix

20 % P. I. 5 T

34 % DRI 8.50 T

42 % HMS 10.50 T

4% MSTB 1.00 T

Total 100 % 25.00 T

Scrap

Type

C Mn S P Si N.Mag Moist. Net

Fe M

P. I.

5Ton

3.5%

175Kg

0.6%

30Kg

0.04%

2Kg

0.15%

7.5Kg

1.5%

75Kg

0.1%

5 Kg

-

-

94.37%

4.72T

DRICB

8.5T

0.12%

10.2Kg

-

-

0.04%

3.4 Kg

0.04%

3.4 Kg

-

-

1.00%

85 Kg

-

-

87.87%

7.47T

HMS

10.5T

0.5%

52.5Kg

0.8%

84Kg

0.025%

2.625Kg

0.025%

2.625Kg

0.4%

42Kg

1.00%

105Kg

- 97.50%

10.28T

MSTB

1.00T

0.45%

4.5 Kg

0.80%

8Kg

0.025%

0.25Kg

0.025%

0.25Kg

0.25%

2.5Kg

1.5%

15Kg

4%

40Kg

93.65%

0.93T

Total

25 T

242.2

Kg

122

Kg

8.28

Kg

13.78Kg 119.5

Kg

210 Kg 40Kg 23.36

Kg

8

Carbon reaction

[C] + [O] {CO}

12 16 28

For 242.20 Kg of C, [O] needed = 323Kg

And forms {CO} = 565.13 Kg i.e. 2.26 % of 25 T scrap mix

Mn reaction

[Mn] + [O] [MnO]

55 16 71

For 122 Kg of Mn, [O] needed = 35.49 Kg

And forms (MnO) = 157.49 Kg i.e. 0.63 % of 25 T scrap mix

S reaction

[S] + (CaO) (CaS) + [O]

32 56 72 16

For 8.28 Kg S, (Cao) Consumed = 14.49 Kg

And [O] liberated = 4.14 Kg

P reaction

2[P] + 5 [O] [P2O5]

62 80 142

For 13.78 Kg P, [O] needed = 17.78 Kg

And (P2O5) formed = 31.56 Kg i.e. 0.13 % of 25 T scrap mix

Si reaction

[Si] + 2[O] [SiO2]

28 32 60For 119.5 Kg Si, [O] needed = 136.57 Kg

9

And SiO2 formed = 256.07 Kg i.e. 1.02% of 25 T scrap mix

%CaO / %Sio2 this ratio is known as V – ratio. The ability of a slag to retain

oxides is generally expressed as the ratio of basic to acid oxides & is represented

as:

V ratio = %CaO / % SiO2

Modified V ratio is given by = CaO% / (SiO2% + P2O5% + Al2O3%)

Basicity is generally defined as ratio of all basic oxides upon all acidic oxides. For

effective refining under basic steel making, basicity always taken more than 2

Basicity = all basic oxides / all acidic oxides

= (%CaO + %MgO ) / (%Al2O3 +% SiO2 + %P2O5)

In practice Basicity is taken as

= (%CaO + %MgO) / (%SiO2 + % P2O5)

Or = %CaO / (%SiO2 + % P2O5)

Or = % Cao / % SiO2

Assuming Basicity = 2.5

Taking

Basicity = (%CaO + %MgO) / (%SiO2 + % P2O5)

2.5 = (%CaO + %MgO) / 1.02 % + 0.13 %

or (%CaO + %MgO) = 2.5 X 1.15 %

= 2.88 %

= 720 kg for 25 T

10

In available Metallurgical Lime we have CaO % = 85 % and MgO % = 4 %

If we consider only the effect of CaO

The amount of metallurgical lime needed = 847kg

If we consider effect of both CaO and MgO

The amount of metallurgical lime needed = 809 Kg

Total amount of O2 gas needed per charge mix = 323 + 35.5 + 17.78 + 136.57 – 4.14

= 508.71 Kg

Volume of O2 needed = 508.71 / 32 X 22.7 at NTP

= 360.86 Nm3

V ratio = % CaO / % SiO2 = 2.88 / 1.02 = 2.82

*Non Magnetic (N. Mag) material s that are present along the charge materials also have

to be considered as acidic oxide.

Hence when we consider the N. Mag material (0.84% for the 25 T charge) in calculation

of Basicity or lime needed then we have

Basicity = (%CaO + %MgO) / (%SiO2 + % P2O5 + % N. Mag)

Or 2.5 = (%CaO + %MgO) / (1.02 + .0.13 + 0.84)

Or (%CaO + %MgO) = 2.5 X 1.99

Or (%CaO + %MgO) = 4.9 % for 25 T charge

= 1225 Kg

Now Met Lime contains 89 % (CaO + MgO)

Then total Met Lime needed = 1400 Kg

Generally the amount of flux needed always kept 100 Kg to 200 Kg more than the

minimum required amount to compensate losses during charging.

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