Course work topic: «Flash smelting of sulfide copper concentrates on matte with heated air blast»

Advantages of melting concentrate in air blast

Melting in a suspended state has a number of advantages:

1) The smelting process is autogenous;
2) obtaining a constant stream of concentrated exhaust gases (16 - 80%), suitable for obtaining liquid sulfur dioxide, sulfuric acid or elemental sulfur;
3) the possibility of full automation of the process;
4) a high degree of desulfurization (70 - 80%), which makes it possible, respectively, to obtain a matte rich in copper content and even blister copper;
5) high degree of sulfur recovery (about 90%) for the plant;
6) increase in labor productivity.


2.Metallrugical calculations


2.1. Flash smelting of sulfide copper concentrates on matte with heated air blast

2.1.1 Calculation of the rational composition of the concentrate

The composition of the copper concentrate, %: Сu -22.0; Fe - 26.0; S-34.0, CaO-2.0; MgO-2.0; SiO2-11.0; Al2O3-2.0; others - 1.0.
We accept that in the concentrate Fe, Cu and S are represented by minerals CuFeS2, CuS, FeS2
We calculate for 100 kg of copper concentrate.
Let us denote the number of kg of minerals contained in 100 kg of copper concentrate through the letters: x-CuFeS2, y-CuS and z-FeS2.
Then the mass of copper contained in CuFeS2 and CuS will be; kg:

63.54x + 63.54y = 20

The mass of iron contained in CuFeS2 and FeS2 will be, kg:

55.85x + 55.85z = 26.1

The mass of sulfur contained in CuFeS2, CuS and FeS2 will be, kg:

64.12x + 32.06y + 64.12z = 33.4

Thus, we have a system of three equations:

63.54x + 63.54y = 20 (3.1)

55.85x + 55.85z = 26.1 (3.2)
64.12x + 32.06y + 64.12z = 33.4 (3.3)

We solve equations (2) and (3) with respect to z

55.85x + 55.85z = 26.1 × 64.12

64.12x + 32.06y + 64.12z = 33.4 x 55.85

64.12∙55.85x + 64.12∙55.85z = 64.12∙26.1

55.85∙64.12x + 55.85∙32.06y + 55.85∙64.12z = 55.85∙33.4

Subtracting equation (2) from equation (3), we obtain

1790.55 y = 1865.39 - 1673.53 = 191.86

Where do we get

y=0.107

Substituting the value of y into equation (1) we get

x = 0.236

Substituting the value of x into equation (2), we get

z = 0.231

Thus, 100 kg of copper concentrate contains minerals:

CuFeS2 - 0.236 kMol, CuS - 0.107 kMol and FeS2 - 0.231 kMol.
Then each mineral contains components:

CuFeS2: Cu - 14.99 kg, Fe - 13.18 kg, S - 15.13 kg.

СuS: Cu - 6.80 kg, S - 3.43 kg

FeS2: Fe - 12.90 kg, S - 14.81 kg

The data obtained are summarized in table 1.

Table 1 - Rational composition of sulfide copper concentrate

Mineral Component	CuFeS2	CuS	FeS2	CaO	MgO	SiO2	Al2O3	Others	Total
Cu	14,99	6,80							21,79
Fe	13,18		12,90						26,08
S	15,13	3,43	14,81						33,37
CaO				2,1					2,1
MgO					1,5				1,5
SiO2						13,8			13,8
Al2O3							1,0		1,0
Others								0,3	0,3
Total	43,3	10,23	27,71	2,1	1,5	13,8	1,0	0,3	100

2.1.2 Dust calculation

We accept the removal of dust from the furnace 10%. We calculate for 100 kg of concentrate
The amount of CuFeS2 carried away by the exhaust gases will be:
39,81∙0,1 = 3,98 kg.
It contains:
Cu – 13,8∙0,1 = 1,38 kg,

Fe – 12,1∙0,1 = 1,21 kg кг,

S – 13,91∙0,1 = 1,39 kg

The amount of CuS carried away by the exhaust gases will be

12,33∙0,1 =1,23 kg.

It contains

Cu – 8,2∙0,1 = 0,82 kg,
S – 4,13∙0,1 = 0,41 kg.

The amount of FeS2 carried away by the exhaust gases will be

29,86∙0.1 = 2,99 kg.
It contains

Fe – 13,9∙0,1 = 1,39 kg,

S – 15,96∙0,1 = 1,60 kg.

Other components of the concentrate will pass into dust in an amount

СаО - 2∙0,1 = 0,2 kg, MgО - 2∙0,1 = 0,2 kg, SiO₂ – 11,0∙0,1 = 1,1 kg, Al₂O₃ - 2∙0,1 = 0,2 kg.

others – 1,0∙0,1 = 0,1 kg.

Table 2 - Quantity and composition of dust from concentrate

Минерал Компонент	CuFeS2	CuS	FeS2	CaO	MgO	SiO2	Al2O3	Others	Total, kg
Cu	1,38	0,82							2,2
Fe	1,21		1,39						2,6
S	1,39	0,41	1,6						3,4
CaO				0,2					0,2
MgO					0,2				0,2
SiO2						1,1			1,1
Al2O3							0,2		0,2
Others								0,1	0,1
Total, kg	3,98	1,23	2,99	0,2	0,2	1,1	0,2	0,1	10

The rational composition of the concentrate minus dust is presented in table 3.

Table 3 - Rational composition of sulfide copper concentrate, taking into account dust

Mineral Component	CuFeS2	CuS	FeS2	CaO	MgO	SiO2	Al2O3	Others	Total, kg
Cu	12,42	7,38							19,8
Fe	10,89		12,51						23,4
S	12,52	3,72	14,36						30,6
CaO				1,8					1,8
MgO					1,8				1,8
SiO2						9,9			9,9
Al2O3							1,8		1,8
Others								0,9	0,9
Total, kg	35,83	11,1	26,87	1,8	1,8	9,9	1,8	0,9	90

2.1.3 Calculation of the rational composition of the matte
During the smelting process, higher sulfides decompose according to the equations
2CuFeS₂ = Cu₂S + 2FeS + 0,5S₂

This creates

Сu₂S - = 15,54 kg.

FeS - = 17,16 kg

S - = 3,13 kg

2CuS = Cu₂S + 0,5S₂

This creates

Cu₂S - = 9,24 kg

S - = 1,86 kg.

FeS₂ = FeS + 0,5S₂

This creates
FeS - = 19,68 kg

S - = 7,18 kg.

Total formed as a result of the dissociation of higher sulfides

Cu₂S – 24,78 kg, FeS -36,84 kg, S – 12,17 kg

We accept that the copper content in the matte is 60.0%, and the extraction of copper in the matte is 94.5%.

Then copper will pass into the matte
19,8 ∙0,954 = 18,89 kg

The mass of the matte will be

= 31,48 kg

The copper content in the matte according to the Mostovich rule is 25%. Then the mass of sulfur in the matte will be

31,48∙0,25 = 7,87 kg

The copper in the matte is in the form of Cu2S. The amount of sulfur in the matte associated with copper will be

= 4,77 kg.
The amount of Cu2S in the matte will be
18,89 + 4,77 = 23,66 kg

Oxygen is always dissolved in copper matte, which is bound with iron in the form of Fe3O4. the amount of oxygen in the matte containing 60% copper is 1.24%. The mass of oxygen in the matte is

0,0124∙31,48 = 0,39 kg.

The amount of iron associated with oxygen in the matte will be

= 1,02 kg

The mass of magnetite in the matte will be

1,02 + 0,39 = 1,41 kg

We accept the content of others in the matte 1%, which is 0.31 kg.

The rest of the matte mass will be represented by FeS, the amount of which will be

31,48 – 23,66 – 1,41 – 0,31= 6,1 kg

The amount of iron associated with FeS will be

= 3,88 kg.

The amount of sulfur in the matte associated with FeS will be

= 2,22 kg
The rational composition of the matte is given in table 4.
Table 4 - Rational composition of the matte

Component	Cu2S	FeS	Fe3O4	Others	Total
					kg	%
Cu	18,89				18,89	60,0
Fe		3,88	1,02		4,9	15,57
S	4,77	2,22			6,99	22,20
O2			0,39		0,39	1,24
Others				0,31	0,31	1,0
Total, kg.	23,66	6,1	1,41	0,31	31,48	100,01

2.1.4 Calculation of the composition and yield of slag without the addition of fluxes

The amount of copper that passes into the slag will be

19,8 -18,89 = 0,91 kg

Copper is present in the slag in the form of Cu2S. The amount of sulfur in the slag associated with Cu2S will be

= 0,23
The amount of Cu2S in the slag will be

0,91 + 0,23 = 1,14 kg

The amount of sulfur passing into the gas phase will be

S_gas.= S_con.- S_dust -S_sl..-S_pcs.= 34.0 - 3.4 - 0.23 -6.99 = 23.38 kg.

The amount of FeS oxidized to oxides

36,84 -6,1 = 30,74 kg

It contains iron

= 19,53 kg
and sulfur


= 11,21 kg
Sulfur passes into the gas phase

12,17 + 11,21 =23,38

The oxidation of FeS in the flame proceeds according to the equation

FeS + 1,5O₂ = FeО + 3SO₂

As a result, FeO is formed.


= 25,12 kg

Part of FeO is oxidized to magnetite according to the equation

6FeО + О₂ = 2Fe₃О₄

We accept that 30% FeO is oxidized to magnetite, which will be

25,12∙0,3 = 7,54 kg FeО

The amount of magnetite formed will be

= 8,1 kg

Of this amount of magnetite, 1.41 kg goes into matte. The rest of the magnetite

8,1 -1,41 = 6,69 kg

turns into sludge.

It contains oxygen


= 1,85 kg.
and iron


= 4,84 kg

FeO will also pass into the slag in the amount

25,12 – 7,54 = 17,58 kg

It contains oxygen

= 3,91 kg

and iron

= 13,67 kg

The composition and amount of smelting slag without flux addition are given in table 5.

Table 5 - Composition and amount of slag without the addition of fluxes

Component	Cu	Fe	S	O	SiO2	CaO	MgO	Al2O3	Others	T
										kg	%
Cu2S	0,91		0,23							1,14
FeO		13,67		3,91						17,58
Fe3O4		4,84		1,85						6,69
SiO2					9,9					9,9
CaO						1,8				1,8
MgO							1,8			1,8
Al2O3								1,8		1,8
Others									0,59	0,59
Total, kg	0,91	18,51	0,23	5,76	9,9	1,8	1,8	1,8	0,59	41,3
%

2.1.5 Calculation of the amount and composition of slag with the addition of fluxes

The recommended content of SiO2 in suspended melting slag is about 30%.
Quartz ore is used as a flux with composition, %: SiO2- 78.4; FeO- 4.9; MgO 6.0, Al2O3 - 6.0 others - 4.7

3,8∙16/55,85 =1,09

Denote by X the mass of added quartz ore.

We draw up a balance for silica, assuming that all silicon ore goes into slag

0,784X + 9,9 = 0,3(41,3 + X)

0,784X -0,3X = 12,39 – 9,9

0,484X = 2,49

X = 5,14 kg

The mass of slag will be

41,3 + 5,14 = 46,44 kg

With flint ore, SiO2 enters the slag

5,14∙0,784 = 4,03 kg

With flint ore, FeO enters the slag

5,14∙0,049 = 0,25 kg

It contains iron

= 0,19 kg
and oxygen


= 0,06 kg.

With silicon ore, the following also enters the slag:

MgO - 5.14∙0.06 = 0.31 kg, Al2O3 - 5.14∙0.06 = 0.31 kg, others - 5.14∙0.047 = 0.24 kg

Thus, the slag will contain oxides:

SiO₂ - 9,9 + 4,03 = 13,93 kg, Fe₃O₄ – 6,69 kg, FeO – 17,58 + 0,25 = 17,83 kg,

(Fe -13,86 kg, О – 3,97 kg)

CaO – 1,8 kg, MgO – 1,8 + 0,31 = 2,11 kg, Al₂O₃ – 1,8 + 0,31 = 2,11 kg,

others – 0,59 + 0,24 = 0,83 kg.

The composition and amount of slag, taking into account the addition of flux, is presented in Table 6.

Table 6 - Composition and amount of slag, taking into account the addition of fluxes

Component	Cu	Fe	S	O	SiO2	CaO	MgO	Al2O3	Others	Total
										kg	%
Cu2S	0,91		0,23							1,14
FeO		13.86		3,97						17,83
Fe3O4		4,84		1,85						6,69
SiO2					13,93					13,93
CaO						1,8				1,8
MgO							2,11			2,11
Al2O3								2,11		2,11
Others									0,83	0,83
Total, kg	0,91	18,7	0,23	5,82	13,93	1,8	2,11	2,11	0,83	46,44
%

2.1.6 Calculation of the amount and composition of dust

We believe that 10% of quartz ore passes into dust. Then it is necessary to introduce quartz flux into the charge


= 5,71 kg

Turns into dust

5,71 – 5,14 = 0,57

quartz ore. It contains

SiO₂ -0,57∙0,784 = 0,45 кг, FeO - 0,57∙0,049 = 0,03 кг (Fe – 0,02 kg, О – 0,01 kg).

MgO – 0,57∙0,06 =0,03 kg, Al₂O₃ – 0,57∙0,06 = 0,03 kg, others –0,57∙0,047= 0,03 kg.

Then the dust contains

SiO₂ – 1,1 + 0,45 = 1,55 kg; FeO 0,03 kg (Fe – 0,02 kg, О – 0,01 kg);

MgO – 0,2 + 0,03 = 0,23 kg; Al₂O₃ – 0,2 + 0,043 = 0,23 kg; others – 0,1 + 0,03 = 0,13 kg

The final dust composition is given in Table 7.

Table 7 - Quantity and composition of dust

Mineral Component	CuFeS2	CuS	FeS2	CaO	MgO	SiO2	Al2O3	FeO	Others	Total, kg
Cu	1,38	0,82								2,2
Fe	1,21		1,39					0,02		2,62
S	1,39	0,41	1,6							3,4
О								0,01		0,01
CaO				0,2						0,2
MgO					0,23					0,23
SiO2						1,55				1,55
Al2O3							0,23			0,23
Others									0,13	0,13
Total, kg	3,98	1,23	2,99	0,2	0,23	1,55	0,23	0,03	0,13	10,57

2.1.7 Calculation of the amount and composition of the gas phase

We assume that sulfur in the gas phase is in the form of SO2. The amount of sulfur transferred to the gas phase will be

34,0 - 3,4 – 0,23 - 6,99 = 23, 38 kg

Then the degree of desulfurization of the melt will be

= 76,4 %.

The amount of oxygen spent on the formation of SO2 will be

= 23,34 kg
The amount of SO2 in the gas phase will be

23,38 + 23,34 = 46,72 kg.

The amount of oxygen spent on the formation of FeO due to the oxidation of FeS will be

= 5,59 kg

The amount of oxygen spent on the oxidation of FeO to Fe3O4 will be

= 0,28 kg.

Total oxygen consumption

23,34 + 5,59 + 0,28 = 29,21 kg

The amount of air nitrogen supplied to the furnace will be

= 97,79 kg.
The amount of air spent for melting will be

29,21 + 97,79 = 127 kg.

For melting in a suspended state, the charge is dried to a moisture content of about 0.1%. Then, with the charge, moisture enters the furnace


= 0,11 kg.
Considering the exhaust gases as an ideal gas, we calculate the volume of gas using the Mendeleev-Clapeyron equation

V = ,

where g – mass of gas, kg;

М – molecular weight of gas, kg
22,4 – volume occupied by 1 kmol of gas, м³;
V – gas volume, м³.

The volume of exhaust gases will be:

SO₂ - = 16,3 м³; N₂ - = 78,23 м³; Н₂О - = 0,14 м³.

The amount and composition of exhaust gases are presented in table 8.

Table 8 - Quantity and composition of exhaust gases

Gases	kg.	м^3	%( voluminous)
SO2	46,72	16,3	17,22
N2	97,79	78,23	82,63
H2O	0,11	0,14	0,15
Total	144,62	94,67	100

2.1.8 Material balance of flash melt

On the basis of the performed calculations, we draw up a material balance for the smelting of copper concentrates in suspension. The material balance is presented in table 9.
Table 9 - Material balance of heat

	Total kg	Cu	Fe	S	O	SiO2	CaO	MgO	Al2O3	H2O	N2	Other
Loaded
Kt	100,11	22,0	26,0	34,0		11,0	2,0	2,0	2,0	0,11		1,0
Ore	5,71		0,21		0,07	4,48		0,34	0,34			0,27
Air	127				29,21						97,79
Total	232,82	22,0	26,21		29,28	15,48	2,0	2,34	2,34	0,11	97,79	1,27
Received
Matte	31,48	18,89	4,9	6,99	0,39							0,31
Slag	46,44	0,91	18.7	0,23	5,82	13,93	1,8	2,11	2,11			0,83
Dust	10,57	2,2	2,62	3,4	0,01	1,55	0,2	0,23	0,23			0,13
Gases	144,62			23,38	23,34					0,11	97,79
Total	233,11	22,0	26,22	34	29,56	15,48	2,0	2,35	2,35	0,11	97,79	1,27

The discrepancy is


= 0,12%.

2.1.9 Calculation of heat balance of melting

2.1.9.1.1 Initial data for calculation
We accept the values of heat capacities for
С_sht. = 0,941 kJ/(kg∙deg.)
С_shl.. = 1,42 kJ/(kg∙deg.)

С_shiht. = 0,84 kJ/(kg∙deg.)

С_dust. = 0,84 kJ/(kg∙deg.)

С_SO2 = 2,298 kJ/(nм³∙ deg.)

С_N2 = 1,423 kJ/(nм³∙ deg.)

С_H2O(gas) = 1,803 kJ/(nм³∙ deg.)

С_H2O(zh) = 4,186 kJ/(kg∙deg.)

С_air = 1,443 kJ/(nм³∙ deg.)

The heat of evaporation of water is 2300 kJ/kg.

The temperature of the materials entering the furnace is 25 °C.

The temperature of the produced matte is 1180 °C.

The temperature of the produced slag is 1250 °C.

Exhaust gas temperature 1300 °C.

Blast heating - 500°C.

2.1.9.2 Articles of the arrival of heat

2.1.9.2.1 Heat of exothermic reactions

FeS + 1,5O₂ = FeO + SO₂ ∆H = - 461,33 kJ

Due to the oxidation of 30.74 kg of FeS, heat will be released

= 161316 kJ

3FeO + O₂ = Fe₃O₄ ∆H = - 322,48 kJ

Due to the oxidation of 7.54 kg of FeO, heat will be released

= 11280 kJ
S₂ + 2O₂ = 2SO₂ ∆H = - 593,8 kJ

Due to the oxidation of 12.17 kg of S₂, heat will be released

= 117797 kJ

2FeO + SiO₂ = 2FeO∙SiO₂ ∆H = - 34,3 kJ

Due to the slagging reaction, 17.83 kg of FeO will release heat

= 4256 kJ.

СаО + SiO₂ = СаО∙ SiO₂ ∆H = - 37,54 kJ

Due to the CaO slagging reaction, heat will be released

= 1339 kJ.

Exothermic reactions generate heat

161316 + 11280 + 117797 + 4256 + 1339 = 295 988 kJ.

2.1.9.2.2 Heat supplied to the furnace with loaded materials

The heat supplied to the furnace with the loaded materials is calculated by the equation

Q = C _i∙g_i∙t_i (26)

where C_i - heat capacity of the 9th component, kJ/(kg∙deg.);

g_i - component weight, kg;
t_i - temperature, ^оС
The heat supplied to the furnace with the charge will be

105,82∙25∙0,84 = 2222 kJ

The heat supplied to the furnace with air will be

127 ∙ 500∙1,343 = 85281 kJ.

The heat supplied to the furnace with loaded materials will be

2222 + 85291 = 87503 kJ

2.1.9.3 Heat consumption

2.1.9.3.1 The heat spent on the decomposition of higher sulfides due to

2CuFeS₂=Cu₂S+2FeS+0,5S₂ ∆H = 100,4 kJ

Due to the dissociation reaction, 35.83 kg CuFeS2 absorbs heat
= 9801 kJ

2CuS = Cu₂S + 0,5S₂ ∆H = 53,56 kJ

Due to the dissociation of 11.1 kg of CuS, heat is absorbed

= 3,109 kJ

FeS₂ = FeS + 0,5S₂ ∆H = 71,18 kJ

Due to the dissociation of 26.8 kg FeS2

= 16115 kJ.

The heat expended on heating and evaporation of moisture will be

4,186∙75∙0,1 + 0,11∙2300 = 288 kJ

Heat is absorbed through endothermic reactions.

9801 + 3109 + 16115 + 288 = 29313 kJ

2.1.9.3.2 Heat lost by melt products discharged from the furnace

Heat lost with matte matte

31,48∙0,941∙1180 = 34955 kJ

The slag heat lost with the slag will be

46,44∙1,42∙1250 = 82431 kJ.

The heat lost with dust is

10,57∙0,84∙1300 = 11542 kJ

The heat lost with the exhaust gases will be

(16,3∙2,298 + 78,23∙1,423 + 0,14∙1,803)∙1300 = 193739 kJ.

Heat loss with melting products will be

34955 + 82431 + 11542 + 193739 = 322667 kJ

The total heat loss will be

29313 + 322667 = 351980 kJ.

We accept that heat loss to the environment will be 5% of the total loss

= 18525 kJ

The heat balance of melting is presented in table 10

Table 10 - Heat balance of the melting process

	The arrival of heat				Heat consumption
	parish article	kJ	%		Expenditure	kJ	%
1	Heat supplied with feed materials	87503		1	Heat carried away by the matte	34955
2	Heat of exothermic reactions	295988		2	Heat carried away by the slag	82431
3				3	Heat carried away by flue gases	193739
4				4	Heat of endothermic reactions	29313
5				5	Heat carried away by dust	11542
6					Losses to the environment	18525
					Unaccounted losses	3985
	Total	383490			Total	379505	Total